[SOLVED] Group Theory For Dummies

[climbhi]

I've become interested in learning about Group Theory. I don't know too much but I see it spring up all over the place and would just like to know what it is about and some of the basics. Could some one please point me in the direction of a good resource that wouldn't be too far over my head? Thanks.

[BoulderHead]

Oops, I thought you said 'group therapy for dummies'. I guess I can't help after all. [:D]

[climbhi]

Originally posted by BoulderHead
Oops, I thought you said 'group therapy for dummies'. I guess I can't help after all. [:D]

Well perhaps I need some of that too...

[rutwig]

Originally posted by climbhi
I've become interested in learning about Group Theory. I don't know too much but I see it spring up all over the place and would just like to know what it is about and some of the basics. Could some one please point me in the direction of a good resource that wouldn't be too far over my head? Thanks.

Good (introductory) references are:

M. A. Armstrong. Groups and symmetry. Springer Verlag 1988.

J. Rotman. An introduction to the theory of groups, Sprnger Verlag 1995.

J. D. Dixon, Problems in Group Theory. New York: Dover, 1973.

R. Mirman. Group Theory: An Intuitive Approach. World Scientific, 1995.

[damgo]

Hmm, I used Dummit&Foote "Abstract Algebra" as an introductory textbook, and found it to be excellent. Though the emphasis is on mathematics rather than the physics applications (Lie groups, representations, etc). The nice thing about pure group theory is it requires basically zero prerequisites.

[chroot]

I'll also recommend the Schaum's Outline of Group Theory. It doesn't specifically cover some of the more physically interesting topics such as groups of 3x3 matrices, but it gives you all of the tools necessary to understand just about any group-theoretical system.

- Warren

[Tom Mattson]

Originally posted by chroot
I'll also recommend the Schaum's Outline of Group Theory.


Yep, and the one called Abstract Algebra, too. All in all, a 30 dollar committment.

[climbhi]

Thanks for the replies, does anyone know of a good online (read free) source? I'm kind of interested in group theory to see how it relates to QM and what not, but also just for pure math. Would the Schaubs outlines work well for both if there is no good free source available?

[rutwig]

Here a book that has been a quite interest source for many physicists/


http://www.cns.gatech.edu/GroupTheory/index.html

[Tom Mattson]

Originally posted by climbhi
Would the Schaubs outlines work well for both if there is no good free source available?

There are many free sources available, but I recommend the Schaum's outlines anyway, because they are loaded with solved examples and exercises with answers.

[marcus]

Originally posted by climbhi
I've become interested in learning about Group Theory. I don't know too much but I see it spring up all over the place and would just like to know what it is about and some of the basics. Could some one please point me in the direction of a good resource that wouldn't be too far over my head? Thanks.

I'm wondering if it would be possible to have an entrylevel workshop here at PF on groups.

I mean a collective teach-each-other tutorial-----no one person doing all the teaching but trading around.

I see Tom and Rutwig and Chroot have posted online resources
and also hardcopy books to buy.

The big question is------is there enough interest?
A secondary question is-----could we stand to type all the subscripts, superscripts, matrices, and greek letters? PF is a great medium for non-hierarchical learning. But the sheer typing of symbols and inability to draw pictures imposes some limits on what one can handle here.

So I am skeptical that a group theory tutorial or workshop would get anywhere.

But just to see how it might go----here is my proposal

Focus on the simplest most classical groups central to basic physics--dimensions 2, 3, 4.

Focus on things like SO(3) the special orthogonal group. ["special" just means det = 1 in this case, think of rotations]

And SU(2) the special unitary group----because of its relation to SO(3) and the pauli spinmatrices. among other things.

And SL(2,C) because of its relation to the Lorentz group.

It seems to me that the goal should be not to snow anybody or discourage anybody----not to show off or try to pull rank on people (as non-PF people sometimes do when discussing math)----but simply to go over the group theory that is most basic and do it in an entrylevel way.

This might not be possible---it might simply not work.

Also it might be tiresome to try to type in matrices---even like the three pauli spinmatrices which are about as simple as 2x2 matrices can get would be sort of tedious to type into PF-style posts.

Anyway I am broaching the idea. Reactions? Better ideas of how to do it?

[Lonewolf]

I like the sound of your idea, Marcus. I'd be interested once my exams are done. I like especially the sound of learning its applications to Physics. We get taught Group Theory, but only in the sense of pure maths.

[marcus]

Originally posted by Lonewolf
I like the sound of your idea, Marcus. I'd be interested once my exams are done. I like especially the sound of learning its applications to Physics. We get taught Group Theory, but only in the sense of pure maths.

If anyone knows good notes on the web that correspond to what Lonewolf is talking about (basic classical group theory with an eye to applications in physics) please post a link.

Lonewolf, this thread may possibly remain dormant until you are thru exams. Depends on how interested the others are. When you or anybody returns I will probably get a notice by email. but to be sure, send me a PM.

[rutwig]

The book by Cvitanovic is one of the links. There are many others, but it should be specified whether one is interested on discrete, continuous (non differentiable) or Lie groups, or even generalizations like Kac-Moody groups, supergroups, etc. Each of the topics is a world in itself.

[marcus]

Originally posted by rutwig
The book by Cvitanovic is one of the links. There are many others, but it should be specified whether one is interested on discrete, continuous (non differentiable) or Lie groups, or even generalizations like Kac-Moody groups, supergroups, etc. Each of the topics is a world in itself.

Rutwig I do not know if you have any interest in LQG or follow it at all but, if you do, then you probably have noted that
a recent result of Olaf Dreyer seems to force a change in the group from SU(2) to SO(3)
Lubos Motl has what seems to be a clearsighted outsiders
perspective on this (not being especially an advocate of LQG)

Have you any comment on this----perhaps the change seems insignificant given that one is a 2-fold cover of the other---or does it have some interesting ramifications?

I will edit this to add a link to Lubos Motl's paper, though
I would not be surprised if you had already noticed it.

http://www.arxiv.org/abs/gr-qc/0212096

[Dimitri Terryn]

A good place for online textbooks is

http://www.math.gatech.edu/~cain/textbooks/onlinebooks.html

I printed out "Abstract Algebra, the basic graduate year" by Prof. Robert Ash, and it looks pretty good so far.

A workshop would be pretty nice, since I had already planned to study some algebra this summer anyway.
I got an introduction this semester, and although most of my fellow physics students hated the abstactness of it all, it grew on me. Seems like a fun game to play.

[rutwig]

Originally posted by marcus
perhaps the change seems insignificant given that one is a 2-fold cover of the other---or does it have some interesting ramifications?


Interesting ramifications should be searched for experimentally, but it is not at all insignificant that the adequate group is not the simply connected universal cover, but some projection of it. With respect to the covering, this would indicate that the system makes no distinction of the covering elements (as happens at the tangent space level), and probably this has some significant consequences.

[marcus]

Originally posted by Lonewolf
I like the sound of your idea, Marcus. I'd be interested once my exams are done. I like especially the sound of learning its applications to Physics. We get taught Group Theory, but only in the sense of pure maths.

Lonewolf, I got your PM that exams are over. I am here but
have been preoccupied with an LQG thread in "theoretical"
forum. The thread is about SO(3) and its Lie algebra
so(3). Good stuff to know. Marsden's introductory treatment is good.

Look at Marsden's Chapter 9 "An introduction to Lie groups" if you want.

for some people who have just posted here, they are waaaay
beyond that entrylevel introduction by Marsden. But if you and I want to start talking it has to be somewhere and the beginning is apt to be a good place.
Besides, Jerry Marsden is a CalTech professor and his approach
connects up to the physics-needs of CalTech students. It doesnt look at all "pure" to me, so you might like it.

Do you find anything in Chapter 9 interesting or whatever?

I will go fetch the link and edit it in here. Really nice of Marsden to put it online.

http://www.cds.caltech.edu/~marsden/bib_src/ms/Book/

[marcus]

Originally posted by rutwig
Interesting ramifications should be searched for experimentally, but it is not at all insignificant that the adequate group is not the simply connected universal cover, but some projection of it. With respect to the covering, this would indicate that the system makes no distinction of the covering elements (as happens at the tangent space level), and probably this has some significant consequences.

rutwig, thanks
hope to hear further---any thoughts you have about this
very interesting switch to SO(3) or news, if you receive any,
about them finding some way, cunning as they are, to switch groups yet again.

[marcus]

On the off chance that we might have a collective learning effort in classical Lie groups here----which might begin at least by being based on Marsden's chapter 9----I have pasted in this extract dealing with the group of rotations. It is a summary of rotation facts made a day or two ago for a thread in "theoretical" forum.
Maybe it is not the perfect thing for this thread but it is a start.

for the moment I am thinking of this very concretely---not at all abstractly---as 3x3 rotation matrices. Anyone else is welcome to take the lead here, but because nothing is happening as yet I will paste in this extract (essentially part of what is covered by Marsden)

Here are some basic facts about SO(3)
**************************************
SO(3) is a compact Lie group of dimension 3.

Its Lie algebra so(3) is the space of real skew-symmetric 3x3 matrices
with bracket [A,B] = AB - BA.

The Lie algebra so(3) can be identified with R3
the 3-tuples of real numbers by a vectorspace isomorphism
called the"hat map"

v = (v1,v2,v3) goes to v-hat, which is a skew-symmetric matrix
meaning its transpose its its NEGATIVE, and you just stash the three numbers into such a matrix like:

+0 -v3 +v2
+v3 +0 -v1
-v2 +v1 +0

v-hat is a matrix and apply it to any vector w and
you get vxw.

Everybody in freshman year got to play with v x w
the cross product of real 3D vectors
and R3 with ordinary vector addition and cross product v x w is kind of the ancestral Lie algebra from whence all the others came.

And the hat-map is a Lie algebra isomorphism

EULER'S THEOREM

Every element A in SO(3) not equal to the identity is a rotation
thru an angle φ about an axis w.

SO SO(3) IS JUST THE WAYS YOU CAN TURN A BALL---it is the group of rotations

THE EIGENVALUE LEMMA is that if A is in SO(3) one of its
eigenvalues has to be equal to 1.
The proof is just to look at the characteristic polynomial which is of degree three and consider cases.

Proof of Euler is just to look at the eigenvector with eigenvalue one----pssst! it is the axis of the rotation. Marsden takes three sentences to prove it.

A CANONICAL MATRIX FORM to write elements of SO(3) in
is

+1 +000 +000
+0 +cosφ -sinφ
+0 +sinφ cosφ

For typography I have to write 0 as +000
to leave space for the cosine and sine under it
maybe someone knows how to write handsomer matrices?

EXPONENTIAL MAP
Let t be a number and w be a vector in R3
Let |w| be the norm of w (sqrt sum of squares)
Let w^ be w-hat, the hat-map image of w in so(3), the Lie algebra. Then:

exp(tw^) is a rotation about axis w by angle t|w|


It is just a recipe to cook up a matrix giving any amount of rotation around any axis you want.

[KillaMarcilla]

Wait, wait, wait, is group theory just that mathematical dealy wherein you count numbers by grouping them in various ways, like if you want to prove there are more numbers between 0 and 1 than there are integers greater than zero?

[chroot]

Originally posted by KillaMarcilla
Wait, wait, wait, is group theory just that mathematical dealy wherein you count numbers by grouping them in various ways, like if you want to prove there are more numbers between 0 and 1 than there are integers greater than zero?
No. Group theory deals with sets of mathematical entities and operations upon those entities.

For example, take the set of real numbers and the addition operation. Together, the set of reals and the addition operation form a group. Groups have the following properties:

1) The result of applying the operator to any two elements of the group is itself an element of the group. (The sum of any two reals is itself a real.)

2) Every group has an identity element, such that the operation of any element with the identity returns that element. (The sum of any real with zero is left unchanged -- zero is the identity.)

3) Every element in a group has an inverse element. (The inverse of 1, for example, is -1.)

4) For any three elements in the group, (A + B) + C is the same as A + (B + C).

Marcus is talking about groups of 3x3 matrices. These groups are given names like SO(3) and so on to reflect the various characteristics that elements of each group share. The operation on these groups is that of matrix multiplication.

- Warren

[KLscilevothma]

Mathematical Groups (teach yourself) by Tony Barnard and Hugh Neill is a good book that introduces basic concepts of group. Topics include properties of group, notations, cyclic groups, isomorphism, etc. There are sufficient examples for beginners to understand and suitable for senior high school students or above.

[marcus]

Does everybody know matrix multiplication
and what a matrix transpose is?

(you get the transpose of a square matrix by flipping it over its main diagonal)

If you dont, please ask. If A is a square matrix the transpose is At.

If anyone has different notation from me they like better I am open to changing notation as long as I can type it easily.

We might have a small informal workshop on matrix groups
right here in the "Groups for Dummies" thread with no fanfare.
It might work, and no harm done if it didnt. But someone besides me has to do the lion's share of the explaining or I will get
too boring and monotonous.

There is a cool kind of matrix whose transpose is equal to its inverse (something you dont normally expect!)

At = A-1


√1/2 -√1/2
√1/2 √1/2

[KillaMarcilla]

Yeah, that's what I meant, chroot

h0 h0, and here I thought Group Theory was some arcane mystery, unknowable to low-level undergraduates like myself

[Lonewolf]

What is the Lie algebra, and how does it relate to its respective Lie group?

[Tom Mattson]

Originally posted by KillaMarcilla
h0 h0, and here I thought Group Theory was some arcane mystery, unknowable to low-level undergraduates like myself

Actually, group theory is one of the few mathematical subjects that has no prerequisites. It is purely axiomatic and logical.

[Tom Mattson]

Originally posted by Lonewolf
What is the Lie algebra, and how does it relate to its respective Lie group?

A Lie algebra is a nonAbelian algebra whose elements ai satisfy the following properties:

1. [ai,ai]=0 (ai commutes with itself.)
2. [aj+ak,ai]=[aj,ai]+[ak,ai] (Linearity of commutator.)
3. [ai,[aj,ak]]+[aj,[ak,ai]]+[ak,[ai,aj]]=0 (Jacobi identity.)

The relation between the Lie algebra and the Lie group is that the elements of the algebra generate the group.

Example:

Consider the Lie algebra of the angular momentum operators in quantum mechanics:

[Ji,Jj]=i(hbar)εijkJk

The elements Ji generate the Lie group of rotations D about a normal vector n through an angle φ as follows:

D(n;φ)=exp(-iJ.nφ/(hbar))

Let me know if you want to go into more detail.

edit: typo

[marcus]

Originally posted by Lonewolf
What is the Lie algebra, and how does it relate to its respective Lie group?

a big question.
I hope others will help answer. gradually filling in the gaps
in the picture.

A Lie group is a group that is a smooth manifold
and each point x in a manifold has a tangent space Tx
and the tangent space at any point is a vector space

That is intuitive I guess----the space of tangent vectors at a point, in the 2D case a tangent plane.

But if the manifold is a group then there is a special element, the identity element of the group. Call it e.

So there is a special tangent space Te of tangent vectors at the identity e.

that is what the Lie algebra is, as a set. But it has a lot of uses and more structure than you expect from just a vector space (with a zero and vector addition and all). There is a way of parlaying the group multiplication (which goes on down in the manifold) up into the tangent space---so that you get an operation up in the tangent space sort of like multiplying two vectors to get a third vector. This "multiplication-like" operation is called "bracket" and is written [A,B] where A,B are vectors in Te the tangent space at the identity. It is not commutative, but then not all forms of multiplication are.

The Lie algebra is kind of an "infinitesimal" version of the group and the group (at least neighborhood of the identity and actually more) can be regenerated from the algebra.

If you lose the group you can grow (at least a piece of) it back just from the tangent space at the identity---the algebra.

Maybe someone else will supply a rigorous definition or some more intuitive insight.

What I would like to do is stop here and look at one simple example of a Lie group and its Lie algebra. Then let someone else take over, if they want.

[marcus]

I just happened to notice that Greg Tom and chroot are browsing the math forum and any of them could give a rigorous def
of a Lie group (I have not given the definition so far) and its Lie algebra.

and that would be a step in the right direction (of collectivizing and getting several persons approaches)

[Tom Mattson]

Originally posted by marcus
I just happened to notice that Greg Tom and chroot are browsing the math forum and any of them could give a rigorous def
of a Lie group (I have not given the definition so far) and its Lie algebra.


Actually, I can't. They did not talk about this stuff in my Abstract Algebra course. I only know about it through my QM courses, which is why I talk in terms of examples. We need people such as Hurkyl, Lethe, SelfAdjoint, etc... to provide the rigorous generalities.


and that would be a step in the right direction (of collectivizing and getting several persons approaches)

I posted mine a few seconds before yours (see above).

[marcus]

it rarely a mistake to look at examples before studying
abstract definitions and my favorite example of a Lie group/algebra
is rotations/skew-symmetric matrices.

Everybody has had linear algebra so probably know
At the transpose of a (real) matrix and
A-1 the matrix inverse
and may also know that an orthogonal matrix
(one that doesnt change the length of vectors
obviously a very valuable interesting kind and it
also does not change their inner product when you apply
it to two vectors)
this very nice kind of matrix is described by
At = A-1

Now those things form a group because if A and B dont
change lengths or inner products then AB will not either
and you can also check the At = A-1
condition for AB.

But they arent a vector space because if you add two A+B
is usually not that kind of matrix any more

Everybody knows the determinant and that detAt = detA
and that detA-1 = 1/detA
So if you look at the At = A-1 condition in that light you will see that there is no possible thing that det A can be except +1 or -1. The matrices with det = +1 form a subgroup.

These are very nice simple useful Lie groups and the question is, what is the Lie algebra. What does the tangent space at the
identity matrix look like?

So you Lonewolf ask "what is the Lie algebra" and I am temporarily turning this question into a very concrete one: "what is the Lie algebra of this particular group of matrices, the orthogonal ones, or the subgroup of them which are simple rotations. We can try to answer that in either 2D or 3D.
Are there any questions so far?

Anybody who wishes is invited to take over explaining and discussing at this point.

[marcus]

Originally posted by Tom
....know about it through my QM courses, which is why I talk in terms of examples.


I posted mine a few seconds before yours (see above).

Glad to see you here
We may need examples far more than rigor
Would invite and encourage examples

[Hurkyl]

We need people such as Hurkyl, Lethe, SelfAdjoint, etc... to provide the rigorous generalities.

Eep!


Paraphrased from my abstract algebra text:

A Lie Algebra is simply a vector space A over a field F equipped with a bilinear operator [,] on A that satisfies [x, x] = 0 and the jacobi identity:

[[x, y], z] + [[y, z], x] + [[z, x], y] = 0


(If F does not have characteristic 2, [x, x] = 0 is equivalent to [x, y] = -[y, x])


I would like to point out that [x, y] is not defined by:

[x, y] = xy - yx

(or various similar definitions); it is merely a bilinear form that satisfies the Jacobi identity and [x, x] = 0.


However, for any associative algebra A, one may define the lie algebra A- by defining the lie bracket as the commutator.


An example where [,] is not a commutator is (if I've done my arithmetic correctly) the real vector space R3 where [x, y] = x * y, where * is the vector cross product.


edit: fixed an omission

[chroot]

Quoting from John Baez' "Gauge Fields, Knots, and Gravity,"

"Lie algebras are a very powerful tool for studying Lie groups. Recall that a Lie group is a manifold that is also a group, such that the group operations are smooth. It turns out that the group structure is almost completely determined by its behavior near the identity. This, in turn, can be described in terms of an operation on the tangent space of the Lie group, called the 'Lie bracket.'

"To be more precise, suppose that G is a Lie group. We define the Lie algebra of G, often written g, to be the tangent space of the identity element of G. This is a vector space with the same dimension of G. A good way to think of Lie algebra elements is as tangent vectors to path in G that start at the identity. An example of this is the physicists' notion of an 'infinitesimal rotation.' If we let [gamma] be the path in SO(3) such that [gamma](t) corresponds to a rotation by the angle t (counterclockwise) about the z axis:
[gamma](t) =

cos t -sin t 0
sin t cos t 0
0 0 1
"Then the tangent vector to [gamma] as it passes through the identity can be calculated by differentiating the components of [gamma](t) and setting t = 0:
[gamma]'(0) =

0 -1 0
1 0 0
0 0 0

This is an element of so(3), the Lie algebra of SO(3). Any such matrix, which is the tangent vector to a path through the identity of SO(3), is a member of so(3).

- Warren

[lethe]

Originally posted by Hurkyl

A Lie Algebra is simply a vector space A over a field F equipped with a bilinear operator [,] on A that satisfies [x, x] = 0 and the jacobi identity:

[[x, y], z] + [[y, z], x] + [[z, x], y] = 0


(If F does not have characteristic 2, [x, x] = 0 is a consequence of the Jacobi identity and may be dropped as an axiom)


whoa! is that true? i m not so sure. i think what you want to say here is:
[x,y]=-[y,x] is a consequence of [x,x]=0, and if the field does not have characteristic 2, then [x,y]=-[y,x] implies [x,x]=0, but not in fields with characteristic 2, so we drop that as an axiom.


I would like to point out that [x, y] is not defined by:

[x, y] = xy - yx

(or various similar definitions); it is merely a bilinear form that satisfies the Jacobi identity and [x, x] = 0.


However, for any associative algebra A, one may define the lie algebra A- by defining the lie bracket as the commutator.


An example where [,] is not a commutator is (if I've done my arithmetic correctly) the real vector space R3 where [x, y] = x * y, where * is the vector cross product.

other examples include the poisson bracket and the lie bracket (well, the lie bracket does turn out t be a commutator, but it is certainly not defined that way).

[rutwig]

You wan a complete synthetic definition of Lie algebra? Here it is: A Lie algebra L is a pair (V,t) formed by an F-module (F being a commutative ring) and an alternated tensor t of mixed type (2,1) satisfying the Jacobi identity.

A special case is F a field.

[rutwig]

Originally posted by Hurkyl
I would like to point out that [x, y] is not defined by:
[x, y] = xy - yx
[/B]

For the so called abstract Lie algebras only the bracket [,] by itself has a meaning, but it can be proven that for any (finite dimensional) Lie algebra L we can find a vector space V such that the elements of L are linear transormations of V, so that the formulae above holds (that is, we can alsays find a faithful linear representation of L). For those interested in details, this is known as the theorem of Ado (1945).

[rutwig]

Originally posted by Tom
A Lie algebra is a nonAbelian algebra whose elements ai satisfy the following properties:

1. [ai,ai]=0 (ai commutes with itself.)
2. [aj+ak,ai]=[aj,ai]+[ak,ai] (Linearity of commutator.)
3. [ai,[aj,ak]]+[aj,[ak,ai]]+[ak,[ai,aj]]=0 (Jacobi identity.)


Just a comment, you can drop the word nonabelian since any vector space is an abelian Lie algebra simply taking the zero bracket. Indeed abelian algebras play a fundamental role in the theory (see for example the Cartan subalgebras). You have also the require bi-linearity, otherwise the result is not necessarily a Lie algebra. The (local!) relation with the Lie groups is expressed by the Campbell-Hausdorff formula (exponentiation of elements). But this is delicate, since not all elements of the Lie group must be the exp of some element of the Lie algebra (the example is well known to you!).

[rutwig]

Finally, you can also use operators to construct Lie algebras. If you take hermitian conjugate operators B, B* (in an infinite dimensional space) with the rule [B,B*]=BB*-B*B you obtain the Heisenberg Lie algebra, which is the basis of all classical analysis of harmonic oscillators and gave rise to the boson formalism used by Schwinger, Holstein and Primakoff in the 40's to analyze angular momentum.

[lethe]

Originally posted by rutwig
But this is delicate, since not all elements of the Lie group must be the exp of some element of the Lie algebra (the example is well known to you!).

what? this is not known to me, i thought that any element of the Lie Group could indeed be obtained by exponentiation of the Lie Algebra. what is the example?

[rutwig]

Originally posted by lethe
what? this is not known to me, i thought that any element of the Lie Group could indeed be obtained by exponentiation of the Lie Algebra. what is the example?

If you have worked only with compact groups, then you will not have observed this; any element is the exponentiation of some element in the Lie algebra. But for noncompact groups this is no longer true, and we have to consider a finite number of elements in the Lie algebra to recover the elements of the group.

Example: show that the element

|-a 0 |
|0 -1/a |

of SL(2,R) cannot be expressed as the exponential of an unique element X of the Lie algebra sl(2,R) if a is different from 1.

[lethe]

Originally posted by rutwig
If you have worked only with compact groups, then you will not have observed this; any element is the exponentiation of some element in the Lie algebra. But for noncompact groups this is no longer true, and we have to consider a finite number of elements in the Lie algebra to recover the elements of the group.

Example: show that the element

|-a 0 |
|0 -1/a |

of SL(2,R) cannot be expressed as the exponential of an unique element X of the Lie algebra sl(2,R) if a is different from 1.

OK, so lemme see. the lie algebra of SL(2,R) is just the set of real 2x2 traceless matrices right? a basis for this algebra is:

[1 0] [0 1] [0 0]
[0 -1], [0 0], [1 0]

right? obviously the matrix you mentioned has to be constructed from the first basis element.

what about
exp(ln a[1 0])
[0 -1]

no wait, that will give me only

[a 0]
[0 1/a]

obviously, i ll never be able to get negative numbers by exponentiating these matrices, so it s impossible, as you say. why is that? what does this have to do with compactness?

[rutwig]

The result is not entirely obvious, but compactness ensures some properties (like existence of invariant integration)
that are not given otherwise. For this case, the key is that for compact (connected) groups any element is conjugate to
an element in a maximal torus (analytic subgroups corresponding to an abelian subalgebra of the Lie algebra).

[Lonewolf]

These are notes taken from Marsden. Some of the proofs have been omitted, but are available in Marsden's text.

The Real General Linear Group

GL(n,R) is defined as GL(n,R) = {A in Rnxn: det(A)!=0}
GL+(n,R) is defined as GL+(n,R) = {A in Rnxn: det(A)>0}
GL-(n,R) is defined as GL-(n,R) = {A in Rnxn: det(A)<0}

where R in the set of real numbers, and Rnxn is the set of real nxn matrices.

GL+(n,R) is the connected component of the identity in GL(n,R), and GL(n,R) has exactly two connected components. Marsden proves this using the real polar decomposition theorem. Following the proof of the the conclusion below is reached.

The real general linear group is a non-compact disconnected n2 dimensional Lie group whose Lie algebra consists of the set of all nxn matrices with the bracket [A,B] = AB-BA.


The Special Linear Group

SL(n,R) is defined as SL(n,R) = {A in GL(n,R): det(A)=1}

R\{0} is a group under multiplication, and det:GL(n,R)->R\{0} is a Lie group homomorphism since det(AB) = det(A)det(B).

The Lie algebra of SL(n,R) consists of the set of nxn matrices with trace 0 and bracket [A,B] = AB-BA.

Since trace B=0 imposes one condition dim[sl(n,R)]=n2-1, where sl(n,R) is the Lie algebra of SL(n,R).

It is useful to introduce the inner product on the Lie algebra gl(n,R) of GL(n,R) <A,B> = trace(ABT). Note that ||A||2 = &Sigma;i,j=1naij2 which shows that the norm on gl(n,R) coincides with the Euclidean norm on Rn2. This norm can be used to show that SL(n,R) is not compact.

Let v1 = (1,0,...,0), v2 = (0,1,...,0),...,vn-1 = (0,...,1,0) and vn = (t,0,...,1) where all vi are members of Rn

Let A = (v1, v2,...,vn) be a matrix in Rnxn. All matrices of this form are elements of SL(n,R) whose norm is equal to &radic;(n+t2) for all t in R. SL(n,R) is not a bounded subset of gl(n,R), and hence SL(n,R) is not compact. SL(n,R) is also connected, but the proof has been left to Marsden due to space constraints.

The section concludes with the following propostition:

The Lie group SL(n,R) is a non-compact connected (n2-1) dimensional Lie group whose Lie algebra sl(n,R) consists of the nxn matrices with trace 0 and bracket [A,B] = AB-BA.

Apologies for any typos/inaccuracies. I'm pretty new to this stuff.

[marcus]

I could not find any typos/inaccuracies
except for the one typo on an unintended extra t in "proposition"
"The section concludes with the following propostition:"
And I could not even find any instance of lack of clarity.
This is great. We might even have a Lie groups workshop
going. If the others, like Chroot and Hurkyl, keep in touch.

I am wondering what the others think would be good to do now.

One could look at what you just said and try to
say intuitively why those things are, in fact, the Lie algebra
of GL(n,R). How gl(n,R) really does correspond to infinitesimal transformations around the identity----and how it is the tangent space at the point on GL(n, R) which is the identity matrix.
Or one could do the same kind of concrete case investigation for SL(n, R) and its Lie algebra. I mean, try out and verify a few special cases and get our hands dirty.

Or, alternatively, we could move on to some more matrix groups like O(n) and SO(n), or begin looking at their complex cousins.

Or, if enough people were involved, we could go in both directions at once. Some could proceed to describe the other classic Lie groups and their algebras while others, like me, cogitate about the very simplest examples.

Let's see what happens. I hope something does.

[marcus]

In a previous post I was going thru a section of marsden
in that pages 283-292 part of chapter 9, and it mentioned
the exponential function defined by the power series
exp(x) = 1 + x + x2/2! +...
and gave a case of where you plug a matrix A in for x
and get a matrix exp(A)

this has always seemed to me like a cool thing to do
and I see it as illustrating a kind of umbilical connection between Lie algebra and Lie group.

The algebra element A is what gets plugged into exp() to give exp(A) which is in the group.

Or in more cagey differential geometry style----exp(tA) for t running from 0 to 1 gives you a curve down in the Lie group (the manifold) which starts out at the identity point and travels along in the manifold and reaches exp(A) as its destination. Indeed exp(tA) for t running forever gives a one-dimensional subgroup--but this is a bit too abstract for this time of morning.

What I always think is so great is that if A is 3x3 skew sym
matrix, meaning AT = -A
then plugging A into that good old exp() power series gives a rotation matrix, one of the SO(3) Lie group.

More wonderful still, exp(A) is the rotation by exactly |v| radians about the vector v = (v1, v2, v3) as axis where A is
given by

+0 -v3 +v2
+v3 +0 -v1
-v2 +v1 +0

any skew symmetric matrix would have such a form for some
v1,v2,v3

And we may be able to convince ourselves of this, or prove it a bit, without much effort, just by looking at the power series in A.

If I stands for the identity matrix,

B = exp(A) = I + A + A2/2! +...

Now consider that since AT = - A, we can take the transpose of this whole power series and it will be as if we put a minus sign in front of A.

BT = exp(A)T = exp(- A)

But multiplying exp(x) and exp( -x) always gives one. When you multiply the two power series there is a bunch of cancelation and it boils down to the identity. So exp (-A) is the matrix INVERSE of exp(A).

BT = exp(A)T = exp(- A) = exp(A)-1 = B-1

BT = B-1 means that B is orthogonal


BTW one reason to think about paths exp(tA) from the identity to the endpoint exp(A) is to see clearly that exp(A) is in the same connected component of the group. O(3) is split into two pieces, one with det = 1 and one with det = -1.

The latter kind turn your shirt inside out as well as rotating it, so they are bad mothers and it is generally safer to work with the det = 1 kind which are called "special" or SO(3).

this curve going t = 0 to 1 shows that exp(A) is in the same connected component as the identity, because how could the curve ever leap the chasm between the two components?
So it shows det A = 1. But that is just mathematical monkeyshines, of course the determinant is one! [;)]

All this stuff can be written with an n sitting in for 3, but
as an inveterate skeptic I often suspect that
dimensions higher than 3 do not exist and prefer to write 3 instead of n. It looks, somehow, more definite and specific that way.

We should check out the elementary fact that [A,B] works with
skew sym matrices A and B! Why not! Maybe later today, unless someone else has already done it.

I will bring along this earlier post with an extract from pages 289-291 of the book
**************************************
SO(3) is a compact Lie group of dimension 3.

Its Lie algebra so(3) is the space of real skew-symmetric 3x3 matrices
with bracket [A,B] = AB - BA.

The Lie algebra so(3) can be identified with R3
the 3-tuples of real numbers by a vectorspace isomorphism
called the"hat map"

v = (v1,v2,v3) goes to v-hat, which is a skew-symmetric matrix
meaning its transpose its its NEGATIVE, and you just stash the three numbers into such a matrix like:

+0 -v3 +v2
+v3 +0 -v1
-v2 +v1 +0

v-hat is a matrix and apply it to any vector w and
you get vxw.

Everybody in freshman year got to play with v x w
the cross product of real 3D vectors
and R3 with ordinary vector addition and cross product v x w is kind of the ancestral Lie algebra from whence all the others came.

And the hat-map is a Lie algebra isomorphism

EULER'S THEOREM

Every element A in SO(3) not equal to the identity is a rotation
thru an angle &phi; about an axis w.

SO SO(3) IS JUST THE WAYS YOU CAN TURN A BALL---it is the group of rotations

THE EIGENVALUE LEMMA is that if A is in SO(3) one of its
eigenvalues has to be equal to 1.
The proof is just to look at the characteristic polynomial which is of degree three and consider cases.

Proof of Euler is just to look at the eigenvector with eigenvalue one----pssst! it is the axis of the rotation. Marsden takes three sentences to prove it.

A CANONICAL MATRIX FORM to write elements of SO(3) in
is

+1 +000 +000
+0 +cos&phi; -sin&phi;
+0 +sin&phi; cos&phi;

For typography I have to write 0 as +000
to leave space for the cosine and sine under it
maybe someone knows how to write handsomer matrices?

EXPONENTIAL MAP
Let t be a number and w be a vector in R3
Let |w| be the norm of w (sqrt sum of squares)
Let w^ be w-hat, the hat-map image of w in so(3), the Lie algebra. Then:

exp(tw^) is a rotation about axis w by angle t|w|

It is just a recipe to cook up a matrix giving any amount of rotation around any axis you want.

[marcus]

sometimes just doing the routine checks is a good way to
get used to something. In the last post I was talking about
so(3) the skewsym matrices that are the Lie algebra of SO(3) the rotations and I said


"We should check out the elementary fact that [A,B] works with
skew sym matrices A and B! Why not! Maybe later today, unless someone else has already done it."

What I mean is just verify the extremely simple fact that
if you have skew sym A,B then the bracket [A,B] is also skew sym!

And there is also the dreaded jacobite identity to verify namely

[[A,B], C] + [[B,C], A] + [[C,A], B] = 0

this terrible formula can only be verified by those who have memorized the alphabet, at least up to C, and
in our culture very young children are made to recite the alphabet to ensure that when they reach maturity they will be able to
verify the Jacobi identity.

It is, you may have noticed the main axiom of abstract Lie algebra.

There are sort of two wrong approaches to anything, (1)purely axiomatic and (2)bloodyminded practical----really have to do both, if one is learning about concrete examples one should occasionally look around and verify that they satisfy the axioms too.

[Hurkyl]

whoa! is that true? i m not so sure. i think what you want to say here is:
[x,y]=-[y,x] is a consequence of [x,x]=0, and if the field does not have characteristic 2, then [x,y]=- [y,x] implies [x,x]=0, but not in fields with characteristic 2, so we drop that as an axiom.

Yes, that's essentially what I meant to say.

[Lonewolf]

I've got a proof for the Lie algebra of skew-symmetric matrices in three dimensions produces another three-dimensional skew-symmetric matrix since that's our focus, but it's pretty simplistic, and unelegant. Maybe someone has a better one.

Let A be defined as

(0 -a -b)
(a 0 -c)
(b c 0)

Let B be defined as

(0 -d -e)
(d 0 -f)
(e f 0)

Let g=-(ad+be), h=-(ad+ce), i=-(be+cf)

Then AB is

( g -bf af)
(-ce h -ae)
( cd -bd i)

And BA is

( g -ce cd)
(-bf h -bd)
( af -ae i)

So, [A,B]=AB-BA=

(0 ce-bf af-cd)
(bf-ce 0 bd-ae)
(cd-af ae-bd 0 )

Which is again a skew-symmetric matrix.

EDIT: Took chroot's advice.

[chroot]

The best way to render matrices here is to put them in a [ code ][ /code ] container, which preserves spacing:

( 0 ce-bf af-cd )
( bf-ce 0 bd-ae )
( cd-af ae-bd 0 )

- Warren

[Hurkyl]

For A and B skew symmetric matrices:

(AB - BA)t = (AB)t - (BA)t
= BtAt - AtBt
= (-B)(-A) - (-A)(-B)
= BA - AB
= -(AB - BA)

So the commutator of any two skew symmetric matrices is again skew symmetric.

In general, for any involution *:

[A, B]* = (AB-BA)*
= (AB)* - (BA)*
= B*A* - A*B*
= [B*, A*]

where [,] is the commutator

edit: fixed a formatting error

[marcus]

two good things just happened.
Lonewolf who is new to groups (background = one course in linear algebra) tackled it and proved it down-in-the-mud
and then Hurkyl proved it elegantly as a special
case of a more general fact that would include
the complex case of skew-Hermitian where you
take transpose and then complex conjugate of the matrix entries
can not restrain a broad grin
because both the dirtyhands approach and the elegant one
are indispensible
great


Originally posted by Hurkyl
For A and B skew symmetric matrices:

(AB - BA)t = (AB)t - (BA)t
= BtAt - AtBt
= (-B)(-A) - (-A)(-B)
= BA - AB
= -(AB - BA)

So the commutator of any two skew symmetric matrices is again skew symmetric.

In general, for any involution *:

[A, B]* = (AB-BA)*
= (AB)* - (BA)*
= B*A* - A*B*
= [B*, A*]

where [,] is the commutator

[lethe]

you know, this thread is turning into a pretty nice lie group/lie algebra thread. there is the differential forms thread. now all we need if for someone to start a representation theory thread, and we ll have all the maths we need to do modern particle physics.

who wants to volunteer?

[chroot]

I would absolutely love a rep theory thread -- especially if we could include both the down-n-dirty and the high-level approaches. I'm resonably competent to talk about Lie groups, but I am lost on representations.

- Warren

[lethe]

Originally posted by chroot
I would absolutely love a rep theory thread -- especially if we could include both the down-n-dirty and the high-level approaches. I'm resonably competent to talk about Lie groups, but I am lost on representations.

- Warren

i m down for the high level part.

[Lonewolf]

Sure, I'll have a go at representation theory. Even if I don't understand it all, I'm sure I'll get something out of it.

[lethe]

Originally posted by Lonewolf
Sure, I'll have a go at representation theory. Even if I don't understand it all, I'm sure I'll get something out of it.

lonewolf-

how much maths do you know? i don t think representation theory is all that hard. hang in there, i m sure we can get through it.

[Lonewolf]

I've covered the basics of group theory, and completed a course in linear algebra to be concluded next academic year. I'm pretty comfortable with the prerequisites you listed in the other thread. I'm willing to learn and I've got four months to fill, so I'm prepared to put some time in.

[lethe]

Originally posted by Lonewolf
I've covered the basics of group theory, and completed a course in linear algebra to be concluded next academic year. I'm pretty comfortable with the prerequisites you listed in the other thread. I'm willing to learn and I've got four months to fill, so I'm prepared to put some time in.
dat s good to hear!

[marcus]

I see several people are interested in group representations
and I'm thinking maybe we can just follow our interests.

I dont remember being part of an online study group and
dont have much idea of what works and what doesnt.

I propose chroot to be our nominal emcee or leader if we need one. But I dont care if we have a leader or are complete anarchy. And if somebody else is leader that is fine too.

Lonewolf defines the prerequisites, as I see it----one course in linear algebra and some time and willingness to work.

Why dont we see if we can get to some target in, say, the representation of some classic Lie group.

Maybe we will run out of gas halfway, but anyway we will have a destination.

What say this for a target-----classify the irreducible representations of SU(2). Can we get there from scratch?
Start with basic definitions and try to touch all the essential bases on the way?

I mention it because that target is highly visible. Maybe Hurkyl, or Chroot of Lethe can suggest a more practical goal.[
Having some goal will determine for us what things we have to cover, so we wont have to decide anything.

It might not matter what order we do things either.
Lethe for example could probably say right now what all the irred. reps of SU(2) are (up to isomorphism)

oops have to go

QUOTE]Originally posted by Lonewolf
I've covered the basics of group theory, and completed a course in linear algebra to be concluded next academic year. I'm pretty comfortable with the prerequisites you listed in the other thread. I'm willing to learn and I've got four months to fill, so I'm prepared to put some time in. [/QUOTE]

[lethe]

Originally posted by marcus

What say this for a target-----classify the irreducible representations of SU(2). Can we get there from scratch?
Start with basic definitions and try to touch all the essential bases on the way?

I mention it because that target is highly visible. Maybe Hurkyl, or Chroot of Lethe can suggest a more practical goal.[
Having some goal will determine for us what things we have to cover, so we wont have to decide anything.

It might not matter what order we do things either.
Lethe for example could probably say right now what all the irred. reps of SU(2) are (up to isomorphism)


a slightly more ambitious goal, that i would like to suggest, is the poincaré group SL(2,C)/Z2. it includes the rotation group as a subgroup (and thus includes all the concepts of SU(2), which would probably be a very good starting place), but it has a less trivial algebra, it is noncompact, so we can address those issues, and not simply connected, so we can also address those issues.

perhaps this is too ambitious. at any rate, SU(2) is a good starting point, and if that ends up being where we finish too, so be it.

[Hurkyl]

I too think going for the representations of SU(2) and SO(3) would be a good first goal, if only because of the importance of those groups in physics. In any case, that's the first goal I had set myself after I read that LQG primer. [:)]

[marcus]

Originally posted by Hurkyl
I too think going for the representations of SU(2) and SO(3) would be a good first goal, if only because of the importance of those groups in physics. In any case, that's the first goal I had set myself after I read that LQG primer. [:)]

Two online books have been mentioned.

Hurkyl I believe you indicated you were using Brian Hall
("An Elementary Introduction to Groups and Reps.")

That is 128 pages and focuses on matrix groups so it works
with a lot of concrete relevant examples. I really like it.

Earlier I was talking about Marsden's Chapter 9, and Lonewolf
extracted some stuff from that source and posted his notes,
I essentially did likewise with another patch of Marsden.

It would be helpful if we all had one online textbook to focus on.

I now think Brian Hall (your preference) is better adapted to people's interests and that maybe I goofed when I suggested Marsden.

I regret possibly causing people to waste time and printer paper printing off that long Chapter 9. I'm personally glad to have it for reference though, not the end of the world. But Brian Hall on balance seems better.

Lets see what theorems he needs to get the representations of SU(2). I mean---work backwards and figure out a route.

Brian hall's chapter 3, especially pp 27-37, seem to me to be grand central station.

chapter 3 is "Lie algebras and the exponential mapping"

He shows how to find the *logarithm* of a matrix
and he proves the useful formula

det exp(A) = exp( trace(A) )

and he proves the "Lie product formula"

and I can honestly say to Lonewolf that there is nothing scary here----nothing (that I can see with my admittedly foggy vision) that is fundamentally hard

(except at one point he uses the Jordan canonical form of a matrix----the fact that you can put it in a specially nice upper triangular form---which is a bit tedious to prove so nobody ever does they just invoke it. just one small snag or catch which we need not belabor)

It seems to me that to get where we want to go the main "base camp" destination is to show Lonewolf (our only novice and thus the most important person in a curious sense) the logarithm map that gets you from the group up into its tangent space (the algebra)
and the exponential map that gets you back down from the tangent space to the group

these are essentially the facts Brian Hall summarizes in the first 10 pages or so of Chapter 3 and then he gives a whole bunch of nice concrete examples illustrating it----pages 37-39.

Hurkyl, I am glad you mentioned Brian Hall's book

arXiv;math-ph/0005032

[marcus]

Have to say, if where we want to go is the representations of SU(2) that we can certainly take a peek at the destination
and it is lovely

just was glancing at Brian Hall's page 71

this being about five pages or so into his chapter 5 "Basic Representation Theory"

So simple!

SU(2) is just a nice kind of 2 x 2 matrices of complex numbers! We always knew that, but suddenly he does the obvious thing and uses a matrix U, or (just a slight variation on the idea) its inverse U-1 to STIR UP polynomials in two complex variables!

We have to be talking to an imaginary novice to define the level of explanation and for better or worse Lonewolf is standing in for that novice. I think this polynomial idea will make sense to him!

If you have a polynomial in two variables z1 and z2,
then you can, before plugging z1 and z2 into the polynomial,
operate on them with a 2 x 2 matrix!

This gives a new polynomial in effect. It is a sweet innocent obvious idea. Why not do this and get new polynomials?

And indeed the polynomials of any given combined degree in two variables are a vector space. So there we already have our group busily working upon some vectorspace and stirring the vectors around.

And to make everything as simple as possible we will consider only homogeneous polynomials of degree m
meaning that in each term the power z1 is raised to and the power z2 is raised to---those two powers add up to m.
It is a "uniformity" condition on the polynomial, all its terms have the same combined degree.

this must be the world's easiest way to come up with an action of SU(2) on an m+1 dimensional vectorspace. Must go back to the days of Kaiser Wilhelm.

a basis of our vectorspace Vm can consist of m+1 singletons like

(z1)2(z2)m-2

the coefficients can be complex numbers, it is a vector space over the complex numbers which may be somewhat less familiar than over the reals but still no big deal.

The official (possibly imaginary) novice may be wondering "what does irreducible mean". Indeed i hope Lonewolf is around and wondering this because we really need someone to explain to.
Well there is a group
and a mapping of the group into the linear operators on a vector space (some method for the group to act on vectors, like this scheme of using matrices to stir up polynomials)

that is called a representation (speaking unrigorously)
and it is irreducible if there is no part of the vectorspace left unstirred.
no subspace of V which is left invariant by the group.
no redundant part of V which doesnt get moved somewhere by at least one element of the group.

if there were an invariant subspace you could factor it out and
so-to-speak "reduce" the representation to a lower dimensional one.
so thats what irreducible means

it looks like these polynomials get pretty thoroughly churned around by preprocessing z1 and z2 with a matrix, but to be quite correct we need to check that they really are and that there is no invariant subspace.

******footnote*****

I think I said this before but just to be fully explicit about the action of the group:


If P(z1,z2) is the old polynomial, then the matrix U acts on it to produce a new polynomial by taking U-1 and acting on (z1, z2) to produce a new pair of complex numbers

(w1, w2) = U-1 (z1,z2)

and then evaluate the polynomial with (w1, w2):

P(U-1 (z1,z2) )

*****************
hope its not unwise to take a peek at the destination
first before trying to see how to get there
especially hope to get comments from Lethe Chroot Hurkyl
on how this should go, which theorems to hit, whether to have an orderly or random progression, whether Brian Hall gives a good focus etc.

[marcus]

Originally posted by lethe
a slightly more ambitious goal, that i would like to suggest, is the poincaré group SL(2,C)/Z2. it includes the rotation group as a subgroup (and thus includes all the concepts of SU(2), which would probably be a very good starting place), but it has a less trivial algebra, it is noncompact, so we can address those issues, and not simply connected, so we can also address those issues.

perhaps this is too ambitious. at any rate, SU(2) is a good starting point, and if that ends up being where we finish too, so be it.

first off, I would love it if you would do a whole bunch of explanation and get us started moving.
I tend to talk to much so I have to shut up and wait.
But I dont want this thread to get cold!

second. I totally agree. SU(2) and SO(3) are good initial targets but if it turns out to be fun to get to them then it would be
great to go on past to Poincare

I am counting (hoping) on you (plural) to explain the exponential map that connects the L.algebra to the L.group, because that seems to be crucial to everything including describing the reps

[marcus]

Hey Lonewolf, is there anything you need explained.

I wish Chroot or Lethe, both of whom could take over,
would take over and move this ahead.
I tend to talk too much and would like to be quiet for a while.

It is a good thread. It should do something.
What are you up to mathwise now its summer vacation?

[Lonewolf]

Please don't slow down the threads on my behalf. I'll be around, just nodding and smiling in the background.

[Lonewolf]

Explaining? Only the exponential map. I can't seem to see how it relates to what it's supposed to...maybe that gets explained further along in the text than I am, or I'm just missing the point.

[marcus]

Originally posted by Lonewolf
Please don't slow down the threads on my behalf. I'll be around, just nodding and smiling in the background.

OK I must have said something wrong and derailed the thread.
I have this fundamental fixed opinion that in any explanation the most important person is the novice and I cannot imagine having a explanation party about groups or lie algebras or anything else without one person who freely confesses to not knowing the subject.

Then you focus with one eye on the target (the theorems you want to get to) and with one eye on the novice

and you try to get the novice to the target destination

and the novice is also partly imaginary----the real one may get bored and go away meanwhiles.

but anyway that is how I imagine it. I cant picture doing groups with just Lethe and Chroot because they both already KNOW groups. Chroot is a tech Stanford student almost to his degree. Lethe is also clearly very capable and knowledgeable.

Dont sit in the background nodding for heavens sake. ASK these people to explain something to you. Well that is how I picture things and that is my advice. But who knows, it may all work out differently.

this is a great fact:

det( exp A) = exp (trace A)

do you know what det is and what trace is and do you know
what the exponential ex map is? I sort of assume so.
But if not then ask those guys and make them work it will be good for their mathematical souls.

[Lonewolf]

Could you elaborate on what you mean by

it is irreducible if there is no part of the vectorspace left unstirred.

and what an invariant subspace is, please?

[marcus]

Originally posted by Lonewolf
Explaining? Only the exponential map. I can't seem to see how it relates to what it's supposed to...maybe that gets explained further along in the text than I am, or I'm just missing the point.

You have had a mathcourse where they said

exp(t) = 1 + t + t2/2! + ......(you can continue this)

If not you will be hurled from a high cliff.

Suppose instead of 1 one puts the n x n identity matrix

and instead of t one puts some n x n matrix A.

At some time in our history someone had this fiendishly clever idea, put a matrix into the series in place of a number. It will converge and give a matrix.

But here is an easy question for YOU Lonewolf.

What if A is a diagonal matrix with say 1/2 all the way down the diagonal

then what is exp (A)?

Dont be reluctant to ask things. Dont wait for it to be "covered later". Any of us may fail to give a coherent answer but ask.

But now Im asking you, can you calculate that nxn, well to be specific call it 3x3, matrix exp(A). Can you write it down.

What is the trace of A
What is the determinant of exp A

If I am poking at you a little it is because I am in the dark about what you know and dont know.

[Hurkyl]

We're supposed to think of a Lie Group as a group of transformations with various properties. One of the more interesting properties is that we can form "one-parameter families" that have the property that:

T0 x = x
Ts Tt x = Ts+t x

We can think of the parameter as being the "size" of the transformation. An example will probably make this clear.


Consider R2, and let T&theta; be rotations around the origin through an angle of &theta;. Then, T0 is the identity transformation, and T&theta; T&phi; x = T&theta;+&phi; x, so rotations form a one-parameter family when parametrized by the angle of rotation.


Since we have this continuous structure, it's natural to extend the ideas of calculus to Lie Groups. So, what if we consider an infinitessimal transformation Tdt in a one-parameter family?

Let's do an example using rotations in R2. Applying rotation T&theta; can be expressed by premultiplying by the matrix:


/ cos &theta; -sin &theta; \
\ sin &theta; cos &theta; /


So what if we plug in an infinitessimal parameter? We get


/ cos d&theta; -sin d&theta; \ = / 1 -d&theta; \
\ sin d&theta; cos d&theta; / \ d&theta; 1 /

= / 1 0 \ + / 0 -1 \ * d&theta;
\ 0 1 / + \ 1 0 /


So the infintessimal rotations are simply infinitessimal translations. This is true in general; we can make locally linear approximations to transformations just like ordinary real functions, such as:

f(x + dx) = f(x) + f'(x) dx

We call the algebra of infinitessimal transformations a Lie Algebra.


The interesting question is how to go the other way. What if we had the matrix


/ 0 -1 \
\ 1 0 /


and we wanted to go the other way to discover this is the derivative of a family of transformations?

Well, integration won't work, so lets take a different approach; let's repeatedly apply our linear approximation. If X is our element from the lie algebra, then (1 + t X) is approximately the transformation we seek Tt. We can improve our approximation by applying the approximation twice, but each time half as long:

(1 + (t/2) X)2

And in general we can break it up into n legs:

(1 + (t/n) X)n

So then we might suppose that:

Tt = limn->&infin; (1 + (tX/n))n

And just like in the ordinary case, this limit evaluates to:

Tt = et X

That's from where the exponential map comes!

You can then verify that the derivitive of Tt at 0 is indeed t X


To summarize, we exponentiate elements of the Lie Algebra (iow apply an infinitessimal transformation an infinite number of times) to yield an elements of the Lie Group.



edit: fixed some hanging formatting tags

[marcus]

my browser draws a blank sometimes and shows boxes so I am
experimenting with typography a bit here. Nice post.
I dont seem able to get the theta to show up inside a "code" area. All I get is a box.

Well that is all right. I can read the box as a theta OK
Strange that theta shows up outside "code" area but not
inside

That is a nice from-first-principles way to introduce the
exponential of matrices.

Can you show

det exp(A) = exp (trace A)

in a similarly down-to-earth way?

I see it easily for diagonal matrices but when I thought about it I had to imagine putting the matrix in a triangular form

[marcus]

Lonewolf your job is to react when people explain something in a way you can understand. stamp feet. make hubub of some kind

You are about to see an example of a Lie algebra.

Hurkyl is about to show you what the L.A. is that belongs to the group of DET = 1 matrices for example SL(2, C).
The L.A. for SL(2,C) is written with lowercase as sl(2, C)

The L.G. of matrices with det = 1 is made by exponential map exp(A) from TRACE ZERO matrices A.

because exp(0) = 1.

So if Hurkyl takes one more step he can characterize the L.A.
of the group of det = 1 matrices.

Actually of any size and over the reals as well as the complexes I think. But just to be specific think of 2x2 matrices.

Lonewolf, do you understand this. Do you like it. I think it is terrific, like sailing on a windy day. L.G. and L.A. are really neat.

Well probably it is 4 AM in the morning in the UK so you cannot answer.

[Lonewolf]

If not you will be hurled from a high cliff.

I guess you don't have to bother coming over here and finding a high cliff then. [;)]

then what is exp (A)?

exp(A) =

(e1/2 0 0)
(0 e1/2 0)
(0 0 e1/2)


What is the trace of A

trace(A) = sum of diagonal entries = a11 + a22 + a33

What is the determinant of exp A

det[exp(A)] = trace(A)

[Lonewolf]

And in general we can break it up into n legs:

(1 + (t/n) X)n



This is pretty much when the penny dropped.

because exp(0) = 1.

This makes sense as well, and I can see where the exponential map is used now. Thanks.

[chroot]

det[exp(A)] = trace(A)
Very good! Except I think you mean:

det[exp(A)] = exp[trace(A)]

Probably a typo..

- Warren

[Lonewolf]

Oops, yeah. I probably should learn to read my posts...

[Hurkyl]

Well that is all right. I can read the box as a theta OK
Strange that theta shows up outside "code" area but not
inside

You're having font issues then. Your default font does indeed have the theta symbol, but the font your browser uses for the code blocks does not have a theta symbol (and replaces it with a box).


This is pretty much when the penny dropped.

Eep! I've never heard that phrase before, is that good or bad?



Can you show

det exp(A) = exp (trace A)

in a similarly down-to-earth way?

Nope. The only ways I know to show it are to diagonalize or to use the same limit approximation as above and the approximation:

det(I + A dt) = 1 + tr(A) dt

which you can verify by noting that all of the off diagonal entries are nearly zero, so the only important contribution is the product of the diagonal entries.


I think there's a really slick "down-to-earth" proof as well. I know the determinant is a measure of how much a transformation scales hypervolumes. (e.g. if the determinant of a 2x2 matrix near a point is 4, then applying the matrix will multiply the areas of figures near that point by 4) I know there's a nice geometrical interpretation of the trace, but I don't remember what it is.

[jeff]

Originally posted by Hurkyl
Nope. The only ways I know to show it are to diagonalize or to use the same limit approximation as above and the approximation:

det(I + A dt) = 1 + tr(A) dt

which you can verify by noting that all of the off diagonal entries are nearly zero, so the only important contribution is the product of the diagonal entries.

All that shows is that the formula holds to good approximation for matrices with elements that are all much less than one.

One correct proof goes as follows:

For any matrix A, there is always a matrix C such that CAC-1 is upper triangular meaning that all elements below the diagonal vanish. The key properties needed for the proof are that the space of upper triangular matrices are closed under matrix multiplication, and their determinants are the product of the elements on their diagonals. The only other thing we use is the invariance of the trace under cyclic permutations of it's arguments so that Tr(CAC-1) = TrA. The proof follows trivially.

[Hurkyl]

The proof to which I was alluding is:

det(eA) = det(limn->&infin;(I + A/n)n)
= limn->&infin; det((I + A/n)n)
= limn->&infin; (det(I + A/n))n
= limn->&infin; (1 + tr(A) / n + O(1 / n2))n
= etr(A)

[lethe]

another proof if you know some topology: diagonalizable matrices are dense in GL(n).

[Lonewolf]

Eep! I've never heard that phrase before, is that good or bad?

It's a good thing. We use it over here to mean the point where somebody realises something. Sorry about that, I thought it was in wider use than it is.

[marcus]

Originally posted by Lonewolf
It's a good thing. We use it over here to mean the point where somebody realises something. Sorry about that, I thought it was in wider use than it is.

I always assumed it was like the coin dropping in a payphone.
Maybe going back to old times when cooking gas was metered
out by coin-operated devices---the penny had to drop for something to turn on.

I have lost track of this thread so much has happened.

Just to review something:
A skewsymmetric means AT = - A
and a skewsymmetric matrix must be zero down the diagonal
so its trace is clearly zero, and another definition:
B orthogonal means BT = B-1

Can you prove that if
A is a skew symmetric matrix then exp(A) is orthogonal and
has det = 1?
I assume you can. It characterizes the Lie algebra "so(3)" that goes with the group SO(3). You may have noticed that they use lowercase "what(...)" to stand for the Lie algebra that goes with the Lie group "WHAT(...)"

Excuse if this is a repeat of something I or someone else said earlier.

[Hurkyl]

SO(3) is defined to be the space of all 3x3 real matrices G such that:

Gt = G-1
det G = 1

So what about its corresponding Lie Algebra so(3)? It is the set of all 3x3 matrices A such that exp(A) is in SO(3).

So how do the constraints on SO(3) translate to constraints on so(3)?

The second condition is easy. If A is in so(3), then:

exp(tr A) = det exp(A) = 1

so tr A must be zero. Conversely, for any matrix A with tr A zero, the second condition will be satisfied.


The first one is conceptually just as simple, but technically trickier. Translated into so(3) it requires:

exp(A)t = exp(A)-1
exp(At) = exp(-A)
*** this step to be explained ***
At = -A

Therefore if A is in so(3) then A must be skew symmetric. And conversely, it is easy to go the other way to see that any skew symmetric matrix A satisfies the first condition.

Therefore, so(3) is precisely the set of 3x3 traceless skew symmetric matrices.


I skipped over a technical detail in the short proof above. If exponents are real numbers then the marked step is easy to justify by taking the logarithm of both sides... however logarithms are only so nice when we're working with real numbers!!! I left that step in my reasoning because you need it when working backwards.

The way to prove it going forwards is to consider:

exp(s At) = exp(-s A)

If A is in so(3), then this must be true for every s, because so(3) forms a real vector space. Now, we differentiate with respect to s to yield:

(At) exp(s At) = (-A) exp(-s A)

Which again must be true for all s. Now, plug in s = 0 to yield:

At = -A

This trick is a handy replacement for taking logarithms!!!


Anyways, we've proven now that so(3) is precisely all 3x3 real traceless skew symmetric matrices. In fact, we can drop "traceless" because real skew symmetric matrices must be traceless.

For matrix algebras we usually define the lie bracket as being the commutator:

[A, B] = AB - BA

I will now do something interesting (to me, anyways); I will prove that so(3) is isomorphic (as a Lie Algebra) to R3 where the lie bracket is the vector cross product!


The first thing to do is find a (vector space) basis for so(3) over R. The most general 3x3 skew symmetric matrix is:


/ 0 a -b \
| -a 0 c |
\ b -c 0 /


Where a, b, and c are any real number. This leads to a natural choice of basis:


/ 0 0 0 \
A = | 0 0 -1 |
\ 0 1 0 /

/ 0 0 1 \
B = | 0 0 0 |
\ -1 0 0 /

/ 0 -1 0 \
C = | 1 0 0 |
\ 0 0 0 /


As an exercise for the reader, you can compute that:
AB - BA = C
BC - CB = A
CA - AC = B

So now I propose the following isomorphism &phi from so(3) to R3:

&phi;(A) = i
&phi;(B) = j
&phi;(C) = k

And this, of course, extends by linearity:

&phi;(aA + bB + cC) = ai + bj + ck


So now let's verify that this is actually an isomorphism:

First, the vector space structure is preserved; &phi is a linear map, and it takes a basis of the three dimensional real vector space so(3) onto a basis of the three dimensional real vector space R3, so &phi; must be a vector space isomorphism.

The only remaining thing to consider is whether &phi preserves lie brackets. We can do so by considering the action on all pairs of basis elements (since the lie bracket is bilinear)

&phi;([A, A]) = &phi(AA - AA) = &phi;(0) = 0 = i * i = [i, i] = [&phi;(A), &phi;(A)]
(and similarly for [B, B] and [C, C])
&phi;([A, B]) = &phi;(AB - BA) = &phi;(C) = k = i * j = [i, j] = [&phi;(A), &phi;(B)]
(and similarly for other mixed pairs)

So we have verified that so(3) and (R3, *) are isomorphic as Lie Algebras! If we so desired, we could then choose (R3, *) as the Lie Algebra associated with SO(3), and define the exponentional map as:

exp(v) = exp(&phi;-1(v))

So, for example:

exp(tk) = rotation of t radians in the x-y plane

[chroot]

Hurkyl:

Great post!

- Warren

[marcus]

Originally posted by chroot
Hurkyl:

Great post!

- Warren

I agree!

[Hurkyl]

Bah, there are no blushing emoticons!!

Thanks guys!


I'm not entirely sure where to go from here, though, since I'm learning it with the rest of you! (so if any of you have things to post, or suggestions on which way we should be studing, feel free to say something! [:)]) But I did talk to one of my coworkers and got a three hour introductory lecture on Lie Groups / Algebras in various contexts, and I think going down the differential geometry route would be productive (and it allows us to keep the representation theory in the representation theory thread!)... I think we are almost at the point where we can derive Maxwellean Electrodynamics as a U(1) gauge theory (which will motivate some differential geometry notions in the process), but I wanted to work out most of the details before introducing that.

Anyways, my coworker did suggest some things to do in the meanwhile; we should finish deriving the Lie algebras for the other standard Lie groups, such as su(2), sl(n; C), so(3, 1)... so I assign that as a homework problem for you guys to do in this thread! [:)]

[Lonewolf]

I think we are almost at the point where we can derive Maxwellean Electrodynamics as a U(1) gauge theory

As a what now?

[Hurkyl]

More talking with him indicates he may have been simplifying quite a bit when he brought up Maxwell EM. I'll let someone else explain what "gauge theory" means in general; I'm presuming I'll understand the ramifications after I work through the EM exercise, but I haven't done that yet. [:)]

[chroot]

Just to help motivate the thread, I'll find su(n).

Lie algebra of U(n)

First, as a reminder, we know that U(n) is the unitary group of n x n matrices. You should program the word 'unitary' into your head so it reminds you of these conditions:

1) Multiplication by unitary matrices preserves the complex inner product: <Ax, Ay> = <x, y> = [sum]i xi* yi, where A is any member of U(n), x and y are any complex vectors, and * connotes complex conjugation.

2) A* = A-1

3) A* A = I

4) |det A| = 1

Now, to find u(n), the Lie algebra of the Lie group U(n), I'm going to follow Brian Hall's work on page 43 of http://arxiv.org/math-ph/0005032

Recall that we can represent any1 member of a matrix Lie group G by an exponentiation of a member of its Lie algebra g. In other words, for all U in U(n), there is a u in u(n) such that:

exp(tu) = U

where exp is the exponential mapping defined above. Thus exp(tu) is a member of U(n) when u is a member of u(n), and t is any real number.

Now, given that U* = U-1 for member of U(n), we can assert that

(exp(tu))* = (exp(tu))-1

Both sides of this equation can be simplified. The left side's conjugation operator can be shown to "fall through" the exponential, and the left side is equivalent to exp(tu*). Similarly, the -1 on the right side falls through, and the right side is equivalent to exp(-tu). (Exercise: it's easy and educational to show that the * and -1 work this way.) We thus have a simple relation:

exp(tu*) = exp(-tu)

As Hall says, if you differentiate this expression with respect to t at t=0, you immediately arrive at the conclusion that

u* = -u

Matrices which have this quality are called "anti-Hermitian." (the "anti" comes from the minus sign.) The set of n x n matrices {u} such that u* = -u is the Lie algebra of U(n).

Now how about su(n)?

Lie algebra of SU(n)

SU(n) is a subgroup of U(n) such that all its members have determinant 1. How does this affect the Lie algebra su(n)?

We only need to invoke one fact, which has been proven above. The fact is:

det(exp(X)) = exp(trace(X))

If X is a member of a Lie algebra, exp(X) is a member of the corresponding Lie group. The determinant of the group member must be the same as e raised to the trace of the Lie algebra member.

In this case, we know that all of the members of SU(n) have det 1, which means that exp(trace(X)) must be 1, which means trace(X) must be zero!

You can probably see now how su(n) must be. Like u(n), su(n) is the set of n x n anti-Hermitian matrices -- but with one additional stipulation: members of su(n) are also traceless.

1You can't represent all group members this way in some groups, as has been pointed out -- but it's true for all the groups studied here.

- Warren

edit: A few very amateurish mistakes. Thanks, lethe, for your help.

[marcus]

The weather's been pretty hot and chroot's derivation of su(n) is really neat and clear so I'm thinking I will just be shamelessly lazy and quote Warren with modifications to get sl(n, C).

I see that he goes along with Brian Hall and others in using lower case to stand for the Lie Algebra of a group written in upper case. So su(n) is the L.A. that belongs to SU(n).

In accord with that notation, sl(n,C) is the L.A. that goes with the group SL(n,C), which is just the n x n complex matrices with det = 1. Unless I am overlooking something, all I have to do is just a trivial change in what Warren already did:

Originally posted by chroot, with minor change for SL(n, C)


Lie algebra of SL(n, C)

SL(n, C) is a subgroup of GL(n, C) such that all its members have determinant 1. How does this affect the Lie algebra sl(n, C)?

We only need to invoke one fact, which has been proven above. The fact is:

det(exp(X)) = exp(trace(X))

If X is a member of a Lie algebra, exp(X) is a member of the corresponding Lie group. The determinant of the group member must be the same as e raised to the trace of the Lie algebra member.

In this case, we know that all of the members of SL(n, C) have det 1, which means that exp(trace(X)) must be 1, which means trace(X) must be zero!
...sl(n, C) is the set of n x n complex matrices but with one additional stipulation: members of sl(n, C) are...traceless.


That didnt seem like any work at all. Even in this heat-wave.
Hurkyl said to give the L.A. of SO(3,1) so maybe i should do that to so as not to look like a slacker. Really like the clarity of both Hurkyl and Chroot style.

I guess Lethe must have raised the "topologically connected" issue. For a rough and ready treatment, I feel like glossing over manifolds and that but it is nice to picture how the det = 0 "surface" slices the GL group into two chunks...

Because "det = 0" matrices, being non-invertible, are not in the group!


...so that only those with det > 0 are in the "connected component of the identity". The one-dimensional subgroups generated by elements of the L.A. are like curves radiating from the identity and they cannot leap the "det = 0" chasm and reach the negative determinant chunk.

Now that I think of it, Lethe is here and he might step in and do SO(3,1) before I attend to it!

[marcus]

Hurkyl has a notion of where to go. I want to follow the hints taking shape here:
***********
....But I did talk to one of my coworkers and got a three hour introductory lecture on Lie Groups / Algebras in various contexts, and I think going down the differential geometry route would be productive (and it allows us to keep the representation theory in the representation theory thread!)... I think we are almost at the point where we can derive Maxwellean Electrodynamics as a U(1) gauge theory (which will motivate some differential geometry notions in the process), but I wanted to work out most of the details before introducing that.

Anyways, my coworker did suggest some things to do in the meanwhile; we should finish deriving the Lie algebras for the other standard Lie groups, such as SU(2), SL(n; C), SO(3, 1)... so I assign that as a homework problem for you guys to do in this thread!
***********
the suggestion is----discuss SO(3,1) and so(3,1). Then back to Hurkyl for an idea about the next step. Lets go with that.

Originally posted by chroot, changed to be about SO(3,1)


Lie algebra of SO(3,1)

SO(3,1) is just the group of Special Relativity that gets you to the moving observer's coordinates---it contains 4x4 real matrices that preserve a special "metric" dx2 + dy2 + dz2 - dt2

to keep the space and time units the same, distance is measured in light-seconds----or anyway time and distance units are made compatible so that c = 1 and I dont have to write ct everywhere and can just write t.

This "metric" is great because light-like vectors have norm zero. So the definition that a matrix in this group takes any vector to one of the same norm means that light-like stays light-like!

All observers, even those in relative motion, agree about what is light-like---the world line of something going that speed. (Another way of saying the grandfather axiom of SR that all agree about the speed of light.)

the (3,1) indicates the 3 plus signs followed by the 1 minus sign in the "metric".

So we implement the grand old axiom of SR by having this special INNER PRODUCT* in our 4D vector

1) Multiplication by SO(3,1) matrices preserves the special inner product: <Ax, Ay> = <x, y> = [sum]*i xiyi, where A is any member of SO(3,1), x and y are any real 4D vectors, and * is a reminder that the last term in the sum gets a minus sign.

2) This asterisk notation is a bit clumsy and what Brian Hall does instead is define a matrix g which is diag(1,1,1,-1).

g looks like the 4 x 4 identity except for one minus sign
BTW notice that g-1 = g
and also that gt = g

and he expresses the condition 1) by saying

At g A = g

[[[[to think about.....express <x,y> as xtg y
express <Ax, Ay> as xt At g A y]]]]

3) Then he manipulates 2) to give
g-1 At g = A-1

.....multiply both sides of 2) on the left by g-1
to give
g-1 At g A= I

then multiply both sides on the right by A-1.....

4) then---ahhhh! the exponential map at last----he writes 3) using a matrix A = expX, and solves for a condition on X


g-1 At g = g-1 exp(Xt) g = exp(g-1 Xt g ) = exp(- X) = A-1

the only way this will happen is if X satisfies the condition
g-1 Xt g = -X

it is something like what we saw before with SO(n) except gussied up with g, so it is not a plain transpose or a simple skew symmetric condition. also the condition is the same as

g Xt g = -X

because g is equal to its inverse.
Better post this and proofread later.

. BTW multiplying by g on right and left like that does not change trace, so as an additional check

trace(X) = trace(g Xt g) = trace( -X) = - trace (X)

showing that trace (X) = 0

so now we know what matrices comprise so(3,1)

they are the ones that satisfy

g Xt g = -X

[Lonewolf]

Not sure how relevant this is to where the thread is going, but I didn’t want people to think I’d given up on it.

The Heisenberg Group

The set of all upper triangular 3x3 matrices with determinant 1 coupled with matrix multiplication forms a group known as the Heisenberg Group, which will be denoted H. The matrices A in H are of the form


(1 a b)
(0 1 c)
(0 0 1)


where a,b,c are real numbers.

If A is in the form above, the inverse of A can be computed directly to be


(1 -a ac-b)
(0 1 -c )
(0 0 1 )


H is thus a subgroup of GL(3:R)

The limit of all matrices in the form of A is again in the form of A. (This bit wasn’t as clear to me as the text indicated. Can someone help?)

The Lie Algebra of the Heisenberg Group

Consider a matrix X such that X is of the form


(0 d e)
(0 0 f)
(0 0 0)


then exp(X) is a member of H.

If W is any matrix such that exp(tW) is of the form of matrix A, then all of the entries of W=d(exp(tW))/dt at t=0 which are on or below the diagonal must be 0, so W is of the form X.

Apologies for the possible lack of clarity. I kinda rushed it.

[Hurkyl]

I don't think I'll have time over the next week or so to prepare anything, so it'd be great if someone else can introduce something (or pose some questions) for a little while!

[marcus]

Originally posted by Hurkyl
I don't think I'll have time over the next week or so to prepare anything, so it'd be great if someone else can introduce something (or pose some questions) for a little while!

Hey Warren, any ideas?
Maybe we should hunker down and wait till
Hurkyl gets back because he seemed to give the
thread some direction. But on the other hand
we dont want to depend on his initiative to the
point that it is a burden! What should we do?

I am thinking about the Lorentz group, or that thing SO(3,1)
I discussed briefly a few days ago.
Lonewolf is our only audience. (in part a fiction, but one must
imagine some listener or reader)
Maybe we should show him explicit forms of matrices implementing the Lorentz
and Poincare groups.

It could be messy but on the other hand these are so
basic to relal speciativity. Do we not owe it to ourselves
to investigate them?

Any particular interests or thoughts about what to do?

[marcus]

If we were Trekies we might call it "the Spock algebra of the Klingon group" or if we were on firstname basis with Sophus Lie and Hendrik Lorentz we would be talking about
"the Sophus algebra of the Hendrik group"
such solemn name droppers... Cant avoid it.

Anyway I just did some scribbling and here it is. Pick any 6 numbers a,b,c,d,e, f
This is a generic matrix in the Lie algebra of SO(3;1):


0 a b c
-a 0 d e
-b -d 0 f
c e f 0


what I did was take a line from preceding post (also copied below)
g-1 Xt g = -X

remember that g is a special diagonal matrix diag(1,1,1,-1)

and multiply on both sides by g to get
Xt g = -gX

that says that X transpose with ritemost colum negged
equals -1 times the original X with its bottom row negged.

This should be really easy to see so I want to make it that way.
Is this enough explanation for our reader? Probably it is.

But if not, let's look at the original X with its bottom row negged



0 a b c
-a 0 d e
-b -d 0 f
-c -e -f 0


And let's look at the transpose with its ritemost column negged



0 -a -b -c
a 0 -d -e
b d 0 -f
c e f 0


And just inspect to see if the first is -1 times the second.
It does seem to be the case.

Multiplying by g on left or right does things either to the
bottom row or the rightmost column, I should have said at the beginning---and otherwise doesnt change the matrix.

Ahah! I see that what I have just done is a homework problem in Brian hall's book. It is exercise #7 on page 51, "write out explicitly the general form of a 4x4 real matrix in so(3;1)



Originally a chroot post but changed to be about SO(3;1)


Lie algebra of SO(3;1)

SO(3;1) is just the group of Special Relativity that gets you to the moving observer's coordinates---it contains 4x4 real matrices that preserve a special "metric" dx2 + dy2 + dz2 - dt2

to keep the space and time units the same, distance is measured in light-seconds----or anyway time and distance units are made compatible so that c = 1 and I dont have to write ct everywhere and can just write t.

1) Multiplication by SO(3;1) matrices preserves the special inner product: <Ax, Ay> = <x, y> = [sum]*i xiyi, where A is any member of SO(3,1), x and y are any real 4D vectors, and * is a reminder that the last term in the sum gets a minus sign.

2) This asterisk notation is a bit clumsy and what Brian Hall does instead is define a matrix g which is diag(1,1,1,-1).

g looks like the 4 x 4 identity except for one minus sign
BTW notice that g-1 = g
and also that gt = g

and he expresses the condition 1) by saying

At g A = g

3) Then he manipulates 2) to give
g-1 At g = A-1

4) then---ahhhh! the exponential map at last----he writes 3) using a matrix A = expX, and solves for a condition on X

g-1 At g = g-1 exp(Xt) g = exp(g-1 Xt g ) = exp(- X) = A-1

the only way this will happen is if X satisfies the condition
g-1 Xt g = -X

it is something like what we saw before with SO(n) except gussied up with g, so it is not a plain transpose or a simple skew symmetric condition. also the condition is the same as

g Xt g = -X

[Hurkyl]

I've been trying to devise a good way to introduce differential manifolds...

(by that I mean that I hate the definition to which I was introduced and I was looking for something that made more intuitive sense!)

I think I have a way to go about it, but it dawned on me that I might be spending a lot of effort over nothing, I should have asked if everyone invovled is comfortable with terms like "differentiable manifold" and "tangent bundle".

[marcus]

Originally posted by Hurkyl
I've been trying to devise a good way to introduce differential manifolds...

(by that I mean that I hate the definition to which I was introduced and I was looking for something that made more intuitive sense!)

I think I have a way to go about it, but it dawned on me that I might be spending a lot of effort over nothing, I should have asked if everyone invovled is comfortable with terms like "differentiable manifold" and "tangent bundle".

I like Marsden's chapter 4 very much
"Manifolds, Vector Fields, and Differential Forms"
pp 121-145 in his book----25 pages
His chapter 9 covers Lie groups and algebras, not too
differently from Brian Hall that we have been using.
So Marsden is describing only the essentials.
I will get the link so you can see if you like it.

Lonewolf and I started reading Marsden's chapter 9 before
we realized Brian Hall was even better. So at least two of us
have some acquaintance with the Marsden book.

We could just ask if anybody had any questions about
Marsden chapter 4----those 25 pages----and if not simply
move on.

On the other hand if you have thought up a better way
to present differential geometry and want listeners, go for it!
Here is the url for Marsden.

http://www.cds.caltech.edu/~marsden/bib_src/ms/Book/

[marcus]

H., I had another look at Marsden.
His chapter 9 is too hard and the book as a whole is
too hard. It is a graduate textbook.
But maybe his short chapter 4 on manifolds, vector
fields and differential forms is not too hard.
a short basic summary. It seems to me OK.
If you agree then perhaps this is a solution.
We dont have to give the definitions because
they are all summarized for us.

We should proceed only where it will give us pleasure,
and at our own pace, being under no obligation to anyone. If Lonewolf is still around we can provide whatever explanations
he asks for so he can keep up with the party. If we decide
it is time to stop we will stop (substantial ground has already
been covered). I shall be happy with whatever you decide.

I am interested to know if there are any matrix group, lie group,
lie algebra, repr. theory topics that you would like to hit.
E.g. sections or chapters of Brian Hall (or propose some other online text).

I am currently struggling to understand a little about spin foams
but can find no direct connection there to this thread.
Baez has a introductory paper gr-qc/9905087

[Hurkyl]

I've been thinking more about my idea of trying to derive Maxwell's equations from the geometry of M4*U(1) (M4=Minowski space)... the way the idea was presented to me, I got the impression it would be an interesting application of lie groups requiring just a minimal amount of differential geometry... but as I've been mulling over what we'd have to do to get there I'm thinking it might actually be an interesting application of differential geometry requiring just a minimal amount of lie groups. [:(]

So basically, I don't know where to go from here!


The way I usually like to learn is to delve a little bit into a subject, then figure out a (possibly almost trivial) concrete example of how the subject can be used to describe "real world" things, and then continue studying deeper into the subject. The problem is I just don't know what "real world" thing we can get to early on. I guess the solution is to just delve deeper into the math before looking back at the real world.

[marcus]

Originally posted by Hurkyl

The way I usually like to learn is to delve a little bit into a subject, then figure out a (possibly almost trivial) concrete example of how the subject can be used to describe "real world" things, and then continue studying deeper into the subject. The problem is I just don't know what "real world" thing we can get to early on. I guess the solution is to just delve deeper into the math before looking back at the real world.

I just happened onto a 3 page online account of
"Representation Theory of SL(2,C)"

It is an appendix in an 8 page paper by Perez Rovelli
"Spin Foam Model for Lorentzian General Relativity"

They lifted it from W. Ruhl (1970) "The Lorentz Group and Harmonic Analysis" and some other classical sources like that.

Baez also reviews SL(2,C) rep theory on page 4 of what I think is a great paper he wrote with Barrett, gr-qc/0101107.
That paper Baez and Barrett "Integrability for Relativistic Spin
Networks" is 22 pages but there is already a good bit of grist for the mill in just the first 4 or 5 pages.

If you have other directions in mind, drop a few hints and I will try to come up with source material.

Oh! we had better not forget to go over the irreps of SU(2)
Do you happen to have an online source? That is easier.
What was I thinking of! irreps of SU(2) naturally come well
before one tries SL(2,C).

think of something nice and simple, my brain is fried from spin foams and 10j symbols

[Lonewolf]

I'm still around. Can someone explain tangent bundles, please. Marsden defines them as the disjoint union of tangent vectors to a manifold M at the points m in M. Am I right in thinking that this gives us a set containing every tangent vector to the manifold, or did I miss something?

[lethe]

Originally posted by Lonewolf
I'm still around. Can someone explain tangent bundles, please. Marsden defines them as the disjoint union of tangent vectors to a manifold M at the points m in M. Am I right in thinking that this gives us a set containing every tangent vector to the manifold, or did I miss something?

that s almost right. the tangent bundle is every tangent vector at any point of the manifold, along with the manifold itself.

the tangent bundle is itself given the structure of a manifold.

[marcus]

Originally posted by Lonewolf
I'm still around. Can someone explain tangent bundles, please. Marsden defines them as the disjoint union of tangent vectors to a manifold M at the points m in M. Am I right in thinking that this gives us a set containing every tangent vector to the manifold, or did I miss something?

Such a good question! Thread would die without a questioner like that---Hurkyl and I, chroot and/or Lethe etc. wouldnt like to just talk to selves. I want to let Hurkyl answer this because will do it in clear orderly reliable manner.

But look, it is a necessary and basic construction! The tangent vectors on a manifold are the most important vectorspacetype things in sight!
And yet each tangentspace at each separate point is different. So at the outset all one has is a flaky disjoint collection of vectorspaces. One HAS to glue it all together into a coherent structure and give it a topology and, if possible, what is even better namely a differential structure.

Imagine a surface with a postagestampsized tangent plane at every point but all totally unrelated to each other. How flaky and awful! But now imagine that with a little thought you can merge all those tangentplanes into a coherent thing-----a dualnatured thing because it is both a linearspace (in each "fiber") and a manifold. Now I am imagining each tangentspace as a one-dimensional token (in reality n-dimensional) and sort of like a hair growing out of the point x in the manifold. All the hairs together making a sort of mat.

And these things generalize----not just tangentspace bundles but higher tensor bundles and Lie algebra bundles. Socalled fiber bundles (general idea). It is a great machine.

A vectorfield is a "section" of the tangent bundle. The graph of a function is a geometrical object and the "graph" of a vectorfield lives in the tangent bundle. A choice of one vector "over" each point x in the manifold. Great way to visualize things.

The problem is how to be rigorous about it!!!! Hurkyl is good at this. You get this great picture but how do you objectify gluing and interrelating all the tangent spaces into a coherent bundle and giving them usable structure.

It turns out to be ridiculously easy. To make a differentiable manifold out of anything you merely need to specify local coordinate charts. The vectorspace has obvious coordinates and around every point in the manifold you have a coordinatized patch so if it is, like, 3D, you have 3 coordinates for the manifold and 3 for the vectorspace. So you have 6 coordinates of a local chart in the bundle.

Charts have to fit together at the overlaps and the teacher wastes some time and chalk showing that the 6D charts for the bundle are smoothly transformable one to the other on overlapping territory------why? because, surprise, the original 3D manifold charts were smoothly compatible on overlaps.

You will see a bit of magic. An innocent looking LOCAL condition
taking almost no time to mention will unexpectedly suffice to make it all coherent. All that is needed is that, at every point, right around that point, the tangent bundle looks like a cartesian product of a patch of manifold and a fixed vectorspace.

what should I do, edit this? delete it? it is attitudinal prep before someone else writes out the definition (if they ever do) of a fiber bundle-----tangent bundle just a special case of fiber bundle
once that is done, can erase this----I dont want to bother editing it since just provisional. Glad yr still around LW

[marcus]

Lethe, chroot, Hurkyl
does everybody know the dodge used to represent the lorentz group on infinite dimensional function spaces?

I sort of suspect you all do.

Let's identify at least the proper lorentz group, the connected component of the identity or whatever, with SL(2,C)
and just look at SL(2,C)

to transform a function f(x) using a 2x2 complex matrix all you need to do is process the x with the matrix first before you feed it to the function

in a generalized sense it is SL(2,C) "acting by translation".

x ----> (ax + b)/(cx + d)

f[x] ----> f[(ax + b)/(cx + d)]

Lately when I see representations of SL(2,C) they mostly involve this action on functions by generalized "translation". So I assume its familiar to y'all.

And the vectorspace of the action is infinite dimensional. There isn't just a discrete set of reps labeled by integers or halfintegers, rather a whole slew labeled by the real line (and maybe another parameter as well).

something to prove here, namely that composing
two maps of the form
x ----> (ax + b)/(cx + d)
results in one of that form
and if you do things in the right order it really gives a group representation
maybe someone should state the definition so we could check details like that?

[Hurkyl]

Doh, and I thought I had weaseled out of explaining fiber bundles!

Disclaimer: the terminology "fiber bundle" is somewhat new to me... although I've known the idea behind them for quite a while. It is possible I have some subtle detail wrong about them. (really, studying this whole subject with rigor is new to me... I just had the good fortune to have gotten partway through Brian Hall's text and another one on differential geometry before this discussion started! It's fortunate that much or even all of differential geometry is ideas we've had since our calculus days... just phrased in nifty ways that make them precise, easy to use, and generalizable)


Let's start with something mundane. Everyone remember studying calculus of a single variable? [:)]

In this context we were primarily concerned with functions in the domain of the real numbers and range in the real numbers. Real numbers have some very nice properties (i.e. they form a differentiable manifold), and we were primarily concerned with how the nice properties of the domain and range interacted with our single variable functions. For example, a continuous function is (essentially) one that preserves the property of nearness in the domain and range. A differentiable function relates the differential structure of the domain to that of the range.

So, our studies began by considering certain functions that map R to R. The "total space" of functions was simply R*R; the set that contains all possible graphs of functions.


Eventually our studies became more sophisticated. We no longer considered x an independant variable and y a dependant variable, but we treated x and y both as fully fledged variables. It became interesting to study the structure of R*R as a differentiable manifold in its own right (though, of course, we didn't call it that!).


These same ideas can be applied to any 2 differentiable manifolds; all of our ideas from calculus still apply (but may be a little tricker) when we're considering functions from, say, R2 to SO(3)! And just like in the single variable case, it pays to also be able to consider the total space as a structure on its own merit.


A fiber bundle is the abstraction that covers the above notions. We have a manifold M (analogous to a domain) and a fiber F (analogous to a range). We consider a "total space" E which looks locally like M*F. More precisely, that means we have a projection mapping &pi; that projects E onto M... so that for any neighborhood U of M, &pi;-1(U) is isomorphic (or isometric or diffeomorphic or whatever) to U*F. That is, the set of all points in E that map onto U must have the same structure as U*F.

Why do we only ask this to hold locally? Why not just take the total space to be equal to U*F? (that would be called a trivial bundle... and incidentally, the local isomorphisms from E to U*F are called trivializations) Well, it helps to consider what appears to be everyone's favorite first nontrivial fiber bundle... the mobius strip!

Recall the procedure for making a mobius strip: you take a long thin rectangle of paper, you loop it around into a cylinder, but then before pasting the ends together so it really is a cylinder, you twist the strip one time, and you get a mobius strip.

But what does this have to do with fiber bundles?

Well, we can describe the strip as being:

S = [0, 10]*[-1, 1]

"Looping" the strip means topologically identifying the two ends of the strip {0}*[-1, 1] and {10}*[-1, 1]. The "domain" (aka base manifold) here is [0, 10] with 0 and 10 identified; this is just the circle S1. The fiber is [-1, 1]. This pasting gives us the cylinder S1*[-1, 1]... a trivial fiber bundle.

But wait! There are two (topologically distinct) ways we can map [-1, 1] onto [-1, 1]; instead of using the identity map f(x)=x, use the map f(x)=-x. This corresponds to the "twist" we use when making a mobius strip. This fiber bundle is (obviously) different from the cylinder. Describing the mobius strip as S1*[-1, 1] does not work anymore...

Presume it does work. Then define f from S1 to [-1, 1] as:

f(x) = 1

This is a constant function, and clearly continuous... but it is not a continuous function on the mobius strip! (please ignore the minor details that would take too much effort to patch this into a perfectly rigorous demonstration)


A "section" is simply a fancy name for a graph of a function! But recall that the total space E cannot always be decomposed into M*F... so "section" only coincides with "graph" locally. More precisely, a section S is simply a surface in the total space E such that for every point x on the manifold M, there is a unique point y on S such that &pi;(y) = x.



For a n-dimensional manifold M, the tangent bundle is a special fiber bundle where the fiber is Rn; for any point x on M, &pi;-1(x) = TMx. In other words, the total space E is simply the collection of the tangent spaces for every point on the manifold. The associated projection map &pi; takes each tangent space onto the point on the manifold to which it is tangent. (alternatively, &pi;-1 inverse takes each point on the manifold to its associated tangent space)


Question: is the tangent bundle a trivial bundle? In other words, is the tangent bundle TM diffeomorphic to M*Rn? (I don't know the answer to this one... I presume it is yes, but I haven't tried to prove or disprove it yet)



Anyways, to summarize:

A fiber bundle is a generalization of the cross product; it permits the resulting structure to "twist" as it goes around the base manifold so that the net result does not globally have the same structure as M*F (though it does locally). A section S is the corresponding generalization of a graph of a function; to each point x on the manifold corresponds a unique point y of S, and the projection onto M of y is simply x. A tangent bundle is simply one where the fibers are the tangent spaces of M.

Lonewolf, you were correct (I believe) in your summary of what a tangent bundle is... but the important thing is that bundles also has properties related to the structures of M and F, be it merely topological, differential, metric, or even a linear structure in the case of vector spaces.


Sorry this isn't as clear as my other ones, I hadn't prepared any nice examples and demonstrations of the concepts. [:(]

[Hurkyl]

I'm not familiar with that construction (as I mentioned, I'm still somewhat new to this subject)... but I have seen that action before, but in the context of complex analysis as the mobius transformation (and, before that, I learned about the roughly equivalent notion of inversions in the Euclidean plane).

[marcus]

this is a quality treatment

I'll try to think about your question. Lethe probably could reassure us on this point, or else knows a counterexample.
<<Question: is the tangent bundle a trivial bundle? In other words, is the tangent bundle TM diffeomorphic to M*Rn? (I don't know the answer to this one... I presume it is yes, but I haven't tried to prove or disprove it yet)
>>
Thanx for doing bundles!


Hey Hurkyl, I found a GREAT site about knots and the Jones
polynomial. Anyone including a highschool kid could quickly learn from this site how to calculate the jones of a a trefoil knot. It is an AMS website and it is classy.

http://www.ams.org/new-in-math/cover/knots2.html

[lethe]

Originally posted by Hurkyl

For example, a continuous function is (essentially) one that preserves the property of nearness in the domain and range.

essentially? this is the exact definition of a continuous mapping: one that takes close points to close points. you just need to find a precise way to formulate a notion of closeness.


Question: is the tangent bundle a trivial bundle? In other words, is the tangent bundle TM diffeomorphic to M*Rn? (I don't know the answer to this one... I presume it is yes, but I haven't tried to prove or disprove it yet)

obviously. any vector space is easily isomorphic and diffeomorphic to Rn. which gives you a vector bundle morphism.

[marcus]

Originally posted by lethe
essentially? this is the exact definition of a continuous mapping: one that takes close points to close points. you just need to find a precise way to formulate a notion of closeness.


obviously. any vector space is easily isomorphic and diffeomorphic to Rn. which gives you a vector bundle morphism.

Merci monsieur Lethe for both comments! I believe Hurkyl is going to point to the next direction for the thread to go, but do you have any suggestions? I feel all of us can be proud of this
congenial and useful thread and would be glad to hear of ideas for things that the thread might do. Tho at moment it is Hurkyl's choice.

BTW Lethe, in that "Knot and Jones" thread I could not see how to introduce the orientation idea without being able to draw better pictures and without sounding pedantic by introducing
lots of words like "righthanded, lefthanded". So what I calculated is only correct up to orientation. Just a first look. I expect you know something about the Jones polynomial and would be happy if you want to edit or emend what I wrote. Feel free if you can improve it.

[Hurkyl]

essentially? this is the exact definition of a continuous mapping: one that takes close points to close points. you just need to find a precise way to formulate a notion of closeness.

Posting late at night saps the urge to prove things that seem obvious. [:)] Preserving nearness is the same as preserving limits (I guess commuting with limits is the right phrase), which is the very definition of a continuous function.



obviously. any vector space is easily isomorphic and diffeomorphic to Rn. which gives you a vector bundle morphism.


I'm not convinced... we can modify the mobius strip construction so the fiber is R, so then the mobius strip is a vector bundle over S1, but is clearly not topologically equivalent to S1*R.

[lethe]

Originally posted by Hurkyl

I'm not convinced... we can modify the mobius strip construction so the fiber is R, so then the mobius strip is a vector bundle over S1, but is clearly not topologically equivalent to S1*R.

ehh... insert the word locally about three places in my post. locally trivial, locally diffeomorphic, etc. then we should be in business

[Hurkyl]

ehh... insert the word locally about three places in my post. locally trivial, locally diffeomorphic, etc. then we should be in business

I knew it was locally trivial; I was wondering about globally... I presume from the last part of jeff's post that it is not globally true in general.

[jeff]

Originally posted by Hurkyl
I presume from the last part of jeff's post that it is not globally true in general.

You are correct sir. For example, no spheres save for S1,3,7 are parallelizable.

[Hurkyl]

Aha, I see!

IIRC, any smooth tangent vector field on the sphere S^2 must contain a zero vector. However, it is trivial to find a smooth section of S^2*R^2 that is nonzero everywhere... therefore T(S^2) cannot be (globally) diffeomorphic to S^2*R^2 and thus is not parallelizable.

[Hurkyl]

Originally posted by jeff
Triviality and principle bundles:

A bundle E is trivial iff it's associated principle bundle P(E) - obtained from E by replacing it's fibre with it's structure group - has a "cross-section", that is, a continuous map s: B &rarr; E satisfying &pi;s(x) = x, x &isin; B. In the case of the mobius strip, if &theta; is a local coordinate on S1, then we must have s(&theta;) = s(&theta;+2&pi;) (for purposes of illustration, we ignore the fact that because each coordinate chart covers less than 2&pi; radians, local coordinates on S1 should not be allowed to exceed this). However, G = Z2 means that P(E) is a double cover of S1, so we can't have s(&theta;) = s(&theta;+2&pi;) unless s jumps discontinuously between corresponding points on the two "branches".


Hrm... Z2 is also the structure group of the cylinder, right? This proof needs to also take into account the twist in the construction of the mobius strip so that P(E) is a connected double cover, right? It seems that you would need to use this fact to prove P(E) is a double cover, so you might as well use this fact by itself to show that s(&theta;) != s(&theta;+2&pi;)

[marcus]

Originally posted by Hurkyl


Question: is the tangent bundle a trivial bundle?



the simplest counterexample is probably the 2D sphere S2 because you can't comb the hair on a billiard ball

It is a famous diff/geom theorem that any vectorfield on the sphere must be zero at at least one point


but if the tangent bundle on the sphere were isomorphic (as bundle) to the cartesian product of the sphere with the plane R2

then one could define a vectorfield or "section" of the bundle by giving every point the same vector (1,0)
and it would map to a never-vanishing vectorfield on the sphere
contradiction
[6)]

I actually put this 2D sphere counterexample in my first post
replying to Hurkyl's where he originally asked the question
but erased it I guess before anyone read it---not feeling completely confident about the definitions

this is where it matters what the definition of a vector bundle morphism is------I theeenk. It ought to act like a linear map upstairs and a diffeo downstairs

havent read jeff's weighty contribution, maybe it says something about this?

[marcus]

Originally posted by jeff
You are correct sir. For example, no spheres save for S1,3,7 are parallelizable.

Ahah! so jeff did make this point! Regret to say I just got back
and have not been keeping up.

[Hurkyl]

Doh! Diffeomorphism is not the right word for a vector bundle morphism. Bad Hurkyl!

[marcus]

Originally posted by Hurkyl
Doh! Diffeomorphism is not the right word for a vector bundle morphism. Bad Hurkyl!

Good Hurkyl!
I liked your treatment of tangent bundles
and was in doubt myself about the definitions which
is why i erased mention of sphere in my initial reply
Am not too concerned with semantics in any case
morphism schmorphism

I trust your judgement about what is a relaxed
not-overly-technical level of discussion and what
would be useful to discuss. Where shall we go next?

Or do we wait till Lonewolf asks another question?

[Hurkyl]

Well, the problem was that I was actually thinking diffeomorphism; I wasn't just using the word because it has "morphism" in it!


As for where to go next, I'm wondering if everyone wanted to stick primarily to lie groups, or if we want expand our goal to study differential geometry in more detail as well.


Anyways, now that we know what the tangent bundle is, I can submit the next homework problem my coworker suggested! (and finally get back to lie groups! [;)])


Suppose M is a differential manifold and f is a morphism of M into itself. The differential structure of M allows us to define a function (*f) on T(M) that acts as a morphism (*f)x from Tx(M) to Tf(x)(M) for every x in M.

Informally, f(x + dx) = f(x) + (*f)x(dx)

More precisely, for any x on M, define (*f)x as follows:

For any v in Tx(M), choose a smooth curve &gamma; through x whose tangent vector at x is v. Then, define (*f)x(v) to be the tangent vector to f(&gamma;) at f(x). (proof that this is well-defined is left to the reader! I've always wanted to say that!)

(Of course, you could do it much more easily by using coordinate charts... but I've been making a conscious effort to avoid using coordinate charts whenever possible because, IMHO, they obscure the geometric meaning behind everything)

An invariant tangent vector field (with respect to a group G of automorphisms of M) is one that is unchanged after applying elements of G. IOW, for a vector field V and a group element g, (*g)(V) = V. Alternatively. (*g)x(V(x)) = V(g(x))


Now, suppose M is a lie group. Since M is a group, we are given a natural class of automorphisms; those of M acting on itself by left multiplication (also by right multiplication)! For an element g of M, define:

Lg : M -> M : h -> gh

That is, Lg is the "left multiplication by g" operator.

Define Rg similarly to be the right multiplication operator.


Finally, let E be the identity element of M.


Problem 1: Prove that there is a one to one correspondence between TE(M) and the set of all tangent vector fields invariant under left multiplications. (called left invariant vector fields)

Problem 2: Prove that right multiplication maps left invariant vector fields to left invariant vector fields.

(there is an exercise 3 that goes with this problem set, but we haven't talked about Adjoint mappings)

[jeff]

Originally posted by Hurkyl
...T(S^2) cannot be (globally) diffeomorphic to S^2*R^2 and thus is not parallelizable.

It's S2 that's not parallelizable implying the stronger statement that T(S2) and S2xR2 aren't homeomorphic.

Originally posted by Hurkyl
Hrm... Z2 is also the structure group of the cylinder, right?

No, trivial bundles have trivial structure groups, that's the whole point.

Originally posted by Hurkyl
This proof needs to also take into account the twist in the construction of the mobius strip so that P(E) is a connected double cover, right? It seems that you would need to use this fact to prove P(E) is a double cover, so you might as well use this fact by itself to show that s(&theta;) != s(&theta;+2&pi;)

Was it not obvious that P(E)'s connectedness was used? Sorry. What I meant was going once round the open set S1 traces a closed arc in P(E) beginning at one point of a fibre and ending at the other point of the same fibre so that s-1 maps a closed set to an open set and so isn't continuous.

[marcus]

Problem 1: Prove that there is a one to one correspondence between TE(M) and the set of all tangent vector fields invariant under left multiplications. (called left invariant vector fields)
---------------------
Both TE(M) and the set of LIVFs are vector spaces and this 1-1 corresp will turn out to be a linear isomorphism (so they are essentially the same as vector spaces)

you already told us about how a mapping f : M--->M has a lift *f up to the tangent space level Tx(M) --->Ty(M) which we now apply to a manifold which is a group G.

and you described the right and left multiplication maps Rg and Lg : G ---> G

So we can use the lifts of those maps, like for example *(Lg)

Now as to Problem 1, for any v in TE(M) let's define a LIVF denoted by Xv
g ---> *(Lg) v

this is a vector field which at a point g in G has a vector which is the image of v by the lift of the left multiplication map that goes from the group identity element to g.

I just need to show that this vector field is left invariant so I study
Xv(h) where h is in G. If I left multiply by g, I get
Xv(gh) and to show left invariance
I have to show this is the same as *(Lg)Xv(h)
This is just your definition of left invariance, shifting around on the group level has to have the same effect as lift-mapping upstairs in the tangent spaces.

But by how Xv was defined in the first place
Xv(gh) = *(Lgh) v
= *(Lg) *(Lh) v ...[[]]
= *(Lg) Xv(h) ...[[[by Xv definition]]]

I think its clear that the correspondence here is linear----adding vectors v and v' in the tangent space at the group identity will correspond to adding left invariant vector fields Xv and Xv' just by the linear way the fields were defined.

All I really have left to do is exhibit the inverse of this map. Given a LIVF, say call it X(g), how do I go back to a tangent vector at the identity. Well it is obvious. Just take X(e), the field's value AT the identity.

footnote, there is that chain rule thing. Lifts preserve the composition of mappings, and specializing that to the case of left multiplication mappings we have that the original mappings
compose groupishly---I'm showing composition of maps denoted by the little o symbol.

Since for any k in G, associativity gives us (gh)k = g(hk) we have
Lgh = Lg o Lh
and that extends to the tangent spaces because of the chain rule
*(Lgh) = *(Lg) o *(Lh)

---------------------------
Problem 2: Prove that right multiplication maps left invariant vector fields to left invariant vector fields.
---------------------------

Well suppose we have vectorfield X(g) which is left invariant and we
apply Rh to it in some sensible way that produces a new vector field Y(g)
A sensible way to define Y(g) might be

Y(g) = *(Rh) X(g h-1)

So let us check if this is left invariant by an action Lk

Is it true that Y(kg) = *(Lk) Y(g)?

Well Y(kg) = *(Rh) X(kg h-1)
= *(Rh) *(Lk) X(g h-1)...[[[by left invariance of X]]]
= *(Lk) *(Rh) X(g h-1) ...[[[commutativity]]]
= *(Lk) Y(g)

which was to be proved.

footnote for any two elements of the group k and h
right and left multiplication by them commute
Lk Rh = Rh Lk
I guess that is obvious k(gh) = (kg)h assoc. and this
commuting business goes upstairs to the lifted right and left multiplications maps
*(Lk) *(Rh) = *(Rh) *(Lk)
so there [;)]


Originally posted by Hurkyl
[B]

Suppose M is a differential manifold and f is a morphism of M into itself. The differential structure of M allows us to define a function (*f) on T(M) that acts as a morphism (*f)x from Tx(M) to Tf(x)(M) for every x in M.

Informally, f(x + dx) = f(x) + (*f)x(dx)

More precisely, for any x on M, define (*f)x as follows:

For any v in Tx(M), choose a smooth curve &gamma; through x whose tangent vector at x is v. Then, define (*f)x(v) to be the tangent vector to f(&gamma;) at f(x). (proof that this is well-defined is left to the reader! I've always wanted to say that!)

(Of course, you could do it much more easily by using coordinate charts... but I've been making a conscious effort to avoid using coordinate charts whenever possible because, IMHO, they obscure the geometric meaning behind everything)

An invariant tangent vector field (with respect to a group G of automorphisms of M) is one that is unchanged after applying elements of G. IOW, for a vector field V and a group element g, (*g)(V) = V. Alternatively. (*g)x(V(x)) = V(g(x))


Now, suppose M is a lie group. Since M is a group, we are given a natural class of automorphisms; those of M acting on itself by left multiplication (also by right multiplication)! For an element g of M, define:

Lg : M -> M : h -> gh

That is, Lg is the "left multiplication by g" operator.

Define Rg similarly to be the right multiplication operator.


Finally, let E be the identity element of M.


Problem 1: Prove that there is a one to one correspondence between TE(M) and the set of all tangent vector fields invariant under left multiplications. (called left invariant vector fields)

Problem 2: Prove that right multiplication maps left invariant vector fields to left invariant vector fields.

(there is an exercise 3 that goes with this problem set, but we haven't talked about Adjoint mappings)

[lethe]

Originally posted by Hurkyl

As for where to go next, I'm wondering if everyone wanted to stick primarily to lie groups, or if we want expand our goal to study differential geometry in more detail as well.


at the risk of sounding self-serving, let me say: yes, continue this conversation, but don t do it in the group theory thread, do it in my differential forms thread!

no, seriously though, don t worry about keeping your conversation "on topic". just let it go where it goes. i like the dynamic of this board a lot.

[Hurkyl]

at the risk of sounding self-serving, let me say: yes, continue this conversation, but don t do it in the group theory thread, do it in my differential forms thread!

Actually, your thread is the main reason I didn't want to go deep into differential forms in this one. [:)]



It's S2 that's not parallelizable implying the stronger statement that T(S2) and S2xR2 aren't homeomorphic.

Yah, I was using (and thinknig) the wrong word. [:(]


No, trivial bundles have trivial structure groups, that's the whole point.

For the cylinder, the principle bundle is S1*Z2, a trivial (and disconnected) one.


Was it not obvious that P(E)'s connectedness was used?

I know you were using it, I was remarking that it was yet to be proven... and the only method I saw for proving it could have itself proved the fact you were using P(E)'s connectedness to prove.


Edit: fixed typo; I meant to have S1 for the base space of the cylinder

[jeff]

Originally posted by Hurkyl
For the cylinder, the principle bundle is S2*Z2, a trivial (and disconnected) one.

No, the cylinder's structure group is trivial so it's principle bundle is just it's base space S1.

Originally posted by Hurkyl
I was remarking that it [P(E) is connected] was yet to be proven

This needs no proof since it's the transition functions that encode topology and P(E) by definition has the same ones as E.

[Hurkyl]

My typo of writing S2 for S1 aside...


What's the definition of a structure group? I had presumed it was the group that preserved the structure of the fiber (i.e. diffeomorphisms for diff. manifolds, isometries for metric spaces, et cetera)... so if I used the same fiber for the cylinder (instead of orienting the fiber) I should have the same structure group.


Spivak's treatment of the mobius strip goes:


Consider, in particular, the Mobius strip as a 1-dimensional vector budle &pi;:E&rarr;S1 over S1. A frame in a 1-dimensional vector space is just a non-zero vector, so F(E) consists of the Mobius strip with the zero-section deleted. This space is connected (cut a paper Mobius strip along the center if you don't believe it); more generally, a vector bundle &pi;:E&rarr;M over a connected space M is orientable if and only if F(E) is disconnected.


F(E) is a principle bundle, so principle bundles aren't always connected spaces. For the cylinder E with the same fiber, F(E) would have to be disconnected.

[jeff]

Originally posted by Hurkyl
My typo of writing S2 for S1 aside...


What's the definition of a structure group? I had presumed it was the group that preserved the structure of the fiber (i.e. diffeomorphisms for diff. manifolds, isometries for metric spaces, et cetera)... so if I used the same fiber for the cylinder (instead of orienting the fiber) I should have the same structure group.


Spivak's treatment of the mobius strip goes:


Consider, in particular, the Mobius strip as a 1-dimensional vector budle &pi;:E&rarr;S1 over S1. A frame in a 1-dimensional vector space is just a non-zero vector, so F(E) consists of the Mobius strip with the zero-section deleted. This space is connected (cut a paper Mobius strip along the center if you don't believe it); more generally, a vector bundle &pi;:E&rarr;M over a connected space M is orientable if and only if F(E) is disconnected.


F(E) is a principle bundle, so principle bundles aren't always connected spaces. For the cylinder E with the same fiber, F(E) would have to be disconnected.

Spivak is right about Gcylinder = Z2. What I tried to do was avoid this by taking fibres to be unoriented line segments instead of vector spaces, but I realize now that they get flipped anyway. See my "Revised overview of fibre bundles" post below for detailed responses to all of your questions. In particular, I show how the structure group is obtained by explicitly constructing it for my cylinder and mobius strip examples. I also construct their principle bundles and that of T(S1). I think my treatment should make the significance of the structure group and the transition functions fairly clear.

[Hurkyl]

If we define AG(H) to be GHG-1 for G and H in a lie group, we can define:

Ad G = *AG

to be the adjoint map on the lie algebra.


Problem 3 is to prove that right multiplication by G on left invariant vector fields is the same as to applying Ad G to the equivalent lie algebra element.



I was trying to hold off to introduce another fact about the adjoint map, but I haven't worked out the proof yet (except for when M is a matrix lie group)...

Ad is a mapping from the lie group G to the group of linear transformations on its lie algebra GL(g). From this we can lift a new map ad from the tangent bundle of G to the tangent bundle of GL(g)... in particular, it maps g to gl(g).

The goal is to prove that the adjoint map ad satisfies the axioms of a lie bracket so that we may define:

[f, g] = (ad f)g

Which justifies our calling the tangent space at the identity (alternatively, the space of all left invariant vector fields) a lie algebra.

[marcus]

this Problem 3 Hurkyl mentioned now seems like an urgent and critical part of the program. Its like Lie algebras are gradually emerging out of the unknown. First the Tangent space of a manifold appears, And then a group that is a manifold.
And then the Tangent space at that group's Identity!

And then we discover that TeG (the tangent space at the group identity element) is linearly isomorphic to the set of all Left Invariant Vector Fields living on the group itself.

At this point then, Hurkyl says what the goal is:

<<The goal is to prove that the adjoint map ad satisfies the axioms of a lie bracket so that we may define:

[f, g] = (ad f)g

Which justifies our calling the tangent space at the identity (alternatively, the space of all left invariant vector fields) a lie algebra. >>

For me this represents the Lie algebra looming up out of nothingness in a kind of natural way as the tangent space at the identity except it is beginning to grow and morph an algebraic structure with a kind of "bracket" operation and "adjoint" map that plain old vectors dont ordinarily have. So to keep it growing and morphing we should (according to Hurkyl) do a Problem 3:

<<Problem 3 is to prove that right multiplication by G on left invariant vector fields is the same as to applying Ad G to the equivalent lie algebra element.>>

Everybody who studies basic Group theory (not just Lie Groups but finite groups) learns that about the most important thing in groups is the "inner automorphism"

g ---> hgh-1

and indeed this is what is used to define socalled "normal" subgroups and that ultimately is how you classify all possible
crystals and symmetries and all possible finite groups and all that jazz.

Hurkyl wants us to look at the lift of "inner automorphism"

Oh, he calls the lift of inner automorphism by h the ADJOINT map using h.

Well OK.

and this is going to engender the Lie bracket and cultivate the algebraic structure on Te

So we better get on with it and do Problem 3

I'm busy now but may have a moment later in afternoon
however anyone who wants should go ahead



Originally posted by Hurkyl
If we define AG(H) to be GHG-1 for G and H in a lie group, we can define:

Ad G = *AG

to be the adjoint map on the lie algebra.


Problem 3 is to prove that right multiplication by G on left invariant vector fields is the same as to applying Ad G to the equivalent lie algebra element.



I was trying to hold off to introduce another fact about the adjoint map, but I haven't worked out the proof yet (except for when M is a matrix lie group)...
The goal is to prove that the adjoint map ad satisfies the axioms of a lie bracket so that we may define:

[f, g] = (ad f)g

Which justifies our calling the tangent space at the identity (alternatively, the space of all left invariant vector fields) a lie algebra.
Ad is a mapping from the lie group G to the group of linear transformations on its lie algebra GL(g). From this we can lift a new map ad from the tangent bundle of G to the tangent bundle of GL(g)... in particular, it maps g to gl(g).

[marcus]

Just to get my bearings, the tangent space at a point is essentially equivalence classes of curves thru that point----two curves being equivalent if taking the derivative along them at the point gives the same answer. There is a kind of convergence of views on this, a few posts ago Hurkyl was saying:

<<...For any v in Tx(M), choose a smooth curve &gamma; through x whose tangent vector at x is v. Then, define (*f)x(v) to be the tangent vector to f(&gamma;) at f(x). ..>>

And IIRC Lethe was defining tangent space in the diff forms thread in a comparable way----the directions of directional derivative
And eg Marsden chapter 4 page 123 says much the same.
Anyway whatever the fine print of the definition says I will consider the tangent space to be equiv classes of curves, because I want to be able to pick a representative of the equiv class and take the derivative along that curve.

So then with &phi; some map M--->M it is easy to define the lift Tx&phi; or *&phi;: Tx --->T&phi;(x).

Given v in Tx pick a representative curve &psi; from the equiv classs and just compose mappings to get a new curve
&phi;(&psi;) passing thru &phi;(x) and take its equiv class which will be a vector belonging to the target tangent space T&phi;(x)

Some people say equiv classes of curves and differentiate along them and other people define tangent vectors in other equivalent ways but it all comes to the same thing.

THE POINT IS YOU ALWAYS HAVE A JACOBI LIE BRACKET. If X and Y are tangent vector fields on a manifold they for any smooth function f there is always an obvious
meaning to the derivatives X[f] and Y[f] which are some new smooth functions on the manifold. So one can do it in either order and define [X,Y] [f] = XY[f] - YX[f].

This seems kind of easy and direct, so where does it get hard if it ever does?

It must be when M turns into a group G as well as a manifold. then we have concepts like "Left Invariant Vector Field" and tangent space not just anywhere but at the identity, and inner automorphisms of the group, and lifting that to the "Adjoint" map which is a kind of stirring around or automorphism of the tangent space at the identity, and so on.

And also, dont forget, we can always go back and fetch the primitive old JACOBI LIE BRACKET which is just switching the order of differentiation w/rt a couple of vectorfields and then we
have something to prove which is that the brack of left invariants is left invariant and that ADJOINT which is a group-type thing gives the same as the jacobi lie bracket and allemand left dosiedo up the middle. Anyway thats how I see it.

So I am going to repeat the first two problems I proved for homework, without proof, just in case they are needed and then
go on to look at adjoint map.


Problem 1: Prove that there is a one to one correspondence between Te(M) and the set of all tangent vector fields invariant under left multiplications. (called left invariant vector fields)
---------------------------
Problem 2: Prove that right multiplication maps left invariant vector fields to left invariant vector fields.
---------------------------

Problem 3 is to prove that right multiplication by g on left invariant vector fields is the same as to applying Ad(g) to the equivalent lie algebra element, i.e to the equivalent tangent vector at the identity.

In other words we have a Left Invariant field X defined on G and there is the one-one correspondence to Te given by the laughably obvious X(e), the value of the field at the identity.
And for any g in G there is the adjoint map Ad(g) which is a way of mucking around with the tangent space at the identity.

And we want to see what doing that Ad(g) corresponds to in the world of Left Invariant vectorfields.


The inner aut map G ---> G is just h--->ghg-1 and
the lift of that is clearly *Lg*Rg-1

Problem 3 says to take a L.I. field X and operate with Ad(g) on X(e)

OK

*Lg*Rg-1 X(e)

*LgX'(eg-1) [[[X' is also left invariant]]]

X'(geg-1) = X'(e)

Darn, I have to go, but I think this is problem 3

have to get back to this and check and maybe edit.

This step *Rg-1 X(e)
corresponded to doing right mult by g to the invariant vector field X and getting an invariant field X'

And I calculated the Ad(g) of X(e)
and it turned out to give the same answer.
However must check this later since I have to go.

[Hurkyl]

Grr, I forgot why I wanted to bring up the differential geometry in the first place! Anyways, I'm kinda stuck on the adjoint thing, so someone want to introduce representations while I try to develop enough of the geometry to continue that track? (I'm probably gonna check out Vol I of Spivak's diff. geom text now too for this thread; so much for my plan to dive right in with curvature)

[marcus]

Originally posted by Hurkyl
Grr, I forgot why I wanted to bring up the differential geometry in the first place! Anyways, I'm kinda stuck on the adjoint thing, so someone want to introduce representations...

I think this means shifting to Brian Hall page 41 and page 68.

Good thing about Hall is no manifolds, no differential geometry, just plain old matrices! A lot of what they want to make happen in great generality and abstraction is just what happens naturally and concretely with matrices.

If you want, I'll discuss Hall pages 41 and 68, and then we would have the option to continue from there if you so choose. On page 41 Hall says:

"The following very important theorem tells us that a Lie group homomorphism between two Lie groups gives rise in a natural way to a map between the corresponding Lie algebras..." Isomorphic groups have isomorphic algebras...

Is this obvious or did you discuss it earlier and I just forgot? Please tell me, before I start proving it, if this is just repetitive or obvious. Here is the statement (Hall's Theorem 3.18)

Let G and H be matrix Lie groups, with Lie algebras g and h. Let &phi; :G --> H be a Lie group homomorphism. Then there exists a unique real linear map &phi;*: g --> h,
such that for all X in g we have

&phi:(exp(X)) = exp (&phi;*(X)).

Moreover this unique real linear map &phi;* has certain properties which I will list, if this has not been covered yet, and the star operation is compatible with the composition of mappings
(&phi; o &psi;)* = &phi;* o &psi;*

Hurkyl I mention this only because you asked someone to temporarily take the initiative going towards representations. You have the baton the moment you want to resume directing the band.

[marcus]

This theorem summarizes some things we have already discussed on this thread like the exponential map and like
one parameter subgroups exp(tX)
the way you actually compute &phi;*(X) is to take the
derivative at t = 0 of &phi;(exp(tX))

This is so obvious! You just use &phi;, since it is a group homomorphism, to map a one-parameter subgroup of one into a one-parameter subgoup of the other-----and an element of the algebra is always the infinitesimal move belonging to some one-parameter subgroup



(Hall's Theorem 3.18, restated with some more detail)

Let G and H be matrix Lie groups, with Lie algebras g and h. Let &phi; :G --> H be a Lie group homomorphism. Then there exists a unique real linear map &phi;*: g --> h,
such that for all X in g we have

&phi;(exp(X)) = exp (&phi;*(X)).

Moreover this unique real linear map &phi;* has certain properties:

1. For all X in g and all A in G,
&phi;*(AXA-1) = &phi;(A) &phi;*(X) &phi;(A-1)

2. For all X, Y in g,
&phi;*(XY - YX) = &phi;*(X) &phi;*(Y) - &phi;*(Y)&phi;*(X)

this is to say that &phi;* commutes with taking brackets or the "adjoint" map, whatever, namely
&phi;*([X,Y]) = [&phi;*(X), &phi;*(Y)]


3. The bedrock fact no matter what anybody says, is that the lifted map takes infinitesimal moves into the corresponding infinitesimals, namely,
&phi;*(X) = d/dt|t = 0 &phi;(exp(tX))

and FINALLY that the star operation is compatible with the composition of mappings
(&phi; o &psi;)* = &phi;* o &psi;*

[Hurkyl]

It makes sense, but isn't entirely obvious. The * here seems to be the same * I introduced in the geometrical context... but we haven't proved much about * in that context either.

I don't mind someone else leading; I'm usually more comfortable playing second fiddle anyways!

Besides, you seem to know the first round of details for representations and I don't, so it'd be better for you to lead that part anyways. [:)]

[marcus]

Originally posted by Hurkyl
It makes sense, but isn't entirely obvious. The * here seems to be the same * I introduced in the geometrical context... but we haven't proved much about * in that context either.

I don't mind someone else leading; I'm usually more comfortable playing second fiddle anyways!

Besides, you seem to know the first round of details for representations and I don't, so it'd be better for you to lead that part anyways. [:)]

You are still stuck with the job leading. I am only interjecting this because you asked for someone to cover for you for a moment.
Dont try to wiggle out. I am even fonder of secondfiddle than you and you really are more generally competent. I am reckless at times but do not mistake that for confidence[;)]
Also I flatly deny knowing whatever you are trying to insinuate that I know. However what I do think is that this thread has to be fun! If it is not we should stop whenever.

Come to think of it, I should make proving properties 1. 2. and 3. mentioned above into homework. When you assigned some things about tangent mappings as homework, earlier, I filled in the details. could you deal with those three properties of &phi;* in some fashion. A line or two of proof or a reference to some page in Hall or whatever seems judicious and perspicacious?

I wonder if Lonewolf is still around and has questions?

OH, ABOUT THE ASTERISK!!! I realize the ambiguity caused by this usage. Brian Hall uses a squiggle tilda over the phi. But I cannot type this. I tried typing various things and they looked too messy and ad hoc. So I finally concluded that I had to use asterisk, EVEN THOUGH you had already used it in a diff geometry context as notation for something else.

[Hurkyl]

I think I see why I'm having difficulties; to take the geometric approach means to work out tons of details that are "obvious" yet nontrivial to prove.


(In the following, all derivatives are to be taken at 0)

Anyways, proofs of properties 1-3. Using the fact &phi;* is linear and properties of the exponential we remember from earlier:

(1)

&phi;*(AXA-1) = (d/dt) exp(t &phi;*(AXA-1))
= (d/dt) exp(&phi;*(tAXA-1))
= (d/dt) &phi;(exp(tAXA-1))
= (d/dt) &phi;(A exp(tX) A-1)
= &phi;(A) (d/dt)&phi;(exp(tX)) &phi;(A-1)
= &phi;(A) (d/dt)exp(&phi;*(tX)) &phi;(A-1)
= &phi;(A) (d/dt)exp(t&phi;*(X)) &phi;(A-1)
= &phi;(A) &phi;*(X) &phi;(A-1)

(3)

&phi;*(X) = (d/dt) exp(t &phi;*(X))
= (d/dt) exp(&phi;*(tX))
= (d/dt) &phi;(exp(tX))

(&phi;&psi;)*(X) = (d/dt) exp(t (&phi;&psi;)*(X))
= (d/dt) exp((&phi;&psi;)*(tX))
= (d/dt) (&phi;&psi;)(exp(tX))
= (d/dt) (&phi;)(&psi;(exp(tX))
= (d/dt) (&phi;)(exp(&psi;*(tX)))
= (d/dt) exp(&phi;* &psi;* (tX))
= (d/dt) exp(t &phi;* &psi;* (X))
= &phi;* &psi;* (X)

(2)'s a little messier, I'll get it tomorrow unless Lonewolf polishes it off in the meanwhile.

Anyways, there is no ambiguity in the use of *; it's the exact same operator in both contexts.

In the first identity in problem (3), notice that exp(tX) is a curve with tangent vector X @ t = 0, and &phi;*(X) is defined to be the tangent vector @ t = 0 to the image of exp(tX) under &phi;... that's precisely how we defined (*&phi;) in the geometric context!

[marcus]

As usual you came through in spades, points 1-3 are proven.
Also you indicate here what is quite true, that we have been chewing over the same material----the exponential map, the logarithm of a matrix (which you defined earlier by a limit as I recall), the one parameter subgroup which is, by golly, a curve, and its derivative or tangent vector at the identity----in various different forms. At least I think we have been doing essentally that for a while. Maybe this theorem 3.18 of Hall can give us a place from which to move onwards.

Originally posted by Hurkyl
...
In the first identity in problem (3), notice that exp(tX) is a curve with tangent vector X @ t = 0, and &phi;*(X) is defined to be the tangent vector @ t = 0 to the image of exp(tX) under &phi;... that's precisely how we defined (*&phi;) in the geometric context!

There are still two parts to theorem 3.18 which I did not ask anyone to prove and I am going to nonchallantly leave them without proof. Anyone who wants can look it up in Hall.

The unproven parts are:
&phi;* exists and is a unique real linear map: g --> h,

and also that (&phi; o &psi;)* = &phi;* o &psi;*

The proof involves stuff we have already been doing lots of, you define phi-star in a by-now-very-familiar way by saying: "take X in g that we want to define phi-star of, and make a one parameter subgroup exp(tX) which you can think of as a curve of matrices in G passing thru the identity matrix
and use phi to MAP THIS WHOLE CURVE into the matrix group H.
and since phi is a smooth group homomorphism the image is a nice smooth curve passing thru the identity in the matrix group H.
And then as destiny decrees you just look at the tangent vector of that curve up in the tangent space of matrices h, and that is some matrix and you call THAT matrix = &phi;*(X)

then you have to check that this map is linear between the two vectorspaces (of matrices) g --> h, which just means trying it out with a scalar multiple rX and with a sum X+Y, and you have to check that it is the unique linear map that commutes with exponentiation namely
&phi;(exp(X)) = exp (&phi;*(X))
each of which little facts Brian Hall proves in one line on page 42 or 43 in case anyone wants to check up.

Now I think we can move on and see where this theorem and the discussion surrounding it have gotten us. In a way all the theorem does is work matrix multiplication into a familiar geometry picture

the geometry picture is two manifolds and a smooth map phi: M--->N that takes point x --->y

and we add just one thing to the picture namely that M and N are now matrix groups and x and y are the group identities (that is identity matrices) and phi is now a homomorphism----it preserves matrix multiplication.

this is just a tiny embellishment of the basic geometry picture and we want to know what happens with the lifted map of the Tangent spaces Tx ---> Ty

It is only natural to ask what happens when the smooth group homomorphism is lifted to the tangentspace level and the answer is this theorem which says that all is as well behaved as one could wish

not only is the thing linear and uniquely defined and consistent with the exponential map and one parameter subgroups (which are curves thru the identity) but we even get a bonus that the
map commutes with a certain "multiplication-like" operation upstairs called the [X,Y].

phi-star doesnt commute with ordinary matrix multiplication, it commutes with bracket. This is how god and nature tell us that we must endow the tangent space at the group identity with an algebraic structure involving the bracket.

We are predestined to do this because IT, the bracket, is what the lift of a group homomorphism preserves and it does not preserve anything else resembling multiplication.

And it is a linear map on tangent spaces so it preserves addition, so it is telling us what a Lie algebra is, namely vectorspace ops plus bracket----and whatever identities the bracket customarily obeys.

well thats one way to look at it. sorry if I have been long-winded.

now we can try a long jump to theorem 5.4 on page 68, which talks about Lie algebra representations, or else in a more relaxed frame of mind we can scope out some of the followup stuff that comes right after this theorem 3.18

oh, theorem 3.34 about the "complexification" of a real Lie algebra seems like a good thing to mention. sometimes we might need to drag in complex numbers to get some matrix diagonalized or solve some polynomial or for whatever reason and there is a regular proceedure for complexifying things when and if that is needed

well that is certainly enough said about theorem 3.18




(Hall's Theorem 3.18, restated with some more detail)

Let G and H be matrix Lie groups, with Lie algebras g and h. Let &phi; :G --> H be a Lie group homomorphism. Then there exists a unique real linear map &phi;*: g --> h,
such that for all X in g we have

&phi;(exp(X)) = exp (&phi;*(X)).

Moreover this unique real linear map &phi;* has certain properties:

1. For all X in g and all A in G,
&phi;*(AXA-1) = &phi;(A) &phi;*(X) &phi;(A-1)

2. For all X, Y in g,
&phi;*(XY - YX) = &phi;*(X) &phi;*(Y) - &phi;*(Y)&phi;*(X)

this is to say that &phi;* commutes with taking brackets or the "adjoint" map, whatever, namely
&phi;*([X,Y]) = [&phi;*(X), &phi;*(Y)]


3. The lifted map takes infinitesimal moves into the corresponding infinitesimals, namely,
&phi;*(X) = d/dt|t = 0 &phi;(exp(tX))

and FINALLY that the star operation is compatible with the composition of mappings
(&phi; o &psi;)* = &phi;* o &psi;*

[Lonewolf]

Sorry about the lack of input, I'm busier than I expected I would be at work. I'm switching to a part time position in two weeks, so I'll be able to input more then. I need to work on this more than I have time for at the moment as a lot of it is completely new to me. I'll work on (2) of theorem 3.18 tonight, and I'll post if I get anywhere.

[Lonewolf]

&Phi;^([X,Y]) = [&Phi;^(X),&Phi;^(Y)]

Using the fact that

[X,Y]=d(exp(tX)*Y*exp(-tX))/dt at t=0

we can define

&Phi;^([X,Y])=&Phi;^(d(exp(tX)*Y*exp(-tX))/dt) at t=0 = d(exp(tX)*Y*exp(-tX))/dt at t=0

since a derivative commutes with a linear transformation.

From (1) of theorem 3.18 we thus obtain

&Phi;^([X,Y]) = d(&Phi;(exp(tX)*&Phi;^Y*&Phi;(exp(-tX)))/dt at t=0 = d(exp(&phi;(X))*&Phi;^Y*exp(-&phi;(X))/dt, at t=0 = [&Phi;(X),&Phi;(Y)]

by our definition of [X,Y]

Hurkyl was right, this was a messy one. Sorry!

[marcus]

Hello LW, no apologies! great you are still on hand despite summer job etc. I believe if OK with you I will edit your post to remove asterisks used as multiplications signs. We are over-using the asterisk round about now---Hurkyl and I have been using it to denote a map lifted up from the manifold or basic group level to the tangent space level. And you are using caret ^ for that!!! So although caret is a good thing to use, asterisk is not a good thing to use for multiplication. Tempest in a teapot and really no confusion, but I will edit your post to conform and hope you are not vexed by my taking the liberty:

Originally posted by Lonewolf
&Phi;^([X,Y]) = [&Phi;^(X),&Phi;^(Y)]

Using the fact that

[X,Y]=d(exp(tX)Yexp(-tX))/dt at t=0

<<<<fact easy to prove by product rule of differential calculus, I will prove it soon unless someone else does>>>>

we can define

&Phi;^([X,Y])=&Phi;^(d(exp(tX)Yexp(-tX))/dt) at t=0 = d&Phi;((exp(tX)Yexp(-tX))/dt at t=0

since a derivative commutes with a linear transformation.

From (1) of theorem 3.18 we thus obtain

&Phi;^([X,Y]) = d(&Phi;(exp(tX)&Phi;^Y&Phi;(exp(-tX)))/dt at t=0 = d(exp(t&Phi;^(X))&Phi;^Yexp(-t&Phi;^(X))/dt, at t=0 = [&Phi;^(X), &Phi;^(Y)]

by our definition of [X,Y]
...

[marcus]

By slogging thru Hall's notation and variants of that we may
hope to eventually see how to deal elegantly with this whole business, anyway here suddenly the ordinary freshman calculus product rule appears like a lighthouse in a fog

Lonewolf says:<<Using the fact that

[X,Y]=d(exp(tX)Yexp(-tX))/dt at t=0 ...>>>

So group (exp(tX)Y) as your function f
and make exp(-tX)) your function g
and calculate d/dt of fg
(fg)' = f'g + f g'


And f' turns out to be XY (because multiplying by Y after you take the derivative is just a linear thing that doesnt disturb differentiation

And g, evaluated at t = 0 is just 1!!!

So f'g is just equal to XY


and how about f g' ?

Well f evaluated at t = 0 is just Y (because exp(0 X) is the identity.)

And g' equals (- X)

So f g' equals - YX

so (fg)' does in fact turn out to be XY - YX = [X, Y]

Its good time and weather for a barbecue right now so I am
taking off for a friend's house, be back later.

[Lonewolf]

I will edit your post to conform and hope you are not vexed by my taking the liberty:

Not at all. I just finished typing the post when I realised I'd unwittingly used an asterix for both multiplication and the map, so ^ seemed the best candidate for a replacement.

[marcus]

Originally posted by Lonewolf
Not at all. I just finished typing the post when I realised I'd unwittingly used an asterix for both multiplication and the map, so ^ seemed the best candidate for a replacement.

Do you see anything that needs clearing up (with the theorem we just discussed or related matters) or shall we see where we can go from here?

there should be stuff we can derive from this "theorem 3.18" of Hall

it has a nice wide-gauge feel to it------saying that group morphisms lift up to tangentspace level and turn into algebra morphisms (linear maps that preserve bracket)

maybe I am not saying this altogether clearly but it seems as if we ought to be able to get some mileage out of the work we have done so far

but also, since we have jumped into this in a very rough and somewhat improvisational way we may have left gaps where you would like more discussion. if so please say

[jeff]

What books are you guys using?

[Hurkyl]

"An Elementary Introduction to Groups and Representations" by Brian C. Hall, @ arXiv:math-ph/0005032 v1


A coworker recommeded I check out "A Comprehensive Introduction to Differential Geometry" volumes I and II by Mike Spivak for the geometric side of the topic (which I have done).

[MathematicalPhysicist]

i hope no one has asked them:
1. what is quantum group theory and with what does it deal with?
2. what are the differences between simple group theory and quantum group theory?

[Lonewolf]

Take a look at

http://www.maths.qmw.ac.uk/~majid/bkintro.html

[marcus]

Originally posted by Lonewolf
Take a look at

http://www.maths.qmw.ac.uk/~majid/bkintro.html

Hello Lonewolf and LoopQG,
I remember looking at some of majid's pages and getting the
impression that he was pushing his book, understandably, and not revealing very much of the subject matter. I may have missed something but I came away dissatisfied.

There is an australian account
http://www-texdev.mpce.mq.edu.au/Quantum/Quantum/Quantum.html
I cannot recommend it, except to say that it tries to be a regular online book about quantum groups. It is not selling anything, but is giving it away free.

I am not recommending that anyone try to learn quantum groups either--since it seems arcane: up in the realms of Category Theory and Hopf Algebras.

But there is a nagging fascination about the subject. There is this parameter "q" which if it is very close to zero the quantum group is almost indistinguishable from a group. And one hears things, like:

In cosmology there is an extremely small number which is
1.3 x 10-123 and is the cosmological constant
(a gossamer-fine energy density thoughout all space) expressed in natural units.
In one of his papers John Baez suggested that if you take q = the cosmological constant and use a quantum group tweaked by q instead of a usual group then something works that wouldnt if you used the usual group.
Tantalizing idea, that something in nature might deviate from being a straightforwards symmetry group by only one part in
10123.

I hate to be a name-dropper but quantum groups come up in the context of Chern-Simons q. field theory. Just another straw in the wind.


On another topic altogether, sometimes people say "quantum group theory" to mean simply ordinary Lie Groups etc. applied to quantum physics! That is "quantum group theory" is just the group theory that one employs in quantum mechanics and the like. These then are true groups---good solid law-abiding citizens of group-dom, just doing their job and helping physics out.

But what the folk in High Abstract Algebra call a "quantum group"
is a different kettle of fish. Those babies dont even have a group inverse---instead they have something that is almost but not quite an inverse called an "antipode". Make sure you still have your wristwatch after you shake hands with one of them.

[Lonewolf]

Make sure you still have your wristwatch after you shake hands with one of them.

Not to mention your internal organs...

Where do you recommend the thread should go from here? Are we at a stage where we can start applying some of what we've covered, or not?

[marcus]

Originally posted by Lonewolf
Not to mention your internal organs...


that had me laughing out loud, unfortunately it does seem
to have an element of truth---"quantum groups" proper does
seem a mathematically quite advanced subject.

Originally posted by Lonewolf
Where do you recommend the thread should go from here? Are we at a stage where we can start applying some of what we've covered, or not?

I defer to Hurkyl. If his job allows him time to think of a possible sortie we could make from here, and he wants to initiate it, then it will happen.

Or, as you have done in the past, you could try asking a specific question...

[Hurkyl]

That's part of my worry too, this is a point in a subject where I like to start applying ideas to some simple problems, but I don't know what to do!!!

I think we can explain what a spinor is, though, at this point, and thus better understand the idea of spin. (I don't know if y'all know this backwards and forwards yet, but I've not seen it rigorously presented) I need a break from the differential geometry aspect anyways, so I'll figure this out. [:)]

Edit: we might need representation theory first for spinors too. [:(]

[marcus]

Originally posted by Hurkyl
...and thus better understand the idea of spin. (I don't know if y'all know this backwards and forwards yet, but I've not seen it rigorously presented)...[:)]

great
take it for granted we dont (know it b.&f. yet) and that we want to

go for it!

page 71 of Hall says why there is an irrep of SU(2)
for every integer m
(on a space of dimension m+1)

only a minor amount of sweat and we have how idea of
spin comes out of irred. reps.

(physicists always divide that number m by 2 and
catalog the irreps in half-integers but allee-samee integers)

sounds good to me

[Hurkyl]

Ok, brief interlude back to differential geometry!

Recall that we were interested in proving that [g, h] = (ad g)(h) satisfied the definition of a Lie bracket.

It finally struck me that I wasn't giving enough emphasis to the group structure of a lie group, and I was trying to be too abstract in the geometrical aspect and wasn't using the calculus we all know and love on Rn!


So let's see how we synthesize these two familiar concepts!


The defining characteristic of a differential manifold is that it is locally diffeomorphic to Rn. Let us select for our Lie Group G a neighborhood U of the identity element E and a diffeomorphism &phi; mapping U to Rn. Since the group operations are continuous, if we focus our attention on points near the identity, we can keep all of our manipulations within U, and thus in the domain of &phi;. Also, I will insist that &phi;(E) = 0.

Now, how do we export the group structure from G to Rn? By taking the axioms of a group and exporting them via &phi;! In particular, I will define the two operations:

f(x, y) = &phi;(&phi;-1(x) * &phi;-1(y))
g(x) = &phi;(&phi;-1(x)-1)

f(x, y) is the Rn interpretation of multiplication, and g(x) is the Rn interpretation of inversion.

We can import the group structure by encoding associativity and identity

f(x, f(y, z)) = f(f(x, y), z)
f(x, gx) = f(gx, x) = 0
f(x, 0) = f(0, x) = x

And now we have moved everything into Rn and can proceed with what we learned from (advanced) calculus text!


Before I proceed, I will have to introduce the notation I will use; I've found in the past that generalizing this notation for scalars for use with vectors has been extraordinarily useful.

f1(a, b) is the differential of f(x, y) at (a, b) where we are holding y constant.


Before I proceed with the proof, first some preliminary results:

f(x, 0) = x : differentiate WRT x
f1(x, 0) dx = dx
f1(x, 0) = I (I for matrix identity)

similarly, f2(0, x) = I

If we differentiate WRT x again, we get:

f11(x, 0) dx = 0 = f22(0, x) dx
f11(x, 0) = 0 = f22(0, x)

(note: I use rank 3 tensors in this proof, such as these second partials, and they worry me because, while I think I know how they behave, I've never used them in this type of proof! Really all I need is that they are 3-dimensional arrays of numbers)

Also, we need to know dg(0):

f(x, gx) = 0 : differentiate WRT x
f1(x, gx) dx + f2(x, gx) dg(x) dx = 0
f1(0, 0) + f2(0, 0) dg(0) = 0
I + I dg(0) = 0
dg(0) = -I


Now, recall how (ad x)(y) was defined: We started with (Ad G)(H) = GHG-1, then we "differentiated" WRT H to get a function (Ad G)(h) acting on the tangent space, and then we differentiate again WRT G to get the function (ad g)(h). Now that we live in Rn, we can actually carry out this operation!

start with

f(f(x, y), gx)

holding x constant, we take the differential @ y = 0, yielding

f1(f(x, 0), gx) f2(x, 0) dy
= f1(x, gx) f2(x, 0) dy

now take the differential @ x = 0, yielding

(f11(0, g0) dx + f12(0, g0) dg(0) dx) f2(0, 0) dy + f1(0, g0) (f21(0, 0) dx) dy

Using the formulae we derived above and the associativity of tensor product:

-f12(0, 0) dx dy + f21(0, 0) dx dy

And using the anticommutativity of differential forms:

f21(0, 0) [dx, dy]

So we see that in Rn-land, (ad g)(h) is simply f21(0, 0) times the commutator of the corresponding differential forms, so it is clear (ad g)(h) satisfies the axioms of a lie bracket!


(I feel like I've skipped over a few too many details, and probably a few opportunities for profound observations, but I just can't see where...)

[Hurkyl]

Grr, I found one of my mistakes! Differential forms are cotangent vectors, not tangent vectors. [:(]

This disturbs me; I've always pictured differential forms as infinitessimal displacements, and that's what tangent vectors are supposed to represent...

Anyways, I'm eager to get onto representations, have you been preparing to post something Marcus, or should I work on that?

[marcus]

Originally posted by Hurkyl


Anyways, I'm eager to get onto representations, have you been preparing to post something Marcus, or should I work on that?

I havent been. And I have been hoping you would start the ball rolling. I am ever ready to try my hand at whatever lemmas checks and homeworks you propose. This has been rather fun so far, so I hope you continue.

(however always remember we are free to drop it anytime for any reason---its not as if we signed a contract! [;)] )

[Lonewolf]

I feel comfortable with the notion of tangent vectors, but I haven't got any references that I can find for cotangent vectors. Any one care to explain, please?

[marcus]

Originally posted by Lonewolf
I feel comfortable with the notion of tangent vectors, but I haven't got any references that I can find for cotangent vectors. Any one care to explain, please?

this is so important that we can do with several explanations, so I will offer one. But I hope to hear Hurkyls account of the same business.

there is a terrible fundamental and easy thing in math called the dual
of any vectorspace
WHENEVER you have any kind of vectorspace or any sort at all
(any set that satisfy those few obvious criteria that vectorspaces have to meet, mainly that there is a sensible way to add two of them and one or two other things like that)
whenever you have ANY vectorspace

then you can define another vectorspace called its dual which is just the linear functions defined on the first one
or, as one says with a certain panache, the linear "functionALs".
If it is a real vector space then a linear functional is just any real-valued function defined on the mother that happens to
be linear

f(x + y) = f(x) + f(y), and all that

I have to go, a friend just telephoned. But anyway I think the
"cotangent" space is just some specialized jargon for the dual of the tangent space------and the jargon is going to snowball: in a couple of seconds we are going to call it the space of "1-forms" too. It is actually exciting because of a bunch of geometrical meanings that emerge and what mathematicians do when they get excited is make up more and more names to call things. You can hear a rising hum of jargon and you know it is going to be a good party.

[Hurkyl]

The simplest example of dual vectors comes from our old friend Rn

Ordinary vectors are considered to be column vectors.
Dual vectors are row vectors.

For example, if I take the gradient of the scalar function f(x), I get a row vector. If I then postmultiply the gradient by an ordinary vector (with ordinary matrix arithmetic), the result is a real number (the directional derivative).


As marcus said, dual vectors are all real-valued linear functions on vectors. Similarly, mod the appropriate isomorphism, vectors are real-valued linear functions on dual vectors.


However, because Rn has the euclidean metric, we can convert freely between vectors and dual vectors, so the difference between the two is often underemphasized or even ignored because we have a nice isomorphism between the two (the transpose map). We even have the audacity to use the transpose map to allow us to write bilinear functions as matrices!


Differential one-forms are dual vectors to tangent vectors (thus we call them cotangent vectors); to put it simply, they tell you how to convert the tangent vector to a curve into a number... for instance, in the (x, y) plane, dx means take the x-coordinate of the tangent vector.

[marcus]

Originally posted by Hurkyl

Differential one-forms are dual vectors to tangent vectors (thus we call them cotangent vectors); to put it simply, they tell you how to convert the tangent vector to a curve into a number... for instance, in the (x, y) plane, dx means take the x-coordinate of the tangent vector.

At this point we could move this part of the conversation over to the "differential forms" thread that Lethe initiated if we wanted, and have two conversations:

one about differential geometry (at basic intro level suitable to our novice)
and one about matrix groups and reps and the like.
we are blessed, after all, with two sticky threads


the one thing I have a problem with in the diff form thread is that Lethe uses codes for symbols which my browser sees as boxes.
When I read a page by Lethe I see a lot of boxes.

I dont want to update my browser because of being a technophobe stick-in-the-mud. I only change habits and software gradually and I am not ready to make a big (for me) change just for one person's posts.

So I would suggest using the symbols that Greg posted and just using capital Lambda for wedge like & Lambda ; makes &Lambda;
and sigma wedge tau is written &sigma;&Lambda;&tau;

but if you dont, and I see boxes, then I will just cope with it somehow---no big deal.

[Hurkyl]

You can have two browsers installed on your computer, ya know. [:D]

Though I think you just need to update your fonts.

[marcus]

I put myself in Lonewolf shoes and I think
well the idea of dual of some vectorspace-----the space of linear functionals on the space----is extremely simple almost idiotically
simple

the puzzling thing is why make such a fuss

novices sometimes have this problem----they understand the idea but are baffled why mathematicians get so excited about it


there actually are reasons


and it is the same in the special case of the tangent space and ITS dual (the socalled cotangent space). Like, why even bother?

but there really are reasons, not only is there nice mathematics that grows out of these things but more urgently a whole bunch of physical concepts ARE linear functionals (and jacked up cousins called multilinear functionals) on the tangent space

Pretty much any physical quantity that has a "per" in its name.

flux is flow per patch of area (which two tangent vectors describe)

charge density is charge per element of volume (which three tangent vectors describe)

wavenumber?---change of phase or number of cycles associated with an infinitesimal move in some direction (which a tangent vector describes)

maybe the magnetic field? it is a linear response to a moving charge---perhaps all these examples are wrong but I believe that correct physical examples would be easy to get.

So suppose you want to be free take physical ideas over onto manifolds---to go places that dont already have an established Euclidean metric like R3. then you dont always have the easy equivalence between row vectors and column vectors. You have to keep track of what is a tangent vector and what is a function OF tangent vectors.

Lethe may have already given motivation for differential forms in the other thread, I havent read it all and dont remember. But anyway linear functions of various kinds built on the tangent space are good for physics and handy to have around.

Subscripts and superscripts make some people break out in hives, but Lethe I seem to recall, was trying to use hypoallergenic notation that avoided the worst infestations of these notational crablice.

[Hurkyl]

I have been trying to avoid use of the idea of coordinate charts in this thread for the very same reason. [:)]


Anyways, on to what a representation is!



To make a representation of a group G is to find some vector space V and define a group action for G on V such that the action is a (invertible) linear transformation. (As Hall puts it, it's a homomorphism from G into GL(V))

The fact we are working with matrix lie groups somewhat obscures the profoundness of this idea; after all a matrix lie group is, by definition, a group acting linearly on a vector space!


IMHO it pays now to think about lie groups in general for a moment. How can we get a general lie group to act linearly on a vector space?

Well, we've already found a way for a lie group to act on its lie algebra (which is a vector space); the adjoint mapping (Ad G)... however this is far too unambitious!

What about the tangent vector fields over a lie group? We know how to act on those by left multiplication! Specifically, if v(x) is a tangent vector field, then:

(Gv)(x) = v(Gx)

this is clearly a linear action, so the lie group action on its tangent vector fields is a representation, and this one is pretty interesting (by interesting I mean that it is more complex than the obvious case)! (Is it clear that the dimension of the space of all tangent vector fields is infinite?)


But this vector space is a little "too big"; this representation is reducible, meaning that there is a subspace of the vector space that is left unchanged by any element of G; in particular, the left invariant vector fields we constructed earlier. However, that's no matter; that's a finite dimensional subspace and if we mod it out, what's left is still interesting!


So we see that all lie groups have interesting representations, but do they have any useful ones?


Well, allow me to construct one! We know that Maxwell's equations are all sphericially symmetric, correct? Any rotation of a solution is another solution. So we know that elements of SO(3) act on the solutions to Maxwell's equations. However, rotations are linear; so we have found that the solutions to Maxwell's equations form a representation of SO(3)!

IIRC this last idea is the original reason Lie Groups were invented! If we can find the symmetry group of a set of differential equations, we know that the solutions to those equations must form a representation of the symmetry group! (is it obvious the symmetries of a DiffEq act linearly? or am I missing something?)


Some rote definitions:

An invariant subspace of a representation of a group G acting on V is a subspace S of V such that GS = S for all G in G.

An invariant subspace S of a vector space V is called trivial if it is all of V or if it is simply the zero vector. It is non-trivial otherwise.

A representation is real if the underlying vector space is a real vector space. Similarly for complex.

A representation is faithful if Gx = Hx for all x implies G = H.

If G acts on two vector spaces V and W, then a morphism between representations is a morphism (linear transformation) from V to W that commutes with group action. That is, &phi;(Gx) = G&phi;(x)

A morphism is an isomorphism if its invertible.


A unitary representation of a group G is one where the vector space is a Hilbert space, the group actions are unitary actions, and strong continuity holds; if the sequence An converges to A when viewed as elements of a lie group, then An converges to A when viewed as unitary operators.

(According to Hall, examples of something that violates strong continuity are difficult to come by)


All of the above holds for representations of a lie algebra as well (except for the unitary representation), except that the lie algebra action doesn't have to be invertible. (it's mapped into gl(V))


Phew, that's a lot to digest, any questions?

[marcus]

Originally posted by Hurkyl


Phew, that's a lot to digest, any questions?

It all seems fine, no questions. Looking forward to whatever comes next.

Originally posted by Hurkyl

All of the above holds for representations of a lie algebra as well (except for the unitary representation), except that the lie algebra action doesn't have to be invertible. (it's mapped into gl(V))


That is, I think I understand what was about groups. I will think about how this all carries over to Lie algebras....right now I dont see any questions about that part either.

Good point about the solutions to a set of equations being a representation of their symmetry. So the crafty physicist tries to understand what all the possible representations of a symmetry group can be as he fervently hopes to avoid ever having to solve systems of partial differential equations.

One time I did a google search with "group representation" and found a John Baez piece (apparently co-written with a character named Oz) which motivated the subject somewhat along these same lines---I dont remember the details but it was entertaining.

Anyway I'm eager to see what comes next.

[Hurkyl]

Baez is a character. [:)] Have you read his GR tutorial?

[marcus]

Originally posted by Hurkyl
Baez is a character. [:)] Have you read his GR tutorial?
He has a tutorial on GR which I have read, called "The Meaning of Einstein's Equation", or something like that. he rewrites the equation in an intuitive form involving the volume of a blob of test particles in free fall. I liked the tutorial and have recommended it. But you may be referring to something else which I havent seen. If so let me know about it---always happy to check out a Baez page.

[Hurkyl]

http://math.ucr.edu/home/baez/gr/gr.html

"Oz and the Wizard"

[marcus]

Originally posted by Hurkyl
http://math.ucr.edu/home/baez/gr/gr.html

"Oz and the Wizard"

thanks, I will take a look
cant say much off topic here because of not wanting
the thread to wander but will start a new thread soon
probably, to let you know what I'm reading---
has to do with representations of *-algebras
for example:
http://arxiv.org/gr-qc/0302059

another paper has a theorem to the effect that
"the Ashtekar-Isham-Lewandowski representation of
the [a certain LQG analog of the Weyl algebra] is irreducible"
http://arxiv.org/gr-qc/0303074

[Hurkyl]

So we've seen some examples of the representations of a group, let's look at some examples of the representations of the lie algebra.

A lie algebra representation is a morphism from a lie algebra g into the algebra of linear transformations on a vector space V. Being a morphism means that it must preserve lie bracket. Specifically,

&phi;([A, B]) = &phi;(A)&phi;(B) - &phi;(B)&phi;(A)

The product on the algebra of linear transformations is, of course, composition.


Just like matrix lie groups, both the trivial space {I} and the matrix lie algebra itself are representations of a matrix lie algebra.

What about generic lie algebras? Well, just like we did with the lie group, we can make the lie algebra act on the tangent vector bundle of the lie group! Because lie groups are parralelizable, we can write the tangent bundle as Gxg, and then for any lie algebra element g and tangent vector field v we have

(gv)(x) = ([nab]gv)(x)

Note this corresponds to the idea of a lie algebra element as an infinitessimal translation; suppose g is the tangent vector to the curve G(s) in G. Then:

gv(x) = ([nab]gv)(x)
= limh&rarr;0 (v(x + hg) - v(x))/h
= limh&rarr;0 (v(x + hg + O(h2)) - v(x))/h
= limh&rarr;0 (v(G(h)x) - v(x))/h
= limh&rarr;0 ((G(h)v)(x) - v(x))/h

so gv is related to Gv in the way we expect; gv is the derivative of Gv wrt G!


In general, for any representation of G acting on a vector space, we can induce a representation of g in the same way:

If g is the tangent vector to G(h):
gv := (d/dh) (G(h)v) @ h = 0


From this, we can actually write g as a directional derivative field! We have:

gv(x) = ((d/dh) (G(h)v) @ h = 0)(x)
= (d/dh) v(G(h)(x)) = dv (d/dh)(G(h)(x))

So at each point x, gv(x) is simply the derivative of v in the direction tangent to G(h)(x) at h = 0.


P.S. how do I make [nab] as a character instead of a smiley?

[marcus]

I understand the notation and can follow this pretty well. It is amazing how much can be done with the available symbols. I understand your use of the @ sign. Am glad to see the nabla (did not know we had [nab]).

I have changed a v to g in a couple of the following equations. Marked them with (*). Is this right?

Originally posted by Hurkyl
...then for any lie algebra element g and tangent vector field v we have

(gv)(x) = ([nab]gv)(x)

Note this corresponds to the idea of a lie algebra element as an infinitessimal translation; suppose g is the tangent vector to the curve G(s) in G. Then:

gv(x) = ([nab]gv)(x)
= limh&rarr;0 (v(x + hg) - v(x))/h &nbsp;&nbsp;(*)
= limh&rarr;0 (v(x + hg + O(h2)) - v(x))/h&nbsp;&nbsp;(*)
= limh&rarr;0 (v(G(h)x) - v(x))/h
= limh&rarr;0 ((G(h)v)(x) - v(x))/h

so gv is related to Gv in the way we expect; gv is the derivative of Gv wrt G!

[Hurkyl]

Lol, yes. Can you tell that I had originally used 'v' for the tangent vector in my scratchwork? [:)]

[marcus]

Originally posted by Hurkyl
Lol, yes. Can you tell that I had originally used 'v' for the tangent vector in my scratchwork? [:)]

In fact i thought it was something like that. BTW should say
that your running a basic group rep sticky in the background
has been personally beneficial for me in several ways, should say thanks sometimes

the main way that comes to mind is that it raises my consciousness of the essence of any quantum theory.
In any quantum theory, it seems to me, the ALGEBRA of
observables, and their more general operator friends, acts
on the HILBERTSPACE of quantum states.

nature seems to smile and beckon when people set things up this way. people get lucky and discover things and publish lots of papers when they set things up this way.

it is down at the level of superstition that we believe this is the right way to do something which maybe we still do not completely understand but nevertheless think we ought to do

so quantum theory of any sort is a theory of operators acting on a vectorspace, usually a C* algebra of operators acting on a hilbert space---that is, a representation theory

[Hurkyl]

Bah, this is gonna be a little terser than I wanted to make it; I need to stop debating in that Zeno's paradox thread, I spend too much time on it.


While representations of groups have to be on vector spaces, groups can act on all sorts of things. For example, SO(3) acts faithfully on any origin-centered sphere in R3. More generally, we can take any representation of G and consider each orbit of the vector space as a set upon which G acts. (if an orbit is a nontrivial subspace, then the action of G on the whole vector space is reducible)

We can build representations out of these sets by considering vector fields on them. So, for example, we have a representation of SO(3) by considering any scalar field on the unit sphere.



We can make representations out of representations. The tensor product of two lie groups is a lie group. The group action is

(a, b) (c, d) = (ac, bd)

and the lie algebra of the product is the product of the lie algebras. If we have a representation of G and a representation of H, then the tensor product of the representations is a representation fo the tensor product of the groups.


Alternatively, we can take two representations of the same group G, and then the tensor product of the representations is another representation of G. The action is given by

G(a, b) = (Ga, Gb)

An interesting example of this is vector fields on R3, with group SO(3). We can pretend (I think) that vector fields are surfaces in the tensor product of R3 with its tangent space (which is just itself), and then elements of SO(3) act by simultaneously rotating the space and rotating the vectors.

[Hurkyl]

Just to let everyone know, I probably won't be able to think about this for a while.

[Lonewolf]

Oh dear. lol. I'm quitting work this Friday, so I'll have time to give the topic the attention it needs. Be glad to hear your input when you return, Hurkyl. Always nice to have two people explaining something.

[marcus]

Originally posted by Lonewolf
Oh dear. lol. I'm quitting work this Friday, so I'll have time to give the topic the attention it needs. Be glad to hear your input when you return, Hurkyl. Always nice to have two people explaining something.

I agree----two or MORE.
Several times Hurkyl assigned "homework problems" or steps in the exposition for the reader to prove and Lonewolf or I would solve stuff or fill in the gaps which is a good kind of dialog. And
Lethe may have gotten in there too.

So Lonewolf, we should consider what kind of holding operation we want during Hurkyl's absence

I could for instance do a chapter on something entirely different but related to the representation theory of groups/algebras

or I could keep on with the basic (Hall's book) expo of matrix
groups and algebras---pretending to be Hurkyl, in effect.

Also LETHE! any comment? ideas of where thread could go?
want to be the substitute schoolteacher while Hurkyl's away?

[Lonewolf]

Do as you see fit. I'm kinda struggling to get to grips with representations, but then again, I haven't given it much attention as of yet. I'd quite like to see "something entirely different". I'm largely using this as a preview of years to come, but I also want to get something solid out of it, if you know what I mean.

[marcus]

Originally posted by Lonewolf
Do as you see fit. I'm kinda struggling to get to grips with representations, ...

Have you seen this story about Representations by John Baez?
-------------"spr" quote-----------------
"Well," said the Wiz, putting down his alembic and scratching his
head, "I'll try to explain it with a minimum of math..."

Oz smiled. "Good. As Feynman said, we don't really understand
anything unless we could explain it to our mother"

At this the Wiz bristled. "Speak for yourself. *My* mother knew
quite a bit about this stuff... she was a high priestess of the
Pythagorean Order... that's how I was born knowing tensor calculus."

"Really?" said Oz, not sure whether the Wiz was having him on.
"But enough of my personal life," said the Wizard impatiently.
"You'll probably disappear any minute now, so let me give you a
quick crash course on this stuff you're wondering about."
He thought a moment, running his fingers through his beard, and
then launched into an explanation:

"You know about various physical quantities: mass, energy, velocity, momentum, angular momentum, and so on. And you probably know that it's useful to keep track of their dimensions - in terms of length, time and mass, for example - at least that's how they teach you dimensional analysis in grade school. But: why is this so useful?"

"Well," said Oz confidently, "it keeps you from making mistakes: it
keeps you from adding apples and oranges and getting potatoes, so to speak. Sometimes you can even guess what the answer has to be just by remembering that the units work out right."

"Exactly!" said the Wiz. "Now, when you did this stuff, did you
ever realize you were doing group representation theory?"
Oz's eyes bulged. "Group... representation... theory? Err, no.
That's wizard talk, that is! I know nothing of that."

"Well, you were. Group representation theory is just the study
of how quantities change when apply various sorts of transformations to them. For example, if I tell you how long something is in rods, and ask you to tell me what its length is in feet, what do you?"

"Why, multiply by sixteen and a half, of course," replied Oz,
instantly losing his terrified demeanor. "Every schoolchild knows that!" "Right; that's how it transforms. But now say I tell you an *area*and ask you to convert it from units where length is measured in rods to units where length is measured in feet. What do you do then?"

"Multiply by sixteen and a half *squared*, of course, area has
units of length squared." "Right: area transforms in a different representation," agreed the Wiz.
"Hey, wait a minute!" said Oz. "You're trying to trick me into
thinking I understand this representation theory business, when I
actually don't!"The Wiz smiled. "No: I'm trying to trick you into realizing you actually DO understand it better than you think! When we change units of length, various physical quantities transform in various ways. We can actually imagine expanding all distances by a factor of 2 and seeing how various quantities change: some would stay the same (like times), some would double (like distances), others would quadruple (like areas), while other would be divided by 8 (like densities). Each of these ways of transforming is called a "representation" - in this case, a representation of the group of dilations."

"Dilations?" asked Oz, his pupils widening in terror.
"Yes, that's just a fancy wizard-word for stretching. Anyway,
when you're solving a physics problem, you know it doesn't make
sense to add a distance and an area, because they transform
differently under dilations, so even if your calculation *happened*
to work out correctly in units of feet, it wouldn't in units of rods."
"Right," said Oz. "But what does this have to do with vectors,
and pseudovectors, and bivectors, and ...."
"Well," said the Wizard, "Just as it's handy to keep track of
how quantities transform under dilations, as a bookkeeping device to keep from making silly mistakes, it's also handy to keep track of how they transform under *rotations*, and other sorts of transformations. Only here the options are more varied. For example, we have quantities like mass or energy, that don't change at all under rotations... we
call these SCALARS."

--------to be continued---------

[marcus]

-----exerpt from sci.physics.research, "spr"-----

Oz nodded, then thought more deeply and got confused. "What do you mean, they don't change under rotations?"

The Wiz glared. "I mean just what I say! See this weight?" With
a wave of his wand, a bang and a puff of smoke, an enormous weight labelled 50 TONS appeared on the floor. Oz held his hand over his eyes and squinted, leaning forward.
"Yes... but you have to realize, everything keeps fading in and out, over there!"

"Well, suppose we rotate it." With another wave of the wand the Wiz conjured up an enormous greenish troll, who grabbed the weight and turned it a bit, and then stared dumbly at it, drool oozing from between his half-open lips. "What's it's mass now?"

Oz rolled his eyes at the enormous expenditure of magic being
wasted on such a simple point. "Why, exactly what it did before!"
"Right!" said the Wiz. He snapped his fingers, and the weight
and the troll disappeared. "Mass doesn't change at all under rotations, so we call it a scalar! On the other hand, something like velocity does! We can measure the velocity of a bullet in some Cartesian coordinate system and get 3 numbers: the x, y and z components."

He pulled out a rusty old flintlock from one of the cabinets and fired it out the window. The glass shattered; the bullet left a trail of smoke, magically labelled by 3 numbers. "If we rotate the experiment and do it again, we get different numbers." He turned....

"Hey, don't point that thing at me!" yelped Oz.

"Okay, hopefully you get the point," said the Wiz. "It's a nuisance
having these windows repaired, after all. The point is, we know a
specific rule for how the numbers change when we do a rotation. Or at least *I* do. Do *you* remember it?"

"Umm, err..." said Oz. "I think maybe I sort of vaguely do, though
not quite. You take the numbers, line them up to form a column, and then you multiply them by a matrix... a square box of numbers... you do this by moving your left finger across the box, while moving your right finger down the column, multiplying the numbers and adding them up as you go... it's rather mysterious, come to think of it!"

"Yes, it's actually rather profound," said the Wiz, smiling. "But
for now, my only point is that for any rotation you... or at least
*I* ... can work out a 3 x 3 matrix which tells us how a velocity
transforms under that rotation. Anything that transforms according to this rule, we call a VECTOR. For example, not only velocity, but also momentum, is a vector."

"Okay," continued the Wiz. "How many other ways are there for
physical quantities to transform under rotations?" Oz thought and thought, but couldn't decide. "In other words," said the Wiz, "How many other REPRESENTATIONS are there of the ROTATION GROUP? This is just wizard-speak for the same
question... I don't expect it to help you just yet... I'm only mentioning it so that when you hear wizards muttering about group representations, you'll have more of a sense of what they're up to."

"Yes," said Oz, "that's helpful already. But - how many ARE there?"

"Lots!" said the Wiz. "But the wonderful thing is, I have a list,
which I keep up here," he said, tapping on his forehead, "of what they all are!"

"Hmm!" said Oz. "Could you, umm, tell me what they all are?" On
second thought, getting a bit scared, he backed off a bit. "Or, at
least some of them?"

"Well, for starters I'll tell you this: every different sort of TENSOR
gives you a different representation of the rotation group. To take the simplest example: the stress tensor."

Oz gulped. "Stress tensor? That's the simplest example? It sounds scary... I always get stressed out when you start talking abstract math, and now you're making me even tenser!"

"It's simple, honest!" said the Wiz. "Take this block of rubber" - with a wave of his hand, one appeared in his palm - "and twist, stretch or squash it however you like." He almost tossed it to Oz, but reconsidered. "Hmm, if you're really in a parallel universe, Oz, that may be risky. I'll do it myself."

He stretched it out and twisted it. "Now, imagine how each tiny piece of this rubber feels stretched, squashed or twisted. We can describe this with numbers, but not with 3 numbers - it takes 6!
In fact, we can arrange them in a 3 x 3 matrix, but it's a symmetric matrix: the entry in the ith row and jth column equals that in the jth row and ith column, so there are only 6 independent entries."

Oz looked puzzled. "Symmetric matrices... symmetric rank-2 tensors -- are those the same thing?"

"Yes," said the Wiz, "for now at least - they transform the same way under rotations, anyway. And that's just the point! You see --"

"Wait! I don't really understand it all yet. Where do we get this
matrix from? What do all the numbers mean?"

"Well," said the Wiz, "I don't really want to get into this now, but
the 3 numbers down the diagonal say how much the rubber is being squashed in the x, y, and z directions... or stretched, if the
number is negative. The other 3 numbers say how much and which way it's being twisted. Hmm. I thought you learnt all this stuff in the general relativity tutorial!"

"Well, maybe I did, Sir - I do remember a "stress-energy tensor",
vaguely, but that was a 4 x 4 matrix, and it had to do with pressure and energy density and..."

The Wiz cut him off impatiently. "Yes, that's another aspect of
the same idea. Back then we were doing SPACETIME, so we had 4 dimensions, but right now we're just doing SPACE, to keep things simple... anyway, the details don't matter here: I was just trying to give you another example of a representation of the rotation group. That is, a physical quantity that doesn't transform like a scalar when you rotate it, and doesn't transform like a vector. The stress tensor is basically a batch of 6 numbers - arranged artistically in a matrix - and there is a rule, which I will not tell you now, for how the stress tensor of this piece of rubber transforms when I rotate it."

"Oh!" said Oz, "Please tell me the rule, please do...."

"NO!" thundered the Wiz. "I can sense your time here is dwindling to a close. I only have time for this: by keeping track of how things transform under ROTATIONS, we can avoid foolish mistakes like adding things that transform differently, so it is profitable to CLASSIFY ALL REPRESENTATIONS OF THE ROTATION GROUP - and every mathematical physicist worth his or her salt knows this classification. It basically amounts to listing all possible sorts of TENSORS, of which the scalars and vectors
are the very simplest kinds."

"But," he continued, "this is just the beginning. You can do even better if you also keep track of how things transform under reflections! For example: angular momentum transforms just like a vector under rotations, but differently when we do reflections. Have you ever looked a moving object in a mirror, and wondered precisely how the velocity of the mirror image is related to that of the original object?"

[marcus]

-------exerpt of Baez post on "spr"----

"Umm, I can't say as I have, though it must be fairly simple."

The Wiz grew even more impatient. "No? What a stunning lack of
curiosity... anyway, do it sometime!

You will then know how a VECTOR transforms under reflections. Then, compare a *spinning* object to its mirror image, and figure out how their angular momenta are related. The rule is different! So we say that angular momentum is a PSEUDOVECTOR! This means that adding velocity and angular momentum is as bad as adding apples and oranges."

"But I already knew that," said Oz. "Velocity and angular momentum have different units!"

"Yes," the Wiz growled, "but even if they DIDN'T, it would STILL
be bad. If I had time, I could invent an example of quantities
with exactly the same units, but one a vector and the other a
pseudovector. But I don't! Or more precisely, *you* don't have
time. Next: what do you get if you take the dot product of two vectors?"

"A scalar!" replied Oz proudly.

"Right! But what if you take the dot product of a vector and a
pseudovector? Like velocity dotted with angular momentum?"

"Umm," said Oz, guessing wildly, "a PSEUDOSCALAR?"

"Right!" said the Wiz. "A scalar doesn't change under reflections;
a pseudoscalar changes sign under reflections."

Oz scratched his head, trying to work it out.

"Anyway," said the Wiz, "I hope you get the pattern: as we consider more and more sorts of transformations - dilations, rotations, reflections -- our classification of physical quantities according to how they transform becomes every more complicated and subtle... but also more POWERFUL, because we can make finer distinctions. This increase our chances that we can figure out the right answer to a physics problem just by writing down the only possibilities that transform correctly!"

"But what about spinors?" asked Oz.

The Wiz sighed. "Ah yes. Well, when we get to quantum mechanics, we need to replace the rotation group by something bigger and better. The reason is that some physical quantities turn out to change when you apply a 360 degree rotation to them! They only come back to where they were after a 720 degree rotation. They're not scalars, or vectors, or any other sort of tensor: to understand them, we need a group that's LIKE the rotation group, but distinguishes between "no rotation at all"
and "a 360 degree rotation about any axis". This new group has 2 elements for each element of the rotation group, so we call it a DOUBLE COVER of the rotation group."

"Is this, umm, US(2)?" asked Oz.

"You mean SU(2)!" replied the Wiz. "Yes. But the real point is
this: as soon as we discover that the rotation group is not sufficiently sophisticated for quantum mechanics, and we have to replace it by some other group, we had better run down to the math department and ask them to tell us all the REPRESENTATIONS of this group, so we can avoid adding apples to oranges in this brave new world!

And if we do, they'll tell us: `well, you've got your scalar representation, and your vector representation, and all your tensors just as before, but now you've got a whole wad of new ones, the simplest being the so-called SPINOR representation...in fact, you've got one for each spin j = 0, 1/2, 1, ...'"

"Spin!" said Oz. "So that's all it is??"

"Yes, for each spin we have a seperate rule saying how things should transform under rotations... or, not really rotations, but these SU(2) transformations, which are a lot like rotations, but a little fancier."

"But where do the Dirac matrices come in to it?" asked Oz.

"Well, for any representation of any group, you need a lot of
matrices to describe the rules for how things transform. You
know how it works for vectors... I hope... and all the other cases
are similar, but fancier. For the spinor representation of SU(2),
it helps to have some 2 x 2 matrices called `Pauli matrices' at your
disposal. But that's just about rotations in SPACE. If we switch
to studying SPACETIME, we also have to know how quantities transform under Lorentz transformations! We switch to a bigger group called SL(2,C), work out its representations, and discover that there's still something called the spinor representation... only now, to calculate with it, we need some 4 x 4 matrices called Dirac matrices. That's for "Dirac spinors", actually. There are also "Weyl spinors," which work differently..."

POOF! All of a sudden, Oz disappeared.

The Wiz sighed. "Just when it was getting interesting!"

He turned back to his alembic, picked it up, and started scraping
it off, muttering to himself as he worked. "Well, I hope he learned at least a *little* before taking off like that...."

-----end of post---

I found this at
http://www.lns.cornell.edu/spr/2002-01/msg0038075.html
the Cornell "spr" archive. But it may be other places on the
web as well. I seem to recall Hurkyl giving an address for
this as well as for the Baez tutorial on relativity.

[Lonewolf]

Lol, madness. He explains it well though. Just checking out the end? in the link you provided. I read part of it when Hurkyl posted, but wasn't focusing completely on it really, so it didn't sink in too well. Thanks for that.

[Lonewolf]

Oh, it seems the end has already been posted. Oops.

[Lonewolf]

SO(3) and Orbital Angular Momentum

Imagine an electron in a spherically symmetric attractive potential of some atom’s nucleus. The wavefunction of the electron can, as we all well know, be characterised by 3 quantum numbers n, l, m that are related to the eigenvalues of conserved operators H L2 and Lz. However, the energy is 2l+1-fold degenerate, depending only on n and l (for an unpure Coulomb potential. In a pure Coulomb potential, the dependency is only on n). The degeneracy may be explained by the spherical symmetry, independent of &theta; and &phi;.

Did I just say ‘spherical symmetry’? Does SO(3) spring to mind? It should. The above explanation for the degeneracy is equivalent to the Schrodinger-Hamiltonian (-hbar2/2m)&nabla;2+V(r) is invariant under ordinary spatial rotations, which is exactly where SO(3) comes into play. Recall SO(3) is the group of ordinary spatial rotations.

The spherical symmetry of the potential V(r) ensures that the orbital angular momentum L is conserved. Instead of using operators to represent Lx, Ly and Lz, we can use matrices. The Li generate the (2l+1)x(2l+1) irreducible representations of SO(3). The dimension 2l+1 is associated with the 2l+1 degenerate states.

This degeneracy is removed by introducing a constant magnetic induction field B, which leads to the Zeeman effect. This magnetic interaction adds to the Schrodinger-Hamiltonian a term which is not invariant under SO(3). This leads us to conclude that B is that magic thing, a symmetry breaking term.

[Hurkyl]

Phew, it seems I'm not drawn away for as long as I thought I'd be!


I guess we should keep heading towards spin, so I'll introduce a class of representations of SU(2). (And it will turn out that any finite dimensional representation of SU(2) will be isomorphic to one of these)


SU(2) is a group of 2x2 invertible complex matrices which means it acts on 2 dimensional complex vectors... we can reinterpret this action as a transformation on pairs of complex numbers.


Now, let's take the space Vm of m-dimensional homogenous complex polynomials in two variables, y and z. That is, the polynomials of the form:


f(y, z) = &Sigma;k=0..m akym-kzk

Note that f(y, z) is an m+1-dimensional vector space with basis {ykzm-k | k in 0 .. m}


I'll leave it as an exercise that, for A in SU(2), the mapping:

(Af)(y, z) -> f(A-1(y, z))

is a linear invertible mapping, and that A(Bf) = (AB)f, which makes Vm an m+1 dimensional complex representation of SU(2).


I'll leave it as homework problem to prove that this is an irreducible representation (since I don't see right away how to prove it! [;)])

[marcus]

Lots of tidbits to check here, among other things


I'll leave it as an exercise that, for A in SU(2), the mapping:

(Af)(y, z) -> f(A-1(y, z))

is a linear invertible mapping, and that A(Bf) = (AB)f...

a kindergarten question (often the best time) is why did they use
f(A-1(y, z)) instead of f(A(y, z))?

Would f(A(y, z)) have worked as well? no because then we would have gotten an unwanted reversal: "A(Bf) = (BA)f..."

so lets prove the Hurkyl-stated fact!

let g(y,z) be temporary notation for f(A-1(y, z))
and let B act on it
(Bg)(y,z) = g(B-1(y, z)) = -> f(A-1(B-1(y, z)))


oops, have to go

[Hurkyl]

Yep, the reversal is important!

Note we know another involution of complex matrix rings; the conjugate transpose. We could also have defined the action

(Af)(y, z) = f(A*(y, z))

For SU(2), of course, this is identical with the representation using the inverse, but for, say, SL(2, C) these would be two different actions (at least superficially)

[marcus]

Now I can resume that HW exercise. I got started on the wrong foot earlier. I have to show A(Bf) = (AB)f. Acting on the polynomial f first by B and then by A gives the same as acting by AB.

I will temporarily use the letter g to stand for the polynomial Bf, the intermediate result gotten by first acting with B on f.

[QUOTE
let g(y,z) be temporary notation for f(B-1(y, z))
and let A act on it
(Ag)(y,z) = g(A-1(y, z)) = f(B-1(A-1(y, z))) = f((B-1A-1(y, z))
by associativity
and there's the fact about matrix multiplication that B-1A-1 = (AB)-1, so
QUOTE]

the above is f((AB)-1(y,z))
which is (AB)f, the result of AB acting on f,
so that shows A(Bf) = (AB)f

[marcus]

About these SU(2) matrices, all novices should now know
what the 4 complex numbers are in the general form of
one
Since we seem to be moving on it would ordinarily be time
to review basic knowledge with a midterm, but perhaps since we go by a sort of anti-academic set of rules we will omit this.

given two complex numbers u and v,
with uu* + vv* = 1,
the general SU(2) matrix is


u v
-v* u*


And if you invert this by swapping on the main diagonal and negging on the other, in the time-honored way when det = 1, then by the Three Graces and the Seven Muses you get


u* -v
v* u


and this is also the complex conjugal flip transpose A* thing
so A-1 = A*

Now the natural naive question to ask is whether any special SU(2) property of the matrix is helping out here. Does the fact that A is SU(2) help make the polynomial f(y,z) map into a new polynomical with is homogeneous with the same total degree????

[marcus]

Originally posted by Hurkyl

I'll leave it as homework problem to prove that this is an irreducible representation (since I don't see right away how to prove it! [;)])

The way the group thread works is if Hurkyl assigns homework we do it.

So as a contribution to classifying the irreducible representations of SU(2) I have made two PF footnotes

http://physicsforums.com/showthread.php?s=&postid=58530#post58530

which is about complexifying Lie algebras and uniquely extending representations of the algebra to its complexified version, and also this one:

http://physicsforums.com/showthread.php?s=&threadid=4671

which is about complexifying the Lie algebra su(2) to get
sl(2, C)

the irreducible reps of sl(2, C) are not dreadfully hard to study
and this will tell us about those of su(2)
which in turn (because reps of the LA correspond to reps of the group)
will tell us about the reps of SU(2)
which will turn out to be on the spaces of homogenous polynomials as we were just now discussing

[marcus]

Hurkyl has described an action of SU(2) on the homogeneous polynomials of total degree m-----an m+1 dimensional complex vectorspace.

How may we show that these are irreducible? And that they are ALL the finite dimensional irreps up to isomorphism.

I think Hall page 72-73 sort of does the first part of this, roughly as follows.

Any rep of the group can be lifted to a rep of the Lie algebra su(2), which can be viewed as a representation over the reals and uniquely extended (see footnote) to a rep of sl(2,C) over the complexes.

Call this unique extension [pi]m
It is a (complex) LA homomorphism from sl(2, C) to the linear operators on the polynomial space Vm

To repeat, Vm is homog. polys with complex coefficients and total degree m, in two variables y,z. as Hurkyl described.

The action with matrix A of sl(2,C) on poly P(y,z) is simply
giving us the new poly P(A-1(y,z))

[pi]m(A) sends P(y,z) to P(A-1(y,z))

NOW WE LOOK AT A BASIS OF sl(2, C), which are TRACE ZERO 2x2 matrices

Following Hall's notation the basis is H, X, and Y where
H = diag(1, -1) = (1,0,0,-1)
X = (0,1,0,0)
Y = (0,0,1,0)
I will use "code" to type these in block form when I edit but here I am just reciting 2x2 matrices left to right and top to bottom as words are written on the page. From the commutation relations we can figure out what the operators [pi]m(H), [pi]m(X) , [pi]m(Y) actually do to the polynomials and show irreducibility directly----no proper invariant subspaces.

Now for the second part, which Hurkyl may be going to prove, we need to show that ANY irreducible rep on ANY finite dimensional V is isomorphic to the polynomial one of the same dimension.

Since we have these three sl(2,C) matrices H,X,Y and know their bracket relations, to study any rep all we need to do is study what it does to these three. Now [pi]m(H) is an operator on V, so it is actually an (m+1)x(m+1) matrix, assuming V has dimension m+1.
And we can diagonalize it! We can find its eigenvalues and eigenvectors!
This approach parallels Hall pages 76-78 where he proves the theorem that any two irreps of sl(2,C) which have the same dimension are equivalent.
Did not mean to hog the exposition like this! It was really Hurkyl's turn to talk but I got carried away. Sorry.

[Hurkyl]

Did not mean to hog the exposition like this! It was really Hurkyl's turn to talk but I got carried away. Sorry.

No problem. Means less work for me. [;)] Besides, I think it's helpful to see different styles of explanations for things, so it's good as long as we don't confuse the audience (which, of course, includes ourselves!


I wonder if it might not be good to explore the "geometry" of SU(2) and SL(2, C); we know lots about SO(3), but I don't really have much intuition for those other two groups, and I'm not confident in what I do have. This would be unnecessary for this thread, but might be helpful! But, of course, it might be entirely the case that it would be far easier to understand the geometry after we've gotten through sl(2, C)'s representations instead of before... EDIT: I guess we're nearly there so might as well finish it up. [:)]


Incidentally, here's an amusing nonrigorous homework problem! At least I found it amusing when I did it. It seems to require ignoring some "obvious" yet nontrivial technical details, so keep that in mind while you do the problem. (this one's just for fun!)

As we know, so(3) and su(2) are isomorphic three-dimensional real algebras. I demonstrated earlier that so(3) was isomorphic to R3 with the cross product for the bracket, via the basis:


/ 0 0 0 \
i = | 0 0 -1 |
\ 0 1 0 /

/ 0 0 1 \
j = | 0 0 0 |
\ -1 0 0 /

/ 0 -1 0 \
k = | 1 0 0 |
\ 0 0 0 /


One can check things like [i, j] = i*j. This basis has the nice property that eit is a t-radian rotation around the R3 vector i, and similarly for j and k.

We know that SO(3) is generated by these three classes of rotations (but is probably nontrival to prove, so don't!)

So what if we take a basis for SU(2)? In particular:


i = 0.5 / i 0 \
\ 0 -i /

j = 0.5 / 0 1 \
\ -1 0 /

k = 0.5 / 0 i \
\ i 0 /


You can check that the bracket again corresponds to the cross product. You can exponentiate these to get elements of SU(2). Recall elements of SU(2) act on su(2) by the adjoint mapping U(X) := UXU-1. One can then check that eti acts as *drumroll*... a rotation of t radians around the vector i!!! We again assume that SU(2) is generated by these three classes of rotations... but the neat part is that the explicit formula for these exponentials makes it clear why SU(2) is a double cover of SO(3). [:)]

[MathematicalPhysicist]

i need to proove (a^-1)^-1=a with group theory.
the text says the proof should be the same as to the proof to b=a^-1
here this proof:
a.b=e (given)
a^-1.(a.b)=a^-1.e
(a^-1.a).b=a^-1.e
e.b=a^-1.e
b=a^-1

i apply it too:
a^-1.a=e (given)
a^-1.(a^-1.a)=a^-1.e
a^-1.e=a^-1.e
a^-1=a^-1

does this what i neede to come to?
btw im new to group theory so be gentle (-:

[Lonewolf]

Not sure that you've explicitly proven it, though I may be wrong. Try to end up with (a-1)-1 = a at the end of your proof.

[MathematicalPhysicist]

that's the problem i cant.

[Lonewolf]

Try letting b = a-1, getting a.(b-1)-1 = a.a, sub a back in for b-1, then you're almost there.

[marcus]

We used to have a table of how to make greek letters using
font = symbol
I have lost track of where it is or what the symbol font looks like
so I am going to have a look at each key in that font

It is handy because & pi ; gives something that doesnt look very much like pi (in the default) and same for gamma (looks like Y in default) and theta (looks like the number 8 in default)


a in symbol font is a
b in symbol font is b
c in symbol font is c
d in symbol font is d
e in symbol font is e
f in symbol font is f
g in symbol font is g

h in symbol font is h
i in symbol font is i
j in symbol font is j
k in symbol font is k
l in symbol font is l
m in symbol font is m
n in symbol font is n


o in symbol font is o
p in symbol font is p
q in symbol font is q
r in symbol font is r
s in symbol font is s
t in symbol font is t
u in symbol font is u

v in symbol font is v
w in symbol font is w
x in symbol font is x
y in symbol font is y
z in symbol font is z

these are mostly intuitive except that the theta is typed
using the letter q and a couple of things like that. here is
a sample in size = 4 and size = 3 to make the greek letters
more legible

a few size 4:
a in symbol font is a
b in symbol font is b
c in symbol font is c
d in symbol font is d
e in symbol font is e
f in symbol font is f
g in symbol font is g

a sampling of size 3:
h in symbol font is h
i in symbol font is i
j in symbol font is j
k in symbol font is k
l in symbol font is l
m in symbol font is m
n in symbol font is n


ordinary size:
o in symbol font is o
p in symbol font is p
q in symbol font is q
r in symbol font is r
s in symbol font is s
t in symbol font is t
u in symbol font is u

v in symbol font is v
w in symbol font is w
x in symbol font is x
y in symbol font is y
z in symbol font is z

[MathematicalPhysicist]

shouldnt this thread should be a sticky one?

[Hurkyl]

On my computer, all of those are just roman letters drawn in a fancy way, not greek letters...

[pdelong]

Good (introductory) references are:

M. A. Armstrong. Groups and symmetry. Springer Verlag 1988.



Would you happen to know where one could get an errata listing for this book? I'm currently in the middle of it and I'm stuck on a problem. I'm pretty sure the reason I'm stuck is because of a misprint, but I just want to be sure.

Thanks.

[Janitor]

I learned (well, sort of :rolleyes: ) my group theory from a moldy-oldie library book written by _____ Hall. It was already old when I was 20, and I imagine it is long out of print.

[ahrkron]

On my computer, all of those are just roman letters drawn in a fancy way, not greek letters...

Same here! :frown: Mozilla on RedHat Linux. I'll check later on a Windows XP.

[ahrkron]

Same thing on XP (Mozilla also).

[meteor]

Groups are fantastic things, I'm recently starting to appreciate its importance, before I considered them rather arbitrary constructions
I'm wondering if the group operation is restricted to some binary operations, or instead, any binary operation is valid. Cause I know that for example addition, multiplication, matrix multiplication, composition of functions,... are binary operations that are permitted like group operations, but is this general?, I mean, can any binary operation serve like a group operation?

[matt grime]

No, by defintion the binary operation must satisfy certain rules, but by the same token any binary operation satisfying those rules is a group operation. "Serving like a group operation" is a nebulous phrase which could mean anything you chose it to mean.

A group is an axiomatic object, anything satisfying those axioms is a group, end of story. If you want a binary operation that isn't a group operation, there's multiplication on the real line - 0 has no inverse. Or addition on the set of multiples of 2 and 5 - that isn't closed, Addition on the strictly positive real numbers has no identity. For failure of asscoiativity I must be more creative: consider the group defined by this table:

* |a b
--------
a|a b
b|a b


That isn't associative by the failure of the latin square principle.

I forgot to emphasize that your question over looks the fact that there needs to be an underlying set the operation is defined on - matrix multiplication is not a group operation on the set of infinite matrices.

[meteor]

Matt,
Yeah, I know that a group has to satisfy certain rules (namely closure, associativity, identity element and existence of the inverse), but my question is: given a group of elements and a binary operation, and then those elements under the binary operation satisfy the rules, then this is considered a group, indepently of the binary operation? I ask this because I do not remember any group where the binary operation is division, or for example the modulus (a mod b), or the Legendre symbol, or many others, but would be nice if such groups could exist
Regards

[matt grime]

The binary operation is part of the group structure, it cannot be a group independently of the operation, hell it can't even be a group.

Given a set there are many ways of putting a group structure on it. There are two groups of order 6 that are not isomorphic.

I think you are misuing the word group in that clause 'given a group of elements'. Do you just mean set or group in its proper definition?

Division would not work for associativity reasons.

And those things you cite (legndgre symbol etc) cannot form group operations - one isn't even a binary operation.

[Janitor]

As Matt points out, division is not associative. Example: what is 20/10/2?

(20/10)/2 = 1,

20/(10/2) = 4.

[matt grime]

I misread what you wrote, they are all binary operations but cannot possibyl be group operations given, firstly you aren't offering a set on which they are to be defined which contradicts the definition that a group is a set with a binary operation, and they fail to define injective maps in that if you fix the first argument there are many (infinitely many) second arguments that will give the same output.

Remember, a group is a set WITH an operation satisfying.... you need both.

[H-bar None]

There is the Dog School of Mathematics of dogpile fame. They have a nice tutorial on group theory.