Group Theory Basics: Where Can I Learn More?

[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 wouldn't 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 don't 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 don't (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 φ 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 φ. Also, I will insist that φ(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 φ! In particular, I will define the two operations:

f(x, y) = φ(φ^-1(x) * φ^-1(y))
g(x) = φ(φ^-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 haven't 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 don't 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 Λ
and sigma wedge tau is written σΛτ

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, you know.

Though I think you just need to update your fonts.

 

[marcus]

why so much fuss about the dual

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 don't already have an established Euclidean metric like R³. then you don't 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 haven't read it all and don't 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, φ(Gx) = Gφ(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 A_n converges to A when viewed as elements of a lie group, then A_n 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 don't 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 don't 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 haven't 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,

φ([A, B]) = φ(A)φ(B) - φ(B)φ(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)
= lim_h→0 (v(x + hg) - v(x))/h
= lim_h→0 (v(x + hg + O(h²)) - v(x))/h
= lim_h→0 (v(G(h)x) - v(x))/h
= lim_h→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)
= lim_h→0 (v(x + hg) - v(x))/h   (*)
= lim_h→0 (v(x + hg + O(h²)) - v(x))/h  (*)
= lim_h→0 (v(G(h)x) - v(x))/h
= lim_h→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 going to 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]

Baez: Oz and the Wizard story about Representations

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]

Baez story about Representations, ctd

-----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 learned 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]

Baez introduces spin concept, end of story

-------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 separate 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 into 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 L_z. 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 θ and φ.

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 (-h_bar²/2m)∇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 L_x, L_y and L_z, we can use matrices. The L_i 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 V_m of m-dimensional homogenous complex polynomials in two variables, y and z. That is, the polynomials of the form:


f(y, z) = Σ_k=0..m a_ky^m-kz^k

Note that f(y, z) is an m+1-dimensional vector space with basis {y^kz^m-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 V_m 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 let's 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

https://www.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:

https://www.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]

the irred reps of SU(2)

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 V_m

To repeat, V_m 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 \
[b]i[/b] = |  0  0 -1 |
    \  0  1  0 /

    /  0  0  1 \
[b]j[/b] = |  0  0  0 |
    \ -1  0  0 /

    /  0 -1  0 \
[b]k[/b] = |  1  0  0 |
    \  0  0  0 /

One can check things like [i, j] = i*j. This basis has the nice property that e^it 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:

[b]i[/b] = 0.5 /  i  0 \
        \  0 -i /

[b]j[/b] = 0.5 /  0  1 \
        \ -1  0 /

[b]k[/b] = 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 e^ti 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]

proove a theorem in group theory

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 I am 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 doesn't 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[/size]
b in symbol font is b[/size]
c in symbol font is c[/size]
d in symbol font is d[/size]
e in symbol font is e[/size]
f in symbol font is f[/size]
g in symbol font is g[/size]

a sampling of size 3:
h in symbol font is h[/size]
i in symbol font is i[/size]
j in symbol font is j[/size]
k in symbol font is k[/size]
l in symbol font is l[/size]
m in symbol font is m[/size]
n in symbol font is n[/size]


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...