Group Theory Basics: Where Can I Learn More?

[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 T_x(M) to T_f(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 T_x(M), choose a smooth curve γ through x whose tangent vector at x is v. Then, define (*f)_x(v) to be the tangent vector to f(γ) 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:

L_g : M -> M : h -> gh

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

Define R_g 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 T_E(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(θ) != s(θ+2π)

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 T_E(M) and the set of all tangent vector fields invariant under left multiplications. (called left invariant vector fields)
---------------------
Both T_E(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 T_x(M) --->T_y(M) which we now apply to a manifold which is a group G.

and you described the right and left multiplication maps R_g and L_g : G ---> G

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

Now as to Problem 1, for any v in T_E(M) let's define a LIVF denoted by X_v
g ---> *(L_g) 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
X_v(h) where h is in G. If I left multiply by g, I get
X_v(gh) and to show left invariance
I have to show this is the same as *(L_g)X_v(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 X_v was defined in the first place
X_v(gh) = *(L_gh) v
= *(L_g) *(L_h) v ...[[[by chain rule]]]
= *(L_g) X_v(h) ...[[[by X_v 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 X_v and X_v' 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
L_gh = L_g o L_h
and that extends to the tangent spaces because of the chain rule
*(L_gh) = *(L_g) o *(L_h)

---------------------------
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 R_h 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) = *(R_h) X(g h^-1)

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

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

Well Y(kg) = *(R_h) X(kg h^-1)
= *(R_h) *(L_k) X(g h^-1)...[[[by left invariance of X]]]
= *(L_k) *(R_h) X(g h^-1) ...[[[commutativity]]]
= *(L_k) 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
L_k R_h = R_h L_k
I guess that is obvious k(gh) = (kg)h assoc. and this
commuting business goes upstairs to the lifted right and left multiplications maps
*(L_k) *(R_h) = *(R_h) *(L_k)
so there

Originally posted by Hurkyl


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 T_x(M) to T_f(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 T_x(M), choose a smooth curve γ through x whose tangent vector at x is v. Then, define (*f)_x(v) to be the tangent vector to f(γ) 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:

L_g : M -> M : h -> gh

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

Define R_g 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 T_E(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 π:E→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 π:E→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 π:E→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 π:E→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 G_cylinder = 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 A_G(H) to be GHG^-1 for G and H in a lie group, we can define:

Ad G = *A_G

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 T_eG (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 don't 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 T_e

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

Ad G = *A_G

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 T_x(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 T_x&phi; or *&phi;: T_x --->T_&phi;(x).

Given v in T_x 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, don't 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 that's 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 T_e(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 T_e 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 *L_g*R_g^-1

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

OK

*L_g*R_g^-1 X(e)

*L_gX'(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 *R_g^-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 going to 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 T_x ---> T_y

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 doesn't 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 that's 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 doesn't 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 realized 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 realized 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]

two questions

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