1. I am trying to get a handle on the the coadjoint representation of a Lie group on its coalgebra. This will become a genuine question, but in order to locate the casual reader in context (and because I am a long winded bastard), it may take a while to get to it. I hope someone can help nonetheless.

For starters, suppose that is a generic group, and where denotes an unspecified but legit. group operation. One may say that acts on from the left. One may also have that acts on from the right i.e. .

We may not assume that , i.e. abelianism for our group.

Now let's call the left action of as . Then by the above, . And let's call the right action .

Now note the following startling fact: one of the group axioms stipulates associativity, i.e. . This implies commutativity of the actions of the form for any group, abelian or not. How weird is that?

No matter, it simply says that the algebra of elements of a group and that of their actions are quite different beasts. This motivates the following.

A representation of a group is its realization as the group of all invertible linear transformations (invertibilty is mandated by the group axioms) on a vector space . This is called the general linear group , so we may have a mapping .

Now since can be viewed as an action on the group , I will find that .

OK, so here's the question: one may also find a representation of on the dual to , that is . Then obviously, for some , I will have that .

So (dropping the argument on ), I am told (Fulton & Harris)that .

Now, I understand why the transpose is required, as is the pullback of the mapping

But why do I need to take the inverse element of my group as an argument on the rep. map? Or is this only true for matrix groups?

Or is it more subtle? Or am I being dumb? Either way, I have more questions.....

2.

3. Originally Posted by Guitarist
I am trying to get a handle on the the coadjoint representation of a Lie group on its coalgebra. This will become a genuine question, but in order to locate the casual reader in context (and because I am a long winded bastard), it may take a while to get to it. I hope someone can help nonetheless.

For starters, suppose that is a generic group, and where denotes an unspecified but legit. group operation. One may say that acts on from the left. One may also have that acts on from the right i.e. .

We may not assume that , i.e. abelianism for our group.

Now let's call the left action of as . Then by the above, . And let's call the right action .

Now note the following startling fact: one of the group axioms stipulates associativity, i.e. . This implies commutativity of the actions of the form for any group, abelian or not. How weird is that?

No matter, it simply says that the algebra of elements of a group and that of their actions are quite different beasts. This motivates the following.

A representation of a group is its realization as the group of all invertible linear transformations (invertibilty is mandated by the group axioms) on a vector space . This is called the general linear group , so we may have a mapping .

Now since can be viewed as an action on the group , I will find that .

OK, so here's the question: one may also find a representation of on the dual to , that is . Then obviously, for some , I will have that .

So (dropping the argument on ), I am told (Fulton & Harris)that .

Now, I understand why the transpose is required, as is the pullback of the mapping

But why do I need to take the inverse element of my group as an argument on the rep. map? Or is this only true for matrix groups?

Or is it more subtle? Or am I being dumb? Either way, I have more questions.....
You need it to make a homomorphism. Using to represent the adjoint (aka transpose) of a linear transformation where indicated (in a dual role, pun intended),

4. Thank for that, it's neat. But, if I read you correctly, you seem to be using your asterisk in two different ways; one for the dual and another for the adjoint.

No harm in that - the great Alexander Kirillov does the same.

But it induces something wonderfully weird, which I shall try to explain (just in case any casual reader might be interested in my pathetic ramblings)

Let be a Lie group. For simplicity I will suppose that is a finite group and also a matrix group. So that in this circumstance I may define the adjoint representation by where is just the tangent space over the identity of .

For aesthetic reason only I will set the image in as . Thus invertibly. Jumping ahead of myself rather, I am going to make the identity , the Lie algebra uniquely associated to the group . So, to re-word: .

Thus I may think of as a (linear?) operator on a finite-dimensional vector space

I now consider the dual: , where again . This called the coadjoint representation. By the above identity, I will have that , so that, again is a linear operator.

Now, a condition on representations is that they respect the natural paring of elements of a vector space and those in its dual. So we must have that ().

So it seems that I can always find some such that .

Holy mother of god!! The linear operator induced by the adjoint representation is adjoint to that induced by the coadjoint representation (and v.v).

Wow (or "wong"?).

PS by edit: I now think this should be . In which case my argument more-or-less fails, since, using the group law, and the fact that is a group homomorphism, then it is not hard to show that

Now I am even loster than before

5. Originally Posted by Guitarist
Thank for that, it's neat. But, if I read you correctly, you seem to be using your asterisk in two different ways; one for the dual and another for the adjoint.

No harm in that - the great Alexander Kirillov does the same.
Kirillov did some good work in identifying the irreducible representations of simply connected nilpotent Lie groups with orbits of the coadjoint representation.

But be warned. Kirillov has written at least one books= that is just filled with errors. When I was in graduate school we did a seminar out of Kirrolov's then new book on representation theory. It essentially ended in a book burning. There were so many serious errors that the book was basically useless. It was so bad that when I briefly talked to a professor who was in that seminar over 30 years ago, he remembered it immediately.

David Vogan, who really is a great mathematician specializing in representation theory, seems to have found similar problems in Kirillov's book on the orbit method.

http://www.ams.org/bull/2005-42-04/S...05-01065-7.pdf

In short, I don't personally trust Kirillov's writings. I find him sloppy with a tendancy to state things that are simply untrue.

If you are interested in Kirillov's theorem for simply connected nilpotent Lie groups, Pukansky's book, Lecons sur les representations des groupes is pretty good. It is, so far as I know, only available in French. Forgive me, I corrected some typos

6. Um. It would appear that A. Kirillov has (had?) a son with the exact same name and (more-or-less) the exact same mathematical interests. Are we talking about the same guy?

No matter for my present purpose.

As an aside:
Pukansky's book, Lecons sur les representations des groupes is pretty good. It is, so far as I know, only available in French.
Then it's a good job that I am pretty fluent in French.

But let's do some mathematics. What do you say to the content of my last?

"Rubbish" I hear you say, as you are wont to.

Wanna discuss it?

7. Originally Posted by Guitarist
Um. It would appear that A. Kirillov has (had?) a son with the exact same name and (more-or-less) the exact same mathematical interests. Are we talking about the same guy?
Alexandre A. Kirillov

I know of only one other A. Kirillov. He did his thesis in combinatorics. I don't know if he is related to A.A. Kirillov.

8. Originally Posted by Guitarist

But let's do some mathematics. What do you say to the content of my last?

"Rubbish" I hear you say, as you are wont to.

Wanna discuss it?
Not rubbish.

However, you do not want to assume that G is finite. When you do that G is also discrete, hence 0-dimensional. That obviates the rest of your post, since you are representing G as linear operators on the associated Lie algebra, the tangent space at the identity. You need to exploit the manifold structure of the Lie group to make any headway.

I am not quite sure where you are heading with this. The theory of representations of Lie groups is big subject, far too big to cover in a forum such as this.

9. Originally Posted by DrRocket
The theory of representations of Lie groups is big subject, far too big to cover in a forum such as this.
So that's that, then.

Shame, really.....

Oh, just for the record, I had meant to say "finite-dimensional" (as a manifold, obviously) Sorry for my error.

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