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.....