# Thread: Deriving the Lie algebra

1. Well, I had thought I was OK with that, but it seems not.

Specifically, on a friend's recommendation, I went to our library and looked in Fulton & Harris: Representation Theory.

There I find that, for the generic Lie group , the map is a nice mapping that fixes the the identity.

This is glaringly obvious, right?

But they, F&H, that is, go on to say this defines a mapping . Now this step I don't quite see.

Clearly is the group of all automorphisms , and, by their (F&H's) construction, then is a Lie group homomorphism.

What I DON'T get is the the leap from to .

Any help out there?

2.

3. Originally Posted by Guitarist
Well, I had thought I was OK with that, but it seems not.

Specifically, on a friend's recommendation, I went to our library and looked in Fulton & Harris: Representation Theory.

There I find that, for the generic Lie group , the map is a nice mapping that fixes the the identity.

This is glaringly obvious, right?

But they, F&H, that is, go on to say this defines a mapping . Now this step I don't quite see.

Clearly is the group of all automorphisms , and, by their (F&H's) construction, then is a Lie group homomorphism.

What I DON'T get is the the leap from to .

Any help out there?
It is

4. Rocket I thank you for your response.

Either you have read the book I referenced, or you are more clever than you appear (own up!) since I made a confusing typo.

The mapping should be

So may I say this? Since, by the group axioms , and for the same reason has inverse, then so does .

The operation is, as you well know, called "conjugation".

So since each conjugates all thusly, I may have that .

Hmm. Still feels like I am missing something, but I can't quite see what.

5. Originally Posted by Guitarist
Rocket I thank you for your response.

Either you have read the book I referenced, or you are more clever than you appear (own up!) since I made a confusing typo.