Inner automorphisms of a lie algebra

Question

I am having trouble under standing why something of the form $\exp(ad x)$ where $ad x$ is nilpotent is a an automorphism of a lie algebra.

As far as I understand the exponential map is mapping from the Lie Algebra to a Lie group. But my understanding does not really make sense as we could be dealing with an abstract lie algebra which apriori need not have a lie group.

Answer 1

Certainly you know that for a Lie algebra $L$ and any element $x\in L$, the map $ad_x$ is an endomorphism of the Lie algebra.

You also know that if one composes two such endomorphisms, or adds them, one gets endomorphisms again.

Now, assuming our Lie algebra lives over a field with characteristic $0$, consider the sum:

$id_L + \dfrac{ad_x}{1!} + \dfrac{ad_x \circ ad_x}{2!} + ... + \dfrac{(ad_x)^{(n)}}{n!} + ...$

with $\cdot^{(n)}$ meaning $n$-fold composition. Now if $ad_x$ is nilpotent, this is actually a finite sum, so we can avoid convergence issues. By the above, it is an endomorphism of $L$.

To show it's an automorphism, find its inverse. Hint: What's the inverse of $e^k$?

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It is the above sum what I would denote $exp(ad_x)$. If the Lie algebra $L$ is given as matrices, so one also has the exponential of matrices available as a map

$exp: L \rightarrow G$

to a certain matrix group, which acts on $L$ via conjugation, then one could alternatively look at the map

$y \mapsto exp(x) \cdot y \cdot exp(x)^{-1}$

and check that it identifies with the map I defined above.

Answer 2

Here $exp$ is the exponential of matrices, $exp:M(n,\mathbb{R})\rightarrow Gl(n,\mathbb{R})$. $ad_x$ is a linear map of the Lie algebra $L$.