Inner automorphisms of a lie algebra 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.
 A: 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 $e^k$?

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.
A: 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 $L$.
