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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.

samlanader
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2 Answers2

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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.

Torsten Schoeneberg
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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 $L$.

Tsemo Aristide
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    I still don't see why this makes $\exp(ad x )$ and automorphism – samlanader Jul 22 '20 at 14:41
  • https://en.wikipedia.org/wiki/Matrix_exponential – Tsemo Aristide Jul 22 '20 at 14:57
  • True but if e.g. our Lie algebra $L = (M_n(\mathbb R), [,])$ then $ad_x \in End(L) \simeq M_{\color{red}{n^{2}}}(\mathbb R)$ hence $exp(ad_x) \in GL_{\color{red}{n^2}}(\mathbb R)$, which can be identified with $Aut(L)$ by viewing $L$ as an $n^2$-dimensional vector space. But since usually one rather identifies $Aut(L)$ with a subset of $GL_{\color{blue}{n}}(\mathbb R)$ acting on $L$ via conjugation as in the second part of my answer, all that can be a bit confusing, which is why I prefer the basis-free approach in the first part of my answer. – Torsten Schoeneberg Jul 22 '20 at 17:01