# The Unapologetic Mathematician

## Automorphisms of Lie Algebras

Sorry for the delay; I’ve had a couple busy days. Here’s Thursday’s promised installment.

An automorphism of a Lie algebra $L$ is, as usual, an invertible homomorphism from $L$ onto itself, and the collection of all such automorphisms forms a group $\mathrm{Aut}(L)$.

One obviously useful class of examples arises when we’re considering a linear Lie algebra $L\subseteq\mathfrak{gl}(V)$. If $g\in\mathrm{GL}(V)$ is an invertible endomorphism of $V$ such that $gLg^{-1}=L$ then the map $x\mapsto gxg^{-1}$ is an automorphism of $L$. Clearly this happens for all $g$ in the cases of $\mathfrak{gl}(V)$ and the special linear Lie algebra $\mathfrak{sl}(V)$ — the latter because the trace is invariant under a change of basis.

Now we’ll specialize to the (usual) case where no multiple of $1\in\mathbb{F}$ is zero, and we consider an $x\in L$ for which $\mathrm{ad}(x)$ is “nilpotent”. That is, there’s some finite $n$ such that $\mathrm{ad}(x)^n=0$ — applying $y\mapsto[x,y]$ sufficiently many times eventually kills off every element of $L$. In this case, we say that $x$ itself is “ad-nilpotent”.

In this case, we can define $\exp(\mathrm{ad}(x))$. How does this work? we use the power series expansion of the exponential: $\displaystyle\exp(\mathrm{ad}(x))=\sum\limits_{k=0}^\infty\frac{\mathrm{ad}(x)^k}{k!}$

We know that this series converges because eventually every term vanishes once $\mathrm{ad}(x)^k=0$.

Now, I say that $\exp(\mathrm{ad}(x))\in\mathrm{Aut}(L)$. In fact, while this case is very useful, all we need from $\mathrm{ad}(x)$ is that it’s a nilpotent derivation $\delta$ of $L$. The product rule for derivations generalizes as: $\displaystyle\frac{\delta^n}{n!}(xy)=\sum\limits_{i=0}^n\frac{1}{i!}\delta^i(x)\frac{1}{(n-i)!}\delta^{n-i}(y)$

So we can write \displaystyle\begin{aligned}\exp(\delta(x))\exp(\delta(y))&=\left(\sum\limits_{i=0}^{n-1}\frac{\delta^i(x)}{i!}\right)\left(\sum\limits_{j=0}^{n-1}\frac{\delta^j(y)}{j!}\right)\\&=\sum\limits_{k=0}^{2n-2}\left(\sum\limits_{i=0}^k\frac{\delta^i(x)}{i!}\frac{\delta^{k-i}(y)}{(k-i)!}\right)\\&=\sum\limits_{k=0}^{2n-2}\frac{\delta^k(xy)}{k!}\\&=\sum\limits_{k=0}^{n-1}\frac{\delta^k(xy)}{k!}\\&=\exp(\delta(xy))\end{aligned}

That is, $\exp{\delta}$ preserves the multiplication of the algebra that $\delta$ is a derivation of. In particular, in terms of the Lie algebra $L$, we find that $\displaystyle[\exp(\delta(x)),\exp(\delta(y))]=\exp(\delta([x,y]))$

Since $\exp(\delta):L\to L$ we conclude that this is an epimorphism of $L$. It’s invertible by the usual formula $\displaystyle(1+\eta)^{-1}=1-\eta+\eta^2-\cdots\pm\eta^{n-1}$

which means it’s an automorphism of $L$.

Just like a derivation of the form $\mathrm{ad}(x)$ is called inner, an automorphism of the form $\exp(\mathrm{ad}(x))$ is called an inner automorphism, and the subgroup $\mathrm{Inn}(L)$ they generate is a normal subgroup of $\mathrm{Aut}(L)$. Specifically, if $\phi\in\mathrm{Aut}(L)$ and $x\in L$ then we can calculate \displaystyle\begin{aligned}\phi(\mathrm{ad}(x)(\phi^{-1}(y)))&=\phi([x,\phi^{-1}(y)])\\&=[\phi(x),y]\\&=\mathrm{ad}(\phi(x))(y)\end{aligned}

and thus $\displaystyle\phi\exp(\mathrm{ad}(x))\phi^{-1}=\exp(\mathrm{ad}(\phi(x)))$

so the conjugate of an inner automorphism is again inner.

August 18, 2012 - Posted by | Algebra, Lie Algebras