The Unapologetic Mathematician

Mathematics for the interested outsider

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

3 Comments »

  1. […] . All we need now is to verify that this is the inverse of , but the expanded Leibniz identity from last time tells us that , thus proving our […]

    Pingback by An Explicit Example « The Unapologetic Mathematician | August 18, 2012 | Reply

  2. […] all . In particular, applying enough times will eventually kill any element of . That is, each is ad-nilpotent. It turns out that the converse is also true, which is the content of Engel’s […]

    Pingback by Engel’s Theorem « The Unapologetic Mathematician | August 22, 2012 | Reply

  3. […] on , and obviously its action on the subalgebra is nilpotent as well. Thus each element of is ad-nilpotent, and Engel’s theorem then tells us that is a nilpotent Lie algebra. Share […]

    Pingback by Flags « The Unapologetic Mathematician | August 25, 2012 | Reply


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