# The Unapologetic Mathematician

## An Explicit Example

Let’s pause and catch our breath with an actual example of some of the things we’ve been talking about. Specifically, we’ll consider $L=\mathfrak{sl}(2,\mathbb{F})$ — the special linear Lie algebra on a two-dimensional vector space. This is a nice example not only because it’s nicely representative of some general phenomena, but also because the algebra itself is three-dimensional, which helps keep clear the distinction between $L$ as a Lie algebra and the adjoint action of $L$ on itself, particularly since these are both thought of in terms of matrix multiplications.

Now, we know a basis for this algebra:

\displaystyle\begin{aligned}x&=\begin{pmatrix}0&1\\ 0&0\end{pmatrix}\\y&=\begin{pmatrix}0&0\\1&0\end{pmatrix}\\h&=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\end{aligned}

which we will take in this order. We want to check each of the brackets of these basis elements:

\displaystyle\begin{aligned}{}[h,x]&=2x\\ [h,y]&=-2y\\ [x,y]&=h\end{aligned}

Writing out each bracket of basis elements as a (unique) linear combination of basis elements specifies the bracket completely, by linearity. We call the coefficients the “structure constants” of $L$, and they determine the algebra up to isomorphism.

Okay, now we want to use this basis of the vector space $L$ and write down matrices for the action of $\mathrm{ad}(x)$ on $L$:

\displaystyle\begin{aligned}\mathrm{ad}(x)&=\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}\\\mathrm{ad}(y)&=\begin{pmatrix}0&0&0\\ 0&0&2\\-1&0&0\end{pmatrix}\\\mathrm{ad}(h)&=\begin{pmatrix}2&0&0\\ 0&-2&0\\ 0&0&0\end{pmatrix}\end{aligned}

Now, both $\mathrm{ad}(x)$ and $\mathrm{ad}(-y)$ are nilpotent. In the case of $x$ we can see that $\mathrm{ad}(x)$ sends the line spanned by $y$ to the line spanned by $h$, the line spanned by $h$ to the line spanned by $x$, and the line spanned by $x$ to zero. So we can calculate the powers:

\displaystyle\begin{aligned}\mathrm{ad}(x)^0&=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\\\mathrm{ad}(x)^1&=\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}\\\mathrm{ad}(x)^2&=\begin{pmatrix}0&-2&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\\\mathrm{ad}(x)^3&=\begin{pmatrix}0&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\end{aligned}

and the exponential:

\displaystyle\begin{aligned}\exp(\mathrm{ad}(x))&=\frac{1}{0!}\mathrm{ad}(x)^0+\frac{1}{1!}\mathrm{ad}(x)^1+\frac{1}{2!}\mathrm{ad}(x)^2\\&=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}+\begin{pmatrix}0&0&-2\\ 0&0&0\\ 0&1&0\end{pmatrix}+\frac{1}{2}\begin{pmatrix}0&-2&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\\&=\begin{pmatrix}1&-1&-2\\ 0&1&0\\ 0&1&1\end{pmatrix}\end{aligned}

Similarly we can calculate the exponential of $\mathrm{ad}(-y)$:

$\displaystyle\exp(\mathrm{ad}(-y))=\begin{pmatrix}1&0&0\\-1&1&-2\\1&0&1\end{pmatrix}$

So now it’s a simple matter to write down the following element of $\mathrm{Int}(L)$:

$\displaystyle\sigma=\exp(\mathrm{ad}(x))\exp(\mathrm{ad}(-y))\exp(\mathrm{ad}(x))=\begin{pmatrix}0&-1&0\\-1&0&0\\ 0&0&-1\end{pmatrix}$

In other words, $\sigma(x)=-y$, $\sigma(y)=-x$, and $\sigma(h)=-h$.

We can also see that $x$ and $-y$ themselves are also nilpotent, as endomorphisms of the vector space $\mathbb{F}^2$. We can calculate their exponentials:

\displaystyle\begin{aligned}\exp(x)&=\begin{pmatrix}1&1\\ 0&1\end{pmatrix}\\\exp(-y)&=\begin{pmatrix}1&0\\-1&1\end{pmatrix}\end{aligned}

and the product:

$\displaystyle s=\exp(x)\exp(-y)\exp(x)=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$

It’s easy to check from here that conjugation by $s$ has the exact same effect as the action of $\sigma$: $sls^{-1}=\sigma(l)$.

This is a very general phenomenon: if $L\subseteq\mathfrak{gl}(V)$ is any linear Lie algebra and $x\in L$ is nilpotent, then conjugation by the exponential of $x$ is the same as applying the exponential of the adoint of $x$.

Indeed, considering $\mathrm{ad}(x)\in\mathrm{End}_\mathbb{F}(\mathrm{End}(V))$, we can write it as

$\displaystyle\mathrm{ad}(x)=\lambda_x+\rho_{-x}$

where $\lambda_x$ and $\rho_x$ are left- and right-multiplication by $x$ in $\mathrm{End}(V)$. Since these two commute with each other and both are nilpotent we can write

\displaystyle\begin{aligned}\exp(\mathrm{ad}(x))&=\exp(\lambda_x+\rho_{-x})\\&=\exp(\lambda_x)\exp(\rho_{-x})\\&=\lambda_{\exp(x)}\rho_{\exp(-x)}\end{aligned}

That is, the action of $\exp(\mathrm{ad}(x))$ is the same as left-multiplication by $\exp(x)$ followed by right-multiplication by $\exp(-x)$. All we need now is to verify that this is the inverse of $\exp(x)$, but the expanded Leibniz identity from last time tells us that $\exp(x)\exp(-x)=\exp(x-x)=\exp(0)=1_V$, thus proving our assertion.

We can also tell at this point that the nilpotency of $x$ and $-y$ and that of $\mathrm{ad}(x)$ and $\mathrm{ad}(-y)$ are not unrelated. Indeed, if $x\in\mathfrak{gl}(V)$ is nilpotent then $\mathrm{ad}(x)\in\mathrm{End}(\mathfrak{gl}(V))$ is, too. Indeed, since $\lambda_x$ and $\rho_x$ are commuting nilpotents, their difference — $\mathrm{ad}(x)=\lambda_x-\rho_x$ — is again nilpotent.

We must be careful to note that the converse is not true. Indeed, $I_V\in\mathfrak{gl}(V)$ is ad-nilpotent, but $I_V$ itself is certainly not nilpotent.

August 18, 2012 Posted by | Algebra, Lie Algebras | 1 Comment

## 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