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.

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

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[...] go back to our explicit example of and look at its Killing form. We first recall our usual [...]

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