The Unapologetic Mathematician

Mathematics for the interested outsider

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 | Algebra, Lie Algebras

1 Comment »

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

    Pingback by Back to the Example « The Unapologetic Mathematician | September 7, 2012 | Reply


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