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.

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

   

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