Posted by: Steven Sam | March 1, 2010

## Nilpotent orbits in graded Lie algebras

In the last post, I described the nilpotent orbits of the classical Lie algebras. I did not prove it, but there a semisimple Lie algebra always has finitely many nilpotent orbits. In this post, I want to use this knowledge to show that certain representations have finitely many orbits. In particular, our goal will be to show that the action of ${\bf GL}({\bf C}^n)$ on ${\bigwedge^3 {\bf C}^n}$ has finitely many orbits if and only if ${n \le 8}$.

First, let’s take care of an easy direction of that claim. By Chevalley’s theorem on constructible sets, all orbits are locally closed, i.e., open in their closure. So if there are finitely many orbits, then there must be an open orbit. Since this orbit has a surjective map from ${\bf GL}({\bf C}^n)$, its dimension must be less than ${n^2}$. But the dimension of this orbit is the same as the dimension of ${\bigwedge^3 {\bf C}^n}$, which is ${\binom{n}{3} = n(n-1)(n-2)/6}$. By inspection, the inequality ${\binom{n}{3} \le n^2}$ is only valid when ${n \le 8}$.

For the reverse direction, that ${n \le 8}$ implies finitely many orbits, we’ll use the following setup due to Vinberg in his paper The Weyl group of a graded Lie algebra. Let ${G}$ be a connected complex semisimple Lie group, and let ${\mathfrak{g}}$ be its Lie algebra. Give ${\mathfrak{g}}$ a ${\mathbf{Z}/m}$-grading, i.e., ${\mathfrak{g} = \bigoplus_{i \in \mathbf{Z}/m} \mathfrak{g}_i}$ such that ${[\mathfrak{g}_i, \mathfrak{g}_j] \subseteq \mathfrak{g}_{i+j}}$. Here ${m}$ could be 0 if we like, but we’ll only use the case ${m>0}$. Since ${\mathfrak{g}_0}$ is a subalgebra of ${\mathfrak{g}}$, there is a connected subgroup ${G_0}$ of ${G}$ whose Lie algebra is ${\mathfrak{g}_0}$. Then ${G_0}$ acts on ${\mathfrak{g}}$ via conjugation, and thanks to the grading, ${G_0}$ acts on each ${\mathfrak{g}_i}$. We’ll be particularly interested in the action of ${G_0}$ on ${\mathfrak{g}_1}$, but everything we say is valid for the other ${\mathfrak{g}_i}$. Our main tool will be the following proposition.

Proposition. The action of ${G_0}$ on the set of nilpotent elements in ${\mathfrak{g}_1}$ has finitely many orbits.

To prove this, we’ll use the following lemma.

Lemma. Let ${U}$ be a vector space, and ${H \subseteq {\bf GL}(U)}$ an algebraic subgroup with Lie algebra ${\mathfrak{h}}$. Let ${L \subseteq H}$ be a closed, connected subgroup of ${H}$ with Lie algebra ${\mathfrak{l}}$, and ${V \subset U}$ be a subspace which is invariant under ${L}$. Suppose that for all ${v \in V}$, one has

$\displaystyle \mathfrak{h} v \cap V = \mathfrak{l} v.$

Then the intersection of an orbit of ${H}$ with the subspace ${V}$ is nonsingular, and each of its connected components is an orbit of ${L}$.

Proof. Pick ${v_0 \in V}$ and ${v \in Hv_0 \cap V}$. Then the tangent space of ${Hv_0 \cap V}$ at ${v}$ is contained in the intersection of the tangent spaces of ${Hv_0}$ and ${V}$ at ${v}$, i.e., ${\mathfrak{h} v \cap V}$, which is by assumption the same as ${\mathfrak{l} v}$. But ${\mathfrak{l} v}$ is the tangent space of ${Lv}$ at ${v}$. Hence the tangent space of ${Hv_0 \cap V}$ at ${v}$ is contained in the tangent space of ${Lv}$ at ${v}$. But we have ${Lv \subset Hv_0 \cap V}$ by assumption that ${V}$ is closed under the action of ${L}$, so in fact we get equality of tangent spaces.

This equality implies that ${Lv}$ is open inside of ${Hv_0 \cap V}$ and that ${v}$ is a nonsingular point (since ${Lv}$ is nonsingular by virtue of being an orbit). Since ${v}$ was arbitrary, we conclude that ${Hv_0 \cap V}$ is nonsingular, and thus ${Lv}$ must be equal to one of its connected components. QED

Proof of Proposition. Let ${U = \mathfrak{g}}$ and ${V = \mathfrak{g}_1}$, and ${H}$ and ${L}$ be the images of ${G}$ and ${G_0}$, respectively, in ${\bf GL}(\mathfrak{g})$ under the action of conjugation. So the hypotheses of the lemma are satisfied. Now we use the fact that ${\mathfrak{g}}$ has finitely many nilpotent orbits under the action of ${G}$. The intersection of each orbit with ${\mathfrak{g}_1}$ has finitely many components (since they are algebraic varieties), so we are done. QED

How do we get interested gradings on ${\mathfrak{g}}$? First, we can pick a Cartan subalgebra so that ${\mathfrak{g}}$ has a root space decomposition ${\mathfrak{g} = \bigoplus_{\alpha \in \Delta} \mathfrak{g}_\alpha}$. Define a nonnegative additive function ${f \colon \Delta \rightarrow \mathbf{N}}$, and set ${\mathfrak{g}_i = \bigoplus_{\alpha,\ f(\alpha) = i} \mathfrak{g}_\alpha}$. In this case, all elements of ${\mathfrak{g}_1}$ will be nilpotent, so the action of ${G_0}$ on ${\mathfrak{g}_1}$ has finitely many orbits by the proposition. Here’s a particularly simple one that we will use. Pick a basis of simple roots ${\{ \alpha_1, \dots, \alpha_n \}}$ and pick one of the roots ${\alpha_i}$. Then define ${f(\alpha_j) = \delta_{i,j}}$ and extend linearly.

We’ll use this table from Wikipedia for Dynkin diagrams.

Let’s apply this to the Lie algebra ${\mathfrak{g} = \mathfrak{so}_{2n}}$, with root system ${D_n}$. Our distinguished simple root ${\alpha}$ will be one of the two branches at the end of the Dynkin diagram. In determining what ${\mathfrak{g}_i}$ is, we just need to know which roots have coefficient ${i}$ for the simple root that we are distinguishing. In the case ${i=0}$, we also get to use the zero roots, so we get the Cartan subalgebra as well. Then ${\mathfrak{g}_0 = \mathfrak{sl}_n \oplus {\bf C}}$: removing the distinguished node gives a Dynkin diagram of type ${A_{n-1}}$, so we get the ${\mathfrak{sl}_n}$, but we also get a 1-dimensional subalgebra ${{\bf C}}$ because the Cartan subalgebra is contained in ${\mathfrak{g}_0}$. To figure out what ${\mathfrak{g}_1}$ is, we just need to figure out the roots of type ${D_n}$ which use the marked node once when written as a linear combination of simple roots. In fact, this representation will be irreducible, so we just need to figure out what the highest weight is (with respect to the ${A_{n-1}}$ subdiagram); the Abelian part ${{\bf C}}$ acts as scalars.

The highest weight ${\alpha = \sum c_i \alpha_i}$ is the root given by labeling all valence 1 nodes with a 1, and the rest with a 2. Which weight is this with respect to the ${A_{n-1}}$ subdiagram? Let ${\omega_1, \dots, \omega_{n-1}}$ be the fundamental weights of ${A_{n-1}}$. Then the coefficient of ${\omega_i}$ in this highest weight will be the Cartan integer ${\alpha(h_i)}$ where the ${h_1, \dots, h_n}$ are the coroots spanning our Cartan subalgebra. This number is obtained by taking ${2c_i}$ and subtracting the ${c_j}$ for all neighbors ${j}$ of ${i}$. Hence we see that the weight is ${\omega_{n-2}}$, so ${\mathfrak{g}_1 = \bigwedge^{n-2} {\bf C}^n \otimes {\bf C}}$ as a representation of ${\mathfrak{g}_0}$. By symmetry, we can replace ${n-2}$ by 2. So we recover the fact that ${{\bf SL}_n} \times {\bf C}$ has finitely many orbits when acting on ${\bigwedge^2 {\bf C}^n}$.

How do we get ${\bigwedge^3 {\bf C}^n}$ to appear as ${\mathfrak{g}_1}$? Take the Lie algebras of type ${\rm E}_n$! These give finite-dimensional Lie algebras if and only if ${n \le 8}$, so we can only do these cases (as it should be, since we proved above that we need ${n \le 8}$.) We take the distinguished root to be the single branch (in the diagram, this is the node at the top). A similar analysis to the previous reveals that ${\mathfrak{g}_0 = \mathfrak{sl}_n \oplus {\bf C}}$ and ${\mathfrak{g}_1 = \bigwedge^3 {\bf C}^n \otimes {\bf C}}$. For reference, the labels for the highest weight in ${\mathfrak{g}_1}$ in each case (just the labels on the bottom row) are:

${n=6}$: 1, 2, 3, 2, 1

${n=7}$: 1, 2, 3, 3, 2, 1

${n=8}$: 1, 2, 3, 3, 3, 2, 1

And that’s just some of the examples that we can do. There are many other diagrams and nodes to work with. This will produce many examples of representations with finitely many orbits, but not all will appear in this way. For more details on this, see the paper Some remarks on nilpotent orbits by Victor Kac.

-Steven