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 on has finitely many orbits if and only if .

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 , its dimension must be less than . But the dimension of this orbit is the same as the dimension of , which is . By inspection, the inequality is only valid when .

For the reverse direction, that 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 be a connected complex semisimple Lie group, and let be its Lie algebra. Give a -grading, i.e., such that . Here could be 0 if we like, but we’ll only use the case . Since is a subalgebra of , there is a connected subgroup of whose Lie algebra is . Then acts on via conjugation, and thanks to the grading, acts on each . We’ll be particularly interested in the action of on , but everything we say is valid for the other . Our main tool will be the following proposition.

**Proposition.** The action of on the set of nilpotent elements in has finitely many orbits.

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

**Lemma.** Let be a vector space, and an algebraic subgroup with Lie algebra . Let be a closed, connected subgroup of with Lie algebra , and be a subspace which is invariant under . Suppose that for all , one has

Then the intersection of an orbit of with the subspace is nonsingular, and each of its connected components is an orbit of .

*Proof*. Pick and . Then the tangent space of at is contained in the intersection of the tangent spaces of and at , i.e., , which is by assumption the same as . But is the tangent space of at . Hence the tangent space of at is contained in the tangent space of at . But we have by assumption that is closed under the action of , so in fact we get equality of tangent spaces.

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

*Proof of Proposition*. Let and , and and be the images of and , respectively, in under the action of conjugation. So the hypotheses of the lemma are satisfied. Now we use the fact that has finitely many nilpotent orbits under the action of . The intersection of each orbit with has finitely many components (since they are algebraic varieties), so we are done. QED

How do we get interested gradings on ? First, we can pick a Cartan subalgebra so that has a root space decomposition . Define a nonnegative additive function , and set . In this case, all elements of will be nilpotent, so the action of on has finitely many orbits by the proposition. Here’s a particularly simple one that we will use. Pick a basis of simple roots and pick one of the roots . Then define and extend linearly.

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

Let’s apply this to the Lie algebra , with root system . Our distinguished simple root will be one of the two branches at the end of the Dynkin diagram. In determining what is, we just need to know which roots have coefficient for the simple root that we are distinguishing. In the case , we also get to use the zero roots, so we get the Cartan subalgebra as well. Then : removing the distinguished node gives a Dynkin diagram of type , so we get the , but we also get a 1-dimensional subalgebra because the Cartan subalgebra is contained in . To figure out what is, we just need to figure out the roots of type 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 subdiagram); the Abelian part acts as scalars.

The highest weight 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 subdiagram? Let be the fundamental weights of . Then the coefficient of in this highest weight will be the Cartan integer where the are the coroots spanning our Cartan subalgebra. This number is obtained by taking and subtracting the for all neighbors of . Hence we see that the weight is , so as a representation of . By symmetry, we can replace by 2. So we recover the fact that has finitely many orbits when acting on .

How do we get to appear as ? Take the Lie algebras of type ! These give finite-dimensional Lie algebras if and only if , so we can only do these cases (as it should be, since we proved above that we need .) 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 and . For reference, the labels for the highest weight in in each case (just the labels on the bottom row) are:

: 1, 2, 3, 2, 1

: 1, 2, 3, 3, 2, 1

: 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

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