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I understand if a partition $\lambda$ has all parts even and all multiplicities even, then the nilpotent orbit corresponding to $\lambda$ splits up into two orbits. By the nilpotent orbit corresponding to $\lambda$, what I mean is the set of all orthogonal nilpotent matrices with Jordan type $\lambda$; by 'splitting' I mean that this set is not a single orbit under the action of $SO_{n}(\mathbb{C})$, but is the union of two orbits (this is a theorem from Collingwood & McGovern - "Nilpotent orbits in Semisimple Lie Algebras").

How can we describe explicitly this ''splitting'' - i.e. since we know the affine coordinate ring of the closure of $\mathcal{O}_{\lambda}$ (this is the set of all orthogonal nilpotent matrices for the type $\lambda$), by having the equation for a matrix to be orthogonal, together with the equations for it to be nilpotent of type $\lambda$, how can we describe algebraically each of the two orbits that it splits up into?

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You're missing some important piece of context here...what splits in what context? Restriction to a subgroup? – Ben Webster Dec 19 '09 at 14:49
Ben: I think I understand the question. See my answer below. – David Speyer Dec 19 '09 at 22:26
Sorry I've tried to clarify my question and add more details. (It is what you thought I meant, David). – Vinoth Dec 20 '09 at 4:59
up vote 2 down vote accepted

OK, I have an answer to the question of how to distinguish different $SO_{2n}$ orbits that have the same Jordan form. I also have a proof that it is correct, but that is much longer, using the ideas in Ben's excellent answer. So, the short answer for now, and the long answer if it is needed.

Let $N$ be a nilpotent matrix acting on a vector space $V$. Define the vector space $W$ to be the subspace of $V$ given by

$$N \ker N^2 + N^2 \ker N^4 + N^3 \ker N^6 + \cdots.$$

Exercise: If $N$ is skew-self-adjoint with respect to an inner product $\langle \ , \ \rangle$, then $W$ is isotropic for $\langle \ , \rangle$.

In terms of Jordan decomposition, $W$ is the "latter half" of each Jordan block.

In particular, suppose all of the Jordan blocks of $N$ have unit size. Then $W$ has dimension $(1/2) \dim V$. It is well known that the Lagrangian grassmannian has two connected components. The orbit of $N$ is determined by which component $W$ is in.

So, how do we determine which component of the Lagrangian grassmannian $W$ is in? First, change bases so that $\langle e_i, e_j \rangle = \delta_{i (j+n)}$, where the indices are modulo $2n$. Consider the $2^n$ subsets $I$ of $\{ 1,2, \ldots, 2n \}$ which contain exactly one of $\{ i, n+i \}$ for each $i$. For half of these $I$, the corresponding Plucker coordinate $p_I(W)$ is zero. Which half? That tells you which component we are in.

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Thanks. I'm slightly confused as to $W$ has that dimension - if the Jordan type is $(1^{2k})$ doesn't that mean that $N$=0, making that subspace the zero subspace? Perhaps you meant that the Jordan type is a partition with all parts & multiplicities even (and only in this case), $W$ has $(1/2) dim V$; my calculations seem to confirm this. Then I understand that if $N$ and $N'$ are conjugate, the conjugacy preserves the component of $W$ that it is in. – Vinoth Dec 22 '09 at 2:16
Next, I guess you mean the Plucker embedding of this Lagrangian Grassmanian inside the projective space of the $n$ -th exterior power, and this projective variety has two components, each of which involves wedge products of $n$ of the $e_j$, involving one of either $e_{i}$ or $e_{n+i}$, and the Plucker coordinate $p_{I}(W)$ refers to the number in front of the wedge product of the $e_i$'s corresponding to $I$. That seems to make sense to me now :), thanks. – Vinoth Dec 22 '09 at 2:24
Just noticed that "unit size" should read "even size". – David Speyer Apr 2 '10 at 13:02

I don't have a full answer, or a good reference, but here's how I think one should answer the question:

By Jacobson-Morozov (see Chriss & Ginzburg for more details), classifying nilpotents in a semi-simple Lie algebra up to conjugacy is basically the same as classifying homomorphisms of $\mathfrak{sl}(2)$ into that Lie algebra, up to conjugacy.

So, a nilpotent orbit in $\mathfrak{so}(n)$ is the same as a representation of $\mathfrak{sl}(2)$ on $\mathbbb{C}^n$ which preserves the standard bilinear form, up to changes of basis preserving that form (EDIT: as David pointed out, I was a bit sloppy about writing this sentence the first time around). I leave classifying those as an exercise to the reader.

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The form should be $\mathfrak{sl}(2)$-invariant, right? – Mariano Suárez-Alvarez Dec 21 '09 at 20:21
Nice answer! You have to be careful about what you mean by isomorphism, though. In the case of $SO_4$, there are two $SL_2$ actions on $\mathbb{C}^4$ which are not $SO_4$-conjugate, but are isomorphic in the sense that there is an automorphism of $\mathbb{C}^4$ carrying one to the other. These are precisely the two orbits that rajamanikkam wants to distinguish! – David Speyer Dec 21 '09 at 21:16
And said autmorphism preserves the bilinear form, I should add. – David Speyer Dec 21 '09 at 21:36

I don't know the answer, but I am posting to make sure I understand the question and give some partial ideas.

As Ben says, the question is hard to follow. I am interpreting it this way:

Consider the cone of nilpotent elements in $\mathfrak{so}_{2n}$. How does it break into $SO_{2n}$ orbits under the adjoint action? How is this related to the decomposition according to Jordan normal form? (Note: the decomposition according to Jordan normal form can be thought of as the $SL_{2n}$ orbits for the nilpotent matrices in $\mathfrak{sl}_{2n}$.)

Some partial thoughts. The Lie algebra $\mathfrak{so}_{2n}$ is skew symmetric matrices. Any skewsymmetric matrix has even rank. Moreover, if $A$ is skewsymmetric, so is every odd power of $A$. So $A$, $A^3$, $A^5$ etcetera all have even rank. This implies that, if $\lambda$ is the partition of $2n$ coming from the Jordan form of $A$, then every even part of $\lambda$ occurs with even multiplicity. So, when $2n=4$, the only partitions that occur are $31$, $22$ and $1111$. So we have determined which Jordan forms appear.

In the particular case $n=4$, we have $\mathfrak{so}_4 \cong \mathfrak{sl}_2 \oplus \mathfrak{sl}_2$. Using this isomorphsim, one checks that $n_1 \oplus n_2$ is nilpotent if and only if $n_1$ and $n_2$ are. We get Jordan form $31$ if both $n_1$ and $n_2$ are nonzero nilpotents; we get $22$ is one is zero and the other isn't; and we get $1111$ if both are nilpotent.

So we see that Jordan form $22$ splits into two orbits, according to which of $n_1$ and $n_2$ is nonzero. Presumably, rajamanikkam knows some theorem which says that this happens whenever all the parts of $\lambda$ are even, and would like to know how to make the theorem explicit.

In general, I know how I would attack this problem. The Jordan form of $A$ fixes the Jordan form of $S_{\mu}(A)$, for any Schur functor $S_{\mu}$. But the $SO_{2n}$ conjugacy class fixes the Jordan form $A$ acting on any $\mathfrak{so}_{2n}$ representation. In particular, the Jordan form of the action of $A$ on the spin representations should give additional data, allowing us to separate distinct $SO_{2n}$ orbits.

I don't know the details of how this works, but I'm sure someone reading this does!

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I think I know what sort of questions I would have to answer to answer your question. This feel like the sort of thing that is easy for an expert, but hard for me, so I'm really hoping David BZ, Ben, or someone like them will fill in the gaps.

Let $D$ denote the Weyl group $S_{n} \ltimes (\mathbb{Z}/2)^{n-1}$.

Looking at Theorem 3.6.9 in Representation Theory and Complex Geometry, by Chriss and Ginzburg:

Irreducible $\mathbb{C}$-representations of $D$ are in bijection with $SO_{2n}$ conjugacy classes of pairs $(x, \chi)$ where $x \in \mathfrak{so}_{2n}$ is nilpotent and $\chi$ is a character of a certain finite group $C(x)$, obeying certain additional conditions.

You, of course, want to describe $SO_{2n}$ conjugacy classes of nilpotent elements, without the extra data of $\chi$. My gut suspects that this corresponds to irreducible $\mathbb{Q}$-representations of $D$, but I can't find a reference.

Question 1: In the Springer variety picture, which object corresponds to $SO_{2n}$ conjugacy classes of nilpotent elements, forgetting the character data?

You are interested in relating $SL_{2n}$ and $SO_{2n}$ conjugacy classes. Presumably, this is related to the maps of groups $D \to S_{2n}$.

Question 2: What operation on representation theory of Weyl groups corresponds to intersecting with a sub-Lie-algebra?

Finally, you want to make this all explicit. So

Question 3: Explain how to convert a representation of $D$ to the Jordan form of the corresponding nilpotent acting on a Spin representation.

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This sounds WAY too complicated to me. If you are trying to answer this question using the Springer correspondence, I suspect you are thinking way too hard. – Ben Webster Dec 21 '09 at 19:57
Thanks for spending so much time thinking about my question, David, and Ben. Your answers are very helpful. I'm still learning about the Springer correspondence- would you suggest that Chriss & Ginzburg is the best place for me to learn it from? Or should I try reading, some paper (perhaps Springer's original paper "A construction of some representations of Weyl groups"?) Chriss & Ginzburg is very good, but for Springer correspondence it seems to focus on SL in Chapter 3, and come back to it at the very end, so I wasn't sure if it was the best place to learn Springer correspondence from. – Vinoth Dec 22 '09 at 1:25

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