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? 



How can we describe the splitting of nilpotent orbit for "very even" partitions in the special orthogonal group?
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](https://mathoverflow.net/questions/9361/how-can-we-describe-the-splitting-of-nilpotent-orbit-for-very-even-partitions-i/9501#9501). 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. 


