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 mathfrakso2n. How does it break into SO2n 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 SL2n orbits for the nilpotent matrices in mathfraksl2n.)
Some partial thoughts. The Lie algebra mathfrakso2n is skew symmetric matrices. Any skewsymmetric matrix has even rank. Moreover, if A is skewsymmetric, so is every odd power of A. So A, A3, A5 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 mathfrakso4congmathfraksl2oplusmathfraksl2. Using this isomorphsim, one checks that n1oplusn2 is nilpotent if and only if n1 and n2 are. We get Jordan form 31 if both n1 and n2 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 n1 and n2 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 Smu(A), for any Schur functor Smu. But the SO2n conjugacy class fixes the Jordan form A acting on any mathfrakso2n representation. In particular, the Jordan form of the action of A on the spin representations should give additional data, allowing us to separate distinct SO2n orbits.
I don't know the details of how this works, but I'm sure someone reading this does!
No comments:
Post a Comment