so let's consider for example the coadjoint partial flag variety of sl(2,reals), as the projective coadjoint orbit of a highest weight vector in the coadjoint representation... perhaps we can get away with using the complexified coadjoint representation here... yes, seems like it... just so i don't have to worry about to what extent stuff like "highest weight vector" makes sense using the uncomplexified coadjoint representation... the group itself remains uncomplexified though... so then we want to try to figure out where, if anywhere, this orbit fits into the picture given by the orbit space of the root system under the kaleidoscope group... the "if anywhere" issue seems significant here... ??....

first let me try to see whether there's some standard "normal form" theorem for linear operators on finite-dimensional vector spaces over a field that i more or less know that's relevant here...

hmm, first let's consider the characteristic polynomial of a 2-by-2 matrix with real entries as an invariant of the matrix...

so given real numbers a,b, consider the affine real algebraic variety of matrixes m=

c d
e f

satisfying m^2 = m*a + b ... or something like that...

actually not too much like that; i was getting mixed up here between merely satisfying a polynomial p and having it as characteristic polynomial... or something...

g h
h j

g+j=1
gj-hh=0

g(1-g)=hh

g-gg=hh

sorry, just trying a few things...

anyway, constraining trace and determinant tends to cut down from 4 dimensions c,d,e,f to 2 ...

c d
e -c

determinant = -(cc+de)

hmmm... setting determinant to zero here as (quadratic) homogeneous constraint...

??so the coadjoint partial flag variety is here the _only_ projective coadjoint orbit that's contact rather than symplectic??? or something??? ???what's going on here???

??hmm, is there something here about "homogeneous poisson ideal" of the enveloping poisson algebra ??? or something???

??so in the split real a-series case, an adjoint orbit is closed under scalar multiplication precisely in case all the eigenvalues of the operator are zero?? ??is that correct??

??so given a linear operator, consider the flag obtained by the kernels of its powers... ?? ...or something...