Timeline for Are nilpotent orbits degenerations of semi-simple orbits ?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2012 at 3:24 | comment | added | Victor Protsak | Alexander, it's not possible to deform e.g. the minimal nilpotent orbit outside of type A while preserving the dimension - it's an interesting calculation for classical Lie algebras. | |
| Jan 14, 2012 at 16:51 | comment | added | Alexander Chervov | Let us look more carefully what happens for t≠0. Equations $A^2=t^2Id$ easy to solve−they define 4 semisimple orbits diag(t,t,t);diag(−t,−t,−t);diag(t,−t,−t);diag(t,t,−t). Orbits of diag(t,t,t);diag(−t,−t,−t) are just points−notveryinteresting. Orbits diag(t,−t,−t);diag(t,t,−t) have the same dimension as rank2 nilpotent orbit(amIright?). So I almost got what I want−a system of equtions−such that fort=0−closure of rank2 nilp.orbit,and for $t\ne 0$ we have semi-simple orbit(s) of the SAME dimension. Some trouble here is that that in semi-simple case not only one orb | |
| Jan 14, 2012 at 16:40 | comment | added | Alexander Chervov | @Jim Let me try to describe the answer to the question in situation of rank 2 orbits. As you suggest. Consider gl(3)=Mat(3,3)=C^9. Let us write a system of 9 equations on matrix A: A^2=t^2Id. We see that for t=0, we get a closure of rank 2 nilpotent orbit. And for $t\ne 0$ we get equations defining semisimple orbits (more precisely uninion of several orbits, see below)... | |
| Jan 14, 2012 at 16:13 | comment | added | Alexander Chervov | @Jim Thank you very much for your answer. Of course I want degeneration preserving dimensions of orbits. However, indeed, thinking of what I asked, I find it might not be good question, since in some sense I can (can I?) achieve degeneration in rather trivial way - just take a curve s(t) in "g" , such that s(t), for t=0 we get some nilpotent element, and for $t\ne 0$ we get semi-simples, SUCH THAT their orbits have the same dimension for all t. I think I can do this, am I right ? (My question required may be more subtle thing, but may be it is not necessary). | |
| Jan 14, 2012 at 16:00 | history | answered | Jim Humphreys | CC BY-SA 3.0 |