bio | website | |
---|---|---|
location | ||
age | ||
visits | member for | 5 years, 5 months |
seen | Jul 19 at 4:37 | |
stats | profile views | 3,306 |
Jul 14 |
comment |
Homological dimension of Joseph quotients
I think Tony doesn't know the answer (and this question is definitely not discussed in his 1976 paper). At first I thought that the homological dimension should always be finite but it now seems to me that for $so(8)$ I have a very roundabout argument which proves that it can't be so... |
Jul 13 |
asked | Homological dimension of Joseph quotients |
Mar 31 |
awarded | Nice Question |
Feb 8 |
awarded | Yearling |
Jan 21 |
asked | Mellin transform of Plancherel measure |
Dec 31 |
awarded | Revival |
Dec 30 |
answered | Definition of Givental $J$-function of cotangent bundle of flag variety |
Dec 30 |
asked | Non-commutative normalization |
Oct 24 |
comment |
Annihillator of the highest weight vector in a finite-dimensional representation
"Degree" actually means weight. In fact, I want something more general. Let $\lambda$ be any weight. Then we can consider the Verma module $M(\lambda)$ and the simple module $L(\lambda)$. We know exactly for which $\lambda$ these are not isomorphic and in what weights they are different for every $\lambda$ (this is given by the Shapovalov determinant formula). I'd like to see this explicitly - namely, for every such weight I'd like to have an explicit (one) element of $M(\lambda)$ which vanishes in $L(\lambda)$. |
Oct 23 |
comment |
Annihillator of the highest weight vector in a finite-dimensional representation
I am not looking for a general answer, I need just some non-trivial elements. For example, I expect that it should be possible to write one canonical element for every degree. |
Oct 23 |
asked | Annihillator of the highest weight vector in a finite-dimensional representation |
Oct 4 |
comment |
Weight multiplicity formulae for $(\mathfrak g,B)$-irreps
I second what Jim says - I doubt that there is anything better than what is given by the Kazhdan-Lusztig conjecture. |
Aug 18 |
accepted | FIltrations on a vector bundle on a curve |
Aug 18 |
comment |
FIltrations on a vector bundle on a curve
Thanks! This was exactly the reference I was looking for. |
Aug 17 |
comment |
FIltrations on a vector bundle on a curve
No, the point is that if $E$ has rank 2 then the claim is that we can find a line subbundle $E_1$ of $E$ such that $deg(E_1)-deg(E/E_1)$ is not too small (I think you can always make it smaller than $g$ in this case) -- this is true even when the degree of $E$ is very negative. |
Aug 17 |
revised |
FIltrations on a vector bundle on a curve
added 19 characters in body |
Aug 16 |
comment |
FIltrations on a vector bundle on a curve
Yes, if you wish (otherwise the question is empty) |
Aug 16 |
comment |
FIltrations on a vector bundle on a curve
Well, it has something to do with Harder-Narasimhan, but it is not quite that. For example, if $n=2$ then the statement is that any rank 2 bundle has a line sub-bundle whose degree is not too small. |
Aug 16 |
asked | FIltrations on a vector bundle on a curve |
Aug 16 |
accepted | A question on non-archimedian Fourier transform |