Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In a well known construction of finite W-algebras, one first constructs a certain nilpotent subalgebra $\mathfrak{m}$ along with a character $\chi:\mathfrak{m}\rightarrow \mathbb{C}$. Then one defines

$$U(\mathfrak{g},e)=(U(\mathfrak{g})/U(\mathfrak{g})\mathfrak{m}_{\chi})^\mathfrak{m}$$

where $\mathfrak{m}_\chi$ is the set of all $m-\chi(m)$ and $\mathfrak{m}$ acts on $U(\mathfrak{g})$ by derivations, extending the adjoint action on $\mathfrak{g}$. Is this the same as

$$U(\mathfrak{g})^{\mathfrak{m}}/(U(\mathfrak{g})\mathfrak{m}_{\chi})^\mathfrak{m}?$$

Of course one can reformulate this question and ask if the following cohomology group vanishes:
$$H^1(\mathfrak{m},U(\mathfrak{g})\mathfrak{m}_{\chi})=0?$$ Maybe this follows from some Lynch style vanishing, but I am not very familiar with these theorems.

share|improve this question
    
Two very minor edits. Maybe lie-algebras would also be a useful tag? –  Jim Humphreys Sep 22 '10 at 16:46
    
Thanks, added lie-algebras tag. –  Jan Weidner Sep 22 '10 at 18:31
add comment

1 Answer

Look at Propositions 5.1 and 5.2 of Gan and Ginzburg's paper Quantization of Slodowy slices. The "reason" behind the vanishing is its identification with algebraic deRham cohomology of an affine space.

share|improve this answer
    
Thanks, unfortunately I do not yet quite see, how to use this in my situation. Do you suggest to show that $U(g)mχ=\mathbb{C}[M]\otimes_\mathbb{C} W$ for some trivial rep $W$ and then conclude $H^i(m,U(g)mχ)=H^i(m,\mathbb{C}[M])\otimes W=0$? Or did I get this completely wrong? –  Jan Weidner Sep 23 '10 at 9:01
    
I suggest you read the paper, where the statements you want are propositions cited, and the proof is given. –  Ben Webster Sep 23 '10 at 20:14
    
I already read this paper, quite thoroughly. What Gan and Ginzburg show in these propositions is that the higher cohomology groups $H^i(m,U(g)/U(g)m_\chi)$ vanish. What I want is that $H^i(M,U(g)m_\chi)$ vanishes. I don't see how their propositions imply my statement, nor how to adapt their proof. Maybe I miss something obvious. –  Jan Weidner Sep 24 '10 at 7:40
1  
Their proof goes as follows. By a spectral sequence argument they show that it is enough to check that $H^i(m,gr(U(g)/U(g)m_\chi))$ vanishes. Now they already showed that there is an equivariant isomorphism $gr(U(g)/U(g)m_\chi)=\mathbb{C}[M]\otimes \mathbb{C}[S]$ where the action on the second factor is trivial. Then they deduce $H^i(m,gr(U(g)/U(g)m_\chi))=H^i(m,\mathbb{C}[M])\otimes \mathbb{C}[S]$. Now as you already remarked $H^i(m,\mathbb{C}[M])$ is the algebraic deRham cohomology of $N\cong\mathbb(A)^n$, which vanishes for $i>0$. –  Jan Weidner Sep 24 '10 at 7:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.