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

My question is about an isomorphism between two varieties of the form discussed in this thread; it is also related to an earlier question of mine.

Let $z_n$ be the standard nilpotent of type $(n,n)$. For $1 \leq k \leq 2n-1$, let $P_k$ be the parabolic subgroup of $GL(2n)$ stabilizing a flag of the form $(0 \subset V_1 \subset \cdots \subset \widehat{V_k} \subset \cdots \subset V_{2n}=V)$. Let $S_n$ denote the Slodowy slice to $z_n$, and let $\pi_k: T^* G/P_k \rightarrow \mathfrak{gl}_{2n}$ be the natural projection. Consider also the Borel $B \subset GL(2n)$, and the Springer map $\pi: T^*G/B \rightarrow \mathfrak{gl}_{2n}$. Then my question is: Why is $\pi_k^{-1}(S_{n}) \simeq \pi^{-1}(S_{n-1})?$

This claim is made on page 10 (immediately after the proof of Lemma $5$) in this paper. I understand why $\pi_k^{-1}(z_n) \simeq \pi^{-1}(z_{n-1})$ (this is the Lemma $5$ I just mentioned). This follows, since if $(0 \subset V_1 \subset \cdots \subset \widehat{V_k} \subset \cdots \subset V_{2n}=V) \in \pi_k^{-1}(z_n)$, it is stabilized by $z_n$, then $z_n V_{k+1}=V_{k-1}$ since $\text{ker}(z_n)$ is $2$-dimensional; and we can map this flag to $(0 \subset V_1 \subset \cdots \subset V_{k-1}=z_n V_{k+1} \subset \cdots \subset z_n V_{2n}) \in \pi^{-1}(z_{n-1})$. A similar argument may be made when we replace $z_n$ by a nilpotent $x \in S_n \cap \text{im}(\pi_k)$, showing that $\pi_k^{-1}(x) \simeq \pi^{-1}(x|_{xV})$ but then the vector space $xV$ is not a fixed vector space, and I am having trouble patching up the isomorphism.

I hope it's ok for me to post this question here (maybe it's better to ask the author of the paper, but I wasn't able to contact the author in this case.)

share|improve this question

1 Answer 1

If you're talking about the Slodowy slice, then you've picked $f_n$ which makes an $\mathfrak{sl}_2$ with Chevalley presentation against $z_n$. If $x\in S_n$, then $xV$ is transverse to $\ker f_n$ (the $-n$-weight space of our $\mathfrak{sl}_2$); if one filters $\mathbb{C}^{2n}$ by things of weight $\leq k$, then $x$ acts with degree $2$, and its action on the associated graded is $z_n$. Thus, we can canonically project $xV$ to the sum of all the other weight spaces. If $x=z_n$, then this is just the identity, and the induced endomorphism is $z_{n-1}$; in general, we arrive at an element of the Slodowy slice to $z_{n-1}$ and the appropriate flag is exactly the one you write.

share|improve this answer

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.