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I have a couple questions regarding the proof of Proposition $3$ (see page $10-11$ of arxiv.org/abs/math/0102039) in Bezrukavnikov's paper "Quasi-exceptional sets and equivariant coherent sheaves on the nilpotent cone" . For simplicity, assume $G$ is simply connected, simple algebraic group (so that $G$ is its own universal cover). Also, $\alpha$ is a simple root, let $P_{\alpha}$ be the associated minimal parabolic and $u_a$ its unipotent radical; $G/B, G/P_{\alpha}$ are the flag variety and a partial flag variety with projection map $p_{\alpha}: G/B \rightarrow G/P_{\alpha}$; $\lambda, \lambda' \in \Lambda$ lie in the weight lattice and satisfy $s_{\alpha}(\lambda) = \lambda - n \alpha, s_{\alpha}(\lambda') = \lambda' - (n-1) \alpha$; and $O_{G/B}(\lambda')$ for instance is the line bundle associated to $\lambda'$. Set $\tilde{\mathcal{N}} = G \times_B \mathfrak{u}$ to be the Springer resolution ($u$ is unipotent radical, as usual).

Question $1$: Set $V_{\lambda'} = p_{\alpha}^* p_{\alpha *} O_{G/B}(\lambda')$.

On the first paragraph of pg $11$ it is claimed that there is a $G$-invariant filtration on $V_{\lambda'}$ with the quotients being $\mathcal{O}_{G/B}(\lambda'-i \alpha)$ for $0 \leq i \leq n-1$; why is this?

Question $2$: Set $\tilde{\mathcal{N}_{\alpha}} = G \times_B u_a$ (I am under the impression this is the same as what Roman defines on pg 11; if not, please correct me). How is the following exact sequence obtained (equation $23$ in the paper)? $0 \rightarrow O_{N'}(\alpha) \rightarrow O_{N'} \rightarrow O_{N'_{\alpha}} \rightarrow 0$

[In the above equation, I have used a ' to replace the tilde, since I cannot get Latex to work.]

Question $3$: Consider the natural projection $\pi': \tilde{\mathcal{N}_{\alpha}} \rightarrow T^{*}(G/P_\alpha)$.

The fibers are clearly projective lines. If $p': \tilde{\mathcal{N}_{\alpha}} \rightarrow G/B$ is the projection, in the last line of the proof (just before Prop $4$), it seems to be claimed that $p'^*(V_{\lambda'}(\lambda - \lambda'-\alpha))$ is isomorphic to a direct sum of copies of $\mathcal{O}_{\mathbb{P}^1}(-1)$ when restricted to any fibre of $\pi'$. Why is this?

EDIT: In all of the above, I had to simplify many of the Latex formulas to make them work (e.g. replace tilde by '). I think everything should be readable now, at the cost of bad notation.

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Does "Roman" refer to Bexrukavnikov? It is usually more convenient for readers if you link to the ArXiv abstract page, rather than the pdf. –  S. Carnahan Aug 24 '11 at 9:41
    
Yep; ok, I've edited it now. –  Vinoth Aug 24 '11 at 10:53

1 Answer 1

up vote 1 down vote accepted

For question 1, let $L_\alpha$ denote the Levi of the parabolic $P_\alpha$; then $L_\alpha$ has derived group $SL_2$. Let $B_\alpha$ denote the Borel of $L_\alpha$ such that $B \cap L_\alpha = B_\alpha$. Note that $V_{\lambda'} = p_\alpha^* O_{G/B}(V_\alpha(\lambda'))$, where $V_\alpha(\lambda')$ is the $L_\alpha$ representation induced from $\lambda'$. That is, $V_\alpha(\lambda') = Ind_{B_\alpha}^{L_\alpha} (\chi_{\lambda'}) = Ind_B^{P_\alpha} (\chi_{\lambda'})$, where $\chi_{\lambda'}$ is the 1-dimensional representation corresponding to the weight $\lambda'$. Since $L_\alpha$ has derived group $SL_2$, the weights of $V_\alpha(\lambda')$ are precisely $\lambda'- i \alpha$, $0 \leq i \leq n-1$. Hence $V_\alpha(\lambda')$ has (considered as a $B$-module now by extending trivially over the unipotent radical of $P_\alpha$) a $B$-equivariant filtration with quotients $\chi_{\lambda' - i \alpha}$, $0 \leq i \leq n-1$, so that $V_{\lambda'}$ has the desired $G$-equivariant filtration.

For question 3, the fibers of $\pi'$ are all isomorphic to $L_\alpha / B_\alpha$, the flag variety of $L_\alpha$; and this flag variety is isomorphic to $\mathbb P^1$. Since $(\lambda - \lambda' - \alpha)(\alpha^\vee) = -1$, under the identification of $L_\alpha / B_\alpha$ with $\mathbb P^1$, we see that the restriction of $p'^* O_{G/B}( \lambda - \lambda' - \alpha )$ to the fibers of $\pi'$ has degree -1, as desired. (I assume you want $p'^* O_{G/B}( \lambda - \lambda' - \alpha )$ here and not $p'^* V_{\lambda'}(\lambda - \lambda' - \alpha)$, since $V_{\lambda'}(\lambda - \lambda' - \alpha)$ is not a bundle on $G/B$).

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