This question is elementary. Let $G$ be a simple algebraic group over $\mathbb{C}$, and let $B$ be a choice of Borel subgroup, with unipotent radical $U$ with Lie algebra $\mathfrak{n}$. Then the Springer resolution of the nilpotent cone is $Z = G \times_B \mathfrak{n}$; it is identified with the cotangent bundle of $G/B$. In "Cohomology and the resolution of the nilpotent variety", Math. Ann. 1976, Hesselink proves that for each $p > 0$, $H^p(Z, \mathcal{O}_Z) = 0$.

My question is as follows: this group is identified with $\oplus_{l \geq 0} H^p(G/B, \text{Sym}^l \mathfrak{n}^\vee)$, so it is equivalent to show the vanishing of each of these summands. Why does the following simple argument not work?

For each $l \geq 0$, $\text{Sym}^l \mathfrak{n}^\vee$ (as a representation of $B$) has a filtration with 1-dimensional graded pieces where $T$ acts with anti-dominant weights (where the positive roots are with respect to $B$). The cohomology of the line bundles associated to these graded pieces vanishes by Kempf's vanishing theorem, so the long exact sequence in sheaf cohomology should imply that the higher cohomology of $\text{Sym}^l \mathfrak{n}^\vee$ also vanishes.

  • $\begingroup$ The typography in the last paragraph got mixed up a little. Aside from that, it's good to keep in mind that most of the machinery makes sense over an algebraically closed field of arbitrary characteristic. This especially includes Kempf's vanishing theorem. Also, many people assume that $B$ corresponds to negative roots in order to avoid some awkwardness about dominant weights. $\endgroup$ – Jim Humphreys Jul 18 '11 at 22:03

The reason your argument doesn't work is because it's not true that $\text{Sym}^l \mathfrak n^\vee$ has a filtration with 1-dimensional graded pieces where $T$ acts with anti-dominant weights. In fact, this is false even in the case $l = 1$. I'll assume, as you do, that $B$ corresponds to the positive roots. Then the weights of $\mathfrak n^\vee$ correspond to the negative roots, and most of the negative roots are NOT anti-dominant. (In fact, even the negative simple roots are not antidominant, unless all components of $G$ are of type $A_1$). This shows why something subtle is going on here: $H^i( G/B, \mathfrak n^\vee ) = 0$ for all $i > 0$, but it is not the case in general that $H^i( G/B, \; \textrm{gr} \; \mathfrak n^\vee ) = 0$ for all $i > 0$.

  • $\begingroup$ I see. Thank you for explaining this to me! $\endgroup$ – user1594 Jul 19 '11 at 20:23

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