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.

`$B$`

corresponds to negative roots in order to avoid some awkwardness about dominant weights. $\endgroup$ – Jim Humphreys Jul 18 '11 at 22:03