Cohomology of Springer resolution - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T10:36:57Z http://mathoverflow.net/feeds/question/70679 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70679/cohomology-of-springer-resolution Cohomology of Springer resolution JT 2011-07-18T21:40:35Z 2011-07-20T16:12:44Z <p>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 $p > 0$, $H^p(Z, \mathcal{O}_Z) = 0$. </p> <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/70679/cohomology-of-springer-resolution/70764#70764 Answer by Chuck Hague for Cohomology of Springer resolution Chuck Hague 2011-07-19T18:48:28Z 2011-07-20T16:12:44Z <p>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 $i > 0$.</p>