Geometric intuition behind perverse coherent sheaves? - MathOverflow

Geometric intuition behind perverse coherent sheaves?

2012-11-26T22:56:17Z

I would like to know an intuition behind perverse coherent sheaves. I am aware that it is induced by a heart of another t-structure on the derived category. Are there any better, probably more geometric, way to understand perverse coherent sheaves?

Just in case, let us recall the definition of perverse coherent sheaves. Let $X$ be a projective threefold with at worst Gorenstein terminal singularities and $f:Y\rightarrow X$ be a crepant resolution. Define a full subcategory $\mathrm{Per}(Y/X) \subset \mathrm{D}(Y)$ consisting of objects $E \in \mathrm{D}(Y)$ satisfying the following three conditions;

$H^i(E)=0$ unless $i=0,-1$,
$R^1f_*H^0(E)=0$ and $R^0f_∗H^{−1}(E)=0$,
$Hom_Y(H^0(E),C)=0$ for any sheaf $C$ on $Y$ satisfying $Rf_∗(C)=0$.

We call the objects of $\mathrm{Per}(Y/X)$ perverse coherent sheaves.

Answer by 36min for Geometric intuition behind perverse coherent sheaves?

2012-11-27T07:36:12Z

This is the definition appear in Bridgeland's paper which shows that flops of smooth 3-folds induces equivalence of derived category of coherent sheaves. From your question I think you know the word "perverse" is kind of related to t-structures.

The main theorem of that paper, indicates that for a flop $Y\to X\leftarrow W$, $Per(Y/X)$ will be send to $Coh(W)$ under that isomorphism. In other words, these objects are sheaves on another scheme which you can construct from the data $Y\to X$! In my opinion that's pretty cool, not "perverse" at all. But as for the name, so be it.

BTW, you don't need $X$ and $Y$ to be three fold in the definition. If you check Bridgeland's paper, most of the time he work with birational morphism such that $Rf_*\mathcal{O}_Y=\mathcal{O}_X$ and fibers have dimension at most 1. (For 3-folds that's just a small resolution.)