2
$\begingroup$

I apologize in advance that this question is probably too basic for MO, but I reckoned I would not get an answer on Math.Stackexchange.

I am starting to learn about perverse sheaves, the decomposition theorem, the equivariant derived category and applications in representation theory (as in the book of Chriss and Ginzburg, for example). What I don't quite understand at the moment is the difference between taking Ext in the derived category and the equivariant derived category. For example, if I let $$ \mu : G/B \to \{pt\} $$ be the projection of a flag variety onto a point and let $C$ denote the constant sheaf on $G/B$ then $$Ext^*(\mu_*C,\mu_*C) \cong H_*(G/B \times G/B) \cong End_k(H_*(G/B))$$ and $$Ext_G^*(\mu_*C,\mu_*C) \cong H_*(G/B \times G/B) \otimes H_*^G(\{pt\})$$ is the NilHecke ring. So let us consider the simplest possible example. Let $C$ now denote the constant sheaf on a point, $G$ a one-dimensional torus $\mathbb{C}^*$. Then, if I take Ext in the derived category, I get $$Ext^*(C,C) = k.$$ This means that $$Ext^m(C,C) = Hom(C,C[m]) = 0$$ for $m \neq 0$. If I now regard $C$ as an object in the equivariant derived category (i.e. more precisely a triple consisting of the constant sheaf on a point, the constant sheaf on the classifying space $BG$ and an isomorphism between the respective pullbacks to $EG$) and take Ext in the equivariant derived category I get $$Ext_G^*(C,C) = k[x],$$ i.e. $$Ext^m_G(C,C) = Hom_G(C,C[m]) = k$$ for $m \geq 0$. My question is: why do we have non-trivial morphisms between $C$ and its shifts $C[m]$ in the equivariant derived category? What are they explicitly?

$\endgroup$
2
$\begingroup$

First of all, you made a mistake: $x$ has degree 2, so your equation is only correct for $m$ even.

As for what the classes are, they're cohomology classes on $\mathbb{CP}^\infty\cong B\mathbb{C}^*$. If you like cellular homology, you can decompose $\mathbb{CP}^\infty$ into cells: there will be one of each even dimension. The cocycle with value 1 on this cell is the class you want.

More generally, you should think of taking an equivariant sheaf on a $G$-variety $X$, and looking at the induced sheaf on $(X\times EG)/G$. So, Ext of the constant sheaf of $X$ with itself is the equivariant cohomology of $X$. That, together with the general nonsense of the 6-functor formalism is the main thing you need to know about equivariant derived categories.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.