MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In their book, Bernstein an Lunts define the equivariant derived category in several ways. One can be expressed as follows:

Let $X$ be a say complex variety with an action by an algebraic group $G$.

Consider the fibration $\pi:D^b(- , \mathbb Z) \rightarrow Var$ which assigns to every variety $T$ the bounded derived category of sheaves of say abelian groups $D^b(T, \mathbb Z)$. Then the equivariant derived category $D_G^b(X,\mathbb Z)$ is defined to be the category of cartesian sections of $\pi$ over the quotient stack $X/G \rightarrow Var$. Informally one can read this as:

"An equivariant sheaf is on a space $X$ is just a map from $X/G$ to the classifying space of sheaves"

Now my questions are:

In what situation is this the correct definition? What properties of $X/G$ and $D^b(T,\mathbb Z)$ make this definition reasonable?

For example:

If we replace $X/G$ by an artin stack $\mathfrak{X}$ does this give a reasonable definition of $D^b(\mathfrak{X}, \mathbb Z)$?

If we replace $D^b(T, \mathbb Z)$ by some other "derived category of sheaves" do we still get a reasonable definition?

For example we could drop the boundedness assumption in one or both directions/allow non finitely generated cohomology sheaves, replace sheaves of abelian groups by quasi-coherent sheaves etc...

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.