4
$\begingroup$

Suppose $M$ is a quasi-projective variety, $G$ is a finite group acting on $M$. Let $X$ be the quotient $M/G$ (we assume $X$ to be singular) and $\pi: M\to X$ be the natural projection.

We have $(\pi)^{G}_{*}\circ \pi^{*}=id$, and in the case when $G$ is acting freely, the two functors are left and right adjoints of each other.

My question is:

1.What is the adjoint relation of the two functors in general?

2.I heard that since $X$ is singular, $L\pi^*$ would not be a map from $D^b(X)$ to $D^{b}(Coh_G(M))$. And this has to do with perfect complexes. Can someone explain this a little more and maybe provide some references?

Thanks!

$\endgroup$

1 Answer 1

4
$\begingroup$

The functors are still adjoint and the same relation holds. For details, see for example http://arxiv.org/pdf/1406.4409.pdf (especially the proof of Theorem 3.0.14) and references therein.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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