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I am reading Kapranov and Vasserot's paper 'Kleinian singularities, derived categories and Hall algebras',where they proved McKay correspondence in dimension 2. In section 1.2, they constructed two functors $\Phi$ and $\Psi$

$\Phi(\mathcal{F})=(Rp_{1*}(p_2^*(\mathcal{F}\otimes\mathcal{O}_{\Sigma}))^G$

$\Psi(\mathcal{G})=Rp_{2*}R\mathcal{Hom}(\mathcal{O}_{\Sigma},p_1^*(\mathcal{G}))$

and claimed that $\Phi$ is the left adjoint of $\Psi$.

What puzzles me is that when one is taking the right adjoint of $Rp_{1*}$ and since $p_1$ is flat, we get $p^*_1(\mathcal{F})\otimes \omega_{p_1}$ under the assumption that $p_1$ is proper. Here neither the properness is assumed nor the relative differential can be seen. I cannot see how the adjunction is justified.

I would greatly appreciate any hint to clear this issue.

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    $\begingroup$ Is not $p_1:\Sigma\to X/\!\!/G$ actually finite? If I am not confusing something, cardinalities of fibers are bounded by the order of $G$. (Btw, at the end of your definition of $\Phi$ must be $)^G$.) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 4, 2016 at 9:07
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    $\begingroup$ As მამუკა ჯიბლაძე says, the properness is not really an issue: You don't need the properness of $p_1$, but only of the restriction of $p_1$ onto $\Sigma$. As for the missing twist by $\omega$, I think you are right, $p_1^*$ should be $p_1^!$. This can be seen already when $G=\{e\}$: then $\Phi=Id$ (the missing $G$-invariants won't matter), but the given formula for $\Psi$ has a non-trivial twist in it. $\endgroup$
    – t3suji
    Commented Dec 4, 2016 at 14:27
  • $\begingroup$ @მამუკაჯიბლაძე I agree with you that the projection from $\Sigma$ to $X//G$ is finite(and hence has a duality), but here $p_1$ denotes the projection from $X//G\times X$ to $X//G$, which is not finite. And thank you for pointing that missing $G$ out, I edited that. $\endgroup$
    – Xuqiang QIN
    Commented Dec 4, 2016 at 15:19
  • $\begingroup$ I believe if you interpret these functors as going through $\Sigma$ instead of the product, everything should work as needed $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 4, 2016 at 15:53
  • $\begingroup$ @t3suji I just realized since in the case we are concerning about, $X//G$ is a crepant resolution of the quotient $X/G$, and has trivial canonical line bundle, and hence the relative bundle for $p_1$ is trivial too. $\endgroup$
    – Xuqiang QIN
    Commented Dec 5, 2016 at 3:40

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