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In Mukai's paper Duality between $D(X)$ and $D(\hat{X})$ with its application to Picard sheaves, Nagoya Math Journal, 1981, there is one sentence that puzzles me.

Let $X$ be an abelian variety over an algebraically closed field, $Y$ be its dual abelian variety. Let $\pi_X:X\times Y\to X$ be the projection, $\pi_Y$ be the other projection. Let $F$ be an $O_X$-flat (i.e., $\pi_X$-flat) $O_{X\times Y}-$module. Define a functor $\Phi:Mod(O_X)\to Mod(O_{X\times Y})$ by $$\Phi(?)= F\otimes \pi_X^*(?).$$ Then $\Phi$ is exact. Define another functor $S_{X\to Y,F}:Mod(O_X)\to Mod(O_Y)$ by $$S_{X\to Y,F}=\pi_{Y,*}\circ\Phi.$$ Then $S_{X\to Y,F}$ is left exact.

On p.154, two lines above Proposition 1.3, it writes that the derived functor of $S_{X\to Y,F}$ is given by $$RS_{X\to Y,F}(?)=R\pi_{Y,*}(F\otimes^L \pi_X^*?):D^-(Mod(O_X))\to D^-(Mod(O_Y)).$$

To simplify, let me take $F=O_{X\times Y}$, then $\Phi=\pi_X^*$. If the statement in last paragraph holds, by https://stacks.math.columbia.edu/tag/015M, $\pi_X^*(I)$ is right acyclic for $\pi_{Y,*}$ for each injective object $I\in Mod(O_X)$. But I cannot see why. Therefore, I have doubts on it.

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  • $\begingroup$ Not to be rude but if you cannot see it, why have doubts on it? This is literally flat base change at work. $\endgroup$ Commented Jan 21, 2023 at 9:22
  • $\begingroup$ Dear @Clamp, Thanks for your comment! First of all, the flat base change theorems that I have seen are about "quasi-coherent" modules. But in Mukai's case, $Mod(O_X)$ is the category of all $O_X$-modules. Secondly, I don't know how flat base change helps this problem when $F$ is NOT $O_{X\times Y}$. For these reasons, I keep my doubts $\endgroup$
    – Doug Liu
    Commented Jan 22, 2023 at 7:58
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    $\begingroup$ @CraniumClamp This particular linguistic point occurred a number of times on MathOverflow already, but “I have doubts” is sometimes used, esp. by non-native English speakers, to convey their misunderstanding rather than their belief that there is a mistake in the reasoning. (Not being a linguistic prescriptivist, I will not endeavor to discuss whether they're “right” or “wrong” to use this phrase in this sense.) $\endgroup$
    – Gro-Tsen
    Commented Jan 23, 2023 at 9:42
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    $\begingroup$ If your doubt is about quasi-coherent $O_X$-modules vs, $O_X$-modules then you are correct and this is incorrect as stated in Mukai's paper (he is aware of this) see mathoverflow.net/questions/243202/… $\endgroup$
    – Hacon
    Commented Jan 27, 2023 at 5:12
  • $\begingroup$ Dear @Hacon, Thanks for letting me know this! $\endgroup$
    – Doug Liu
    Commented Jan 27, 2023 at 14:06

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