Let $X,Y$ be smooth varieties defined over $k$. Suppose $P$ is a coherent sheaf on $X \times Y$ flat over $X$, considering the Fourier-Mukai transform $$\Phi_P : D^{b}(X) \to D^{b}(Y)$$ which is defined as $$F \mapsto q_* (P \otimes p^{*}F).$$

Suppose $k(x)$ is the skyscraper sheaf on the closed point $x \in X$, and $k(x) \cong k$. Then how to justify the following claim: $$\Phi(k(x)) \cong P_x$$ where $P_x := P \mid_{ \{ x \} \times Y}$ is considered as a sheaf on $Y$ via the second projection: $\{x\} \times Y \to Y$ ?

my difficulties comes at two places: (1)All the definition here are in the derived sense, so I have no idea of compute them efficiently. (2) It seems $P \mid_{ \{ x \} \times Y}$ is a pullback rather than pushforward which is what the claim looks like.

Moreover, where is the condition $P$ is flat used in the calculation?


$p^*$ for smooth $p$ is exact, so $p^*$ of a skyscraper sheaf in the derived and regular senses are identical. So now we just have a pullback of a skyscraper sheaf - the structure sheaf of a single fiber and zero elsewhere.

We want to check that the higher tensor products vanish, so that the derived tensor product is just the regular tensor product, which is of course the sheaf $P_x$ on $x \times Y$ and $0$ elsewhere.

This is where flatness comes in. It ensures that the tensor product is exact, so its higher derived functors vanish.

So the tensor product is just the pushforward of $P_x$ along the inclusion $Y \to X \times Y$ defined by $x$. Since this is a closed immersion the higher pushforwards vanish. If we pushforward again to $Y$, this is the same as the pushforward along the composition morphism, which is the identity, so we just end up with $P_x$.

  • $\begingroup$ Thank you! Could you explain a few places in your argument which I did not follow. (1) First, $p$ is smooth implies $p^*$ is exact, could you let me know the reference of this result? (2) If $p^*$ is exact, why the image of sheaf is a sheaf? (3) You wrote:"Since this is a closed immersion the higher pushforwards vanish", I did not see why. I appreciate your time! $\endgroup$ – Li Yutong Mar 14 '13 at 1:22
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    $\begingroup$ (1) It actually follows from flatness. Just check it locally on affine things, where $p^*$ is just tensoring - but you're tensoring with an flat thing, so it's exact. (2) Because the higher derived functors vanish. The derived category is just a way of measuring the higher derived functors. No higher derived functors, just a regular sheaf. (3) This follows from affineness. Again check it locally on affine things. Now pushforward is just taking an $R/I$-module and viewing it as an $R$-module. Clearly exact. So no higher derived functors. $\endgroup$ – Will Sawin Mar 14 '13 at 1:48
  • $\begingroup$ That makes a lot sense, thank you again! $\endgroup$ – Li Yutong Mar 15 '13 at 13:51

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