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Suppose $X$ and $Y$ are schemes of finite type over a field, and let $f: X\rightarrow Y$ be a morphism. Let $\Gamma_f$ be the closed subscheme of $X\times Y$, then the first example of the Fourier-Mukai transform says that $$f_*()=p_Y{_*}(p_X^*()\bigotimes_{\mathcal{O}_{X\times Y}}\mathcal{O}_{\Gamma_f}),$$ similarly there is an expression for $f^*$. Has anyone checked this before? I am having some difficulties (at least on some commutative algebra) in verifying them. Does anyone know why those formulas are true?

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I did my best in the typing and tried more than five times. However, the preview and the actual post just are different. Sigh... –  Wayne Nov 16 '09 at 19:00
    
I fixed it with backquotes, to what you intended in the first draft. The backquotes trick disables the italics interpretation of the underscores. You can change it more if you don't like the formula. –  Greg Kuperberg Nov 16 '09 at 19:07
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You are missing hypotheses I think - you at least want Y to be separated (so that the graph is a closed immersion) and probably for X to be reduced. Also as pointed out below unless you meant for things to be derived one usually wouldn't call this a Fourier-Mukai transform I don't think? –  Greg Stevenson Nov 16 '09 at 20:12

2 Answers 2

You should use the fact that $p_X^*(-)\otimes_{\mathcal{O}_{X\times Y}}\mathcal{O}_{\Gamma_f}\cong i_*(p_X|_{\Gamma_f})^*(-)$ where $i:\Gamma_f\to X\times Y$ is the inclusion. Then the statment becomes much easier.

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Indeed, as follows from the comments below, maps between schemes provide examples of Fourier-Mukai transform, most famous example being a similar map with additional twisting by a bundle in $A\times \hat A$ for an Abelian variety $A$.

Anyway, since the restriction $p'_X:\Gamma_f\to X$ is actually an isomorphism (the inverse is $x\mapsto (x, f(x))$) and the composition $p_Y\circ {p'_X}^{-1}: X \to \Gamma_f \to Y$ is exactly $f$, the statement you have written is actually equivalent to $f_* () = f_*()$. Thus there is no hard commutative algebra stuff.

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The term Fourier-Mukai transform is used more widely now than to just referring to the original example of an abelian variety and its dual. –  Greg Stevenson Nov 16 '09 at 19:59
    
Indeed, it now refers to different examples, but to my knowledge they are mostly about some derived category A being equivalent to some derived category B, not an arbitrary map. –  Ilya Nikokoshev Nov 16 '09 at 20:02
    
Any morphism of varieties gives Fourier-Mukai transforms, they're only equivalences in nice situations, but any one gives an $f_*$ as the transform of the graph. –  Charles Siegel Nov 16 '09 at 20:08
    
Okay, okay, I said "in my opinion" and I was wrong. –  Ilya Nikokoshev Nov 16 '09 at 20:08
    
I actually meant in the derived setting - maybe one could call the ones which aren't equivalences integral transforms but I think one can use either term. –  Greg Stevenson Nov 16 '09 at 20:10

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