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I am a beginner at derived categories and I'm looking for a proof of the following fact:

If $X$ and $Y$ are smooth projective curves such that $D^b(Coh\,X)$ is equivalent to $D^b(Coh\,Y)$ then $X$ and $Y$ are isomorphic.

Can anyone give me the explanation of this fact or provide a good reference for this?

As I understand, the crucial point here is that any object in $D^b(Coh \, X)$ splits as a sum of its cohomology sheaves but I don't know how to finish the proof.

Thank you.

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2 Answers 2

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The non-elliptic case is easier. By a theorem of Bondal-Orlov, see [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 4.11]:

Theorem. Let $X$ and $Y$ be smooth projective varieties with equivalent derived category $D^b(X)$ and $D^b(Y)$. If the (anti)-canonical bundle of $X$ is ample, then $X$ and $Y$ are isomorphic.

For elliptic curve, we need a generalized theorem by Kawamata, See [Kawamata, D-equivalence and K-equivalence] or [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 6.15]:

Theorem. Let $X$ and $Y$ be smooth projective varieties with equivalent derived category $D^b(X)$ and $D^b(Y)$. Then the (anti)-canonical bundle of $X$ is nef if and only if the (anti)-canonical bundle of $Y$ is nef.

This theorem implies if $X$ is elliptic curve, the so is $Y$.

By [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 5.39], the hodge structuces of $X$ and $Y$ are the same, which yeilds that $X$ and $Y$ are isomorphic.

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I don't know of a uniform way to prove it for all curves. For elliptic curves it follows from Hodge theory and for the rest it's a consequence of the Bondal-Orlov theorem. It's all explained in Huybrechts's excellent book on Fourier-Mukai transforms.

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  • $\begingroup$ Can you get a uniform proof using Torelli's theorem? It says that if two curves have the same integral Hodge structure (including Poincare duality) then they are isomorphic. $\endgroup$
    – Will Sawin
    Aug 2, 2014 at 0:13
  • $\begingroup$ @WillSawin: that's an interesting thought. I don't know enough of Hodge theory to answer that. I hope someone does though! $\endgroup$ Aug 2, 2014 at 0:38

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