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Let $A$ be an abelian variety over $\mathbb{C}$, $L$ be a very ample line bundle on $A$, then the dual abelian variety is $\hat{A} \cong A/K(L)$ with $K(L)$ the kernel of surjective morphism $A \to Pic^0(A)$. Let $P$ be the Poincare sheaf on $A \times \hat{A}$. It is known that $\Phi_{A \to \hat{A}}^{P}$ (i.e.the Fourier-Muaki functor with kernel $P$ ) is a derived equivalence between $D^b(A)$ and $D^b(\hat{A})$.

Suppose $H \subseteq K(L)$ is a subgroup, and let $\tilde{A} = A/H$. Do we still have derived equivalence between $D^b(A)$ and $D^b(\tilde{A})$?

I feel that one could similarly define a Poincare sheaf $\tilde{P}$ on $A \times \tilde{A}$, but I am not sure if this gives the equivalence.

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The Poincare bundle and the associated FM transform are independent of the polarization $L$. – Piotr Achinger Dec 12 '13 at 20:48
This would imply that $D^b(A)= D^b(A')$ for any $A'$ isogenous to $A$, by taking a sufficiently large tensor power of $A$. That seems wrong, but I don't know a derived category invariant that would prove it. – Will Sawin Dec 12 '13 at 22:29
up vote 7 down vote accepted

No. As Will Sawin indicates, every finite subgroup $H$ of $A$ is contained in $K(L)$ for some very ample line bundle on $A$: indeed, let $L_1$ be your favorite very ample line bundle on $A$, and let $n = \# H$; then $H \subset A[n] \subset K(nL_1)$. Thus you are asking whether every abelian variety $B$ which is isogenous to $A$ is a Fourier-Mukai partner of $A$.

This is not true: in order for complex abelian varieties to be Fourier-Mukai partners, it is necessary but not sufficient that they be isogenous. Indeed, Proposition 5.1 of

MR1827500 (2002a:14017) Bridgeland, Tom Maciocia, Antony Complex surfaces with equivalent derived categories. Math. Z. 236 (2001), no. 4, 677–697.

shows that any complex abelian surface has only finitely many Fourier-Mukai partners. On the other hand, it is shown in

Hosono, Shinobu; Lian, Bong H.; Oguiso, Keiji; Yau, Shing-Tung Kummer structures on K3 surface: an old question of T. Shioda. Duke Math. J. 120 (2003), no. 3, 635–647.

that for every positive integer $n$, there is a complex abelian surface having precisely $n$ Fourier-Mukai partners. Via Hodge theory the question is reduced to the study of positive definite integral quadratic forms....which is quite nontrivial and interesting in its own right, of course.

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Thank you! May I ask one more question: If $H \subset A$ is a finite group, does there always exist an ample line bundle $L$ on $A$, such that $K(L) \cong H$? – Li Yutong Dec 13 '13 at 3:40
No, because $A/K(L) \cong A^{\vee}$, whereas for "most" finite subgroups $H$ of $A$, $A/H$ is not isomorphic to $A^{\vee}$. For instance, if $A$ is an elliptic curve without complex multiplication and $H$ has squarefree order $n > 1$, then $A/H$ cannot be isomorphic to $A$ since $A$ has no endomorphisms of order $n$. – Pete L. Clark Dec 13 '13 at 4:46

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