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.



Fourier-Mukai transform for abelian varieties
**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



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


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



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> 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.
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> 


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.


