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In my case, I have an abelian surface $A$ of (2,8)-polarization, and I have some finite group (may not be abelian group) $G$ acting on $A$ without fixed point. I want to understand when there is a Hodge isometry $$\phi: H^{2*}(A/G,\mathbb{Z}) \to H^{2*}(A,\mathbb{Z}),$$ where $\phi$ is the map induced from the natrual morphism $A \to A/G$.

For any abelian surface $X$, the dimension of $H^k(X,\mathbb{Z})$ is ${4 \choose k}$, hence it is enough to show $\phi$ is injective.

The background for this question is: according to Orlov, Mukai that $D(A),D(A/G)$ are derived equivalent iff there exists such Hodge isometry. I don't know if the polarization type is relevent or not, but that is what I have in my case.

Any comments/suggestions/references are very welcome!!!

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  • $\begingroup$ What you ask is not clear at all. Is $\phi=p^*$, where $p:A\rightarrow A/G$ is the quotient map? Is $G$ abelian? What is the role of the polarization in your question? $\endgroup$
    – abx
    Commented Jan 12, 2014 at 6:54
  • $\begingroup$ (1)$\phi = p^*$ (2) $G$ may not be abelian (3) I don't know what is the role of the polarization in my problem, but that's the (only) information I know about the abelian surface. $\endgroup$
    – Li Yutong
    Commented Jan 12, 2014 at 9:40
  • $\begingroup$ $p^*$ is not an isometry unless $G$ is trivial. I suggest that you begin by a good book on the subject, for instance Barth et al. $\endgroup$
    – abx
    Commented Jan 12, 2014 at 13:05
  • $\begingroup$ This is certainly not true - abelian variety and its dual are derived equivalent and hence when $G=K(L)$ we should have the isometry by Orlov & Mukai... $\endgroup$
    – Li Yutong
    Commented Jan 13, 2014 at 11:47
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    $\begingroup$ It is true -- it is just an elementary fact about cohomology of surfaces. There might be an isometry, but it is certainly not $p^*$. $\endgroup$
    – abx
    Commented Jan 13, 2014 at 12:26

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