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Let $A$ be an abelian variety over $\overline{\mathbb{Q}}$. We work up to isogeny (i.e., Hom sets are tensored with $\mathbb{Q}$). I am looking for a reference (and ideally, a short explanation) for the following statement (or, if wrong, for its correct version):

Claim: Let $\phi:A \to A^*$ be a polarization (in particular, a symmetric isogeny). Then $\phi$ induces a non-degenerate alternating pairing $\langle-,-\rangle:H_1(A,\mathbb{Q}) \times H_1(A,\mathbb{Q}) \to \mathbb{Q}$ on the degree $1$ singular homology of $A$.

Such a pairing is equivalent to an anti-symmetric isomorphism $\psi:H_1(A,\mathbb{Q}) \to H_1(A,\mathbb{Q})^*$. The thing that puzzles me is that $\phi^*=\phi$ but $\psi^*=-\psi$. This might have to do with the way in which $A^{**}$ is identified with $A$ or with the way in which the pairing is defined, but I am not sure: if the functor $(-)^{**}$ on abelian varieties is isomorphic to the identity, and if the functor $H_1(-,\mathbb{Q})$ from abelian varieties to $\mathbb{Q}$-vector spaces is faithful and exact and preserves duality, and if $\psi$ is the image of $\phi$ under $H_1(-,\mathbb{Q})$, then how can the symmetric morphism $\phi$ turn into an anti-symmetric morphism $\psi$? Where am I doing something wrong?

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    $\begingroup$ The polarization corresponds to a hermitian form on $H^1(A,\Bbb{C})$, whose imaginary part is an integer-valued skew-symmetric form. You'll find this in any book on Abelian varieties, for instance Mumford's. $\endgroup$
    – abx
    Commented Jul 1, 2022 at 12:08
  • $\begingroup$ what do you mean $\phi^*=\phi$? the domain of one morphism is $A$ and the other is $A^*$! Any way the definition is like this: a polarisation gives you an ample line bundle $\lambda$ on $A$, this gives you an element of $H^2(A,\mathbb{Q})$ and $H^2(A,\mathbb{Q})$ is the second exterior power of $H^1(A,\mathbb{Q})$ e.g the space of antisymmetric forms on H^1. $\endgroup$
    – ali
    Commented Jul 1, 2022 at 12:44
  • $\begingroup$ Are you familiar with the Koszul sign rule ? In brief, an alternating pairing on $H_1$ is in fact what you would expect from a symmetric construction. Compare with the cup product in the cohomology of a space $X$: it is induced by the diagonal $X \to X \times X$, so it should clearly be commutative as the diagonal is $S_2$-invariant, but in fact it satisfies $\alpha \beta = (-1)^{\vert \alpha \vert \vert \beta \vert} \beta\alpha$ and in particular classes in odd degree anticommute. This type of commutativity twisted by the Koszul rule is in fact the natural one! $\endgroup$
    – Dan Petersen
    Commented Jul 2, 2022 at 9:28
  • $\begingroup$ @ali If $\phi:A \to B$ is a map of abelian varieties, the dual map is $\phi^*:B^* \to A^*$. If $B=A^*$, then $B^*=A^{**} \cong A$ and $\phi^*$ is again in $\mathrm{Hom}(A,A^*)$. $\endgroup$
    – 57Jimmy
    Commented Jul 2, 2022 at 20:14
  • $\begingroup$ @DanPetersen But in this pairing there is no multiplication involved, right? So where does the Koszul sign rule come into play? $\endgroup$
    – 57Jimmy
    Commented Jul 2, 2022 at 20:29

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