0
$\begingroup$

Let $A$ be a complex abelian variety of dimension $d$. Let $\omega_1, \ldots, \omega_j \in H^0(A, \Omega^1_A)$ be linearly independent (so $j \leq d$) and consider $\gamma_1, \ldots, \gamma_j \in H_1(A(\mathbb{C}), \mathbb{Q})$. Look at the matrix

$M=(\int_{\gamma_i} \omega_j)$

Is it possible to express the determinant of $M$ as the integral of $\omega_1 \wedge \ldots \wedge \omega_j \in H^0(A, \Omega^j_A)$ against some element of $H_j(A(\mathbb{C}), \mathbb{Q})$ constructed out of $\gamma_1, \ldots, \gamma_j$?

$\endgroup$
9
  • $\begingroup$ What if the dimension of $X$ is less than $n$? Do you want to assume that $X$ is a complex torus and $\omega_1,\dots,\omega_n$ form a basis? $\endgroup$ May 27, 2014 at 13:59
  • $\begingroup$ In that case both sides are zero, no? But you are right that the case I have in mind is $X$ an abelian variety of dimension $n$ and $\omega_i$ a basis of $H^0(X, \Omega^1_X)$ $\endgroup$
    – detted92
    May 27, 2014 at 14:01
  • $\begingroup$ "In that case both sides are zero, no?" Why? Certainly the integral you write is zero, but that does not imply that the determinant is zero. That is why I raised my objection. $\endgroup$ May 27, 2014 at 14:04
  • $\begingroup$ What is $\gamma _1\cup\ldots \cup \gamma _n$? $\endgroup$
    – abx
    May 27, 2014 at 14:04
  • 1
    $\begingroup$ You can't take the cup-product of homology classes! Cup-product is an operation on cohomology, so the cycle you're integrating over is nonsensical as soon as $n > 1$. $\endgroup$ May 27, 2014 at 14:26

1 Answer 1

1
$\begingroup$

You can form the $j$-fold self product, $A^j$, together with its addition morphism to $A$, $$\Sigma:A^j \to A.$$ Via Künneth, you have an inclusion $$\bigotimes_{i=1}^j H_1(A;\mathbb{Q}) \hookrightarrow H_j(A^j;\mathbb{Q}).$$ Now take the image of the cycle $\gamma_1\otimes \dots \otimes \gamma_j$ in $H_j(A^j;\mathbb{Q})$, and then take the pushforward via the morphism $\Sigma$ (proper and locally a fiber bundle) to get a class $\gamma\in H_j(A;\mathbb{Q})$. Similarly, given de Rham differentials, $$\omega_1,\dots,\omega_j \in H^{1,0}(A),$$ you can form $\omega= \omega_1\wedge \dots \wedge \omega_j$. Consider $\int_\gamma \omega$, or equivalently, the pairing, $$\int_\gamma \omega = \langle \Sigma^*(\omega_1\wedge \dots \wedge \omega_j),\gamma_1\otimes \dots \otimes \gamma_j\rangle.$$
Then you can ask whether or not, $$ \int_\gamma \omega = \text{det}\left[\int_{\gamma_\alpha} \omega_\beta\right]_{1\leq \alpha,\beta \leq j}?$$

Of course, ultimately, this has nothing to do with the differentials $\omega_\beta$ being contained in $H^{1,0}(A)$. Really this is a question about the singular cohomology of $A$: how does the Hopf algebra structure induced by addition behave? That question answers itself: the cohomology is a Hopf algebra. More precisely, it is the free exterior algebra on $H^1(A;\mathbb{Q})$ (with the usual trace on the top exterior power) equipped with its standard structure of cocommutative Hopf algebra. In particular, considered as an element in $H^1(A;\mathbb{Q})\otimes H^1(A;\mathbb{Q})$, $\Delta(\omega_\beta)$ equals $\omega_\beta\otimes 1 + 1\otimes \omega_\beta$, where $\Delta$ is pullback via addition, $$m:A\times A \to A,$$ together with the Künneth isomorphism, $$H^1(A\times A;\mathbb{Q}) = \left(H^1(A;\mathbb{Q})\otimes \mathbb{Q}\right) \oplus \left(\mathbb{Q}\otimes H^1(A;\mathbb{Q}) \right).$$ By my computation, this does imply that the Künneth component in $H^1(A;\mathbb{Q})\otimes \dots \otimes H^1(A;\mathbb{Q})$ of $\Sigma^*(\omega_1\wedge \dots \wedge \omega_j)$ is equal to $$\sum_{\sigma\in{\mathfrak{S}_j}} \text{sgn}(\sigma) \omega_{\sigma(1)}\otimes \dots \otimes \omega_{\sigma(j)}$$ (it would be wise for you to double-check that computation). Thus the pairing against $\gamma_1\otimes \dots \otimes \gamma_j$ equals $$\int_\gamma \omega = \sum_{\sigma\in{\mathfrak{S}_j}} \text{sgn}(\sigma) \langle \omega_{\sigma(1)}\otimes \dots \otimes \omega_{\sigma(j)}, \gamma_1\otimes \dots \otimes\gamma_j \rangle = \text{det}\left[\int_{\gamma_\alpha} \omega_\beta\right]_{1\leq \alpha,\beta \leq j}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.