Let $X$ be a smooth complex projective variety. Let $\omega_1, \ldots, \omega_n$ be elements of $H^0(X, \Omega^1_X)$ and $\gamma_1, \ldots, \gamma_n \in H_1(X(\mathbb{C}), \mathbb{Q})$. Consider the matrix given by the integrals $M=(\int_{\gamma_i} \omega_j)$ Is it true that $\det(M)=\int_{\gamma_1 \cup \cdots \cup \gamma_n} \omega_1 \wedge \cdots \wedge \omega_n$? If so could anybody provide a proof or a reference?

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 $\gamma_1, \ldots, \gamma_j$?

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 $\gamma_1, \ldots, \gamma_j$?

Let $X$ be a smooth complex projective variety. Let $\omega_1, \ldots, \omega_n$ be elements of $H^0(X, \Omega^1_X)$ and $\gamma_1, \ldots, \gamma_n \in H_1(X(\mathbb{C}), \mathbb{Q})$. Consider the matrix given by the integrals

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

Is it true that $\det(M)=\int_{\gamma_1 \cup \cdots \cup \gamma_n} \omega_1 \wedge \cdots \wedge \omega_n$?

If so could anybody provide a proof or a reference?

Let $X$ be a smooth complex projective variety. Let $\omega_1, \ldots, \omega_n$ be elements of $H^0(X, \Omega^1_X)$ and $\gamma_1, \ldots, \gamma_n \in H_1(X(\mathbb{C}), \mathbb{Q})$. Consider the matrix given by the integrals $M=(\int_{\gamma_i} \omega_j)$ Is it true that $\det(M)=\int_{\gamma_1 \cup \cdots \cup \gamma_n} \omega_1 \wedge \cdots \wedge \omega_n$? If so could anybody provide a proof or a reference?

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 $\gamma_1, \ldots, \gamma_j$?