$T$ is torus. $H_1(T,Z)\to \operatorname{Hom}\Omega^1(T),C)$ is injection.

Let $T=\frac{V}{\Lambda}$ be a torus where $V$ is a complex vector space of dimension $n$ and $\Lambda$ is a rank $2n$ lattice in $V$. $\Omega^1(T)$ is the space of holomorphic 1-forms of $T$. $dim_C(\Omega^1(T))=n$.

Poincare duality says $H_1(T)\cong H_1(T)^\star$ where $\star$ indicates duality. However $H_1(T)^\star$ is isomorphic to de Rham group of rank $2n$ over $Z$.(Correct me if I am wrong here.) The $(\Omega^1(T))^\star$ has rank $2n$ over $R$. $H_1(T)^\star$ is full rank lattice in $(\Omega^1(T))^\star$.

$\textbf{Q:}$ Should $(\Omega^1(T))^\star$ isomorphic to de Rham group as I want to see there is an embedding $H_1(T) \to(\Omega^1(T))^\star$ by integrating closed curves of $H_1(T)$? And what is the reason that $H_1(T,Z)\to (\Omega^1(T))^\star$ is an embedding?(I knew it is Poincaré duality but I need to apply de Rham isomorphism somewhere.)

Let $T = \mathbb C^n / \Lambda$ be the complex torus, and let $\{ z_1, \dots, z_n \}$ be a complex coordinates for the $\mathbb C^n$. Then $\Omega^1(T)$ is the $\mathbb C$-vector space generated by the one-forms $\{ dz_1, \dots dz_n \}$.

[To see that this is true, observe that every holomorphic 1-form on $T$ can be written as $g_1(z_1, \dots, z_n) dz_1 + \dots g_n(z_1, \dots, z_n) dz_n$ for holomorphic functions $g_1, \dots, g_n$ on $\mathbb C^n$ that are invariant under translations by lattice vectors in $\Lambda$. By Liouville's theorem for several complex variables, the functions $g_1, \dots, g_n$ must be constant.]

Now suppose that $\vec\lambda_1, \dots, \vec\lambda_{2n}$ is a set of vectors in $\mathbb C^n$ that generate the lattice $\Lambda$ over $\mathbb Z$. Each vector $\vec\lambda_i \in \mathbb C^n$ corresponds to a generator $C_i$ of $H_1(T, \mathbb Z)$: to be more specific, this cycle $C_i$ is represented by the straight-line segment in $\mathbb C^n$ joining the zero vector to $\vec\lambda_i$, which descends to a closed cycle in the quotient space $\mathbb C^n / \Lambda$.

Finally, the cycle $C_i$ acts on one-forms in $\Omega^1(T)$ by integration; that is, the cycle $C_i$ corresponds to the linear functional in $(\Omega^1(T))^\star$ that sends each $\omega \in \Omega^1(T)$ to the integral $\int_{C_i} \omega$. In particular, $C_i$ sends the basis vector $dz_j \in \Omega^1(T)$ to $\int_{C_i} dz_j$, and this is equal to the $j$th coordinate of the lattice vector $\lambda_i$.

Since no two $\lambda_i$'s have identical coordinates, this shows that no two elements of $H_1(T, \mathbb Z)$ act identically on $\Omega^1(T)$. In other words, the map $H_1(T, \mathbb Z) \to (\Omega^1(T))^\star$ is injective.

• Thanks for the answer. Maybe this is a dumb question. Is poincare duality used here? It seems that argument does not involve poincare duality explicitly here. – user45765 Jun 16 '18 at 19:10
• @user45765 I did wonder about that! It's easy to see that $\Omega^1(T)$ is the same as $H^{1,0}(T)$, which embeds in $H^1(T)$ by standard Hodge theory. We also know that de Rham's theorem says that the natural map $H_1(T) \to H^1(T)^\star$ is an isomorphism. But I can't see from this that the composition $H_1(T) \to H^1(T)^\star \to H^{1,0}(T)$ is injective. I suspect I'm missing something really obvious, but I'm too tired today... – Kenny Wong Jun 16 '18 at 19:15
• Thanks a lot. I will look this part in Voisin then. – user45765 Jun 16 '18 at 19:16
• and by the way, in my comment above, $H_1(T)$ should be $H_1(T, \mathbb C)$ (not $H_1(T, \mathbb Z)$), and so on... – Kenny Wong Jun 16 '18 at 19:25