Albanese varieties are described here, and provide motivation for this question, although are not central to it.
Let $X$ be an algebraic variety and $\Omega^1$ the space of everywhere regular differential $1$-forms on $X$. In the construction of Albanese varieties, one defines the map $\phi: H_1(X,\mathbb{Z})\rightarrow (\Omega^1)^*$ given by $\phi: \gamma\mapsto (\omega\mapsto \int_\gamma \omega)$ for each $\gamma\in H_1(X,\mathbb{Z})$ and $\omega\in \Omega^1$. It is stated that $\phi(H_1(X,\mathbb{Z}))$ is a lattice in $(\Omega^1)^*$. Why is this the case?
Recall that being a lattice implies $\phi(H_1(X,\mathbb{Z}))\otimes_\mathbb{Z} \mathbb{R}\simeq (\Omega^1)^*$.
Note: some sources such as chapter 5 of Beauville note that this is due to Hodge theory, but I don't see how
Best Answer
It suffices to show that the map $H_1(X,\mathbb R)\to(\Omega_X^1)^*$ is surjective. Using Serre duality we have $(\Omega_X^1)^*=H^0(X,\Omega_X^1)^* = H^n(X,\Omega^{n-1}_X).$ Using Poincare duality we have a map $$\tag{$*$} H_1(X,\mathbb R)\cong H^{2n-1}(X,\mathbb R)\to H^{2n-1}(X,\mathbb C) = H^n(X,\Omega^{n-1}_X)\oplus H^{n-1}(X,\Omega^n_X)$$ where the last equality uses the Hodge decomposition. Now since the Hodge decomposition satisfies symmetry under conjugation, we must have that, inside $H^{2n-1}(X,\mathbb C),$ $$H^{2n-1}(X,\mathbb R)=\{\alpha+\overline\alpha\mid \alpha\in H^n(X,\Omega^{n-1}_X)\}.$$ Under this description the map $H_1(X,\mathbb R)\to (\Omega_X^1)^*$ is just the map $(*)$ composed with projection onto the first factor and is clearly surjective.