Albanese map for complex compact manifold

Question

Let $X$ be a compact Kahler manifold. To it, it is possible to associate a complex torus $Alb(X)$ with a map $$alb: X \to Alb(X) .$$

In class, our teacher claimed the image of this map generates the whole torus, but I'm not able neither to prove this result nor to find it on our textbooks.

Answer 1

As I mentioned in the comment, it is proved in Voisin's Hodge Theory and Complex Algebraic Geometry, Volumn I, Lemma 12.11. I will just paraphrase the proof here.

Let $k\ge 1$ and denote $X^k=X\times\cdots \times X$, the $k$-the Cartesian product. Let $alb^k$ be the map $$alb^k:X^k\to Alb(X):=H^0(X,\Omega^1)^*/H_1(X,\mathbb Z)$$ $$(x_1,...,x_k)\mapsto (\omega\mapsto \sum_{i=1}^k\int_{x_0}^{x_i}\omega).$$

Theorem: The map $alb^k:X^k\to Alb(X)$ is surjective for $k>>0$.

Proof: By proper mapping theorem, it suffices to show that $alb^k$ is submersive at some point $\vec{x}=(x_1,...,x_k)$. Dualize the tangent map, it reduces to show injectivity between cotangent spaces

$$T^*_{alb^k(\vec{x})}Alb(X)\to T^*_{x_1}X\times\cdots \times T^*_{x_k}X.\tag{1}\label{1}$$ at some point $alb^k(\vec{x})$. Since a torus $V/\Lambda$ has trivial (co)tagent bundle, the cotangent space at any of its point is canonically identified with $V^*$. Therefore, the LHS of $(\ref{1})$ is identified with $H^0(X,\Omega^1)$ and the map $(\ref{1})$ is just restriction. Now, it is clear that one can choose integer $k$ and points $x_1,...,x_k$ realizing the injectivity. $\square$

Remark 1: When $\dim X=1$, this is just Jacobi inversion theorem for compact Riemann surfaces. Moreover, $Alb(X)$ coincides with Jacobi variety $Jac(X)$ and one can choose $k=\text{genus}(X)$ to generate the torus. (See Griffiths & Harris p.237 for an elementary proof, without using proper mapping theorem.)

Remark 2: An alternative way to prove this theorem is to apply universality property of Albanese map $alb:X\to Alb(X)$, i.e., for any holomorphic map $f:X\to T$ to some complex torus with $f(x_0)=0$, there exists a unique morphism of complex tori $$g:Alb(X)\to T$$ such that $f=g\circ alb$.

On the other hand, one can show that the analytic subvariety generated by $alb(X)$ in $Alb(X)$ is a subtorus, which still satisfies the universal property. Now, uniqueness guarantees that it has to be the whole $Alb(X)$. (See Birkenhake-Lange's Complex Abelian Varieties, Prop. 11.11.8).

Answer 2

Let $A$ be the set of analytic maps from $X$ to a complex torus, for $h\in A,h:X\to \Bbb{C}^m/L$ then $dh(x)_i$ is an holomorphic one-form on $X$, let $V$ be the complex subspace of $\Omega_X$ generated by all those $dh(x)_i,h\in A$.

$\Omega_X$ thus $V$ is finite dimensional because $\pi_1(X,x_0)$ is finitely generated and if $\omega$ integrates to $0$ on every closed-loop then $\int_{x_0}^x \omega$ is holomorphic $X\to \Bbb{C}$ thus it attains its maximum modulus at some $x\in X$, thus it is constant and $\omega=0$.

Take a basis $V=\sum_{j=1}^d \omega_j \Bbb{C}$, fix some $x_0\in X$, let $\Lambda = \{ (\int_\gamma\omega_1,\ldots,\int_\gamma \omega_d)\in \Bbb{C}^d,\gamma \in \pi_1(X,x_0)\}$ then $$Alb(X)=\Bbb{C}^d/\Lambda, \qquad F: X\to Alb(X),\qquad F(x)=(\int_{x_0}^x\omega_1,\ldots,\int_{x_0}^x\omega_d)$$ $\Lambda$ is a discrete subgroup of $\Bbb{C}^d$ because any $h\in A,h:X\to \Bbb{C}^m/L$ factors as $h=h(x_0)+H\circ F+L$ with $H$ linear $\Bbb{C}^d\to \Bbb{C}^m$ (where $H(y)_i=\sum_j c_{ij} y_j, dh(x)_i = \sum_j c_{ij}\omega_j$) and with enough different $h$ the obtained $H$ is injective. $\Bbb{R}\Lambda=\Bbb{C}^d$ for the same reason.

Thus $\Lambda$ is a lattice and $Alb(X)$ is a complex torus.

Every $h\in A$ factors uniquely through $F$. By a dimensionality argument (the image of $\mathrm{x}\in X^d \to \sum_{k=1}^d F(\mathrm{x}_k)\in Alb(X)$ contains a $d$-dimensional complex ball) any point of $Alb(X)$ is a $\Bbb{Z}$-linear combination of some $F(\mathrm{x}_k)$.