Complex Geometry – Second Homology Group of Kähler Surfaces

Question

Let $M$ be a closed Kähler surface, i.e., Kähler manifold of complex dimension $2$.

Is it true that for any element $\alpha \in H_2(M,\mathbb Z)$, there exists a nonsingular holomorphic curve $\Sigma$ of $M$ such that $[\Sigma]=\alpha$?

Answer 1

This is not true, since Hodge structure provides obstruction.

By the Poincare duality, we have $H_2(M,{\mathbb Z})\cong H^2(M, {\mathbb Z})$. Now the cohomology of a compact Kahler manifold has Hodge structure. So $$ H^2(M,{\mathbb C}) = H^2(M,{\mathbb Z})\otimes {\mathbb C}\cong H^{2,0}(M) \oplus H^{1, 1}(M)\oplus H^{0, 2}(M). $$

The cohomology class of an integral algebraic cycle, in particular of a holomorphic curve, belongs to the Hodge class, that is, $H^{1, 1}(M)\cap H^2(M, {\mathbb Z})$.

Therefore, in particular, a class in $(H^{2, 0}(M)\oplus H^{0, 2}(M))\cap H^2(M, {\mathbb Z})$, and its dual in $H_2(M, {\mathbb Z})$, can't be represented by a holomorphic curve.

Here is a concrete example. Let $M=E_1\times E_2$ be the product of two elliptic curves, where both $E_i={\mathbb C}/({\mathbb Z}+{\mathbb Z}\sqrt{-1})$, that is, $\tau_i=\sqrt{-1}$. Let $a_i={\mathbb R}/{\mathbb Z}$, and $b_i={\mathbb R}\sqrt{-1}/{\mathbb Z}\sqrt{-1}$ be the generators of $H_1(E_i, {\mathbb Z})$.

By the Kunneth formula, $$ H_2(M,{\mathbb Z})=\bigoplus_{i+j=2} H_i(E_1,{\mathbb Z}))\otimes H_j(E_2,{\mathbb Z}). $$ Let $z_i=x_i+\sqrt{-1} y_i$ be the complex coordiates on the ${\mathbb C}$ covering $E_i$. Then $dz_i\in H^{1, 0}(E_i)$, and $$ dz_1\wedge dz_2\in H^{2, 0}(M). $$ We compute \begin{align} dz_1\wedge dz_2&=(dx_1+\sqrt{-1}dy_1)\wedge (dx_2+\sqrt{-1}dy_2)\\ &=(dx_1\wedge dx_2 - dy_1\wedge dy_2)+\sqrt{-1}(dx_1\wedge dy_2+dy_1\wedge dx_2). \end{align} The real and imaginary parts are Poincare dual to $$ a_1\times a_2-b_1\times b_2,\ \ a_1\times b_2+b_1\times a_2\in H_2(M,{\mathbb Z}). $$ The elements in this rank 2 lattice in $H_2(M,{\mathbb Z})$ (whose total rank is 6) can't be represented by an integral algebraic cycle since it is Poinare dual to elements in $H^{2, 0}(M)\oplus H^{0, 2}(M)$.