Is every Kahler form on a Stein manifold exact outside a compact set?

Question

Is it true that on a smooth Stein manifold (or smooth affine variety or smooth complete intersection in $\mathbb{C}^{n}$), every Kahler form is exact outside a compact set?

Answer 1

This seems extremely false to me: I can't think of any example of a positive dimensional Stein manifold $W$ and a compact subset $K$ such that $H^2(W) \to H^2(W \setminus K)$ has nontrivial kernel. That means there should be an abundance of counter-examples: Just take any Kahler form representing a nontrivial class in $H^2(W)$.

In any case, here is a specific counter-example. Take $W = (\mathbb{C}^*)^2$ with coordinates $z_1$ and $z_2$. Take a $(1,1)$ form of the form: $$\omega := a \frac{dz_1 \wedge d\overline{z_1}}{z_1 \overline{z_1}} + b \frac{dz_1 \wedge d\overline{z_2}}{z_1 \overline{z_2}} + c \frac{dz_2 \wedge d\overline{z_1}}{z_2 \overline{z_1}} + d \frac{dz_2 \wedge d\overline{z_2}}{z_2 \overline{z_2}}$$

This is obviously closed. If I haven't dropped any signs, it is Kahler if and only if $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ is positive definite Hermitian. (It is easiest to do this computation after setting $z_i = e^{w_i}$, so $d z_i/z_i = d w_i$.)

Now, let $T$ be the torus $|z_i|=r_i$. Irrespective of what $r_1$ and $r_2$ are, $\int_T \omega = (b-c) (4 \pi^2)$. (Again, if I haven't dropped any signs.) So, if we take $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ a positive definite Hermitian matrix with $b-c \neq 0$, then we will have a Kahler form $\omega$ for which $\int_T \omega \neq 0$. For example, $\left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) = \left( \begin{smallmatrix} 2 & i \\ -i & 2 \end{smallmatrix} \right)$ would do.

Now, let $K$ be any compact set. For $(r_1, r_2)$ large enough, the torus $T$ is disjoint from $K$. So we have exhibited a Kahler form on $W$ such that, for any $K$, there is a $2$-cycle $T$ in $W \setminus K$ with $\int_{T} \omega \neq 0$. So $\omega$ is not exact on any $W \setminus K$.