MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 9 down vote accepted

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 $W \setminus K$.

share|cite|improve this answer
I think I have corrected all the sign errors now. – David Speyer Feb 26 '11 at 15:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.