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Hello, i still have a question about positive closed currents. In particular i know that if $X$ is a compact complex manifold and $T$ is a positive closed current of bidegree $(1,1)$ such that its cohomology class is zero then is itself zero. Now, is it possible that is trivial, but is still true if the bidegree is greater than $1$ ? thanks in advance.

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Take any positive $(1,1)$-form $\omega$ on $X$ and let $T$ be a positive $(p,p)$-current. Then, the trace measure $$ \sigma_T=\frac{1}{2^{n-p}(n-p)!}T\wedge\omega^{n-p} $$ is a positive measure on $X$ which dominates the mass measure $||T||$ of $T$. In particular, if $\sigma_T$ has vanishing total mass then it is zero, and if it is zero then $||T||$ and hence $T$ is zero.

If $T$ and $\omega$ are closed (thus, in particular $X$ must be Kähler), then the total mass $\sigma_T(X)=\frac{1}{2^{n-p}(n-p)!}\int_XT\wedge\omega^{n-p}\ge 0$ depends only on the cohomology classes of $T$ and $\omega$. In particular, if $T$ is zero in cohomology, then the total mass of the trace measure of $T$ is zero and hence $T$ is zero.

Therefore the answer to your question is yes, provided the manifold $X$ is Kähler.

On the other hand, the answer is no in general, even for $(1,1)$-currents. Here is a counterexample:

Take $X=(\mathbb C^2\setminus\{0\})/\mathbb Z$, where $\mathbb Z$ acts by homotheties, to be the Hopf surface. It is topologically $S^1\times S^3$, hence by Künneth formula $b_2(X)=0$ (in particular $X$ is not Kähler). The image of the two (punctured) axes of $\mathbb C^2$ by the projection are two elliptic curves on $X$. Take as $(1,1)$-current the current of integration over one of these two elliptic curves: it is then non-zero closed and positive but since there is no non-trivial $H^2$-cohomology, it is also exact.

Your statement about $(1,1)$-current holds instead always true if you look exactness in Bott-Chern cohomology! In this case, indeed, your current is a $\partial\bar\partial$ of a function which is plurisubharmonic, since your current is positive by assumption. The conclusion follows by the maximum principle.

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thanks so much. so the assumption to be Kahler is necessary. thank you very clear. –  alike Jan 11 '12 at 15:41
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The assumption to be Kähler is not necessary. Indeed, there are non-Kähler manifolds in which the dd-bar lemma holds (e.g. Moishezon manifolds), so that you can use the same arguments. –  Henri Jan 11 '12 at 15:46
    
Indeed... Kähler is sufficient, not necessary! What Henri says is of course about the last part of my answer, that is about $(1,1)$-currents. A more general class of manifolds for which the $\partial\bar\partial$-lemma holds are the manifolds of the Fujiki class $(\mathcal C)$ , that is compact complex manifolds bimeromorphic to Kähler ones. –  diverietti Jan 11 '12 at 16:27
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A theorem of Harvey and Lawson (Inventiones 1983), says that a compact manifold $X$ of dimension $n$ is non-Kahler if and only if it supports a non-zero, positive current $T$ of bidimension $(1,1)$ (hence bidegree $(n-1,n-1)$) which is the $(n-1,n-1)$ component of a $d$-exact current. The current $T$ is $\partial\bar\partial$-closed (it doesn't have to be $d$-closed), so your claim is almost equivalent to $X$ being Kahler.

For manifolds in the Fujiki class ${\mathcal C}$, one can show that $T$ can be chosen to be $d$-closed: $X$ is Fujiki but not Kahler if and only if it supports a non-zero, positive current of bidimension $(1,1)$ which is $i\partial\bar\partial$-exact.

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