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.



Cohomology class of a current 
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.


