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.
Take any positive $(1,1)$form $\omega$ on $X$ and let $T$ be a positive $(p,p)$current. Then, the trace measure $$ \sigma_T=\frac{1}{2^{np}(np)!}T\wedge\omega^{np} $$ 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 $\sigma_T(X)=\frac{1}{2^{np}(np)!}\int_XT\wedge\omega^{np}\ge 0$ 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 nonzero closed and positive but since there is no nontrivial $H^2$cohomology, it is also exact. Your statement about $(1,1)$current holds instead always true if you look exactness in BottChern 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. 


A theorem of Harvey and Lawson (Inventiones 1983), says that a compact manifold $X$ of dimension $n$ is nonKahler if and only if it supports a nonzero, positive current $T$ of bidimension $(1,1)$ (hence bidegree $(n1,n1)$) which is the $(n1,n1)$ 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 nonzero, positive current of bidimension $(1,1)$ which is $i\partial\bar\partial$exact. 

