Let $M$ be a compact Kähler manifold. Recall that (by definition) a cohomology class in $H^2(M, \mathbb{R})$ is called positive (negative) if it can be represented by a closed positive (negative) $(1,1)$ form.
Why is this well defined? In other words, is it ever possible to have a positive and a negative form both representing the same cohomology class?
Even though I suspect that the answer may be elementary, I would appreciate any help on this.
Thank you.
No, it's not possible for a cohomology class on a compact complex manifold to be represented by both a positive $(1,1)$-form and a negative one. Here's a proof.
Let $M$ be a compact complex manifold of (complex) dimension $n$, and suppose $\alpha_+$ and $\alpha_-$ are cohomologous closed $(1,1)$-forms, with $\alpha_+$ positive and $\alpha_-$ negative. Then the $(1,1)$-form $\omega = \alpha_+ - \alpha_-$ is positive and exact.
Because $\omega$ is positive and closed, it's a Kähler form. On a Kähler manifold, $\omega^n = n! dV$, where $\omega^n$ is the $n$-fold wedge product of $\omega$ with itself, and $dV$ is the volume form of the associated Riemannian metric; and therefore $\int_M \omega^n >0$.
On the other hand, because $\omega$ is exact, there is a $1$-form $\beta$ such that $\omega = d\beta$. It follows that $\omega^n = d( \beta\wedge \omega^{n-1})$, so $\omega^n$ is exact. It then follows from Stokes's theorem that $\int_M \omega^n = 0$, a contradiction.