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Let $(M,J)$ be a complex manifold. We say that $\omega\in H^2(X;\mathbb{R})$ is Kahler if and only if $\omega$ is closed, positive [i.e. $\omega(v,Jv)>0, \forall v\neq0$], and a $(1,1)$-form [i.e. $\omega(Jv,Jw)=\omega(v,w), \forall v,w$ ].

We denote by $K_M$ the set of all Kahler classes on $M$.

Now consider $(N,J')$ some complex manifold and its corresponding Kahler cone $K_N$.

What is known about $K_{M\times N}$ versus $K_M$ and $K_N$?

Feel free to add additional assumptions on $M$ and $N$. I am only curious to find out what the known theorems are.

For instance: there is a natural map from $K_{M}\times K_N$ to $K_{M\times N}$, sending $(\omega,\eta)$ to $$\omega\times 1+1\times\eta.$$

What can be said about this map? Does the cone generated by its image equal $K_{M\times N}$?

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Even if M and N are (some high genus) curves, this is an open question! – Mohammad F. Tehrani Jun 5 '13 at 16:42
Look at "positivity in algebraic geometry"; couple of examples are discussed there. – Mohammad F. Tehrani Jun 5 '13 at 16:44
Think first about tori $M=\mathbb{C}^n/\Lambda$ and $N=\mathbb{C}^m/\Lambda'$, where every Kahler class has a constant representative, and you can compute everything explicitly. You will see that in this case $K_{M\times N}$ is much larger than the cone generated by $K_M\times K_N$, essentially because of Kunneth's formula. – YangMills Jun 6 '13 at 21:26

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