Let $M$ and $N$ be Kähler manifolds. I know that the cartesian product $M\times N$ is a kahler manifold. I was wondering which form has the Kähler form on $M\times N$. Here's what I thought:
Let $\omega_M$ and $\omega_n$ be the kahler forms respectively on $M$ and on $N$, and let $h^M$ and $h^N$ be the associated hermitian forms on $M$ and on $N$. Then for every $(p,q)\in M\times N$ i define the hermitian form $h_{(p,q)}$ on $T_{(p,q)}(M\times N)\simeq T_p(M)\oplus T_q(N)$ as $h_{(p,q)}(\frac{\partial}{\partial x_i^M},\frac{\partial}{\partial x_j^N})=0$, $h_{(p,q)}(\frac{\partial}{\partial x_i^M},\frac{\partial}{\partial x_j^M})=h^M_{p}(\frac{\partial}{\partial x_i^M},\frac{\partial}{\partial x_j^M})$ and $h_{(p,q)}(\frac{\partial}{\partial x_i^N},\frac{\partial}{\partial x_j^N})=h^N_{p}(\frac{\partial}{\partial x_i^N},\frac{\partial}{\partial x_j^N})$ i.e. $h$ is the pullback of the hermitian forms on $M$ and $N$.
Then taking $a_{ij}$ coefficients of the hermitian form $h$ and $z_i$ coordinates on $M\times N$, the form
$\omega=\frac{i}{2}\sum_{i,j}a_{ij}dz_i\wedge d{\overline{z}}_{j}$
is closed and thus Kähler
Am I right?