TITLE: Hodge numbers of a cartesian product of copies of $\mathbb{C}P^1$
QUESTION [8 upvotes]: I wonder if some works have been done in the context of cohomology space of projective complex manifolds. Specifically I want to study the Hodge diagrams of $\mathbb{C}P^1\times\mathbb{C}P^1$ and $\mathbb{C}P^1\times\mathbb{C}P^1\times\mathbb{C}P^1$. A reference would be very helpful to get started.

REPLY [12 votes]: There is an analogue of the Künneth Theorem for Dolbeault cohomology. It can be found on page $105$ of Principles of Algebraic Geometry by Griffiths and Harris.
If $M$ and $N$ are compact complex manifolds, then we have the following equality of Hodge numbers
$$h^{u, v}(M\times N) = \sum_{\substack{p\, +\, r\, =\, u\\ q\, +\, s\, =\, v}}h^{p,q}(M)h^{r,s}(N).$$
For $\mathbb{CP}^1\times\mathbb{CP}^1$, you should obtain the following Hodge diamond:
\begin{matrix}
  &   & 1 &   &  \\
  & 0 &   & 0 &  \\
0 &   & 2 &   & 0\\
  & 0 &   & 0 &  \\
  &   & 1 &   &
\end{matrix}
For $\mathbb{CP}^1\times\mathbb{CP}^1\times\mathbb{CP}^1$, the Hodge diamond is
\begin{matrix}
  &   &   & 1 &   &   &  \\
  &   & 0 &   & 0 &   &  \\
  & 0 &   & 3 &   & 0 &  \\
0 &   & 0 &   & 0 &   & 0\\
  & 0 &   & 3 &   & 0 &  \\
  &   & 0 &   & 0 &   &  \\
  &   &   & 1 &   &   &
\end{matrix}
You can prove by induction that $h^{k,k}((\mathbb{CP}^1)^n) = \displaystyle\binom{n}{k}$ and for $p \neq q$, $h^{p,q}((\mathbb{CP}^1)^n) = 0$.