Hodge numbers of base of fibration Let $X$ be a smooth complex manifold which is a fibration over a complex base manifold $B$. Supposing we know the details of the fibration, is it possible from knowledge of the Hodge numbers of $X$ to extract information about the Hodge numbers of $B$?
 A: Let me expand my comment. I need to assume that $X$ and $B$ are algebraic, compact and smooth. Then, the Hodge numbers of $X$ coincide with the Hodge numbers of $F \times B$.
For see this, consider the Grothendieck ring of varieties $K_0(\text{Var}/\Bbb C)$. This is the ring generated by varieties over $\Bbb C$, quotiented by the "scissor relation" $[X] = [Y] + [X \backslash Y]$ where $Y \subset X$ is closed. The product is defined as $[X] \cdot [Y] = [X \times Y]$.
Now we have the following theorem : 

Theorem : There is a ring morphism : $P_{Hodge} : K_0(\text{Var}/\Bbb C) \to \Bbb Z[u^{\pm 1},v^{\pm 1}]$ such that if $X$ is smooth and proper then $ P_{Hodge}(X) = \sum_{p,q} h^{p,q}u^pv^q$

In particular, this means that for compute $P_{Hodge}$ you can cut and glue as you like as long as you finish with proper smooth varieties. For example, $P_{Hodge}(\Bbb P^1 \times \Bbb P^1) = P_{Hodge}(\Bbb A^1 \times \Bbb A^1) + 1 + uv + uv = (uv)^2 + 2uv + 1$ which indeed coincide with the usual Hodge number for $\Bbb P^1 \times \Bbb P^1$. 
Now it is clear that in $K_0(\text{Var}/\Bbb C)$ we have $[X] = [B][F]$ if $X$ is a fibration with base $B$ and fiber $F$, so the claims follows. 
This gives you all the informations you can have, with the fact that $$ h^{p,q}(X \times Y) = \sum_{u+v = p, r+s = q}h^{u,r}(X)h^{v,s}(Y)$$
Edit : In fact, for the more general case of a fibration one can also carry similar computations, for simplicity let me assume that $X \to B$ is a fibration with finitely many singular fibers $F_1, \dots, F_n$. Then we have the equality $[X] - \sum_i ([F_i] - [F])= [B][F]$ where $F$ is the generic fiber, and this can gives you information about the Hodge numbers of $B$ assuming that you can compute everything else. 
