If the fiber and the base are Kähler manifolds is the total space also Kähler? Let $E$ be a compact manifold and consider the fiber bundle
$$F\to E\to B$$
Assume that $F$ and $B$ are Kähler manifolds. Is $E$ also a Kähler  manifold?
 A: Let $H$ be the standard Hopf surface $(\mathbb{C}^2\setminus\{(0,0)\})/\mathbb{Z}$ where the $\mathbb{Z}$-action is generated by the map $(z_1, z_2) \mapsto (2z_1, 2z_2)$. There is a holomorphic submersion $\pi : H \to \mathbb{CP}^1$ given by $[(z_1, z_2)] \mapsto [z_1, z_2]$, so $\pi$ is a holomorphic fibre bundle. The fiber of $\pi$ over $[1, 0]$ is $C := \{[(w, 0)] : w \in \mathbb{C}^*\} \cong \mathbb{C}^*/\mathbb{Z}$ where the $\mathbb{Z}$-action is given by $w \mapsto 2w$. This is a one-dimensional compact complex submanifold of $H$, namely a torus. So $\pi : H \to \mathbb{CP}^1$ is a holomorphic fibre bundle with fibre a torus. 
Note that tori and $\mathbb{CP}^1$ are Kähler manifolds, but $H$ is not because it is diffeomorphic to $S^1\times S^3$ and $H^2(S^1\times S^3; \mathbb{Z}) = 0$. Topologically, we have taken the product of the standard Hopf fibration $S^1 \to S^3 \to S^2$ with a circle to obtain $S^1\times S^1 \to S^1\times S^3 \to S^2$.

More generally, if $F$ and $B$ are manifolds which admit Kähler metrics, and $F \to E \to B$ is a fibre bundle, then $E$ may not even be orientable. For example, the Klein bottle $K$ is an $S^1$-bundle over $S^1$ so there is a fibration $S^1\times S^1 \to K\times K \to S^1\times S^1$.
A: The book Algebraic Geometry II, part 2 Algebraic Surfaces by Shafarevich and Iskovskikh answers this question in section 14.3 (p. 241) with the example of a specific type of Hopf surface
$$ (\mathbf C^2\smallsetminus 0) / \langle T\rangle$$
where
$$T(z_1,z_2) = (\alpha z_1, \alpha z_2)$$
 for some $$|\alpha|<1$$
as it maps to $\mathbf P^1$ via
$$(z_1,z_2)\mapsto (z_1,z_2)$$
with fibres elliptic curves $E$.
