If the base and fiber of a fiber bundle are complex manifolds, is the total space a complex manifold?

Question

Let $\pi: E\rightarrow B$ be a fiber bundle with fibers $F$. If $B$ and $F$ are complex manifolds, does this imply $E$ is also a complex manifold?

Answer 1

Initially, your question seemed to ask: if the total space and fibre were complex manifolds, is the base a complex manifold? The answer is no, one interesting example arises from twistor geometry. There is a fibre bundle

$$\mathbb{CP}^1 \to \mathbb{CP}^3 \to S^4$$

called the Penrose fibration. Although the fibre and the total space are complex manifolds, the base is not (it doesn't even admit an almost complex structure).

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Your actual question was: if the fibre and the base are complex manifolds, is the total space a complex manifold? Again the answer is no.

The Klein bottle $K$ is a fibre bundle with $S^1$ fibre and $S^1$ base, so there is a fibre bundle

$$S^1\times S^1 \to K\times K \to S^1\times S^1.$$

Although $S^1\times S^1$ admits a complex structure, $K\times K$ does not (complex manifolds are orientable).