The total space of an $S^2$ bundle over a closed surface is a ruled surface

Question

In complex geometry, a complex surface $S$ is called a ruled surface if there is a holomorphic map $\pi:S\to C$ to a complex curve $C$ such that each fiber is $\Bbb CP^1$. Typical examples are $\Bbb P(L\oplus \epsilon)$ where $L\to C$ is a holomorphic line bundle and $\epsilon\to C$ is a trivial complex line bundle.

Is it true that for a smooth $S^2$-bundle $\pi:E\to B$ where $B$ is a Riemann surface that $E$ is a ruled surface?

Answer 1

I assume what you're asking is if, given any smooth, real $S^2$-bundle $\pi:E\to B$ where $B$ has a complex structure, there exists a complex structure on $E$ such that $\pi$ is a holomorphic $\mathbb{C}P^1$-bundle.

The answer is no. There are nonorientable $S^2$ bundles over Riemann surfaces, which carry no complex structure at all. One example is to start with the nonorientable $S^2$ bundle over $S^1$ (the mapping torus of the antipodal map), which I'll denote $\pi:E\to S^1$ and multiply by $S^1$ to obtain $\hat{\pi}:E\times S^1\to T^2$. The base space can be given a complex structure, but the total space cannot.

Answer 2

Question: "Is it true that for a smooth S2-bundle π:E→B where B is a Riemann surface that E is a ruled surface?"

Answer: Let $k$ be the real numbers and $K$ the complex numbers.

It is well known that if $C:=\mathbb{P}^1_K$ is the complex projective line and $S^2_k$ the real 2-sphere, there is an isomorphism of real smooth manifolds $\phi:C_k \cong S^2_k$ where $C_k$ is the projective line viewed as a real smooth surface. The isomorphism $\phi$ may be constructed using projection from the north and south pole of $S^2_k$.

Whenever you have a locally trivial fibration

$$\pi: E \rightarrow B$$

where $B$ is a (compact) Riemann Surface, $E$ a complex projective manifold and where the fibers of $\pi$ are complex projective lines, it follows there is a rank two locally trivial sheaf $F$ on $B$ with $E \cong \mathbb{P}(E^*)$ (this is proved in Hartshorne exercise II.7.10).

If you have a smooth (locally trivial) $S^2_k$-bundle

$$\pi: E \rightarrow B$$

with fibers $\psi_b:\pi^{-1}(b) \cong S^2_k$, it follows each fiber is isomorphic as real smooth manifolds to $C_k$. For $\pi$ to be a ruled surface you must put extra conditions on the isomorphisms $\psi_b$ and the map $\pi$. For $E$ to be a complex manifold you need an "integrable almost complex structure" (an endomorphism of the tangent bundle $T_E$)

$$J: T_E \rightarrow T_E$$

with $J^2=-1$.

https://en.wikipedia.org/wiki/Almost_complex_manifold