# Non-Kähler complex structure on $S^2 \times T^4$

Consider $$M = S^2 \times T^4$$. Then we can construct a non-Kähler complex structure as follows. Let $$L$$ be a line bundle over $$\mathbb{P}^1$$ such that there are two sections $$s_1, s_2 \in H^0 (L)$$ without common zeros. Thus, $$\sigma = (s_1, s_2)$$ is a nowhere-vanishing section of the vector bundle $$W = L \oplus L$$. The Hamiltonian quaternions act on $$\sigma$$ giving four holomorphic sections $$\sigma_1 = \sigma, \sigma_2 = i \sigma, \sigma_3 = j \sigma, \sigma_4 = k \sigma$$. Then the quotient $$X$$ of $$W$$ by the $$\mathbb{Z}^4$$-action, acting fiberwise by translations $$v \mapsto v + \sum a_i \sigma_i$$, is a compact complex three-fold. By calculating the canonical divisor, $$X$$ is Kähler if and only if $$W$$ is trivial.

I am wondering if it is the only construction, i.e., if we endow $$M$$ with an arbitrary complex structure, then whether the universal covering is a holomorphic vector bundle $$W$$ over $$\mathbb P^1$$?

• Another example to consider is the twistor space of $T^4$ with a flat metric. Nov 21, 2023 at 9:31
• Your construction always gives rise to a complex manifold $M$ which is a holomorphic principal bundle over $\mathbb{CP}^1$. In particular, the family of complex tori over $\mathbb{CP}^1$ is isotrivial. As Michael suggests, you can take the twistor space, then the family will not be isotrivial, so it does not come from this construction. It would be interesting to ask if there exists a complex structure on $M$ such that $M$ does not fibre over $\mathbb{CP}^1$. Nov 28, 2023 at 17:11