Let $M$, $N$ be $n$-dimensional real manifolds. Does $M\times N$ admits a complex structure? If not, are there known condidtions ensuring that $M\times N$ admits a complex structure?
['https://mathoverflow.net/questions/346122', 'https://mathoverflow.net', 'https://mathoverflow.net/users/147073/']
I'm going to take $M$ and $N$ to be connected. Note that they cannot have boundary. Furthermore, as complex manifolds are orientable, $M$ and $N$ must also be orientable.

If $n = 1$, then $M, N \in \{\mathbb{R}, S^1\}$. Note that $\mathbb{R}\times\mathbb{R}$ has a complex structure, in fact, precisely two: $\mathbb{D}$ and $\mathbb{C}$. In addition, $\mathbb{R}\times S^1 = S^1\times \mathbb{R}$ and $S^1\times S^1$ have uncountably many complex structures (annuli and tori respectively).

If $n = 2$, then $M$ and $N$ are orientable surfaces. These always admit complex structures, and therefore, so do their products.

Already for $n = 3$, we do not have a complete answer. In this case, the manifold $M\times N$ always admits an almost complex structure, but it is not clear when it admits an integrable one (i.e. a complex structure). This is a major open problem in complex geometry:

> 
> If a manifold admits an almost complex structure, does it admit a complex structure?
> 
> 
> 

The answer is known to be false in real dimension four, but completely open in higher dimensions.

For $n = 4$, the manifold $M\times N$ may not even admit an almost complex structure. For example, $M = N = S^4$. Theorem 1 of [this paper](https://www.sciencedirect.com/science/article/pii/0040938370900315) of Heaps completely characterises which closed eight-manifolds admit an almost complex structure. In principle, this could be used to determine what $M$ and $N$ could be, at least in the compact case.
[EDIT: I started to write this answer, and then realized it doesn't really answers the question - I leave it here anyway, just in case]

Let $X=M\times N$, and let $E=\pi\_M^\*TM$ and $F=\pi\_N^\*TN$. Note that $TX=E\oplus F$. 
As suggested in Ben McKay's answer, you could first look for the existence of an almost complex structure $J\in End(TX)$, $J^2=-Id$, for which both projections $\pi\_M$ and $\pi\_N$ have totally real fibers. 

This may exist or not. It does for instance under the assumption that there exists an isomorphism of vector bundles $\phi:E\to F$. In that case, define
$$
J:=\left(\matrix{0 &\phi^{-1} \\ -\phi & 0}\right)
$$
In this situation the eigenbundle $T^{1,0}X$ corresponding to $+i$ is isomorphic to $E^{\mathbb{C}}$ in the following way: 
$$
E^{\mathbb{C}}\to T^{1,0}X\quad;\quad v\mapsto v+i\phi(v)
$$
You may want to look for conditions on $\phi$ for $J$ to be integrable.