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?
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$\begingroup$ It's not a complex manifold... better ask whether it admits a complex structure / a structure of complex manifold. $\endgroup$– YCorCommented Nov 15, 2019 at 20:31
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3$\begingroup$ In general, the answer is no. In particular, if either $M$ or $N$ are non-orientable, so is their product, which rules out the existence of a complex structure. I'm not sure that there are any known non-trivial conditions which ensure that $M \times N$ admits a complex structure. $\endgroup$– Gabe KCommented Nov 15, 2019 at 21:01
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2$\begingroup$ You are probably thinking of the statement that if $V$ is an $n$-dimensional real vector space, then $V\oplus V$ acquires a complex structure via the matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$. If so, then first consider the fact that if $W$ is another $n$-dimensional real vector space, then one needs to pick an isomorphism between $V$ and $W$ to have such a complex structure. Mimicking this, one could ask whether $M\times M$ has a complex structure. $\endgroup$– KapilCommented Nov 15, 2019 at 21:28
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3 Answers
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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 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.
answered Nov 15, 2019 at 22:38
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$\begingroup$ Thank you! when n=4, why is there not almost complex structure? We can define $J :T_{x}M\rightarrow T_{y}N$ and $J :T_{y}N\rightarrow T_{y}M$ such that J^{2}=-Id? Then J is an almost complex structure. Is it true? $\endgroup$– lidingCommented Nov 16, 2019 at 1:50
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1$\begingroup$ That can work at a point, but you cannot get an almost complex structure on $M\times N$ that way. Note that $T(M\times N) \cong \pi_1^*TM\oplus\pi_2^*TN$ and $\pi_1^*TM \not\cong \pi_2^*TN$ so there is no such map $J : \pi_1^*TM \to \pi_2^*TN$. $\endgroup$ Commented Nov 16, 2019 at 5:29
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$\begingroup$ I checked the definition of almost complex structure. I think almost complex structures only need to be defined point by point. $\endgroup$– lidingCommented Nov 16, 2019 at 6:59
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$\begingroup$ It needs to be smoothly varying, your construction won't be. $\endgroup$ Commented Nov 16, 2019 at 13:56
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$\begingroup$ In my first comment, I should have said that $\pi_1^*TM$ and $\pi_2^*TN$ are not isomorphic unless they are trivial. $\endgroup$ Commented Nov 16, 2019 at 18:11
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I assume you are asking if $M \times N$ admits a complex structure. I believe that there is no result known on this question, although we might have something to say about the existence of an almost complex structure for which both factors are totally real. Once again we can quote Gromov: Page 30. "How much do we gain in global understanding of a compact $(V, J)$ by assuming that the structure $J$ is integrable (i.e. complex)? It seems nothing at all: there is no single result concerning all compact complex manifolds".
answered Nov 15, 2019 at 20:36
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[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.
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$\begingroup$ Since $E$ is trivial on $N$ and $F$ is trivial on $N$, if they are isomorphic, they are both trivial, so $M$ and $N$ have trivial tangent bundle. It then follows that there is an almost complex structure. For example, any 3-manifolds $M$ and $N$ have trivial tangent bundle, so the resulting $M \times N$ has an almost complex structure with two totally real fibrations. $\endgroup$ Commented Nov 16, 2019 at 13:36