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Let $M$ be an even dimensional Reimannian manifold with trivial tangent bundle $TM$. Then there is a global orthonormal basis $\{e_1,e_2,...,e_{2n-1},e_{2n}\}$ of $TM$.

Define a bundle map $J :TM \to TM$ which sends each $e_{2k-1}$ to $e_{2k}$ and $e_{2k}$ to $-e_{2k-1}$ for $k=1,...,n$. Then, does this define a complex structure on $M$? Is it well-defined?

Ref

Complex structure v.s. conformal structure in more than 1 complex dimension

Proof of equivalence of conformal and complex structures on a Riemann surface.

Edit for understanding the answer.

First, we show that $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ has a trivial tangent bundle and to do so, it suffices to show that

(I) the Euler numbr of $M$ is zero and

(II) $M$ is a spin manifold.

Consider the (I).

For any triangularizable toplogical space $X$ and $Y$, there are simple algebraic relations between their Euler numbers.

$\chi(X \#Y) = \chi(X ) +\chi( Y) -2,$

$\chi(X \times Y) = \chi(X ) \times \chi( Y). $

Using these formulas, we can calculate the Euler number of the manifold $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ as follows.

$\chi(M) = \chi (S^2)\chi( S^2)+\chi(S^1)\chi( S^3)+\chi(S^1)\chi( S^3) -4.$

Because $\chi (S^2)=2$ and $\chi (S^1)=0$ and $\chi (S^3)=1-0+0-1=0$, we obtain

$\chi(M) = 2 \times 2 +0 \times \chi( S^3)+0 \times \chi( S^3) -4=0.$

Thus, $\chi(M)=0$ and thus (I) is O.K.

Next, consider the (II).

According to Albanese's comment, the connected sum of spin manifolds is spin.

Thus we show that the spaces $S^2\times S^2$, $S^1\times S^3$ and $S^1\times S^3$ are spin. To poove this, for example, we show $w_2(S^1\times S^3)=0$.

According to problem 4-A of page 54 in Milnor-Stasheff,

$w_2(S^1\times S^3)=w_2(S^1) \times w_0(S^3) + w_1(S^1) \times w_1(S^3) + w_0(S^1) \times w_2(S^3) =0$

where we use the total Stiefel-Whitney class of sphere is 1, namely, $w(s^d)=1$ (page 42 example 1 in Milnor Stasheff).

Jean Billie
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    The connected sum of spin manifolds is spin, so the fact that $M$ is spin follows from the fact that $S^2\times S^2$ and $S^1\times S^3$ are. – Michael Albanese Dec 31 '19 at 03:40

2 Answers2

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First of all, the fact that $M$ is Riemannian plays no role here.

The endomorphism $J$ is indeed well-defined and as $J\circ J = -\operatorname{id}_{TM}$, it is an almost complex structure on $M$. However, it need not be an integrable almost complex structure. That is, it may not be an almost complex structure induced by a complex manifold. For example, $M = (S^2\times S^2)\#(S^1\times S^3)\#(S^1\times S^3)$ has trivial tangent bundle and hence has an almost complex structure as you describe, but it does not admit a complex structure.

In real dimension two, every almost complex structure is integrable, see this question. In real dimension four and above, there are examples of almost complex structures which are not integrable. In real dimension four, there are manifolds (such as $M$) which admit almost complex structures, none of which are integrable. It is an open problem whether such manifolds exist in real dimension six and above.

Michael Albanese
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  • The example seems interesting but it is difficult for me. Please let me give a hint,key words, materials to verify why the manifold has trivial tangent bundle and it does not admit any integral complex structure. I have no experience to consider the tangent bundle of the connected sum. Also I understand the Reimannian metric is redundant ( though, the almost complex structure is compatible with the metric?). – Jean Billie Dec 10 '19 at 11:48
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    The fact that $M$ is parallelisable is a direct consequence of the Dold-Whitney Theorem which states two orientable real rank four bundles over a compact four-manifold are isomorphic if and only if they have the same second Stiefel-Whitney class $w_2$, first Pontryagin class $p_1$, and Euler class $e$. In particular, a compact four-manifold is parallelisable if and only if it is spin manifold with signature and Euler characteristic zero. The fact that $M$ does not admit a complex structure combines the following two facts: 1. If a compact complex manifold has fundamental group a free group ... – Michael Albanese Dec 10 '19 at 11:57
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    ... on a positive number of generators, then it does not admit a Kähler metric; 2. A compact complex surface admits a Kähler metric if and only if its first Betti number is even. So, if $M$ in my answer were to admit a complex structure, it would not admit a Kähler metric by the first point, but that is impossible by the second point. Therefore $M$ cannot admit a complex structure. – Michael Albanese Dec 10 '19 at 11:59
  • I need a time to understand. Thank you. – Jean Billie Dec 10 '19 at 12:08
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I would estimate yes in the sense each 2n is mapped one-to-one in the sense each +2n may be mapped onto the pure reals for the real part of the complex mapping and each of the -2n may be mapped onto the pure imaginary numbers of the imaginary part of the complex system. I am sorry here if will not do. I am an economist, not a mathematician. Sincerely, Richard Anthony Baum

Richard Anthony Baum
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