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I would like to check, whether the product space $X = \mathbb{S}^{n} \times \mathbb{S}^{m}$ admits an almost complex structure for odd $m,n$.

For example, if $m=1$ and $n=3$, then $X = \mathbb{S}^{1} \times \mathbb{S}^{3}$ -- in this case one can construct an almost complex structure as follows:

(1) Since $\mathbb{S}^{k}$ is parallelizable for $k = 1, 3, 7$, there exist linearly independendent sections $e_{1}$, and $e_{1}', e_{2}', e_{3}'$ coresspondingly. By lifting up those vector fields to the globals sections of $T(\mathbb{S}^{1} \times \mathbb{S}^{3})$ we obtain a paralelization of the tangent bundle of a product.

(2) Thus, we calculate the Lie bracket and seek for an endomorphism $J: T_{p}{(S^{1} \times S^{3})} \rightarrow T_{p}{(S^{1} \times S^{3})}$ of the tangent space that satisfies the $J^{2} = -I$.

(3) Moreover, we can easily check the integrability condition in order to establish, whether the almost complex structure lifts to a complex one.

Is it possible to deal with the general case somehow, i.e. that is not restricted by parallelization property?

Michael Albanese
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hyperkahler
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    Note that $\mathbb S^1\times \mathbb S^{2n-1}$ is complex: those are the Hopf surface. –  Dec 16 '17 at 04:50
  • Any product $S^{2n+1}×S^{2m+1}$ of odd spheres admits a complex structure, as the total space of a holomorphic $S^1\times S^1$ fiber bundle over $CP^n \times CP^m$ (see "Calabi--Eckmann manifolds"). – Aleksandar Milivojević Nov 05 '18 at 15:08

1 Answers1

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Let $n$ and $m$ be odd.

As $\chi(S^n) = 0$, the manifold $S^n$ has a nowhere-vanishing vector field, and hence $TS^n \cong E\oplus\varepsilon^1$ for some rank $n - 1$ vector bundle $E$. Now note that

\begin{align*} T(S^n\times S^m) &\cong \pi_1^*(TS^n)\oplus\pi_2^*(TS^m)\\ &\cong \pi_1^*(E\oplus\varepsilon^1)\oplus\pi_2^*(TS^m)\\ &\cong \pi_1^*E\oplus\pi_1^*\varepsilon^1\oplus\pi_2^*(TS^m)\\ &\cong \pi_1^*E\oplus\varepsilon^1\oplus\pi_2^*(TS^m)\\ &\cong \pi_1^*E\oplus\pi_2^*\varepsilon^1\oplus\pi_2^*(TS^m)\\ &\cong \pi_1^*E\oplus\pi_2^*(\varepsilon^1\oplus TS^m)\\ &\cong \pi_1^*E\oplus\pi_2^*(\varepsilon^{m+1})\\ &\cong \pi_1^*E\oplus\varepsilon^{m+1}\\ &\cong \pi_1^*E\oplus\varepsilon^2\oplus\varepsilon^{m-1}\\ &\cong \pi_1^*E\oplus\pi_1^*\varepsilon^2\oplus\varepsilon^{m-1}\\ &\cong \pi_1^*(E\oplus\varepsilon^2)\oplus\varepsilon^{m-1}\\ &\cong \pi_1^*(E\oplus\varepsilon^1\oplus\varepsilon^1)\oplus\varepsilon^{m-1}\\ &\cong \pi_1^*(TS^n\oplus\varepsilon^1)\oplus\varepsilon^{m-1}\\ &\cong \pi_1^*(\varepsilon^{n+1})\oplus\varepsilon^{m-1}\\ &\cong \varepsilon^{n+1}\oplus\varepsilon^{m-1}\\ &\cong \varepsilon^{n+m} \end{align*}

so $S^n\times S^m$ is parallelisable. More generally, a product of two or more spheres is parallelisable if and only if at least one of them has odd dimension.

As $S^n\times S^m$ is a parallelisable manifold of even dimension, it admits almost complex structures. As pointed out in the comments, such manifolds actually admit complex structures. Note however that there are parallelisable manifolds of even dimension which do not admit complex structures, for example $(S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$.

Michael Albanese
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