Consider the product of $2n$ two-spheres $X_n=(S^2)^{2n}$. This manifold admits an orientation preserving involution that preserves the product structure and acts as the (orientation reversing) central symmetry for each $S^2$. Is it possible to say for which $n$ the quotient manifold $X_n/\mathbb Z_2$ admits an almost complex structure? I believe that if such $n$ exists then $n>1$, i.e. $S^2\times S^2/\mathbb Z_2$ is not almost complex (for the defined action).
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$\begingroup$ Hmm. I would have said that $X_1/\mathbb{Z}_2$ is almost complex. Since this is an orientable $4$-manifold, the only obstruction is that $w_2$ is the reduction of an integral class. There is a fibration $X_1\to X_1/\mathbb{Z}_2\to B\mathbb{Z}_2$, and $w_2(X_1/\mathbb{Z}_2)$ pulls back to $w_2(X_1)=0$, so comes up from the nonzero class in $H^2(B\mathbb{Z}_2;\mathbb{Z}_2)$, which is an integral reduction. $\endgroup$– Mark GrantCommented Dec 12, 2018 at 14:27
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3$\begingroup$ For $n = 1$ we have $X_1 = \operatorname{Gr}^+(2, 4)$ and $X_1/\mathbb{Z}_2 = \operatorname{Gr}(2, 4)$ which does not admit an almost complex structure. $\endgroup$– Michael AlbaneseCommented Dec 12, 2018 at 14:59
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7$\begingroup$ @MarkGrant: That is the first obstruction, but not the only one for a four-manifold. You also need the integral lift $c$ of $w_2$ to satisfy $c^2 = 2\chi + 3\sigma$. Every closed orientable four-manifold has an integral lift of $w_2$ (i.e. they are spin${}^c$), but they do not always admit almost complex structures. $\endgroup$– Michael AlbaneseCommented Dec 12, 2018 at 15:03
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7$\begingroup$ @MarkGrant: Theorem 1 of this paper by Heaps completely answers the question of when an eight-manifold has an almost complex structure, but it is significantly more complicated. $\endgroup$– Michael AlbaneseCommented Dec 12, 2018 at 15:10
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3$\begingroup$ Neat. Checking these should be do-able, for someone with sufficient motivation and time on their hands. Note that these manifolds are projective product spaces, and Don Davis has calculated their cohomology and Steenrod operations here: arxiv.org/abs/0908.0525 $\endgroup$– Mark GrantCommented Dec 12, 2018 at 15:16
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1 Answer
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Let $Y_n = X_n/\mathbb{Z}_2$.
If $G$ is a finite group acting freely on a manifold $M$, and $\pi : M \to M/G$ denotes the quotient map, then $\pi^* : H^*(M/G; \mathbb{Q}) \to H^*(M; \mathbb{Q})$ is injective; moreover, the image is $H^*(M; \mathbb{Q})^G$.
Involution acts on $H^2(S^2; \mathbb{Q})$ by $-1$, so by the Künneth Theorem, $\mathbb{Z}_2$ acts on $H^{2l}(X_n; \mathbb{Q})$ by $(-1)^l$. Therefore $H^{4k + 2}(Y_n; \mathbb{Q}) \cong H^{4k + 2}(X_n; \mathbb{Q})^{\mathbb{Z}_2} = 0$ and hence $H^{4k + 2}(Y_n; \mathbb{Z})$ is a torsion group. In particular, if $Y_n$ admits an almost complex structure, the odd Chern classes must be torsion.
A product of spheres is stably parallelisable, so $p_i(TX_n) = 0$. As $\pi : X_n \to Y_n$ is a covering map, we have $\pi^*TY_n \cong TX_n$, so $\pi^*p_i(TY_n) = p_i(\pi^*TY_n) = p_i(TX_n) = 0$. Therefore $p_i(TY_n) = 0 \in H^{4i}(Y_n; \mathbb{Q})$ and hence $p_i(TY_n) \in H^{4i}(Y_n; \mathbb{Z})$ is torsion.
If $Y_n$ admits an almost complex structure, then
$$p_i(TY_n) = 2(-1)^ic_{2i}(TY_n) +\ \text{terms involving lower Chern classes}.$$
Together with the fact that the odd Chern classes of $Y_n$ are torsion, it follows inductively from the above formula that all of the Chern classes of $Y_n$ are torsion. In particular, $c_{2n}(Y_n) = 0$ as $H^{4n}(Y_n; \mathbb{Z}) \cong \mathbb{Z}$ is torsion-free. But this is a contradiction as
$$\langle c_{2n}(Y_n), [Y_n]\rangle = \langle e(Y_n), [Y_n]\rangle = \chi(Y_n) = \frac{1}{2}\chi(X_n) = \frac{1}{2}2^{2n} = 2^{2n-1} \neq 0.$$
Therefore $Y_n = X_n/\mathbb{Z}_2$ does not admit an almost complex structure for any $n$.
Although this is not part of your question, it is worth noting that this implies that $\operatorname{Gr}(2, 4)^n$ does not admit an almost complex structure for any $n$ (because $\operatorname{Gr}(2, 4)^n$ is covered by $Y_n$).
answered Jan 9, 2019 at 5:59