# An almost complex structure on $S^2\times …\times S^2 / \mathbb{Z_2}$

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).

• 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. – Mark Grant Dec 12 '18 at 14:27
• 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. – Michael Albanese Dec 12 '18 at 14:59
• @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. – Michael Albanese Dec 12 '18 at 15:03
• @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. – Michael Albanese Dec 12 '18 at 15:10
• 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 – Mark Grant Dec 12 '18 at 15:16

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 $$Y_n$$).