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).



An almost complex structure on $S^2\times ...\times S^2 / \mathbb{Z_2}$
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 $Y\_n$).


