Orientable Double Cover of $\mathbb{R}\mathbb{P}^2 \times \mathbb{R}\mathbb{P}^2$

Question

I'm trying to understand the orientable double cover of manifolds and a friend posed a question about $M = \mathbb{R}\mathbb{P}^2 \times \mathbb{R}\mathbb{P}^2$.

First, here is some set up. The fundamental group of $M$ is $\mathbb{Z}_2 \times \mathbb{Z}_2$ and so the number of covering spaces for $M$ is equal to the number of conjugacy classes of subgroups in $\pi_1(M)$. Being abelian, there are 4 subgroups which correspond to the universal cover $S^2 \times S^2$, $S^2 \times \mathbb{R}\mathbb{P}^2$, $\mathbb{R}\mathbb{P}^2 \times S^2$ and $\widehat{M} = S^2 \times S^2/\langle (1,1) \rangle$ where $\langle (1,1) \rangle$ is the "diagonal" order 2 subgroup.

The question: What are the de Rham groups of $\widehat{M}$?

We know that since $M$ is non-orientable and compact, $\widehat{M}$ should be connected and so $H^0_{dR}(\widehat{M}) = \mathbb{R} = H^4_{dR}(\widehat{M})$ by Poincare duality. But I don't know how to compute $H^1, H^2$. I was thinking of using the only other tool that I know which is the Mayer-Vietoris theorem but I don't know what two open subsets to choose. And Kunneth's Formula doesn't seem to apply because the quotient by group action complicates the product.

Answer 1

You can see $\hat M$ has a flat bundle (suspension) of $S^2$ over $\mathbb{R}P^2$ and apply the Serre spectral sequence which gives $H^1(\hat M)=H^1(\mathbb{R}P^2,H^0(S^2))=H^1(\mathbb{R}P^2,\mathbb{R})$ and $H^2(\hat M)=H^2(\mathbb{R}P^2,H^0(\mathbb{R}))=H^2(\mathbb{R}P^2,\mathbb{R})$.

Answer 2

The forms on $M/G$ correspond to forms on $M$ invariant under $G$. Now, if you have a complex of vector spaces with an action of a finite group $G$, and you consider the subcomplex of invariant elements, the cohomology of the complex of invariants will be isomorphic to the invariants of the cohomology. That is, if $A$ is your complex of vector spaces, we have a morphism of complexes $A^G\to A$ so a morphism of the cohomology that maps $H(A^G)$ to $H(A)^G$.

Injective: let $a$ $G$ invariant, such that $a = d a'$. Then $a = d( \frac{1}{|G|} \sum g a')$, the image of an invariant cochain.

Surjective: Let $\hat a \in H(A)^G$. Then $g a - a = d( b_g)$ for some $b_g$. Then $a = \frac{1}{|G|}(\sum g a) - d(\frac{1}{|G|}\sum b_g)$.

Now, we know that the cohomology of the complex of invariant forms is isomorphic to the invariants of the cohomology. On $H^2(S^2)\otimes H^2(S^2)$ the map $-1\otimes -1$ acts trivially, so the invariants is the full space, $\mathbb{R}$.

On $H^2(S^2\times S^2)= H^2(S^2)\otimes \mathbb{R}\oplus \mathbb{R}\otimes H^2(S^2)$, the map $(-1,-1)$ acts by $-1$ on each term. So we get $0$. No other non-zero term except $H^0$.

Answer 3

We can use the fact that $H^1_{dR}(\widehat{M}) \cong H_1(\widehat{M}, \mathbb{R})^* = \text{Hom}(H_1(\widehat{M}),\mathbb{R})$ and since $H_1(\widehat{M},\mathbb{R}) = H_1(\widehat{M},\mathbb{Z}) \otimes_\mathbb{Z} \mathbb{R} = 0$, $H^1_{dR}(\widehat{M}) = 0 = H^3_{dR}(\widehat{M})$ by Poincare duality. We can then compute $H^2$ by looking at the Euler characteristic. In general, we know that $\chi(M \times N) = \chi(M)\chi(N)$ and if $M$ is a $k$-fold cover of $N$, then $\chi(M) = k \chi(N)$. Thus, we know that $\chi(\widehat{M}) = b_2 + 2 = 2\chi(\mathbb{R}\mathbb{P}^2)\chi(\mathbb{R}\mathbb{P}^2)$. So we compute the Euler characteristic of $\mathbb{R}\mathbb{P}^2$. $\mathbb{R}\mathbb{P}^2$ is covered twice by $S^2$ and $\chi(S^2) = 2 = 2\chi(\mathbb{R}\mathbb{P}^2)$. So $\chi(\mathbb{R}\mathbb{P}^2) = 1 \Rightarrow b_2 + 2 = 2$. Thus, $H^2_{dR}(\widehat{M}) = 0$.