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If I have two non-orientable connected manifolds such that their orientable double covers are homeomorphic, can anything be said about the manifolds? Are they homeomorphic?

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up vote 4 down vote accepted

Here's a slightly easier example than mland's.

Let $X = S^2\times S^4$ Then $X$ is the universal (orientation) cover of both $\mathbb{R}P^2\times S^4$ and $S^2\times \mathbb{R}P^4$. By the Kunneth formula, $H_2(\mathbb{R}P^2\times S^4) \cong 0$ while $H_2(S^2\times \mathbb{R}P^4) = \mathbb{Z}$.

Somewhat relatedly, (if you want a lower dimensional example), $T^2\times S^2$ covers both $K\times S^2$ and $T^2\times \mathbb{R}P^2$ where $K$ denotes the Klein bottle. In this case, the fundamental group of the first is nonabelian while the fundamental group of the second is abelian.

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Does anyone know if $4$ is the lowest possible dimension of a counterexample? –  Jason DeVito Feb 22 '13 at 18:45
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They are in general not homeomorphic. Here is a counterexample. Consider the manifold $X = S^2 \times S^2 \times S^2$. We define two free $C_2$ actions on X, where $C_2$ is the cyclic group of order $2$. The first is $(x,y,z) \sim (-x,y,z)$ and the second is $(x,y,z) \sim (-x,-y,-z)$. You obtain two quotients with are on the one hand $\mathbb{R}P^2\times S^2 \times S^2$ and on the other hand some $6$-manifold $M$.

Since both actions are orientation reversing you see that both quotients are non-orientable. Moreover since $X$ is simply-connected both quotients have $C_2$ as fundamental group. This directly implies that $X$ must be in both cases the orientation double cover (since there is only one connected 2-sheated covering over the quotients).

Now we claim that $M$ and $\mathbb{R}P^2\times S^2 \times S^2$ are not homeomorphic. This you can prove using rational cohomology. For $\mathbb{R}P^2\times S^2\times S^2$ you can do this by Künneth, and for $M$ I at the moment do not know of a better reason than using the Cartan-Leray spectral sequence to show that $H^*(M;\mathbb{Q}) \cong H^*(X;\mathbb{Q})^{C_2}$ where the last means the $C_2$-invariants of the cohomology of $X$ by the induced $C_2$-action. Now you can distinguish the two manifolds by H^2 and H^4.

It would be interesting to explicitly identify the manifold $M$ if this is possible.

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Thank you for answering. I'm working out the details of the counterexample. –  Chu Feb 21 '13 at 3:43
    
You are welcome. Were you able to check all details? –  mland Feb 22 '13 at 14:38
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