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I've been studying homology groups, and this question is stumping me:

Prove there can be no compact manifold $X$ without boundary whose homology groups are $$H_i(X) = \left\{ \begin{array}{ll} \mathbb{Z} & i = 0 \\ \mathbb{Z}_3 & i=1 \\ 0 & i = 2 \\ \mathbb{Z}_2 & i=3 \\ 0 & i\geq 4 \end{array} \right.$$

I tried creating a chain complex in order to look at differential maps, and $H_2(X) = 0$ helps gives injectivity to one of the maps, but I'm not seeing how to prove no such manifold can exist.

phenomenalwoman4
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1 Answers1

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Denote our space by $M$.

Note that the fundamental group cannot have any subgroup of index two since this would constitute a nontrivial map $\pi_1M \rightarrow \mathbb{Z}/2$ which must factor through the abelianization of $\pi_1M$, i.e. $\mathbb{Z}/3$. But every map $\mathbb{Z}/3 \rightarrow \mathbb{Z}/2$ is trivial.

Whence $M$ has no nontrivial double covers. If $M$ were a manifold, this would imply that the orientation double cover was trivial, whence $M$ would be orientable- but this is clearly false because the homology groups do not satisfy Poincaré duality for any possible dimension for the manifold.

Dylan Wilson
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    What a brilliant answer, Dylan! I'm sorry I can only upvote you once, but I'm sure other users will show you their appreciation. I'll try to remember the trick that a manifold with $H_1$ or $\pi_1$ of finite odd cardinality is orientable. – Georges Elencwajg Aug 16 '13 at 21:10
  • haha thanks Georges :) – Dylan Wilson Aug 17 '13 at 08:11