11
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 5
    $\begingroup$ What general facts do you know about the homology of compact manifolds without boundary? $\endgroup$ – Qiaochu Yuan Aug 15 '13 at 21:09
  • 1
    $\begingroup$ I can't solve this problem (the space is not orientable, hence Poincaré duality with $\mathbb Z$-coefficients does not apply) and I would be very gratuful to @Qiaochu or someone else to post a solution. $\endgroup$ – Georges Elencwajg Aug 16 '13 at 8:32
  • $\begingroup$ Oops. My mistake. $\endgroup$ – Qiaochu Yuan Aug 16 '13 at 8:37
18
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – Georges Elencwajg Aug 16 '13 at 21:10
  • $\begingroup$ haha thanks Georges :) $\endgroup$ – Dylan Wilson Aug 17 '13 at 8:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.