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This is an example from a question sheet (non-assessed) of a university class. If M is a non-orientable, closed, connected 3 manifold, prove $H_1(M;\mathbb{Z})$ is an infinite group. I know that since $M$ is non-orientable it follows that $H_3(M;\mathbb{Z}) = 0$ and since $M$ is connected $H_0(M;\mathbb{Z}) = \mathbb{Z}$. I want to apply a combination of Poincare duality and Universal coefficient theorem to try and find a contradiction if $H_1(M;\mathbb{Z})$ has no free part. However poincare duality doesn't apply here so I'm stuck.

I would appreciate any hints as to how to proceed.

EHH
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1 Answers1

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We can use the fact that the Euler characteristic of any odd dimensional closed manifold is $0$.

Given this, we have $0= \mbox{rk}H_0 - \mbox{rk}H_1 + \mbox{rk}H_2 - \mbox{rk}H_3$.

$M$ is connected so $\mbox{rk}H_0 = 1$. $M$ is non-orientable so $\mbox{rk}H_3 = 0$. If $H_1$ is finite, then $\mbox{rk}H_1 = 0$, so we get $$0 = 1-0+\mbox{rk}H_2 - 0$$ which implies that $\mbox{rk}H_2$ is negative which is impossible.

So $H_1$ must be infinite.

Dan Rust
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    So if we use the facts listed above and the fact that the homology will be trivial for n>3 we get the sum 0 = 1 - H1 + H2 (where Hi is the no. of Z summands in the ith homology group) and so H1 = 1+H2 >= 1 and so H1 is infinite. Why did you reccomend rational coefficients? Thank you for your hint as well :) – EHH Mar 17 '14 at 17:30
  • @EdwardHart I was imagining something like "Suppose $H_1$ is finite, then over $\mathbb{Q}$ we have $H_1=0$... hence contradiction". I suppose the Euler characteristic only sees non-torsion components anyway so it makes little difference. – Dan Rust Mar 17 '14 at 18:13
  • Yeah I guess either works. Thanks for your help! – EHH Mar 17 '14 at 18:42
  • @EdwardHart No problem. As a corollary, this also leads to a quick proof that $\pi_1(M)$ must be infinite as well. – Dan Rust Mar 17 '14 at 18:45
  • So this just follows directly from the fact that H1 is the abelianization of the fundamental group right? – EHH Mar 21 '14 at 15:50
  • Yep. If $\pi_1$ is finite than its abelianisation would also be finite which can't be the case. – Dan Rust Mar 21 '14 at 15:54
  • Now that OP has worked out the details in the comments, would you consider expanding your hint into a full answer? I can even do it for you if you don't want to bother. As you can see other people have the same question, and a hint-only answer runs contrary to the "knowledge repository" aspect of the site. – Najib Idrissi Nov 28 '15 at 08:13
  • @NajibIdrissi Good point. Edited. – Dan Rust Nov 29 '15 at 17:21