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The questions asks us to show that if $M$ is a closed orientable 4-manifold such that $H_2(M)$ is rank $1$, then $M$ does not admit a free action of $\mathbb{Z}/2$.

My attempt has been to suppose $M$ has a free action of $\mathbb{Z}/2$. So there is a homeomorphism $\phi:M\rightarrow M$ satisfying $\phi^2=\textrm{id}$. I've looked at the duality isomorphism $H^2(M)\rightarrow H_2(M)$ and played around with identities like $$ \phi_{\ast}(\phi^{\ast}\alpha\cap[M])=\alpha\cap\phi_{\ast}[M]=\pm\alpha\cap[M] $$ trying to get at some kind of contradiction. But I suspect I need to incorporate the freeness assumption for the action $\mathbb{Z}/2$. I understand that this action will be properly discontinuous and so if $M$ were path connected then the quotient map $M\rightarrow M/(\mathbb{Z}/2)$ would be a covering space and $M/(\mathbb{Z}/2)$ would inherit the structure of a manifold. The trouble is that $M$ isn't necessarily path-connected. Can anyone suggest a way of procedding?

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  • $\begingroup$ Every topological manifold is locally path-connected (being locally homeomorphic to $\mathbb{R}^n$). Does it help in something? $\endgroup$ – Rodrigo Dias Jun 21 '19 at 2:22
  • $\begingroup$ Assume $M=\bigcup _{i\in I} X_i$, where $\{X_i\}$ are all the connected components of $M$. Then $\mathbb{Z}=H_2(M)=\oplus_{i\in I} H_2(X_i)$, so exist exactly an $i$ such that $\mathbb{Z}=H_2(X_i)$ and $\phi(X_i)= X_i$. $\endgroup$ – Bonbon Jun 21 '19 at 2:46
  • $\begingroup$ Now we can just consider the case $M=X_i$ and assume that $M$ is path-connected. $\endgroup$ – Bonbon Jun 21 '19 at 2:48
  • $\begingroup$ Thank you, yes a reduction like this is perfect! $\endgroup$ – user683708 Jun 21 '19 at 3:13
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Let $\phi\colon M\to M$ be a homeomorphism with $\phi^2=\text{id}$. You can use the Lefschetz fixed-point theorem: If $$\Lambda_\phi=\sum_{k\geq0}(-1)^k \text{Tr}(\phi_*|_{H_k(M;\mathbb{Q})})$$ is non-zero then $\phi$ has a fixed-point. Now we have $H_k(M;\mathbb{Q})=H_k(M;\mathbb{Z})\otimes \mathbb{Q}$ and by Poincaré duality together with universal coefficients $H_k(M;\mathbb{Q})\cong H_{4-k}(M;\mathbb{Q})$. The map $\phi$ induces involutions on all these vector spaces, so it is diagonalizable with eigenvalues $\pm1$. With these information you can calculate $\Lambda_\phi\mod 2$ and show that it is non-zero.

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  • $\begingroup$ Great, thanks so much.I'll have to read up on Lefschetz fixed-point theorem. $\endgroup$ – user683708 Jun 21 '19 at 3:16
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A free action of $\mathbb{Z}/2$ leads to a covering map $M \rightarrow M/(\mathbb{Z}/2)$. For an n-sheeted covering map $X \rightarrow Y$, $\chi (X)=n\chi(Y)$. In this case $n=2$ and $\chi(M)$ is odd, by Poincare duality, which yields a contradiction.

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    $\begingroup$ Why is $\chi(M)=1$? I only see that $\chi(M)$ is odd by Poincaré duality (which still gives you the contradiction). $\endgroup$ – P R Jun 21 '19 at 3:06
  • $\begingroup$ I was counting in $\mathbb{Z}/2$ (not really I just got my signs mixed up). $\endgroup$ – Connor Malin Jun 21 '19 at 3:14
  • $\begingroup$ Thanks a lot, this is a very nice solution. $\endgroup$ – user683708 Jun 21 '19 at 3:15

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