# Closed orientable 4-manifolds with $H_2(M)\cong \mathbb{Z}$ do not admit free actions of $\mathbb{Z}/2$

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?

• Every topological manifold is locally path-connected (being locally homeomorphic to $\mathbb{R}^n$). Does it help in something? – Rodrigo Dias Jun 21 '19 at 2:22
• 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 $\phi(X_i)= X_i$. – Bonbon Jun 21 '19 at 2:46
• Now we can just consider the case $M=X_i$ and assume that $M$ is path-connected. – Bonbon Jun 21 '19 at 2:48
• Thank you, yes a reduction like this is perfect! – user683708 Jun 21 '19 at 3:13

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