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Prove that there doesn't exist a free $\mathbb{Z}_2\times \mathbb{Z}_2$ action on $S^n$.

I know that an action of a group $G$ on a space $X$ is a homomorphism from $G$ to the group $Homeo(X)$ of homeomorphisms $X\to X$, and the action is free if the homeomorphism corresponding to each nontrivial element of $G$ has no fixed points.

And $\mathbb{Z}_2=\{-1,1\}$ can act freely on $S^n$ via $(1,x)\mapsto x$ and $(-1,x)\mapsto -x$. I also know that $S^n/\mathbb{Z}_2\cong \mathbb{R}P^n$

But how to analyze the action of $\mathbb{Z}_2\times \mathbb{Z}_2$? I got a little confused and don't know where to start. Any helps will be highly appreciated!

Added:

I have learned some homology theory and I got some results based on that. But I haven't learned the cohomology theory. I would like to show you some of the results I've got, and I hope to find a method that doesn't go beyond homology. Any enlightening ideas will be highly appreciated!

  1. The only nontrivial group that can act freely on $S^{2n}$ is $\mathbb{Z}_2$

It's known that if $f:S^n\to S^n$ has no fixed point, then $f$ is homotopic to the antipodal map whose degree is $(-1)^{n+1}$. Define $d:G\to \{-1,1\}$ by taking $g\in G$ to the degree of the homeomorphism derived by $g$. Then we know $$G\cong G/\ker d \cong \text{Im } d \le \mathbb{Z}_2$$

  1. $\mathbb{Z}_2\times \mathbb{Z}_2$ can't act freely on $S^1$.

It's known that if $G$ is a finite group, $Y$ is path connected and locally path connected, then the quotient map $p:Y\to Y/G$ is a normal covering map, and G is the group of deck transformations of $Y\to Y/G$ and $$G\cong \pi_1(Y/G)/p_*(\pi_1(Y))$$ For $S^1$ we know that $S^1/G$ is homeomorphic to $S^1$. Hence G must be cyclic.

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    $\begingroup$ Well, I know an answer that is not very straightforward but is beautiful to me. Existence of free action implies the existence of free resolution of $\mathbb{Z}$ over the group ring of $\mathbb{Z}/2\times\mathbb{Z}/2.$ But for elementary abelian group it implies cyclicity. As I've said it's not straightforward and requires some insight into group cohomology. You can look it up in Brown's book on cohomology of groups. $\endgroup$ – Gregory G Dec 8 '19 at 10:37
  • $\begingroup$ @GregoryG Thanks for your words! But that's a little out of my knowledge, and I wonder whether we can solve it with homology since that's what I already learned. I will also go into the chapter you mentioned soon cause I really want to know such a method. $\endgroup$ – Chiquita Dec 10 '19 at 22:43
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I'm not sure where I first encountered (a version of) this proof, but it's definitely not original.

For ease of writing, I will write $G = \mathbb{Z}_k\oplus\mathbb{Z}_k$. Note that if $G$ acts freely, then any subgroup of $G$ acts freely. Hence, we may assume $k$ is prime.

Assume for a contradiction that $G$ acts freely on $S^n$ with $n$ odd. You have already handled the case $n = 1$, so we will assume $n\geq 3$. Because $n$ is odd, a simple application of Lefschetz shows that $M:=S^n/G$ is a closed orientable manifold. Further, because $n > 1$, $G\cong \pi_1(M)\cong H_1(M)\cong H_1(M;\mathbb{Z}_k)$. Poincare duality now forces $H^{n-1}(M;\mathbb{Z}_k)\cong G$.

Now, $\pi_k(M) = 0$ for $k = 2,..., n-1$. So, we can turn $M$ into an Eilenburg-Maclane space $K(G,1)$ by attaching cells of dimension $n+1$ or higher. In particular, by uniqueness (up to homotopy) of a $K(G,1)$, we have $H^\ast(M)\cong H^\ast(K(G,1))$ for $\ast\leq n-1$.

Thus, $H^{n-1}(K(G,1);\mathbb{Z}_k)\cong G$.

On the other hand, a model for $K(G,1)$ is a product of 2 infinite dimensional lens spaces $L_k$. Each of these has $H^\ast(L_k;\mathbb{Z}_k)\cong Z_k$ for all $\ast$.

Now, because $k$ is prime, $\mathbb{Z}_k$ is a field, so the Kunneth theorem is particularly nice: $H^{n-1}(K(G,1); \mathbb{Z}_k)\cong \bigoplus_{s=0}^{n-1} H^s(L_k;\mathbb{Z}_k)\otimes H^{n-1-s}(L_k;\mathbb{Z}_k)$. Since each cohomology group is $\mathbb{Z}_k$ and $\mathbb{Z}_k\otimes \mathbb{Z}_k\cong \mathbb{Z}_k$, so $H^{n-1}(K(G,1);\mathbb{Z}_k)\cong \bigoplus_{s=0}^{n-1} \mathbb{Z}_k \cong \mathbb{Z}_k^n$.

Thus, we conclude $$\mathbb{Z}_k^2 \cong G\cong H^{n-1}(M;\mathbb{Z}_k)\cong H^{n-1}(K(G,1);\mathbb{Z}_k)\cong \mathbb{Z}_k^n$$

In other words, $n=2$. Since $n$ was the (odd!) dimensions of our sphere, we have a contradiction.

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    $\begingroup$ Rereading the OP's request, I now see the "homology only" part of it. Well, good news! Because I'm using field coefficients, everything I wrote above applies if you just lower all the indices from superscrips to subscripts. $\endgroup$ – Jason DeVito Dec 11 '19 at 15:07
  • $\begingroup$ Thanks for your words! But I don't get your point that "just lower all the indices from superscrips to subscripts." If I don't use cohomology, won't I just get a Kunneth formula as $H_{1}(K(G,1); \mathbb{Z}_k)\cong \bigoplus_{i+j=1} H_i(L_k;\mathbb{Z}_k)\otimes H_j(L_k;\mathbb{Z}_k)$ which doesn't work? $\endgroup$ – Chiquita Dec 11 '19 at 18:19
  • $\begingroup$ You still want to use $H_{n-1}$ on the left and $i+j = n-1$ on the right. $\endgroup$ – Jason DeVito Dec 11 '19 at 18:21
  • $\begingroup$ But we don't have $H_{n-1}(K(G,1),\mathbb{Z}_k)\cong G$, where is the contradiction? $\endgroup$ – Chiquita Dec 11 '19 at 18:24
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    $\begingroup$ @Zero: It's not stupidity! I mean exactly what you wrote in the first sentence. The way to compute $H_{n-1}(M;\mathbb{Z}_k)$ is via Poincare duality, where I'm thinking of the original Betti number (the $\mathbb{Z}_k$ Betti numbers) formulation of Poincare duality (with doesn't use cohomology). See en.wikipedia.org/wiki/Poincar%C3%A9_duality $\endgroup$ – Jason DeVito Dec 11 '19 at 21:08

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