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I am attemping to find groups which acts freely on $S^2\times S^2$.

$S^2\times S^2$ is Hausdorff, and $G$ is finite. So if $G$ acts freely on $S^2\times S^2$, Hatcher's book says it implies $G$ gives a covering space action.

Use the notation in Hatcher's book, so $S^2\times S^2\to S^2\times S^2/G$ is a covering space, and the covering space is $|G|$ sheeted. As $S^2\times S^2$ has Euler characteristic $4$, (from an exercise in Hatcher we have)$|G|\cdot\chi(S^2\times S^2/G)=\chi(S^2\times S^2)=4$, this shows that $|G|$ is a factor of $4$. Then $G$ could only be $\{e\},\Bbb Z_2,\Bbb Z_4,\Bbb Z_2\times \Bbb Z_2$.

I think I can define an action for $\{e\},\Bbb Z_2,\Bbb Z_2\times \Bbb Z_2$. But I have no idea define an action for $\Bbb Z_4$. May I please ask how may I define an action for $\Bbb Z_4$ on $S^2\times S^2$? Thanks!

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  • $\begingroup$ @AnubhavMukherjee I think the trivial group which gives the trivial map also satisfies the definition of free action, may I please ask why the action of the trivial group cannot be free? For $\Bbb Z_2$ I could define $e$ is the identity map and $1$ is the antipodal map. And for $\Bbb Z_2\times \Bbb Z_2$ I think I can take one $\Bbb Z_2$ to act on one $S^2$, and the other $\Bbb Z_2$ acts on the other $S^2$. Is that correct? How can I prove that $\Bbb Z_4$ cannot act freely on $S^2\times S^2$? $\endgroup$ – PropositionX Oct 20 '17 at 11:40
  • $\begingroup$ Let me try it: Suppose we have an action of $\Bbb Z_4$, if the generator of $\Bbb Z_4$ corresponds to a map $f$, then we require $f^2$ is the antipodal map(I am not sure if the only map which apply twice to give the identity map is the antipodal map, so I am not sure for this step), so $deg(f^2)=deg(f)deg(f)=deg(antipodal)=-1$, which implies $deg(f)=\pm i$, but this is impossible. So we cannot have a free action of $\Bbb Z_4$. $\endgroup$ – PropositionX Oct 20 '17 at 11:47
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Observe that $\mathbb Z_4$ cannot act freely and orientation preserving way on $S^2\times S^2$.

proof: If not then, since its a a finite group, the free action is a covering action. So $M=S^2\times S^2 /\mathbb Z_4$ is a oriented 4 manifold. Now $H_1(M)=H^1(M)=\mathbb Z_4= H_3(M)$. [The last equality comes from Poincare Dulatily]. So $\chi(M)=\sum (-1)^i rank(H_i)\geq2.$ But $\chi(S^2\times S^2)=4$ and order of the group action implies $\chi(M)=1$. So contradiction.

Here is a Free $\mathbb Z_4$ action on $S^2\times S^2$ as $(x,y)\mapsto (-y,x)$.

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    $\begingroup$ Do you know of another way to describe the quotient $(S^2\times S^2)/\mathbb{Z}_4$? $\endgroup$ – Michael Albanese Jan 16 at 22:51
  • $\begingroup$ That's an interesting question,let me think. $\endgroup$ – Anubhav Mukherjee Jan 17 at 1:54
  • $\begingroup$ @MichaelAlbanese I think it is most likely true that any self diffeomorphism of S^2xS^2 sends one generator sphere to the other or fixed them. It is certainly true that the homology classes of S^2 x pt and pt x S^2 must be preserved or switched since you can easily check they are the only homology classes with self-intersection 0. And now if we can prove that in the complement of this, we can always extends the diffeomorphism, then we are done. You have any thoughts? $\endgroup$ – Anubhav Mukherjee Jan 17 at 14:45
  • $\begingroup$ I don't think that's necessarily true. Even if it were, I wouldn't know how that could be used to identify $(S^2\times S^2)/\mathbb{Z}_4$. $\endgroup$ – Michael Albanese Jan 18 at 9:02

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