# Finite groups $G$'s which acts freely on $S^2\times S^2$

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!

• @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 $S^2\times S^2$? – PropositionX Oct 20 '17 at 11:40
• 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 $\Bbb Z_4$. – PropositionX Oct 20 '17 at 11:47

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 $(x,y)\mapsto (-y,x)$.
• Do you know of another way to describe the quotient $(S^2\times S^2)/\mathbb{Z}_4$? – Michael Albanese Jan 16 at 22:51
• I don't think that's necessarily true. Even if it were, I wouldn't know how that could be used to identify $(S^2\times S^2)/\mathbb{Z}_4$. – Michael Albanese Jan 18 at 9:02