I think it is true that there is no free-action of $(\mathbb{Z}_2)^3$ on product of $\mathbb{S}^{m}$ and $\mathbb{CP}^n(n$ is odd). I don't know how to prove it. A detailed proof will be very much helpful.

  • 1
    $\begingroup$ This is true for $m=1,n=1$. Closed $3$-manifolds covered by $\mathbb{R} \times S^{2}$ fall into 5 homeomorphism classes, and $(\mathbb{Z}_{2})^{3}$ does not embed into $\pi_{1}$ or any of them. See page 457 of homepages.warwick.ac.uk/~masgar/Teach/2012_MA4J2/geometry.pdf. The case $m=1, n \geq 1$, can probably treated similarly, although it seems quite different to the general case. $\endgroup$ – Nick L Apr 10 '18 at 9:22

Here's a proof that works for all even $m$ and $n\geq 2$ and shows a stronger result: There is no freely acting group $G$ containing three non-identity elements $a,b,c$ for which $ab, ac,bc,abc$ are all non-identity elements. (So, we do not assume $a,b,c$ commute).

I suspect one would have to work much harder for odd $m$, because the corresponding fact about $(\mathbb{Z}_2)^2$ actions on $S^m$ is much harder when $m$ is odd.

Each element in $G$ induces an action on the cohomology ring of $M:=S^m\times\mathbb{C}P^n$. By Kunneth, the cohomology ring of $M$ is given by $$H^\ast(M)\cong\mathbb{Z}[x,y]/x^2 = y^{n+1} = 0$$ with $|x| = m$ and $|y| = 2$.

Because $H^\ast(M)$ is generated (as a ring) by $x$ and $y$, any automorphism is determined by where it sends $y$ and where it sends $x$.

Lemma: If $m$ is even and $n\geq 2$, then any such automorphism maps $x$ to $\pm x$ and $y$ to $\pm y$.

Proof: Assume first $m\geq 4$. Then $y\in H^2(M)$ generates, so any automorphism must map $y$ to $\pm y$.

Now, we know $x, y^{m/2}\in H^m(M)$. If $y^{m/2} = 0$, then $x$ generates $H^m(M)$, so $x$ maps to $\pm x$. If $y^{m/2}\neq 0$, then notice that the equation $0 = (ax + by^{m/2})^2 = 2ab xy^{m/2} + b^2y^m$ forces $a= 0$ or $b=0$ because $xy^{m/2}\neq 0$. Thus, the are 2 ``lines'' (multiples of $x$ and of $y^{m/2}$) are ring theoretically characterized by squaring to $0$. Since an automorphism must map $x$ to a primitive element, it must be sent to $\pm x$ or $\pm y$. But $y$ already maps to $\pm y$, and an automorphism is injective, so $x$ must map to $\pm x$. This concludes the case $m\geq 4$.

If $m = 2$ (and $n\geq 2$) then the only solutions to $0 = (ax + by)^2 = 2abxy + b^2 y^2$ have $b=0$. Thus, $x\in H^2(M)$ is characterized ring theoretically (up to sign) by being a primitive element which squares to $0$, so $x$ must map to $\pm x$. Now, the equation $0 = (ax + by)^{n+1} = (n+1) ab^n x y^n$ (using the fact that $y^{n+1} = x^2 = 0$ and that $x$ and $y$ commute) implies $a=0$ or $b=0$. Since $x$ maps to $\pm x$, then just as in the case of $m\geq 4$, this implies $y$ maps to $\pm y$.


Now we use Lefschetz: if an automorphim of $H^\ast(M)$ maps $x$ to $x$ and $y$ to $y$, then its Lefschetz number is non-trivial, so this map has a fixed point. If there are three elements $a,b,c\in G$ with $ab,ac,bc,abc$ all non-identity then we have a contradiction as follows. There are only three choices of mapping $x\mapsto \pm x$ and $y\mapsto \pm y$ which are not the identity map. If, say, $a$ and $b$ induced the same map, then $ab$ induces the trivial map on $x$ and $y$. This implies the non-identity transformation $a\circ b$ on $M$ has fixed points, contradicting freeness. So, we may assume $a,b,c$ all generated different maps on $x$ and $y$. Then $abc$ acts as the identity on $H^\ast(M)$, so has fixed point.

All this still leaves the case of $m=2, n=1$, when $M = S^2\times S^2$. We can still prove the $\mathbb{Z}_2^3$ can't act freely, but it takes more work: there are new automorphisms of $H^2(M)$ coming from swapping $x$ and $y$.

The calculation $0 = (ax + by)^2$ still implies the the two lines (multiples of $x$ and multiples of $y$) are characterized ring theoretically, so $x$ and $y$ map to elements of $\{\pm x, \pm y\}$. It follows that, up to changing signs, swapping $x$ and $y$ is the only new thing we need to worry about. Of these new automorphisms, only 2 have 0 Lefschetz number: $x \mapsto -y, y \mapsto x$ and $x\mapsto y, y\mapsto -x$. Both of these have order 4, so if every element of $G$ has order $2$, then these can't happen, and so we reduce to the case where $x\mapsto \pm x$ and $y\mapsto \pm y$.

  • $\begingroup$ Does the case $S^{2} \times S^{2}$ not follow from the fact that $\chi(S^{2} \times S^{2}) = 4$? $\endgroup$ – Nick L Apr 10 '18 at 18:21
  • $\begingroup$ @NIck: You're right! No group of order greater than $4$ can act freely on $S^2\times S^2$ for that reason. Much nicer than my arugment - thanks! $\endgroup$ – Jason DeVito Apr 10 '18 at 18:45
  • $\begingroup$ @JasonDeVito Thank you so much. It will be very helpful if you give some hint how to solve it when $m$ is odd ? $\endgroup$ – Shivani Sengupta Apr 11 '18 at 6:01

This is not a complete solution, but it puts a relatively strong restriction on the possible $n$ and $m$ for which this can happen.

Suppose that $m$ is even and a group of order $8$ acts freely on $S^{m} \times \mathbb{CP}^{n}$. We will show that $$ n =3 (\mod 4). $$

Firstly, note that since $S^{m} \times \mathbb{CP}^{n}$ is compact the action is automatically properly discontinuous so the quotient is a manifold and the quotient map is a covering map.

Now, $8| \chi (S^{m} \times \mathbb{CP}^{n}) = 2 (n+1)*$ since the quotient has a degree 8 cover by $S^{m} \times \mathbb{CP}^{n}$ (recall that Euler characteristic is multiplicative for covers), therefore $4| (n+1)$ which proves the claim.

  • since Euler characteristic is multiplicative on products.

Here is the $m$ odd case with one major gap. I don't know how to fill it - does anyone else?

We will assume $m \geq 3$ since $m = 1$ has been handled in the comments by Nick L.

Consider the principal $S^1$ bundle $N\xrightarrow{\pi} M$ where $N:=S^m\times S^{2n+1}$ and $M:= S^m\times \mathbb{C}P^n$.

Lemma: Let $X$ be a reasonable topological space and consider a map $f:X\rightarrow M$. This map has a lift $\tilde{f}:X\rightarrow N$ iff $f^\ast:H^2(M)\rightarrow H^2(X)$ is the $0$ map.

Proof: Because $H^2(N) = 0$, if $f = \pi\circ \tilde{f}$, then $f^\ast$ factors through the $0$ map on $H^2$. So the condition on $H^2$ is definitely necessary.

Sufficieincy is more fun. The bundle $\pi$ is classified by a map $\phi:M\rightarrow BS^1 = \mathbb{C}P^2$, which also classifies $H^2(M)$. If $f^\ast$ is $0$ on $H^2$, it then follows that the pull back bundle $f^\ast N$ is a trivial bundle, $f^\ast N\cong X\times S^1$.

Of course, $f^\ast N = \{(x,n):f(x) = \pi(n)\}$ comes equipped with a natural map into $N$: projection onto the second factor.

Then $\tilde{f}$ is the composition $X\rightarrow X\times S^1\cong f^\ast N\rightarrow N$. $\square$

We will use this lemma in the case where $X = N\times G$, with $G$ a finite group acting freely on $M$.

Then the map $f: X\rightarrow M$ given by $f(n,g)= g\ast\pi(n)$ induces the $0$ map on $H^2$, because $H^2(X) = 0$. By the lemma, there is a lift $\tilde{f}:X\rightarrow N$.

Is this lift a group action?

I don't know the answer. But for the rest of the post, I will assume it is a group action. Then note that the projection map is equivariant. Indeed, writing $g \tilde{f}(n)$ for $\tilde{f}(n,g)$, the fact that $\tilde{f}$ is a lift means $\pi(g\tilde{f}(n)) = g\pi(n)$.

Proposition: This group action is free.

Proof: Suppose $g$ fixes $n$. Then $\pi(n) = \pi(g n) = g \pi(n)$ so $g$ fixes $\pi(n)$. Since the action of $G$ on $M$ is assumed to be free, $n$ is the identity element.$\square$

So we have a free action of $G$ on $S^m\times S^{2n+1}$.

If $G$ contains $(\mathbb{Z}_p)^3$, we are done, by an Annals paper of Adem and Smith which can be accessed here.

  • $\begingroup$ Thank you for your nice idea. I have some problem related to group action on a space which uses spectral sequence. May I contact you through mail. If you can give some idea it will be very helpful. $\endgroup$ – Shivani Sengupta Apr 13 '18 at 14:00
  • $\begingroup$ @Sneha: I prefer if you would just ask a question on MSE and the direct my attention to it (say, by commenting here.) I should also add that I am comfortable with compact connected Lie groups, and less so with finite groups. $\endgroup$ – Jason DeVito Apr 13 '18 at 18:10

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