Even dimensional indecomposable simply-connected closed manifolds with a free action of $\mathbb{Z}/4$

Question

Proposition 2.29 of Hatcher's Algebraic Topology asserts that $\mathbb{Z}/2$ is the only nontrivial group that can act freely on even dimensional spheres.

My question: Which even dimensional indecomposable simply-connected closed manifolds admit a free action of $\mathbb{Z}/4$?

Note that the only simply-connected closed 2-manifold is the 2-sphere, but in dimensions four and higher there are simply-connected, closed manifolds which are not homotopy equivalent to an $n$-sphere.

Of course, $\mathbb{Z}/4$ acts freely on odd dimensional spheres, so it also acts freely on the product of an odd dimensional sphere and another odd dimensional simply-connected closed manifold. Do you have more nontrivial examples?

Thank you!

Answer 1

Just in case you aren't happy with fiber bundles (as they are, perhaps, "too close" to being decomposable), here's a whole infinite family of simply connected $4$-manifolds which are not fiber bundles and which admit a free $\mathbb{Z}_4$ action.

(Note that $4$ is the minimal dimension where you can find such an example since a simply connected $2$-manifold is automatically $S^2$).

To that end, let $M = S^2\times S^2$ and let $\phi:M\rightarrow M$ with $\phi(x,y) = (-y,x)$ be the order $4$ fixed point free homeomorphism in Jyrki's answer. Let $n,s\in S^2$ denote the north and south poles.

Pick $N$ to be any simply connected $4$ manifold which is not homeomorphic to $S^4$. Note that this implies the second Betti number $b_2(N)\geq 1$. Connect sum 4 copies of $N$ to $M$ at each of the points $(n,n), (n,s), (s,n),$ and $(s,s)$. Call the resulting $4$-manifold $N'$.

Notice that $\phi$ extends to an order $4$ fixed point free homeomoprhism of $N'$ by permuting around the copies of $N\setminus\{Ball\}$.

Now, there are compact $4$-manifolds with arbitrarily large second Betti number. For example, one can take connect sum of $\mathbb{C}P^2$ with itself any number of times. Since $b_2(N') = 4b_2(N) + 2 \geq 6$, which follows, for example, from Meyer-Vietoris, we have an infinite family of $N'$ up to homotopy.

Finally, why are none of the $N'$ bundles? Well, if it's a bundle of the form $S^1\rightarrow N'\rightarrow B^3$, then the long exact sequence (LES) in homotopy groups, together with the fact that $\pi_1(N')$ is trivial implies $\pi_1(B^3)$ is trivial. Hence, $B^3$ is a sphere (Poincare' conjecture), so $\pi_2(B^3)$ is trivial. Then the LES again tells us $\pi_2(N') = 0$. This contradicts the fact that $b_2(N') \geq 6$.

What if it's a bundle of the form $F^2\rightarrow N' \rightarrow B^2$? Well, as above, we must have $\pi_1(B^2) = 0$, so $B^2 = S^2$. Since $\pi_2(F^2)$ is either $\mathbb{Z}$ or trivial, this shows $\pi_2(N')$ is generated by at most $2$ elements. So, $b_2(N') \leq 2$, a contradiciton.

Finally, if it's a bundle $F^3\rightarrow N'\rightarrow S^1$, then the LES shows that $F^3$ has infinitely many components. This contradicts the fact that $N'$ is compact.

Answer 2

Proffering $S^3\times S^3$ as an example.

We can identify $S^3$ with $SU(2)$, i.e. complex matrices of the form $$ \left(\begin{array}{rr}z_1&z_2\\-\overline{z_2}&\overline{z_1}\end{array}\right), $$ where we add the constraint $|z_1|^2+|z_2|^2=1$. These can, in turn, be identified with unit quaternions. Anyway the diagonal matrix $D=diag(i,-i)$ is obviously of order four, and multiplication by it gives a free action of $C_4$ on $S^3$.

We then let it act diagonally on $S^3\times S^3$.

Observe that $\pi_1(S^3)$ is trivial, so the same holds for $S^3\times S^3$ as well.

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As a 4-dimensional example I suggest $S^2\times S^2$. Taking advantage of the antipodal mapping $\overline{x}\mapsto-\overline{x}$ we can construct a fixed-point-free homeomorphism of order four $\phi$ by the recipe $$ \phi(\overline{x},\overline{y})=(\overline{y},-\overline{x}). $$ The composition square $\phi\circ\phi$ simply maps both components to their antipodes, so it cannot have fixed points. Therefore neither do the first or the third powers (a fixed point of either would also be a fixed point for the square). Again, the space is also simply connected.

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Qiaochu Yuan observed that $SU(3)$ is an example that is not a product of lower dimensional spaces. It is a fiber bundle over $S^5$ with fibers $S^3$ though. I'm too ignorant to tell whether we can also construct lower dimensional examples as fiber bundles. It's been a while, and I never built up a large enough personal collection of example manifolds to refer to in the first place :-(