Does a manifold which bounds always admit a free involution?

Question

If a closed smooth manifold $M$ admits a smooth free involution $T$, then it bounds. In fact, the mapping cylinder of the quotient map $M \to M/T$ is the manifold whose boundary is $M$. Is the converse true? If not, then could someone give an example of a closed smooth manifold which bounds but does not admit any free involution.

Answer 1

The answer is no. In your example, $M$ is the boundary of a twisted $I$-bundle over another manifold. That's the level of generality in which there's an if and only if statement.

If you want a closed smooth manifold which bounds, but which does not admit a free involution (you can go further and say it does not admit any involution, regardless of having a fixed point set or not) then take a compact orientable hyperbolic 3-manifold that has no symmetries. These exist, since Sadayoshi Kojima proved there is a hyperbolic 3-manifold whose isometry group (therefore the group of homotopy-classes of homotopy equivalences by Mostow rigidity) is any finite group. But all compact 3-manifolds are the boundary of some 4-manifold, this is the Dehn-Lickorish theorem.

Answer 2

Consider $M = \mathbb{CP}^2\#\mathbb{CP}^2$. All of its Stiefel-Whitney numbers vanish, so $M$ is unorientedly nullcobordant; more generally, $X\# X$ always bounds.

We have $H^2(M; \mathbb{Z}) \cong H^2(\mathbb{CP}^2; \mathbb{Z})\oplus H^2(\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}a\oplus \mathbb{Z}b$, and $a^2 = b^2$ is a generator for $H^4(M; \mathbb{Z})$ while $ab = 0$. Now let $f : M \to M$ be a continuous map, and suppose $f^*a = ma + nb$, then

$$f^*(a^2) = (f^*a)^2 = (ma + nb)^2 = m^2a^2 + n^2b^2 = (m^2 + n^2)a^2$$ so $\deg f \geq 0$. In particular, if $f : M \to M$ is a homeomorphism, it must be orientation preserving.

Suppose $f : M \to M$ is a free involution. Then $N := M/\mathbb{Z}_2$ is a compact connected four-manifold which is orientable because $f$ preserves orientation, so $b_0(N) = b_4(N) = 1$. Furthermore, $N$ has fundamental group $\pi_1(N) = \mathbb{Z}_2$, so $b_1(N) = 0$ and hence $b_3(N) = 0$ by Poincaré duality. As the Euler characteristic is multiplicative under coverings, we see that $\chi(N) = \frac{1}{2}\chi(M) = \frac{1}{2}(4) = 2$ so $b_2(N) = 0$. On the other hand, the signature is also multiplicative under coverings, so $\sigma(N) = \frac{1}{2}\sigma(M) = \frac{1}{2}(2) = 1$; this is impossible as $b_2(N) = 0$. Therefore $M$ does not admit a free involution.

More generally, this argument can be used to show that $k\mathbb{CP}^2$ does not admit a fixed-point free homeomorphism of finite order. Therefore $k\mathbb{CP}^2$ is not the covering space of any manifold other than itself.