Does a fixed-point free "homotopy involution" imply that a manifold bounds?

2011-08-18T15:08:46Z

Let $M^n$ be a closed (compact, connected, without boundary) smooth manifold. It is known that if there exists a fixed point free involution $\tau:M \rightarrow M$, then M bounds. That is, there exists a compact manifold $W^{n+1}$ such that $\partial W = M$.

But now suppose $\tau$ is only a "homotopy involution". That is $\tau^2$ is only homotopic to the identity on $M$ rather than equal to the identity. Can we say that $M$ bounds?

For some reason I feel this statement is not true..., but I have not been able to construct a counterexample yet. For a counterexample, maybe an aspherical, nonbounding manifold would be the best candidate?

On a related question, what if we say that $\tau^2$ is isotopic to the identity on M. Then does M bound?

Thanks, I appreciate any responses.

User scott van thuong - MathOverflow

Does a fixed-point free "homotopy involution" imply that a manifold bounds?