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.

Answer by Bruno Martelli for Does a fixed-point free "homotopy involution" imply that a manifold bounds?

2011-08-18T19:19:57Z

A manifold with zero Euler characteristic admits a nowhere-vanishing vector field, which generates a one-parameter group of diffeomorphisms that are (smoothly) isotopic to the identity. A sufficiently small element $\tau$ is fixed-point free since the vector field does not vanish and the manifold is compact.

There are manifolds with zero Euler characteristic that do not bound, for instance the unoriented cobordism group in dimension 5 is not trivial, see the Wikipedia page on cobordisms.

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

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

Answer by Bruno Martelli for Does a fixed-point free "homotopy involution" imply that a manifold bounds?