# Does a fixed-point free “homotopy involution” imply that a manifold bounds?

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.

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Off the top of my head, the reason $M$ bounds in the involution case is that its the boundary of the mapping cylinder of the quotient map $M \to M/\tau$. But that's not a very homotopy-friendly argument. Perhaps you can instead directly argue all the Stiefel-Whitney numbers are zero, and see in that argument if you really need $\tau$ to be an involution. – Ryan Budney Aug 18 '11 at 17:01
A Stiefel-Whitney number argument. Claim: a double cover $f : N \to M$ bounds. Proof: S-W classes of N are pulled back from M, so $<w_I(M), [M]> = <w_I(N), f_*[M]>$ but $f_*[M] = 2[N]=0$. – Oscar Randal-Williams Aug 18 '11 at 18:35
The argument for the involution map is not so easy, see: www.maths.ed.ac.uk/~aar/papers/browfram.pdf – Igor Rivin Aug 18 '11 at 19:57
@Igor: It's not clear to me how the Brown paper relates. Another way to state my argument above is that $M/\tau$ has $M$ as a $2:1$-cover, and $2:1$-covers are the boundaries of $I$-bundles. – Ryan Budney Aug 18 '11 at 20:17

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.
Nice example. I suppose this indicates that to make the question interesting you'd have to put some kind of homotopy non-triviality condition on $\tau$. – Ryan Budney Aug 18 '11 at 20:53