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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.

flag	asked Aug 18 2011 at 15:08 Scott Van Thuong 63●3

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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 2011 at 17:01

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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 2011 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 2011 at 19:57

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@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 2011 at 20:17

Bruno Martelli · Accepted Answer · 2011-08-18 19:19:57Z UTC

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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.

answered Aug 18 2011 at 19:19

Bruno Martelli
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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 2011 at 20:53

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+1. On the other hand, maybe such a thing should not be considered a "homotopy involution," as it is isotopic to the identity itself. – Richard Kent Aug 19 2011 at 4:25

Good point. The question is then still open with this natural interpretation of "homotopy involution". – Bruno Martelli Aug 23 2011 at 10:18

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

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