Has anyone seen the following result in the literature? I've asked a few experts but so far I've come up with nothing.

Given a manifold M and an open cover {U_i} of M, we want to see how families of diffeomorphisms of M can be adapted to {U_i}. We will think of families of diffeomorphisms as generators of C_*(Diff(M)), where C_*() denotes singular chains.

Def: A k-parameter family of diffeomorphisms f: P^k \times M -> M is supported on V \subset M if, for all y not in V, we have f(p, y) = f(q, y) for all p, q \in P. In other words, f is independent of the parameters P outside of V.

Define A_k \subset C_k(Diff(M)) to be the subcomplex generated by all k-parameter families (k-chains) of diffeomorphisms f such that f is supported on a union of at most k of the U_i's, and such that (inductively) the boundary of f is in A_{k-1}.

Claim: A_* is homotopy equivalent to C_*(Diff(M)).

There is a similar result if we replace Diff(M) with Maps(M -> T), where T is some topological space. It is used in the proof of the claim in this question.

  • $\begingroup$ Maybe it's worth saying that we know the result is true, because we've written down a long and complicated proof, but that we'd be much happier if someone else has already done this or if there's a better explanation. $\endgroup$ – Scott Morrison Oct 23 '09 at 16:13

This is not a full answer, but there is a once-well-known technique in geometric topology that could help clarify this result. Kirby and Siebenmann, in their old work on triangulability of manifolds, have a technique called handle straightening. It is a very nice way to push around pieces of a diffeomorphism without disturbing other pieces.

To use this technique, give the manifold M a handle decomposition (which in your case should refine your open cover) and collar all of the handles. Then you can isotop the 0-handles with without disturbing the 1-handles. How? You should shrink back the higher handles along the collars of the 0-handles, then isotop the 0-handles in a region that is that contains the original 0-handles but is disjoint from the trimmed higher handles. Then you can extend the higher handles back again so that they attach to the 0-handles. You can repeat this for the k-handles in turn, provided that the requested isotopy of the k-handles is the identity in a neighborhood of the (k-1)-skeleton.

I don't know if your whole result is standard, but this part, if you can use it in your construction, is standard.

  • $\begingroup$ Thanks Greg. It's not immediately clear to me how to apply this technique to our problem, but does have a similar flavor. $\endgroup$ – Kevin Walker Nov 6 '09 at 17:43
  • $\begingroup$ For convenience, let's let M be compact. The it's easy to reduce to the case that all of the motions are smaller than the handles, and the Jacobian matrices are all close to the identity. I'm not sure, but it then may not be hard to factor the isotopy into isotopies internal to each handle. $\endgroup$ – Greg Kuperberg Nov 6 '09 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.