Mapping Class Group of Contractible Spaces

[WWGD]

Hi all,

Isn't the mapping class group of a contractible space trivial (or, if we consider isotopy, {+/-Id})?

Since every map from a contractible space is (homotopically)trivial.

 

[homeomorphic]

I think you have to specify what you mean by the mapping class group. Not that I don't know what it is, do the maps have to fix the boundary?

Generally, homotopic might not imply isotopic, although that's generally true for homeomorphisms of surfaces from what I remember (possibly with a couple exceptions?).

 

[WWGD]

Thanks, I mean the space of self -homeomorphisms, up to
isotopy :http://en.wikipedia.org/wiki/Mapping_class_group

 

[homeomorphic]

Orientation-reversing homeomorphism of a 2-disk to itself is homotopic to the identity, but not isotopic. An isotopy has to be a homeomorphism at every point in time, so it should map boundary to boundary and preserve the degree of the map on the boundary. But, as you say, everything is homotopic because it's contractible.

 

[WWGD]

homeomorphic ,
This is a related question: let K be a knot in $\mathbb R^3$ , and let h: be a homeo (a relative of yours ;) ) of $\mathbb R^3$ to itself . Is h(K) ~K (as knots, i.e., they are isotopic)? I say yes (up to orientation), since MCG($\mathbb R^3$)= {Id, -Id}, so that h(K)~Id or h(K)~ -Id(K). Do you agree?

 

[homeomorphic]

Seems like that should be true, but how do you know that's the mapping class group of R^3?

There's only one homotopy class of maps, but I'm not 100% sure about isotopies.

In any case, Gordon-Lueke does imply that the knots are isotopic.

 

[WWGD]

But any two isotopies are homotopies, so we only need to consider {+/- Id}. And isotopies preserve orientation, so we're done.

 

[homeomorphic]

But any two isotopies are homotopies, so we only need to consider {+/- Id}. And isotopies preserve orientation, so we're done.

Not following. Isotopies are homotopies, but if you want to establish what the isotopy classes are, you can't use homotopies. Sure, anything is homotopic to the those, but we want isotopic.

 

[WWGD]

Sorry, I skipped a few steps :we use Alexander's trick, since $\mathbb R^n,D^n$ are isomorphic. Then, every two homeos. that agree on the boundary (a condition on elements of the MCG) are isotopic .

 

[homeomorphic]

Sorry, I skipped a few steps :we use Alexander's trick, since Rn,Dn are isomorphic. Then, every two homeos. that agree on the boundary (a condition on elements of the MCG) are isotopic.

Still not following. D^n usually includes the boundary, but R^n has no boundary. For D^n, I agree now that the mapping class group is trivial--incidentally, I just remembered, the mapping class group is usually defined to be orientation-preserving homeos, if there's an orientation.

A contractible space is a pretty general thing, though. No orientation, no -Id, no Alexander's trick. So, isotopy classes of self-homeomorphisms aren't so clear in general.

 

[WWGD]

But the doesn't the isotopy between maps on D^n restrict to one between the respective interiors ? Specially since the boundary is sent to the boundary in each embedding in the path? Related :any two contractible subspaces of the same space are homotopic (an equiv. rel. ). Are they also isotopic?

 

[homeomorphic]

But the doesn't the isotopy between maps on D^n restrict to one between the respective interiors ? Specially since the boundary is sent to the boundary in each embedding in the path?

The problem is that you are starting with a map on the interior, not on the whole D^n. You'd have to be able to extend it. I can't think of a counter-example off the top of my head, but it doesn't seem like you should be able to extend an arbitrary homeo of the open disk to the closed disk.

Related :any two contractible subspaces of the same space are homotopic (an equiv. rel. ). Are they also isotopic?

I don't have a very good feel for isotopy equivalence, but I don't think so.