Degree of multiplication map of a topological group

[Jamma]

Hi. I'm trying to find the degree of the map of f(g,h)=g.h (i.e. multiplication in g) for fixed g. It is a map G-->G (if we fix g).

We can assign a degree to this map for any topological group for which the last non-zero homology group is Z and proceed like we do for the degree of a map between n-spheres.

My intuition is that it should obviously be 1. In fact, it must be +1 or -1 since it is invertible. I have also noted that the set of elements H={g in G|deg(f(g,-))=1} must be a normal subgroup of G and also that G/H would then be the group of order 2 if there were elements with degree -1; reasonably strong conditions.

I can't seem to finish it off though and feel that I've probably just missed something very obvious. Maybe top. groups with final homology group Z are path connected meaning that
H:[0,1]xG-->G
H(t,x)=q(t).x
(where q(t) is a path between e (identity element) and g)
would be a homotopy between the identity map and any mult. map in question, showing the degree to be 1.

Any thoughts to nudge me in the right direction?

 

[Hurkyl]

It's not clear which part you're stuck on -- is it:

(1) You haven't managed to prove the group is path-connected

(2) Assuming it's path connected, translation by g really is homotopic to the identity


If the former, then it would seem the thing to do would be to assume otherwise, and look at the path-connected components...

(Though I confess I am fairly uncomfortable with dealing with connected-but-not-path-connected spaces or other pathologies relating to connectedness)

 

[Jamma]

I haven't managed to prove that the group is path connected, that was a thought that I had. When this is known, the way I continued would be a proof that the degree of the maps are 1.

I wasn't sure if the statement I made was true. Not all topological groups are path connected but I thought that maybe if the last homology group was Z this may be true, although I'm thinking it probably isn't. As I said, it was an idea of how to proceed.

I'll quickly examine the path-components are report back ;)

 

[Jamma]

Hmm not quite seeing how looking at the path components is going to help; couldn't we have a separate path component of elements who's mult' map is degree -1?

I'm not that familiar with topological groups yet so I do not know elementary results about them as such.

 

[Jamma]

By the homotopy, we can see that we have path components of elements in H and others not. Also there must be an equal number of such components as a degree -1 element would send path components to opposite ones since path components -> path components under continuous maps and also every point is mapped to since this is an isomorphism.

So I've deduced there are an equal number of path components...

At least I seem to be narrowing it down

 

[Hurkyl]

If you have several path connected components, what kind of structure do the homology groups have?

 

[Jamma]

Well considered as a topological space, we could just consider the space as a disjoint union of these other path components; and if we had one with Z as its final homology group and others with 0, we could still get something with elements with degree -1 elements.

Maybe being a topological group reduces the generality here however, I will examine this, thanks for the suggestions.

 

[Hurkyl]

Maybe being a topological group reduces the generality here however,

Yep -- the most important aspect is that any points in the group looks just like every other point, because multiplication is a homeomorphism.

Spoiler

Or more specifically, all of the connected components look the same...

 

[Jamma]

Ahha, I think I've got it, thanks a lot!