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Degree of multiplication map of a topological group

  1. May 6, 2009 #1
    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
    (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?
  2. jcsd
  3. May 6, 2009 #2


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    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)
  4. May 6, 2009 #3
    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 ;)
  5. May 6, 2009 #4
    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.
  6. May 6, 2009 #5
    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... :uhh:

    At least I seem to be narrowing it down :smile:
  7. May 7, 2009 #6


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    If you have several path connected components, what kind of structure do the homology groups have?
  8. May 7, 2009 #7
    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.
    Last edited: May 7, 2009
  9. May 8, 2009 #8


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

    Or more specifically, all of the connected components look the same....
  10. May 8, 2009 #9
    Ahha, I think I've got it, thanks a lot!
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