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).(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Degree of multiplication map of a topological group

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