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Question on equivariant maps

  1. Jan 16, 2016 #1
    Hello,

    I have a doubt about equivariant maps in the context of group theory. In particular, if we consider an automorphism of a group G, we would have f(g.h)=f(g).f(h)

    I would expect f to be also an equivariant map, but from the definition it wouldn't seem so, because one should have f(g.h)=g.f(h)

    Can anyone clarify this issue?
     
  2. jcsd
  3. Jan 16, 2016 #2

    Ssnow

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    If the automorphism is also equivariant ## f(g\cdot h)=f(g)\cdot f(h)=gf(1_{G})f(h)=g\cdot f(h)##
     
  4. Jan 16, 2016 #3
    Doesn't that basically prove that the only equivariant automorphism is the identity map f(g)=g ?
     
  5. Jan 16, 2016 #4

    Ssnow

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    If you want that your automorphism will be equivariant you must include this in the definition ...
     
  6. Jan 19, 2016 #5

    mathwonk

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    i never heard of an equivariant map of a group to itself. this concept to me relates to maps of sets on which the group acts. of course a group does act on itself by translation, but it seems this action is not compatible with any homomophism except the identity.
     
  7. Jan 19, 2016 #6

    lavinia

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    If ##f## is an automorphism of a group then one can define an action by ##g.h## = ##f(g)h##. Since ##f(gh) = f(g)f(h)##, ##f(gh) = g.f(h)##.

    More generally groups act on sets. An action satisfies the formal rule: ##(gh).x = g.(h.x)##. This rule says that the action is a homomorphism of ##G## into the group of bijections of ##S##.

    ##G \rightarrow ## Bijections##(S)##

    If ##S## is a group then the action may not only be a group of bijections but it may also be a group of automorphisms. A standard example is the action of a group on itself by conjugation.

    In this case,

    ##G \rightarrow## Automorphisms##(S)##

    This is the sense in which an action is usually thought of as being a homomorphism.
     
    Last edited: Jan 20, 2016
  8. Jan 20, 2016 #7

    mathwonk

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    I am puzzled Lavinia. It seems to me the fact that f(gh) = g.f(h) does not say f is equivariant for the action g.h = f(g)h. Rather I think one needs that f(g.h) = g.f(h), which would require that f(f(g)h) = f(g)f(h), which is usually false....???

    a nice example of an equivariant map in real life is an odd function in calculus like sin, i.e. the group of order 2 acts on the reals by the non trivial element sending x to -x, so equivariance means that f(-x) = -f(x).
     
  9. Jan 20, 2016 #8

    lavinia

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    Maybe you are right. If one defines the action,g.h, to be f(g)h then (kg).h = f(kg)h = f(k)f(g)h = k.(f(g)h) = k.(g.h). If g = id then f(g) = id since f is a homomorphism. What am I missing? Oh. This meant g.f(h) is multiplication by g not by f(g) in order to be equivariant. So f(g.h) would have to be f(f(g).h) which usually does not work. I misunderstood.

    I agree the other does not work.
     
    Last edited: Jan 20, 2016
  10. Jan 22, 2016 #9

    Ssnow

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    yes, with this definition is equivariant.
     
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