Understanding Equivariant Maps in Group Theory

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Discussion Overview

The discussion revolves around the concept of equivariant maps in group theory, particularly in relation to automorphisms of groups. Participants explore the definitions and implications of equivariance, questioning whether automorphisms can be considered equivariant under certain conditions.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions whether an automorphism of a group can be considered equivariant, noting the difference between the definitions of automorphisms and equivariant maps.
  • Another participant suggests that if an automorphism is equivariant, it leads to the conclusion that the only equivariant automorphism is the identity map.
  • Some participants argue that the definition of equivariance must be explicitly included when discussing automorphisms.
  • There is a perspective that equivariant maps are typically related to actions on sets rather than groups acting on themselves, with the identity being a notable exception.
  • One participant introduces a specific action defined by an automorphism and discusses the implications of this action on the concept of equivariance.
  • Another participant expresses confusion about the relationship between the definitions and provides a real-world example of an odd function as an equivariant map, illustrating the concept with a specific case.
  • Some participants acknowledge misunderstandings regarding the definitions and the implications of equivariance in the context of group actions.
  • A later reply confirms that with a specific definition of action, the map can indeed be considered equivariant.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between automorphisms and equivariant maps, with no consensus reached on whether automorphisms can be classified as equivariant under the discussed definitions. The discussion remains unresolved regarding the implications of these definitions.

Contextual Notes

Participants highlight the need for clarity in definitions and the potential for misunderstandings regarding the conditions under which automorphisms may or may not be equivariant.

mnb96
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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?
 
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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)##
 
Ssnow said:
If the automorphism is also equivariant ## f(g\cdot h)=f(g)\cdot f(h)=gf(1_{G})f(h)=g\cdot f(h)##

Doesn't that basically prove that the only equivariant automorphism is the identity map f(g)=g ?
 
If you want that your automorphism will be equivariant you must include this in the definition ...
 
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.
 
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:
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).
 
mathwonk said:
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).

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:
yes, with this definition is equivariant.
 

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