# Question on equivariant maps

1. Jan 16, 2016

### mnb96

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. Jan 16, 2016

### Ssnow

If the automorphism is also equivariant $f(g\cdot h)=f(g)\cdot f(h)=gf(1_{G})f(h)=g\cdot f(h)$

3. Jan 16, 2016

### mnb96

Doesn't that basically prove that the only equivariant automorphism is the identity map f(g)=g ?

4. Jan 16, 2016

### Ssnow

If you want that your automorphism will be equivariant you must include this in the definition ...

5. Jan 19, 2016

### mathwonk

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.

6. Jan 19, 2016

### lavinia

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
7. Jan 20, 2016

### mathwonk

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).

8. Jan 20, 2016

### lavinia

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
9. Jan 22, 2016

### Ssnow

yes, with this definition is equivariant.