A question about the definition of equivariant map

If $$S$$ is a set of functions from $$X$$ to $$Y$$ then I can consider the action of a group $$G$$ on $$S$$ via its action on $$X$$ and $$Y$$ by the formula

$$(g \cdot f)(x) = g \cdot f(g^{-1} \cdot x),$$

So we are considering left actions both on $$X$$ and $$Y$$.

Then, the definition of equivariant map pops out but I don't really understand how it is related to previous statement.

An function $$f: X \rightarrow Y$$ is equivariant if it satisfies

$$f(g \cdot x) = g \cdot f(x) \, \, \, \, \forall g \in G.$$

What's happening here? Are we assuming that the group $$G$$ acts trivially on $$Y$$? How to get this definition from the previous statement?

• I don't follow. It seems like the first thing is giving a construction so that a set of functions might have a $G$-action assuming both domain and codomain do. The second is a defintion of an equivariant map, that a single function may or may not respect. Why should there be a connection? Is there some claim that when giving $S$ this $G$-action that all elments of $S$ are now suddenly equivariant? Nov 15, 2021 at 18:47
• A priori the definition of equivariant map doesn't have anything to do with the induced action of $G$ in $S$. It is just a notion of action-preservingness of a function. Nov 15, 2021 at 18:54
• Oh, i thought about this a bit more and I think there's a connection that can be stated. Under that specific action of $G$ in $S$, a function will be equivariant if and only if it is $G$-invariant, if I'm not mistaken... Nov 15, 2021 at 19:00
• Just to make things clear, I’ve read what I’ve reported here from this book math.ens.fr/~benoist/refs/Dolgachev.pdf (p. 1) where these definitions are put very close together Nov 15, 2021 at 21:16

So, recapitulating, we have actions of $$G$$ on $$X$$ and $$Y$$. This induces an action of $$G$$ on $$S=\{f:X\longrightarrow Y\}$$ by setting $$(g\cdot f)(x)=g\cdot f(g^{-1}\cdot x)$$ On the other hand, a function $$f\in S$$ is said to be equivariant provided $$f(g\cdot x)=g\cdot f(x) \quad \forall g\in G$$ Taking $$x=g^{-1}\cdot x^{*}$$ in this last expression yields $$f(x^{*})=g\cdot f(g^{-1}\cdot x^{*})=(g\cdot f)(x^{*})$$ Thus, a function $$f\in S$$ will be equivariant if and only if $$f=g\cdot f$$ for all $$g\in G$$, i.e., if it is $$G$$-invariant under the above defined action of $$G$$ on $$S$$.

The two definitions are a priori independent, the only thing you can deduce is that for the induced action $$(g,f)\mapsto g*f$$ we don't have $$(g*f)(x)=f(g\cdot x)$$ in general.
Instead, by the first definition, we have $$(g*f)(x)\ =\ g\cdot f(g^{-1}\cdot x)\,.$$ Note that the second definition doesn't say anything about the induced action of the first definition.

However, as pointed out in the comments, there's a connection between these definitions, namely

A map $$f:X\to Y$$ is equivariant iff $$g* f=f$$ for every $$g\in G$$.

Indeed, if $$f$$ is equivariant, then $$(g* f)(x)=g\cdot f(g^{-1}\cdot x)=f(g\cdot g^{-1}\cdot x)=f(x)\,.$$ And if $$f$$ is stabilized by $$G$$, then in particular for any $$g^{-1}\in G$$ we have $$f=g^{-1}* f$$, so $$f(x)=(g^{-1}* f)(x)=g^{-1}\cdot f(g\cdot x)\implies g\cdot f(x)=f(g\cdot x)\,.$$

• why if $f$ is equivariant, we have $(g \cdot f)(x) = g \cdot f(g^{-1} \cdot x)$? That is not the definition.. In the definition we are supposing $g^{-1} = e$ ? I'm a bit confused about that Nov 16, 2021 at 12:07
• Yes, that is the definition, namely the definition of the $G$-action on functions $X\to Y$ (def 1). As I wrote, the definition of being equivariant has (a priori) nothing to do with the above mentioned induced action. The action $g\cdot f$ is defined for every function $f:X\to Y$. Some of these functions (notably just the fixed points of this action) satisfy the additional property of def.2 and they are called 'equivariant'. Nov 16, 2021 at 12:13
• I edited my answer to distinguish the action of def.1 (now denoted by $*$). I hope it's clearer now. Nov 16, 2021 at 12:23
• I just don't understand why that $g^{-1}$ appears that is not present at all in the definition of equivariant function. Nov 16, 2021 at 12:41
• It only appears if you want to combine the two. My statement wants to connect the two notions, and observe that both definitions are used. Nov 16, 2021 at 12:45