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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?

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  • $\begingroup$ 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? $\endgroup$
    – Randall
    Nov 15, 2021 at 18:47
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    $\begingroup$ 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. $\endgroup$
    – maikel
    Nov 15, 2021 at 18:54
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    $\begingroup$ 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... $\endgroup$
    – maikel
    Nov 15, 2021 at 19:00
  • $\begingroup$ 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 $\endgroup$
    – James Arten
    Nov 15, 2021 at 21:16

2 Answers 2

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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 $S$.

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

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  • $\begingroup$ 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 $\endgroup$
    – James Arten
    Nov 16, 2021 at 12:07
  • $\begingroup$ 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'. $\endgroup$
    – Berci
    Nov 16, 2021 at 12:13
  • $\begingroup$ I edited my answer to distinguish the action of def.1 (now denoted by $*$). I hope it's clearer now. $\endgroup$
    – Berci
    Nov 16, 2021 at 12:23
  • $\begingroup$ I just don't understand why that $g^{-1}$ appears that is not present at all in the definition of equivariant function. $\endgroup$
    – James Arten
    Nov 16, 2021 at 12:41
  • $\begingroup$ 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. $\endgroup$
    – Berci
    Nov 16, 2021 at 12:45

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