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⋅f)(x)=g⋅f(g−1⋅x), So we are considering left actions both on X and Y. Then, the definition of equivariant map pops out but I dont

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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⋅f)(x)=g⋅f(g−1⋅x),  So we are considering left actions both on X and Y.   Then, the definition of equivariant map pops out but I dont

Elois Puryear · Accepted Answer

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

SlabydouluS62 · Accepted Answer

The two definitions are a priori independent, the only thing you can deduce is that for the induced action (g,f)→g⋅f we don&#039;t have (g⋅f)(x)=f(g⋅x) in general.Instead, by the first definition, we have(g⋅f)(x)=g⋅f(g−1⋅x).Note that the second definition doesn&#039;t say anything about the induced action of the first definition.However, as pointed out in the comments, there&#039;s a connection between these definitions, namelyA map f:X→Y is equivariant ifg⋅f=ffor every g∈G.Indeed, if f is equivariant, then(g⋅f)(x)=g⋅f(g−1⋅x)=f(g⋅g−1⋅x)=f(x)And if f is stabilized by G, then in particular for anyg−1∈G we have f=g−1⋅f, sof(x)=(g−1⋅f)(x)=g−1⋅f(g⋅x)⇒g⋅f(x)=f(g⋅x)

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 acti

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