Group action definition

Question

The Wikipedia for group action and several posts here have what seems to me to be two, slightly different definitions for a group action.$\ $ Let G be a group acting on space X.$\ $ These two definitions are:

1. A group homomorphism $\varphi$ from the group to the automorphism group of the space. $$\varphi : G\ \longrightarrow \text{Aut}(X)$$

2. A map that takes an element of the group and an element of the space, and returns an element of the space. $$\alpha : G\, \times X\ \longrightarrow X$$

Which of these is the the group action?$\ $ Is it a function of one argument or two?$\ $ What is the return type? Does it return a function (automorphism) or an element of the space?$\ $ The first definition is in line with what I think of as a representation, ex. from a symmetry group C$_\text{2v} \longrightarrow $ GL(V) that returns a matrix, rather than a vector.

I (think) I understand the basic idea of what is going on, but I'd like to know what the actual mathematical definition is.

Thanks in advance.

Answer 1

The definitions end up being more or less the same in the sense that if you have a map $f: G \times X \to X$ satisfying the second definiton, there exists a group homomorphism (induced by $f$) satisfying the first definition (and vice versa).

To show you how this might work, suppose you have $f: G \times X \to X$. For each $g \in G$, you can define a map $\phi_g: X \to X$ by letting $\phi_g(x) = f(g,x)$. This will be a bijection since $\phi_g^{-1}$ is precisely $\phi_{g^{-1}}$. Then, the homomorphism $G \to \text{Aut}(X)$ is the one sending $g \mapsto \phi_g$. If instead you start with a homomorphism $f: G \to \text{Aut}(X)$ and let $f_g := f(g)$, we can define a map $\alpha: G \times X \to X$ by letting $\alpha(g,x) = f_g(x)$. Hence, there is a one-to-one correspondence between maps $G \times X \to X$ (which satisfy other conditions as mentioned in the comments) and homomorphisms $G \to \text{Aut}(X)$

Though, in my experience, the 2nd definition you mentioned is a more conventional starting place for defining group actions. If you look in many classic Algebra books like the one by Dummit and Foote, they will define a group action as a map $G \times X \to X$ satisfying various conditions. However, as you learn more about Abstract Algebra, the difference between these two notions becomes miniscule, and you will start to use both notions of a group action interchangeably.

Answer 2

There's a more general setting for this equivalence. Let $G$ be a group and $X$ a set; consider:

D1) There is a map: $\cdot:G\times X\to X$, $(g,x)\mapsto g\cdot x$, such that:

• $e\cdot x=x$, for every $x\in X$;
• $g\cdot(h\cdot x)=(gh)\cdot x$, for every $g,h\in G, x\in X$.

D2) There is a group homomorphism $\varphi\colon G\to\operatorname{Sym}(X)$.

Claim. D1) $\iff$ D2).

Proof.

• D1) $\Longrightarrow$ D2). Let's define $\varphi(g)(x):=g\cdot x$; then (injectivity):

$$\varphi(g)(x)=\varphi(g)(y)\Rightarrow g\cdot x=g\cdot y\Rightarrow g^{-1}\cdot(g\cdot x)=g^{-1}(g\cdot y)\stackrel{\text{(D1)}}{\Longrightarrow}x=y$$

and (surjectivity), for every $y\in X$, $y\stackrel{\text{(D1)}}{=}\varphi(g)(g^{-1}\cdot y)$. Therefore, indeed $\varphi(g)\in\operatorname{Sym}(X)$ for every $g\in G$. Now, for every $g,h\in G,x\in X$:

\begin{alignat}{1} \varphi(gh)(x) &= (gh)\cdot x \\ &\stackrel{\text{(D1)}}{=} g\cdot(h\cdot x) \\ &= \varphi(g)(\varphi(h)(x)) \\ &= (\varphi(g)\varphi(h))(x) \\ \end{alignat}

whence $\varphi(gh)=\varphi(g)\varphi(h)$.

• D2) $\Longrightarrow$ D1). Let's define $g\cdot x:=\varphi(g)(x)$; then, for every $x\in X$:

$$e\cdot x=\varphi(e)(x)\stackrel{\text{(D2)}}{=}Id_X(x)=x$$

Moreover, for every $g,h\in G,x\in X$:

$$(gh)\cdot x =\varphi(gh)(x)\stackrel{\text{(D2)}}{=}(\varphi(g)\varphi(h))(x)=\varphi(g)(\varphi(h)(x))=g\cdot(h\cdot x)$$

$\Box$

Therefore, feel free to choose any between D1 and D2 as definition of group action.

Answer 3

What you are seeing here is a special case of the very general concept of currying: For three sets $A, B, C$ there is a natural bijection between maps $A\times B\to C$ and maps $A\to \operatorname{Maps}(B,C)$, where $\operatorname{Maps}(B,C)$ denotes the set of maps $B\to C$.

The bijection is rather trivial and works as follows: Given a map $f\colon A\times B\to C$ we can define for each $a\in A$ the map $f_a\colon B\to C$ given by $f_a(b)=f(a,b)$. This defines a map $A\to\operatorname{Maps}(B,C)$ by $a\mapsto f_a$. Conversely, from a map $g\colon A\to\operatorname{Maps}(B,C)$ you can always define a map $A\times B\to C$ by $(a,b)\mapsto \big(g(a)\big)(b)$.

These two constructions are inverse to each other and hence establish a bijection between maps $A\times B\to C$ and maps $A\to\operatorname{Maps}(B,C)$.

In your case $A=G$ and $B=C=X$ so that maps $G\times X\to X$ correspond to maps $G\to\operatorname{Maps}(X,X)$. However, you are only interested in the subset of those maps that actually map into $\operatorname{Sym}(X)$ (which is the subset of $\operatorname{Maps}(X,X)$ consisting of only the bijections) and have the property of being a group homomorphism. In the setting of maps $G\times X\to X, (g,x)\mapsto g\cdot x$, these criteria translate to the group action axioms $1\cdot x=x$ and $(gh)\cdot x=g\cdot(h\cdot x)$ for all $g,h\in G$ and $x\in X$.