What is the difference between a group action and a group homomorphism

Question

In most of the literature I have read for Differential geometry and Lie groups, group actions have been introduced. In some of them "Introduction to smooth manifolds" by Lee they get introduced after (Lie) group homomorphisms. Other authors such as Khesin and Wendt in "The geometry of infinite-dimensional groups" directly define a (transformation) group as a set $G$ of transformations of some other set $M$ so that with composition of maps it is a (abstract) group. Then something like group actions is not even really introduced. And it is referred to V.I Arnold, that one always should think of a group as a transformation of some set. In no book I read, there is precisely said what the differences between a group action and a group homomorphism between an (abstract) group and some specific transformation group is. At the level of groups a group action seams to be just a useful notation for transformation groups. But when looking at groups with more structure, it can be something different whether a map $\triangleright: G×M→M$ is (jointly) continuous or a map $G→T(M)$ is continuous with $T(M)≔\{f_g:M→M\mid f_g(x)≔\triangleright(g,x)\}$. Same for Lie groups, where $T(M)$ might not be a manifold in the same sense. What are other differences between a group action and a group homomorphism to some transformation group, where it is not only different notation, i.e. what are benefits of introducing the definition of a (left) group action instead of just talking about the (transformation) group that is given by the induced group homomorphism? Same for right group actions and group antihomomorphisms.

That is a copy of What is the difference between a group action and a group acting on a set?

Answer 1

A group action is always (something you can view as) a group homomorphism -- namely a homomorphism whose codomain is the group of permutations of the set the action acts on.

But not all homomorphisms are actions in this sense -- consider for example the determinant as a homomorphism from the general linear group to $\mathbb R^\times$. There the codomain is not a permutation group, so the determinant is not a group action.

Sure, we can make an action on $\mathbb R$ by declaring that a matrix acts on a number by multiplying it with the determinant of the matrix. But that multiplication is not usually implied when we just say "determinant", so the determinant itself is not a group action.

Answer 2

Here is my take. If you are just interested in group actions on sets, then there is effectively no difference between group actions on such sets and homomorphisms to permutation/automorphism groups of such sets. But the sources that you are reading are about group actions that have extra structure (as you noted in your question). For concreteness, suppose that $X$ is a topological space and $G$ is a topological group. An arbitrary homomorphism $h: G\to \operatorname{Homeo}(X)$ will not define a continuous group action of $G$ on $X$. One way to handle this is to (suitably) topologize $\operatorname{Homeo}(X)$ so that it becomes a topological group and then require $h$ to be continuous. This takes quite a bit of effort and, most likely, you will end up not using this topology on $\operatorname{Homeo}(X)$ at all! The situation is similar if $G$ is a Lie group and $X$ is a smooth manifold, or maybe a complex manifold and a holomorphic group action, or $G$ is an algebraic group and $X$ is algebraic variety, etc. (Actually, for algebraic varieties the situation is even worse since the automorphism group can be "too big", as it happens for instance if $X=\mathbb C^2$.) For most applications that I know, one does not need to topologize the automorphism group of the target space and the formalism of group actions suffices. This formalism is quite general: You work in some category $C$ so that both $G$ and $X$ are objects of that category, then a (left!) $G$-action is a morphism (assuming that $C$ has products), $G\times X\to X$ satisfying the homomorphism property that you already know.

Answer 3

Q: "What is the difference between a group action and a group homomorphism?"

A: Given an abstract group $G$ and a set $M$: To give a left group action $\sigma: G \times M \rightarrow M$ is equivalent to give a map of groups $\rho: G \rightarrow Aut(M)$ from $G$ into the group of automorphisms $Aut(M)$ of the set $M$.

Given any left action $\sigma$ you get a map of groups

$$\rho_{\sigma}: G \rightarrow Aut(M) $$

defined by

$$\rho_{\sigma}(g)(m):=\sigma(g,m).$$

Conversely, given a map of groups $\rho: G \rightarrow Aut(M)$ you get a left action $\sigma_{\rho}: G \times M \rightarrow M$ defined by

$$\rho_{\sigma}(g,m):=\rho(g)(m).$$

This gives a 1-1 correspondence between left actions and maps of groups.

C: "The second comment in the referd post is confusing for me. Does he want to say that a group action does not need to be a group acting on a set, where the answer you already gave is, the image of the group homomorphism induced by the action always is a group acting on a set (i.e., a transformation group)? And on the other side, every transformation group can be given the notation of a left group action. So group actions and Transformation groups are the same. Or does he want to say that the set the group is acting on can have more structure and there can rise differences? – Epod Commented 2 days ago"

A: "@Epod - the other answer claims the group action $det(−):GL_k(W) \rightarrow k^*$ is not a group action but this is wrong. We may view $k^∗≅GL_k(V)$ where $dim_k(V)=1$ and then we get a map of groups $det(−):GL_k(W)\rightarrow GL_k(V):=Aut_k(V)$. – hm2020"

Note: You may study Lie groups, linear algebraic groups or group schemes - these are "groups with additional structure". Then you want your representation $\rho$ to be a map of Lie groups, linear algebraic groups or group schemes.

In the case of linear algebraic groups and group schemes, you may speak of group-actions

$$\sigma: G \times_k X \rightarrow X$$

where $X$ is a scheme (or an algebraic variety over a field $k$) and not a finite dimensional vector space. In the case of schemes you may ask for a corresponding map

$$\rho_{\sigma}: G \rightarrow Aut_k(X).$$

And this is a problem: the construction of $Aut_k(X)$ is problematic since the algebra of functions $k[X]$ (in the case when $X$ is affine) is an infinite dimensional vector space over $k$. In the case when $dim_k(W)$ is finite we may speak of $GL_k(W)$ since the determinant is a polynomial function. When $dim_k(W)$ is infinite there is no obvious analogue of the determinant function. You may define $Aut_k(k[X])$ as the group of $k$-linear endomorphisms $f:k[X] \rightarrow k[X]$ with an inverse $f^{-1} \in Aut_k(k[X])$. But $Aut_k(k[X])$ is not a group-scheme (or linear algebraic group) in general.

Note: If $k$ is a ring and $B$ is a $k$-algebra that is finitely generated and projective as $k$-module, it follows $Aut_k(B)$ has the struture as group-scheme in the language of Demazure/Gabriel (the SGA3 books). But when $B$ is not finitely generated there is no such construction.