Please compare the following two definitions:

Definition 1

A representation of a topological group $G$ in a vector space ${\mathbb{V}}$ over complex numbers is a continuous map $$ A\, :\quad G\times {\mathbb{V}}\longrightarrow{\mathbb{V}}\;\,,\qquad (g\,,\;v)\longmapsto A(g)\,v\;\,,\;\;\;\;\; g\in G\,,\;\;v\in{\mathbb{V}}\;\; \label{1}\tag{1} $$ with the following properties:

(1) $\;\; A\,$ is a group action;

(2) $\;\; A\,$ is linear, i.e. $$ A(g)\,(\alpha\, v+\beta)=\alpha\, A(g)\, v+ A(g)\, w\quad \mbox{for}\quad\forall g\in G\,,\;\; v,\; w\in {\mathbb{V}}\,,\;\; \alpha\in{\mathbb{C}}\,\;. $$

Definition 2

A representation of a topological group $G$ in a vector space ${\mathbb{V}}$ is a homomorphism of $ G$ into the group of invertible linear transformations of ${\mathbb{V}}$: $$ A\,:\quad G\longrightarrow GL({\mathbb{V}})\;\,. \label{2}\tag{2} $$

$$ \, $$ Definition 2 $\;\;\Longrightarrow\;\;$ Definition 1,

Indeed, the $A$ from Definition 2 is linear and ensures $\,(g\,,\;v)\longmapsto A(g)\,v\,$.

Also, Definition 2 says that $A$ is a homomorphism $-$ which guarantees the continuity of $\,(g\,,\;v)\longmapsto A(g)\,v\;$ in Definition 1.


To show that Definition 1 $\;\Longrightarrow\;$ Definition 2, we must demonstrate that Definition 1 ensures \eqref{2} being a homomorphism.

How to do this?

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    $\begingroup$ Are you assuming that $\Bbb{V}$ is a vector space over the real numbers, or maybe the complex numbers? (If not then neither of your definitions gives enough information about the topologies involved.) $\endgroup$
    – Rob Arthan
    Commented Mar 1, 2021 at 1:01
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    $\begingroup$ Definition 2 is not equivalent to Definition 1 unless we put a topology on $\text{GL}(V)$ and insist that the map be continuous. In particular, I disagree with the implication that you wrote. $\endgroup$
    – hunter
    Commented Mar 1, 2021 at 1:03
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    $\begingroup$ I think you should try to separate your thoughts about the algebraic content of your problem and its topological content. You don't need any topological reasoning to show that the algebraic part of Definition 1 implies the algebraic part of Definition 2. $\endgroup$
    – Rob Arthan
    Commented Mar 1, 2021 at 1:08
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    $\begingroup$ @RobArthan I see that I am confusing homomorphism with homeomorphism! $\endgroup$ Commented Mar 1, 2021 at 1:21
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    $\begingroup$ The algebraic property of being a group homomorphism has nothing to do with topology. I think you may be mixing up "homomorphism" with "homeomorphism" (an easy mistake to make, in this context). $\endgroup$
    – Rob Arthan
    Commented Mar 1, 2021 at 1:24

1 Answer 1


Actually, a group action by $G$ on a set $X$ is equivalent to a group homomorphism $\psi: G\to \operatorname{Aut}(X)$, where in this case automorphisms are bijections of the set with itself. Indeed, the group action can be thought of as $\alpha:G\times X\to X$ such that $\alpha(hg,x)=\alpha(h,\alpha(g,x))$ etc. This is equivalent to $\psi(hg)=\psi(h)\circ \psi(g)$ where $\psi(g)(x)=\alpha(g,x)$. $\alpha(e_G,x)=x$ for all $x\in X$ is equivalent to $\psi(e_G)=\operatorname{Id}_X$. These are good facts to check carefully yourself as an exercise in the definitions (also usually in any algebra text).

So, definition (1) already gives us a group homomorphism $G\to \operatorname{Aut}^{\operatorname{Sets}}(V)$. A priori, the codomain on the right is self-bijections. However, the requirement that $G$ act linearly means that $G$ maps into $\operatorname{GL}(V)\subsetneq \operatorname{Aut}^{\operatorname{Sets}}(V)$. The continuity of the action stipulated in definition (1) then implies that as a map $G\to \operatorname{GL}(V)$ it is continuous, where $\operatorname{GL}(V)$ is viewed as a topological group.

  • $\begingroup$ I agree that the continuity of action in Definition 1 makes the map $G\longrightarrow GL(V)$ continuous. Does it also make it homomorphism? $\endgroup$ Commented Mar 1, 2021 at 1:18
  • $\begingroup$ The continuity is unrelated to it being a homomorphism in some sense. As a stupid example, take $\Bbb{C}^\times$ viewed as $G=\operatorname{GL}(1,\Bbb{C})$. Then we can cook up all sorts of continuous maps of groups into $G$. This does not mean the actions are linear. For instance, we could send $S^1$ to $\{z:\lvert z\rvert=2\}$. This does not send the identity of $S^1$ to the identity element of $G$, but it sure is continuous. It is really the algebraic data of being an action that guarantee that we have a homomorphism. $\endgroup$ Commented Mar 1, 2021 at 1:18
  • $\begingroup$ I see that I am confusing homomorphism with homeomorphism. Thank you! $\endgroup$ Commented Mar 1, 2021 at 1:22
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    $\begingroup$ The question here though is whether or not continuity guarantees homomorphism which is very false. $\endgroup$ Commented Mar 1, 2021 at 1:25
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    $\begingroup$ Great. I should mention that in my longer comment above that I made a mistake: the map fails to be a group homomorphism but is valued in linear automorphisms of $\Bbb{C}$. $\endgroup$ Commented Mar 1, 2021 at 1:54

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