What is the intuition behind the definition of the action of a group on a set? Why do we call a homomorphism $G\to\mathcal S(X)$ a "representation"?

271 Views Asked by At

Let $G$ be a group acting on a set $X$. For all $ g\in G $, we consider the application \begin{align*} \varphi_{g}: &~ X \longrightarrow X \\ &~ x \longrightarrow g.x \end{align*} It is clear that $\varphi_{gh}=\varphi_g\circ\varphi_h$, $\forall~ g,h \in G,$ $\varphi_e=\text{Id}_X$ ( $e$ the neutral element of $G$ ), and $\varphi_g \circ \varphi_{g^{-1}}=\varphi_{g^{-1}} \circ \varphi_g $, so $\varphi_g$ is bijective, for all $g\in G$. i.e: $\varphi \in \mathcal S(X)$, $\forall~ g\in G$, and the application: \begin{align*}\Phi :&~ G \longrightarrow \mathcal S(X) \\ &~g\longrightarrow \varphi_g\end{align*} is a group homomorphism that we call a representation of $G$ in $\mathcal S(X).$

I didn't understand what this definition is for, I'm looking for the intuition or the idea behind it. Why do we call the application $\Phi$ by this name: "representation", it is only a homomorphism? if anyone has any ideas or comments that he can add, I will be very grateful.

1

There are 1 best solutions below

6
On BEST ANSWER

Put in general terms, a representation of an abstract object (group, ring, etc.) is a way of thinking about that object acting on a certain other thing we can describe in concrete terms. The most natural candidates for the other thing are sets (so that group elements act as permutations of those objects) or vector spaces (so that group elements act as matrices).

It's quite hard to understand dihedral groups abstractly, just using muliplication tables, but easy to understand them as symmetries of an $n$-gon via a matrix representation (i.e., as $2\times 2$ matrices over $\mathbb{R}$), or as permutations of the vertices of the $n$-gon.

So one aspect of group actions is that they can be used to more easily study the group. They can also be used to study the object being acted upon. For example, a cubic equation with real coefficients must have a real root. Why? Because complex conjugation permutes the roots, and it therefore swaps them in pairs. But there are an odd number of roots, so (at least) one is fixed. That's a trivial application of group actions.