Here is what I understand a group representation to be:

A (linear) representation of a group $G$ in a vector space $V$ over a field $\mathbb{F}$ is a homomorphism $\sigma: G\to GL(V)$.

And there are plenty of nice examples, such as the one where we map each element in $S_n$ to its associated permutation matrix in $GL(\mathbb{R}^3)$.

Later on in a book I am reading, they give an example that is of specific importance, but the notation loses me completely. I have reread it several times over the weekend and it is like a complete mental brick wall. Here is what they say:

Let $G$ be a finite group of order $m$. Let $V$ be a vector space over a field $\mathbb{F}$ having basis $\{e_h:h\in G\}$. The left regular representation $\sigma_r$ of $G$ in $V$ is defined as:

$$\sigma_r(g):e_h\mapsto e_{gh}\;\;\forall g\in G$$

I find this horribly confusing. Why is our representation taking an element of $V$ to another element of $V$? Isn't it supposed to take an element of $G$ to an operator in $GL(V)$? Why is there suddenly an inherent dependence on another element of $G$ (specifically $h$), when there wasn't any such thing in the previous example (about $S_n$, above)? Is "$\sigma_r(g)$" its own map, and every such element of $G$ has its own associated map? If so, then shouldn't it say $\sigma_r(g):h\mapsto e_{gh}$? But still, where did $GL(V)$ go?


What is an operator in $\mathrm{GL}(V)$? It is a thing that takes an element of $V$ to another element of $V$.

Thus, if $\sigma_r$ takes an element of $G$ to an element of $\mathrm{GL}(V)$, then $\sigma_r(g)$ is a thing that takes an element of $V$ to an element of $V$.

What is one of the ways of specifying a linear operator on $V$? By specifying what it does on a basis.

What is the basis for $V$ we are given? It is the collection of $e_h$, indexed by $h \in G$.

To specify an arbitrary basis element, we use a variable $h$ from $G$ to indicate the basis element.

Thus, we need to say what $\sigma_r(g)$ does to $e_h$, for all $g$ and for all $h$.

  • $\begingroup$ This did it. Thank you. I thought $\sigma_r(g)$ was the representation. I now see my tiny error, which turned into a which misunderstanding. Am I correct in saying the notation is incredibly subtle? Also, thank you again so much. I will accept as soon as I am allowed. $\endgroup$ – The Count May 1 '17 at 15:51
  • 2
    $\begingroup$ @TheCount: Function valued functions always seem to give people a hard time when first encountered. It is, in my opinion, straightforward and fairly explicit when you're used to it, the trick is getting used to it. $\endgroup$ – Hurkyl May 1 '17 at 15:53
  • $\begingroup$ Honestly, it is as though you just cleared miles of fog from my mind. I can explain it perfectly now, and just did to my office mate. $\endgroup$ – The Count May 1 '17 at 15:53

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