3
$\begingroup$

Reading chapter 2.4 of linear representations of finite groups by Serre, can someone create an example for me to understand what's going on?

Let the irreducible characters be denoted by $\chi_i$ and their degrees are $n_i$. Let the order of $G$ be $|G|$ and let $V$ be a vector space of dimension $|G|$ with a basis $\left\lbrace e_t \right\rbrace$ for $t \in G$. Define the linear map of $\rho_s$ from $V$ into $V$ as $\rho_s(e_t) = e_{st}$, which is called the regular representation R. Its degree is equal to the order of $G$.

Proposition 5

The character $r_G$ of the regular representation $R$ is given by \begin{align*} r_G(1) &= |G| \\ r_G(s) &= 0 \qquad \text{if } s\neq 1 \end{align*}

Proof

If $s=1$, we have $$ Tr(\rho_s)=Tr(1)=\dim(R)=| G| $$ If $s\neq 1$, then $st\neq t$ and $\rho_s(e_t) = e_{st} \neq e_t$ for all $t$ so all diagonal terms of the matrix for $\rho_s$ are zero.

Corollary 6

Every irreducible representation $W_i$ is contained in the regular representation with multiplicity equal to its degree $n_i$.

Proof

From theorem 2.3.3, this number equates to $(r_G|\chi_i)$ $$ (r_G|\chi_i)=\frac{1}{\,|G|} \sum_{s \in G} r_G(s)\chi_i(s)^* = \chi_i(1)^* = n_i $$

Remark: Hence we notice, there is only a finite number of irreducible representations.

Corollary 7

(i) The degrees $n_i$ satisfy $\sum_{i=1} n_i^2 =|G|$

(ii) If $s \in G$ is different from $1$, we have $\sum_{i=1} n_i \chi_i(s) =0$

Proof

By corollary 2.4.2, we have $r_G(s) = \sum n_i \chi_i(s)$ for all $s \in G$. For $(i)$ take $s=1$ and for $(ii)$ take $s\neq1$. \end{proof}

Corollary 8 The matrix entries of the unitary irreducible representations form an orthogonal basis for the set of all functions on $G$.

Proof

We know that the matrix entries are all orthogonal as functions. There are $\sum n_i^2=|G|$ of them, and this is the dimension of the vector space of all functions.

$\endgroup$
  • 2
    $\begingroup$ My suggestion is to try to work this all out by hand for $G = S_3$, the symmetric group on 3 letters. $\endgroup$ – Dane May 9 at 13:55
4
+50
$\begingroup$

As @Dane suggested, lets consider the group $G = S_3$ the symmetric group on $3$ letters $1,2,3$. We will consider propositions $5,6$ here. Let $\mathsf{k}$ be some field. Consider the six dimensional $\mathsf{k}$-vector space $V$ with basis $e_0$, $e_{(12)}$, $e_{(23)}$, $e_{(13)}$, $e_{(123)}$, $e_{(132)}$. All we have done here is indexed the basis by the elements of $S_3$. We then define a representation $\rho$ of $G$ by $\rho(g)(e_{h}) =e_{gh}$. So for example

$$ \rho((12))e_{(23)} = e_{(132)}, \quad \rho((13))e_{(23)} = e_{(123)}, \quad \rho((12))e_{(12)} = e_0. $$

Let's explicitly calculate the matrix of $\rho(12)$ with respect to the ordered basis $e_0$, $e_{(12)}$, $e_{(23)}$, $e_{(13)}$, $e_{(123)}$, $e_{(132)}$. Note that

$$ \begin{array}{c | c c } g & e_0 & e_{(12)} & e_{(23)} & e_{(13)} & e_{(123)} & e_{(132)} \\ \hline \rho((12))e_{g} & e_{(12)} & e_0 & e_{(132)} & e_{(123)} & e_{(13)} & e_{(23)} \end{array} $$

and so

$$ \rho((12)) = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\end{pmatrix} $$

so note that $\chi((12)) = 0$, where $\chi = \operatorname{trace} \circ \rho$. This is what is called $r_G$ in your post. You might notice that

$$ \chi(g) = \left| \left\{ h \in S_3 \mid gh = h \right\} \right| = \left|\operatorname{Fix}_{g}(S_3) \right|. $$

Hence indeed

$$ \chi(g) = \left\{ \begin{array}{ll} \left| G \right| = 6, & \text{if} \ g = e_0, \\ 0, & \text{otherwise} \end{array} \right. $$

Now we have a representation of $S_3$ given by the sign representation. That is $\rho_{\operatorname{sign}} : S_3 \to \mathsf{k}^{\times}$ such that

$$ \rho_{\operatorname{sign}}(g) = \left\{ \begin{array}{ll} 1, & \text{if} \ g \ \text{is a product of an even number of transpositions,} \\ -1, & \text{if} \ g \ \text{is a product of an odd number of transpositions.} \end{array} \right. $$

Now we will find this a subrepresentation of $V$. Consider the vector

$$ v_{\operatorname{sign}} = e_0 - e_{(12)} - e_{(23)} - e_{(13)} + e_{(123)} + e_{(132)} \in V. $$

and define $V_{\operatorname{sign}} = \mathsf{k} \cdot v_{\operatorname{sign}}$, then I claim that this is a subrepresentation. But it is easy to check that $S_3$ acts on $v_{\operatorname{sign}}$ by $\pm 1$ so this is indeed a subrepresentation and whats more its easy to check that this is indeed the sign representation. It is also easy to check that this is the unique subrepresentation of $V$ isomorphic to the sign representation.

$\endgroup$
  • $\begingroup$ what is meant by $|Fix S_3(g)|$ apart from this, everything else is well explained. $\endgroup$ – johnny May 13 at 9:49
  • $\begingroup$ Oh sorry this is standard notation to do with group actions. If you have a group $G$ (left) acting on a set $S$, (so to every $g \in G$, we can associate some bijective function $f_g : S \to S$ such that $f_{h} \circ f_g = f_{hg}$, and $f_{e} = \operatorname{Id}_S$) then if $T$ is some subset of $G$ then $\operatorname{Fix}_G\left( T \right)$ is the set of $g \in G$ such that $f_g(t) = t$ for all $t \in T$. I used the wrong notation above. I'll change it to what I meant. $\endgroup$ – Adam Higgins May 13 at 9:57
  • $\begingroup$ Rereading this, where you wrote $\chi((12)) = 0$, where $\chi =$ trace $\circ \rho$ shouldn't it read $\chi =$ trace of $\rho$? $\endgroup$ – johnny Jul 15 at 12:44
  • $\begingroup$ Well, either is fine. You can view $\operatorname{trace}$ as a function justifies the function composition notation. $\endgroup$ – Adam Higgins Jul 15 at 13:03
  • $\begingroup$ Can you expand on the part where you considered sign representation? From "Now we have a representation of $S_3$ given by the sign representation..." $\endgroup$ – johnny Jul 15 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.