2.10. Lecture 10

In this lecture we will finally define the last of the machinery needed to finish the proof of Theorem 2.8.7, and then we will finish said proof.

2.10.1. Universal projections

Recall that for any representation $(\pi,V)$ , the sum of matrices $\frac{1}{|G|}\sum_{g\in G}\pi(g)$ acts on $V$ as the orthogonal projection onto $V^{G}$ , the subspace of vectors in $V$ that are $G$ -invariant, i.e. the subrepresentation of $(\pi,V)$ that is isomorphic to $\operatorname{dim}(V^{G})$ copies of the irreducible trivial representation. We now generalise this. To start with, recall that for any element of the group algebra $\sum_{g\in G}z_{g}g\in{\mathbb{C}}G$ , we define

\pi\left(\sum_{g\in G}z_{g}g\right):=\sum_{g\in G}z_{g}\pi(g).

The space of ${\mathbb{C}}$ -valued functions on $G$ is isomorphic to ${\mathbb{C}}G$ ; one possible choice of isomorphism is

f\mapsto\frac{1}{|G|}\sum_{g\in G}\overline{f(g)}g.

Composition then lets us define $\pi(f)$ :

\pi(f):=\frac{1}{|G|}\sum_{g\in G}\overline{f(g)}\pi(g).

Lemma 2.10.1.

If $f\in CF(G)$ , then $\pi(f)\in\operatorname{Hom}_{G}(V)$ .

Proof.

This is Problem 2.10.3 below. ∎

Proposition 2.10.2.

Let $f\in CF(G)$ and $(\rho,W)$ be an irreducible representation of $G$ . Then

\rho(f)=\frac{1}{\operatorname{dim}(W)}\langle\chi_{\rho},f\rangle_{G}\,{\,% \mathrm{Id}}.

Proof.

By Schur’s Lemma 1.4.1, $\operatorname{Hom}_{G}(W)={\mathbb{C}}\,{\,\mathrm{Id}}$ . For any $f\in CF(G)$ , Lemma 2.10.1 then gives $\rho(f)=\lambda{\,\mathrm{Id}}$ for some $\lambda\in{\mathbb{C}}$ . We compute this $\lambda$ by taking the trace of $\rho(f)$ :

\lambda\operatorname{dim}(W)=\operatorname{tr}\big{(}\rho(f)\big{)}=\frac{1}{|% G|}\sum_{g\in G}\overline{f(g)}\operatorname{tr}\big{(}\rho(g)\big{)}=\frac{1}% {|G|}\sum_{g\in G}\overline{f(g)}\chi_{\rho}(g)=\langle\chi_{\rho},f\rangle.

∎

Proof of Theorem 2.8.7, continued:.

We want to prove that $\{\chi_{\rho}\}_{\rho\in\mathrm{Irr}(G)}$ spans the space $CF(G)$ . To do this, we assume that $f\in CF(G)$ is orthogonal to all $\chi_{\rho}$ ; and need to show that then $f=0$ .

Given any representation $(\pi,V)$ , we may decompose $(\pi,V)$ into irreducible subrepresentations:

(\pi,V)=(\pi,W_{1})\oplus\ldots\oplus(\pi,W_{r}).

Now, $\pi(f)$ preserves each irreducible component, hence $\pi(f)=\sum_{i}\pi(f)|_{W_{i}}$ . By Proposition 2.10.2, $\pi(f)|_{W_{i}}=\frac{\langle\chi_{\rho_{i}},f\rangle_{G}}{\operatorname{dim}(% W_{i})}{\,\mathrm{Id}}|_{W_{i}}$ , where $\pi|_{W_{i}}\cong\rho_{i}$ . By assumption, $\langle\chi_{\rho_{i}},f\rangle_{G}=0$ for all $i$ , hence $\pi(f)=0$

Thus: $\pi(f)=0$ for any representation $(\pi,V)$ of $G$ . In particular, considering the regular representation $(\lambda,{\mathbb{C}}G)$ , we have

{\bf 0}=\lambda(f)(e)=\frac{1}{|G|}\sum_{g\in G}\overline{f(g)}\lambda(g)e=% \frac{1}{|G|}\sum_{g\in G}\overline{f(g)}g.

The set $\{g\}_{g\in G}$ forms a basis of ${\mathbb{C}}G$ , so the linear combination is only zero if and only if $\overline{f(g)}=0$ for all $g\in G$ , i.e. $f=0$ . ∎

We collect two more useful facts regarding maps $\pi(f)$ :

Corollary 2.10.3.

If $(\pi,V)$ and $(\rho,W)$ are two irreducible representations, then

\pi(\chi_{\rho})=\begin{cases}\frac{1}{\operatorname{dim}(V)}{\,\mathrm{Id}}% \quad&\mathrm{if}\;(\pi,V)\cong(\rho,W),\\ 0\quad&\mathrm{otherwise}.\end{cases}

Proof.

This follows directly from Proposition 2.10.2 and Theorem 2.8.7. ∎

Corollary 2.10.4.

Let $(\pi,V)$ be a representation of $G$ with decomposition into irreducible subrepresentations

(\pi,V)=(\pi,W_{1})\oplus(\pi,W_{2})\oplus\ldots\oplus(\pi,W_{r}).

For each $\rho\in\mathrm{Irr}(G)$ , define

(\pi,V_{\rho}):=\bigoplus_{\underset{\pi|_{W_{i}}\cong\rho}{i}}(\pi,W_{i}).

Then $\operatorname{dim}(\rho)\,\pi(\chi_{\rho})$ is a projection onto $V_{\rho}$ .

Proof.

This follows directly from the previous corollary. ∎

The subrepresentation $(\pi,V_{\rho})$ is called the $\rho$ -isotopic component of $\pi$ , and $\operatorname{dim}(\rho)\pi(\chi_{\rho})$ is called the $\rho$ -isotopic projector. Observe that since $\operatorname{dim}(\rho)\,\pi(\chi_{\rho})$ is independent of the choice of decomposition of $(\pi,V)$ into irreducibles, $(\pi,V_{\rho})$ must also be independent of this choice. This implies that the only choices in the decomposition are given by the different decompositions of each $(\pi,V_{\rho})$ into copies of $\rho$ .

Example 2.10.5.

We consider the permutation representation $(\pi,{\mathbb{C}}^{3})$ of $S^{3}$ , and compute the matrices of the isotopic projectors for all the irreducible representations of $S_{3}$ :

\pi(\chi_{{\mathrm{triv}}})=\frac{1}{6}\big{(}\pi(e)+\pi(12)+\pi(13)+\pi(23)+% \pi(123)+\pi(132)\big{)}=\frac{1}{6}\left(\begin{matrix}2&2&2\\ 2&2&2\\ 2&2&2\end{matrix}\right).

\pi(\chi_{\mathrm{sgn}})=\frac{1}{6}\big{(}\pi(e)-\pi(12)-\pi(13)-\pi(23)+\pi(% 123)+\pi(132)\big{)}=\frac{1}{6}\left(\begin{matrix}0&0&0\\ 0&0&0\\ 0&0&0\end{matrix}\right).

2\,\pi(\chi_{(\pi,W_{0})})=\frac{1}{3}\big{(}2\pi(e)-\pi(123)-\pi(132)\big{)}=% \frac{1}{3}\left(\begin{matrix}2&-1&-1\\ -1&2&-1\\ -1&-1&2\end{matrix}\right).

2.10.2. Example: the character table of $S_{4}$ .

Example 2.10.6.

We determine the character table of $S_{4}$ . First, we have the 1-dimensional trivial and sign representations:

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr{\mathrm{triv}}&1&1&1&1&1\\ \mathrm{sgn}&1&-1&1&1&-1\end{array}

Next, we have the 3-dimensional irreducible permutation representation $(\pi,W_{0})$ of $S_{4}$ on $\{1,2,3,4\}$ , whose character $\chi_{\pi}$ satisfies

\chi_{\pi}(g)=\#\{\text{fixed points of $g$}\}-1.

Notice that we also have an operation on representations known as twisting: if $(\rho,V)$ is any irreducible representation of $G$ and $\psi$ is a 1-dimensional character of $G$ , then we can define a new representation, $\psi\rho$ , of $G$ on $V$ by the formula $(\psi\rho)(g)=\psi(g)\rho(g)$ . It has character $\psi\chi_{\rho}$ . In this case, we can look at the representation $(\mathrm{sgn}\pi,W_{0})$ and see that it will also be irreducible:

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\pi&3&1&-1&0&-1\\ \mathrm{sgn}\pi&3&-1&-1&0&1\\ \end{array}

Since $\|\chi_{\pi}\|_{S_{4}}^{2}=\|\mathrm{sgn}\chi_{\pi}\|_{S_{4}}^{2}=1$ , these are both irreducible. The characters of the two representations are different, hence they are non-isomorphic.

There is one more row of the table to find. This can be done using orthonormality with the first column, this is Corollary 2.8.8. If we recall that the dimension must be a positive integer we can test candidate cases and determine that it must be $2$ . Thus we obtain that the full character table is

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr{\mathrm{triv}}&1&1&1&1&1\\ \mathrm{sgn}&1&-1&1&1&-1\\ \pi&3&1&-1&0&-1\\ \mathrm{sgn}\pi&3&-1&-1&0&1\\ \rho&2&0&2&-1&0\end{array}

Notice that we constructed the character of the final representation without constructing the representation itself!

2.10.3. Exercises

Problem 58. Prove Lemma 2.10.1.

Problem 59. Let $(\pi,V)$ and $(\rho,U)$ be two representations of a group $G$ . Show that $(\pi,V)\cong(\rho,U)$ if and only if

\operatorname{dim}\operatorname{Hom}_{G}(V,W)=\operatorname{dim}\operatorname{% Hom}_{G}(U,W)

for all representations $(\sigma,W)$ of $G$ .

Problem 60. Let $(\pi,V)$ be a representation of a group $G$ and $H$ a subgroup of $G$ . Show that

\chi_{\pi|_{H}}(h)=\chi_{\pi}(h)\qquad\forall\,h\in H.

Problem 61. Find the character table of $A_{4}$ . Decompose the restriction of each irreducible representation of $S_{4}$ to $A_{4}$ into irreducibles.

Remark: When we say “decompose a representation into irreducibles”, we simply mean find out how many times each irreducible representation occurs in the decomposition. This is in contrast to if we were to ask for a “decomposition into irreducible subrepresentations”, which involves finding irreducible subrepresentations of the given representation and showing that their direct sum is the whole representation - this is a much harder task.

Problem 62. Decompose ${\mathbb{C}}[Q_{8}]$ into irreducible isotopic components.

Problem 63. Let $X$ be a finite set on which a group $G$ acts. Consider the permutation representation $(\pi,{\mathbb{C}}(X))$ of $G$ on the free vector space of $X$ .

(a)

Show that $\operatorname{dim}({\mathbb{C}}(X)^{G})$ (recall that ${\mathbb{C}}(X)^{G}=\{{\bf v}\in{\mathbb{C}}(X)\,|\,\pi(g){\bf v}={\bf v}\,% \forall g\in G\}$ ) is equal to the number of $G$ -orbits in $X$ .
(b)

Compute $\langle\chi_{\pi},\chi_{{\,\mathrm{Id}}}\rangle_{G}$ , and use it to show Burnside’s lemma: the number of $G$ -orbits in $X$ is equal to

$\frac{1}{|G|}\sum_{g\in G}\#\{x\in X\,|\,g\cdot x=x\},$

i.e. the average number of fixed points of the group elements.

2.10. Lecture 10

2.10.1. Universal projections

Lemma 2.10.1.

Proof.

Proposition 2.10.2.

Proof.

Proof of Theorem 2.8.7, continued:.

Corollary 2.10.3.

Proof.

Corollary 2.10.4.

Proof.

Example 2.10.5.

2.10.2. Example: the character table of S4S_{4}.

Example 2.10.6.

2.10.3. Exercises

2.10.2. Example: the character table of $S_{4}$ .