Lecture 10

Solution: Let $h\in G$. Note that conjugation by $h$ is a bijection, i.e. $G=\{hgh^{-1}\,|\,g\in G\}$. We will also make use of the fact that if $f$ is a class function $f(g)=f(hgh^{-1})$, for all $g\in G$. Then

As $h$ was arbitrary we have that $\pi(f)$ is in $\operatorname{Hom}_{G}(V)$.

Solution: We have

Thus, if $\operatorname{dim}\operatorname{Hom}_{G}(V,W)=\operatorname{dim}\operatorname{Hom}_{G}(U,W)$ for all $(\sigma,W)$, we have $\chi_{\pi}=\chi_{\rho}$, hence $(\pi,V)\cong(\rho,U)$ by Theorem 29.

On the other hand, if $(\pi,V)\cong(\rho,U)$, then $\chi_{\pi}=\chi_{\rho}$, hence

Solution: Since $\pi|_{H}(h)=\pi(h)$ for all $h\in H$, $\operatorname{tr}\pi|_{H}(h)=\operatorname{tr}\pi(h)$, hence $\chi_{\pi|_{H}}=\chi_{\pi}$.

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.

Solution: We start by determining the conjugacy classes of $A_{4}$. Recall the following:

Thus, the conjugacy classes of $A_{4}$ are

By Theorem 19, we then know that there are four irreducible representations of $S_{4}$, and the squares of their dimensions sum up to $12$. Recall the character table of $S_{4}$:

The trivial representation is of course an irreducible representation, and we see that the character of $(\pi|_{A_{4}},W_{0})$ has norm one:

hence $(\pi|_{A_{4}},W_{0})$ is irreducible. Since $\operatorname{sgn}=1$ on $A_{4}$, restricting $\operatorname{sgn}\,\pi$ to $A_{4}$ gives the same representation. We thus have

By checking the dimensions, we see that we are missing two one-dimensional representations $(\psi_{1},{\mathbb{C}})$ and $(\psi_{2},{\mathbb{C}})$. All the elements of ${\mathcal{D}}_{2}$ have order two, hence the characters there must have order two in ${\mathbb{C}}^{\times}$ (i.e. they are $\pm 1)$, and all the elements of ${\mathcal{D}}_{3}$ and ${\mathcal{D}}_{4}$ have order three, hence the characters must be third roots of unity. Writing $\omega=e^{2\pi i/3}$, we get

We now use column orthogonality with the first column to see that both $\pm 1$S must in fact be $+1$s, and

We then get that $j_{2}=-j_{1}$, $k_{2}=-k_{1}$, and $j_{1},k_{1}=\pm 1$. Column orthogonality between the last two columns gives

so there are two solutions (each giving one of the characters $\psi_{1},\psi_{2}$): $j_{1}=\pm 1$. The completed table reads

The sign representation on $S_{4}$ restricted to $A_{4}$ is simply the trivial representation, while $(\pi,W_{0})$ and $(\operatorname{sgn}\,\pi,W_{0})$ restricted to $A_{4}$ are both the irreducible three-dimensional irreducible representation.

It remains to decompose the restriction of the two-dimensional irreducible representation $(\rho,V)$ to $A_{4}$. Since $A_{4}$ has no two-dimensional irreducibles, this must be the direct sum of two one-dimensional representations. We compute the inner product of $\chi_{\rho|_{A_{4}}}$ with $\chi_{\psi_{1}}$ and $\chi_{\psi_{2}}$:

hence $(\rho|_{A_{4}},V)=(\psi_{1},{\mathbb{C}})\oplus(\psi_{2},{\mathbb{C}})$.

Solution: The character table of $Q_{8}$ reads

Thus

For each irreducible representation $(\sigma,V)$, the isotopic component of $\sigma$ in $(\lambda,{\mathbb{C}}(Q_{8}))$ is the image of the isotopic projector $\lambda_{\sigma}$. For the four one-dimensional representations, we compute $\lambda_{\sigma}(\mathbf{1})$:

For the two-dimensional representation, the $(\pi,{\mathbb{C}}^{2})$-isotopic component is

Given any of the canonical basis elements $g\in Q_{8}$, we have

hence the matrix of $I-\lambda_{\pi}$ w.r.t. the basis $\{\mathbf{1},\mathbf{-1},\mathbf{i},\mathbf{-i},\mathbf{j},\mathbf{-j},\mathbf{k},\mathbf{-k}\}$ is

The kernel of this is then spanned by $\mathbf{1}-(\mathbf{-1})$, $\mathbf{i}-(\mathbf{-i})$, $\mathbf{j}-(\mathbf{-j})$, $\mathbf{k}-(\mathbf{-k})$, giving

1. (a)
   ​

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

2. (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.

Solution:

1. (a)
   ​

   We write $X=X_{1}\sqcup X_{2}\sqcup\ldots\sqcup X_{d}$, where each $X_{i}$ is a $G$-orbit in $X$. For each $1\leq i\leq d$, let $y_{i}=\sum_{x\in X_{i}}x$. Clearly $g\cdot y_{i}=y_{i}$, for all $g\in G$ so each $y_{i}$ is in ${\mathbb{C}}(X)^{G}$. Also note that, as the orbits are disjoint each $y_{i}$ is disjoint and the set $\{y_{i}\}_{1\leq i\leq d}$ is linearly independent. To show it spans ${\mathbb{C}}(X)^{G}$ let $\sum_{x\in X}z_{x}x\in{\mathbb{C}}(X)^{G}$ so $g\cdot\sum_{x\in X}z_{x}x=\sum_{x\in X}z_{x}x$ for all $g\in G$. We can partition this into $G$-orbits to get

   $$g\cdot\sum_{X_{i}\subseteq X}\sum_{x\in X_{i}}z_{x}x=\sum_{X_{i}\subseteq X}\sum_{x\in X_{i}}z_{x}x$$

   and thus for each orbit we have $g\cdot\sum_{x\in X_{i}}z_{x}x=\sum_{x\in X_{i}}z_{x}x$ and for each $x\in X_{i}$, $z_{x}=z_{g^{-}1}\cdot x$. Thus all the $z_{x}$ are the same for all $x\in X_{i}$ and $\sum_{x\in X_{i}}z_{x}x\in{\mathbb{C}}y_{i}$. The set $\{y_{i}\}_{1\leq i\leq d}$ therefore spans ${\mathbb{C}}(X)^{G}$ and hence is a basis so $\operatorname{dim}{\mathbb{C}}(X)^{G}=d$.

   Functional method: We write $X=X_{1}\sqcup X_{2}\sqcup\ldots\sqcup X_{d}$, where each $X_{i}$ is a $G$-orbit in $X$. For each $i=1,\ldots,d$, let $f_{i}\in V$ be the function defined through

   $$f_{i}(x)=\begin{cases}1\quad&\mathrm{if\;}x\in X_{i}\\ 0\quad&\mathrm{otherwise.}\end{cases}$$

   Noting that $x\in X_{i}\Leftrightarrow g^{-1}\cdot x\in X_{i}\;\forall g\in G$, we see that each $f_{i}\in V^{G}$. Furthermore, since the support of $f_{i}$ is $X_{i}$, and the $X_{i}$ are all disjoint, the set $\{f_{i}\}_{i=1,\ldots,d}$ is linearly independent. We show that it spans $V^{G}$: given $f\in V^{G}$, we have $f(x)=[\pi(g)f](x)=f(g^{-1}\cdot x)$ for all $g\in G$, $x\in X$. In particular, $f$ is constant on each $X_{i}$, allowing us to write

   $$f=(f|_{X_{1}})f_{1}+(f|_{X_{2}})f_{2}+\ldots+(f|_{X_{d}})f_{d}.$$

   The set $\{f_{i}\}_{i=1,\ldots,d}$ is therefore a basis of $V^{G}$, giving $\operatorname{dim}V^{G}=d$, which is the number of $G$-orbits in $X$.

2. (b)
   ​

   By Lemma 9.4, we have

   $$\langle\chi_{\pi},\chi_{{\,\mathrm{Id}}}\rangle_{G}=\frac{1}{|G|}\sum_{g\in G}\chi_{\pi}(g)=\operatorname{dim}V^{G}=d=\#(G\text{-orbits in }X).$$

   From Problem 46,

   $$\chi_{\pi}(g)=\#\{x\in X\,|\,g\cdot x=x\},$$

   which, when substituted into the above identity, gives

   $$\frac{1}{|G|}\sum_{g\in G}\#\{x\in X\,|\,g\cdot x=x\}=\#(G\text{-orbits in }X),$$

   i.e. Burnside’s lemma.