Lecture 7 ‣ Representation theory exercise solutions for Michaelmas

Problem 43. Let $(\pi,W_{0})$ denote the usual $n-1$ dimensional irreducible subrepresentation of the permutation representation of $S_{n}$ on ${\mathbb{C}}^{n}$ . Compute the character $\chi_{(\pi,W_{0})}$ of this subrepresentation.

Hint: combine Example 7.7 with Proposition 7.9 (iii).

Solution: We saw in class that the character of the $1$ -dimensional trivial representation $({\,\mathrm{Id}},{\mathbb{C}})$ of a group $G$ is given by

\chi_{({\,\mathrm{Id}},{\mathbb{C}})}(g)=1\qquad\forall g\in G.

We also saw that for the permutation representation $(\pi,{\mathbb{C}}^{n})$ of $S_{n}$ , we have

\chi_{(\pi,{\mathbb{C}}^{n})}(\sigma)=\#\big{\{}j\in\{1,\ldots,n\}\,|\,\sigma(% j)=j\big{\}}\qquad\forall\sigma\in S_{n}.

Since $(\pi,{\mathbb{C}}^{n})=(\pi,{\mathbb{C}}(1,1,\ldots,1))\oplus(\pi,W_{0})$ , by Proposition 7.9 iii),

\chi_{(\pi,{\mathbb{C}}^{n})}=\chi_{(\pi,(1,\ldots,1))}+\chi_{(\pi,W_{0})}

Now, $(\pi,(1,\ldots,1))$ is isomorphic to $({\,\mathrm{Id}},{\mathbb{C}})$ , so $\chi_{(\pi,(1,\ldots,1))}(\sigma)=1$ for all $\sigma\in S_{n}$ , giving

\chi_{(\pi,W_{0})}(\sigma)=\chi_{(\pi,{\mathbb{C}}^{n})}(\sigma)-1=\#\big{\{}j% \in\{1,\ldots,n\}\,|\,\sigma(j)=j\big{\}}-1\qquad\forall\sigma\in S_{n}.

Problem 44. Let $g\in G$ be such that $g$ and $g^{-1}$ are both in the same conjugacy class. Show that $\chi_{\pi}(g)$ is real-valued for any representation $\pi$ of $G$ .

Solution: Let $(\pi,V)$ be a representation of $G$ . If $g$ and $g^{-1}$ are in the same conjugacy class of $G$ then there exists a $h\in G$ such that $hgh^{-1}=g^{-1}$ . Thus, by Proposition 7.9(ii), $\chi_{\pi}(g)=\chi_{\pi}(g^{-1})$ . However, Proposition 7.9(iv), gives us that $\chi_{\pi}(g^{-1})=\overline{\chi_{\pi}(g)}$ and thus $\chi_{\pi}(g)=\overline{\chi_{\pi}(g)}$ . We therefore have that $\chi_{\pi}(g)$ is real valued and as $(\pi,V)$ was arbitrary this is true for all representations of $G$ .

Problem 45. Let $(\pi,V)$ be a representation of group $G$ . Show that

(a)

$|\chi_{\pi}(g)|\leq\operatorname{dim}(\pi)\quad\forall g\in G$ .
(b)

$\ker\pi=\chi_{\pi}^{-1}\big{(}\{\operatorname{dim}(\pi)\}\big{)}.$

Solution:

(a)

Let $\{\lambda_{i}\}_{i=1,\ldots,\operatorname{dim}(\pi)}$ be the eigenvalues of $\pi(g)$ . Then

$|\chi_{\pi}(g)|=|\lambda_{1}+\ldots+\lambda_{\operatorname{dim}(\pi)}|\leq|% \lambda_{1}|+\ldots+|\lambda_{\operatorname{dim}(\pi)}|=\operatorname{dim}(\pi)$

(since each $\lambda_{i}$ is a root of unity).
(b)

If $g\in\ker\pi$ , then $\chi_{\pi}(g)=\operatorname{tr}(\pi(g))=\operatorname{tr}({\,\mathrm{Id}})=% \operatorname{dim}(\pi)$ . On the other hand, if $\chi_{\pi}(g)=\operatorname{dim}(\pi)$ , we then have

$\chi_{\pi}(g)=\lambda_{1}+\ldots+\lambda_{\operatorname{dim}(\pi)}=|\lambda_{1% }|+\ldots+|\lambda_{\operatorname{dim}(\pi)}|.$

However, since equality in the triangle inequality is only obtained for complex numbers in the ${\mathbb{R}}$ -span of each other, we must then have $\lambda_{1}=\lambda_{2}=\ldots=\lambda_{\operatorname{dim}(\pi)}=1$ , hence $\pi(g)={\,\mathrm{Id}}$ .

Problem 46. Let the finite group $G$ act on the finite set $X$ , and consider the representation $(\pi,{\mathbb{C}}(X))$ as in Definition 1.9. Show that

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

Solution: Let ${\mathcal{B}}=\{x\}_{x\in X}$ denote the basis of ${\mathbb{C}}(X)$ . Note that $\pi(g)x=g\cdot x$ . Choosing some ordering of the basis elements, and writing down the coordinate vectors of the $\pi(g){x_{i}}$ gives:

[\pi(g){x_{i}}]_{{\mathcal{B}}}=\left(\begin{matrix}0\\ \vdots\\ 0\\ 1\\ 0\\ \vdots\\ 0\end{matrix}\right),

where the one is in row $j$ iff $g\cdot x_{i}=x_{j}$ . Now, as seen previously, we have

\chi_{\pi}(g)=\operatorname{tr}\pi(g)=\operatorname{tr}[\pi(g)]_{{\mathcal{B}}% }=\operatorname{tr}\left(\begin{matrix}\vdots&&\vdots\\ [\pi(g){x_{1}}]_{{\mathcal{B}}}&\cdots&[\pi(g){x_{n}}]_{{\mathcal{B}}}\\ \vdots&&\vdots\end{matrix}\right);

and each column has a one on the diagonal iff $g\cdot x_{i}=x_{i}$ , and a zero otherwise. Thus

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