Lecture 8

Solution:

1. (a)
   ​

   $S_{3}$: The character table reads

   	$e$	$(1\;2)$	$(1\;2\;3)$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1
   $(\mathrm{sgn},{\mathbb{C}})$	1	-1	1
   $(\pi,W_{0})$,	2	$0$	-1

   The conjugacy classes have sizes as follows $[e]$: 1, $[(1\;2)]$: 3, $[(1\;2\;3)]$: 2. The norms of the columns are computed:

   $$1^{2}+1^{2}+2^{2}=6=\frac{6}{1},\quad 1^{2}+(-1)^{2}+0^{2}=2=\frac{6}{3},\quad 1^{2}+1^{2}+(-1)^{2}=3=\frac{6}{2},$$

   and the inner products between different columns are easily seen to be zero:

   $${\mathcal{C}}_{1}\cdot{\mathcal{C}}_{2}=1\cdot 1+1\cdot(-1)+2\cdot 0=0,\quad{\mathcal{C}}_{1}\cdot{\mathcal{C}}_{3}=1\cdot 1+1\cdot 1+2\cdot(-1)=0,\quad{\mathcal{C}}_{2}\cdot{\mathcal{C}}_{3}=1\cdot 1+(-1)\cdot 1+0\cdot(-1)=0.$$

   We now compute the norms of the rows (with the weights given by the size of each conjugacy class):

   $$\frac{1}{6}(1\cdot 1^{2}+3\cdot 1^{2}+2\cdot 1^{2})=\frac{6}{6}=1,\quad\frac{1}{6}(1\cdot 1^{2}+3\cdot(-1)^{2}+2\cdot 1^{2})=\frac{6}{6}=1,\frac{1}{6}(1\cdot 2^{2}+3\cdot 0^{2}+2\cdot(-1)^{2})=\frac{6}{6}=1.$$

   Row orthogonality is also easily shown:

   	$\displaystyle\langle\chi_{{\,\mathrm{Id}}},\chi_{\operatorname{sgn}}\rangle_{G}=\frac{1}{6}\big{(}1(1)(1)+3(1)(-1)+2(1)(1)\big{)}=0$
   	$\displaystyle\langle\chi_{{\,\mathrm{Id}}},\chi_{\pi}\rangle_{G}=\frac{1}{6}\big{(}1(1)(2)+3(1)(0)+2(1)(-1)\big{)}=0$
   	$\displaystyle\langle\chi_{\operatorname{sgn}},\chi_{\pi}\rangle_{G}=\frac{1}{6}\big{(}1(1)(2)+3(-1)(0)+2(1)(-1)\big{)}=0.$

2. (b)
   ​

   $C_{5}$. Let $\omega=e^{2\pi i/5}$. The character table then reads

   	$0$	1	2	3	4
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1	1
   $(\chi_{1},{\mathbb{C}})$	$1$	$\omega$	$\omega^{2}$	$\omega^{3}$	$\omega^{4}$
   $(\chi_{2},{\mathbb{C}})$	$1$	$\omega^{2}$	$\omega^{4}$	$\omega$	$\omega^{3}$
   $(\chi_{3},{\mathbb{C}})$	$1$	$\omega^{3}$	$\omega$	$\omega^{4}$	$\omega^{2}$
   $(\chi_{4},{\mathbb{C}})$	$1$	$\omega^{4}$	$\omega^{3}$	$\omega^{2}$	$\omega$

   We observe that the first row and the first column are identical, and both are of unit length:

   $$\frac{1}{5}(1^{2}+1^{2}+1^{2}+1^{2}+1^{2})=1.$$

   The remaining rows and columns are identical (up to permutation), thus they also all have norm one:

   $$\frac{1}{5}(|1|^{2}+|\omega|^{2}+|\omega^{2}|^{2}+|\omega^{3}|^{2}+|\omega^{4}|^{2})=\frac{1}{5}(1^{2}+1^{2}+1^{2}+1^{2}+1^{2})=1.$$

   The inner product of these rows/columns with the first row/column is zero, since

   $$1+\omega+\omega^{2}+\omega^{3}+\omega^{4}=0.$$

   It remains to show row/column orthogonality. Firstly, the inner product of one of the “non-trivial” rows/columns with the first row/column gives

   $$1\cdot 1+\omega\cdot 1+\omega^{2}\cdot 1+\omega^{3}\cdot 1+\omega^{4}\cdot 1=\frac{1-\omega^{5}}{1-\omega}=0.$$

   For the inner product of two distinct non-trivial rows/columns the relevant sum is of the form (for some $j\in{\mathbb{N}}\setminus\{1\}$)

   	$\displaystyle 1\cdot 1+\omega\cdot\overline{\omega^{j}}+\omega^{2}\cdot\overline{(\omega^{2})^{j}}+\omega^{3}\cdot\overline{(\omega^{3})^{j}}+\omega^{4}\cdot\overline{(\omega^{4})^{j}}$
   	$\displaystyle=1+\omega^{1-j}+\omega^{(1-j)2}+\omega^{(1-j)3}+\omega^{(1-j)4}$
   	$\displaystyle=\frac{1-\omega^{5(1-j)}}{1-\omega}=0.$

3. (c)
   ​

   The character table of $D_{4}$ reads:

   size:	1	1	2	2	2
   	$e$	$r^{2}$	$r$	$s$	$sr$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1	1
   $(\pi_{+-},{\mathbb{C}})$	$1$	$1$	$1$	$-1$	$-1$
   $(\pi_{-+},{\mathbb{C}})$	$1$	$1$	$-1$	$1$	$-1$
   $(\pi_{--},{\mathbb{C}})$	$1$	$1$	$-1$	$-1$	$1$
   $(\rho,{\mathbb{C}}^{2})$	$2$	$-2$	$0$	$0$	$0$

   Thus the norm of the any of the first four rows equals

   $$\frac{1}{8}(1^{2}+1^{2}+2\big{(}(\pm 1)^{2}+(\pm 1)^{2}+(\pm 1)^{2}\big{)}=1,$$

   and the norm of the final row is

   $$\frac{1}{8}(2^{2}+(\pm 2)^{2}+2(0^{2}+0^{2}+0^{2}))=1.$$

   Similarly, the norm of the first two columns equals

   $$1^{2}+1^{2}+1^{2}+1^{2}+(\pm 2)^{2}=\frac{8}{1},$$

   whereas the norm of any of the last four columns equals

   $$1^{2}+(\pm 1)^{2}+(\pm 1)^{2}+(\pm 1)^{2}=\frac{8}{2}.$$

   The inner product of the first two columns equals

   $$1\cdot 1+1\cdot 1+1\cdot 1+1\cdot 1+2\cdot(-2)=0,$$

   the inner product of one of the first two columns with any of the last three is

   $$1+1-1-1+(\pm 2))\cdot 0=0,$$

   and the inner product of any two distinct columns out of the last three is

   $$1\cdot 1+(-1)\cdot 1+1\cdot(-1)+(-1)\cdot(-1)+0\cdot 0=0.$$

   The inner product of the last row with any of the first four equals

   $$\frac{1}{8}\big{(}2\cdot 1+(-2)\cdot 1+2(0\cdot(\pm 1)+0\cdot(\pm 1)+0\cdot(\pm 1))\big{)}=0,$$

   and the inner product of the first row with any of rows two to four equals

   $$\frac{1}{8}\big{(}1\cdot 1+1\cdot 1+2(1\cdot 1+1\cdot(-1)+1\cdot(-1))\big{)}=0.$$

   Finally, the inner product of two distinct rows from two to four equals

   $$\frac{1}{8}\big{(}1\cdot 1+1\cdot 1+2(1\cdot(-1)+(-1)\cdot 1+(-1)\cdot(-1))\big{)}=0.$$

Solution:

1. (a)
   ​

   This is an Abelian group, so every element is its own conjugacy class. The irreducible representations are all one-dimensional. Let $\omega=e^{2\pi i/9}$. Then each irreducible representation is of the form $(\psi_{j},{\mathbb{C}})$, where $\psi_{j}(k)=\omega^{jk}$. The table character table then reads

   	$0$	$1$	$2$	$3$	$4$	5	6	7	8
   $({\,\mathrm{Id}},{\mathbb{C}})$	$1$	$1$	$1$	$1$	$1$	1	1	1	1
   $(\psi_{1},{\mathbb{C}})$	$1$	$\omega$	$\omega^{2}$	$\omega^{3}$	$\omega^{4}$	$\omega^{5}$	$\omega^{6}$	$\omega^{7}$	$\omega^{8}$
   $(\psi_{2},{\mathbb{C}})$	$1$	$\omega^{2}$	$\omega^{4}$	$\omega^{6}$	$\omega^{8}$	$\omega$	$\omega^{3}$	$\omega^{5}$	$\omega^{7}$
   $(\psi_{3},{\mathbb{C}})$	$1$	$\omega^{3}$	$\omega^{6}$	$1$	$\omega^{3}$	$\omega^{6}$	$1$	$\omega^{3}$	$\omega^{6}$
   $(\psi_{4},{\mathbb{C}})$	$1$	$\omega^{4}$	$\omega^{8}$	$\omega^{3}$	$\omega^{4}$	$\omega^{5}$	$\omega^{6}$	$\omega^{7}$	$\omega^{8}$
   $(\psi_{5},{\mathbb{C}})$	$1$	$\omega^{5}$	$\omega$	$\omega^{6}$	$\omega^{2}$	$\omega^{7}$	$\omega^{3}$	$\omega^{8}$	$\omega^{4}$
   $(\psi_{6},{\mathbb{C}})$	$1$	$\omega^{6}$	$\omega^{3}$	$1$	$\omega^{6}$	$\omega^{3}$	$1$	$\omega^{6}$	$\omega^{3}$
   $(\psi_{7},{\mathbb{C}})$	$1$	$\omega^{7}$	$\omega^{5}$	$\omega^{3}$	$\omega$	$\omega^{8}$	$\omega^{6}$	$\omega^{4}$	$\omega^{2}$
   $(\psi_{8},{\mathbb{C}})$	$1$	$\omega^{8}$	$\omega^{7}$	$\omega^{6}$	$\omega^{5}$	$\omega^{4}$	$\omega^{3}$	$\omega^{2}$	$\omega$

2. (b)
   ​

   Also an Abelian group, with nine conjugacy classes. The irreducible representations factor as a product of two irreducible representations of $C_{3}$, i.e. they are given as $(\psi_{j,k},{\mathbb{C}})$, where $\psi_{j,k}(a,b)=\omega^{aj+bk}$, where $\omega=e^{2\pi i/3}$. The character table reads

   	$(0,0)$	$(1,0)$	$(2,0)$	$(0,1)$	$(1,1)$	$(2,1)$	$(0,2)$	$(1,2)$	$(2,2)$
   $({\,\mathrm{Id}},{\mathbb{C}})$	$1$	$1$	$1$	$1$	$1$	1	1	1	1
   $(\psi_{1,0},{\mathbb{C}})$	$1$	$\omega$	$\omega^{2}$	$1$	$\omega$	$\omega^{2}$	$1$	$\omega$	$\omega^{2}$
   $(\psi_{2,0},{\mathbb{C}})$	$1$	$\omega^{2}$	$\omega$	$1$	$\omega^{2}$	$\omega$	$1$	$\omega^{2}$	$\omega$
   $(\psi_{0,1},{\mathbb{C}})$	$1$	$1$	$1$	$\omega$	$\omega$	$\omega$	$\omega^{2}$	$\omega^{2}$	$\omega^{2}$
   $(\psi_{1,1},{\mathbb{C}})$	$1$	$\omega$	$\omega^{2}$	$\omega$	$\omega^{2}$	$1$	$\omega^{2}$	$1$	$\omega$
   $(\psi_{2,1},{\mathbb{C}})$	$1$	$\omega^{2}$	$\omega$	$\omega$	$1$	$\omega^{2}$	$\omega^{2}$	$\omega$	$1$
   $(\psi_{0,2},{\mathbb{C}})$	$1$	$1$	$1$	$\omega^{2}$	$\omega^{2}$	$\omega^{2}$	$\omega$	$\omega$	$\omega$
   $(\psi_{1,2},{\mathbb{C}})$	$1$	$\omega$	$\omega^{2}$	$\omega^{2}$	$1$	$\omega$	$\omega$	$\omega^{2}$	$1$
   $(\psi_{2,2},{\mathbb{C}})$	$1$	$\omega^{2}$	$\omega$	$\omega^{2}$	$\omega$	$1$	$\omega$	$1$	$\omega^{2}$

3. (c)
   ​

   We have a complete classification of the irreducible representations from Lecture 4. The conjugacy classes are given in Problem 5 Prerequisites: there are two one-dimensional representations and to irreducible two-dimensional representations. Filling in the table gives

   size:	1	2	2	5
   	$e$	$r$	$r^{2}$	$s$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1
   $(\epsilon,{\mathbb{C}})$	$1$	$1$	$1$	$-1$
   $(\rho_{1},{\mathbb{C}}^{2})$	$2$	$2\cos({\textstyle\frac{2\pi}{5}})$	$2\cos({\textstyle\frac{4\pi}{5}})$	$0$
   $(\rho_{2},{\mathbb{C}}^{2})$	$2$	$2\cos({\textstyle\frac{4\pi}{5}})$	$2\cos({\textstyle\frac{2\pi}{5}})$	0

4. (d)
   ​

   The conjugacy classes were computed in Problem 3 Prerequisites, and the irreducible representations in Problem 22 1. Putting these together gives the character table:

   size:	1	1	2	2	2
   	$1$	-1	$i$	$j$	$k$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1	1
   $(\pi_{i},{\mathbb{C}})$	$1$	$1$	$1$	$-1$	$-1$
   $(\pi_{j},{\mathbb{C}})$	$1$	$1$	$-1$	$1$	$-1$
   $(\pi_{k},{\mathbb{C}})$	$1$	$1$	$-1$	$-1$	$1$
   $(\pi,{\mathbb{C}}^{2})$	$2$	$-2$	$0$	$0$	$0$

   Funny observation: this is the “same” as the table of $D_{4}$!

5. (e)
   ​

   We combine the result of Problem 5 Prerequisites with the classification of irreducible representations from Lecture 4. It will be convenient to split into two cases:

   Even $n$:
   

   size:	1	1	2	2	$\cdots$	2	$n/2$	$n/2$
   	$e$	$r^{n/2}$	$r$	$r^{2}$	$\ldots$	$r^{n/2-1}$	$s$	$sr$
   $({\,\mathrm{Id}},{\mathbb{C}})$	$1$	$1$	$1$	$1$	$\ldots$	$1$	$1$	$1$
   $(\pi_{+-},{\mathbb{C}})$	$1$	$1$	$1$	$1$	$\ldots$	$1$	$-1$	$1$
   $(\pi_{-+},{\mathbb{C}})$	$1$	$(-1)^{n/2}$	$-1$	$1$	$\ldots$	$(-1)^{n/2-1}$	$1$	$-1$
   $(\pi_{--},{\mathbb{C}})$	$1$	$(-1)^{n/2}$	$-1$	$1$	$\ldots$	$(-1)^{n/2-1}$	$-1$	$1$
   $(\rho_{1},{\mathbb{C}}^{2})$	$2$	$2\cos(1\pi)$	$2\cos(1\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot 1\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos(({\textstyle\frac{n}{2}}-1)\cdot 1\cdot{\textstyle\frac{2\pi}{n}})$	$0$	$0$
   $(\rho_{2},{\mathbb{C}}^{2})$	$2$	$2\cos(2\pi)$	$2\cos(2\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot 2\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos(({\textstyle\frac{n}{2}}-1)\cdot 2\cdot{\textstyle\frac{2\pi}{n}})$	$0$	$0$
   $(\rho_{3},{\mathbb{C}}^{2})$	$2$	$2\cos(3\pi)$	$2\cos(3\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot 3\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos(({\textstyle\frac{n}{2}}-1)\cdot 3\cdot{\textstyle\frac{2\pi}{n}})$	$0$	$0$
   $\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$
   $(\rho_{n/2-1},{\mathbb{C}}^{2})$	$2$	$2\cos\big{(}({\textstyle\frac{n}{2}}-1)\pi\big{)}$	$2\cos\big{(}({\textstyle\frac{n}{2}}-1)\cdot{\textstyle\frac{2\pi}{n}}\big{)}$	$2\cos(2\cdot({\textstyle\frac{n}{2}}-1)\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos(({\textstyle\frac{n}{2}}-1)^{2}\cdot{\textstyle\frac{2\pi}{n}})$	$0$	$0$

   
   

   Odd $n$:
   

   size:	1	2	2	$\cdots$	2	$n$
   	$e$	$r$	$r^{2}$	$\ldots$	$r^{(n-1)/2}$	$s$
   $({\,\mathrm{Id}},{\mathbb{C}})$	$1$	$1$	$1$	$\ldots$	$1$	$1$
   $(\epsilon,{\mathbb{C}})$	$1$	$1$	$1$	$\ldots$	$1$	$-1$
   $(\rho_{1},{\mathbb{C}}^{2})$	$2$	$2\cos(1\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot 1\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos({\textstyle\frac{n-1}{2}}\cdot 1\cdot{\textstyle\frac{2\pi}{n}})$	$0$
   $(\rho_{2},{\mathbb{C}}^{2})$	$2$	$2\cos(2\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot 2\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos({\textstyle\frac{n-1}{2}}\cdot 2\cdot{\textstyle\frac{2\pi}{n}})$	$0$
   $(\rho_{3},{\mathbb{C}}^{2})$	$2$	$2\cos(3\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot 3\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos({\textstyle\frac{n-1}{2}}\cdot 3\cdot{\textstyle\frac{2\pi}{n}})$	$0$
   $\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$
   $(\rho_{{\textstyle\frac{n-1}{2}}},{\mathbb{C}}^{2})$	$2$	$2\cos({\textstyle\frac{n-1}{2}}\cdot{\textstyle\frac{2\pi}{n}})$	$2\cos(2\cdot{\textstyle\frac{n-1}{2}}\cdot{\textstyle\frac{2\pi}{n}})$	$\ldots$	$2\cos({\textstyle\frac{n-1}{2}}\cdot{\textstyle\frac{n-1}{2}}\cdot{\textstyle\frac{2\pi}{n}})$	$0$

Solution: Consider the matrix

By Theorem 8.7, the matrix $M$ is unitary, i.e.

hence

Consider the matrix $T$, whose entries are those of the character table. We now note that

hence