Lecture 12

Solution: Let $a_{i}\in{\mathbb{C}}$, ${\bf v}_{i}^{(1)},{\bf v}_{i}^{(2)}\in V$. For $g\in G$ and $\sum_{i}a_{i}\big{(}{\bf v}_{i}^{(1)}\otimes{\bf v}_{i}^{(2)}-{\bf v}_{i}^{(1)}\otimes{\bf v}_{i}^{(2)}\big{)}\in U_{1}$, we have

hence $(\pi,\otimes\pi,U_{1})$ is a subrepresentation. Similarly, given $\sum_{i}a_{i}\big{(}{\bf v}_{i}^{(1)}\otimes{\bf v}_{i}^{(2)}+{\bf v}_{i}^{(1)}\otimes{\bf v}_{i}^{(2)}\big{)}\in U_{2}$, we have

hence $(\pi,\otimes\pi,U_{2})$ is a subrepresentation.

Note that $U_{1}\cup U_{2}=V\otimes V$ as

for all ${\bf v}_{1},{\bf v}_{2}\in V$. If ${\bf e}_{1},\ldots{\bf e}_{n}$ is a basis of $V$ it is clear $\{{\bf e}_{i}\otimes{\bf e}_{j}-{\bf e}_{j}\otimes{\bf e}_{i}\,|\,1\leq i\leq n,1\leq j<i\}$ and $\{{\bf e}_{i}\otimes{\bf e}_{j}+{\bf e}_{j}\otimes{\bf e}_{i}\,|\,1\leq i\leq n,1\leq j\leq i\}$ are bases of $U_{1}$ and $U_{2}$ respectively. Thus $\operatorname{dim}(v\otimes V)=n^{2}=\operatorname{dim}(U_{1})+\operatorname{dim}(U_{2})$ and $V\otimes V=U_{1}\oplus U_{2}$.

Let $\phi$ be the map

which we extend linearly. This map is clearly a bijection as it maps a basis to a basis thus these vector spaces are isomorphic. For any $g\in G$ and $v_{1},v_{2}\in V$ we have

so $\phi$ is a $G$-homomorphism and they are isomorphic representations.

In a very similar fashion we define the map

and show that the representations are isomorphic (omitted). Thus

.

Solution:

1. (a)
   ​

   We compute

   	$\displaystyle\mathrm{Sym}^{2}\pi(g)b_{1}^{2}=(\pi(g)b_{1})(\pi(g)b_{1})=(ab_{1}+cb_{2})(ab_{1}+cb_{2})=a^{2}b_{1}^{2}+2acb_{1}b_{2}+c^{2}b_{2}^{2}$
   	$\displaystyle\mathrm{Sym}^{2}\pi(g)b_{1}b_{2}=(\pi(g)b_{1})(\pi(g)b_{2})=(ab_{1}+cb_{2})(bb_{1}+db_{2})=abb_{1}^{2}+(ad+bc)b_{1}b_{2}+cdb_{2}^{2}$
   	$\displaystyle\mathrm{Sym}^{2}\pi(g)b_{2}^{2}=(\pi(g)b_{2})(\pi(g)b_{2})=(bb_{1}+db_{2})(bb_{1}+db_{2})=b^{2}b_{1}^{2}+2bdb_{1}b_{2}+d^{2}b_{2}^{2},$

   so the matrix of $\mathrm{Sym}^{2}\pi(g)$ is

   $$\left(\begin{matrix}a^{2}&ab&b^{2}\\ 2ac&ad+bc&2bd\\ c^{2}&cd&d^{2}\end{matrix}\right).$$
2. (b)
   ​

   Similarly to (a),

   $$\pi\wedge\pi(g)(b_{1}\wedge b_{2})=(\pi(g)b_{1})\wedge(\pi(g)b_{2})=(ab_{1}+cb_{2})\wedge(bb_{1}+db_{2})=ab(b_{1}\wedge b_{1})+ad(b_{1}\wedge b_{2})+cb(b_{2}\wedge b_{1})+cd(b_{2}\wedge b_{2}).$$

   Now, $b_{1}\wedge b_{1}=b_{2}\wedge b_{2}=0$, and $b_{2}\wedge b_{1}=-b_{1}\wedge b_{2}$, hence

   $$\pi\wedge\pi(g)(b_{1}\wedge b_{2})=(ad-bc)(b_{1}\wedge b_{2}).$$

Solution: The character table of $S_{5}$ reads (see Lecture 14)

The formulas for the character of $\mathrm{Sym}^{2}\sigma$ and $\sigma\wedge\sigma$ are given by $\frac{1}{2}\big{(}\chi_{\sigma}(g)^{2}\pm\chi_{\sigma}(g^{2})\big{)}$, so we need to compute $\chi_{\sigma}(g^{2})$ for $g$ in each of the conjugacy classes of $S_{5}$:

these formulas hold for both $\sigma=\rho$ and $\sigma=\mathrm{sgn}\rho$. Similarly, $\chi_{\rho}^{2}=\chi_{\mathrm{sgn}\rho}^{2}$, so WLOG we may assume that $\sigma=\rho$. Thus, using the above to compute the characters:

Starting with $\mathrm{Sym}^{2}\rho$, we can see directly from the character table that $\langle\chi_{\mathrm{Sym}^{2}\rho},\chi_{{\,\mathrm{Id}}}\rangle>0$; to make this more precise, we compute

We can also see that $(\pi,W_{0})$ is a subrepresentation:

and $(\rho,V)$:

and $\mathrm{sgn}\rho$:

The sum of the dimensions of these four irreducible representations is 15, hence

In a similar fashion, one can see directly from the character table that $\langle\chi_{\rho\wedge\rho},\chi_{\pi\wedge\pi}\rangle>0$ and $\langle\chi_{\rho\wedge\rho},\chi_{\mathrm{sgn}\pi}\rangle>0$, and a more explicit computation verifies this:

hence

1. (a)
   ​

   Show that $(\operatorname{Sym}^{2}\pi,\operatorname{Sym}^{2}{\mathbb{C}}^{4})$ has a unique subrepresentation isomorphic to $(\rho,V)$.

2. (b)
   ​

   Use the $\rho$-isotopic projector to find that subrepresentation.

Solution:

1. (1)
   ​

   The character of $(\pi,{\mathbb{C}}^{4})$ is given by

   size:	1	6	3	8	6
   	$e$	$(1\;2)$	$(1\;2)(3\;4)$	$(1\;2\;3)$	$(1\;2\;3\;4)$
   $(\pi,{\mathbb{C}}^{4})$	4	2	0	1	0

   Using the formula $\chi_{\mathrm{Sym}^{2}\pi}(g)=\frac{1}{2}\big{(}\chi_{\pi}(g)^{2}+\chi_{\pi}(g^{2})\big{)}$, hence

   size:	1	6	3	8	6
   	$e$	$(1\;2)$	$(1\;2)(3\;4)$	$(1\;2\;3)$	$(1\;2\;3\;4)$
   $\mathrm{Sym}^{2}\pi$	10	4	2	1	0

   The character of $(\rho,V)$ is given by

   size:	1	6	3	8	6
   	$e$	$(1\;2)$	$(1\;2)(3\;4)$	$(1\;2\;3)$	$(1\;2\;3\;4)$
   $(\rho,V)$	2	0	2	-1	0

   Hence, computing the inner product of $\chi_{\mathrm{Sym}^{2}\pi}$ with $\chi_{\rho}$:

   $$\langle\chi_{\mathrm{Sym}^{2}\pi},\chi_{\rho}\rangle_{S_{4}}=\frac{1}{24}\big{(}1\cdot 10\cdot 2+6\cdot 4\cdot 0+3\cdot 2\cdot 2+8\cdot 1\cdot(-1)+6\cdot 0\cdot 0\big{)}=1,$$

   there is therefore exactly one subrepresentation of $(\operatorname{Sym}^{2}\pi,\operatorname{Sym}^{2}{\mathbb{C}}^{4})$ that is isomorphic to $(\rho,V)$.

2. (2)
   ​

   The $\rho$-isotopic projector is given by

   $$\mathrm{Sym}^{2}\pi_{\rho}=\frac{2}{24}\sum_{g\in S_{4}}\overline{\chi_{\rho}(g)}\mathrm{Sym}^{2}\pi(g)=\frac{1}{12}\left(2{\,\mathrm{Id}}+2\left(\sum_{g\in[(1\;2)(3\;4)]}\mathrm{Sym}^{2}\pi(g)\right)-\left(\sum_{g\in[(1\;2\;3)]}\mathrm{Sym}^{2}\pi(g)\right)\right).$$

   Since $V$ has dimension two, the subrepresentation we are searching for will be spanned by any two linearly independent vectors in $\mathrm{Sym}^{2}\pi_{\rho}(\mathrm{Sym}^{2}{\mathbb{C}}^{4})$. We will succesively apply $\mathrm{Sym}^{2}\pi_{\rho}$ to the standard basis vectors of $\mathrm{Sym}^{2}{\mathbb{C}}^{4}$ until we have two non co-linear vectors;

   $$\mathrm{Sym}^{2}\pi_{\rho}e_{1}^{2}=\frac{1}{12}\left(2e_{1}^{2}+2\big{(}e_{2}^{2}+e_{3}^{2}+e_{4}^{2}\big{)}-\big{(}e_{2}^{2}+e_{3}^{2}+e_{2}^{2}+e_{4}^{2}+e_{3}^{2}+e_{4}^{2}+e_{1}^{2}+e_{1}^{2}\big{)}\right)=0$$

   (recall that

   $$[(1\;2)(3\;4)]=\{(1\;2)(3\;4),\;(1\;3)(2\;4),\;(1\;4)(2\;3)\},$$

   and

   $$[(1\;2\;3)]=\{(1\;2\;3),\;(1\;3\;2),\;(1\;2\;4),\;(1\;4\;2),\;(1\;3\;4),\;(1\;4\;3),\;(2\;3\;4),\;(2\;4\;3)\},$$

   the sums are carried out in the order listed here.) We proceed with other basis vectors:

   	$\displaystyle\mathrm{Sym}^{2}\pi_{\rho}e_{1}e_{2}$	$\displaystyle=\frac{1}{12}\left(2e_{1}e_{2}+2\big{(}e_{1}e_{2}+e_{3}e_{4}+e_{3}e_{4}\big{)}-\big{(}e_{2}e_{3}+e_{1}e_{3}+e_{2}e_{4}+e_{1}e_{4}+e_{2}e_{3}+e_{2}e_{4}+e_{1}e_{3}+e_{1}e_{4}\big{)}\right)$
   		$\displaystyle=\frac{1}{12}\left(4e_{1}e_{2}+4e_{3}e_{4}-2e_{1}e_{3}-2e_{1}e_{4}-2e_{2}e_{3}-2e_{2}e_{4}\right)=b_{1}.$

   	$\displaystyle\mathrm{Sym}^{2}\pi_{\rho}e_{1}e_{3}$	$\displaystyle=\frac{1}{12}\left(2e_{1}e_{3}+2\big{(}e_{2}e_{4}+e_{1}e_{3}+e_{2}e_{4}\big{)}-\big{(}e_{1}e_{2}+e_{2}e_{3}+e_{2}e_{3}+e_{3}e_{4}+e_{3}e_{4}+e_{1}e_{4}+e_{1}e_{4}+e_{1}e_{2}\big{)}\right)$
   		$\displaystyle=\frac{1}{12}\left(4e_{1}e_{3}+4e_{2}e_{4}-2e_{1}e_{2}-2e_{1}e_{4}-2e_{2}e_{3}-2e_{3}e_{4}\right)=b_{2};$

   this is not in ${\mathbb{C}}b_{1}$, hence $(\mathrm{Sym}^{2}\pi,\mathrm{span}\{b_{1},b_{2}\})$ is the subrepresentation we are looking for.

Solution:

1. (1)
   ​

   Since $x$ has order $7$, so must $x^{2},x^{3},\ldots,x^{6}$. hence each of these is in either $C_{7A}$ or $C_{7_{B}}$. In particular, the pigeonhole principle gives that at least two of $\{x,x^{2},x^{3}\}$ are conjugate to each other. We consider the three possibilities: (i) $x,x^{2}$ are conjugate: we are done, as this is what we are trying to show (ii) $x,x^{3}$ are conjugate. Writing $x^{3}=axa^{-1}$, we have $x^{2}=x^{9}=x^{3}\cdot x^{3}\cdot x^{3}=ax^{3}a^{-1}=a^{2}xa^{-2}$, hence $x$ and $x^{2}$ are also conjugate. (iii) $x^{2}$ and $x^{3}$ are conjugate. Writing $x^{3}=bx^{2}b^{-1}$, we then have $x=x^{15}=(x^{3})^{5}=(bx^{2}b^{-1})^{5}=b^{5}x^{10}b^{-5}=b^{5}x^{3}b^{-5}=b^{6}x^{2}b^{-6}$, i.e. $x$ and $x^{2}$ are also conjugate.

2. (2)
   ​

   The trivial representation is of course irrecducible. By Proposition 31 (Lecture 12, cf. Problem 54), $(\pi_{1}^{*},V^{*})$ is also irreducible, with character $\overline{\chi_{\pi_{1}}}$. We have now found three rows of the character table:

   $$\begin{array}[]{|c|c|c|c|c|c|c|}\hline\cr\text{size:}&1&21&56&42&24&24\\ \hline\cr\text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr({\,\mathrm{Id}},{\mathbb{C}})&1&1&1&1&1&1\\ \hline\cr(\pi_{1},V)&3&-1&0&1&\frac{-1+\sqrt{-7}}{2}&\frac{-1-\sqrt{-7}}{2}\\ \hline\cr(\pi_{1}^{*},V^{*})&3&-1&0&1&\frac{-1-\sqrt{-7}}{2}&\frac{-1+\sqrt{-7}}{2}\\ \end{array}$$

   We now consider the representations $(\pi_{1}\wedge\pi_{1},\bigwedge^{2}V)$ and $\mathrm{Sym}^{2}\pi,\mathrm{Sym}^{2}V$. Recall that

   $$\chi_{\pi_{1}\wedge\pi_{1}}(g)=\frac{1}{2}(\chi_{\pi_{1}}(g)^{2}-\chi_{\pi_{1}}(g^{2})),\qquad\chi_{\mathrm{Sym}^{2}\pi}(g)=\frac{1}{2}(\chi_{\pi_{1}}(g)^{2}+\chi_{\pi_{1}}(g^{2}))$$

   If $x$ has order two, then $x^{2}=e$, if $x$ has order three, so does $x^{2}$, if $x$ has order four, then $x^{2}$ has order two, and from part (1), we know that if $x$ is in $C_{7A}$ or $C_{7B}$, then so is $x^{2}$. This lets us compute:

   $$\begin{array}[]{c|cccccc}\text{size:}&1&21&56&42&24&24\\ \text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr\pi_{1}\wedge\pi_{1}&3&-1&0&1&\frac{-1-\sqrt{-7}}{2}&\frac{-1+\sqrt{-7}}{2}\\ \hline\cr\mathrm{Sym}^{2}\pi_{1}&6&2&0&0&-1&-1\\ \end{array}$$

   From this we see that $\pi_{1}\wedge\pi_{1}$ is simply $\pi_{1}^{*}$, and that $\mathrm{Sym}^{2}\pi_{1}$ is a new irreducible representation.

   We are now missing two irreducible representations. Recalling the difference of squares rule, we consider the representation $(\pi_{1},V)\otimes(\pi_{1}^{*},V^{*})$, which has character $\chi_{\pi_{1}}\overline{\chi_{\pi_{1}}}$:

   $$\begin{array}[]{c|cccccc}\text{size:}&1&21&56&42&24&24\\ \text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr\pi_{1}\otimes\pi_{1}^{*}&9&1&0&1&2&2\end{array}$$

   Since all the entries in the table are non-negative and the representation has positive dimension, we must have $\langle\chi_{\pi_{1}\otimes\pi_{1}^{*}},\chi_{{\,\mathrm{Id}}}\rangle>0$, so this is not irreducible. However, when we compute this inner product, we get

   $$\langle\chi_{\pi_{1}\otimes\pi_{1}^{*}},\chi_{{\,\mathrm{Id}}}\rangle=\frac{1}{168}\left(9+21+42+24\cdot 2+24\cdot 2\right)=1,$$

   hence

   $$(\pi_{1},V)\otimes(\pi_{1}^{*},V^{*})\cong({\,\mathrm{Id}},{\mathbb{C}})\oplus(\rho,W).$$

   Subtracting off the trivial character from $\chi_{\pi_{1}\otimes\pi_{1}^{*}}$ gives

   $$\begin{array}[]{c|cccccc}\text{size:}&1&21&56&42&24&24\\ \text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr(\rho,W)&8&0&-1&0&1&1\end{array}$$

   Computing the the norm of $\chi_{\rho}$ gives

   $$\|\chi_{\rho}\|^{2}=\frac{1}{168}\big{(}1\cdot 64+56\cdot 1+24\cdot 1+24\cdot 1\big{)}=1,$$

   which shows that $\rho$ is irreducible! We are now missing just one irreducible representation:

   $$\begin{array}[]{|c|c|c|c|c|c|c|}\hline\cr\text{size:}&1&21&56&42&24&24\\ \hline\cr\text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr({\,\mathrm{Id}},{\mathbb{C}})&1&1&1&1&1&1\\ \hline\cr(\pi_{1},V)&3&-1&0&1&\frac{-1+\sqrt{-7}}{2}&\frac{-1-\sqrt{-7}}{2}\\ \hline\cr(\pi_{1}^{*},V^{*})&3&-1&0&1&\frac{-1-\sqrt{-7}}{2}&\frac{-1+\sqrt{-7}}{2}\\ \hline\cr\mathrm{Sym}^{2}\pi_{1}&6&2&0&0&-1&-1\\ \hline\cr(\rho,W)&8&0&-1&0&1&1\\ \end{array}$$

   We then immediately know that this representation $(\sigma,U)$ must have dimension $7$. Column orthogonality then gives

   $$\begin{array}[]{c|cccccc}\text{size:}&1&21&56&42&24&24\\ \text{class}&C_{1}&C_{2}&C_{3}&C_{4}&C_{7A}&C_{7B}\\ \hline\cr(\sigma,U)&7&-1&1&-1&0&0\end{array}$$