Lecture 3

Solution:

1. (a)
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   Let $T\in\operatorname{Hom}_{G}(V,W)$ be invertible. If $v\in V$ is an eigenvector for $\pi(g),$ i.e. $\pi(g)v=\lambda v$, then

   $$\rho(g)Tv=T\pi(g)v=\lambda Tv,$$

   i.e. $\lambda$ is an eigenvalue for $\rho(g)$. This shows that any eigenvalue of $\pi(g)$ is also an eigenvalue of $\rho(g)$. The same argument with $T^{-1}$ in place of $T$ that any eigenvalues of $\rho(g)$ is also an eigenvalue of $\pi(g).$

2. (b)
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   First we note that, by part (a), $({\,\mathrm{Id}},{\mathbb{C}})$ and $(\epsilon,{\mathbb{C}})$ are not isomorphic because the eigenvalues corresponding to $s$, $1$ and $-1$ respectively, are not the same. It is also clear that both 1-dimensional representations are not isomorphic to the 2-dimensional representations. Note that, for each $1\leq k<n/2$, the eigenvalues of $\rho_{k}(r)$ are $2\pi ik/{n}$ and $-2\pi ik/n$, which are distinct for distinct values of $k$. Thus each $\rho_{k}$ is in its own isomorphism class and everything is the table is non-isomorphic.

Solution: We have an additional two 1-dimensional representations so the table is:

The proof that this is a complete list follows the same argument as the proof of Theorem 3.2, but differs slightly in two places. In Case 1 we note that we exclude $k=n/2$ because then we would have $\lambda=\lambda^{-1}$. In Case 2 we have a further two cases: the case when $\lambda=1$, which is the same as in Theorem 3.2, and the case when $\lambda=-1$. Now

but this is still in ${\mathbb{C}}({\bf v}+{\bf w})$ so $({\bf v}+{\bf w})$ spans a subrepresentation of $V$. If ${\bf v}+{\bf w}\neq 0$ then, as $\pi(s)({\bf v}+{\bf w})={\bf v}+{\bf w}$ we have the representation $\pi_{1}$ with $V={\mathbb{C}}({\bf v}+{\bf w})$. Otherwise, $\pi(s){\bf v}={\bf w}=-{\bf v}$, and we get the representation $\pi_{2}$.

The argument that these are all non isomorphic is the same as in Problem 25(b).

Solution: Since $S_{3}\equiv D_{3}$ (see Problem 6), we know that $S_{3}$ has two one-dimensional irreducible representations and one two-dimensional irreducible representation. From class, we saw that $(\pi,W_{0})$ is irreducible, and we know that $({\,\mathrm{Id}},{\mathbb{C}})$ is the one-dimensional trivial representation, so it remains to show that the sign map defines a representation, and also that it is not isomorphic to the trivial representations. Firstly, recall that the sign of a permutation $\sigma$ equals $-1$ raised to the number of transpositions $(i\,j)$ needed to express $\sigma$. In particular, $\mathrm{sgn}(1\,2)=-1$, which is order two, hence $(\mathrm{sgn},{\mathbb{C}})$ cannot be isomorphic to ${\,\mathrm{Id}},{\mathbb{C}})$. It remains to verify that the sign map is a homomorphism; this is a standard fact from group theory: if $\tau$ is written as $N(\tau)$ transpositions and $\sigma$ is written as $N(\sigma)$ transpositions, then $\sigma\tau$ can be written as $N(\sigma)+N(\tau)$ transpositions, hence $\mathrm{sgn}(\tau)\mathrm{sgn}(\sigma)=(-1)^{N(\tau)}(-1)^{N(\sigma)}=(-1)^{N(\tau)+N(\sigma)}=\operatorname{sgn}(\tau\sigma)$.

Solution: Let $(\pi,V)$ be an irreducible representation of $Q_{8}=\{\pm 1,\pm i,\pm j,\pm k\}$. We choose an eigenvector $v$ for $\pi(i)$:

where $\lambda^{4}=1$. Define also $w=\pi(j)v$. We claim defining $U=\mathrm{span}\{v,w\}$ gives a subrepresentation $(\pi,U)$ of $(\pi,V)$. Note that $\{i,j\}$ is a generating set for $Q_{8}$, so to check that $U$ is a subrepresentation, it suffices to check that $\pi(i)w\in U$ and $\pi(j)w\in U$ (since by definition $\pi(i)v,\pi(j)v\in U$). We compute:

Thus $(\pi,U)$ is a subrepresentation of $(\pi,V)$, and hence ($V$ is assumed to be irreducible) $U=V$. We now consider two different cases:

• Case 1:
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  $w\in{\mathbb{C}}v$, $(\pi,V)$ is then one-dimensional. Since $v=\pi(1)v=\pi(-1)^{2}v$, we have two possibilities: either $\pi(-1)v=v$ or $\pi(-1)v=-v$.

  If $\pi(-1)=-1$, then writing $w=\pi(j)v=\mu v$, from the group relations, we must have

  $$-1=\pi(-1)=\pi(i)^{2}=\lambda^{2}=\pi(j)^{2}=\mu^{2}=\pi(k)^{2}=\pi(ij)^{2}=\lambda^{2}\mu^{2},$$

  i.e. $-1=\lambda^{2}=\mu^{2}=\lambda^{2}\mu^{2}$, which is not possible! Hence, we must have $\pi(-1)=1$. This gives $\lambda,\mu\in\{\pm 1\}$, and one sees that each of these four cases satisfies the above equation, giving four isomorphism classes of one-dimensional representations:
  

  Rep	$\pi(\pm 1)$	$\pi(\pm i)$	$\pi(\pm j)$	$\pi(\pm k)$
  $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1	1
  $(\pi_{i},{\mathbb{C}})$	1	1	-1	-1
  $(\pi_{j},{\mathbb{C}})$	1	-1	1	-1
  $(\pi_{k},{\mathbb{C}})$	1	-1	-1	1

• Case 2:
  ​

  $w\not\in{\mathbb{C}}v$. Writing down the matrices of $\pi(i)$ and $\pi(j)$ with respect to the basis $v,w$ gives

  $$\pi(i)=\left(\begin{matrix}\lambda&0\\ 0&\lambda^{-1}\end{matrix}\right),\quad\pi(j)=\left(\begin{matrix}0&1\\ \lambda^{2}&0\end{matrix}\right).$$

  Since $\pi(i)^{4}={\,\mathrm{Id}}$, we see that $\lambda$ must be a fourth root of unity. If $\lambda=\pm 1,$ then $\pi(i)=\pm{\,\mathrm{Id}}$, which commutes with $\pi(j)$, allowing us to simultaneously diagonalise the matrices, i.e decomposing $(\pi,V)$ as $(\pi,{\mathbb{C}}(u+v))\oplus(\pi,{\mathbb{C}}(u-v))$. This contradicts the assumption that $V$ was irreducible! Hence $\lambda=\pm i$; both choices give representations that are isomorphic to each other (with the isomorphism given by swapping the basis elements). Moreover, they are irreducible, as the two distinct eigenvalues of $\pi(i)$ do not agree with any of those for the one-dimensional representations. Thus, $Q_{8}$ has a single two-dimensional irreducible representation with operators as follows:
  

  Rep	$\pi(\pm 1)$	$\pi(\pm i)$	$\pi(\pm j)$	$\pi(\pm k)$
  $(\pi,{\mathbb{C}}^{2})$	$\pm{\,\mathrm{Id}}$	$\pm\left(\begin{matrix}i&0\\ 0&-i\end{matrix}\right)$	$\pm\left(\begin{matrix}0&1\\ -1&0\end{matrix}\right)$	$\pm\left(\begin{matrix}0&-i\\ i&0\end{matrix}\right)$

  (here the “$\pm$” signs match, i.e. $\pi(-1)=-{\,\mathrm{Id}}$, etc.).