Lecture 1

We elaborate on two ways of checking where a map $\pi:G\rightarrow\mathrm{GL}(V)$ is actually a representation. This point was discussed a bit in Lecture 1, but since it will be used frequently throughout the course, I would like to emphasise this a bit further:

1. (i)
   ​

   Assume that a rule or formula is given for $\pi(g)$ for every $g\in G$. In order to verify that $(\pi,V)$ really is a representation, one needs to either check or show that $\pi(gh)=\pi(g)\pi(h)$ for all $g,h\in G$.

2. (ii)
   ​

   Frequently, we will just give the operators $\pi(g_{1}),\ldots\pi(g_{k})$ for a set of generators $g_{1},\ldots,g_{k}$ of a group $G$. Since every element of $G$ can be written as a word in $g_{1},\ldots,g_{k}$, if $\pi$ is required to be a homomorphism, then there is no choice in the values of $\pi(g)$ for the other group elements $g$: $\pi(g)$ must be the product of the $\pi(g_{i})$s corresponding to the decomposition of $g$ into a product of $g_{i}$s. In order for this procedure to give a well-defined homomorphism $\pi$ on all of $G$, it is both necessary and sufficient that the $\pi(g_{i})$s satisfy the same relations as the $g_{i}$s. (This is a general fact about any group homomorphism, not just ones into $\mathrm{GL}(V)$).

Compare the phrasing of problems 12 and 13 below!

Solution: The operators $\pi(\sigma)$ are clearly linear (it is stated in the problem that they are in $\operatorname{GL}_{3}({\mathbb{C}})$), so it remains to check that $\pi$ is a homomorphism. We compute the action of a group product on a vector:

showing that $\pi$ is indeed a homomorphism.

Solution: To check that the operators define a representation of $G$, we simply to need to verify that $\rho(r)$ and $\rho(s)$ satisfy the same relations as $r$ and $s$. We check this directly:

and

For the other group relation, we have

and

which are equal.

Solution:

1. (a)
   ​

   First note that $\pi((1))={\,\mathrm{Id}}\in{\mathbb{C}}^{3}$. We then find the matrices for a generating set of $S_{3}$. We have

   $$\pi((12)){\bf e}_{1}={\bf e}_{2},\ \pi((12)){\bf e}_{2}={\bf e}_{1},\ \pi((12)){\bf e}_{3}={\bf e}_{3},$$

   and

   $$\pi((23)){\bf e}_{1}={\bf e}_{1},\ \pi((23)){\bf e}_{2}={\bf e}_{3},\ \pi((23)){\bf e}_{3}={\bf e}_{2}.$$

   This gives us

   $$\pi((12))=\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}\text{ and }\pi((23))=\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}.$$

   As $\pi$ is a homomorphism we can compute the remaining matrices by taking the product of the matrices corresponding to each generator. Thus

   $$\pi((123))=\pi((12))\pi((23))=\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}=\begin{pmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{pmatrix}.$$

   By a similar argument we get

   $$\pi((132))=\begin{pmatrix}0&1&0\\ 0&0&1\\ 1&0&0\end{pmatrix}\text{ and }\pi((13))=\begin{pmatrix}0&0&1\\ 0&1&0\\ 1&0&0\end{pmatrix}.$$
2. (b)
   ​

   First we note that for any $\sigma\in S_{3}$ we have

   $$\pi(\sigma)({\bf e}_{1}+{\bf e}_{2}+{\bf e}_{3})={\bf e}_{1}+{\bf e}_{2}+{\bf e}_{3},$$

   so we pick ${\bf v}_{1}={\bf e}_{1}+{\bf e}_{2}+{\bf e}_{3}$ to be our first basis element. Furthermore if we have $\pi(\sigma){\bf v}={\bf v}_{1}$ for some $\sigma\in S_{3}$ then ${\bf v}={\bf v}_{1}$ so our matrices will all take the form

   $$\begin{pmatrix}1&0&0\\ 0&?&?\\ 0&?&?\end{pmatrix}.$$

   We then have a choice for our other two basis vectors. Let ${\bf v}_{2}={\bf e}_{1}-{\bf e}_{2}$ and ${\bf v}_{3}={\bf e}_{1}-{\bf e}_{3}$. Then $\{{\bf v}_{1},{\bf v}_{2},{\bf v}_{3}\}$ clearly forms a basis of ${\mathbb{C}}^{3}$ so we just need to compute each of the matrices. We proceed in a similar way to part (a), so

   $$\pi((12))=\begin{pmatrix}1&0&0\\ 0&-1&-1\\ 0&0&1\end{pmatrix}\text{ and }\pi((23))=\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}.$$

   Thus the remaining matrices are

   $$\pi((123))=\begin{pmatrix}1&0&0\\ 0&-1&-1\\ 0&1&0\end{pmatrix}\text{, }\pi((132))=\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&-1&-1\end{pmatrix}\text{ and }\pi((13))=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&-1&-1\end{pmatrix}.$$

Solution: Given vector spaces $V,W$, and any $S\in\operatorname{Hom}(V)$, $T\in\operatorname{Hom}(W)$, the map $S\oplus T:V\oplus W\rightarrow V\oplus W$ defined by

is in $\operatorname{Hom}(V\oplus W)$, since for all $\alpha,\beta\in{\mathbb{C}}$ and $v_{1},v_{2}\in V,w_{1},w_{2}\in W$,

Moreover, if $S\in\operatorname{GL}(V)$ and $T\in\operatorname{GL}(W)$, then $S\oplus T\in\operatorname{GL}(V\oplus W),$ as

Applying this to $\pi\oplus\rho(g)=\pi(g)\oplus\rho(g)$, we have $\pi\oplus\rho(g)\in\operatorname{GL}(V\oplus W),$ so it remains to show that $g\mapsto\pi\oplus\rho(g)$ is a homomorphism. We compute the image of a product: for all $g,h\in G$, $v\in V$, $w\in W$, we have

where we used that $\pi$ and $\rho$ are homomorphisms for the second equality. This shows that $\pi\oplus\rho$ is a homomorphism.

Solution: Since $h\in H<G$, and $\pi:G\rightarrow\operatorname{GL}(V),$ we have that $\pi|_{H}(h)=\pi(h)\in\operatorname{GL}(V)$ for all $h\in H$. It remains to show that $\pi|_{H}$ is a homomorphism. For any $h_{1},h_{2}\in H$, we have

(the third equality holding since $\pi$ is a homomorphism for $G$).

a) Show that $(\pi,V)$ is a representation of $G$, where for every $g\in G$ and $f\in V$, $\pi(g)f\in V$ is defined via the formula

b) Show that $(\widetilde{\pi},V)$ is a not a representation of $G$, where for every $g\in G$ and $f\in V$, $\widetilde{\pi}(g)f\in V$ is defined via the formula

Solution: We first verify that $\pi(g)$ is linear: given $\alpha,\beta\in{\mathbb{C}}$, $f_{1},f_{2}\in V$, and $x\in X$,

Next we show that $\pi$ is a homomorphism: given $g,h\in G$, $f\in V$, and $x\in X$, we have

For part (b), we simply observe that the above calculation gives $\widetilde{\pi}(gh)=\widetilde{\pi}(h)\widetilde{\pi}(g)$.

Prop. 1.11 Let $G$ be a finite group and $(\pi,{\mathbb{C}})$ a representation of $G$. Show that for every $g\in G$, there exists $n_{g}\in\{0,1,2,\ldots|G|-1\}$ such that

Solution: Since $\operatorname{GL}({\mathbb{C}})={\mathbb{C}}^{\times}$, for each $g\in G$, $\pi(g)=z_{g}\in{\mathbb{C}}^{\times}$. By Lagrange’s theorem, $g^{|G|}=e,$ hence

hence $z_{g}$ is a $|G|$-th root of unity, and may therefore be written as $e^{2\pi ij/|G|}$ for some $j\in\{1,\ldots,|G|\}$.

Solution: Let $(\pi,{\mathbb{C}}v)$ be a one-dimensional vector space. Thus $\pi(r)v=\lambda_{r}v$ and $\pi(s)=\lambda_{s}v$ for some $\lambda_{r},\lambda_{s}\in{\mathbb{C}}$. These numbers must satisfy the group relations, i.e.

Since $\lambda_{s}^{2}=1$, we then get that $\lambda_{r}\lambda_{s}\lambda_{r}\lambda_{s}=1\Rightarrow\lambda_{r}^{2}=1$, hence $\lambda_{r},\lambda_{s}\in\{\pm 1\}.$ Thus, for any $(i,j)\in{\mathbb{Z}}_{2}\oplus{\mathbb{Z}}_{2}$, $(\pi_{(i,j)},{\mathbb{C}}v)$ is a representation, where

is a representation. (This gives four isomporhism classes of one-dimensional representations of $D_{4}$.)