1 Representation theory of finite groups 1.1 Definitions and first examples 1.3 Example: dihedral groups

1.2 Formalities

1.2.1 Subrepresentations and irreducibility

Definition 1.14.

A subrepresentation of a representation $(\rho,V)$ of $G$ is a subspace $W\subset V$ such that $\rho(g)w\in W$ for all $w\in W$ .

Then $(\rho_{W},W)$ is also a representation of $G,$ where $\rho_{W}(g)w=\rho(g)w$ .

This has a block matrix interpretation: if we choose a basis for $W$ and extend it to a basis for $V$ , then the matrix of $\rho(g)$ will have the following block matrix form:

\left(\begin{array}[]{c|c}\rho_{W}(g)&\star\\\hline\cr\mathbf{0}&\rho_{V/W}(g).\end{array}\right).

Here $\rho_{V/W}$ is also a representation of $G$ , the quotient representation.

Definition 1.15.

A representation $V$ of $G$ is irreducible if it is nonzero and has no subrepresentations except for $\{0\}$ and $V$ itself.

Example 1.16.

Let $V$ be the permutation representation of $S_{4}$ acting on the faces of the cube. Remember that we can see a vector in $V$ as a way of writing a complex number on each face. Then we can spot three subrepresentations:

•

the subrepresentation where the numbers on all faces are equal
•

the subrepresentation where the numbers on opposite faces sum to zero
•

the subrepresentation where the numbers on opposite faces are equal and all the labels sum to zero.

Note that in lectures we did the related example of edges of the tetrahedron.

Definition 1.17.

Suppose that $(\rho,V)$ is a representation of $G$ and that $W_{1},W_{2}$ are two subrepresentations such that every element of $V$ can be written uniquely as the sum of an element of $W_{1}$ and an element of $W_{2}$ . Then we say that $V$ is the (internal) direct sum of $W_{1}$ and $W_{2}$ and write $V=W_{1}\oplus W_{2}$ .

This also has a block matrix interpretation: if we choose bases for $W_{1}$ and $W_{2}$ , then the matrix of $\rho(g)$ will have the following block matrix form:

\left(\begin{array}[]{c|c}\rho_{1}(g)&\mathbf{0}\\\hline\cr\mathbf{0}&\rho_{2}(g).\end{array}\right).

We can generalize this definition to a finite number of subrepresentations $W_{1},\ldots,W_{k}$ . If $V$ is the direct sum of subrepresentations that are irreducible, then we will say that $V$ is decomposable. We will see that complex representations of finite groups are always decomposable.

Example 1.18.

Consider the permutation representation of $S_{3}$ . This is the representation on $V=\mathbb{C}^{3}$ for which $\sigma(e_{i})=e_{\sigma(i)}$ where $e_{1},e_{2},e_{3}$ are the standard basis vectors. It is not irreducible!

Let $W_{0}=\left\langle e_{1}+e_{2}+e_{3}\right\rangle$ . Then $\sigma(e_{1}+e_{2}+e_{3})=e_{1}+e_{2}+e_{3}$ for all $\sigma,$ so $W_{0}$ is a subrepresentation (and the action of $G$ is trivial).

Let $W_{1}=\{(x,y,z):x+y+z=0.\}$ . Then since $S_{3}$ permutes the coordinates of a vector, it doesn’t change their sum; so $\sigma(W_{1})\subset W_{1},$ and $W_{1}$ is a subrepresentation (of dimension two). Since $W_{0}\cap W_{1}=\{0\}$ and their dimensions add to $\dim V=3,$ we have $V=W_{0}\oplus W_{1}$ .

I claim that $W_{1}$ is irreducible. Indeed, suppose $U\subset W_{1}$ is a nonzero subrepresentation; we have to show that $U=W_{1}$ . Let $(x,y,z)\in U$ be nonzero. As $x=y=z$ can’t happen, we can apply an element of $G$ to permute the coordinates so that $x\neq y$ . Then applying $(12),$ we have $(y,x,z)\in U$ . Taking the difference, $(x-y,y-x,0)\in U$ ; scaling, $(1,-1,0)\in U$ . Applying $(23),$ we have $(1,0,-1)\in U$ . But these vectors span $W_{1}$ (e.g. because they are linearly independent and $\dim W_{1}=2$ ), so $U=W_{1}$ as required.

Example 1.19.

If $(\rho,V)$ is a finite-dimensional representation of $G=\mathbb{Z},$ with $T=\rho(1),$ then $T$ has an eigenvector $v$ which spans a one-dimensional subrepresentation of $V$ . Thus $V$ is reducible unless $\dim V=1$ . The irreducible subrepresentations of $V$ are exactly the lines spanned by $T$ -eigenvectors, and $V$ is decomposable if and only if it has a basis of $T$ -eigenvectors. This is equivalent to $T$ being diagonalizable. As you know from Algebra II (or Linear Algebra I?) this is not always the case! For example,

T=\begin{pmatrix}1&1\\0&1\end{pmatrix}.

The next definition is optional: we will avoid using quotient representations.

Definition 1.20.

Suppose that $(\rho,V)$ is a representation of $G$ and that $W\subset V$ is a subrepresentation. Then the quotient vector space $V/W$ has a representation $\rho_{V/W}$ of $G$ defined by

\rho_{V/W}(g)\overline{v}=\overline{\rho(g)v}.

Remember that $\overline{v}$ is the coset $v+W$ . You should check that this is well-defined, i.e. that if $\overline{v}=\overline{v}_{1}$ then $\overline{\rho(g)v}=\overline{\rho(g)v_{1}}$ .

1.2.2 Homomorphisms and isomorphisms

Definition 1.21.

Suppose that $(\rho,V)$ and $(\sigma,W)$ are representations of $G$ . Then a $G$ -homomorphism (or homomorphism of representations of $G,$ or map of representations of $G,$ or if we are being lazy just a homomorphism) $V\rightarrow W$ is a linear map $\phi:V\rightarrow W$ such that

\phi(\rho(g)v)=\sigma(g)\phi(v)

for all $v\in V,$ $g\in G$ .

In other words, $\phi$ ’commutes’ with the action of $G$ : $g\phi(v)=\phi(gv)$ . We write $\operatorname{Hom}_{G}(V,W)$ for the (vector space) of $G$ -homomorphisms from $V$ to $W$ .

There is another word that is sometimes used for $G$ -homomorphism: intertwiner, or $G$ -intertwiner.

Definition 1.22.

A $G$ -isomorphism (or just an isomorphism) is a bijective $G$ -homomorphism.

If there is a $G$ -isomorphism $V\rightarrow W$ then we write $V\cong W$ .

Lemma 1.23.

Suppose that $V$ and $W$ are representations of $G$ .

1.

If $T\in\operatorname{Hom}_{G}(V,W)$ is an isomorphism, then $T^{-1}\in\operatorname{Hom}_{G}(W,V)$ .
2.

Suppose $\dim V=\dim W=n,$ and choose bases for them, so that $\rho_{V}(g)$ and $\rho_{W}(g)$ are (invertible) $n\times n$ matrices. Then $\rho_{V}\cong\rho_{W}$ if and only if there exists $T\in\operatorname{GL}_{n}(\mathbb{C})$ such that

$T\rho_{V}(g)T^{-1}=\rho_{W}(g)$

for all $g\in G$ .

Proof.

Exercise. ∎

Lemma 1.24.

Given $\phi\in\operatorname{Hom}_{G}(V,W),$ then $\ker(\phi)\subset V$ and $\operatorname{im}(\phi)\subset W$ are subrepresentations.

Proof.

We know that they are subspaces, so we just have to show that they are preserved by the action of $G$ .

For the kernel: suppose that $v\in\ker(\phi)$ and $g\in G$ . Then $\phi(v)=0,$ and

\phi(g\cdot v)=g\cdot\phi(v)=g\cdot 0=0,

so $g\cdot v\in\ker(\phi)$ . So $\ker(\phi)$ is $G$ -stable as required.

For the image: suppose $w\in\operatorname{im}(\phi)$ . Then $w=\phi(v)$ for some $v\in V$ . Then

g\cdot w=g\cdot\phi(v)=\phi(g\cdot v)

is also in the image of $\phi$ . ∎

Example 1.25.

Let $G=D_{3}\cong S_{3}$ where the isomorphism takes $r\mapsto(123)$ and $s\mapsto(23)$ . Let $(\rho,V)$ be the permutation representation of $G,$ so in the basis $e_{1},e_{2},e_{3}$ we have

\rho(r)=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},\rho(s)=\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}.

Then let $G$ act on the equilateral triangle centred at 0 with vertex at $(0,1),$ giving a two dimensional representation $\sigma$ on $W=\mathbb{C}^{2}$ such that

\sigma(r)=\begin{pmatrix}1/2&-\sqrt{3}/2\\\sqrt{3}/2&1/2\end{pmatrix},\sigma(s)=\begin{pmatrix}-1&0\\0&1\end{pmatrix}.

Let $v_{1},$ $v_{2},$ $v_{3}$ be the vertices labeled anticlockwise from the top and let $T:V\rightarrow W$ be the linear map taking $e_{i}$ to $v_{i}$ . I claim that $T$ is a $G$ -homomorphism.

The matrix of $T$ is

\begin{pmatrix}0&-\sqrt{3}/2&\sqrt{3}/2\\1&-1/2&-1/2\end{pmatrix}

and now we just have to check that

\begin{pmatrix}0&-\sqrt{3}/2&\sqrt{3}/2\\1&-1/2&-1/2\end{pmatrix}\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}=\begin{pmatrix}-1/2&-\sqrt{3}/2\\\sqrt{3}/2&-1/2\end{pmatrix}\begin{pmatrix}0&-\sqrt{3}/2&\sqrt{3}/2\\1&-1/2&-1/2\end{pmatrix},

which is true, and a similar equation coming from $s$ .

In fact we could see this without calculation; by the way we defined the isomorphism $D_{3}\cong S_{3},$ we have that

\sigma(g)T(e_{i})=\sigma(g)v_{i}=v_{g\cdot i}=T(e_{g\cdot i})=T(\rho(g)e_{i})

for all $g\in S_{3}$ .

The kernel of the homomorphism $T$ is the subspace spanned by $e_{1}+e_{2}+e_{3},$ and in fact $T$ defines an isomorphism from the subrepresentation

\{(a,b,c)\in V:a+b+c=0\}\subset V

to $W$ .

Remark 1.26.

If $V_{1}$ and $V_{2}$ are representations of $G$ then we may form their (external) direct sum with underlying vector space $V_{1}\oplus V_{2}$ such that

g(v_{1},v_{2})=(gv_{1},gv_{2})

for all $g\in G,$ $v_{1}\in V_{1},$ $v_{2}\in V_{2}$ . If $V_{1}$ and $V_{2}$ happen to be subrepresentations of some other representation $V,$ then saying $V$ is the direct sum of $V_{1}$ and $V_{2}$ is the same as saying that the map

	$\displaystyle V_{1}\oplus V_{2}$	$\displaystyle\rightarrow V$
	$\displaystyle(v_{1},v_{2})$	$\displaystyle\mapsto v_{1}+v_{2}$

is an isomorphism.

1.2.3 Change of group

We haven’t yet covered this section; we will come back to it.

Definition 1.27.

If $H$ is a subgroup of $G$ and $(\rho,V)$ is a representation of $G,$ then we get a representation $(\rho|_{H},V),$ the restriction of $\rho$ to $G,$ which is given by

\rho|_{H}(h)=\rho(h)

for $h\in H$ .

This representation is also written $V|_{H}$ , $\operatorname{Res}^{G}_{H}\rho$ , or $\operatorname{Res}^{G}_{H}V$ .

If $\rho$ is an irreducible representation of $G$ , then $\rho|_{H}$ need not be. As an extreme example, if $H=\{e\}$ then $\rho|_{H}$ is just $\dim(V)$ copies of the trivial representation!

Example 1.28.

Let $\rho$ be the irreducible two-dimensional representation of $S_{3},$ realised as the space

V=\{(x,y,z):x+y+z=0\}

of the permutation representation. Let $H=\left\langle 123\right\rangle\subset S_{3}$ . Then $\rho|_{H}$ is reducible (as it must be: all irreducible representations of $H\cong C_{3}$ are one-dimensional). Indeed, let $v=(1,\omega,\omega^{2})$ and $w=(1,\omega^{2},\omega)$ where $\omega=e^{2\pi i/3}$ . Then $\left\langle v\right\rangle$ and $\left\langle w\right\rangle$ are $H$ -subrepresentations of $V$ with $V=\left\langle v\right\rangle\oplus\left\langle w\right\rangle$ .

The action of $H$ on $\left\langle v\right\rangle$ is through the character taking $(123)$ to $\omega^{-1},$ and the action on $\left\langle w\right\rangle$ is through the character $(123)\mapsto\omega$ .

Exercise: what is the restriction of $\rho$ to $\{e,(12)\}$ ?

If $K$ is a normal subgroup of $G$ , then $G/K$ is a group. If $(\rho,V)$ is a representation of $G/K$ then we define the lift (or inflation) $\tilde{\rho}$ of $\rho$ to be the homomorphism $G\rightarrow\operatorname{GL}(V)$ defined by $\tilde{\rho}(g)=\rho(gK)$ :

\tilde{\rho}:G\twoheadrightarrow G/K\xrightarrow{\rho}\operatorname{GL}(V).

Then $(\tilde{\rho},V)$ is a representation of $G$ .

Example 1.29.

Take $G=S_{4}$ and take $K=\{e,(12)(34),(13)(24),(14)(23)\}$ . Then there is an isomorphism

G/K\cong S_{3}

defined as follows: label the three non-identity elements of $K$ as $\{a,b,c\}$ . Then $G$ acts by conjugation on this set of three elements, giving us a homomorphism $G\to S_{3}$ . Explicitly, we have

(12)\mapsto(bc)

and

(123)\mapsto(acb).

Then the homomorphism is surjective with kernel $K$ (check these statements!) and so defines an isomorphism

G/K\cong S_{3}.

If $(\rho,V)$ is any representation of $G$ , then its kernel is

\ker(\rho)=\{g\in G:\rho(g)=I\}.

This is just the kernel of the homomorphism $\rho:G\rightarrow\operatorname{GL}(V)$ , and so it is a normal subgroup. If the kernel of $\rho$ is trivial, then we say that $\rho$ is faithful. In this case, $\rho$ determines an embedding of $G$ inside $\operatorname{GL}_{n}(\mathbb{C})$ !

Lemma 1.30.

If $K\subset G$ is a normal subgroup, and $\rho$ is a representation of $G,$ then the following are equivalent:

1.

$\ker(\rho)$ contains $K$ ;
2.

$\rho$ is isomorphic to the inflation of a representation of $G/K$ .

Proof.

Exercise! ∎

Corollary 1.31.

If $\rho$ is a representation of $G/K$ then $\tilde{\rho}$ is irreducible if and only if $\rho$ is.

Proof.

We prove the equivalent statement that $\tilde{\rho}$ is reducible if and only if $\rho$ is. If $\rho$ is reducible, so has a nonzero proper subrepresentation $W,$ then so is $\tilde{\rho}$ since $W$ is also a subrepresentation of $\tilde{\rho}$ . Conversely, if $\tilde{\rho}$ has a nonzero proper subrepresentation $\sigma,$ then $K\subset\ker(\rho)\subset\ker(\sigma)$ and so, by the lemma, $\sigma=\tilde{\rho}_{1}$ for a representation $\rho_{1}$ of $G/K$ that is then a nonzero proper subrepresentation of $\rho$ . ∎