Representation Theory IV 2 Representation theory of finite groups 4 Induced representations

3 Character theory

3.1 Characters

Throughout this section, $G$ is a finite group and $V$ is a finite dimensional complex vector space.

Definition 3.1.

Let $(\rho,V)$ be a finite-dimensional complex representation of $G$ . The character of $(\rho,V)$ is the function $\chi_{\rho}:G\rightarrow\mathbb{C}$ defined by

\chi_{\rho}(g)=\operatorname{tr}(\rho(g)).

We might also write $\chi_{V}$ instead of $\chi_{\rho}$ .

Remark 3.2.

It is not totally obvious that the process ‘Start with an element $\rho(g)\in\operatorname{GL}(V)$ , choose a basis for $V$ , write $\rho(g)$ as a matrix using that basis, and take the trace’ gives a number that is independent of the choice of basis. One way to see this is to use the change of basis formula and the fact that $\operatorname{tr}PMP^{-1}=\operatorname{tr}M$ .

The same argument shows:

Lemma 3.3.

Isomorphic representations have the same character.

It seems that we throw away a lot of information when we pass to the character; the remarkable thing is that, in fact, the character completely determines the representation. Moreover, there is a lot of structure to the characters and it is often possible to find all of the characters of a group, even when it is not clear how to construct the representations!

Example 3.4.

If $\chi:G\rightarrow\mathbb{C}^{\times}$ is a one-dimensional representation, then it is its own character (the trace of a scalar is just itself).

Example 3.5.

Let $\rho$ be the irreducible two-dimensional representation of $S_{3}$ and let $\chi$ be its character. Then

\chi(e)=2\qquad\chi((12))=\chi((23))=\chi((31))=0\qquad\chi((123))=\chi((132))% =-1.

Lemma 3.6.

If $(\rho,V)$ is a representation of $G$ , then $\chi_{\rho}(e)=\dim\rho$ .

Proof.

Exercise. ∎

We call $\dim\rho$ the degree (or the dimension) of the character $\chi_{\rho}$ .

Lemma 3.7.

If $V$ and $W$ are representations with characters $\chi$ and $\psi$ , then $V\oplus W$ has character $\chi+\psi$ .

Proof.

Use block matrices. ∎

Lemma 3.8.

If $\chi$ is the character of a representation $V$ of $G$ , then

\chi(g^{-1})=\overline{\chi(g)}.

Proof.

As $g$ has finite order, say $m$ , we can find a basis such that $\rho(g)$ is diagonal with eigenvalues $\lambda_{1},\ldots,\lambda_{n}$ , and the $\lambda_{i}$ are $m$th roots of unity. Then $\rho(g)^{-1}$ is diagonal with eigenvalues $\lambda^{-1}_{1},\ldots,\lambda^{-1}_{n}$ , which are $\overline{\lambda}_{1},\ldots,\overline{\lambda}_{n}$ , giving the result. ∎

Remark 3.9.

It follows that, if $g$ is conjugate to $g^{-1}$ , then $\chi(g)$ is real for every character $\chi$ .

Lemma 3.10.

Let $\chi$ be the character of a representation $\rho$ of $G$ . If $g$ and $h$ are in the same conjugacy class of $G$ , then

\chi(g)=\chi(h).

Proof.

Let $h=xgx^{-1}$ for some $x\in G$ . Then

\chi(h)=\operatorname{tr}\rho(xgx^{-1})=\operatorname{tr}\left(\rho(x)\rho(g)% \rho(x)^{-1}\right)=\operatorname{tr}\rho(g)=\chi(g)

since conjugate matrices have the same trace. ∎

Definition 3.11.

A class function is a function $G\rightarrow\mathbb{C}$ that is constant on conjugacy classes.

The previous lemma then says that the character of a representation is a class function. We often organise the information into a character table. This has columns labeled by the conjugacy classes of $G$ , and rows labeled by the irreducible representations. The entries are the values of the characters of the irreducible representations on elements of the conjugacy class.

We usually write $\mathbbm{1}$ for the character of the trivial representation.

Example 3.12.

Here is the character table of $S_{3}$ . We label each column by a representative element of the conjugacy class. It is also common, as here, to write the number of elements in the conjugacy class in the second row.

\begin{array}[h]{c|rrr}\text{Class}&e&(12)&(123)\\ \text{size}&1&3&2\\ \hline\cr\mathbf{1}&1&1&1\\ \epsilon&1&-1&1\\ \rho&2&0&-1\end{array}

Example 3.13.

Let $G=C_{n}$ and $\omega=e^{2\pi i/n}$ . Then we can write down the character table of $C_{n}$ ; I will do this for $n=5$ for concreteness. Since $G$ is abelian, all conjugacy classes are singletons so I will omit the second row.

\begin{array}[h]{c|rrrrrrr}\text{Class}&e&g&g^{2}&g^{3}&g^{4}\\ \hline\cr\mathbbm{1}&1&1&1&1&1\\ \chi&1&\omega&\omega^{2}&\omega^{3}&\omega^{4}\\ \chi^{2}&1&\omega^{2}&\omega^{4}&\omega&\omega^{3}\\ \chi^{3}&1&\omega^{3}&\omega&\omega^{4}&\omega^{2}\\ \chi^{4}&1&\omega^{4}&\omega^{3}&\omega^{2}&\omega\end{array}

Exercise 3.14.

Let $G$ act on a finite set $X$ , and let $\rho$ be the permutation representation. Then its character $\chi$ is given by

\chi(g)=|\mathrm{Fix}(g)|,

the number of fixed points of $g$ .

Exercise 3.15.

What is the character of the regular representation?

3.2 Orthogonality of characters

Definition 3.16.

If $\chi$ and $\psi$ are two class functions on $G$ , then their inner product is

\left\langle\chi,\psi\right\rangle=\frac{1}{|G|}\sum_{g\in G}\overline{\chi(g)% }\psi(g).

You could define this for any old functions on $G$ , but we will only use it for class functions. Note that this is a Hermitian inner product on the space of class functions of $G$ , whose dimension is the number of conjugacy classes of $G$ .

Over the next few section, we will prove the following theorem-

Theorem 3.17.

Let $G$ be a finite group with conjugacy classes $\mathcal{C}_{1},\ldots,\mathcal{C}_{r}$ . If $\chi$ is a class function, write $\chi(\mathcal{C}_{i})=\chi(g)$ for any $g\in\mathcal{C}_{i}$ .

1.

The irreducible characters are orthonormal with respect to the inner product defined above. This means that, if $\chi$ and $\psi$ are irreducible characters, then

$\frac{1}{|G|}\sum_{g\in G}\overline{\chi(g)}\psi(g)=\begin{cases}1&\text{if $% \chi=\psi$}\\ 0&\text{otherwise.}\end{cases}$

We can also see this inner product as the standard inner product of the rows of the character table with the entries weighted by $\sqrt{\frac{|\mathcal{C}_{i}|}{|G|}}$ :

$\left\langle\chi,\psi\right\rangle=\frac{1}{|G|}\sum_{i=1}^{r}|\mathcal{C}_{i}% |\overline{\chi(\mathcal{C}_{i})}\psi(\mathcal{C}_{i}).$
2.

The number of irreducible representations is equal to the number of conjugacy classes. In other words, the character table is square.
3.

The columns of the character table are orthonormal with respect to the weighted inner product from part₍₁₎. That is, if $\mathcal{C}_{i}$ , $\mathcal{C}_{j}$ are distinct conjugacy classes, then

$\sum_{\chi}\overline{\chi(\mathcal{C}_{i})}\chi(\mathcal{C}_{j})=0$

while if they are the same conjugacy class then

$\sum_{\chi}\overline{\chi(\mathcal{C}_{i})}\chi(\mathcal{C}_{i})=\frac{|G|}{|% \mathcal{C}_{i}|}.$

To complement this, we have:

Theorem 3.18.

Two irreducible representations of $G$ are isomorphic if and only if they have the same character.

In other words, each row of the character table corresponds to exactly one isomorphism class of irreducible representation.

Theorem 3.19.

1.

A representation is determined up to isomorphism by its character.
2.

If $V$ is a representation with character $\chi$ , then $V$ is irreducible if and only if $\left\langle\chi,\chi\right\rangle=1$ .

Proof.

Let the irreducible representations be $V_{1},\ldots,V_{r}$ and let $\chi_{1},\ldots,\chi_{r}$ be their characters (distinct, by Theorem 3.18 above). By Maschke’s Theorem, any representation $V$ with character $\chi$ can be written as $\bigoplus V_{i}^{a_{i}}$ for some integers $a_{i}\geq 0$ . Its character is then $\bigoplus a_{i}\chi_{i}$ . We then have, for each $i$ ,

\left\langle\chi,\chi_{i}\right\rangle=\left\langle\sum a_{i}\chi_{i},\chi_{i}% \right\rangle=a_{i}\left\langle\chi_{i},\chi_{i}\right\rangle=a_{i}

since $\left\langle\chi_{j},\chi_{i}\right\rangle=0$ for all $j\neq i$ . Thus the integers $a_{i}$ are determined by $c h i$ , the character of $V$ . But the integers $a_{i}$ determine $V$ up to isomorphism so we are done.

For the second part, note that

\left\langle\chi,\chi\right\rangle=\sum a_{i}^{2}.

Since the $a_{i}$ are nonnegative integers, this holds if and only if exactly one of the $a_{i}$ is equal to one and the rest are zero. This means that $V$ is irreducible. ∎

3.3 Example: S₄

Example 3.20.

We determine the character table of $S_{4}$ . First, we have the trivial representation $\mathbbm{1}$ and the sign character $\epsilon$ :

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\mathbbm{1}&1&1&1&1&1\\ \epsilon&1&-1&1&1&-1.\end{array}

Next, we consider the permutation representation $V$ of $S_{4}$ on $\{1,2,3,4\}$ . This contains a copy of the trivial representation, so we write $V=\mathbbm{1}\oplus W$ for some representation $W$ , whose character $\chi$ satisfies

\chi(g)=\#\{\text{fixed points of $g$}\}-1.

Notice that we also have an operation on representations known as twisting: if $(\rho,V)$ is an irreducible representation of $G$ with character $\chi$ and $\psi$ is a one-dimensional character of $G$ , then we can define a new representation, $\rho\psi$ , of $G$ on $V$ by the formula $(\rho\psi)(g)=\rho(g)\psi(g)$ . It has character $\chi\psi$ . In this case, we can look at $\chi\epsilon$ and see that it will also be the character of an irreducible representation.

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\chi&3&1&-1&0&-1\\ \chi\epsilon&3&-1&-1&0&1\\ \end{array}

Since $\left\langle\chi,\chi\right\rangle=\frac{1}{24}(3^{2}+6\times 1^{2}+3\times(-1% )^{2}+8\times 0^{2}+6\times(-1)^{2})=1$ , $\chi$ is irreducible. Note also that $\chi\otimes\epsilon$ is different to $\chi$ , and also irreducible.

There is one more row of the table to find. This can be done using orthonormality with the first column, remembering that the dimension must be a positive integer (in this case, it must be two): we obtain that the full character table is

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\mathbbm{1}&1&1&1&1&1\\ \epsilon&1&-1&1&1&-1\\ \chi&3&1&-1&0&-1\\ \chi\epsilon&3&-1&-1&0&1\\ \psi&2&0&2&-1&0\end{array}

Notice that we constructed the character of the final representation without constructing the representation itself!

3.3.1 Characters of lifts

Recall that if $K\subset G$ is a normal subgroup then we can lift representations of $G/K$ to representations of $G$ , as in Section 2.2.3.

Lemma 3.21.

If $\chi$ is the character of a representation of $G/K$ and $\tilde{\chi}$ is the character of its lift to $G$ , then

\tilde{\chi}(g)=\chi(gK)

for all $g\in G$ .

Proof.

Immediate from the definition. ∎

Example 3.22.

In exercise 2.29, we explained how to lift the irreducible two-dimensional representation of $S_{3}$ to a representation of $S_{4}$ . Then its character agrees with that called $\psi$ in the character table of $S_{4}$ .

We can recognise the kernel of a representation using its character.

Proposition 3.23.

Let $\rho$ be a representation of $G$ with character $\chi$ and dimension $d$ . Then

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

Proof.

It is clear that $g\in\ker(\rho)$ then $\chi(g)=\operatorname{tr}I=d$ . We leave the converse as an exercise. ∎

3.3.2 Decomposing a representation

Example 3.24.

We apply the method of Theorem 3.19 to decompose a naturally occurring representation of $S_{4}$ . Let

X=\{S\subset\{1,2,3,4\}:|S|=2\}.

Then $S_{4}$ acts on $X$ . If we think of $S_{4}$ as the symmetry group of the tetrahedron with vertices labeled 1, 2, 3, 4, then this is ‘the same as’ the action on the set of edges (each edge being determined by a set of two vertices).

Let $(\rho,V)$ be the permutation representation obtained from $X$ and let $\chi_{\rho}$ be its character. Then

\chi_{\rho}(g)=|X^{g}|=|\left\{\{x,y\}\subset\{1,2,3,4\}:\{gx,gy\}=\{x,y\},x% \neq y\right\}|.

We have:

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\chi_{\rho}&6&2&2&0&0\end{array}

Then we compute $\left\langle\chi_{\rho},\mathbf{1}\right\rangle=\left\langle\chi_{\rho},\chi% \right\rangle=\left\langle\chi_{\rho},\psi\right\rangle=1$ , $\left\langle\chi_{\rho},\epsilon\right\rangle=\left\langle\chi_{\rho},\chi% \epsilon\right\rangle=0$ . It follows that

\chi_{\rho}=\mathbbm{1}+\chi+\psi.

\rho\cong W_{0}\oplus W_{1}\oplus W_{2}

where $W_{0}=\mathbb{C}$ is the trivial representation, $W_{1}$ is isomorphic to the 3-D representation of $S_{4}$ with character $\chi$ , and $W_{2}$ is isomorphic to the unique 2-D representation of $S_{4}$ (which has character $\psi$ ).

Exercise 3.25.

In the previous example, find $W_{0}$ , $W_{1}$ , $W_{2}$ as subspaces of $V$ .

Remark 3.26.

The problem of finding irreducible representations of $G$ such that $V$ is isomorphic to their direct sum is called “decomposing $V$ into irreducibles representations (or irreps, or irreducibles)”. It can be solved with character theory as in example 3.24.

The problem of finding irreducible subrepresentations of $V$ such that it is their (internal) direct sum is called “decomposing $V$ into irreducible subrepresentations” and can require more understanding of the nature of the representation $V$ . Exercise 3.25 is an example of this.

3.4 Inner products and homomorphisms

In this and the next section we complete the proof of character orthogonality!

If $(\rho,V)$ and $(\sigma,W)$ are two complex representations of a finite group $G$ , then we can define a representation $\operatorname{Hom}(\rho,\sigma)$ on the vector space

\operatorname{Hom}(V,W)=\{\text{linear maps $T:V\rightarrow W$}\}

with $G$ -action ’by conjugation’:

(g\cdot T)(v)=\sigma(g)T(\rho(g)^{-1}v).

Lemma 3.27.

1.

If $\rho$ and $\sigma$ have characters $\chi$ and $\psi$ , then $\operatorname{Hom}(\rho,\sigma)$ has character $\overline{\chi}\sigma$ .
2.

We have

$\operatorname{Hom}_{G}(V,W)=\operatorname{Hom}(V,W)^{G};$

in other words, the $G$ -homomorphisms are the $G$ -fixed points of $\operatorname{Hom}(V,W)$ .

Proof.

1.

Let $g\in G$ and let $v_{1},\ldots,v_{n}$ and $w_{1},\ldots,w_{m}$ be bases of $V$ and $W$ consisting of eigenvectors for $\rho(g)$ and $\sigma(g)$ , with $\lambda_{i}$ the eigenvalue attached to $v_{i}$ and $\mu_{i}$ that attached to $w_{i}$ . Let $T_{ij}\in\operatorname{Hom}(V,W)$ send $v_{i}$ to $w_{j}$ and all other basis vectors to zero; in matrix terms, this is the matrix with 1 in column $i$ , row $j$ , and 0 everywhere else. Then the $T_{ij}$ are a basis for $\operatorname{Hom}(V,W)$ . Moreover, recalling that the $\lambda_{i}$ are roots of unity, we have

$gT_{ij}=\lambda_{i}^{-1}\mu_{j}T_{ij}=\overline{\lambda_{i}}\mu_{j}T_{ij}.$

So the trace of $\operatorname{Hom}(\rho,\sigma)(g)$ is

$\sum_{i,j}\overline{\lambda_{i}}\mu_{j}=\overline{\sum_{i}\lambda_{i}}\sum_{j}% \mu_{j}=\overline{\chi}(g)\psi(g)$

as required.
2.

If $T\in\operatorname{Hom}(V,W)$ , then $gT=T$ means

$\sigma(g)T(\rho(g)^{-1}v)=T(v)$

for all $v$ . Replacing $v$ by $w=\rho(g)v$ , we get

$\sigma(g)T(w)=T(\rho(g)w)$

for all $w\in W$ . So $T\in\operatorname{Hom}(V,W)^{G}$ if and only if this holds for all $g$ and all $w$ , i.e. if and only if $T$ is a $G$ -homomorphism.∎

The key theoretical result is:

Theorem 3.28.

If $V$ and $W$ are two representations of $G$ with characters $\chi$ and $\psi$ respectively, then

\left\langle\chi,\psi\right\rangle=\dim\operatorname{Hom}_{G}(V,W).

Proof.

Step 1: We first consider the case that $(\rho,V)$ is the trivial representation. Then we have to prove that

\dim W^{G}=\frac{1}{|G|}\sum_{g\in G}\psi(g).

Note that the operator

\pi=\frac{1}{|G|}\sum_{g\in G}\rho(g)

maps $W$ to $W^{G}$ and is the identity on $W^{G}$ . This means that $\pi$ must be a projection onto $W^{G}$ . It follows that, in some basis, its matrix is diagonal with $\dim W^{G}$ ‘1’s and the remaining entries ‘0’,¹¹ 1 See also problem [[prob-projection]]. so

	$\displaystyle\dim W^{G}$	$\displaystyle=\operatorname{tr}(\pi)$
		$\displaystyle=\operatorname{tr}\left(\frac{1}{\|G\|}\sum_{g\in G}\rho(g)\right)$
		$\displaystyle=\frac{1}{\|G\|}\sum_{g\in G}\psi(g)$

as required.

Step 2: Note that $\operatorname{Hom}_{G}(V,W)=\operatorname{Hom}(V,W)^{G}$ . The character of $\operatorname{Hom}(V,W)$ is $\overline{\chi}\psi$ . So, by the first part,

\dim\operatorname{Hom}_{G}(V,W)=\frac{1}{|G|}\sum_{g\in G}\overline{\chi}(g)% \psi(g)

which is exactly what we have to show. ∎

Corollary 3.29.

Suppose that $V$ and $W$ are irreducible representations with characters $\chi$ and $\psi$ . Then

\left\langle\chi,\psi\right\rangle=\begin{cases}1&\text{if $V\cong W$}\\ 0&\text{otherwise}.\end{cases}

Proof.

This is immediate from Theorem 3.28 and Schur’s lemma. ∎

We have therefore proved part (1) of Theorem 3.17, as well as Theorem 3.18 and Theorem 3.19. In other words, we know that representations are determined by their characters and that the rows of the character table are orthonormal (with an appropriately weighted inner product). We now only have to show that the character table is square.

3.5 Universal projections

The map $\pi$ constructed in step 1 of the proof of Theorem 3.28 has an interesting feature: for every representation $V$ , the map $\pi$ is a $G$ -homomorphism $V\rightarrow V^{G}$ . We can generalise this substantially.

Let $\alpha:G\rightarrow\mathbb{C}$ be any class function. Then define

\pi_{\alpha}=\frac{1}{|G|}\sum_{g\in G}\overline{\alpha}(g)[g]\in\mathbb{C}[G].

For every representation $(\rho,V)$ , $\pi_{\alpha}$ acts on $V$ , and we also call this $\pi_{\alpha}$ :

\pi_{\alpha}(v)=\frac{1}{|G|}\sum_{g\in G}\overline{\alpha}(g)\rho(g).

If we take $\alpha$ to be the constant function with value 1, then $\pi_{\alpha}$ is the projection $\pi:V\rightarrow V^{G}$ .

Lemma 3.30.

If $V$ is any representation, then $\pi_{\alpha}:V\rightarrow V$ is a $G$ -homomorphism.

Proof.

Let $g\in G$ . Then we have

	$\displaystyle\pi_{\alpha}(\rho(g)v)$	$\displaystyle=\frac{1}{\|G\|}\sum_{h\in G}\overline{\alpha}(h)\rho(h)\rho(g)v$
		$\displaystyle=\rho(g)\frac{1}{\|G\|}\sum_{h\in G}\overline{\alpha}(h)\rho(g^{-1}% hg)v$
		$\displaystyle=\rho(g)\frac{1}{\|G\|}\sum_{h\in G}\overline{\alpha}(g^{-1}hg)\rho% (g^{-1}hg)v$
as $\alpha$ is a class function
		$\displaystyle=\rho(g)\pi_{\alpha}(v)$
replacing $h$ by $g^{-1}hg$ . ∎

Since $\pi_{\alpha}$ is a $G$ -homomorphism $V\rightarrow V$ for every $V$ , we could call it a universal $G$ -homomorphism.

What happens when we take $V$ to be an irreducible representation?

Proposition 3.31.

Let $\alpha$ be a class function and let $V$ be an irreducible representation with character $\chi$ . Then $\pi_{\alpha}$ acts as the scalar

\frac{1}{\dim(V)}\left\langle\alpha,\chi\right\rangle

on $V$ .

In particular, if $\psi$ is the character of an irreducible representation $W$ , then $\pi_{\psi}$ acts as $\frac{1}{\dim W}$ on $W$ and as 0 on all other irreducible representations.

Proof.

If $V$ is irreducible, then by Schur’s lemma $\pi_{\alpha}\in\operatorname{Hom}_{G}(V,V)$ must be a scalar. But its trace is

\operatorname{tr}\pi_{\alpha}=\frac{1}{|G|}\sum_{g\in G}\overline{\alpha}(g)% \chi(g)=\left\langle\alpha,\chi\right\rangle.

Since the trace of a scalar matrix is just the dimension times that scalar, we get the proposition. ∎

Recall now that, if $\rho$ is an irreducible representation of $G$ and $V$ is another representation of $G$ , then $V(\rho)$ denotes the subrepresentation of $V$ generated by all the subrepresentations of $V$ isomorphic to $\rho$ .

Corollary 3.32.

Let $\rho$ be an irreducible representation of $G$ with character $\chi$ and dimension $d$ . Then the operator $d\pi_{\psi}$ acts, on any $G$ -representation $V$ , as the $G$ -equivariant projection

V\rightarrow V(\rho).

Proof.

We have $V\cong\bigoplus_{\sigma\in\mathrm{Irr}(G)}V(\sigma)$ by Maschke’s Theorem. By the proposition, $d\pi_{\psi}$ kills $V(\sigma)$ for $\sigma\not\cong\rho$ , and acts as the identity on $V(\rho)$ , as required. ∎

So $\dim(\rho)\pi_{\psi}$ projects any representation $V$ onto its $\rho$ -isotypic component. In particular, if $V$ has a unique subrepresentation isomorphic to $\rho$ , then $\dim(\rho)\pi_{\psi}$ is a projection onto that subrepresentation.

Corollary 3.33.

The irreducible characters are a basis for the space of class functions.

Proof.

We already know that they are linearly independent (since they are mutually orthogonal vectors in a Hermitian inner product space), so we have to show that they span. Let $C$ be the space of class functions, and let $X$ be the subspace spanned by the irreducible characters. Then $C=X\oplus X^{\perp}$ , and we must show $X^{\perp}=0$ . Suppose that $\alpha\in X^{\perp}$ , so $\left\langle\chi,\alpha\right\rangle=0$ for every irreducible character $\chi$ .

We consider the operator $\pi_{\alpha}=\frac{1}{|G|}\sum_{g\in G}\overline{\alpha}(g)[g]$ . This acts as (a multiple of)

\left\langle\alpha,\chi\right\rangle=\left\langle\chi,\alpha\right\rangle=0

on every irreducible representation $V$ with character $\chi$ . As every representation is a direct sum of irreducibles, we see that $\pi_{\alpha}$ acts as zero on all representations of $V$ .

However, considering the action of $\pi_{\alpha}$ on the regular representation $\mathbb{C}[G]$ , we see that

\pi_{\alpha}([e])=\frac{1}{|G|}\sum_{g\in G}\overline{\alpha}(g)[g].

Since this must be zero, we have $\alpha(g)=0$ for all $g\in G$ , as required. ∎

Corollary 3.34.

Parts (2) and (3) of Theorem 3.17 are true.

Proof.

Part (2) follows from the obvious fact that the dimension of the space of class functions is the number of conjugacy classes, together with Corollary 3.33.

For part (3), consider the matrix obtained by multiplying each entry of the character table by $\sqrt{|\mathcal{C}_{i}|/|G|}$ , where $\mathcal{C}_{i}$ is the conjugacy class labeling the column. Then row orthogonality says that this is a (square) unitary matrix. But the transpose of a unitary matrix is unitary, which gives the column orthogonality. ∎

Example 3.35.

We give an example of how to use the projection operators. Let $G=S_{4}$ , and let $(\sigma,V)$ be the permutation representation on the set of eight vertices of the cube. Its character $\theta$ is given below.

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\theta&8&0&0&2&0\\ \chi&3&1&-1&0&-1.\end{array}

Then we compute $\left\langle\theta,\theta\right\rangle=4$ , and (by inspection) that $\chi=\mathbbm{1}\oplus\epsilon\oplus\rho\oplus\rho\epsilon$ where $\rho$ is the standard three-dimensional representation of $S_{4}$ whose character $\chi$ is also shown in the table. We label the vertices as in Figure 2, where the labels are shown in red: the labels on the top face are $1,2,3,4$ running counterclockwise, and the label on the vertex opposite $i$ is $i+4$ , and look for a subspace of $V$ isomorphic to $\rho$ .

Refer to caption — Figure 2: Vertices of the cube

To do this we use the projection operator $\dim(\rho)\pi_{\chi}=\frac{3}{24}\sum_{g\in S_{4}}\overline{\chi}(g)\sigma(g)$ . Applying this to $e_{1}$ , we obtain:

\frac{3}{24}(3e_{1}+e_{2}+e_{4}+e_{7}+3e_{5}-e_{3}-e_{6}-e_{8}-2e_{2}-2e_{4}-2% e_{7})

which simplifies to (writing $f_{i}=e_{i}+e_{i+4}$ )

\frac{1}{8}(3f_{1}-f_{2}-f_{3}-f_{4}).

This is illustrated in green in Figure 2. Starting with a different vector $e_{i}$ will permute the roles of the $f_{i}$ , and we see that the image of $\pi_{\chi}$ is the subspace of $V$ where the coefficients of opposite vertices are equal, and the sum of all the coefficients is zero. This is therefore the subspace of $V$ isomorphic to $\rho$ !

Exercise 3.36.

Complete this example by finding the other irreducible subrepresentations of $V$ .

3.6 Linear algebra constructions

We construct new representations from old in various ways, using linear algebra. We continue to work over $k=\mathbb{C}$ throughout, though many of these constructions work over any field.

3.6.1 The dual representation

Recall that we earlier defined, for any $G$ -representations $V$ and $W$ , a $G$ -representation on the space $\operatorname{Hom}(V,W)$ of linear maps from $V$ to $W$ . It has character $\overline{\chi}_{V}\chi_{W}$ .

Definition 3.37.

If $V$ is any vector space, then the dual space of $V$ is

V^{*}=\operatorname{Hom}(V,\mathbb{C}).

We have $\dim V^{*}=\dim V$ . To see this, given a basis $v_{1},\dots,v_{n}$ of $V$ , we have the dual basis $v_{1}^{*},\dots,v_{n}^{*}$ of $V^{*}$ given by

v_{i}^{*}(v_{j})=\delta_{ij}=\begin{cases}1&\text{if $i=j$}\\ 0&\text{otherwise.}\end{cases}

There is a bilinear map $V^{*}\times V\rightarrow\mathbb{C},(\phi,v)\mapsto\phi(v)$ . The choice of basis identifies $V$ with $\mathbb{C}^{n}$ (as column vectors) and then the dual basis realizes $V^{*}$ as $1\times n$ matrices, with the above pairing being the usual matrix/dot product.

If $V$ has a $G$ -representation $\rho$ , then we take $\mathbb{C}$ to have the trivial representation and get an action $\rho^{*}$ of $G$ on $V^{*}$ defined by $\rho^{*}(g)(\phi)=\phi(\rho(g)^{-1}v)$ . From the formula for the character of $\operatorname{Hom}(V,W)$ , we see

\chi_{V^{*}}=\overline{\chi}_{V}.

If the matrix of $\rho(g)$ with respect to some basis is $A$ , then the matrix of $\rho(g)$ with respect to the dual basis is $(A^{T})^{-1}$ , the inverse of the transpose of $A$ .

3.6.2 Tensor products

Let $V, W$ be two vector spaces. Then the tensor product

V\otimes W

is the $\mathbb{C}$ -vector space generated by the symbols $v\otimes w$ for $v\in V$ and $w\in W$ , with the “bilinear” relations

	$\displaystyle\lambda(v\otimes w)$	$\displaystyle=(\lambda v)\otimes w=v\otimes(\lambda w);$
	$\displaystyle(v+v^{\prime})\otimes w$	$\displaystyle=v\otimes w+v^{\prime}\otimes w;$
	$\displaystyle v\otimes(w+w^{\prime})$	$\displaystyle=v\otimes w+v\otimes w^{\prime}.$

Remark 3.38.

Rigorously, we are taking the (infinite dimensional) vector space with a basis element $v\otimes w$ for every $v\in V$ and $w\in W$ and then forming its quotient by the (infinite dimensional) subspace generated by vectors of the form $(v+v^{\prime})\otimes w-v\otimes w-v^{\prime}\otimes w$ and by similar expressions corresponding to the other relations; this quotient is then finite dimensional, as the proposition below shows.

Proposition 3.39.

Let $e_{1},\ldots,e_{n}$ be a basis of $V$ and $f_{1},\ldots,f_{m}$ be a basis of $W$ . Then

\{e_{i}\otimes f_{j}:i=1,\ldots,n,j=1,\ldots,m\}

is a basis of $V\otimes W$ .

Proof.

Omitted.∎

In particular, the dimension of the tensor product is the product of the dimensions of the vector spaces:

\dim(V\otimes W)=\dim V\dim W.

Contrast the direct sum, which has dimension the sum of the dimensions of the vector spaces.

Exercise 3.40.

It is not true that every vector in $V\otimes W$ is of the form $v\otimes w$ . For example, if $V=W$ is two-dimensional with basis $e, f$ then $e\otimes e+f\otimes f\in V\otimes V$ cannot be written in this form.

Then next remark is non-examinable.

Remark 3.41.

Tensor products can be difficult to get used to. Perhaps the most important principle for understanding them is the following:

A linear map $V\otimes W\rightarrow U$ is the same as a bilinear map $V\times W\rightarrow U$ .

The dictionary as follows: given a linear map $\phi:V\otimes W\rightarrow U$ , we define a bilinear map $V\times W\rightarrow U$ by sending $(v,w)$ to $\phi(v\otimes w)$ . The bilinearity is then a consequence of the relations that hold in the tensor product. Conversely, given a bilinear map $\psi:V\times W\rightarrow U$ , define a linear map $V\otimes W\rightarrow U$ by sending $v\otimes w$ to $\psi(v,w)$ . We have to check that this is well-defined, i.e. that the bilinear relations are respected, and this is equivalent to the bilinearity of $\psi$ .

This all seems very abstract, but it is useful: defining bilinear maps is ‘easy’! In fact, tensor products ‘in real life’ often arise in situations where you have a number that depends on the choice of two vectors (in $V$ , say); then this dependency can be expressed as a linear map $V\otimes V\rightarrow\mathbb{C}$ .

All that said, for this course it will not be important to have such a deep theoretical understanding of tensor products as long as you are able to do calculations with them and are willing to take Proposition 3.39 on trust.

Naturally, we can consider more factors $V_{1}\otimes V_{2}\otimes\cdots\otimes V_{r}$ , with linearity in each slot.

Now, if $V$ and $W$ are both representations of $G$ then $V\otimes W$ becomes a representation via

g(v\otimes w)=gv\otimes gw.

We also write $\rho_{V}\otimes\rho_{W}$ for this representation. If we have bases $e_{1},\ldots,e_{n}$ of $V$ and $f_{1},\ldots,f_{m}$ of $W$ , with respect to which the matrices of $g$ acting on $V$ and $W$ are $A$ and $B$ , and we order the resulting basis of $V\otimes W$ as

e_{1}\otimes f_{1},e_{1}\otimes f_{2},\ldots,e_{1}\otimes f_{m},e_{2}\otimes f% _{1},\ldots,

then the matrix of $g$ on $V\otimes W$ is $A\otimes B$ where this is the block matrix

\begin{pmatrix}a_{11}B&a_{12}B&\ldots\\ a_{21}B&a_{22}B&\ldots\\ \vdots&\vdots&\ddots.\end{pmatrix}.

Proposition 3.42.

If $\rho_{V}$ and $\rho_{W}$ have characters $\chi_{V}$ and $\chi_{W}$ , then

\chi_{V\otimes W}=\chi_{V}\chi_{W}.

Proof.

Let $g\in G$ and choose bases $e_{1},\ldots,e_{n}$ and $f_{1},\ldots,f_{m}$ such that $ge_{i}=\lambda_{i}e_{i}$ and $gf_{j}=\mu_{j}f_{j}$ . Then

g(e_{i}\otimes f_{j})=\lambda_{i}\mu_{j}e_{i}\otimes f_{j},

so with respect to the basis $\{e_{i}\otimes f_{j}\}$ of $V\otimes W$ , $g$ acts diagonally with entries $\lambda_{i}\mu_{j}$ . So

\chi_{V\otimes W}(g)=\sum\lambda_{i}\mu_{j}=\left(\sum\lambda_{i}\right)\left(% \sum\mu_{j}\right)=\chi_{V}(g)\chi_{W}(g).\qed

The tensor product generalises the ’twisting’ construction earlier. If $V$ is any vector space then $V\otimes\mathbb{C}$ is isomorphic to $V$ via the map $v\otimes\lambda\mapsto\lambda v$ . If $(\chi,\mathbb{C})$ is a $1$ -dimensional representation and $(\rho,V)$ is any representation of $G$ , then $\rho\otimes\chi$ is a representation acting on $V\otimes\mathbb{C}$ . We have

(\rho\otimes\chi)(g)(v\otimes\lambda)=(\rho(g)v)\otimes(\chi(g)\lambda)\mapsto% \chi(g)\rho(g)\lambda v

via the above isomorphism so that

\rho\otimes\chi\cong\chi\rho.

Note that if $\rho$ is irreducible, so is $\chi\otimes\rho$ . Furthermore, $\chi\otimes\rho$ might or might not be isomorphic to $\rho$ .

Lemma 3.43.

Let $V$ and $W$ be two finite-dimensional representations of a group $G$ . Then

V^{*}\otimes W\cong\operatorname{Hom}(V,W)

as $G$ -modules.

Proof.

The best way to prove this is to show that the map $\phi\otimes w\mapsto T$ , where $T(v)=\phi(v)w$ , is a $G$ -isomorphism. This is straightforward but a bit technical.

Another proof which works in our situation is simply to observe that both sides have character $\overline{\chi}_{V}\chi_{W}$ . ∎

3.6.3 Symmetric and alternating powers

The symmetric square $\operatorname{Sym}^{2}(V)$ of $V$ is the vector space spanned by symbols $vv^{\prime}$ subject to the bilinear relations above and, additionally,

vv^{\prime}=v^{\prime}v

for all $v,v^{\prime}\in V$ .

Remark 3.44.

Formally, $\operatorname{Sym}^{2}(V)$ is the quotient of $V\otimes V$ by the subspace spanned by all elements of the form $v\otimes v^{\prime}-v^{\prime}\otimes v$ , and then $vv^{\prime}=v^{\prime}v$ is the image of $v\otimes v^{\prime}$ in $\operatorname{Sym}^{2}V$ .

Proposition 3.45.

Given a basis $e_{1},\dots,e_{n}$ of $V$ , the $e_{i}e_{j}$ with $i\leq j$ are a basis of $\operatorname{Sym}^{2}(V)$ . ∎

Hence

\dim\operatorname{Sym}^{2}(V)=\frac{n(n+1)}{2}.

The alternating square $\sideset{}{{}^{2}}{\bigwedge}(V)$ of $V$ is spanned by elements of the form $v\wedge v^{\prime}$ subject to the bilinear relations above and, additionally,

v\wedge v^{\prime}=-v^{\prime}\wedge v

for all $v,v^{\prime}\in V$ .

Remark 3.46.

Formally, it is the quotient of $V\otimes V$ by the subspace spanned by all elements of the form

v\otimes v^{\prime}+v^{\prime}\otimes v,

and then $v\wedge v^{\prime}=-v^{\prime}\wedge v$ is the image of $v\otimes v^{\prime}$ in $\Lambda^{2}(V)$ .

If $V$ and $W$ are representations of $G$ , we define actions of $G$ on these spaces as for the tensor product.

We can define linear maps $\operatorname{Sym}^{2}V\rightarrow V\otimes V$ and $\sideset{}{{}^{2}}{\bigwedge}(V)\rightarrow V\otimes V$ sending $vv^{\prime}\mapsto v\otimes v^{\prime}+v^{\prime}\otimes v$ and $v\wedge v^{\prime}\mapsto v\otimes v^{\prime}-v^{\prime}\otimes v$ . Any $v\otimes w\in V\otimes V$ can be written

v\otimes w=\frac{1}{2}\left((v\otimes w+w\otimes v)+(v\otimes w-w\otimes v)\right)

and this shows that

V\otimes V\cong\operatorname{Sym}^{2}(V)\oplus\sideset{}{{}^{2}}{\bigwedge}(V).

In fact this decomposition holds as $G$ -representations.

Remark 3.47.

The space $V\otimes V$ has an involution²² 2 Map whose square is the identity.

\sigma:v\otimes w\mapsto w\otimes v.

As $\sigma^{2}=I$ , its eigenvalues are $\pm 1$ . The decomposition above is the eigenspace decomposition for $\sigma$ : $\operatorname{Sym}^{2}V$ is the $(+1)$ -eigenspace, $\Lambda^{2}(V)$ the $(-1)$ -eigenspace.

Proposition 3.48.

If $(\rho,V)$ has character $\chi$ , then

\chi_{\operatorname{Sym}^{2}V}(g)=\frac{1}{2}\left(\chi(g)^{2}+\chi(g^{2}))\right)

and

\chi_{\Lambda^{2}V}(g)=\frac{1}{2}\left(\chi(g)^{2}-\chi(g^{2}))\right).

Proof.

If $\rho(g)$ has eigenvalues $\lambda_{1},\ldots\lambda_{n}$ , then diagonalise it as usual to get an eigenvector basis $v_{1},\ldots,v_{n}$ . Using the basis $v_{i}\wedge v_{j}$ of $\Lambda^{2}(V)$ you find that

\chi_{\Lambda^{2}V}(g)=\sum_{i<j}\lambda_{i}\lambda_{j}.

This is

\frac{1}{2}\left(\left(\sum\lambda_{i}\right)^{2}-\left(\sum\lambda_{i}^{2}% \right)\right)

as required.

The proof for the symmetric square is similar, or use the decomposition of $V\otimes V$ . ∎

Remark 3.49.

One can also define spaces $\operatorname{Sym}^{k}(V)$ and $\Lambda^{k}(V)$ for any $k$ , the latter vanishing if $k>\dim V$ . The former is spanned by expressions $v_{1}v_{2}\ldots v_{k}$ where ‘the order doesn’t matter’, while the latter is spanned by expressions $v_{1}\wedge v_{2}\wedge\ldots\wedge v_{k}$ where ‘switching two vectors introduces a minus sign’.

A particular special case is $k=\dim(V)$ . In this case, $\Lambda^{k}(V)$ is exactly one dimensional (it is easy to see it is spanned by $e_{1}\wedge\ldots\wedge e_{k}$ for $e_{1},\ldots,e_{k}$ any basis of $V$ , showing that the dimension is at most one, and there is an injective map to $V^{\otimes k}$ given by

v_{1}\wedge\ldots\wedge v_{k}\mapsto\sum_{\sigma\in S_{n}}\epsilon(\sigma)e_{1% }\otimes\ldots\otimes e_{k}

which shows that the dimension is at least one).

If $T:V\rightarrow V$ is any linear map, then we get a linear map $\Lambda^{k}(T):\Lambda^{k}(V)\rightarrow\Lambda^{k}(V)$ by setting $T(v_{1}\wedge\ldots\wedge v_{k})=T(v_{1})\wedge\ldots\wedge T(v_{k})$ . Since it is a map from a one-dimensional vector space to itself, $\Lambda^{k}(T)$ is just multiplication by some scalar. This scalar is exactly the determinant of $T$ !

3.6.4 Matrices in dimension 2

Suppose that $(\rho,V)$ is a representation of $G$ and that $\dim V=2$ , with $e_{1},e_{2}$ being a basis of $V$ . Let $g\in G$ , and let

\rho(g)=\begin{pmatrix}a&b\\ c&d\end{pmatrix}

be the matrix of $\rho(g)$ in this basis. We compute the matrices of $\Lambda^{2}\rho(g)$ and $\operatorname{Sym}^{2}\rho(g)$ .

The space $\Lambda^{2}V$ is one-dimensional, with basis vector $e_{1}\wedge e_{2}$ . Then

	$\displaystyle g\cdot(e_{1}\wedge e_{2})$	$\displaystyle=(ge_{1})\wedge(ge_{2})$
		$\displaystyle=(ae_{1}+ce_{2})\wedge(be_{1}+de_{2})$
		$\displaystyle=(ad-bc)e_{1}\wedge e_{2}$
		$\displaystyle=\det(\rho(g))e_{1}\wedge e_{2}$

using that $e_{1}\wedge e_{1}=e_{2}\wedge e_{2}=0$ and $e_{1}\wedge e_{2}=-e_{2}\wedge e_{1}$ . We see that

\operatorname{Sym}^{2}\rho\cong\det\rho.

The space $\operatorname{Sym}^{2}V$ is three-dimensional with basis $e_{1}^{2},e_{1}e_{2},e_{2}^{2}$ , and

	$\displaystyle g(e_{1}^{2})$	$\displaystyle=(ae_{1}+ce_{2})^{2}$
		$\displaystyle=a^{2}e_{1}^{2}+2ace_{1}e_{2}+c^{2}e_{2}^{2}$
	$\displaystyle g(e_{1}e_{2})$	$\displaystyle=abe_{1}^{2}+(ad+bc)e_{1}e_{2}+cde_{2}^{2}$
	$\displaystyle g(e_{2}^{2})$	$\displaystyle=b^{2}e_{1}^{2}+2bde_{1}e_{2}+d^{2}e_{2}^{2}$

whence the matrix of $\operatorname{Sym}^{2}\rho(g)$ (in this basis) is

\begin{pmatrix}a^{2}&ab&b^{2}\\ 2ac&ad+bc&2bd\\ c^{2}&cd&d^{2}\end{pmatrix}.

3.7 The character table of S₅

Let $G=S_{5}$ . We have the trivial representation $\mathbbm{1}$ , the sign representation $\epsilon$ , and the permutation representation $V\cong\mathbbm{1}\oplus W$ , and its twist, as before. So we can start off the character table:

\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\mathbbm{1}&1&1&1&1&1&1&1\\ \epsilon&1&-1&1&1&-1&-1&1\\ \chi&4&2&0&1&-1&0&-1\\ \chi\epsilon&4&-2&0&1&1&0&-1\end{array}

We then try $\Lambda^{2}W$ , which has character as shown (sadly, this is equal to its twist by $\epsilon$ ). This is an irreducible character.

\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\Lambda^{2}\chi&6&0&-2&0&0&0&1\end{array}

We can also try $\operatorname{Sym}^{2}W$ , which has character below; it isn’t irreducible.

\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\operatorname{Sym}^{2}\chi&10&4&2&1&1&0&0\end{array}

By taking inner products with the characters we’ve already found, we see that

\operatorname{Sym}^{2}\chi=\mathbbm{1}\oplus\chi\oplus\psi

where $\psi$ is an irreducible character. We get one more from twisting $\psi$ .

\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\psi&5&1&1&-1&1&-1&0\\ \psi\epsilon&5&-1&1&-1&-1&1&0\end{array}

This gives all of the irreducible characters, which we assemble into Table 2.

\begin{array}[]{c|ccccccc|}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ &1&10&15&20&20&30&24\\ \hline\cr\mathbf{1}&1&1&1&1&1&1&1\\ \epsilon&1&-1&1&1&-1&-1&1\\ \chi&4&2&0&1&-1&0&-1\\ \chi\epsilon&4&-2&0&1&1&0&-1\\ \Lambda^{2}\chi&6&0&-2&0&0&0&1\\ \psi&5&1&1&-1&1&-1&0\\ \psi\epsilon&5&-1&1&-1&-1&1&0\\ \end{array}

Table 2: Character table of

S_{5}

Question 3.50.

Find a more explicit description of the representation with character $\psi$ .

3 Character theory

3.1 Characters

Definition 3.1.

Remark 3.2.

Lemma 3.3.

Example 3.4.

Example 3.5.

Lemma 3.6.

Proof.

Lemma 3.7.

Proof.

Lemma 3.8.

Proof.

Remark 3.9.

Lemma 3.10.

Proof.

Definition 3.11.

Example 3.12.

Example 3.13.

Exercise 3.14.

Exercise 3.15.

3.2 Orthogonality of characters

Definition 3.16.

Theorem 3.17.

Theorem 3.18.

Theorem 3.19.

Proof.

3.3 Example: S4

Example 3.20.

3.3.1 Characters of lifts

Lemma 3.21.

Proof.

Example 3.22.

Proposition 3.23.

Proof.

3.3.2 Decomposing a representation

Example 3.24.

Exercise 3.25.

Remark 3.26.

3.4 Inner products and homomorphisms

Lemma 3.27.

Proof.

Theorem 3.28.

Proof.

Corollary 3.29.

Proof.

3.5 Universal projections

Lemma 3.30.

Proof.

Proposition 3.31.

Proof.

Corollary 3.32.

Proof.

Corollary 3.33.

Proof.

Corollary 3.34.

Proof.

Example 3.35.

Exercise 3.36.

3.6 Linear algebra constructions

3.6.1 The dual representation

Definition 3.37.

3.6.2 Tensor products

Remark 3.38.

Proposition 3.39.

Proof.

Exercise 3.40.

Remark 3.41.

Proposition 3.42.

Proof.

Lemma 3.43.

Proof.

3.6.3 Symmetric and alternating powers

Remark 3.44.

Proposition 3.45.

Remark 3.46.

Remark 3.47.

Proposition 3.48.

Proof.

Remark 3.49.

3.3 Example: S₄

3.7 The character table of S₅