Representation Theory

Representation theory is a way of studying groups by letting them act on vector spaces. The point is that linear algebra gives us tools that are very concrete: matrices, eigenvalues, traces, dimensions, and decompositions.

In this chapter we follow the beginning of Serre’s Linear Representations of Finite Groups. We will work with finite groups and finite-dimensional complex vector spaces. This is already a rich setting, and it is the setting where the basic structure theorems have their cleanest form.

Definitions and Examples

Let $G$ be a group and let $V$ be a finite-dimensional vector space over $\mathbb{C}$.

Definition

A (complex) representation of $G$ in $V$ is a group homomorphism

$$\rho :G\to \mathrm{GL}(V).$$

As a reminder, $\mathrm{GL}(V)$ is the group of linear automorphisms of $V$ called the general linear group, or if $\dim V=d$, the group of invertible $d\times d$ matrices.

In other words, a representation of $G$ is a pair $(V,\rho )$, which associates to each element $g\in G$ an invertible linear map $\rho (g)$ in such a way that

$$\rho (gt)=\rho (g)\rho (t)\text{for }g,t\in G.$$

After choosing a basis of $V$, each $\rho (g)$ becomes an invertible matrix. In this way, we are trying to represent the abstract group $G$ by matrices. Most people will abuse notation supress $\rho$ from the notation when its understood, and say that $V$ is a representation of $G$.

Example

Let $V$ be any finite-dimensional complex vector space. The rule $\rho (g)=I_V$ for every $g\in G$ defines a representation of $G$. This is called the trivial representation.

Example

Suppose $\dim V=1$. In this case, a representation is a map

$$\rho :G\to {\mathbb{C}}^{\times }.$$

Since $G$ is finite, every element $g\in G$ has finite order. If $g^n=e_G$, then

$$\rho (g{)}^n=\rho (g^n)=\rho (e_G)=1.$$

Thus $\rho (g)$ is a root of unity for every $g\in G$.

Exercise

This exercise will complete a classification of one-dimensional representations of a finite group $G$.

1. Show that $\rho$ uniquely factors through the abelianization $G^{\text{ab}}=G/[G,G]$. (Hint: look at previous exercises…)
2. Show that if $G$ is a cyclic group, then the one-dimensional representations of $G$ are in bijection with the $n$-th roots of unity, where $n=|G|$.
3. Use the classification of finite abelian groups to describe all one-dimensional representations of $G$.

Example

The sign representation of $S_n$ is the one-dimensional representation

$$\mathrm{sgn}:S_n\to \{\pm 1\}\subseteq {\mathbb{C}}^{\times }.$$

Even permutations act by multiplication by $1$, and odd permutations act by multiplication by $-1$. This is a representation because if you multiply two even permutations, you get an even permutation, and if you multiply two odd permutations, you get an even permutation. If you multiply an even and an odd permutation, you get an odd permutation. In all cases, the sign of the product is the product of the signs.

Example

Recall that $S_n$ acts naturally on the set $X=\{1,2,\dots ,n\}$ by permutating the elements. There is an extremely natural representation here on the $n$-dimensional vector space ${\mathbb{C}}^n$. Let $e_1,\dots ,e_n$ be the standard basis of ${\mathbb{C}}^n$, one for each element in $X$. For each permutation $\sigma \in S_n$, define

$$\rho (\sigma )(e_i)=e_{\sigma (i)}.$$

So for example, if $n=3$ and $\sigma =(123)$, then

$$\rho (\sigma )=\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right].$$

We call this the permutation representation associated with $S_n\curvearrowright X$.

Example S_n permutation representation extends to a general construction for any group action $G\curvearrowright X$ on a finite set $X$.

Definition

Let $V$ be the complex vector space with basis $(e_x{)}_{x\in X}$. For each $g\in G$, define the permutation representation of $G\curvearrowright X$ by

$$\rho (g)(e_x)=e_{g\cdot x}\text{for all }g\in G,x\in X.$$

Since the action satisfies $g(tx)=(gt)x$, the maps $\rho (g)$ satisfy $\rho (gt)=\rho (g)\rho (t)$.

Recall that every finite group $G$ acts on itself by left multiplication. This gives a permutation representation of $G$ on a vector space of dimension $|G|$ with basis $(e_g{)}_{g\in G}$. This is called the regular representation of $G$.

Example

If $G=S_3$, then the regular representation is 6-dimensional, with one basis vector for each permutation in $S_3$. Order the basis as

$$e_{\mathrm{id}},e_{(12)},e_{(13)},e_{(23)},e_{(123)},e_{(132)}.$$

In this basis, the matrix for $\rho ((123))$ is

$$\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 1\\ 0& 0& 0& 1& 0& 0\\ 0& 1& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\\ 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\end{array}\right].$$

Example

If $G=\mathbb{Z}/3\mathbb{Z}=\{\overline{0},\overline{1},\overline{2}\}$, then the regular representation of $G$ is a representation in a three-dimensional vector space with basis $e_{\overline{0}},e_{\overline{1}},e_{\overline{2}}$, and the group elements act as follows:

$$\begin{array}{rlrlrl}\rho (\overline{0})(e_{\overline{0}})& =e_{\overline{0}},& \rho (\overline{0})(e_{\overline{1}})& =e_{\overline{1}},& \rho (\overline{0})(e_{\overline{2}})& =e_{\overline{2}},\\ \rho (\overline{1})(e_{\overline{0}})& =e_{\overline{1}},& \rho (\overline{1})(e_{\overline{1}})& =e_{\overline{2}},& \rho (\overline{1})(e_{\overline{2}})& =e_{\overline{0}},\\ \rho (\overline{2})(e_{\overline{0}})& =e_{\overline{2}},& \rho (\overline{2})(e_{\overline{1}})& =e_{\overline{0}},& \rho (\overline{2})(e_{\overline{2}})& =e_{\overline{1}}.\end{array}$$

So as a matrix, $\rho (\overline{1})$ is the matrix

$$\left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right].$$

Note that this is the same matrix as the natural permutation representation of $(123)\in S_3$, after identifying $\overline{0},\overline{1},\overline{2}$ with $1,2,3$.

Exercise

Let $G$ be a finite group. The group algebra $\mathbb{C}[G]$ is the complex vector space with basis $(e_g{)}_{g\in G}$. Define multiplication on basis vectors by

$$e_g{e}_h=e_{gh}$$

and extend this rule bilinearly, so that

$$\left(\sum _{g\in G}{a}_g{e}_g\right)\left(\sum _{h\in G}{b}_h{e}_h\right)=\sum _{g,h\in G}{a}_g{b}_h{e}_{gh}.$$

1. Show that this multiplication is associative and has identity element $e_{e_G}$.
2. Show that $\mathbb{C}[G]$ is commutative if and only if $G$ is abelian.
3. For each $s\in G$, left multiplication by $e_s$ sends $e_g$ to $e_{sg}$. Explain why this recovers the regular representation described above.

As usual, we have a notion for when objects are the isomorphic to each other. As usual, it means that there is a homomorphism that respects the structure of the objects.

Definition

Two representations $(V,\rho )$ and $(W,\tau )$ of $G$ are called isomorphic if there is an isomorphism of vector spaces $f:V\to W$ such that

$$\tau (g)\circ f=f\circ \rho (g)$$

for all $g\in G$. In other words, the following diagram commutes for all $g\in G$:

People call such a map $f$ a $G$-equivariant linear map. More generally, a homomorphism from one $G$-representation to another is a homomorphism respecting the $G$-action in the same way.

Isomorphic representations are the same representation written in different coordinates. If bases are chosen for $V$ and $W$, then the condition above says that the matrices for one representation are obtained from the matrices for the other by a single change of basis.

Remark

There is a pattern here that appears throughout modern mathematics. Once we decide what kinds of objects we want to study, we also have to decide what the structure-preserving maps between them should be. For representations of $G$, the objects are representations, and the structure-preserving maps are $G$-equivariant linear maps. The isomorphisms are exactly the invertible $G$-equivariant linear maps. The subject that studies this general pattern of objects and morphisms is called category theory.

Subrepresentations and irreducibility

A representation is a vector space together with a compatible action of $G$. Therefore the right analogue of a subgroup is not just a subset, but a linear subspace that is preserved by the action.

Definition

Let $(V,\rho )$ be a representation of $G$. A subspace $W\subseteq V$ is called a subrepresentation if

$$\rho (g)(W)\subseteq W$$

for every $g\in G$. Sometimes we call $W$ G-stable or invariant under $G$.

This is what is needed in order for the restriction

$${\rho }_W(g)={\rho (g)|}_W:W\to W$$

defines a representation of $G$ on $W$. Since $\rho (g)$ is invertible, the condition $\rho (g)(W)\subseteq W$ is equivalent to $\rho (g)(W)=W$ for every $g\in G$.

Example

Let $G$ act on a finite set $X$, and let $V$ be the associated permutation representation with basis $(e_x{)}_{x\in X}$. The vector

$$u=\sum _{x\in X}{e}_x$$

is fixed by every element of $G$, because each $s\in G$ only permutes the basis vectors. Therefore the line $\mathbb{C}u$ is a subrepresentation, and $G$ acts trivially on it.

Lemma

Let $f:V\to W$ be a $G$-equivariant linear map between representations of $G$. Then $\ker (f)$ is a subrepresentation of $V$, and $\mathrm{im}(f)$ is a subrepresentation of $W$.

Proof

Let $v\in \ker (f)$, so $f(v)=0$. Then for any $g\in G$,

$$f(\rho (g)(v))=\tau (g)(f(v))=\tau (g)(0)=0,$$

so $\rho (g)(v)\in \ker (f)$. Therefore $\ker (f)$ is a subrepresentation of $V$.

Let $w\in \mathrm{im}(f)$, so $w=f(v)$ for some $v\in V$. Then for any $g\in G$,

$$\tau (g)(w)=\tau (g)(f(v))=f(\rho (g)(v))\in \mathrm{im}(f),$$

so $\mathrm{im}(f)$ is a subrepresentation of $W$.

Example

Every representation $V$ has two automatic subrepresentations:

$$0\text{and}V.$$

These are called the trivial subrepresentations.

Definition

A nonzero representation $V$ of $G$ is called irreducible if its only subrepresentations are $0$ and $V$.

Later, Maschke’s theorem will tell us that every finite-dimensional complex representation of a finite group can be built as a direct sum of irreducible representations. Irreducible representations are therefore the basic building blocks of representation theory. They are the “simple groups” of representation theory.

Example

Every one-dimensional representation is irreducible. Indeed, a one-dimensional vector space has only two subspaces: $0$ and the whole space.

Example

If $G$ is nontrivial, then the regular representation of $G$ is not irreducible. The line

$$\mathbb{C}\left(\sum _{t\in G}{e}_t\right)$$

is a one-dimensional subrepresentation, and it is not the whole regular representation because $\dim V=|G|>1$.

Proposition

All irreducible representations of $\mathbb{Z}/n\mathbb{Z}$ are one-dimensional.

Proof

Let $\gamma$ be a generator for $\mathbb{Z}/n\mathbb{Z}$, and let $V$ be an irreducible representation. Let $v$ be an eigenvector for $\gamma$, so $\gamma v=\lambda v$. Consider the complex line $\mathbb{C}v$; it is preserved by $\gamma$ and by all of its powers and thus $\mathbb{C}v$ is preserved by every group element. Therefore its a subrepresentation of $V$ and must be irreducible since its $1$-dimensional.

There are also multiple ways to create representations from existing ones. The most basic is to take their direct sum.

Definition

Let $V_1$ and $V_2$ be representations of $G$. Their direct sum representation is the representation on $V_1\oplus V_2$ defined by

$$s(v_1,v_2)=(s{v}_1,s{v}_2)$$

for every $s\in G$, $v_1\in V_1$, and $v_2\in V_2$.

Example

Let us return to the natural action of $S_n$ on the set $\{1,\dots ,n\}$. The corresponding permutation representation is the representation on ${\mathbb{C}}^n$ with standard basis $e_1,\dots ,e_n$ given by

$$\rho (\sigma )(e_i)=e_{\sigma (i)}.$$

There is a one-dimensional subspace

$$L=\mathrm{span}\{e_1+\cdots +e_n\}$$

on which every permutation acts trivially. There is also an $(n-1)$-dimensional subspace

$$W=\left\{a_1{e}_1+\cdots +a_n{e}_n\mid a_1+\cdots +a_n=0\right\}.$$

This subspace is preserved by every permutation, and the resulting representation on $W$ is called the standard representation of $S_n$.

In fact,

$${\mathbb{C}}^n=L\oplus W.$$

To see this, write $v=a_1{e}_1+\cdots +a_n{e}_n$ and let

$$\overline{a}=\frac{a_1+\cdots +a_n}{n}.$$

Then

$$v=\overline{a}(e_1+\cdots +e_n)+\sum _{i=1}^n(a_i-\overline{a})e_i,$$

where the first term lies in $L$ and the second lies in $W$.

Exercise

This exercise shows that for $n\ge 2$, the standard representation of $S_n$ is irreducible. Suppose $U$ is a nonzero subrepresentation of the standard representation

1. Let $v=a_1{e}_1+\cdots a_n{e}_n$ be a nonzero vector in $U$. Show that there must be $i,j$ such that $a_i\ne a_j$.
2. Show that the transposition $\tau =(ij)$ satisfies $v-\tau v\in U$. Conclude that $e_p-e_q\in U$ for any $p\ne q$.
3. Show that the vectors $e_p-e_q$ for $p\ne q$ span the standard representation. Conclude that $U$ is the whole standard representation.

Schur’s Lemma and Maschke’s Theorem

We now prove the first structural theorems about finite-dimensional complex representations. Maschke’s theorem says that every representation can be broken apart into irreducible pieces. Schur’s lemma says that maps between irreducible representations are very rigid.

More precisely, Maschke’s theorem says that over $\mathbb{C}$, every finite-dimensional representation is assembled from irreducible ones by direct sums. The key point is that every subrepresentation has an invariant complement.

Invariant complements

Let $V$ be a finite-dimensional complex representation of a finite group $G$, and let $W\subseteq V$ be a subrepresentation. Then there is a subrepresentation $W^{\prime }\subseteq V$ such that

$$V=W\oplus W^{\prime }.$$

Proof

Start by choosing any linear complement $U$ to $W$, so that $V=W\oplus U$ as vector spaces. Equivalently, choose a linear projection $p:V\to W$ onto $W$ — this chooses a complement $U=\ker (p)$.

The projection $p$ does not have to respect the $G$-action. We fix this by averaging it over the group. Define

$$P(v)=\frac{1}{|G|}\sum _{s\in G}sp(s^{-1}v).$$

Since $W$ is a subrepresentation, each term $sp(s^{-1}v)$ lies in $W$, so $P(v)\in W$.

First, $P$ is still a projection onto $W$. If $w\in W$, then $s^{-1}w\in W$, so $p(s^{-1}w)=s^{-1}w$. Therefore

$$P(w)=\frac{1}{|G|}\sum _{s\in G}s(s^{-1}w)=\frac{1}{|G|}\sum _{s\in G}w=w.$$

Next, $P$ is $G$-equivariant. Let $t\in G$. Then

$$\begin{array}{rl}P(tv)& =\frac{1}{|G|}\sum _{s\in G}sp(s^{-1}tv).\end{array}$$

As $s$ runs through $G$, we may write $s=tr$ with $r$ running through $G$. Thus

$$\begin{array}{rl}P(tv)& =\frac{1}{|G|}\sum _{r\in G}trp(r^{-1}v)\\ & =t\left(\frac{1}{|G|}\sum _{r\in G}rp(r^{-1}v)\right)\\ & =tP(v).\end{array}$$

Now let

$$W^{\prime }=\ker (P).$$

Since $P$ is $G$-equivariant, $W^{\prime }$ is a subrepresentation of $V$. Finally, every $v\in V$ can be written as

$$v=P(v)+(v-P(v)).$$

The first term lies in $W$. The second lies in $\ker (P)$, because $P$ is the identity on $W$ and $P(v)\in W$. Also, $W\cap \ker (P)=0$, since if $w\in W$ and $P(w)=0$, then $w=0$. Hence

$$V=W\oplus W^{\prime }.$$

Maschke's theorem

Let $G$ be a finite group. Every finite-dimensional complex representation of $G$ is a direct sum of irreducible subrepresentations.

Proof

We use induction on $\dim V$. If $\dim V=0$, there is nothing to prove. If $V$ is irreducible, then $V$ is already a direct sum with one irreducible summand.

Otherwise, $V$ has a nonzero proper subrepresentation $W$. By Lemma Invariant complements, there is a subrepresentation $W^{\prime }$ such that

$$V=W\oplus W^{\prime }.$$

Both $W$ and $W^{\prime }$ have dimension smaller than $\dim V$. By induction, each of them is a direct sum of irreducible subrepresentations. Therefore $V$ is also a direct sum of irreducible subrepresentations.

Maschke’s theorem is the structural foundation of finite representation theory over $\mathbb{C}$. It tells us that the main problem is to understand the irreducible representations and then understand how often each one appears inside a given representation. Characters will give us an efficient way to answer that multiplicity question.

Now we turn to Schur’s lemma, which says that maps between irreducible representations are very rigid. In particular, if two irreducible representations are not isomorphic, then there are no nonzero maps between them. If they are isomorphic, then the only maps from one to the other are scalar multiples of the isomorphism.

Schur's lemma

Let $V$ and $W$ be irreducible complex representations of $G$. Any nonzero $G$-equivariant map $T:V\to W$ is an isomorphism.

In particular, if $V$ is irreducible, then every $G$-equivariant linear map $T:V\to V$ is scalar multiplication by some complex number.

Proof

By Lemma equivariant kernel image subrepresentations, $\ker (T)$ is a subrepresentation of $V$, and $\mathrm{im}(T)$ is a subrepresentation of $W$. Since $V$ is irreducible, $\ker (T)$ is either $0$ or $V$. If $\ker (T)=V$, then $T=0$. If $\ker (T)=0$, then $T$ is injective.

Similarly, since $W$ is irreducible, $\mathrm{im}(T)$ is either $0$ or $W$. If $T\ne 0$, then $\mathrm{im}(T)\ne 0$, so $\mathrm{im}(T)=W$. Therefore every nonzero $G$-equivariant map $T:V\to W$ is both injective and surjective, hence an isomorphism.

Now suppose $V=W$, and let $T:V\to V$ be $G$-equivariant. Since $V$ is a finite-dimensional complex vector space, $T$ has an eigenvalue $\lambda \in \mathbb{C}$. The map

$$T-\lambda I_V$$

is still $G$-equivariant. It is not injective, because it has a nonzero eigenvector in its kernel. By the first part of the theorem, it cannot be a nonzero map. Therefore $T=\lambda I_V$.

Corollary

Let $V$ and $W$ be irreducible complex representations of $G$. Then

$${\mathrm{Hom}}_G(V,W)=\{\begin{array}{ll}0,& V\ncong W\\ \mathbb{C},& V\cong W.\end{array}$$

Intuitively, Maschke’s theorems say that all representations break apart into irreducible pieces, and then Schur’s lemma says that any maps between representations do not “mix” these irreducible pieces. In the next section, we will use this to find the irreducible decomposition of a representation.

Characters

Reminder: the trace of a matrix is the sum of its diagonal entries. Equivalently, the trace is the sum of the eigenvalues of the matrix or the second coefficient of the characteristic polynomial. This is independent of the choice of basis, so it is a well-defined invariant of a linear transformation. Recall that $\mathrm{Tr}(AB)=\mathrm{Tr}(BA)$ and $\mathrm{Tr}(A+B)=\mathrm{Tr}(A)+\mathrm{Tr}(B)$.

Definition

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

$${\chi }_V(s)=\mathrm{Tr}(\rho (s)).$$

The character records one number for each group element: the trace of the linear transformation by which that element acts. One immediate thing is that if two representations are isomorphic, they have the same character.

Lemma

Isomorphic representations have the same character.

Proof

If $V$ and $W$ are isomorphic representations, then $W=PV{P}^{-1}$ for some invertible matrix $P$. That is, it is the same linear transformation written in a different basis. Therefore $\mathrm{Tr}(W)=\mathrm{Tr}(PV{P}^{-1})=\mathrm{Tr}(V)$, so the characters are the same.

Remark

The observation that $\chi (uv)=\chi (vu)$ is equivalent to $\chi (uv{u}^{-1})=\chi (v)$ is called being a class function. In other words, the character is constant on conjugacy classes.

As we will show, the important deeper point is that the converse is also true: if two representations have the same character, then they are isomorphic. Therefore we can reduce the study of representations to that of their characters, and in general, a finite-dimensional representation is specified by a set of $|G|$ numbers.

Example

If $V$ is the trivial representation of dimension $n$, then every $s\in G$ acts as the identity map $I_V$. Therefore

$${\chi }_V(s)=\mathrm{Tr}(I_V)=n$$

for every $s\in G$.

Example

Let $G$ act on a finite set $X$, and let $V$ be the associated permutation representation with basis $(e_x{)}_{x\in X}$. For $s\in G$, the matrix of $\rho (s)$ has a $1$ in the diagonal entry corresponding to $x$ exactly when $sx=x$, and has $0$ in that diagonal entry otherwise. Therefore

$${\chi }_V(s)=|\{x\in X\mid sx=x\}|.$$

So the character of a permutation representation counts fixed points.

Example

Let $V$ be the regular representation of $G$. This is the permutation representation associated to the action of $G$ on itself by left multiplication. If $s=e_G$, then every element of $G$ is fixed, so

$${\chi }_V(e_G)=|G|.$$

If $s\ne e_G$, then left multiplication by $s$ has no fixed points: $st=t$ would imply $s=e_G$. Hence

$${\chi }_V(s)=0\text{for }s\ne e_G.$$

Thus the character of the regular representation is

$${\chi }_V(s)=\{\begin{array}{ll}|G|& \text{if }s=e_G,\\ 0& \text{if }s\ne e_G.\end{array}$$

Example

Let $S_n$ act on ${\mathbb{C}}^n$ by permuting the standard basis. Let $P$ be this permutation representation, let $L$ be the trivial subrepresentation spanned by $e_1+\cdots +e_n$, and let $W$ be the standard representation. We showed earlier that

$$P=L\oplus W.$$

The character of $P$ counts fixed points, and the character of $L$ is identically $1$. Therefore

$${\chi }_W(\sigma )=|\{i\in \{1,\dots ,n\}\mid \sigma (i)=i\}|-1.$$

Thus the character of the standard representation of $S_n$ is “number of fixed points minus one.”

Example

So far, we’ve introduced three different irreducible representations of $S_3$: the trivial representation, the sign representation, and the standard representation. Lets compute their characters.

$$\begin{array}{ccccccc}{\chi }_V(g)& e& (12)& (13)& (23)& (123)& (132)\\ {\chi }_{\mathrm{triv}}& 1& 1& 1& 1& 1& 1\\ {\chi }_{\mathrm{sgn}}& 1& -1& -1& -1& 1& 1\\ {\chi }_{\mathrm{std}}& 2& 0& 0& 0& -1& -1\end{array}$$

We will soon be able to show that these are all the irreducible representations of $S_3$. This is called the character table for $S_3$.

Here are the important properties about characters and character tables:

1. The value of ${\chi }_V(g)$ only depends on the conjugacy class of $g$.

2. The number of rows equals the number of conjugacy classes of $G$.

3. The sum of the absolute squares of any row, summed over all group elements, equals $|G|$.

4. The “dot product” of any row with itself is 1.

5. The “dot product” of any two different rows, summed over all group elements, is $0$.

Most of these we will prove later.

Proposition

If $V$ and $W$ are representations of $G$, then ${\chi }_{V\oplus W}={\chi }_V+{\chi }_W$.

Proof

With respect to bases adapted to the direct sum $V\oplus W$, the matrix for the action of $s\in G$ has block form

$$\left[\begin{array}{cc}{\rho }_V(s)& 0\\ 0& {\rho }_W(s)\end{array}\right].$$

The trace of a block diagonal matrix is the sum of the traces of the diagonal blocks. Therefore

$${\chi }_{V\oplus W}(s)={\chi }_V(s)+{\chi }_W(s)$$

for every $s\in G$.

Lemma

If ${\chi }_V$ is a character, then $\chi (s^{-1})=\overline{\chi (s)}$ for every $s\in G$.

Proof

Since $s$ has finite order, the eigenvalues of $\rho (s)$ are roots of unity. Therefore the eigenvalues of $\rho (s^{-1})$ are the complex conjugates of the eigenvalues of $\rho (s)$. Since the trace is the sum of the eigenvalues, we have

$${\chi }_V(s^{-1})=\mathrm{Tr}(\rho (s^{-1}))=\overline{\mathrm{Tr}(\rho (s))}=\overline{{\chi }_V(s)}.$$

Characters and class functions

Recall from the previous section that characters are class functions on $G$. We can define an inner product on the vector space of all class functions on $G$.

Definition

If $f_1,f_2:G\to \mathbb{C}$ are functions, define

$$⟨f_1,f_2⟩=\frac{1}{|G|}\sum _{s\in G}{f}_1(s)\overline{f_2(s)}.$$

This is the usual Hermitian inner product on the vector space of functions $G\to \mathbb{C}$, normalized by the factor $|G|$.

Remark

If $f_1$ and $f_2$ are class functions, then the inner product can be computed by summing over conjugacy classes. If $C_1,\dots ,C_r$ are the conjugacy classes of $G$ and $c_i\in C_i$ is a representative, then

$$⟨f_1,f_2⟩=\frac{1}{|G|}\sum _{i=1}^r|C_i|f_1(c_i)\overline{f_2(c_i)}.$$

This is one reason conjugacy classes become the columns of a character table.

We now prove the main orthogonality theorem for irreducible characters. The proof is a little complicated, but combining the result with Schur’s lemma gives a very clean and useful result.

Proposition

Let $V$ and $W$ be finite-dimensional complex representations of $G$. Then

$$⟨{\chi }_V,{\chi }_W⟩=\dim {\mathrm{Hom}}_G(W,V).$$

Proof

Let $E=\mathrm{Hom}(W,V)$. We make $E$ into a representation of $G$ by

$$(sT)(w)=sT(s^{-1}w),$$

for $T\in \mathrm{Hom}(W,V)$ and $w\in W$. Its character is

$${\chi }_E(s)={\chi }_V(s){\chi }_W(s^{-1}).$$

Indeed, after choosing bases, the action of $s$ on $E$ has the form

$$T⟼AT{B}^{-1},$$

where $A$ is the matrix of $s$ on $V$, and $B$ is the matrix of $s$ on $W$. A direct calculation on matrix units $E_{ij}$ shows that the trace of the map $T\mapsto AT{B}^{-1}$ is $\mathrm{Tr}(A)\mathrm{Tr}(B^{-1})$. This proves the displayed formula for ${\chi }_E(s)$.

Now average the action of $G$ on $E$. Define

$$P:E\to E,P(T)=\frac{1}{|G|}\sum _{s\in G}sT.$$

This is a projection onto the fixed subspace $E^G$. Indeed, if $U\in E$, then for any $t\in G$,

$$tP(U)=\frac{1}{|G|}\sum _{s\in G}tsU=\frac{1}{|G|}\sum _{r\in G}rU=P(U),$$

so $P(U)\in E^G$. If $U\in E^G$, then

$$P(U)=\frac{1}{|G|}\sum _{s\in G}U=U.$$

Thus $P$ is a projection onto $E^G$, and therefore $\mathrm{Tr}(P)=\dim E^G$ since $\mathrm{Tr}$ of a projection is the dimension of its image. By linearity of trace,

$$\begin{array}{rl}\mathrm{Tr}(P)& =\frac{1}{|G|}\sum _{s\in G}{\chi }_E(s)\\ & =\frac{1}{|G|}\sum _{s\in G}{\chi }_V(s)\overline{{\chi }_W(s)}\\ & =⟨{\chi }_V,{\chi }_W⟩.\end{array}$$

Finally, the fixed subspace $E^G$ is exactly ${\mathrm{Hom}}_G(W,V)$. Indeed, $T\in E^G$ means

$$sT(s^{-1}w)=T(w)$$

for every $s\in G$ and $w\in W$. Replacing $w$ by $sw$ gives

$$sT(w)=T(sw),$$

which is exactly the condition that $T$ is $G$-equivariant. Thus

$$⟨{\chi }_V,{\chi }_W⟩=\mathrm{Tr}(P)=\dim E^G=\dim {\mathrm{Hom}}_G(W,V).$$

The proof of Proposition character inner product hom gives an intuitive picture of inner products between characters. The inner product between characters is the quantity

$$\frac{1}{|G|}\sum _{s\in G}{\chi }_V(s)\overline{{\chi }_W(s)},$$

which is is an average correlation between the two representations, or how much they have in common. This is the same as the dimension of the space of $G$-equivariant maps between the two representations. By adding Schur’s lemma, we get the following orthogonality theorem for irreducible characters.

Orthogonality of irreducible characters

Let $V$ and $W$ be irreducible complex representations of $G$. Then

$$⟨{\chi }_V,{\chi }_W⟩=\{\begin{array}{ll}1& \text{if }V\cong W,\\ 0& \text{if }V\ncong W.\end{array}$$

Proof

By Proposition character inner product hom,

$$⟨{\chi }_V,{\chi }_W⟩=\dim {\mathrm{Hom}}_G(W,V).$$

By Schur’s lemma, ${\mathrm{Hom}}_G(W,V)=0$ if $V$ and $W$ are not isomorphic. If $V\cong W$, then ${\mathrm{Hom}}_G(W,V)$ is one-dimensional: after choosing one isomorphism $W\to V$, every other $G$-equivariant map differs from it by a scalar. This proves the theorem.

This theorem says that irreducible characters form an orthonormal set in the space of class functions. In particular, they are linearly independent. But in fact, they form an orthonormal basis for the space of class functions.

Completeness of irreducible characters

The irreducible characters of $G$ form an orthonormal basis for the vector space of class functions on $G$. In particular, the number of irreducible complex representations of $G$ is equal to the number of conjugacy classes of $G$.

We will not prove this here because it requires a little more machinery I don’t want to introduce.¹ As a direct consequence, we can compute the decomposition of the regular representation into irreducible representations, which gives a very useful formula for the order of a finite group.

Corollary

Let $V_1,\dots ,V_r$ be the irreducible complex representations of $G$. Then the regular representation decomposes as

$$V_{\mathrm{reg}}\cong (\dim V_1)V_1\oplus \cdots \oplus (\dim V_r)V_r.$$

Consequently,

$$|G|=\sum _{i=1}^r(\dim V_i{)}^2.$$

Proof

Let ${\chi }_{\mathrm{reg}}$ be the character of the regular representation. We computed earlier that

$${\chi }_{\mathrm{reg}}(e)=|G|,{\chi }_{\mathrm{reg}}(s)=0\text{ for }s\ne e.$$

Therefore the multiplicity of $V_i$ in the regular representation is

$$\begin{array}{rl}⟨{\chi }_{\mathrm{reg}},{\chi }_{V_i}⟩& =\frac{1}{|G|}|G|{\chi }_{V_i}(e)\\ & =\dim V_i.\end{array}$$

This proves the decomposition. Taking dimensions of both sides gives the result.

Now suppose a finite-dimensional complex representation $V$ decomposes as

$$V\cong m_1{V}_1\oplus \cdots \oplus m_r{V}_r,$$

where $V_1,\dots ,V_r$ are pairwise nonisomorphic irreducible representations, and where $m_i{V}_i$ means a direct sum of $m_i$ copies of $V_i$. Maschke’s theorem guarantees that such a decomposition exists after collecting isomorphic irreducible summands together. The integers $m_i$ are called the multiplicities of the irreducible representations in $V$.

Multiplicity formula

With notation as above,

$$m_i=⟨{\chi }_V,{\chi }_{V_i}⟩.$$

Proof

By additivity of characters under direct sums,

$${\chi }_V=m_1{\chi }_{V_1}+\cdots +m_r{\chi }_{V_r}.$$

Taking the inner product with ${\chi }_{V_i}$ gives

$$\begin{array}{rl}⟨{\chi }_V,{\chi }_{V_i}⟩& =\sum _{j=1}^r{m}_j⟨{\chi }_{V_j},{\chi }_{V_i}⟩\\ & =m_i,\end{array}$$

by the orthogonality of irreducible characters.

Corollary

Two finite-dimensional complex representations of $G$ have the same character if and only if they are isomorphic.

Proof

If $V\cong W$, then ${\chi }_V={\chi }_W$. Conversely, suppose ${\chi }_V={\chi }_W$. Write

$$V\cong m_1{V}_1\oplus \cdots \oplus m_r{V}_r\text{and}W\cong n_1{V}_1\oplus \cdots \oplus n_r{V}_r,$$

where $V_1,\dots ,V_r$ are the irreducible representations that occur in either decomposition. The multiplicity formula gives

$$m_i=⟨{\chi }_V,{\chi }_{V_i}⟩=⟨{\chi }_W,{\chi }_{V_i}⟩=n_i$$

for every $i$. Thus $V$ and $W$ have the same irreducible summands with the same multiplicities, so $V\cong W$.

Remark

This is the first major computational payoff of characters. To find how many times an irreducible representation $V_i$ occurs inside $V$, we do not have to find all of the subrepresentations by hand. We compute one inner product of class functions.

Burnside's lemma revisited

Let $G$ act on a finite set $X$, and let ${\chi }_{\mathrm{perm}}$ be the character of the associated permutation representation. Then

$$⟨{\chi }_{\mathrm{perm}},1⟩=\frac{1}{|G|}\sum _{s\in G}|\{x\in X\mid sx=x\}|=|X/G|.$$

Proof

The first equality is just the definition of the inner product together with the fact that ${\chi }_{\mathrm{perm}}(s)$ counts the fixed points of $s$. The last expression is Burnside’s lemma.

We can also see the representation-theoretic meaning directly. The inner product $⟨{\chi }_{\mathrm{perm}},1⟩$ is the multiplicity of the trivial representation inside the permutation representation. The fixed subspace consists of the vectors

$$\sum _{x\in X}{a}_x{e}_x$$

whose coefficients are constant on each orbit. Therefore the dimension of this fixed subspace is the number of orbits of the action.

Computing Character Tables

As we’ve seen, a character table is a compact way to record the irreducible characters of a finite group. The columns are indexed by conjugacy classes of $G$. This makes sense because characters are class functions. The rows are indexed by the irreducible characters of $G$. If $C_1,\dots ,C_r$ are the conjugacy classes and $c_i\in C_i$ is a representative, then the character table records the values $\chi (c_i)$ as $\chi$ runs through the irreducible characters.

Remark

The choice of representative $c_i$ does not matter, since characters are constant on conjugacy classes. In a character table we usually label a column by a convenient representative, such as $(12)$ for the class of all transpositions in $S_n$.

The row orthogonality relation says that distinct irreducible rows are orthogonal using the weighted inner product

$$⟨\chi ,\psi ⟩=\frac{1}{|G|}\sum _{i=1}^r|C_i|\chi (c_i)\overline{\psi (c_i)}.$$

Thus each irreducible row has norm $1$, and two different irreducible rows have inner product $0$.

There is also a column orthogonality relation for a complete character table. Let $g$ and $h$ be elements of a finite group $G$, and let $V_1,\dots ,V_r$ be irreps of $G$. Then

$$\sum _{i=1}^r{\chi }_{V_i}(g)\overline{{\chi }_{V_i}(h)}=\{\begin{array}{ll}|C_G(g)|=|G|/|C_g|& \text{if }g,h\text{ are conjugate},\\ 0& \text{otherwise}.\end{array}$$

So the columns are orthogonal too, but with a different normalization.

Exercise

Let $C_n=⟨a⟩$ be a cyclic group of order $n$. Write the character table of $C_n$.

Example

Let us return to the symmetric group $S_3$. The conjugacy classes of $S_3$ are:

$$\{e\},\{(12),(13),(23)\},\{(123),(132)\}.$$

Their sizes are $1$, $3$, and $2$. We can rewrite the character table with columns as conjugacy classes:

$$\begin{array}{cccc}& e& (12)& (123)\\ \text{class size}& 1& 3& 2\\ {\chi }_{\mathrm{triv}}& 1& 1& 1\\ {\chi }_{\mathrm{sgn}}& 1& -1& 1\\ {\chi }_{\mathrm{std}}& 2& 0& -1\end{array}$$

Since $S_3$ has three conjugacy classes, Theorem Completeness of irreducible characters says that there are three irreducible characters. Thus this is the complete character table of $S_3$.

Now let $R$ be the regular representation of $S_3$. From our earlier computation of the character of the regular representation,

$${\chi }_R=(6,0,0)$$

on these three conjugacy classes. The multiplicities are

$$⟨{\chi }_R,{\chi }_{\mathrm{triv}}⟩=1,⟨{\chi }_R,{\chi }_{\mathrm{sgn}}⟩=1,⟨{\chi }_R,{\chi }_{\mathrm{std}}⟩=2.$$

Therefore

$$R\cong V_{\mathrm{triv}}\oplus V_{\mathrm{sgn}}\oplus 2V_{\mathrm{std}}.$$

The dimensions check out: $1+1+2\cdot 2=6$, which is the dimension of the regular representation.

Exercise

Let $S_4$ act on the set $X$ of all two-element subsets of $\{1,2,3,4\}$. Let $P$ be the corresponding permutation representation.

1. Compute the character ${\chi }_P$ on the five conjugacy classes

$$e,(12),(12)(34),(123),(1234).$$

2. Compute

$$⟨{\chi }_P,{\chi }_{\mathrm{triv}}⟩\text{and}⟨{\chi }_P,{\chi }_{\mathrm{std}}⟩.$$

Conclude that $P$ contains one copy of the trivial representation and one copy of the standard representation.

3. By Maschke’s theorem, write $P\cong V_{\mathrm{triv}}\oplus V_{\mathrm{std}}\oplus U$ for some representation $U$. Compute ${\chi }_U$ and show that it is irreducible. What is the dimension of $U$?

Example

Let us compute the character table of $S_4$. By Theorem symmetric group conjugacy cycle type, conjugacy classes in $S_n$ are determined by cycle type. The conjugacy classes of $S_4$ are represented by

$$e,(12),(12)(34),(123),(1234),$$

and their sizes are

$$1,6,3,8,6.$$

Since there are five conjugacy classes, there are five irreducible characters.

We immediately have the trivial character and the sign character. The dimension formula from Corollary regular representation decomposition tells us that if the remaining dimensions are $d_1,d_2,d_3$, then

$$d_1^2+d_2^2+d_3^2=24-1^2-1^2=22.$$

The only possibility is

$$\{d_1,d_2,d_3\}=\{2,3,3\}.$$

One of the 3-dimensional irreducible representations is the standard representation of $S_4$. Its character is the number of fixed points minus one:

$${\chi }_{\mathrm{std}}=(3,1,-1,0,-1).$$

Exercise s4 two subsets character gives the remaining 2-dimensional irreducible character

$${\chi }_2=(2,0,2,-1,0).$$

We only need the last irreducible character, which we can compute via the orthogonality relations. That is, we can use the row orthogonality relations and linear algebra to solve for the last row, or some of the column orthogonality relations. We get

$${\chi }_{\mathrm{sgn}\cdot \mathrm{std}}=(3,-1,-1,0,1).$$

Note that this is the character ${\chi }_{\mathrm{std}}$ multiplied entrywise by the sign character ${\chi }_{\mathrm{sgn}}$.

Thus the character table of $S_4$ is

$$\begin{array}{cccccc}{S}_4& e& (12)& (12)(34)& (123)& (1234)\\ \text{class size}& 1& 6& 3& 8& 6\\ {\chi }_{\mathrm{triv}}& 1& 1& 1& 1& 1\\ {\chi }_{\mathrm{sgn}}& 1& -1& 1& 1& -1\\ {\chi }_{\mathrm{std}}& 3& 1& -1& 0& -1\\ {\chi }_2& 2& 0& 2& -1& 0\\ {\chi }_{\mathrm{sgn}\cdot \mathrm{std}}& 3& -1& -1& 0& 1\end{array}$$

One quick check is

$$1^2+1^2+3^2+3^2+2^2=24=|S_4|.$$

Exercise

Show that if $V$ and $W$ are irreps of $G$ with $\dim W=1$, then ${\chi }_V{\chi }_W$ is the character of an irreducible representation (where multiplication happens entrywise). We didn’t have time in the course, but this is the character of the tensor product $V\otimes W$. This is the “product” analog of the direct sum construction and has dimension $\dim V\cdot \dim W$.

Exercise

Let $W$ be the standard representation of $S_4$ and let $E=\mathrm{End}(W)=\mathrm{Hom}(W,W)$.

1. Show that $\sigma \cdot T=\sigma T{\sigma }^{-1}$ defines a representation of $S_4$ on $E$.
2. Show that $\dim E=9$ and compute the character of $E$. (Hint: For the latter, look at the proof that $⟨{\chi }_V,{\chi }_W⟩=\dim {\mathrm{Hom}}_G(W,V)$.)
3. Decompose $E$ into irreducible representations of $S_4$.

Example

Let $D_4=⟨r,s\mid r^4=e,s^2=e,srs=r^{-1}⟩$ be the symmetry group of a square. Its conjugacy classes are

$$\{e\},\{r^2\},\{r,r^3\},\{s,r^{2}s\},\{rs,r^{3}s\}.$$

Thus $D_4$ has five irreducible characters.

Using the earlier exercise about one-dimensional representations, the one-dimensional characters of $D_4$ are exactly the homomorphisms $D_4^{\mathrm{ab}}\to {\mathbb{C}}^{\times }$. Recall the presentation

$$D_4=⟨r,s\mid r^4=e,s^2=e,srs=r^{-1}⟩.$$

The abelianization is what you get by forcing $s$ and $r$ to commute, thus $srs=r^{-1}$ becomes $r=r^{-1}$, so $r^2=e$. Thus

$$D_4^{\mathrm{ab}}=⟨r,s\mid r^2=e,s^2=e,rs=sr⟩\cong C_2\times C_2.$$

Therefore there are four one-dimensional characters, which we can denote by ${\chi }_{++}$,${\chi }_{+-}$,${\chi }_{-+}$, and ${\chi }_{--}$, where the signs indicate the values of the character on $r$ and $s$.

This leaves one last 2-dimensional character, which we could find via the orthogonality relations. However, its very natural: consider the 2-dimensional geometric representation coming from the action of $D_4$ on the square in the plane. In that representation, the identity has trace $2$, the half-turn $r^2$ has trace $-2$, and every quarter-turn or reflection has trace $0$. Therefore the character table is

$$\begin{array}{cccccc}{D}_4& e& r^2& r& s& rs\\ \text{class size}& 1& 1& 2& 2& 2\\ {\chi }_{++}& 1& 1& 1& 1& 1\\ {\chi }_{+-}& 1& 1& 1& -1& -1\\ {\chi }_{-+}& 1& 1& -1& 1& -1\\ {\chi }_{--}& 1& 1& -1& -1& 1\\ {\chi }_{\mathrm{geom}}& 2& -2& 0& 0& 0\end{array}$$

The dimensions again check out:

$$1^2+1^2+1^2+1^2+2^2=8=|D_4|.$$

Exercise

This exercise computes the character table of the quaternion group

$$Q_8=\{\pm 1,\pm i,\pm j,\pm k\}.$$

1. Show that the conjugacy classes of $Q_8$ are

$$\{1\},\{-1\},\{\pm i\},\{\pm j\},\{\pm k\}.$$

2. Show that the abelianization of $Q_8$ is $Q_8/\{\pm 1\}\cong C_2\times C_2$. Use this to find four one-dimensional characters of $Q_8$.
3. Use orthogonality with the trivial character to show that this final character is

$$(2,-2,0,0,0)$$

on the conjugacy classes above.

4. Write the full character table of $Q_8$ and compare it with the table of $D_4$. What does this show about character tables as invariants of groups?

Footnotes

1. One way to prove this is through the group algebra $\mathbb{C}[G]$. The center of $\mathbb{C}[G]$ has a basis given by the conjugacy-class sums ${\sum }_{g\in C}g$, so its dimension is the number of conjugacy classes of $G$. On the other hand, Schur’s lemma implies that a central element acts by a scalar on each irreducible representation, so the center has one independent scalar parameter for each irreducible representation. Thus the number of irreducible representations equals the number of conjugacy classes. ↩