Representation Theory

Groups and algebras studied through the matrices they become. Start with the symmetries of a triangle and end with quivers, which have no group at all: Schur and Maschke, characters, the Fourier transform as the abelian case, induction, Young diagrams, the first Lie algebra $\mathfrak{sl}(2)$ , and Gabriel's theorem. Every module has numbered examples and a long list of exercises, with the important ones starred.

Follows P. Etingof, O. Golberg, S. Hensel, T. Liu, A. Schwendner, D. Vaintrob and E. Yudovina, Introduction to Representation Theory (lecture notes, 2011; AMS Student Mathematical Library 59), Chapters 1–5, with the Lie material from B. C. Hall, An Elementary Introduction to Groups and Representations (arXiv:math-ph/0005032, 2000), Chapter 5. Section and result numbers refer to those texts. The layout follows Bogachev's Measure Theory: numbered examples with their proofs, then many exercises with hints. ★ marks the exercises worth doing first, and each list can be filtered to those. The examples and exercises are written for this page. For how this subject connects to the others, see the roadmap.

InteractiveEtingof et al. §§1.1–1.3, 1.7; Hall Ch. 5 §§1–3

Representations: groups acting by matrices

A representation turns each element of a group into an invertible matrix, so that multiplying in the group becomes multiplying matrices. You then study the group with linear algebra, where every tool is available: eigenvalues, traces, invariant subspaces, bases.

The idea

Representation (Etingof §1.3). A representation of a group

G

on a vector space

V

over a field

k

is a homomorphism

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

: each

g

acts by an invertible linear map, with

\rho(gh)=\rho(g)\rho(h)

and

\rho(e)=I

. We often write

gv

for

\rho(g)v

. Its dimension is

\dim V

The algebra point of view (Etingof §§1.2–1.3). A representation of an associative algebra

A

is a homomorphism

A\to\mathrm{End}(V)

, also called an

A

-module. A representation of

G

is the same thing as a representation of the group algebra

k[G]

, whose basis is the elements of

G

and whose product extends the group law linearly (Exercise 1.1). This is why the subject can study groups, Lie algebras and quivers with one set of tools.

Subrepresentations, irreducible, indecomposable. A subspace

U\subseteq V

with

gU\subseteq U

for all

g

is a subrepresentation; then

V/U

is a representation too.

V\ne0

is irreducible if its only subrepresentations are

0

and

V

, and indecomposable if it is not a direct sum of two nonzero subrepresentations. Irreducible implies indecomposable; Example 1.3 shows the converse fails.

Intertwiners. A homomorphism of representations

T:V\to W

is a linear map with

T(gv)=g\,T(v)

for all

g,v

. Two representations are isomorphic if there is an invertible intertwiner: the same matrices in different bases. Classifying representations means classifying them up to isomorphism.

The two basic problems (Etingof §1.1). Classify the irreducible representations, and classify the indecomposable ones. For finite groups over

\mathbb C

these coincide (module 2), and the answer is a finite list read off from characters (module 3). For other algebras, such as

\mathbb C[x]

or a quiver, the indecomposables are the harder and more interesting question (module 8).

Why Lie groups keep appearing. A Lie group is a group of symmetries that varies continuously: rotations

\mathrm{SO}(3)

, unitary matrices

\mathrm{SU}(n)

, the circle. Whenever a problem has such a symmetry, its space of solutions carries a representation, and splitting it into irreducible pieces is how the problem gets solved: Fourier series are the circle acting on functions on the circle (module 4), spherical harmonics are

\mathrm{SO}(3)

acting on functions on the sphere (module 7), and the energy levels of the hydrogen atom and the quark model are both decompositions of this kind. That is why physics and analysis send so many people to Lie groups through representation theory. But it runs both ways and neither owns the other. Representation theory also studies finite groups (modules 1–6) and objects with no group at all (module 8); Lie groups also have their own geometry, topology and structure theory, which have little to do with representations.

Explore: three representations of the triangle's symmetries

The faint triangle is the starting position, the solid one its image. The six symmetries form the group

S_3

Symmetry g

Then apply h (to see ρ(h)ρ(g) = ρ(hg))

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Representation: A rule that assigns a matrix to every element of a group, so that doing two group operations in a row corresponds to multiplying their matrices.
Group algebra: Formal combinations of group elements with coefficients, multiplied using the group law. Matrix pictures of the group are exactly the matrix pictures of this bigger object.
Subrepresentation: A subspace that every group element maps back into itself, like the line of constant vectors when the group permutes coordinates.
Irreducible: Describes a nonzero representation with no invariant subspaces except zero and the whole space: it can't be cut into smaller pieces at all.
Indecomposable: Describes a representation that can't be split as a direct sum of two smaller invariant pieces, although it may still contain a smaller invariant piece with no invariant complement.
Intertwiner: A linear map between two representations that commutes with the group action: acting by $g$ and then mapping gives the same result as mapping and then acting.
Permutation representation: The representation you get when a group shuffles a finite set: each element becomes the matrix that shuffles the basis vectors the same way.

Examples

Example 1.1.

Every group has the trivial representation $\rho(g)=1$ on $k$ . The symmetric group $S_n$ also has the sign representation $\rho(\sigma)=\mathrm{sgn}(\sigma)\in\{\pm1\}$ , a different 1-dimensional representation when $n\ge2$ .

Example 1.2 (permutation representation).

$S_n$ acts on $k^n$ by permuting coordinates: $\sigma e_i=e_{\sigma(i)}$ . The line spanned by $(1,\dots,1)$ is a subrepresentation (a copy of the trivial one), and so is the hyperplane $V_0=\{x:\sum x_i=0\}$ , the standard representation of dimension $n-1$ . Over $\mathbb C$ (or any field of characteristic not dividing $n$ ), $k^n=k(1,\dots,1)\oplus V_0$ . In the figure, the $3\times3$ matrices are this representation of $S_3$ , and the $2\times2$ ones are $V_0$ in a rotated basis.

Example 1.3 (indecomposable but not irreducible).

Let $\mathbb Z$ act on $\mathbb C^2$ by $\rho(n)=\begin{pmatrix}1&n\\0&1\end{pmatrix}$ . The line $\mathbb Ce_1$ is invariant, so $\rho$ is not irreducible. An invariant line is an eigenline of $\rho(1)$ , and $\rho(1)$ has only the eigenline $\mathbb Ce_1$ . So no invariant complement exists, and $\rho$ is indecomposable.

Example 1.4 (the field matters).

$\mathbb Z/n$ acts on $\mathbb R^2$ with the generator rotating by $2\pi/n$ . For $n\ge3$ a rotation has no real eigenvector, so this real representation is irreducible. Over $\mathbb C$ the same matrix has eigenvectors $(1,\mp i)$ with eigenvalues $e^{\pm2\pi i/n}$ , and the representation splits into two 1-dimensional ones.

Example 1.5 (functions on a set).

If $G$ acts on a set $X$ , it acts on functions $f:X\to k$ by $(gf)(x)=f(g^{-1}x)$ ; the inverse makes $g(hf)=(gh)f$ (Check your understanding, question 2). For finite $X$ this is the permutation representation again. For $X$ a sphere and $G=\mathrm{SO}(3)$ , it is the representation whose irreducible pieces are the spherical harmonics. Most representations in nature arise this way.

Example 1.6 (the regular representation).

$G$ acting on itself by left multiplication gives the permutation representation on $k[G]$ , of dimension $|G|$ . It will turn out to contain every irreducible representation of a finite group, each as many times as its dimension (module 3).

Example 1.7 (one operator).

A representation of the polynomial algebra $\mathbb C[x]$ is a vector space with one linear operator $T=\rho(x)$ . Isomorphism is similarity of operators, so the classification of finite-dimensional representations is the Jordan normal form: the indecomposables are the Jordan blocks $J_m(\lambda)$ , and the irreducibles are only the $1\times1$ blocks. Here the two basic problems have visibly different answers.

Check your understanding

The representation $n\mapsto\begin{pmatrix}1&n\\0&1\end{pmatrix}$ of $\mathbb Z$ on $\mathbb C^2$ is:

irreducible.The line spanned by $e_1$ is invariant.
indecomposable but not irreducible.
a direct sum of two 1-dimensional representations.That would need two independent invariant lines, i.e. two independent eigenvectors of $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ .

Example 1.3: one invariant line, and no invariant complement.

Why does the action on functions use $(gf)(x)=f(g^{-1}x)$ rather than $f(gx)$ ?

So that each $\rho(g)$ is invertible.Both formulas give invertible maps.
So that $g(hf)=(gh)f$ ; with $f(gx)$ the order of multiplication comes out reversed.
It makes no difference.Compute $g(hf)$ both ways for a non-abelian group.

With

f(gx)

g(hf)(x)=(hf)(gx)=f(hgx)

, which is

(hg)f

. The inverse fixes the order:

g(hf)(x)=(hf)(g^{-1}x)=f(h^{-1}g^{-1}x)=f((gh)^{-1}x)

The rotation-by- $90^\circ$ representation of $\mathbb Z/4$ on $\mathbb C^2$ is:

irreducible, as it is over $\mathbb R$ .Over $\mathbb C$ the rotation matrix has eigenvectors.
a direct sum of two 1-dimensional representations, with the generator acting by $i$ and by $-i$ .
indecomposable but not irreducible.The rotation is diagonalizable over $\mathbb C$ .

Example 1.4: eigenvalues

e^{\pm i\pi/2}=\pm i

Exercises

Exercise 1.1Group representations are group algebra modules

Show that representations of $G$ on $V$ correspond exactly to representations of the group algebra $k[G]$ on $V$ , and that intertwiners are the same on both sides.

Hint

Extend $\rho$ linearly; in the other direction, restrict to the basis elements $g$ .

Solution

Given $\rho:G\to\mathrm{GL}(V)$ , define $\tilde\rho(\sum a_gg)=\sum a_g\rho(g)$ . It is linear, sends $e$ to $I$ , and is multiplicative on basis elements, hence on all of $k[G]$ by bilinearity. Conversely, an algebra homomorphism $\tilde\rho:k[G]\to\mathrm{End}(V)$ gives $\rho(g)=\tilde\rho(g)$ , which is invertible with inverse $\tilde\rho(g^{-1})$ , and multiplicative. A linear map commutes with every $\rho(g)$ exactly when it commutes with every linear combination of them.

Exercise 1.2The standard representation is irreducible

Let $n\ge2$ and $V_0=\{x\in\mathbb C^n:\sum x_i=0\}$ (Example 1.2). Show that $V_0$ is irreducible.

Hint

Take a nonzero $w$ in an invariant subspace $W$ , and compare $w$ with $(ij)w$ .

Solution

Let $0\ne W\subseteq V_0$ be invariant and $w\in W$ , $w\ne0$ . Not all coordinates of $w$ are equal (they would be a common $c$ with $nc=0$ , so $c=0$ ). Pick $w_i\ne w_j$ . Then $w-(ij)w=(w_i-w_j)(e_i-e_j)\in W$ , so $e_i-e_j\in W$ . Applying permutations, every $e_k-e_l$ is in $W$ , and these span $V_0$ . So $W=V_0$ .

Exercise 1.3A non-split extension

In Example 1.3, show that the quotient $\mathbb C^2/\mathbb Ce_1$ is the trivial representation, so $\rho$ is "trivial on top of trivial" without being trivial.

Hint

$\rho(n)e_2=e_2+ne_1\equiv e_2$ modulo $e_1$ .

Exercise 1.4Representations of

\mathbb Z/2

(1) Over $\mathbb C$ , show every finite-dimensional representation of $\mathbb Z/2=\{e,g\}$ is $V_+\oplus V_-$ , where $g$ acts by $+1$ on $V_+$ and $-1$ on $V_-$ . (2) Over a field of characteristic 2, show that $g\mapsto\begin{pmatrix}1&1\\0&1\end{pmatrix}$ is a representation that does not split.

Hint

(1) $v=\frac{v+gv}2+\frac{v-gv}2$ . (2) In characteristic 2, $\begin{pmatrix}1&1\\0&1\end{pmatrix}^2=\begin{pmatrix}1&2\\0&1\end{pmatrix}=I$ .

Solution

(1) The maps $P_\pm=\frac12(I\pm\rho(g))$ satisfy $P_\pm^2=P_\pm$ , $P_++P_-=I$ , $P_+P_-=0$ , and commute with $\rho(g)$ , so $V=\mathrm{im}P_+\oplus\mathrm{im}P_-$ with $g=\pm1$ on the pieces. (2) The square is the identity, so this is a representation. Its only eigenline is $ke_1$ , as in Example 1.3, so it has no invariant complement. Dividing by 2 in (1) is exactly what fails: this is the first sign of Maschke's theorem (module 2).

Exercise 1.5The dual representation

Show that $\rho^*(g)=\rho(g^{-1})^\top$ defines a representation on $V^*$ , and that $g\mapsto\rho(g)^\top$ in general does not.

Hint

Transposition reverses products: $(AB)^\top=B^\top A^\top$ . The inverse reverses them back.

Exercise 1.6Kernels and images of intertwiners

Show that the kernel and the image of an intertwiner $T:V\to W$ are subrepresentations of $V$ and $W$ . (This is the whole engine of Schur's lemma in module 2.)

Hint

If $Tv=0$ then $T(gv)=gTv=0$ .

Solution

If $Tv=0$ , then $T(gv)=g\,Tv=0$ , so $gv\in\ker T$ . If $w=Tv$ , then $gw=g\,Tv=T(gv)\in\mathrm{im}\,T$ .

Exercise 1.7Invariants of a permutation representation

$G$ acts on a finite set $X$ . Show that the invariant vectors in the permutation representation $k[X]$ are the functions constant on orbits, so $\dim k[X]^G$ is the number of orbits.

Hint

$f$ is invariant when $f(g^{-1}x)=f(x)$ for all $g,x$ .

Solution

$f$ is invariant iff $f(g^{-1}x)=f(x)$ for all $g,x$ , i.e. $f$ takes one value on each orbit. The indicator functions of the orbits form a basis of these functions. Combined with module 3 this gives Burnside's counting lemma (Example 3.7).

Exercise 1.8One-dimensional representations

Show that every 1-dimensional representation of $G$ is trivial on the commutator subgroup $[G,G]$ , so 1-dimensional representations of $G$ are those of $G/[G,G]$ . Given that $[S_n,S_n]=A_n$ for $n\ge2$ , deduce that the trivial and sign representations are the only 1-dimensional representations of $S_n$ over $\mathbb C$ .

Hint

$\mathrm{GL}_1(k)=k^\times$ is abelian, so $\rho(ghg^{-1}h^{-1})=1$ . $S_n/A_n\cong\mathbb Z/2$ has two characters.

Exercise 1.9Same matrices, different representations

Show that the 2-dimensional representations $n\mapsto\begin{pmatrix}1&n\\0&1\end{pmatrix}$ and $n\mapsto I$ of $\mathbb Z$ have the same characteristic polynomial for every $n$ but are not isomorphic. (For finite groups over $\mathbb C$ this cannot happen: module 3.)

Hint

An isomorphism would conjugate $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ to $I$ .

InteractiveEtingof et al. Prop. 1.16, Cor. 1.17, §§3.1, 3.6; Hall Ch. 5 §§5, 7

Schur's lemma and Maschke's theorem

Two short results carry most of the theory of finite groups. Schur's lemma says there are almost no maps between irreducible representations. Maschke's theorem says every representation of a finite group over $\mathbb C$ is a direct sum of irreducible ones. Together they reduce everything to a list of irreducibles and their multiplicities.

The idea

Schur's lemma (Etingof Prop. 1.16, Cor. 1.17). Let

V,W

be irreducible. Every nonzero intertwiner

V\to W

is an isomorphism. If

k

is algebraically closed (for example

\mathbb C

) and

V

is finite-dimensional, every intertwiner

V\to V

is a scalar multiple of the identity.

The proof is two lines (Exercise 2.1): kernels and images of intertwiners are subrepresentations, and an eigenvalue $\lambda$ of $T:V\to V$ makes $T-\lambda I$ an intertwiner that is not injective, hence zero.

Maschke's theorem (Etingof Thm. 3.1). If

G

is finite and the characteristic of

k

does not divide

|G|

, every subrepresentation

W\subseteq V

has an invariant complement. So every finite-dimensional representation is a direct sum of irreducible ones.

The proof averages over the group. Take any linear projection $P$ of $V$ onto $W$ , and replace it by $\bar P=\frac1{|G|}\sum_g\rho(g)P\rho(g)^{-1}$ . This is still a projection onto $W$ , and it is an intertwiner, so its kernel is an invariant complement (Exercise 2.3). The division by $|G|$ is where the characteristic enters.

The unitary trick (Etingof §3.6; Hall Ch. 5 §5). Over

\mathbb C

, average an inner product instead:

\langle u,v\rangle'=\frac1{|G|}\sum_g\langle gu,gv\rangle

is an inner product for which every

\rho(g)

is unitary, and the orthogonal complement of an invariant subspace is invariant. The figure shows this for

\mathbb Z/3

. For a compact group such as

\mathrm{SU}(2)

, replace the sum by an integral against Haar measure: that is Hall's proof of complete reducibility for compact Lie groups.

When it fails. In characteristic

p\mid|G|

(Exercise 1.4(2)), and for infinite groups such as

\mathbb Z

(Example 1.3), representations need not split. Modular representation theory and the theory of indecomposables study exactly these cases.

Explore: averaging an inner product

The generator acts by a rotation by

120^\circ

seen in skewed coordinates,

A=SRS^{-1}

. Drag the point

v

: its orbit

v,Av,A^2v

leaves the circle through

v

but always lies on the averaged ellipse .

Skew of the coordinates

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Schur's lemma: Between two irreducible pieces, a map that respects the action is either zero or an isomorphism; from an irreducible piece to itself over the complex numbers, it can only be multiplication by a number.
Maschke's theorem: For a finite group, as long as you can divide by its order, every invariant subspace has an invariant complement, so every representation breaks into irreducible pieces.
Complete reducibility: The property of breaking completely into a direct sum of irreducible pieces.
Averaging over the group: Replacing any object (a projection, an inner product) by the mean of all its transforms under the group, which produces something the group leaves unchanged.
Unitary representation: One in which every group element preserves lengths and angles for some fixed inner product. Then the perpendicular space to an invariant subspace is invariant too.
Central character: The scalar by which each element of the centre of the group acts on an irreducible complex representation.

Examples

Example 2.1 (cyclic groups).

Every irreducible complex representation of $\mathbb Z/n$ is 1-dimensional (Exercise 2.2), so it is determined by the image of the generator, an $n$ th root of unity. The irreducibles are $\chi_j(m)=\omega^{jm}$ with $\omega=e^{2\pi i/n}$ , $j=0,\dots,n-1$ : exactly $n$ of them.

Example 2.2 (Schur needs

\mathbb C

The rotation representation of $\mathbb Z/4$ on $\mathbb R^2$ is irreducible over $\mathbb R$ , but every matrix $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ commutes with the rotation. So its intertwiners form a copy of $\mathbb C$ , not just the real scalars. Over $\mathbb R$ , Schur's lemma only says that the intertwiners form a division algebra (Exercise 2.7).

Example 2.3 (Maschke needs the characteristic).

Over $\mathbb F_p$ , $\mathbb Z/p$ acts on $\mathbb F_p^2$ by $m\mapsto\begin{pmatrix}1&m\\0&1\end{pmatrix}$ , well defined because $p\equiv0$ . As in Example 1.3, $\mathbb F_pe_1$ is invariant with no invariant complement.

Example 2.4 (an averaged projection).

In the permutation representation of $S_3$ on $\mathbb C^3$ , start from the non-invariant projection $P(x)=(x_1,x_1,x_1)$ onto the line $\mathbb C(1,1,1)$ . Averaging gives $\bar P(x)=\frac{x_1+x_2+x_3}3(1,1,1)$ , the orthogonal projection, whose kernel is the standard representation $V_0$ .

Example 2.5 (central characters).

In the dihedral group $D_4$ of the square, the rotation $r^2$ by $180^\circ$ is central. On the 2-dimensional representation by symmetries of the square it acts by $-I$ , and on each of the four 1-dimensional ones by $+1$ . Schur's lemma predicts exactly this: central elements act by scalars on irreducibles.

Example 2.6 (a compact group).

The circle $\mathrm{SO}(2)$ acts on continuous functions on the circle by rotation. Averaging with respect to arc length (Haar measure) makes the action unitary for $\langle f,g\rangle=\frac1{2\pi}\int f\bar g$ , and the irreducible pieces are the lines spanned by $e^{ik\theta}$ . Decomposing a function into them is its Fourier series (module 4).

Check your understanding

$V$ is an irreducible complex representation of a finite group, and $T:V\to V$ commutes with every $\rho(g)$ . Then:

$T=\lambda I$ for some $\lambda\in\mathbb C$ .
$T$ is invertible. $T=0$ commutes with everything.
$T$ is diagonalizable but may have several eigenvalues.Each eigenspace of $T$ would be a subrepresentation.

Schur's lemma over an algebraically closed field.

In the averaging proof of Maschke's theorem, which step needs the characteristic of $k$ not to divide $|G|$ ?

Checking that $\bar P$ commutes with $\rho(g)$ .That is reindexing the sum, valid in any characteristic.
Checking that $\bar P$ is the identity on $W$ : the sum of $|G|$ identity maps must be divided by $|G|$ .
Choosing a linear projection onto $W$ .That exists over any field.

For

w\in W

, each term

\rho(g)P\rho(g)^{-1}w=w

, so

\sum_g\rho(g)P\rho(g)^{-1}w=|G|\,w

, and you must divide by

|G|

Is every finite-dimensional complex representation of $\mathbb Z$ a direct sum of irreducible ones?

Yes, by Maschke's theorem.Maschke's theorem is for finite groups.
No: Example 1.3 is a counterexample.

A representation of

\mathbb Z

is one invertible matrix, and it splits completely exactly when that matrix is diagonalizable.

Exercises

Exercise 2.1Prove Schur's lemma

Prove both statements of Schur's lemma.

Hint

Use Exercise 1.6. For the second part, take an eigenvalue of $T$ .

Solution

Let $T:V\to W$ be a nonzero intertwiner. $\ker T$ is a subrepresentation of $V$ , not all of $V$ , so it is $0$ ; $\mathrm{im}\,T$ is a nonzero subrepresentation of $W$ , so it is $W$ . Hence $T$ is an isomorphism. For $T:V\to V$ over an algebraically closed field with $\dim V<\infty$ , $T$ has an eigenvalue $\lambda$ . Then $T-\lambda I$ is an intertwiner with nonzero kernel, so by the first part it is $0$ .

Exercise 2.2Abelian groups have 1-dimensional irreducibles

Show that every finite-dimensional irreducible complex representation of an abelian group is 1-dimensional.

Hint

Each $\rho(h)$ commutes with all $\rho(g)$ .

Solution

Since $G$ is abelian, each $\rho(h)$ is an intertwiner $V\to V$ , so by Schur's lemma $\rho(h)=\lambda_hI$ . Then every subspace is invariant, and irreducibility forces $\dim V=1$ .

Exercise 2.3Prove Maschke's theorem

Let $P$ be a linear projection of $V$ onto an invariant subspace $W$ , and $\bar P=\frac1{|G|}\sum_g\rho(g)P\rho(g)^{-1}$ . Show that $\bar P$ is an intertwiner, that $\bar P|_W=\mathrm{id}$ , that $\mathrm{im}\,\bar P=W$ , and conclude that $\ker\bar P$ is an invariant complement of $W$ .

Hint

For the intertwining property, substitute $g\to hg$ in the sum.

Solution

$\rho(h)\bar P\rho(h)^{-1}=\frac1{|G|}\sum_g\rho(hg)P\rho(hg)^{-1}=\bar P$ . For $w\in W$ , $\rho(g)^{-1}w\in W$ , so $P\rho(g)^{-1}w=\rho(g)^{-1}w$ and each term gives $w$ ; hence $\bar Pw=w$ . Each term maps $V$ into $\rho(g)W=W$ , so $\mathrm{im}\,\bar P\subseteq W$ , with equality by the previous step. So $\bar P$ is a projection onto $W$ , $V=W\oplus\ker\bar P$ , and $\ker\bar P$ is invariant by Exercise 1.6.

Exercise 2.4The unitary trick

For a finite group acting on $\mathbb C^n$ , show that $\langle u,v\rangle'=\frac1{|G|}\sum_g\langle gu,gv\rangle$ is an inner product, that every $\rho(h)$ is unitary for it, and that the $\langle\cdot,\cdot\rangle'$ -orthogonal complement of an invariant subspace is invariant.

Hint

Positivity: each term is $\ge0$ and the $g=e$ term is $>0$ . Invariance: reindex by $g\to gh$ .

Exercise 2.5Two-dimensional representations of

\mathbb Z/2

over

\mathbb F_2

Classify the 2-dimensional representations of $\mathbb Z/2$ over $\mathbb F_2$ up to isomorphism.

Hint

$\rho(g)^2=I$ means $(\rho(g)-I)^2=0$ in characteristic 2. So $\rho(g)-I$ is either $0$ or a nilpotent of rank 1. Answer: two classes, $I$ and $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ .

Exercise 2.6Irreducibles are quotients of the regular representation

Let $V$ be irreducible and $v\in V$ nonzero. Show that $a\mapsto av$ is a surjective intertwiner $k[G]\to V$ . Deduce $\dim V\le|G|$ for finite $G$ .

Hint

Its image is a nonzero subrepresentation.

Solution

$h(av)=(ha)v$ , so the map intertwines left multiplication on $k[G]$ with the action on $V$ . Its image contains $v=ev\ne0$ and is a subrepresentation (Exercise 1.6), hence all of $V$ . So $\dim V\le\dim k[G]=|G|$ .

Exercise 2.7Schur's lemma over

\mathbb R

Show that the intertwiners of an irreducible real representation with itself form a division algebra. Using Frobenius's theorem (every finite-dimensional real division algebra is $\mathbb R$ , $\mathbb C$ or the quaternions $\mathbb H$ ), find a representation for each case.

Hint

A nonzero intertwiner is invertible by the first part of Schur's lemma. Examples: the trivial representation ( $\mathbb R$ ); rotation by $90^\circ$ of $\mathbb Z/4$ on $\mathbb R^2$ ( $\mathbb C$ , Example 2.2); the quaternion group $Q_8$ acting on $\mathbb H\cong\mathbb R^4$ by left multiplication (right multiplications by $\mathbb H$ commute with it).

Exercise 2.8Faithful irreducibles force a cyclic centre

Show that the centre $Z(G)$ of a finite group acts by scalars on each irreducible complex representation. Deduce that if $G$ has a faithful irreducible complex representation, then $Z(G)$ is cyclic.

Hint

Schur's lemma gives a homomorphism $Z(G)\to\mathbb C^\times$ , injective if $\rho$ is faithful. Finite subgroups of $\mathbb C^\times$ are cyclic.

CalculatorEtingof et al. §§3.2–3.9

Characters and orthogonality

The trace of each matrix, one number per group element, turns out to determine a representation of a finite group completely. The irreducible characters are orthonormal, so decomposing a representation is computing inner products, the way you find coordinates in an orthonormal basis.

The idea

Character (Etingof §3.2).

\chi_V(g)=\mathrm{tr}\,\rho(g)

. It is constant on conjugacy classes (

\mathrm{tr}(ABA^{-1})=\mathrm{tr}B

), so it is a class function, and

\chi_V(e)=\dim V

. Characters add over direct sums and multiply over tensor products:

\chi_{V\oplus W}=\chi_V+\chi_W

\chi_{V\otimes W}=\chi_V\chi_W

\chi_{V^*}=\overline{\chi_V}

Orthogonality (Etingof Thm. 3.8). For complex representations of a finite group, with

\langle\chi,\psi\rangle=\frac1{|G|}\sum_g\chi(g)\overline{\psi(g)}

\langle\chi_V,\chi_W\rangle=\dim\mathrm{Hom}_G(W,V).

So the irreducible characters are orthonormal, the multiplicity of an irreducible

V_i

V

\langle\chi_V,\chi_i\rangle

, and

V

is irreducible exactly when

\langle\chi_V,\chi_V\rangle=1

. Since multiplicities determine

V

(Maschke), two representations with the same character are isomorphic.

Counting (Etingof §§3.2, 3.5, 4.4). The irreducible characters form a basis of the class functions, so the number of irreducible representations equals the number of conjugacy classes. The regular representation contains

V_i

exactly

d_i=\dim V_i

times, so

\sum_id_i^2=|G|

. Each

d_i

divides

|G|

(Frobenius divisibility).

The character table. Rows are irreducible characters, columns conjugacy classes. Rows are orthonormal when each column is weighted by its class size, and the columns are orthogonal too:

\sum_i\chi_i(g)\overline{\chi_i(h)}=|C_G(g)|

g,h

are conjugate and

0

otherwise (Exercise 3.9).

Explore: a character table workbench

Group

Representation to decompose

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Character: The list of traces of a representation's matrices, one number per group element. For finite groups over the complex numbers it pins the representation down completely.
Class function: A function on a group that takes the same value on elements that are conjugate to each other.
Conjugacy class: All the elements you can get from one element $g$ as $hgh^{-1}$ : the elements that "do the same thing in a different frame".
Character table: A square grid with one row for each irreducible representation and one column for each class of conjugate elements, filled with traces.
Multiplicity: How many times a given irreducible piece occurs when a representation is broken up; computed as an inner product of traces.
Regular representation: The group acting on formal combinations of its own elements by multiplying on the left. Every irreducible piece appears in it as many times as its dimension.

Examples

Example 3.1 (the table of

S_3

Classes: $\{e\}$ , the three transpositions, the two 3-cycles, so there are three irreducibles, and $d_1^2+d_2^2+d_3^2=6$ forces dimensions $1,1,2$ . The trivial and sign characters are $(1,1,1)$ and $(1,-1,1)$ . The 2-dimensional one, $V_0$ from Example 1.2, is the permutation character $(3,1,0)$ minus the trivial one: $(2,0,-1)$ .

Example 3.2 (decomposing by inner products).

For $S_3$ , $\chi_{V_0}^2=(4,0,1)$ . Its inner products with the three rows are $\frac16(4+0+2)=1$ , $\frac16(4-0+2)=1$ and $\frac16(8+0-2)=1$ , so $V_0\otimes V_0\cong\mathbf 1\oplus\mathrm{sgn}\oplus V_0$ : the symmetric square minus its trivial part, and the alternating square.

Example 3.3 (the regular representation).

$\chi_{\mathrm{reg}}(e)=|G|$ and $\chi_{\mathrm{reg}}(g)=0$ otherwise (left multiplication by $g\ne e$ fixes no element). So $\langle\chi_{\mathrm{reg}},\chi_i\rangle=\frac1{|G|}|G|d_i=d_i$ . For $S_3$ : $\mathrm{reg}=\mathbf 1\oplus\mathrm{sgn}\oplus V_0^{\oplus2}$ , dimension $1+1+4=6$ .

Example 3.4 (

D_4

and

Q_8

The dihedral group of the square and the quaternion group both have order 8, five classes, irreducible dimensions $1,1,1,1,2$ , and literally the same character table (try both in the workbench). They are not isomorphic: $D_4$ has five elements of order 2, $Q_8$ only one. The character table does not determine the group.

Example 3.5 (

A_4

and complex values).

$A_4$ has classes $\{e\}$ , the three double transpositions, and two classes of four 3-cycles each ( $(123)$ is not conjugate to $(132)$ inside $A_4$ ). Its quotient by the Klein four-group is $\mathbb Z/3$ , giving characters $(1,1,\omega,\omega^2)$ and $(1,1,\omega^2,\omega)$ , and the rotations of a tetrahedron give a 3-dimensional irreducible with character $(3,-1,0,0)$ . Then $1+1+1+9=12$ .

Example 3.6 (an irreducibility test).

The rotation representation of $A_4$ has $\langle\chi,\chi\rangle=\frac1{12}(9+3\cdot1+0+0)=1$ , so it is irreducible, with no need to look for invariant subspaces.

Example 3.7 (Burnside's counting lemma).

By Exercise 1.7 and orthogonality, the number of orbits of $G$ on a finite set $X$ is $\langle\chi_{k[X]},\mathbf 1\rangle=\frac1{|G|}\sum_g|\mathrm{Fix}(g)|$ . The rotations of a square acting on the $2^4$ black-and-white colourings of its corners fix $16,2,4,2$ colourings, so there are $\frac{24}4=6$ colourings up to rotation.

Check your understanding

A representation has $\langle\chi,\chi\rangle=3$ . Then it is:

a sum of three non-isomorphic irreducibles, each appearing once.
three copies of one irreducible.That gives $\langle\chi,\chi\rangle=3^2=9$ .
irreducible of dimension 3.Irreducible means $\langle\chi,\chi\rangle=1$ , whatever the dimension.

\langle\chi,\chi\rangle=\sum m_i^2

, and

3=1+1+1

is the only way to write 3 as a sum of squares of positive integers.

How many irreducible complex representations does $S_4$ have?

4Count the cycle types of permutations of four letters.
5
24That is $\sum d_i^2$ , not the number of irreducibles.

The cycle types

1^4,\ 2\,1^2,\ 2^2,\ 3\,1,\ 4

are the five conjugacy classes, with irreducible dimensions

1,1,2,3,3

A non-abelian group of order 8 has how many 1-dimensional representations?

8Then all irreducibles would be 1-dimensional and the group abelian (Exercise 3.10).
4
2Then $8-2=6$ would have to be a sum of squares of dimensions $\ge2$ dividing 8.

8=1+1+1+1+4

is the only option with some

d_i>1

and every

d_i

dividing 8.

Exercises

Exercise 3.1Inverses and complex conjugates

For a finite group over $\mathbb C$ , show that $\chi(g^{-1})=\overline{\chi(g)}$ and $|\chi(g)|\le\chi(e)$ .

Hint

$\rho(g)^{|G|}=I$ , so $\rho(g)$ is diagonalizable with eigenvalues that are roots of unity.

Solution

$\rho(g)$ satisfies $x^{|G|}-1=0$ , which has distinct roots, so it is diagonalizable with eigenvalues $\lambda_j$ , $|\lambda_j|=1$ . Then $\chi(g^{-1})=\sum\lambda_j^{-1}=\sum\bar\lambda_j=\overline{\chi(g)}$ , and $|\chi(g)|\le\sum|\lambda_j|=\dim V=\chi(e)$ .

Exercise 3.2Prove orthogonality

(1) Show that $\frac1{|G|}\sum_g\rho(g)$ is a projection of $V$ onto the invariants $V^G$ , so $\dim V^G=\frac1{|G|}\sum_g\chi_V(g)$ . (2) Apply (1) to $\mathrm{Hom}(W,V)\cong W^*\otimes V$ to prove $\langle\chi_V,\chi_W\rangle=\dim\mathrm{Hom}_G(W,V)$ .

Hint

The invariants of $\mathrm{Hom}(W,V)$ under $g\cdot T=\rho_V(g)T\rho_W(g)^{-1}$ are the intertwiners.

Solution

(1) $e=\frac1{|G|}\sum_g\rho(g)$ satisfies $\rho(h)e=e$ , so its image is invariant, it is the identity on $V^G$ , and $e^2=e$ . The trace of a projection is the dimension of its image. (2) $G$ acts on $\mathrm{Hom}(W,V)$ by $g\cdot T=\rho_V(g)T\rho_W(g)^{-1}$ , with fixed points $\mathrm{Hom}_G(W,V)$ and character $\chi_V(g)\chi_W(g^{-1})=\chi_V(g)\overline{\chi_W(g)}$ by Exercise 3.1. Apply (1).

Exercise 3.3The irreducibility test

Show that $V$ is irreducible if and only if $\langle\chi_V,\chi_V\rangle=1$ , and that two representations with the same character are isomorphic.

Hint

Write $V=\bigoplus V_i^{m_i}$ and use orthonormality of the $\chi_i$ .

Solution

By Maschke, $V\cong\bigoplus V_i^{m_i}$ , so $\chi_V=\sum m_i\chi_i$ and $\langle\chi_V,\chi_V\rangle=\sum m_i^2$ (by Exercise 3.2 and Schur's lemma, $\langle\chi_i,\chi_j\rangle=\delta_{ij}$ ). This is 1 exactly when one $m_i$ is 1 and the rest are 0. The multiplicities $m_i=\langle\chi_V,\chi_i\rangle$ are read off from the character, and they determine $V$ .

Exercise 3.4The two groups of order 4

Write down the character tables of $\mathbb Z/4$ and $\mathbb Z/2\times\mathbb Z/2$ , and show they cannot be made equal by reordering rows and columns.

Hint

$\mathbb Z/4$ has characters with values $\pm i$ ; every character of $\mathbb Z/2\times\mathbb Z/2$ takes values $\pm1$ . So, unlike Example 3.4, here the tables tell the groups apart.

Exercise 3.5Sum of squares of dimensions

Prove that the regular representation contains each irreducible $V_i$ exactly $d_i$ times, and deduce $\sum_id_i^2=|G|$ .

Hint

Example 3.3.

Solution

Left multiplication by $g\ne e$ moves every basis element, so $\chi_{\mathrm{reg}}(g)=0$ , while $\chi_{\mathrm{reg}}(e)=|G|$ . The multiplicity is $\langle\chi_{\mathrm{reg}},\chi_i\rangle=\frac1{|G|}|G|\,\overline{\chi_i(e)}=d_i$ . Comparing dimensions, $|G|=\sum_id_i\cdot d_i$ .

Exercise 3.6Build the table of

S_4

Construct the character table of $S_4$ (classes $e$ , $(12)$ , $(12)(34)$ , $(123)$ , $(1234)$ , of sizes $1,6,3,8,6$ ) from: the trivial and sign characters; the standard representation (permutation character minus trivial); its tensor product with sign; and the 2-dimensional representation pulled back along $S_4\to S_3$ (the action on the three ways of pairing up four letters).

Hint

Rows: $(1,1,1,1,1)$ , $(1,-1,1,1,-1)$ , $(2,0,2,-1,0)$ , $(3,1,-1,0,-1)$ , $(3,-1,-1,0,1)$ . Check each with the irreducibility test and against the workbench.

Exercise 3.7A tensor square in

S_4

Let $V$ be the 3-dimensional standard representation of $S_4$ . Decompose $V\otimes V$ .

Hint

$\chi^2=(9,1,1,0,1)$ . The inner products give $V\otimes V\cong\mathbf 1\oplus V_{(2,2)}\oplus V\oplus(V\otimes\mathrm{sgn})$ , dimensions $1+2+3+3$ . Compare with "standard ⊗ standard" in the workbench.

Exercise 3.8Counting necklaces

Using Example 3.7, count the necklaces of 6 beads in 2 colours up to rotation.

Hint

Rotation by $k$ places fixes $2^{\gcd(k,6)}$ colourings: $\frac16(64+2+4+8+4+2)=14$ .

Exercise 3.9Column orthogonality

Show that $\sum_i\chi_i(g)\overline{\chi_i(h)}$ equals $|C_G(g)|$ if $g$ and $h$ are conjugate and $0$ otherwise.

Hint

The matrix $U_{ic}=\sqrt{|c|/|G|}\,\chi_i(c)$ , over irreducibles $i$ and classes $c$ , is square with orthonormal rows, hence unitary, so its columns are orthonormal too. Use $|G|/|c|=|C_G(g)|$ for $g\in c$ .

Exercise 3.10All irreducibles 1-dimensional means abelian

Show that if every irreducible complex representation of a finite group $G$ is 1-dimensional, then $G$ is abelian.

Hint

Then the regular representation is a sum of 1-dimensional pieces: in a suitable basis all $\rho_{\mathrm{reg}}(g)$ are diagonal, so they commute. The regular representation is faithful.

InteractiveEtingof et al. §3 (the abelian case); Grafakos, Classical Fourier Analysis, §3.1

Abelian groups and the Fourier transform

For an abelian group every irreducible representation is 1-dimensional, and breaking the space of functions on the group into those pieces is exactly the Fourier transform. The discrete Fourier transform, Fourier series and the Fourier integral are the same theorem for three different groups.

The idea

Characters of an abelian group. A character of a finite abelian group

G

is a homomorphism

\chi:G\to\mathbb C^\times

; by Exercise 2.2 of the representation theory page these are all its irreducible representations. They form a group

\hat G

under pointwise multiplication, the dual group, and

\hat G\cong G

(Exercise 4.4). For

G=\mathbb Z/n

\chi_k(x)=e^{2\pi ikx/n}

k=0,\dots,n-1

Fourier transform and inversion. For

f:G\to\mathbb C

put

\hat f(\chi)=\frac1{|G|}\sum_xf(x)\overline{\chi(x)}

. Since the characters are an orthonormal basis of functions on

G

for

\langle f,g\rangle=\frac1{|G|}\sum f\bar g

(Exercise 4.1),

f=\sum_{\chi\in\hat G}\hat f(\chi)\,\chi,\qquad \frac1{|G|}\sum_x|f(x)|^2=\sum_\chi|\hat f(\chi)|^2\quad\text{(Plancherel)}.

In the language of representations: functions on

G

are the regular representation, and

\hat f(\chi)\chi

is the component of

f

in the copy of

\chi

Convolution becomes multiplication. With

(f*g)(x)=\sum_yf(y)g(x-y)

, one has

\widehat{f*g}=|G|\,\hat f\,\hat g

(Exercise 4.2). Translation-invariant linear operators (convolutions, circulant matrices) are intertwiners of the regular representation, so by Schur's lemma each acts on each character by a scalar: the Fourier transform diagonalizes all of them at once.

Three groups, one theorem. For the circle

\mathbb T=\mathbb R/2\pi\mathbb Z

the continuous characters are

e^{ik\theta}

k\in\mathbb Z

, and the transform is the sequence of Fourier coefficients (Grafakos §3.1). For

\mathbb R

they are

e^{2\pi ix\xi}

\xi\in\mathbb R

, and the transform is the Fourier integral. Sums become integrals against Haar measure; the algebra is the same. For non-abelian compact groups the irreducibles are no longer 1-dimensional, and the Peter–Weyl theorem replaces characters by matrix entries of irreducible representations.

Explore: a signal and its Fourier transform on $\mathbb Z/n$

Signal f

n Then convolve with a box of width

f

(drag the bars) , and

f*b

for a box

b

of the chosen width

|\hat f(\chi_k)|

and

|\widehat{f*b}(\chi_k)|=n|\hat f||\hat b|

, for

k=0,\dots,n-1

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Dual group: The collection of all "pure frequencies" of an abelian group (homomorphisms to the unit circle), which is itself a group under pointwise multiplication.
Fourier transform: Writing a function on a group as a combination of pure frequencies, and recording how much of each frequency it contains.
Plancherel identity: The total energy of a signal equals the total energy of its frequency components (with the right normalisation): nothing is lost or gained by changing to frequencies.
Convolution: Sliding one function along another and adding up the products: a weighted moving average. In frequencies it becomes plain multiplication.
Circulant matrix: A square matrix in which each row is the previous row shifted one place to the right, wrapping around. All of them share the same eigenvectors, the pure frequencies.
Peter–Weyl theorem: The version of Fourier analysis for compact groups that need not be abelian: functions are expanded using the matrix entries of irreducible representations instead of single frequencies.

Examples

Example 4.1 (

\mathbb Z/2

The characters are $\chi_0=(1,1)$ and $\chi_1=(1,-1)$ , so $\hat f(\chi_0)=\frac{f(0)+f(1)}2$ and $\hat f(\chi_1)=\frac{f(0)-f(1)}2$ : the average and half the difference: the splitting of a representation of $\mathbb Z/2$ into the parts where the generator acts by $+1$ and by $-1$ .

Example 4.2 (the DFT matrix).

For $\mathbb Z/n$ the matrix $F_{kx}=\frac1n\omega^{-kx}$ , $\omega=e^{2\pi i/n}$ , sends $f$ to $\hat f$ . Orthogonality of characters says $\sqrt nF$ is unitary. The fast Fourier transform computes it in $O(n\log n)$ steps instead of $O(n^2)$ , by splitting $\mathbb Z/2m$ into even and odd residues.

Example 4.3 (a spike and a constant).

The spike $\delta_0$ has $\hat\delta_0(\chi)=\frac1{|G|}$ for every $\chi$ : it contains every frequency equally. The constant 1 has $\hat1=\delta_{\chi_0}$ . Concentrated on one side means spread out on the other (Exercise 4.7). Try "spike" in the figure.

Example 4.4 (moving averages).

Convolving with a box $b=\frac1w\mathbf 1_{\{0,\dots,w-1\}}$ replaces $f$ by its average over a window. Its transform $n\hat b(\chi_k)=\frac1w\sum_{j<w}\omega^{-jk}$ is close to 1 for small $k$ and small for most large $k$ , and vanishes exactly at the $k\ne0$ with $n\mid kw$ : averaging keeps low frequencies and damps high ones.

Example 4.5 (Boolean functions).

For $G=(\mathbb Z/2)^n$ , written multiplicatively as $\{\pm1\}^n$ , the characters are $\chi_S(x)=\prod_{i\in S}x_i$ for $S\subseteq\{1,\dots,n\}$ . The transform is the Walsh–Hadamard transform, the basic tool of the analysis of Boolean functions. For example $\mathrm{Maj}(x_1,x_2,x_3)=\frac12(x_1+x_2+x_3)-\frac12x_1x_2x_3$ .

Example 4.6 (Fourier series).

On the circle, $\hat f(k)=\frac1{2\pi}\int_0^{2\pi}f(\theta)e^{-ik\theta}\,d\theta$ , and Plancherel reads $\frac1{2\pi}\int|f|^2=\sum_k|\hat f(k)|^2$ . The finite theory has no convergence questions; on the circle, whether and in what sense $\sum\hat f(k)e^{ik\theta}$ converges back to $f$ is the subject of Grafakos Chapters 3 and 4.

Check your understanding

Why does the discrete Fourier transform diagonalize every circulant matrix?

Circulant matrices commute with translation, i.e. they intertwine the regular representation of $\mathbb Z/n$ , and each character spans its own 1-dimensional piece, so each is an eigenvector.
Because circulant matrices are symmetric.Most are not, e.g. the shift itself.
Because the DFT matrix is unitary.Unitarity alone doesn't make it diagonalize anything in particular.

The pieces of the regular representation of an abelian group are distinct 1-dimensional representations, so an intertwiner must map each into itself, acting by a scalar (Exercise 4.5).

How many characters does $\mathbb Z/2\times\mathbb Z/4$ have?

6That is $2+4$ ; characters of a product are products of characters.
8
4An abelian group has as many characters as elements.

Every element is its own conjugacy class, so there are

|G|=8

irreducibles, all 1-dimensional:

\chi(a,b)=(-1)^{ja}i^{kb}

Exercises

Exercise 4.1Characters are an orthonormal basis

For a finite abelian group, show directly that distinct characters are orthonormal for $\langle f,g\rangle=\frac1{|G|}\sum f\bar g$ , that there are $|G|$ of them, and hence that Fourier inversion holds.

Hint

If $\chi\ne\psi$ , then $\chi\bar\psi$ is a nontrivial character $\eta$ ; pick $a$ with $\eta(a)\ne1$ and compare $\sum_x\eta(x)$ with $\sum_x\eta(ax)$ . For the count, use $\sum d_i^2=|G|$ with all $d_i=1$ .

Solution

$\langle\chi,\chi\rangle=1$ since $|\chi|=1$ . If $\eta=\chi\bar\psi$ is nontrivial, $S=\sum_x\eta(x)$ satisfies $S=\sum_x\eta(ax)=\eta(a)S$ with $\eta(a)\ne1$ , so $S=0$ . By Schur's lemma (Exercise 2.2 on the representation theory page) all irreducibles are characters, and $\sum d_i^2=|G|$ gives $|G|$ of them. $|G|$ orthonormal functions in a $|G|$ -dimensional space are a basis, and the coefficient of $\chi$ in $f$ is $\langle f,\chi\rangle=\hat f(\chi)$ .

Exercise 4.2The convolution theorem

With the conventions above, prove $\widehat{f*g}=|G|\,\hat f\,\hat g$ .

Hint

Substitute $x=y+z$ and use $\chi(y+z)=\chi(y)\chi(z)$ .

Solution

$\widehat{f*g}(\chi)=\frac1{|G|}\sum_{x,y}f(y)g(x-y)\overline{\chi(x)}=\frac1{|G|}\sum_yf(y)\overline{\chi(y)}\sum_zg(z)\overline{\chi(z)}=\frac1{|G|}\cdot|G|\hat f(\chi)\cdot|G|\hat g(\chi)$ .

Exercise 4.3Plancherel

Prove $\sum_x|f(x)|^2=|G|\sum_\chi|\hat f(\chi)|^2$ , and the polarized form for $\sum_xf(x)\overline{g(x)}$ .

Hint

Expand $f=\sum\hat f(\chi)\chi$ and use orthonormality.

Solution

$\sum_x|f(x)|^2=\sum_{\chi,\psi}\hat f(\chi)\overline{\hat f(\psi)}\sum_x\chi(x)\overline{\psi(x)}=|G|\sum_\chi|\hat f(\chi)|^2$ , since the inner sum is $|G|\delta_{\chi\psi}$ . The same computation with $g$ in the second slot gives $\sum f\bar g=|G|\sum\hat f\,\overline{\hat g}$ .

Exercise 4.4The dual group

Show that $\hat G$ is a group under pointwise multiplication, that the dual group of $\mathbb Z/n$ is isomorphic to $\mathbb Z/n$ , and that $\widehat{G\times H}\cong\hat G\times\hat H$ . Deduce $\hat G\cong G$ for every finite abelian group (using the structure theorem). Note that the isomorphism is not canonical, while $G\cong\hat{\hat G}$ is.

Hint

A character of $\mathbb Z/n$ is determined by $\chi(1)$ , an $n$ th root of unity.

Exercise 4.5Eigenvalues of circulant matrices

Let $C_{xy}=c(x-y)$ on $\mathbb Z/n$ . Show $C\chi_k=n\hat c(\chi_k)\,\chi_k$ .

Hint

$(C\chi)(x)=\sum_yc(x-y)\chi(y)$ ; substitute $z=x-y$ .

Solution

$(C\chi)(x)=\sum_zc(z)\chi(x-z)=\chi(x)\sum_zc(z)\overline{\chi(z)}=n\hat c(\chi)\,\chi(x)$ , using $\chi(-z)=\overline{\chi(z)}$ . So $C$ is $c*\cdot$ , and Exercise 4.2 is the same statement.

Exercise 4.6Majority on three bits

Verify the expansion of $\mathrm{Maj}$ in Example 4.5 by computing $\hat f(S)=\frac18\sum_xf(x)\chi_S(x)$ for every $S\subseteq\{1,2,3\}$ .

Hint

By symmetry the coefficients depend only on $|S|$ ; even-size sets give 0 because $\mathrm{Maj}(-x)=-\mathrm{Maj}(x)$ .

Exercise 4.7An uncertainty principle

Show that for $f\ne0$ on a finite abelian group, $|\mathrm{supp}\,f|\cdot|\mathrm{supp}\,\hat f|\ge|G|$ . When is it an equality?

Hint

$\lVert f\rVert_\infty\le\sum_\chi|\hat f(\chi)|\le|\mathrm{supp}\,\hat f|^{1/2}\big(\sum|\hat f|^2\big)^{1/2}$ , then Plancherel and $\sum|f|^2\le|\mathrm{supp}\,f|\,\lVert f\rVert_\infty^2$ . Equality holds for indicator functions of subgroups (and their translates and modulations).

Exercise 4.8Characters of the circle

Show that the continuous homomorphisms $\mathbb R/2\pi\mathbb Z\to\mathbb C^\times$ are $\theta\mapsto e^{ik\theta}$ , $k\in\mathbb Z$ .

Hint

A continuous homomorphism $\phi:\mathbb R\to\mathbb C^\times$ is $t\mapsto e^{ct}$ (smooth it by integrating: $\int_0^\varepsilon\phi(t+s)ds=\phi(t)\int_0^\varepsilon\phi(s)ds$ ). Being $2\pi$ -periodic forces $c=ik$ .

CalculatorEtingof et al. §§4.8–4.11

Induced representations and Frobenius reciprocity

Restriction takes a representation of $G$ to one of a subgroup $H$ . Induction goes the other way, building a representation of $G$ from one of $H$ . Most irreducible representations are found this way, and Frobenius reciprocity says the two operations are adjoint, so multiplicities can be computed on whichever side is easier.

The idea

Induction (Etingof §4.8). For

H\le G

and a representation

W

H

\mathrm{Ind}_H^GW=k[G]\otimes_{k[H]}W

. Concretely, choose coset representatives

t_1,\dots,t_r

G/H

; then

\mathrm{Ind}W=\bigoplus_it_iW

, and

g

sends

t_iw

t_j(hw)

where

gt_i=t_jh

. So

\dim\mathrm{Ind}W=[G:H]\dim W

: "

W

copied onto every coset, with

G

shuffling the copies".

Character of an induced representation (Etingof §4.8).

\chi_{\mathrm{Ind}W}(g)=\frac1{|H|}\sum_{\substack{t\in G\\ tgt^{-1}\in H}}\chi_W(tgt^{-1}).

Only elements conjugate into

H

can have a nonzero value.

Frobenius reciprocity (Etingof §4.10).

\mathrm{Hom}_G(\mathrm{Ind}_H^GW,V)\cong\mathrm{Hom}_H(W,\mathrm{Res}_HV)

. In characters,

\langle\chi_{\mathrm{Ind}W},\chi_V\rangle_G=\langle\chi_W,\chi_V|_H\rangle_H

: the number of times

V

occurs in the induced representation is the number of times

W

occurs in the restriction of

V

Permutation representations are induced.

\mathrm{Ind}_H^G\mathbf 1=k[G/H]

, the permutation representation on cosets. Every transitive

G

-set is

G/H

for a stabilizer

H

, so every permutation representation is a sum of these.

Explore: inducing from a cyclic subgroup

Group G

Subgroup H = ⟨g⟩

Character ψ of H

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Restriction: Forgetting most of a group and remembering only how a subgroup acts.
Induced representation: A representation of a big group built from one of a subgroup, by placing a copy on each coset and letting the big group shuffle the copies.
Frobenius reciprocity: Building up from a subgroup and cutting down to it are mirror images: how often an irreducible appears in the built-up one equals how often the original appears in its cut-down version.
Coset representatives: One chosen element from each translate $gH$ of a subgroup, used as labels for the copies when building a representation of the whole group.
Monomial representation: One built from a 1-dimensional representation of a subgroup; in a suitable basis every matrix has exactly one nonzero entry in each row and column.

Examples

Example 5.1.

$\mathrm{Ind}_{\{e\}}^G\mathbf 1$ is the regular representation, and $\mathrm{Ind}_G^GW=W$ .

Example 5.2 (from

\mathbb Z/3

S_3

Let $H=\{e,(123),(132)\}$ and $\psi((123))=\omega$ . The formula gives $\chi_{\mathrm{Ind}\psi}=(2,0,\omega+\omega^2)=(2,0,-1)$ , the 2-dimensional irreducible. (For a 3-cycle $g$ , three elements $t$ give $tgt^{-1}=(123)$ and three give $(132)$ , so the value is $\frac13(3\omega+3\omega^2)$ .)

Example 5.3 (from

\mathbb Z/2

S_3

$H=\{e,(12)\}$ , $\psi$ the nontrivial character. At $(12)$ : the $t$ with $t(12)t^{-1}=(12)$ form $C_G((12))$ , of size 2, so the value is $\frac12\cdot2\cdot(-1)=-1$ . So $\chi=(3,-1,0)=\mathrm{sgn}+\chi_{V_0}$ .

Example 5.4 (reciprocity in action).

In Example 5.2, $V_0$ appears once in $\mathrm{Ind}\psi$ . On the other side, $V_0$ restricted to $\mathbb Z/3$ is $\omega\oplus\omega^2$ (the rotation of Example 1.4), which contains $\psi$ once. The calculator shows both sides for every choice.

Example 5.5 (

Q_8

is monomial).

Inducing $\psi(i)=i$ from $\langle i\rangle\cong\mathbb Z/4$ to $Q_8$ gives the character $(2,-2,0,0,0)$ : the 2-dimensional irreducible. In the basis $1,j$ of the induced representation, $i$ acts by $\mathrm{diag}(i,-i)$ and $j$ by $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ : one nonzero entry per row and column.

Example 5.6 (from

A_4

S_4

Induce $\psi=(1,1,\omega,\omega^2)$ from the index-2 subgroup $A_4$ . Conjugation by an odd permutation swaps the two classes of 3-cycles of $A_4$ , so $\chi_{\mathrm{Ind}\psi}$ is $\psi+\psi^{\text{swapped}}$ on $A_4$ and 0 off it: $(2,0,2,-1,0)$ , the 2-dimensional irreducible of $S_4$ .

Check your understanding

$H\le G$ has index 5 and $W$ has dimension 3. What is $\dim\mathrm{Ind}_H^GW$ ?

3One copy of $W$ for each coset.
8Dimensions multiply, they don't add.
15

\dim\mathrm{Ind}W=[G:H]\dim W

$H\ne G$ . Can $\mathrm{Ind}_H^G\mathbf 1$ be irreducible?

Yes, if $H$ is normal.Look for the trivial representation inside it.
No: by reciprocity it contains the trivial representation once, and its dimension is $[G:H]>1$ .

\langle\mathrm{Ind}\mathbf 1,\mathbf 1\rangle_G=\langle\mathbf 1,\mathbf 1\rangle_H=1

: the constants on

G/H

Exercises

Exercise 5.1The induced character formula

From the description $\mathrm{Ind}W=\bigoplus_it_iW$ , prove the character formula above.

Hint

$g$ maps the block $t_iW$ to $t_jW$ with $gt_i=t_jh$ ; only blocks with $j=i$ contribute to the trace, i.e. $t_i^{-1}gt_i\in H$ .

Solution

The block $t_iW$ is mapped into itself exactly when $t_i^{-1}gt_i\in H$ , and then acts there as $t_i^{-1}gt_i$ acts on $W$ . So $\chi(g)=\sum_{i:\,t_i^{-1}gt_i\in H}\chi_W(t_i^{-1}gt_i)$ . Replacing $t_i$ by $t_ih$ changes $t_i^{-1}gt_i$ by conjugation in $H$ , which doesn't change $\chi_W$ ; so summing over all $t\in G$ counts each coset $|H|$ times. With $t$ in place of $t^{-1}$ , this is the stated formula.

Exercise 5.2Prove Frobenius reciprocity for characters

Prove $\langle\mathrm{Ind}\,\psi,\chi\rangle_G=\langle\psi,\mathrm{Res}\,\chi\rangle_H$ for a class function $\psi$ on $H$ and $\chi$ on $G$ .

Hint

Insert the formula and substitute $h=tgt^{-1}$ .

Solution

Extend $\psi$ by 0 outside $H$ and write $\dot\psi$ . Then $\langle\mathrm{Ind}\,\psi,\chi\rangle_G=\frac1{|G||H|}\sum_{g,t}\dot\psi(tgt^{-1})\overline{\chi(g)}$ . For fixed $t$ , $h=tgt^{-1}$ runs over $G$ and $\chi(g)=\chi(h)$ . So the sum is $\frac1{|G||H|}\cdot|G|\sum_{h\in H}\psi(h)\overline{\chi(h)}=\langle\psi,\mathrm{Res}\,\chi\rangle_H$ .

Exercise 5.3Induction in stages

For $K\le H\le G$ , show $\mathrm{Ind}_H^G\mathrm{Ind}_K^HW\cong\mathrm{Ind}_K^GW$ .

Hint

$k[G]\otimes_{k[H]}(k[H]\otimes_{k[K]}W)\cong k[G]\otimes_{k[K]}W$ . Or compare characters using Exercise 5.2 twice and the fact that restriction is transitive.

Exercise 5.4The projection formula

Show $\mathrm{Ind}_H^G(W\otimes\mathrm{Res}\,V)\cong(\mathrm{Ind}_H^GW)\otimes V$ .

Hint

Compare characters: $\chi_V(tgt^{-1})=\chi_V(g)$ .

Solution

$\chi_{\mathrm{Ind}(W\otimes\mathrm{Res}V)}(g)=\frac1{|H|}\sum_{tgt^{-1}\in H}\chi_W(tgt^{-1})\chi_V(tgt^{-1})=\chi_V(g)\cdot\chi_{\mathrm{Ind}W}(g)$ , since $\chi_V$ is a class function on $G$ . Equal characters give isomorphic representations (Exercise 3.3).

Exercise 5.5Every irreducible comes from a subgroup

Show that every irreducible $V$ of $G$ is a constituent of $\mathrm{Ind}_H^GW$ for some irreducible $W$ of $H$ , for any subgroup $H$ .

Hint

Take $W$ inside $\mathrm{Res}\,V$ and apply reciprocity.

Exercise 5.6Inducing from an index-2 subgroup

Let $[G:H]=2$ and $W$ irreducible for $H$ . Let $W^s$ be $W$ twisted by conjugation by some $s\notin H$ . Show that $\mathrm{Ind}W$ is irreducible if $W^s\not\cong W$ , and is a sum of two non-isomorphic irreducibles if $W^s\cong W$ . Check it against Example 5.6.

Hint

$\mathrm{Res}\,\mathrm{Ind}W\cong W\oplus W^s$ (Mackey), so $\langle\mathrm{Ind}\chi,\mathrm{Ind}\chi\rangle=\langle\chi,\chi+\chi^s\rangle_H\in\{1,2\}$ .

Exercise 5.7Artin's theorem for

S_3

Artin's theorem (Etingof §4.25) says every character is a rational combination of characters induced from cyclic subgroups. Write the trivial character of $S_3$ this way.

Hint

$\mathbf 1=\mathrm{Ind}_{C_2}^{S_3}\mathbf 1+\frac12\mathrm{Ind}_{C_3}^{S_3}\mathbf 1-\frac12\mathrm{Ind}_{\{e\}}^{S_3}\mathbf 1$ : check $(3,1,0)+\frac12(2,0,2)-\frac12(6,0,0)=(1,1,1)$ . Integer coefficients are not always possible (that needs elementary subgroups, Brauer's theorem).

InteractiveEtingof et al. §§4.12–4.17

The symmetric group and Young diagrams

For $S_n$ the abstract count "as many irreducibles as conjugacy classes" becomes concrete: both are indexed by partitions of $n$ , drawn as Young diagrams. The dimension of each irreducible is a product formula you can compute by hand.

The idea

Classes and partitions. Two permutations are conjugate exactly when they have the same cycle type (Exercise 6.1). Cycle types of

S_n

are partitions

\lambda=(\lambda_1\ge\lambda_2\ge\dots)

n

, drawn as Young diagrams:

\lambda_1

boxes in the first row,

\lambda_2

in the second, and so on.

Irreducibles of $S_n$ (Etingof Thm. 4.36). For each partition

\lambda

there is an irreducible

V_\lambda=\mathbb C[S_n]c_\lambda

, where the Young symmetrizer

c_\lambda=a_\lambda b_\lambda

symmetrizes over the rows of a filled diagram and antisymmetrizes over its columns. These are pairwise non-isomorphic and exhaust the irreducibles.

V_{(n)}

is trivial,

V_{(1^n)}

is the sign,

V_{(n-1,1)}

is the standard representation, and transposing the diagram corresponds to tensoring with the sign.

Hook length formula (Etingof §4.17). The hook of a box is the box itself, the boxes to its right and the boxes below it. Then

\dim V_\lambda=\frac{n!}{\prod_{\text{boxes}}\text{hook length}},

which is also the number of standard Young tableaux of shape

\lambda

(fillings with

1,\dots,n

increasing along rows and down columns).

Branching rule. Restricted to

S_{n-1}

V_\lambda

is the direct sum of the

V_\mu

for the diagrams

\mu

obtained by removing one corner box from

\lambda

, each once.

Explore: hook lengths

Click an empty cell to grow the diagram until it reaches that cell; click a box to remove it with everything to its right and below. The number in each box is its hook length.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Partition: A way of writing a whole number as a sum of positive whole numbers listed from largest to smallest, like 4 = 2 + 1 + 1.
Young diagram: Rows of boxes, left-justified, each row no longer than the one above: a picture of a way of splitting a number into decreasing parts.
Hook length: For one box in the picture: count it, everything to its right in its row, and everything below it in its column.
Standard Young tableau: A filling of the boxes with the numbers 1 to $n$ , each used once, increasing to the right along rows and downwards along columns.
Young symmetrizer: An element of the group algebra that symmetrizes over the rows of a filled diagram and antisymmetrizes over its columns; it cuts out one irreducible piece.
Branching rule: When you forget the last letter and look only at permutations of the first $n-1$ , an irreducible splits into the pieces obtained by removing one corner box.

Examples

Example 6.1 (

S_3

Partitions $(3)$ , $(2,1)$ , $(1,1,1)$ . Hooks of $(2,1)$ : $3,1,1$ , so $\dim=6/3=2$ . Dimensions $1,2,1$ , matching Example 3.1.

Example 6.2 (

S_4

$(3,1)$ has hooks $4,2,1,1$ : $24/8=3$ . $(2,2)$ has hooks $3,2,2,1$ : $24/12=2$ . The five dimensions are $1,3,2,3,1$ and $1+9+4+9+1=24$ .

Example 6.3 (tableaux).

The standard tableaux of shape $(2,1)$ have first row $1\,2$ and second row $3$ , or first row $1\,3$ and second row $2$ : two of them, matching $\dim V_{(2,1)}=2$ .

Example 6.4 (branching).

$(2,2)$ has one corner, and removing it gives $(2,1)$ : the 2-dimensional irreducible of $S_4$ restricts to the 2-dimensional irreducible of $S_3$ . $(3,1)$ has two corners, giving $(3)$ and $(2,1)$ : $3=1+2$ , matching the permutation representation of $S_3$ inside the standard representation of $S_4$ .

Example 6.5 (

S_5

Seven partitions, with dimensions $1,4,5,6,5,4,1$ for $(5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1),(1^5)$ , and $1+16+25+36+25+16+1=120$ .

Example 6.6 (exterior powers).

The hook $(n-k,1^k)$ gives $\Lambda^kV_{(n-1,1)}$ , of dimension $\binom{n-1}k$ (Exercise 6.4). So the exterior powers of the standard representation are all irreducible.

Check your understanding

How many irreducible representations does $S_5$ have?

5Count partitions of 5, not the number 5.
7
120That is $\sum\dim^2$ .

The partitions of 5:

5,\ 41,\ 32,\ 311,\ 221,\ 2111,\ 11111

$V_\lambda\otimes\mathrm{sgn}$ is:

$V_{\lambda'}$ , where $\lambda'$ is the transposed diagram.
$V_\lambda$ again.Only when $\lambda$ is its own transpose, like $(2,1)$ .
reducible.Tensoring with a 1-dimensional representation preserves irreducibility.

Swapping rows and columns swaps the symmetrizing and antisymmetrizing parts of the Young symmetrizer. Hook lengths are unchanged by transposing, so the dimensions agree.

Exercises

Exercise 6.1Conjugacy classes are cycle types

Show that $\sigma(a_1\,a_2\,\cdots\,a_k)\sigma^{-1}=(\sigma(a_1)\,\sigma(a_2)\,\cdots\,\sigma(a_k))$ , and deduce that two permutations are conjugate exactly when they have the same cycle type.

Hint

Check where both sides send $\sigma(a_i)$ . For the converse, line up the cycles of two permutations of the same type and define $\sigma$ by matching entries.

Solution

$\sigma c\sigma^{-1}$ sends $\sigma(a_i)$ to $\sigma(c(a_i))=\sigma(a_{i+1})$ , and fixes $\sigma(b)$ whenever $c$ fixes $b$ . Applying this to each cycle, conjugation preserves cycle type. Conversely, if $\pi$ and $\pi'$ have the same type, write their cycles one above the other with equal lengths aligned, and let $\sigma$ send each entry of $\pi$ 's list to the entry below it. Then $\sigma\pi\sigma^{-1}=\pi'$ .

Exercise 6.2The standard representation by hooks

Compute $\dim V_{(n-1,1)}$ by the hook length formula.

Hint

The corner box has hook $n$ ; the rest of the first row has hooks $n-2,\dots,1$ ; the box in the second row has hook 1.

Solution

$\prod\text{hooks}=n\cdot(n-2)!\cdot1$ , so $\dim=\frac{n!}{n(n-2)!}=n-1$ , the dimension of the standard representation $V_0$ of Example 1.2.

Exercise 6.3All irreducibles of

S_6

List the 11 partitions of 6 with their dimensions, and check $\sum\dim^2=720$ . Use the figure to check your hook lengths.

Hint

Dimensions: $(6)\,1,\ (5,1)\,5,\ (4,2)\,9,\ (4,1,1)\,10,\ (3,3)\,5,\ (3,2,1)\,16,\ (3,1,1,1)\,10,\ (2,2,2)\,5,\ (2,2,1,1)\,9,\ (2,1^4)\,5,\ (1^6)\,1$ .

Exercise 6.4Hooks and binomial coefficients

Show $\dim V_{(n-k,1^k)}=\binom{n-1}k$ .

Hint

Corner hook $n$ ; the rest of the first row $n-k-1,\dots,1$ ; the column below $k,\dots,1$ .

Solution

$\prod\text{hooks}=n\,(n-k-1)!\,k!$ , so $\dim=\frac{n!}{n\,(n-k-1)!\,k!}=\frac{(n-1)!}{k!\,(n-1-k)!}=\binom{n-1}k$ , which is $\dim\Lambda^kV_{(n-1,1)}$ .

Exercise 6.5Branching and dimensions

Show that the branching rule implies $\dim V_\lambda=\sum_\mu\dim V_\mu$ over the $\mu$ obtained by removing a corner. Check it for $\lambda=(3,2)$ and $\lambda=(3,2,1)$ .

Hint

$(3,2)$ : $5=2+3$ from $(2,2)$ and $(3,1)$ . $(3,2,1)$ : $16=5+6+5$ from $(2,2,1)$ , $(3,1,1)$ , $(3,2)$ . The same recursion counts standard tableaux by the position of $n$ .

Exercise 6.6Transposes and the sign

For $S_4$ , show that the characters of $V_{(3,1)}\otimes\mathrm{sgn}$ and $V_{(2,1,1)}$ agree, given that $V_{(2,1,1)}=\Lambda^2V_{(3,1)}$ (Example 6.6).

Hint

$\chi_{\Lambda^2V}(g)=\frac12(\chi(g)^2-\chi(g^2))$ . With $\chi=(3,1,-1,0,-1)$ and $g^2$ of cycle types $e,e,e,(123),(12)(34)$ , you get $(3,-1,-1,0,1)$ .

Exercise 6.7Characters of

S_n

are integers

Show that every character value of $S_n$ is an integer, using: $g$ and $g^k$ are conjugate when $\gcd(k,\mathrm{ord}\,g)=1$ , and a rational number that is an algebraic integer is an integer.

Hint

$\chi(g)$ is a sum of $m$ th roots of unity, $m=\mathrm{ord}\,g$ . Each Galois automorphism of $\mathbb Q(e^{2\pi i/m})$ sends $\chi(g)$ to $\chi(g^k)$ .

Solution

Raising to a power $k$ coprime to the order of $g$ preserves cycle type, so $g^k$ is conjugate to $g$ and $\chi(g^k)=\chi(g)$ . The automorphism $\zeta\mapsto\zeta^k$ of $\mathbb Q(\zeta)$ , $\zeta=e^{2\pi i/m}$ , sends $\chi(g)=\sum\zeta^{a_j}$ to $\sum\zeta^{ka_j}=\chi(g^k)=\chi(g)$ . So $\chi(g)$ is fixed by the whole Galois group, hence rational. It is a sum of roots of unity, hence an algebraic integer, hence an integer.

Exercise 6.8Counting standard tableaux by hand

List all standard Young tableaux of shape $(3,2)$ and check there are 5.

Hint

The first row starts with 1, and 5 sits in a corner. The first rows are $123,\ 124,\ 125,\ 134,\ 135$ .

InteractiveEtingof et al. §§1.9, 1.13–1.14; Hall Ch. 5 §§4, 6, 8–9

sl(2) and SU(2): the first Lie example

A Lie group is too big to work with element by element, so you differentiate: a representation of the group gives a representation of its Lie algebra, a few matrices satisfying commutation relations. For $\mathfrak{sl}(2)$ the whole classification follows from one observation, that $e$ and $f$ move weights up and down by 2.

The idea

From groups to Lie algebras (Hall Ch. 5 §§1, 8). If

\Pi:G\to\mathrm{GL}(V)

is a smooth representation of a matrix Lie group, then

\pi(X)=\frac{d}{dt}\Pi(e^{tX})\big|_{t=0}

is a representation of its Lie algebra: linear, with

\pi([X,Y])=[\pi(X),\pi(Y)]

. If

G

is connected,

\Pi

is determined by

\pi

; if

G

is also simply connected, every

\pi

comes from some

\Pi

\mathrm{SU}(2)

is simply connected and

\mathfrak{su}(2)\otimes\mathbb C=\mathfrak{sl}(2,\mathbb C)

, so its representations are the representations of

\mathfrak{sl}(2)

$\mathfrak{sl}(2)$ (Etingof §1.9). Trace-zero

2\times2

matrices, with basis

e=\begin{pmatrix}0&1\\0&0\end{pmatrix}

f=\begin{pmatrix}0&0\\1&0\end{pmatrix}

h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}

and relations

[h,e]=2e

[h,f]=-2f

[e,f]=h

. A representation is three matrices satisfying these relations.

Weights move by 2. If

hv=\lambda v

then

h(ev)=(\lambda+2)ev

and

h(fv)=(\lambda-2)fv

(Exercise 7.1). In a finite-dimensional representation

h

has finitely many eigenvalues, so some eigenvector

v

has

ev=0

: a highest weight vector.

Classification (Etingof Thm. 1.1, §1.14; Hall Ch. 5 §4). For each

m=0,1,2,\dots

there is exactly one irreducible representation

V_m

of dimension

m+1

. It has basis

v_k=f^kv_0

k=0,\dots,m

, with

hv_k=(m-2k)v_k,\qquad fv_k=v_{k+1},\qquad ev_k=k(m-k+1)v_{k-1}.

Realized on homogeneous polynomials of degree

m

x,y

e=x\partial_y

f=y\partial_x

h=x\partial_x-y\partial_y

. Every finite-dimensional representation is a direct sum of these (Weyl; for

\mathrm{SU}(2)

, Hall's averaging over the compact group).

Clebsch–Gordan (Hall Ch. 5 §6).

V_m\otimes V_n\cong V_{m+n}\oplus V_{m+n-2}\oplus\dots\oplus V_{|m-n|}

. Weights of a tensor product add, so you can peel off highest weights one at a time (Exercise 7.4).

Group versus Lie algebra (Hall Ch. 5 §9).

\mathrm{SU}(2)\to\mathrm{SO}(3)

is a double cover with kernel

\pm I

, and

-I=\exp(\pi i h)

acts on

V_m

(-1)^m

. So only

V_m

with

m

even are representations of

\mathrm{SO}(3)

: in physics language, integer spin

m/2

. The half-integer spins exist only for

\mathrm{SU}(2)

Explore: weight diagrams and Clebsch–Gordan

m (first factor) n (second factor; 0 = no tensor)

Each dot is a weight vector, placed at its eigenvalue of

h

. With

n=0

the arrows show

e

(up by 2) and

f

(down by 2) with their coefficients. With

n>0

, the weights of

V_m\otimes V_n

are stacked and coloured by the irreducible piece they belong to.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Lie algebra: The infinitesimal version of a continuous group: the space of directions you can move in from the identity, with a bracket recording how badly two such motions fail to commute.
Weight: An eigenvalue of the diagonal element $h$ on a vector; the label that the raising and lowering operators shift by 2.
Highest weight vector: A vector that the raising operator kills. Every irreducible piece is generated from one by applying the lowering operator repeatedly.
Raising and lowering operators: The two off-diagonal elements, which move a vector up or down the ladder of eigenvalues by two steps.
Clebsch–Gordan decomposition: The rule for splitting the product of two irreducible pieces: dimensions $m+1$ and $n+1$ give one piece of each size from $m+n+1$ down to $|m-n|+1$ , in steps of two.
Casimir element: A particular quadratic combination of the basis elements that commutes with everything, so it acts on each irreducible piece by a single number that tells the pieces apart.
Double cover: A two-to-one map of groups that is locally an isomorphism, like the map from unit quaternions to rotations of space, under which $q$ and $-q$ give the same rotation.

Examples

Example 7.1.

$V_0$ is the trivial representation and $V_1=\mathbb C^2$ is the defining one, with weights $\pm1$ on $e_1,e_2$ .

Example 7.2 (the adjoint representation).

$\mathfrak{sl}(2)$ acts on itself by $\mathrm{ad}(X)Y=[X,Y]$ . The weights of $h$ on $e,h,f$ are $2,0,-2$ , and $e$ is a highest weight vector, so this is $V_2$ . Under $\mathrm{SU}(2)\to\mathrm{SO}(3)$ it is the rotation representation on $\mathbb R^3\otimes\mathbb C$ .

Example 7.3 (polynomials).

On degree-2 polynomials with basis $x^2,xy,y^2$ : $h$ has weights $2,0,-2$ ; $f=y\partial_x$ sends $x^2\mapsto2xy\mapsto2y^2$ ; $e=x\partial_y$ sends $y^2\mapsto2xy\mapsto2x^2$ . Up to rescaling the basis, these are the formulas for $V_2$ .

Example 7.4 (two spins).

$V_1\otimes V_1$ has weights $2,0,0,-2$ , so it is $V_2\oplus V_0$ : the symmetric tensors (triplet) and the antisymmetric one $e_1\otimes e_2-e_2\otimes e_1$ (singlet).

Example 7.5 (the Casimir).

$C=ef+fe+\frac12h^2$ commutes with $e,f,h$ (Exercise 7.5), so it is a scalar on each $V_m$ . On $v_0$ : $efv_0=[e,f]v_0+fev_0=mv_0$ and $fev_0=0$ , so $C=m+\frac{m^2}2=\frac{m(m+2)}2$ . In physics this is $2j(j+1)$ for spin $j=m/2$ .

Example 7.6 (spherical harmonics).

$\mathrm{SO}(3)$ acts on homogeneous harmonic polynomials of degree $\ell$ in $x,y,z$ , a space of dimension $2\ell+1$ . It is irreducible, isomorphic to $V_{2\ell}$ . Restricted to the sphere these are the spherical harmonics, and the decomposition of functions on the sphere into them is the $\mathrm{SO}(3)$ analogue of Fourier series.

Check your understanding

What are the weights of $V_3$ ?

$0,1,2,3$ Weights are symmetric about 0.
$3,1,-1,-3$
$3,2,1,0,-1,-2,-3$ Weights move in steps of 2, and $\dim V_3=4$ .

m-2k

for

k=0,\dots,m

$V_2\otimes V_1\cong$

$V_3\oplus V_1$
$V_3\oplus V_2\oplus V_1$ Dimensions: $3\cdot2=6$ , but $4+3+2=9$ .
$V_2\oplus V_1\oplus V_0$ The highest weight is $2+1=3$ .

Clebsch–Gordan with

m=2

n=1

V_3\oplus V_1

, dimensions

4+2=6

Why is $V_1$ not a representation of $\mathrm{SO}(3)$ ?

$-I\in\mathrm{SU}(2)$ maps to the identity rotation but acts on $V_1$ by $-1$ .
It is not irreducible.It is irreducible.
Its dimension is even.Dimension isn't the issue by itself; the kernel of $\mathrm{SU}(2)\to\mathrm{SO}(3)$ is.

A representation of

\mathrm{SU}(2)

factors through

\mathrm{SO}(3)

exactly when

-I

acts trivially, i.e.

m

even.

Exercises

Exercise 7.1Raising and lowering

Show that if $hv=\lambda v$ , then $h(ev)=(\lambda+2)ev$ and $h(fv)=(\lambda-2)fv$ .

Hint

$hev=[h,e]v+ehv$ .

Solution

$h(ev)=[h,e]v+e(hv)=2ev+\lambda ev$ . Similarly $h(fv)=-2fv+\lambda fv$ .

Exercise 7.2Classify the irreducibles of

\mathfrak{sl}(2)

Let $V$ be a finite-dimensional irreducible representation. Take a highest weight vector $v_0$ of weight $\lambda$ and put $v_k=f^kv_0$ . Show $ev_k=k(\lambda-k+1)v_{k-1}$ , that the nonzero $v_k$ span $V$ , and that $\lambda=m$ is a non-negative integer with $\dim V=m+1$ .

Hint

Induct on $k$ with $ev_k=efv_{k-1}=[e,f]v_{k-1}+fev_{k-1}$ . Let $m$ be the last index with $v_m\ne0$ and apply the formula to $v_{m+1}=0$ .

Solution

$ev_k=hv_{k-1}+fev_{k-1}=(\lambda-2k+2)v_{k-1}+(k-1)(\lambda-k+2)v_{k-1}=k(\lambda-k+1)v_{k-1}$ . The nonzero $v_k$ have distinct weights, so they are independent, and their span is invariant under $e,f,h$ ; by irreducibility it is $V$ . If $v_m\ne0$ and $v_{m+1}=0$ , then $0=ev_{m+1}=(m+1)(\lambda-m)v_m$ , so $\lambda=m$ . Hence $\dim V=m+1$ , and all the formulas are forced, which also proves uniqueness. Existence: the polynomial model.

Exercise 7.3The polynomial model

Check that $e=x\partial_y$ , $f=y\partial_x$ , $h=x\partial_x-y\partial_y$ satisfy the $\mathfrak{sl}(2)$ relations on polynomials in $x,y$ , and that degree- $m$ polynomials form a copy of $V_m$ .

Hint

Compute on monomials: $h(x^ay^b)=(a-b)x^ay^b$ , $e(x^ay^b)=b\,x^{a+1}y^{b-1}$ . The monomial $x^m$ is a highest weight vector of weight $m$ .

Exercise 7.4Prove Clebsch–Gordan

Assuming complete reducibility, prove $V_m\otimes V_n\cong\bigoplus_{k=0}^{\min(m,n)}V_{m+n-2k}$ .

Hint

The weights of a tensor product are the sums of weights. A representation is determined by its weight multiplicities, since $V_j$ contributes one vector to each weight $j,j-2,\dots,-j$ .

Solution

$h$ acts on $v_i\otimes w_j$ by $(m-2i)+(n-2j)$ . The multiplicity of weight $m+n-2s$ is the number of $(i,j)$ with $i+j=s$ , $0\le i\le m$ , $0\le j\le n$ : it increases by 1 from $s=0$ up to $s=\min(m,n)$ , then stays constant until the symmetric point. The number of copies of $V_{m+n-2s}$ equals (multiplicity of weight $m+n-2s$ ) minus (multiplicity of weight $m+n-2s+2$ ), which is 1 for $0\le s\le\min(m,n)$ and 0 beyond.

Exercise 7.5The Casimir is central

Show that $C=ef+fe+\frac12h^2$ commutes with $e$ , $f$ and $h$ , and conclude that it acts on $V_m$ by $\frac{m(m+2)}2$ .

Hint

Use $[X,YZ]=[X,Y]Z+Y[X,Z]$ . Then Schur's lemma and Example 7.5.

Exercise 7.6Representations of

\mathrm{SO}(3)

Show that $-I=\exp(i\pi h)$ in $\mathrm{SU}(2)$ , that it acts on $V_m$ by $(-1)^m$ , and deduce that the irreducible representations of $\mathrm{SO}(3)$ are exactly the $V_m$ with $m$ even, of odd dimension.

Hint

$\exp(i\pi h)=\mathrm{diag}(e^{i\pi},e^{-i\pi})$ . On $V_m$ , $h$ has eigenvalues $m-2k$ , and $e^{i\pi(m-2k)}=(-1)^m$ .

Exercise 7.7No nontrivial 1-dimensional representations

Show that every 1-dimensional representation of $\mathfrak{sl}(2)$ is zero. (Compare groups, whose 1-dimensional representations factor through the commutator quotient: here $\mathfrak{sl}(2)=[\mathfrak{sl}(2),\mathfrak{sl}(2)]$ .)

Hint

In dimension 1 all brackets vanish, so $h=[e,f]=0$ and $2e=[h,e]=0$ .

Exercise 7.8The adjoint representation is irreducible

Show that the adjoint representation of $\mathfrak{sl}(2)$ is irreducible, so $\mathfrak{sl}(2)$ has no ideals except $0$ and itself (it is simple).

Hint

A nonzero invariant subspace contains a weight vector ( $h$ acts diagonalizably), and from $e$ , $h$ or $f$ you reach the others by brackets.

CalculatorEtingof et al. §§1.8, 5.1–5.5

Beyond groups: quivers and Gabriel's theorem

A quiver is a directed graph, and a representation puts a vector space at each vertex and a linear map on each arrow. There is no group anywhere, yet the classification of indecomposables brings back exactly the Dynkin diagrams that classify simple Lie algebras. This is the clearest sign that representation theory is a subject in its own right, not a chapter of Lie theory.

The idea

Quiver representations (Etingof §1.8). A quiver

Q

is a finite directed graph. A representation assigns a vector space

V_i

to each vertex and a linear map

V_i\to V_j

to each arrow

i\to j

. Morphisms are families of maps

V_i\to W_i

that make every square commute. Representations of

Q

are the same as modules over its path algebra

kQ

, whose basis is the directed paths (Exercise 8.2). The dimension vector is

d=(\dim V_i)

Finite type and the Tits form.

Q

has finite type if it has finitely many indecomposable representations up to isomorphism. The quadratic form

q(d)=\sum_id_i^2-\sum_{\text{edges } i-j}d_id_j

depends only on the underlying graph, not on the orientation.

Gabriel's theorem (Etingof Thm. 1.2, §5.5). A connected quiver has finite type if and only if its underlying graph is one of the Dynkin diagrams

A_n

D_n

E_6

E_7

E_8

, which happens if and only if

q

is positive definite. In that case

V\mapsto\dim V

is a bijection from indecomposables to the positive roots, the vectors

d\ge0

with

q(d)=1

. The answer does not depend on the orientation of the arrows.

Why the Lie algebra answer appears. The same diagrams classify the simply-laced simple Lie algebras, and the positive roots of

q

are the positive roots of that Lie algebra. Etingof's proof uses reflection functors, which move representations between orientations and act on dimension vectors by the Weyl group. Nobody puts a Lie group in; it comes out.

Explore: the Tits form and positive roots

The graph, labelled with the dimension vector of one indecomposable representation (a positive root). Step through them with the buttons.

Graph

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Quiver: A finite directed graph, possibly with loops and repeated arrows, used as a blueprint: put a vector space at each dot and a linear map along each arrow.
Path algebra: The algebra whose basis is the directed walks in a graph, multiplied by joining walks end to start (and zero when they don't meet).
Dimension vector: The list of dimensions of the spaces placed at the dots.
Finite type: Describes a graph that admits only finitely many essentially different unsplittable arrangements of spaces and maps.
Tits form: The quadratic form "sum of squares of the dimensions, minus the product across each edge". Being positive definite decides whether a graph has only finitely many unsplittable arrangements.
Dynkin diagram: One of the graphs on the list A, D, E: a line, a line with a fork at one end, or one of three special trees with six, seven or eight dots.
Positive root: A list of non-negative whole numbers, one per dot, on which the quadratic form equals 1; for the A, D, E graphs these are exactly the dimension lists of the unsplittable arrangements.

Examples

Example 8.1 (

A_1

One vertex, no arrows: a representation is a vector space, and the only indecomposable is $k$ . The positive roots of $q(d)=d^2$ are $d=1$ .

Example 8.2 (

A_2

: one linear map).

$\bullet\to\bullet$ : a representation is a linear map $A:V\to W$ . Choosing bases with $A=\begin{pmatrix}I_r&0\\0&0\end{pmatrix}$ splits it into $r$ copies of $k\xrightarrow{1}k$ , plus copies of $k\to0$ and $0\to k$ . Three indecomposables, dimension vectors $(1,1),(1,0),(0,1)$ , and those are the solutions of $d_1^2+d_2^2-d_1d_2=1$ with $d\ge0$ .

Example 8.3 (

A_n

For a line of $n$ vertices, in any orientation, the indecomposables are the "intervals": $k$ at vertices $i,\dots,j$ with identity maps between them and $0$ elsewhere. There are $\binom{n+1}2$ of them.

Example 8.4 (one loop).

A single vertex with a loop: a representation is a vector space with an operator, so the indecomposables are the Jordan blocks $J_m(\lambda)$ (Example 1.7), infinitely many in each dimension. Here $q(d)=d^2-d^2=0$ , and the loop is not on the Dynkin list.

Example 8.5 (Kronecker quiver).

Two vertices joined by two arrows: pairs of maps $A,B:V\to W$ , the classical theory of matrix pencils $A+tB$ . The representations $k\rightrightarrows k$ with maps $(1,\lambda)$ are pairwise non-isomorphic, and $q(1,1)=1+1-2=0$ . This graph is $\tilde A_1$ , the first of the "extended" diagrams just beyond finite type.

Example 8.6 (

D_4

A centre joined to three outer vertices has 12 indecomposables. The largest has dimension vector 2 at the centre and 1 at each outer vertex: $q=4+3-6=1$ . It is three lines in a plane, the representation in which "three subspaces in general position" lives.

Check your understanding

How many indecomposable representations does a quiver of type $A_4$ have?

4Those are only the simple ones, one per vertex.
10
Infinitely many. $A_4$ is a Dynkin diagram.

Intervals in a line of 4 vertices:

\binom52=10

Does reversing some arrows of a quiver change whether it has finite type?

Yes, in general.Look at Gabriel's theorem again.
No: finite type depends only on the underlying graph.

The Tits form ignores orientation, and reflection functors match indecomposables between orientations.

Exercises

Exercise 8.1Representations of

A_2

Prove the classification in Example 8.2: every representation of $\bullet\to\bullet$ is $(k\xrightarrow1k)^r\oplus(k\to0)^a\oplus(0\to k)^b$ , with $r$ , $a$ , $b$ determined by the representation.

Hint

Rank normal form. The invariants are $r=\mathrm{rank}A$ , $a=\dim\ker A$ , $b=\dim\mathrm{coker}A$ .

Solution

Choose a basis of $V$ whose last $a$ vectors span $\ker A$ ; the images of the first $r$ are independent in $W$ , and extend them to a basis of $W$ . In these bases $A=\begin{pmatrix}I_r&0\\0&0\end{pmatrix}$ , which is the stated direct sum. An isomorphism of representations is a pair of invertible maps with $QA=A'P$ , which preserves rank, kernel dimension and cokernel dimension.

Exercise 8.2Quiver representations are path algebra modules

Show that a representation of $Q$ is the same as a module over the path algebra $kQ$ .

Hint

The length-zero paths $e_i$ are orthogonal idempotents summing to 1; put $V_i=e_iM$ . An arrow $a:i\to j$ gives a map $V_i\to V_j$ .

Solution

Given a module $M$ , $1=\sum e_i$ with $e_ie_j=\delta_{ij}e_i$ gives $M=\bigoplus e_iM$ . An arrow $a$ from $i$ to $j$ satisfies $a=e_jae_i$ , so $m\mapsto am$ maps $e_iM$ to $e_jM$ . Conversely, a representation defines the action of a path as the composite of its maps, and of $e_i$ as the projection onto $V_i$ . Module maps commuting with all $e_i$ and arrows are exactly morphisms of representations.

Exercise 8.3Intervals are roots

For $A_n$ , show that the indicator vector of an interval $\{i,\dots,j\}$ has $q=1$ , and that every $d\ge0$ with $q(d)=1$ is of this form.

Hint

$q(d)=\frac12\big(d_1^2+\sum_{i}(d_i-d_{i+1})^2+d_n^2\big)$ . This is a sum of squares that must equal 2 after doubling.

Exercise 8.4The extended diagrams are not of finite type

Show that $q$ is not positive definite for a cycle ( $\tilde A_n$ ) and for the star with four leaves ( $\tilde D_4$ ), by finding $d\ne0$ with $q(d)=0$ .

Hint

Cycle: $d=(1,\dots,1)$ . Star: 2 at the centre and 1 on each leaf.

Solution

For a cycle of $n+1$ vertices with all $d_i=1$ : $q=(n+1)-(n+1)=0$ . For $\tilde D_4$ : $q=2^2+4\cdot1^2-4\cdot(2\cdot1)=8-8=0$ . In both cases $q(td)=0$ for all $t$ , and one can build a one-parameter family of pairwise non-isomorphic indecomposables of dimension vector $d$ , as in Example 8.5.

Exercise 8.5Jordan blocks

For the one-loop quiver, show that $J_n(\lambda)$ and $J_n(\mu)$ are non-isomorphic for $\lambda\ne\mu$ , so there are infinitely many indecomposables of each dimension.

Hint

Isomorphic representations have similar operators, hence the same eigenvalues.

Exercise 8.6The Kronecker family

Show that the representations $k\rightrightarrows k$ with maps $(1,\lambda)$ , $\lambda\in k$ , together with $(0,1)$ , are pairwise non-isomorphic and indecomposable.

Hint

An isomorphism is a pair of nonzero scalars $(p,q)$ with $q\cdot1=1\cdot p$ and $q\lambda=\mu p$ , so $p=q$ and $\lambda=\mu$ . They are indecomposable because the dimension vector $(1,1)$ can only split as $(1,0)+(0,1)$ , which would make both maps zero.

Exercise 8.7The roots of

D_4

List the 12 positive roots of $D_4$ and check them in the figure.

Hint

With the centre listed first: the four simple roots; $1$ at the centre with any non-empty set of leaves (7 vectors); and $(2;1,1,1)$ . That's $4+7+1=12$ .

Exercise 8.8Cycles and high degree rule out finite type

Show that if a connected graph contains a cycle, or a vertex of degree at least 4, then it is not of finite type.

Hint

Extending a representation of a subgraph by zero keeps it indecomposable and keeps non-isomorphic ones non-isomorphic. Then use Exercise 8.4 (a vertex of degree at least 4 gives a $\tilde D_4$ inside).

Representation Theory

The idea

Explore: three representations of the triangle's symmetries

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Exercises

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Explore: averaging an inner product

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Explore: a character table workbench

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Explore: a signal and its Fourier transform on ℤ/n\mathbb Z/n

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Explore: weight diagrams and Clebsch–Gordan

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The idea

Explore: the Tits form and positive roots

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Explore: a signal and its Fourier transform on $\mathbb Z/n$