# Representation Theory Primer

### Basic definitions, Schur's lemma, and Fourier analysis on groups

As a general warning, these primers are mainly here just to set up notations and introduce basic definitions and facts that I will refer to in my main posts, so they are a bit lacking in motivation.

The following is adapted from Chapter 2 of my thesis, which is an introduction to the representation theory of Lie groups and their associated homogeneous spaces.

A group homomorphism between two groups is a map $\varphi:G \to G'$ that respects the group composition law: \begin{aligned} \varphi(g_1)\varphi(g_2) &= \varphi(g_1 g_2) \end{aligned} for all $g_1,g_2 \in G$. This implies, among other things, that if $e$ is the identity element of $G$, then $\varphi(e)$ is the identity element of $G'$, and $\varphi(g)^{-1} = \varphi(g^{-1})$. The kernel $\ker \varphi$ of a homomorphism $\varphi$ is the set of elements of $G$ that are sent to the identity element.

Given a subgroup $K$ of $G$, a left coset of $K$ consists of elements of the form $gK$ for some $g \in G$. $g_1$ and $g_2$ belong to the same coset if $g_1 g_2^{-1}$ is in $K$. Two left cosets are either the same or disjoint, so $G$ is partitioned into left cosets. The left coset space $G/K$ is the set of all left cosets. The right coset space $K\backslash G$ is the set of all right cosets, which are defined similarly but with right multiplication by $g^{-1}$. The left and right cosets coincide if and only if $K$ is a normal subgroup i.e. $K$ is invariant under conjugation $gKg^{-1}$ by every $g \in G$. (In that case, $G/K \simeq K\backslash G$ is also a group.)

Groups are things that act reversibly. A group action of $G$ on a set $X$ is the permutation \begin{aligned} G \times X &\to X \\ (g,x) &\mapsto g\cdot x \end{aligned} respecting the group structure \begin{aligned} ex &= x, & g_1 (g_2 x) &= (g_1 g_2)x \end{aligned} for all $x \in X$. In other words, a group action is a group homomorphism from $G$ to $S_{|X|}$, the symmetric group over the set $X$.

Equivariant map

Representation theory begins when a group acts as linear operators (endomorphisms) on a vector space. We say that $\rep$ is a representation of $G$ if it is a homomorphism from $G$ to $\gr{GL}{V}$, the group of all invertible matrices on $V$, \begin{aligned} \rep: G \to \gr{GL}{V}. \end{aligned} Linear representations occur naturally in quantum theory as quantum systems are described by Hilbert spaces. But from a mathematical point of view, the philosophy of representation theory is to “linearize” group actions

A linear representation is an instance of a group action on a vector space. A $G$-homogeneous space $X$ is a set with a transitive action of $G$: for any two elements $x,y \in X$, there is some $g \in G$ such that $y = gx$.

A representation of a group $G$ is a homomorphism from $G$ to $\gr{GL}{V}$, the group of all invertible matrices on $V$, \begin{aligned} \rep: G \to \gr{GL}{V}. \end{aligned} We also say that $(\rep,V)$ is a $G$-representation. When no confusion arises, we also call the vector space $V$ itself a representation. A representation $\rep$ is said to be faithful if the map $\rho$ is one-one. A subrepresentation is a subspace of $V$ stable under $G$ i.e. closed under every $\rep(g)$. An irreducible representation or irrep for short is one that has no nontrivial subrepresentation.

A reducible representation may not decompose as a direct sum of subrepresentations. This is true even for a single matrix: any given matrix has at least one eigenvector in $\mathbb{C}$ but it may not be diagonalizable. As an example of an indecomposable representation, the group of integers with addition as group multiplication $(\mathbb{Z},+)$ has a two-dimensional representation with the generator 1 \begin{aligned} \rep(e) &= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},\end{aligned} which cannot be diagonalized. A representation $V$ is completely reducible if for any subrepresentation $W \subset V$, there is a complementary subrepresentation $W'\subset V$ such that $V \simeq W \oplus W'$.

As every vector space comes with the dual space, every representation comes with the dual representation. Recall that the dual space $V^*$ of a complex vector space is the vector space of all linear maps from $V$ to $\mathbb{C}$. Since $V^*$ and $V$ have the same dimension, they are isomorphic as vector spaces $V^* \simeq V$ but not naturally. (We do not assume the Hermitian inner product structure for a moment.) Nevertheless, if we pick an ordered basis $\{ \ket{v_j} \}$, an isomorphism amounts to the transposition—simply flipping the ket $\ket{v_j}$ to the bra $\bra{v_j}$. The dual representation $(\rep^*,V^*)$ of a representation $(\rep,V)$ can be defined in several ways 2, which can be quite confusing. We follow Fulton and Harris 3 and define the right action $\bra{u} \rep^*(g)$ so that \begin{aligned} (\bra{u} \rep^* (g) ) (\rep(g) \ket{v}) &= \braket{u|v}, \end{aligned} for all $\ket{u},\ket{v} \in V$ and $g \in G$. (The inverse is necessary to make $\rep^*$ a group homomorphism.) Note that if we want to turn the right action to the left action on $\bra{u}$, the matrix representation of $\rep^*(g)$ is given by the transpose $\rep^T(g^{-1})$ so that \begin{aligned} \bra{u}\rep^*(g) &= \bra{\rep^T(g^{-1})u} = \bra{\rep^{-T}(g)u}. \end{aligned}

A representation is unitary (or unitarizable) if it is equivalent to a representation in which every $\rep(g)$ is a unitary matrix, \begin{aligned} \rep\dgg (g) \rep(g) &= \id, \end{aligned} where $\rep\dgg (g)$ is the entry-wise complex conjugate transpose of $\rep(g)$. For a unitary representation, the right-action version of the dual representation coincides with the Hermitian dual representation $\rho\dgg (g)$, while the left-action version coincides with the complex conjugate representation. High energy physicists like to use the latter, indicating the dual representation by an overbar.

## Intertwiners and Schur’s lemma

For complex vector spaces $V$ and $W$, define $\text{Hom}(V,W)$ to be the vector space of linear maps (linear homomorphisms) from $V$ to $W$. When $V = W$, this space is also denoted by $\text{End}(V)$ (endomorphisms). Naturally, $\text{Hom}(V,W) \simeq W \otimes V^*$. When $V$ and $W$ are endowed with $G$-representations $\rep_1$ and $\rep_2$, respectively, an intertwiner between $\rep_1$ and $\rep_2$ is defined as a linear map $\varphi$ that makes the diagram below commutative for any $g\in G$.

In other words, $\varphi$ commutes with the $G$-action: \begin{aligned} \varphi \rep_1 (g) &= \rep_2 (g) \varphi.\end{aligned} The set of all such intertwiners forms a subspace of $\text{Hom}(V,W)$ denoted by $\text{Hom}_G (V,W)$ or, again, $\text{End}_G (V)$ when $V \simeq W$. Two representations $V$ and $W$ of $G$ are said to be equivalent \begin{aligned} V \stackrel{G}{\simeq} W\end{aligned} when $\text{Hom}_G (V,W)$ contains an invertible intertwiner. In this case, they are related just by a change of basis.

Schur proved a collection of elementary but very powerful observations that, in an algebraically closed field $K$ (such as $\mathbb{C}$), intertwiners between irreps behave like the Kronecker delta: \begin{aligned} \text{Hom}_G (V,W) \simeq \begin{cases} K, & V \simeq W, \\ 0, & V \not\simeq W. \end{cases} \end{aligned}

Lemma (Schur):

1. Every intertwiner between irreps of $G$ is either an isomorphism or zero,

2. For a finite-dimensional irrep $V$ in an algebraically closed field $K$, $\text{End}_G (V) = K$. That is, every intertwiner is proportional to the identity operator $\id$, with the proportionality constant in the base field $K$.

Proof. The first observation is proved by noting that the image and the kernel of an intertwiner are subrepresentations. If either is nontrivial, then the irrep has a nontrivial subrepresentation, which is a contradiction.

For the second observation, let $\varphi \in \text{End}_G (V)$ and $\lambda$ be an eigenvalue (which exists because $K$ is algebraically closed). Consider $\varphi - \lambda \id$. It is also in $\text{Hom}_G (V)$ because $\id$ commutes with every operator on $V$. But $\ker (\varphi - \lambda \id)$ is a subrepresentation. Therefore $\varphi - \lambda \id$ must be the zero map i.e. $\varphi = \lambda \id$. $\square$

The assumption that $K$ is algebraically closed is necessary. Consider a representation of $\mathbb{Z}_4$ on $\mathbb{R}^2$ as discrete rotations with the generator \begin{aligned} \rep(e) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.\end{aligned} It is irreducible because $\rep(e)$ cannot be diagonalized over $\mathbb{R}$. But every $\rep(g)$ commutes with matrices of the form $a\id + b\rep(e)$ for some $a,b \in \mathbb{R}$, a two-dimensional real vector space.

An easy corollary is that every complex irrep of an abelian group is one-dimensional because $\rep(g_1)$ and $\rep(g_2)$ commute for every $g_1,g_2 \in G$. So they are all proportional to the identity operator.

Another consequence of Schur’s lemma is the “orthogonality of matrix elements”.

Theorem: Let $G$ be a finite group and $d_{\lambda}$ be the dimension of an irrep $\rep_{\lambda}$ of $G$. \begin{aligned} \frac{1}{|G|} \sum_g \rep_{\lambda}(g)_{jk} \left( \rep_{\lambda'}(g)_{mn} \right)^* &= \frac{1}{d_{\lambda}} \delta_{\lambda \lambda'} \delta_{jm} \delta_{kn} \end{aligned} That is, the matrix elements $(d_{\lambda} / |G|)^{1/2} \rep_{\lambda}(g)_{jk}$ of irreps are orthonormal as functions over $G$.

Proof. For any linear map $A:V_{\lambda'} \to V_{\lambda}$, its twirl \begin{aligned} \frac{1}{|G|} \sum_{g \in G} \rep_{\lambda}(g) A \rep_{\lambda'}(g^{-1}) \end{aligned} is an intertwiner between the two irreps. Therefore, by Schur’s lemma, it is either proportional to the identity or the zero operator. Setting $A = \ketbra{k}{n}$, \begin{aligned} \frac{1}{|G|} \sum_{g \in G} \rep_{\lambda}(g) \ketbra{k}{n} \rep_{\lambda'}(g^{-1}) &= N \delta_{\lambda \lambda'} \id_{d_{\lambda}}, \end{aligned} Taking the trace gives $N = \delta_{kn}/d_{\lambda}$ and taking the $j,m$ matrix element gives the desired result. $\square$

## Every finite-dimensional representation of a compact group is unitary and completely reducible

Any finite-dimensional representation of a compact group, possessing a Haar measure, is unitary. A measure $dg$ on $G$ is a Haar measure when it is left-invariant i.e. for any integrable function $f$, \begin{aligned} \int dg_1 f(g_2 g_1) &= \int dg_1 f(g_1),\end{aligned} and normalized, \begin{aligned} \int dg &= 1.\end{aligned} For a compact group, it is unique and also right-invariant \begin{aligned} \int dg_1 f(g_1 g_2) &= \int dg_1 f(g_1).\end{aligned} (Of course, when $G$ is finite this is just a sum.) Armed with the Haar measure, if $\rep(g)$ is not unitary under some sesquilinear inner product $\braket{v|w}$, redefine the inner product to be the average $\int dg \braket{\rep(g)v|\rep(g)w}$. This new inner product (which amounts to a change of basis) can be seen to be invariant under the $G$-action: \begin{aligned} \int dg_1 \braket{\rep^{-T}(g_1)v|\rep^*(g_2) \rep(g_2)|\rep(g_1)w} &= \int dg_1 \braket{\rep^{-T}(g_2 g_1)v|\rep(g_2 g_1)w} \\ &= \int dg \braket{\rep^{-T}(g)v|\rep(g)w}. \end{aligned} where we have used the left-invariance of the Haar measure. Thus unitarity is not an additional assumption when we deal with compact groups. The existence of an invariant inner product also provides an easy proof that every finite-dimensional representation is completely reducible. Given a subrepresentation $W$ of $V$, the orthogonal complement $W^{\perp}$ is also stable under $G$. Therefore, $V \simeq W \oplus W^{\perp}$.

## Isotypic decomposition

Let $\hat{G}$ be the collection of all inequivalent irreps of $G$. A completely reducible representation (by definition) decomposes into the orthogonal direct sum of irreps $V_{\lambda}$ \begin{aligned} V &\stackrel{G}{\simeq} \bigoplus_{\lambda \in \hat{G}} \bigoplus^{n_{\lambda}} V_{\lambda},\end{aligned} each with (possibly zero) multiplicity $n_{\lambda}$. A decomposition is said to be multiplicity-free if every $n_{\lambda}$ is either 0 or 1. By Schur’s lemma, \begin{aligned} \text{Hom}_G (V_{\lambda},V) &= \bigoplus^{n_{\lambda}} \text{Hom}_G ( V_{\lambda}, V_{\lambda} ) = \mathbb{C}^{n_{\lambda}}.\end{aligned} $\mathbb{C}^{n_{\lambda}}$ is called the multiplicity space where $n_{\lambda} = \dim \text{Hom}_G (V_{\lambda},V)$. Putting these together, we obtain the isotypic decomposition of $V$: \begin{aligned} V &\stackrel{G}{\simeq} \bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes \mathbb{C}^{n_{\lambda}} \stackrel{G}{\simeq} \bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes \text{Hom}_G (V_{\lambda},V).\end{aligned} An important special case is when $V$ is a tensor product of irreps. $V$ may not be irreducible and we have the Clebsch-Gordan decomposition \begin{aligned} V_{\mu} \otimes V_{\nu} &\stackrel{G}{\simeq} \bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes \mathbb{C}^{n^{\lambda}_{\mu\nu}}. \end{aligned} The collection of $\lambda$ that appears in the direct sum is called the Clebsch-Gordan series, and the overlap between vectors in $V_{\mu} \otimes V_{\nu}$ and vectors in $V_{\lambda}$ are Clebsch-Gordan coefficients. When $G$ is a unitary group, the $n^{\lambda}_{\mu\nu}$ are called Richardson-Littlewood coefficients.

Try pandoc! pandoc –from latex –to markdown from

A of $G$ is a map to a set $X$ respecting the group structure

for all $x \in X$. A representation is an instance of a group action on a vector space. A $X$ is a set with a transitive action of $G$: for any two elements $x,y \in X$, there is some $g \in G$ such that $y = gx$.

Given a subgroup $K$ of $G$, a of $K$ consists of elements of the form $gK$ for some $g \in G$. $g_1$ and $g_2$ belong to the same coset if $g_1 g_2^{-1}$ is in $K$. Two left cosets are either the same or disjoint, so $G$ is partitioned into left cosets. The $G/K$ is the set of all left cosets. The $K\backslash G$ is the set of all right cosets, which are defined similarly but with right multiplication by $g^{-1}$. The left and right cosets coincide if and only if $K$ is a normal subgroup i.e. $K$ is invariant under conjugation $gKg^{-1}$ by every $g \in G$. (In that case, $G/K \simeq K\backslash G$ is also a group.)

When $K \subset G$ is a closed subgroup, $G/K$ is a $G$-homogeneous space. Conversely, any homogeneous space can be constructed as a coset space $G/G_p$, where $G_p = \{g \in G | gx = x\}$ is the fixing an element $x$ in some set $X$. Since we will be dealing only with compact and semisimple Lie groups (), the following result is useful:

to
A group action of $G$ is a map to a set $X$ \begin{aligned} G \times X &\to X \\ (g,x) &\mapsto g\cdot x\end{aligned} respecting the group structure \begin{aligned} ex &= x, & g_1 (g_2 x) &= (g_1 g_2)x\end{aligned} for all $x \in X$. A representation is an instance of a group action on a vector space. A $G$-homogeneous space $X$ is a set with a transitive action of $G$: for any two elements $x,y \in X$, there is some $g \in G$ such that $y = gx$.

Given a subgroup $K$ of $G$, a left coset of $K$ consists of elements of the form $gK$ for some $g \in G$. $g_1$ and $g_2$ belong to the same coset if $g_1 g_2^{-1}$ is in $K$. Two left cosets are either the same or disjoint, so $G$ is partitioned into left cosets. The left coset space $G/K$ is the set of all left cosets. The right coset space $K\backslash G$ is the set of all right cosets, which are defined similarly but with right multiplication by $g^{-1}$. The left and right cosets coincide if and only if $K$ is a normal subgroup i.e. $K$ is invariant under conjugation $gKg^{-1}$ by every $g \in G$. (In that case, $G/K \simeq K\backslash G$ is also a group.)

When $K \subset G$ is a closed subgroup, $G/K$ is a $G$-homogeneous space. Conversely, any homogeneous space can be constructed as a coset space $G/G_p$, where $G_p = \{g \in G | gx = x\}$ is the stabilizer subgroup fixing an element $x$ in some set $X$. Since we will be dealing only with compact and semisimple Lie groups (), the following result is useful:

[[thm:closed-if-compact-semisimple]]{#thm:closed-if-compact-semisimple label=“thm:closed-if-compact-semisimple”} [@Berndt §4.2] If $G$ is compact or simply connected and the Lie algebra of $K$ is semisimple, then $K$ is closed in $G$.

## Fourier analysis on groups

For a finite group $G$, define an orthonormal basis $\{\ket{g}|g\in G\}$. Its complex span, the group algebra $\mathbb{C}[G]$, is an algebra (a vector space in which two vectors can be multiplied) with multiplication $*$ inherited from group multiplication $\ket{g_1}*\ket{g_2} = \ket{g_1g_2}$: \begin{aligned} \left( \sum_{g_1} f_{g_1} \ket{g_1} \right) * \left( \sum_{g_2} h_{g_2} \ket{g_2} \right) &= \sum_{g_1,g_2} f_{g_1} h_{g_1^{-1} g} \ket{g}.\end{aligned} Here the expression on the right is a discrete analogue of the convolution $(f*h)(x) = \int dy f(x)h(y-x)$. $\mathbb{C}[G]$ can also be naturally thought of as a representation of $G$ by left multiplication $\rep_L(g_1) \ket{g_2} = \ket{g_1 g_2}$, called the left regular representation $(\rep_L,\mathbb{C}[G])$, or right multiplication $\rep_R(g_1)\ket{g_2} = \ket{g_2 g_1^{-1}}$, called the right regular representation $(\rep_R,\mathbb{C}[G])$. When $G$ is infinite, we can interpret $\mathbb{C}[G]$ to be the convolution algebra of functions on $G$, provided that we agree on what we mean by a function. The standard choice is for $\mathbb{C}[G]$ to be $L^2(G)$, the space of square-integrable functions. The group action on a function is dual to the action on group algebra, \begin{aligned} \rep(g)f(x) &= f(g^{-1} x),\end{aligned} $x \in G$, since a function is a linear map from $G$ to $\mathbb{C}$.

A central result in representation theory is the isotypic decomposition of $\mathbb{C}[G]$: \begin{aligned} \mathbb{C}[G] &\stackrel{G}{\simeq} \bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes \text{Hom}_G (V_{\lambda},\mathbb{C}[G]) \stackrel{G}{\simeq} \bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes \mathbb{C}^{\dim V_{\lambda}}. \end{aligned} In words, every irrep appear as many times as its dimension in the regular representation. Since the left and right representations commute, $\mathbb{C}[G]$ can also be thought of as a representation of $G \times G$. Under this $G \times G$-action, $\mathbb{C}[G]$ decomposes in the multiplicity-free manner: \begin{aligned} \label{eq:regular decomposition} \mathbb{C}[G] \stackrel{G \times G}{\simeq} \bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes V_{\lambda}^* \stackrel{G \times G}{\simeq} \bigoplus_{\lambda \in \hat{G}} \text{End}(V_{\lambda}), \end{aligned} where the representation $(\rep_L \otimes \id, G\times G)$ acts on $V_{\lambda}$ and $(\id \otimes \rep_R, G\times G)$ acts on $V_{\lambda}^*$. The result is sometimes a part of so-called Maschke’s theorem and can be proved entirely in the language of semisimple algebras 4. Another route (which I took in my thesis) is via induced representations. 5. Yet another route, probably the simplest, is to use character theory.

The unitary change of basis from $\{\ket{g}\}$ to an orthonormal basis on the right hand side of is the Fourier transform. Its explicit matrix form, given an orthonormal basis $\{\ket{\lambda,j,k}|1\le j,k\le d_{\lambda}\}$ for each $V_{\lambda} \otimes V_{\lambda}^*$, is \begin{aligned} \label{Fourier transform:unitary} U_{\text{FT}} &= \sum_{g \in G} \sum_{\lambda \in \hat{G}} \sum_{j,k=1}^{d_{\lambda}} \sqrt{\frac{d_{\lambda}}{|G|}} \rep_{\lambda}(g)_{jk} \ketbra{\lambda,j,k}{g} = \sum_{g\in G} \ketbra{\tilde{g}}{g}, \end{aligned} $\ket{\tilde{g}}$ being the Fourier transform of the discrete delta function $\ket{g}$: \begin{aligned} \label{Fourier transform} \ket{\tilde{g}} &= \sum_{\lambda \in \hat{G}} \sum_{j,k=1}^{d_{\lambda}} \sqrt{\frac{d_{\lambda}}{|G|}} \rep_{\lambda}(g)_{jk} \ket{\lambda,j,k}. \end{aligned} (Note the choice of the constant $1/\sqrt{|G|}$, akin to using $1/(2\pi)$ in the continuous Fourier transform, is necessary for $U_{\text{FT}}$ to be unitary).

For the cyclic group $\mathbb{Z}_n$, this is just the discrete Fourier transform \begin{aligned} \label{Fourier transform:Z_n} U_{\text{FT}} = \frac{1}{\sqrt{n}} \sum_{x,y=0}^{n-1} e^{2\pi ixy/n} \ketbra{y}{x}. \end{aligned} More generally, for any abelian group, \begin{aligned} \label{Fourier transform:abelian} \ket{\tilde{g}} &= \frac{1}{\sqrt{|G|}} \sum_{\lambda \in \hat{G}} \chi_{\lambda}(g) \ket{\lambda}, \end{aligned} where $\chi_{\lambda}(g)$ is a one-dimensional irrep of $G$. The Fourier transform on an abelian group is much more well-behaved than the general case because the collection of all distinct irreps $\hat{G}$ also comes equipped with the abelian group structure. \begin{aligned} \chi^{-1}(g) &= \chi(g^{-1}) = \chi^* (g) \\ \chi(g_1)\chi(g_2) &= \chi(g_1 g_2). \end{aligned} In this case, $\hat{G}$ is called the dual group. The compact-discrete duality states that $\hat{G}$ is compact if $G$ is discrete and vice versa. Familiar dual pairs are $G = \hat{G} = \mathbb{Z}_d$, $G = \hat{G} = \mathbb{R}$, and $G = \gr{U}{1}$ dual to $\hat{G} = \mathbb{Z}$.

For compact groups, we have the celebrated

Theorem (Peter-Weyl): For a compact Lie group $G$ and $V_{\lambda}$ its irrep, the orthogonal direct sum $\bigoplus_{\lambda \in \hat{G}} V_{\lambda} \otimes V_{\lambda}^*$ is dense in $L^2(G)$ with respect to the supremum norm $||f||_{\infty} = \sup |f(g)|$. Moreover, every $V_{\lambda}$ is finite-dimensional.

In other words, every function in $L^2(G)$ can be approximated arbitrarily closely in the supremum norm by a finite linear combinations of the matrix elements $\{ \sqrt{d_{\lambda}} \rep_{\lambda}(g)_{jk} \}$. Thus, the matrix elements are not only orthonormal but also complete. This is a generalization of Fourier analysis and orthogonal polynomials and special functions to arbitrary groups.

1. Not to be confused with physicists’ (infinitesimal) “generators" of a Lie group which are Lie algebra elements

2. Richard Borcherds, MATH261 Lecture notes, UC Berkeley

3. Page 4 of William Fulton and Joe Harris, Representation Theory: A First Course, Springer, 1999

4. Pavel Etingof et al., Introduction to Representation Theory, AMS, 2011

5. Santiago Chaves Aguilar, A Survey on Representation Theory Math REU 2015

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