[Math] Understanding induced representations

Representation theory in the language of categories; how tensor operators are actually intertwiners and Wigner-Eckart theorem is Schur’s lemma; induction as an adjoint to the much more trivial restriction functor which restricts representations of a group to a subgroup

created: 17 Sep 2019; modified: 22 Jan 2021; status: draft

Preliminaries: Group representations

The standard way to introduce a representation of a group $G$ is as a pair $(\rho,V)$ of a vector space $V$ and a group homomorphism $\rho: G \to \GL(V)$. That is, for $g_1,g_2 \in G$, $\rho(g_1)$ and $\rho(g_2)$ are invertible matrices over a vector space $V$ such that the group structure is respected: $\rho(g_1)\rho(g_2) = \rho(g_1 g_2)$. Group homomorphisms are examples of ubiquitous structure-preserving maps or morphisms between mathematical objects. Morphisms for sets are arbitrary total functions (i.e. defined for all inputs) between sets. For vector spaces, we have linear operators. Not only algebraic objects have morphisms, of course. For example, continuous maps and smooth maps are morphisms, respectively, for topological spaces and smooth manifolds. (In particular, homeomorphisms are isomorphisms between topological spaces, and diffeomorphisms are isomorphisms between smooth manifolds.)

Once we have objects and morphisms, we have a category. Formally, a category $\mathcal{C}$ is a class ¹ of objects and morphisms between them such that there is an identity morphism $1\circ f = f\circ 1 = f$ and the composition of morphisms is associative: $(f \circ g) \circ h = f \circ (g \circ h)$. It should be emphasized that all the information contained in the definition is about morphisms, whereas the objects can be anything. (Perhaps the lesson here is that morphisms are as important, if not more, than the objects we want to study.) What would be the morphisms in a category of $G$-representations? They are $G$-equivariant maps or intertwiners. Let $(\rho,V)$ and $(\sigma,W)$ be $G$-representations. A linear map $\varphi:V\to W$ is $G$-equivariant if it does not matter if one applies a transformation $g$ on the vector space first or $\varphi$ first. That is, $\varphi\circ\rho(g) = \sigma(g)\circ\varphi$ for all $g \in G$. Pictorially, we say that the following diagram “commutes”.

When $\rho=\sigma$, this just says that $\varphi$ commutes with $\rho(g)$ for all $g \in G$. This is a pretty natural definition (and should be formalizable as a natural transformation in category theory) but can be made more obvious by thinking of $G$-representations as $k[G]$-modules.

In terms of modules

The notion of a module generalizes that of a vector space by extending multiplication by scalars to elements of a ring (such as matrices). Given a ring $R$ with an identity, a (left) $R$-module is an abelian group $M$ representing “vector addition”, and “scalar multiplication” $R \times M \to M$ a ring homomorphism from $R$ to the endomorphism ring $\End(M)$ that also respects the group addition; in equations,

\[\begin{align} \begin{split} 1m &= m, \\ (r+s)m &= rm + sm, \\ (rs)m &= r(sm), \\ r(m+n) &= rm + rn, \end{split} \end{align}\]

for all $r,s \in R$ and $m,n \in M$. A right $R$-module or even a left-right $(S,R)$ bi-module is similarly defined but with the $R$-multiplication from the right and $S$-multiplication from the left. (Left and right $R$-modules may not be the same.) The ring $k[G]$ is the matrix ring formally spanned by $g \in G$ over a field $k$ with the convolution product * inherited from the group multiplication:

\[\begin{align} \left(\sum_{g_1}f_{g_1} g_1 \right) * \left(\sum_{g_2}h_{g_2} g_2 \right) = \sum_{g_1,g} f_{g_1} h_{g_1^{-1}g} g. \end{align}\]

In fact, $k[G]$ is not just a ring but also an algebra (because it is a vector space) called the group algebra. Given a $G$-representation, the linear span of $\rho(g)$ gives a $k[G]$-module. Conversely, given a $k[G]$-module, there is a vector space $V$ over $k$ (because $k\id \in k[G]$), on which the $G$-action gives a group homomorphism $G \to \GL(V)$, that is, a representation. In this language, intertwiners are linear maps that commute with multiplication by scalars in $k[G]$; they are $k[G]$-linear maps, generalizing the fact that linear operators (all of which commute with scalars) are morphisms in the category of vector spaces.

Schur’s lemma

Intertwiners between irreps are completely characterized by Schur’s lemma, a very basic (easily proved using only a little linear algebra) but powerful result (and what inspired the title of this blog). Denote the space of intertwiners between $G$-representations $V$ and $W$ by $\Hom_G(V,W)$ (and $\End_G(V)$ if $V=W$).

Lemma 1. (Schur’s lemma)

• (Burnside) Every intertwiner between irreps is either an isomorphism or zero.
• (Schur) For $k$ algebraically closed, every intertwiner is of the form $k\id$. That is, $\End_G(V) = k$.

The lemma can also be summarized as follows: when $k$ is algebraically closed, intertwiners between irreps behave like the Kronecker delta

\[\begin{align} \dim\Hom_G(V,W) = \begin{cases} 1, V \cong W, \\ 0, V \;\unicode{0x2246}\; W. \end{cases} \end{align}\]

[[The assumption that $k$ is algebraically closed is necessary. Consider a representation of $\Z_4$ on $\R^2$ as discrete rotations with the generator \(\begin{equation} \rho(1) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. \end{equation}\) It is irreducible because $\rho(1)$ cannot be diagonalized over $\R$. But every $\rho(g)$ commutes with matrices of the form $a\id + b\rho(1)$ for some $a,b \in \R$, a two-dimensional real vector space.]]

Digression: Tensor operators

I claim that one does not really understand the concept of tensor operators in physics until one understands intertwiners. Why? Because tensor operators are, by definition, intertwiners. This can be seen by unpacking the seemingly circular definition of tensor operators as “operators that act like tensors”. For concreteness, let us think about one of the simplest non-trivial cases of a vector operator. Let $(R,\R^3)$ be the defining representation of $\SO(3)$ and $\mathbf{r} \in \R^3$. A vector operator $T(\mathbf{r})$ has components $T_j \coloneqq T(\mathbf{r}_j):V\to W, j=1,2,3$ that are linear maps between irreps $(\rho,V),(\sigma,W)$ of $\SO(3)$ (that is, $T_j \in\Hom(V,W) = V^* \otimes W$) such that the $G$-action on $\Hom(V,W)$ (a superoperator $\sigma\dgg(R) \odot \rho(R)$ where $\odot$ is a placeholder for an argument) induces a rotation of $\mathbf{r}$: ²

\[\begin{align} \sigma\dgg(R)T(\mathbf{r}_j)\rho(R) = R_j{}^k T(\mathbf{r}_k). \end{align}\]

Writing this in the coördinate-independent form

\[\begin{align} \sigma\dgg(R)T(\mathbf{r})\rho(R) = T(R\mathbf{r}), \end{align}\]

one can see that $T$ is nothing but a linear map that makes the following diagram commutes!

By the first part of Schur’s lemma, $T$ is an isomorphism between $(R,\R^3)$ and its copy within $\Hom(V,W)$ if it exists (see remark below). This justifies labeling the components of the image $T_j$ the same way we label the components of a vector $\mathbf{r} \in \R^3$. More generally, $\R^3$ can be replaced by any irrep $U$ of $\SO(3)$.

[[Why can’t there be more than one copy of $(R,\R^3)$ in $\Hom(V,W)$? Because every tensor product $V\otimes W$ of irreducible $\SO(3)$-representations decomposes without multiplicity. Wigner gave a characterization of some of these so-called simply reducible groups which was later generalized by Mackey.]]

Does anyone remember what the celebrated Wigner-Eckart theorem says in the form usually taught in a quantum mechanics class? Definitely not me. But by recognizing tensor operators as intertwiners, the theorem is just Schur’s lemma in disguise, specialized to intertwiners between $U$ and $\Hom(V,W)$; the basis-independence of the “reduced matrix element” $\av{j’,m’\lVert T_k\ \rVert j,m}$ is just the statement that the tensor operator is proportional to the identity $T = c\id$ in the block isomorphic to the irrep $U$.

Induced representations

Given a representation of $G$, we can always restrict it to a representation of a subgroup $H \subset G$, but what about the inverse process of constructing a $G$-representation out of an arbitrary $H$-representation? The language of categories and modules will prove useful here. Restriction is a functor from the category of $G$-representations to the category of $H$-representations. Given categories $\mathcal C$ and $\mathcal D$, a (covariant) functor $\mathcal{F}:\mathcal C\to \mathcal D$ assigns to every object in $\mathcal C$ an object in $\mathcal D$ and to every morphism in $\mathcal C$ a morphism in $\mathcal D$:

\[\begin{align} \mathcal{F}:\Hom_{\mathcal C}(A,B) \to \Hom_{\mathcal D}(\mathcal{F}(A),\mathcal{F}(B)), \end{align}\]

where $A,B\in\mathcal C$, such that the identity and the composition is preserved:

\[\begin{align} \mathcal{F}(1_{\mathcal C}) = 1_{\mathcal D}, && \mathcal{F}(f\circ g) = \mathcal{F}(f) \circ \mathcal{F}(g). \end{align}\]

Restriction belongs to a simple kind of functors, forgetful functors that simply forget information, in this case the structure of $G$. Restriction therefore cannot have a strict inverse. An insight from category theory is that we can construct the induction functor as a kind of “weak inverse” to the restriction functor called an adjoint. More precisely, induction should be define as a left adjoint (so named because $\mathrm{Ind}$ is in the left slot below) to the restriction functor $\mathrm{Res}$ implicitly defined by Frobenius reciprocity:

\[\begin{align}\label{eq:left-adj} \Hom_G(\mathrm{Ind}_L\,V,W) \cong \Hom_H(V,\mathrm{Res}\,W) \end{align}\]

or a right adjoint defined similarly

\[\begin{align} \Hom_G(V,\mathrm{Ind}_R\,W) \cong \Hom_H(\mathrm{Res}\,V,W) \end{align}\]

(People call them “adjoints” because they look similar to the adjoints in linear algebra: $\langle x,Ay\rangle = \langle A\dgg x,y\rangle$.) The functorial viewpoint makes it clear that Frobenius reciprocity is not merely a property but the raison d’être of induced representation. That induction is defined to satisfy Frobenius reciprocity (the adjoint relation) might be lost if we give a definition of an induced representation first and then prove the adjunction. Not only will it be unclear why the definition given is the correct one, but the definition may also be different depending on which version (left or right) of Frobenius reciprocity we are trying to prove. For finite groups, the right adjoint is isomorphic to the left adjoint (but not canonically), which is equal to

\[\begin{align} \mathrm{Ind}_L W = \bigoplus_{k \in G/H} kW, \end{align}\]

whereas for Lie groups, we end up with two inequivalent notions of induction: ordinary and compact induction

\[\begin{align} \mathrm{Ind}_R W &= \{f:G\to W|f(gh)=\sigma(h^{-1})f(g) \} \\ \mathrm{Ind}_L W &= \{f:G\to W|f(gh)=\sigma(h^{-1})f(g), f \textrm{ has compact support on } G/H \}. \end{align}\]

These definitions can be proven directly to satisfy the correct Frobenius reciprocity, Chaves REU 2015, but we’ll show more general def

Adjoint functors

Example: The simplest adjoint functors

What does an adjoint functor look like in a category with only one morphism? Such a morphism $x\to y$ can be though of as a comparison $x\le y$, a preorder if you will ³. A functor between two preorders $\mathcal C$ and $\mathcal D$ is a monotone function if it is order-preserving: \(a \le_{\mathcal C} b \implies f(a) \le_{\mathcal D} f(b).\) A pair of monotone functions $f$ and $g$ is an adjoint pair if there is a one-to-one correspondence \(f(a) \le_{\mathcal D} x \iff a \le_{\mathcal C} g(x)\) ⁴. Suppose, for instance, that $\mathcal A = \mathcal B = \mathbb{N}$, the natural numbers. The monotone function $x \to 2x$ has no strict inverse because $y/2$ may not be an integer, but it does have right and left adjoints: the ceiling $\lceil x/2 \rceil$ and the floor $\lfloor x/2 \rfloor$ respectively. (Verify this.) Adjunction generalizes this one-to-one correspondence.

Example: Free functors

Many useful constructions, like those of induced representations, are adjoint functors of a forgetful functor. The free functor $\mathcal F:$ from the category of all sets to the category of all groups sends any set to its free group, the group of all “words”, non-commutative product constructed from the alphabet in $S$ and their inverses i.e. $S$ generates $\mathcal F(S)$ without any constraint. The free functor is defined by the universal property ⁵ that giving a function $f$ from $S$ to (the underlying set of) an arbitrary group $G$ is equivalent to giving a unique group homomorphism $\varphi$ from the free group $\mathcal F(S)$ to $G$. For example, the empty set gives the trivial group with one element. (The empty word is the identity.) In other words, $\mathcal F(S)$ can simulate any interaction between $S$ and an arbitrary group $G$ and does so optimally (the uniqueness of $\varphi$).

The free functor $\mathcal F$ should be an adjoint of the forgetful functor $\mathcal R$ that maps a group to its underlying set. This is because every group homomorphism $\varphi :F(S) \to G$ is completely determined if we know what it does to each generator of $\mathcal F(S)$. Since the generators are just elements of $S$, what $\varphi$ really acts on is the underlying set of the group. Thus, we have the following commutative diagram

which says that the free functor is a left adjoint of the forgetful functor: \(\begin{align} \Hom_{\mathbf{Grp}}(\mathcal F(S),G) \cong \Hom_{\mathbf{Set}}(S,\mathcal R(G)). \end{align}\)

Beware that having an adjoint on one side does not imply having the adjoint on the other side, as the forgetful functor from Grp to Set illustrates. Having the right adjoint would imply the following commutative diagram.

But this can’t happen when $S=\emptyset$ because there is no (total) function to an empty set, whereas there is always a (unique) group homomorphism to $\mathcal F(\emptyset)$ the trivial group with one element. (It is the terminal object in Grp).

Extension of scalars

[[To Rewrite]]

Having seen examples of free functors, let us go back to our original goal of making sense of induction. In the language of modules, induction corresponds to an extension of scalars, similar to how a real vector space $V$ can be complexified to $V^{\C} = V \otimes_{\R} \C$. We begin by looking for an adjunction involving tensor products.

A hint comes from a trivial-looking identity given three finite-dimensional vector spaces $M,N,P$

\[\begin{align} \Hom (P\otimes M,N) = \Hom (M,\Hom(P,N)), \end{align}\]

which expresses on both sides $P^* \otimes M^* \otimes N$ without using the dual. This makes the tensoring with $P$ and the “hom” map $\Hom(P,\odot)$ look like an adjoint pair, the tensoring being the left adjoint and hom being the right adjoint. Interestingly, this adjunction continues to hold when the vector spaces, which are $k$-modules, are replaced by general modules. This is called the tensor-hom adjunction

\[\begin{align} \Hom_S (P\otimes_R M,N) = \Hom_R (M,\Hom_S(P,N)). \end{align}\]

To unpack its meaning, we have to properly define what $M,N,P$ and these tensor products between modules are. The tensor product $M \otimes_R N$ of a right $R$-module $M$ and a left-$R$-module $N$ is the vector space $M \otimes N$ quotient by the relation $mr \otimes n - m \otimes rn = 0$ for every $r \in R$. (“$\otimes$ commutes with $R$.”)

[[A tensor product of two nonzero modules can turn out to be zero. For example, $\Z/2\Z \otimes_{\Z} \Z/3\Z = 0$ because $1m\otimes_{\Z} n = (3-2)m\otimes_{\Z} n = m\otimes_{\Z} 3n - 2m\otimes_{\Z} n = 0$ vanishes identically for any $m \in \Z/2\Z$ and $n \in \Z/3\Z$. More generally, the same conclusion can be reached for $\Z/a\Z \otimes_{\Z} \Z/b\Z$ when $a$ and $b$ are coprime since we can always find integers $x$ and $y$ such that $ax+by=1$.]]

With a $(S,R)$-bi-module $P$ at our disposal, we can turn an $R$-module into an $S$-module and vice versa. Let $M$ be a left $R$-module and $N$ be a left $S$-module.

• $P \otimes_R M$ is a left $S$-module.
• $\Hom_S (P,N)$ is a left $R$-module.

$\Hom_S (P,N)$ makes sense because both $P$ and $N$ are left $S$-modules. The obvious $R$-action \(r\varphi(p) = \varphi(pr)\) turns $\Hom_S (P,N)$ into a left $R$-module. One can verify that the action satisfies the axioms for modules, for example,

\[\begin{align} (r_1 r_2) \varphi(p) = \varphi(p (r_1 r_2)) = r_2 \varphi(p r_1) = r_1 (r_2 \varphi(r)). \end{align}\]

Now we can give a proof of the tensor-Hom adjunction

\[\begin{align} \Hom_S (P\otimes_R M,N) = \Hom_R (M,\Hom_S(P,N)), \end{align}\]

the only thing complicated about which is the bookkeeping.

Proof.

$\blacksquare$ In one direction, given an $S$-module homomorphism $\varphi$ from $P \otimes_R M$ to $N$, we want to define a corresponding $R$-module homomorphism from $M$ to $\Hom_S (P,N)$. By linearity, $\varphi$ is determined by its values on all rank-one tensors

\[\begin{align} \varphi(p \otimes_R m) \in n. \end{align}\]

So let us define

\[\begin{align} \psi_m(\odot) \coloneqq \varphi(\odot \otimes_R m). \end{align}\]

The $S$-linearity of $\psi_m$ follows immediately from the $S$-linearity of $\varphi$. The $R$-linearity of $\psi$ is not hard to show either:

\[\begin{align} \begin{split} \psi_{rm_1 + m_2}&(p) = \varphi(p \otimes_R (rm_1 + m_2)) \\ &= \psi_{m_1}(pr) + \psi_{m_2}(p) = r\psi_{m_1}(p) + \psi_{m_2}(p). \end{split} \end{align}\]

(The last equality is just the previously defined left $R$-action on $\Hom_S (P,N)$).

Conversely, given an $R$-module homomorphism $\psi$ from $M$ to $\Hom_S (P,N)$, define

\[\begin{align} \varphi(\odot_1 \otimes_R \odot_2) \coloneqq \psi_{\odot_2} (\odot_1{}_{}) \end{align}\]

and extend by linearity. $\varphi$ can be verified to be $B$-linear. Finally, since the construction of $\psi$ from $\varphi$ and vice versa is an inverse of each other, the tensor-hom adjunction is established. $\Box$

Tensor-Hom adjunction

I claim that the restriction of scalars from $R$ to a subring $S\subset R$ have both left and right adjoints. Again, $M$ is a left $R$-module and $N$ is a left $S$-module.

• Thinking of $R$ as an $(R,S)$-bi-module, $\mathrm{Res} N \coloneqq \Hom_R (R,N)$ gives a left $S$-module. Then by the tensor-hom adjunction \(\Hom_S (N,\Hom_R (R,M)) \cong \Hom_R (R \otimes_S N,M),\)

  we have that the left adjoint of restriction is $\mathrm{Ind}_L N = R \otimes_S N$, often called the extension of scalars.

• Thinking of $R$ as a $(S,R)$-bi-module, $\mathrm{Res} M \coloneqq R \otimes_R M$ gives a left $S$-module. Then by the tensor-hom adjunction \(\Hom_S (R \otimes_R M,N) \cong \Hom_R (M,\Hom_S (R,N)),\)

  we have that the right adjoint of restriction is $\mathrm{Ind}_R N = \mathrm{Hom}_S (R,N)$ called the co-extension of scalars to distinguish it from the left adjoint.

Setting $R = k[G]$ and $S = k[H]$ gives both versions of Frobenius reciprocity (12,13):

\[\begin{align} \Hom_G(k[G]\otimes_{k[H]} V,W) &\cong \Hom_H(V,\mathrm{Res}\,W),\\ \Hom_G(V,\Hom_{k[H]}(k[G],W)) &\cong \Hom_H(\mathrm{Res}\,V,W). \end{align}\]

We may call the former induced representation of $V$ and the latter co-induced representation of $W$. Clearly, $k[G]\otimes_{k[H]} V$ in (37) is the same as $\mathrm{Ind}L W = \bigoplus{k \in G/H} kW$. \bigoplus_{k\in G/H} kW

There is much more that can be said about induced representations. (How the right and left adjoints for finite groups are isomorphic. How they are in general not isomorphic for Lie groups. The geometric interpretation of adjoint representation as a homogeneous vector bundle.) But those would be topics for another day.

1. The word “class” cannot be replaced by “set”. For example, there is no set of all sets (Russell’s paradox). ↩

2. We think of R as both a rotation matrix (the defining representation of SO(3)) and an abstract group element. We also assume the Einstein summation convention. ↩

3. A partial order is a preorder that is also “antisymmetric”: a≤b and b≤a implies a=b. Also for the curious, the product in this category is the infimum and the coproduct is the supremum. ↩

4. The example is taken from John Baez’ online Applied Category Theory Course. See also one of Qiaochu Yuan’s answers on Math Stack Exchange ↩

5. I recommend reading this blog post by Jeremy Kun. Teaser: “There are only two universal properties and they are that of being initial and final.” ↩