Notes on the Wedderburn decomposition

Wildon's Weblog 2024-09-10

Let $G$ be a finite group and let $\mathrm{Irr}(G)$ denote the set of irreducible representations of $G$ up to isomorphism. Wedderburn’s Theorem is that there is an algebra isomorphism

$\mathbb{C}[G] \cong \bigoplus_{V \in \mathrm{Irr}(G)} \mathrm{Mat}_V(\mathbb{C}).$

where $\mathbb{C}[G]$ is the complex group algebra of $G$ . Since any imaginable proof of the theorem shows that composing with the projection to $\mathrm{Mat}_V(G)$ gives the action of $\mathbb{C}[G]$ on $V$ , I’m going to include that as part of the statement. Thus Wedderburn’s Theorem implies Jacobson’s Density Theorem for $\mathbb{C}[G]$ -modules, namely that the matrices representing the action $G$ on $V$ span $\mathrm{Mat}_V(\mathbb{C})$ .

Incidentally, this corollary, that irreducible representations of finite groups are, after removing a thin disguise, simply matrix algebras acting on column vectors often strikes me as a little surprising: are irreducible representations really so boring? I think the answer is ‘yes they are, from the ring-theoretic perspective’. Consider that many of the interesting problems in representation theory concern the decomposition of big representations into irreducibles: for instance, for the symmetric group, Foulkes’ Conjecture in characteristic zero, or the decomposition matrix problem in prime characteristic. For such problems we care very much about the particular label of the irreducible representations (these come from the group structure), not just their dimension (which we have seen is entirely ring-theoretic information).

I know of two logical classes of proofs of Wedderburn’s Theorem: ‘two shot’ proofs that first prove the corollary for the group action, and then deduce the theorem, and ‘one shot’ proofs that lay the theorem cold in one (possibly somewhat extended) blow. We give specimens of both proofs here and end with a corollary on the automorphism group of group algebras, showing to what extent (not very far) the Wedderburn decomposition is canonical. Both proofs need the following lemma.

Decomposing the regular character of $G$

By character orthogonality, the regular character of $\mathbb{C}[G]$ defined by $\pi(g) = |G|[g = \mathrm{id}_G]$ decomposes as $\pi = \sum_{V \in \mathrm{Irr}(G)} (\dim V) \chi_V$ . Hence as a left $\mathbb{C}[G]$ -module,

$\mathbb{C}[G] \cong \bigoplus_{V \in \mathrm{Irr}(G)} V^{\oplus \dim V} \qquad (\star)$

A one shot proof

A fine example of the first logical class is the following (very standard) proof. By Schur’s lemma applied to $(\star)$ we get

$\begin{aligned} \mathrm{End}_{\mathbb{C}[G]} \mathbb{C}[G] &\cong \bigoplus_{V \in \mathrm{Irr}(G)} \mathrm{End}_{\mathbb{C}[G]} (V^{\oplus \dim V}) \\ &\cong \bigoplus_{V \in \mathrm{Irr}(G)} \mathrm{Mat}_{\dim V}(\mathbb{C})\end{aligned}$

To illustrate the second isomorphism, consider the case $\dim V = 2$ and let $v \in V$ be a any non-zero vector. Let $V'$ be isomorphic to $V$ by an isomorphism sending $v$ to $v'$ . Clearly any $\mathbb{C}[G]$ endomorphism of $V \oplus V'$ is determined by its effect on $v$ and $v'$ , and less clearly, but I hope you’ll see that it follows from yet more Schur’s Lemma, it is of the form $v \mapsto \alpha v + \gamma v'$ and $v' \mapsto \beta v + \delta v'$ for some $\alpha,\beta,\gamma,\delta \in \mathbb{C}$ , corresponding to the matrix

$\left( \begin{matrix} \alpha & \gamma \\ \beta & \delta \end{matrix} \right).$

To complete the proof we use that any $\mathbb{C}[G]$ -endomorphism of $\mathbb{C}[G]$ is of the form $x \mapsto xg$ for a unique $g \in G$ (determined as the image of $\mathrm{id}_G$ ) and so $\mathrm{End}_{\mathbb{C}[G]}\mathbb{C}[G] \cong \mathbb{C}[G]^{\mathrm{op}}$ , the opposite algebra to $G$ . This gives the Wedderburn isomorphism, but with an inconvenient ‘opposite’, which can be removed by noting that $X \mapsto X^{\mathrm{tr}}$ is an isomorphism between $\mathrm{Mat}_d(\mathbb{C})$ and $\mathrm{Mat}_d(\mathbb{C})^{\mathrm{op}}$ .

Critical remarks

The strength of this proof to my mind is that it has just one idea: ‘look at the endomorphism algebra of $\mathbb{C}[G]$ as a module for itself’. After that, it’s just the remorseless application of Schur’s Lemma. The one idea requires is well motivated by experience with computing with representations. For instance, one very important link between character theory and the theory of permutation groups is that if $\mathbb{C}[\Omega]$ is the permutation representation of $G$ acting on a set $\Omega$ , with canonical basis $\{ e_\omega : \omega \in \Omega \}$ , then $\dim \mathrm{End}_{\mathbb{C}[G]}\mathbb{C}[\Omega] = 2$ and

$\mathbb{C}[\Omega] = \bigl\langle \sum_{\omega \in \Omega} e_\omega \bigr\rangle \oplus V$

for an irreducible representation $V$ if and only if the action of $G$ on $\Omega$ is $2$ -transitive.

The weaknesses are, in my view: (1) it meets the criterion to be a ‘one shot’ proof, but the fuss with the opposite algebra lengthens the proof; (2) the map $\mathbb{C}[G] \rightarrow \mathrm{Mat}_{\dim V}(\mathbb{C})$ it gives comes not from $G$ acting on $V$ (as we expected) but from $G$ acting on $V^{\oplus \dim V}$ , in a way that commutes with the diagonal left action of $\mathbb{C}[G]$ . Taken together (1) and (2) make the isomorphism far less explicit than need be the case.

Digression on Yoneda’s Lemma

The appearance of the opposite algebra can be predicted: I’d like to digress briefly to show that it’s just a manifestation of the ‘opposite’ that appears in Yoneda’s Lemma. Stated for covariant functors, a special case of Yoneda’s Lemma states that natural transformation between the representable functors $\mathrm{Hom}_\mathcal{C}(x, -)$ and $\mathrm{Hom}_\mathcal{C}(y, -)$ are in bijection with $\mathrm{Hom}_\mathcal{C}(y, x)$ ; given such a morphism $g_{yx} : y \rightarrow x$ , the map $\mathrm{Hom}_\mathcal{C}(x, -) \rightarrow \mathrm{Hom}_\mathcal{C}(y, -)$ is defined by $f \mapsto f \circ g_{yx}$ , composing from right-to-left. (What other source of natural maps between these hom-spaces could there possibly be beyond morphisms from $y$ to $x$ ?) This makes the Yoneda embedding $x \mapsto \mathrm{Hom}_\mathcal{C}(x, -)$ a contravariant functor from $\mathcal{C}$ to $\mathbf{Set}^\mathrm{op}$ . (Slogan: ‘pre-composition is contravariant’.)

Cayley’s Theorem is the special case where $\mathcal{C}$ is a group (a category with one object and all morphisms invertible): the Yoneda embedding is then the map $G \mapsto \mathrm{Sym}_G$ defined by $g \mapsto \theta_g$ where $\theta_g(x) = xg$ . Note that $g' g$ which may be read as ‘do $g$ then $g'$ ‘ maps to $\theta_{g g'} = \theta_{g'} \circ \theta_{g}$ — we check:

$\begin{aligned} (\theta_{g'} \circ \theta_{g})(x) &= \theta_{g'} \bigl( \theta_g(x) \bigr) \\ & \quad = \theta_{g'} (xg) = (xg)g' = x(gg') = \theta_{g g'}(x). \end{aligned}$

Thus the Yoneda embedding is not a group homomorphism, but becomes one if we regard the codomain as $\mathrm{Sym}_G^{\mathrm{op}}$ . The proof above that $\mathrm{End}_{\mathbb{C}[G]}\mathbb{C}[G] \cong \mathbb{C}[G]^{\mathrm{op}}$ is almost the same ‘dancing around Yoneda’ but this time with a version of Yoneda’s Lemma for categories enriched over $\mathbb{C}$ -vector spaces.

Incidentally, the Yoneda embedding becomes covariant if we instead take natural transformations from $\mathrm{Hom}_\mathcal{C}(-, x)$ to $\mathrm{Hom}_\mathcal{C}(_, y)$ (‘post-composition is covariant’), but the ‘opposite’ pops up in another place, as these hom-into functors are contravariant, or, in the more modern language, ordinary covariant functors but defined on $\mathcal{C}^\mathrm{op}$ .

Finally, Yoneda’s Lemma is often stated as

$\mathrm{Nat}(\mathrm{Hom}_\mathcal{C}(x, -), F) \cong F(x)$

‘natural transformations from representable functors are representable’; observe that when $F = \mathrm{Hom}_\mathcal{C}(y, -)$ , we have $F(x) = \mathrm{Hom}_\mathcal{C}(y,x)$ , so this is indeed a generalization of the result as I stated it.

A two shot proof

In a ‘two shot proof’ we must first prove Jacobsen’s Density Theorem for $\mathbb{C}[G]$ -modules. The proof below closely following Tommaso Scognamiglio elegant proof on MathStackExchange. As motivation, recall that a representation of a finite group is irreducible if and only if it is cyclic, generated by any non-zero vector. Given this, to prove density, it is very natural to show that $\mathrm{Mat}(V)$ is an irreducible representation: but since (supposing the conclusion) we know that as a left $\mathbb{C}[G]$ -module

$\mathrm{Mat}(V) \cong V\, \oplus \stackrel{\dim V}{\cdots}\! \oplus\, V$

is not irreducible, we are going to have to consider a bigger group.

Density

There is a canonical $\mathbb{C}$ -linear isomorphism $\mathrm{End}(V) \cong V \otimes V^\star$ under which $v \otimes \theta$ maps to the rank $1$ endomorphism defined by $x \mapsto v \theta(x)$ . Recall that $V^\star$ is a representation of $G$ (acting on the left as usual) by $\rho_{V^\star}(g) \theta = \theta \circ \rho_V(g^{-1})$ . The matrix giving the action of $g$ is then $\rho_V(g^{-1})^\mathrm{tr}$ , the transpose appearing because if $\theta_1, \ldots, \theta_n$ is the dual basis to $v_1, \ldots, v_n$ then from

$\rho_V(g^{-1}) v_k = \sum_{j=1}^n \rho_V(g^{-1})_{jk} v_j \qquad (\dagger)$

we get $(\theta_j \circ g^{-1})(v_k) = \rho(g^{-1})_{jk}$ , and so $\rho_{V^\star}(g)\theta_j = \sum_k \rho(g^{-1})_{jk} \theta_k$ , which then requires a transpose to agree with the matrix convention in $(\dagger)$ . Thus $V \otimes V^\star$ is a representation of $G \times G$ with action defined by linear extension of

$(f, g)(v \otimes \theta) = \rho_V(f) v \otimes \rho_{V^\star}(g)\theta$

in which $(f, g) \in G \times G$ acts with matrix $\rho_V(f) \otimes \rho_V(g^{-1})^\mathrm{tr}$ . In particular $\chi_{\mathrm{End}(V)} (f, g) = \chi_V(f) \chi_V(g^{-1}) = \chi_V(f) \overline{\chi_V(g)}$ . Hence

$\begin{aligned} \langle \chi_{\mathrm{End}(V)}, \chi_{\mathrm{End}(V)} \rangle_{G \times G} &= \frac{1}{|G|^2} \sum_{(f,g) \in G \times G} |\chi_V(f)|^2 |\chi_V(g)|^2 \\ &= \bigl( \frac{1}{|G|} \sum_{g \in G} |\chi_V(g)|^2 \bigr) \\ &= 1. \end{aligned}$

showing that $\chi_{\mathrm{End}(V)}$ is an irreducible character, as required.

Proof of Wedderburn’s Theorem

By density, for each irreducible representation $V \in \mathrm{Irr}(G)$ , the induced map $\rho_V : \mathbb{C}[G] \rightarrow \mathrm{Mat}_{\dim V}(\mathbb{C})$ is surjective Hence $\bigoplus_{V \in \mathrm{Irr}(G)} \rho_V : \mathbb{C}[G] \rightarrow \bigoplus_{V\in \mathrm{Irr}(G)} \mathrm{Mat}_{\dim V}(\mathbb{C})$

is a surjective algebra homomorphism. But by the decomposition of the regular representation seen in $(\star)$ , the dimensions on each side are $|G|$ . Hence this map is an algebra isomorphism, as required.

Critical remarks

A strength of this proof is that you can swap in your own preferred proof of the density theorem. For instance, to prove the generalization to group algebras over an algebraically closed finite field that

$\frac{\mathbb{F}[G]}{\mathrm{rad}\ \mathbb{F}[G]} \cong \bigoplus_{V \in \mathrm{Irr}(G)} \mathrm{Mat}_{\dim V}(\mathbb{F})$

one can instead prove the density theorem for irreducible representations over an arbitrary field, and deduce that the map is injective because (either by definition, or an easy equivalent to the definition) an element of $\mathbb{F}[G]$ acting as zero on every irreducible $\mathbb{F}[G]$ -module lies in $\mathrm{rad}\ \mathbb{F}[G]$ . The only disadvantage to my mind is that it’s less suitable for beginners: Jacobson’s Density Theorem is not an obvious result, and although I like Scognamiglio’s proof very much, the use of the $G \times G$ action (on the left, but with one factor twisted over to the right, hence the fiddliness with transposition) could be off-putting.

Why ‘density’? A unitary example

The Weyl algebra $\langle \hat{x}, \hat{p} : \hat{x}\hat{p} - \hat{p}\hat{x} = i \hbar \rangle$ models the position and momentum operators in quantum mechanics. The relation above is the algebraic formulation of the uncertainty principle that there can be no simultaneous eigenstate for both the position $\hat{x}$ and momenum $\hat{p}$ operators. In the canonical interpretation of (time independent) wave functions as elements of $L^2(\mathbb{R})$ , i.e. square integrable complex valued functions on $\mathbb{R}$ , there is a representation

$\begin{aligned} \hat{x} \psi(x) &= x \psi(x) \\ \hat{p} \psi(x) &= -i\hbar \frac{\mathrm{d} \psi(x)}{\mathrm{d} x}\end{aligned}$

defined on the dense subspace of $L^2(\mathbb{R})$ of smooth functions in which the Weyl algebra generators act as self-adjoint operators. Exponentiating these operators we obtain $\exp(i t \hat{x})$ and $\exp(i t \hat{p})$ defined by

$\begin{aligned} \exp(i t \hat{x}) \psi(x) &= \exp(itx) \psi{x} \\ \exp(i t \hat{p}) \psi(x) &= \exp (\hbar t \frac{\mathrm{d}}{\mathrm{d}x})\psi(x) = \psi(x + \hbar t) \end{aligned}$

which generate an irreducible unitary representation of the Heisenberg group by multiplication and translation operators. (The second equation is essentially Taylor’s Theorem, and any worries about its applicability can be assuaged by muttering ‘dense subspace’.) The Stone–von Neumann theorem is that any irreducible representation of the Heisenberg group is unitarily equivalent to this representation An excellent question on MathStackexchange asks whether, whether these maps generate a dense subalgebra of $\mathrm{End}(L^2(\mathbb{R}))$ , suitably interpreted. The answer, perhaps surprisingly, is ‘no’ with respect to the operator norm, but ‘yes’ with the strong or weak operator topology; the proofs of this appeals to von Neumann’s double commutatant theorem, which is in turn is a generalization of the density theorem.

Pauli matrices

Continuing with the quantum theme, the density theorem for the Lie group $\mathrm{SU}_2(\mathbb{C})$ has an important consequence for quantum error correction. This Lie group is the transformation group of a single qubit, represented by a quantum state

$| \psi \rangle \in \bigl\langle | 0 \rangle, | 1 \rangle \bigr\rangle_\mathbb{C}.$

Here $|0\rangle$ and $|1\rangle$ are the orthonormal $Z$ -basis of a two-dimensional Hilbert space $\mathcal{H}$ ; the term ‘ $Z$ -basis’ is used because they are eigenvectors for the $Z$ Pauli matrix, defined below

$X = \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), \quad Y = \left( \begin{matrix} 0 & -i \\ i & 0 \end{matrix} \right), \quad Z = \left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right).$ The Pauli matrices are self-conjugate unitary operators in $\mathrm{SU}_2(\mathbb{C})$ whose linear span is $\mathrm{Mat}_2(\mathbb{C})$ . This verified the conclusion of Jacobson’s Density Theorem applied to the natural representation of $\mathrm{SU}_2(\mathbb{C})$ acting on $\mathcal{H}$ . More generally, given $c \in \{I,X,Y,Z\}^n$ , let

$P_c = P_{c_1} \otimes \cdots P_{c_n}$

where $P_I = I$ , $P_X = X$ , $P_Y = Y$ , $P_Z = Z$ . Then since $\mathcal{H}^{\otimes n}$ is an $n$ -fold tensor product of irreducible representations, it is an irreducible representation of $\mathrm{SU}_2(\mathbb{C})^n$ and so the $4^n$ matrices $P_c$ for $c \in \{I,X,Y,Z\}^n$ span the space $\mathrm{End}(\mathcal{H}^{\otimes n})$ . Of course this can also be seen directly. As a small check, note the endomorphism space has dimension $2^n \times 2^n$ .

Quantum error correction

In one important model for quantum error correction, we assume that the three Pauli matrices model the three possible errors that can occur to a logical state $|\psi \rangle_L \in \mathcal{H}^{\otimes n}$ in a quantum computer. To justify this, imagine that there is a huge further Hilbert space, $\mathcal{E}$ which models the environment. (But no imagination is necessary, it’s where we live.) At the start of the computation, the logical state and environment are unentangled, and so of the form

$|\psi \rangle_L |\epsilon \rangle_E$

for some $|\epsilon \rangle_E \in \mathcal{E}$ . After a short amount of time, this state has evolved unitarily, by an operator in the unitary group $U(\mathcal{H}^{\otimes n} \otimes \mathcal{E})$ . As seen at the end of the previous subsection, the density theorem implies that we may express this operator as

$\sum_{c \in \{I,X,Y,Z\}^n} P_c \otimes E_c$

for some unique endomorphisms $E_c \in \mathrm{End}(\mathcal{E})$ . Hence the new quantum state is

$\sum_{c \in \{I,X,Y,Z\}^n} P_c |\psi \rangle_L E_c |\epsilon \rangle.$

In quantum error correction, we get information about the error by entangling the logical state with a further ancilla state, using stabilisers of the code containing $|0\rangle_L$ to specify the entangling circuit. Let us suppose that the code containing $|\psi\rangle_L$ is a CSS code $\mathcal{C}$ and we do full Steane style extraction for both $X$ – and $Z$ -errors. The relevant ancilla state is then the tensor product of a plus logical and a zero logical. After the ancilla interaction, and assuming no further errors have occured, the quantum state is

$\sum_{c \in \{I, X, Y, Z\}^n} P_c |\psi \rangle_L E_c |\epsilon \rangle P_{c(X)} |+\rangle_L P_{c(Z)} |0\rangle_L$

where

$c(X)_i = \begin{cases} I & c_i = I \\ X & c_i = X \\ X & c_i = Y \\ I & c_i = Z \end{cases} \quad c(Z)_i = \begin{cases} I & c_i = I \\ I & c_i = X \\ Z & c_i = Y \\ Z & c_i = Z \end{cases}$

The ancilla states

$P_{c(X)} |+\rangle_L P_{c(Z)} |0\rangle_L$

and

$P_{c'(X)} |+\rangle_L P_{c'(Z)} |0\rangle_L$

are equal if and only if $P_{c(X)}P_{c'(X)}$ is an $X$ -stabiliser of $|+\rangle_L$ , and $P_{c(Z)}P_{c'(Z)}$ is a $Z$ -stabiliser of $|0\rangle_L$ . Equivalently, these products must map $\mathcal{C}$ to itself. Note this is the key place where orthogonality enters: we can’t control the environment, so there is no chance it appears there! Therefore, after measuring the ancilla state, we learn $c(X)$ and $c(Z)$ up to the relevant stabilisers, and moreover, the new quantum state is

$P_{c} |\psi \rangle_L E_c |\epsilon \rangle$

for some $c$ such that $P_c$ is congruent modulo stabiliers of $\mathcal{C}$ to $P_{c(X)} P_{c(Z)}$ . (If $Q$ is the matrix of $Z$ -stabilisers, i.e. $Z^v |\psi\rangle = |\psi\rangle$ for all $|\psi\rangle \in \mathcal{C}$ then $\mathrm{Stab} |+\rangle_L = \bigl\{ X^u : u \in \langle Q \rangle^\perp \bigr\}$ so learning $c(X)$ up to the relevant stabilisers is equivalent to learning the syndrome $Q u$ , where in the product we think of $u \in \mathbb{F}_2^n$ as a column vector. Thus, as one should expect, syndrome information about $X$ -errors comes from the $Z$ -stabilisers.) Note the environment state is now unentangled with the logical state and the logical state can be corrected by applying the self-inverse unitary map $P_{c(X)} P_{c(Z)}$ (which is equal to $P_c$ modulo stabilisers), giving the final state $|\psi \rangle_L E_c|\epsilon \rangle$ .

The automorphism group of $\mathbb{C}[G]$

Suppose that $\tau$ is an algebra automorphism of $\mathrm{Mat}_d(\mathbb{C})$ . Let $V$ be the natural $d$ -dimensional irreducible representation. Then twisting the action by $\tau$ we get a new representation $V^\tau$ in which $\mathrm{Mat}_d(\mathbb{C})$ acts by $X \cdot v = X^\tau v$ . Since $\mathrm{Mat}_d(\mathbb{C})$ has a unique irreducible representation up to isomorphism, we have $V^\tau \cong V$ . Hence, by definition of isomorphism of representations, there exists an invertible matrix $T$ such that $X^\tau = T^{-1}X T$ for all $X \in \mathrm{Mat}_d(\mathbb{C})$ . Therefore $\tau$ is conjugation by $T$ . We have proved the special case of the Skolem–Noether theorem that all automorphisms of $\mathrm{Mat}_d(\mathbb{C})$ are inner, and, moreover, that the automorphism group is $\mathrm{PGL}_d(\mathbb{C})$ . Since the block decomposition in the Wedderburn decomposition is unique: the block $\mathrm{Mat}(V)$ being given by multiplication by the central primitive idempotent

$e_V = \frac{\chi_V(1)}{|G|} \sum_{g \in G} \chi_V(g^{-1}) g,$

it is an easy corollary that

$\mathrm{Aut} \ \mathbb{C}[G] \cong \prod_{V \in \mathrm{Irr}(G)} \mathrm{PGL}_{\dim V}(\mathbb{C}).$

In particular, the Wedderburn decomposition is unique if and only if all irreducible representations have dimension $1$ , or equivalently, if and only if $G$ is abelian.