Math 202B: Lecture 18

In this Lecture, we develop some further purely linear-algebraic background which we will use to study unitary representations of finite groups. Most of this will probably be familiar from Math 202A albeit stated in a possibly different way.

Let V and W be Hilbert spaces of positive, finite dimension. We denote the scalar product in V by \langle \cdot,\cdot \rangle_V and the scalar product in W by \langle \cdot,\cdot \rangle_W. Furthermore, let Y \subset V be an orthonormal basis of V, and let X \subset W be an orthonormal basis of W.

Proposition 18.1. For each T \in \mathrm{Hom}(V,W), there exists a unique T^* \in \mathrm{Hom}(W,V), called the adjoint of T, such that

\langle w,Tv \rangle_W = \langle T^*w,v \rangle_V, \quad v\in V,\ w \in W.

Problem 18.1. Prove Proposition 18.1.

The adjoint operation leads to a natural scalar product on the vector space \mathrm{Hom}(V,W).

Definition 18.2. The Frobenius scalar product on \mathrm{Hom}(V,W) is defined by

\langle S,T \rangle = \mathrm{Tr}\, S^*T, \quad S,T \in \mathrm{Hom}(V,W).

Note that the Frobenius scalar product \langle \cdot,\cdot \rangle carries no subscript, differentiating it from the scalar products \langle \cdot,\cdot \rangle_V and \langle \cdot,\cdot\rangle_W in V and W.

Proposition 18.2. The Frobenius scalar product is indeed a scalar product on \mathrm{Hom}(V,W).

Proof: We have

\langle S,T \rangle = \sum\limits_{y \in Y} \langle y,S^*Ty\rangle_V = \sum\limits_{y \in Y} \langle Sy,Ty \rangle_W.

Linearity in the second slot is easily verified:

\langle S,\beta_1T_1 +\beta_2T_2\rangle = \beta_1\langle S,T_1\rangle+\beta_2\langle S,T_2\rangle,

as is \langle T,S \rangle = \overline{\langle S,T\rangle}. As for positive definiteness, we have

\langle T,T \rangle = \sum\limits_{y \in Y} \langle Ty,Ty\rangle_W = \sum\limits_{y \in Y} \|Ty\|_W^2,

and this sum of nonnegative numbers is zero if and only if all its terms are zero, which happens if and only if Ty=0_W for all y \in Y, which is the case if and only if T \in \mathrm{Hom}(V,W) is the zero transformation.

-QED

Now that \mathrm{Hom}(V,W) has been promoted from a vector space to a Hilbert space, we seek an orthonormal basis in this space. Let Y \subset V and X\subset W be orthonormal bases of the underlying spaces which define \mathrm{Hom}(V,W).

Definition 18.3. The elementary transformations in \mathrm{Hom}(V,W) relative to X,Y are defined by

E_{xy}y' =x\langle y,y'\rangle_V, \quad x \in X,\ y,y'\in Y.

Note that if V=W and X=Y, then the elementary transformations E_{xy} \in \mathrm{Hom}(V,V)=\mathcal{L}(V) are the elementary operators in \mathcal{L}(V) discussed previously. However if V and W are different Hilbert spaces, the transformations E_{xy} \in \mathrm{Hom}(V,W) and E_{yx} \in \mathrm{Hom}(W,V) belong to different vector spaces.

Proposition 18.3. The elementary transformations satisfy E_{xy}^* = E_{yx}.

Proof: Let (x,y) \in X \times Y \subset W \times V, and let E_{xy} \in \mathrm{Hom}(V,W) be the corresponding elementary transformation. For any (x',y') \in X \times Y, we have

\langle E_{xy}y',x'\rangle_W = \langle x\langle y,y'\rangle_V,x'\rangle_W = \langle x,x'\rangle_W \langle y,y'\rangle_V

and

\langle y',E_{yx}x'\rangle_V = \langle y',y\langle x,x'\rangle_W\rangle_V=\langle x,x'\rangle_W \langle y',y\rangle_V.

-QED

Proposition 18.4. The set \{E_{xy} \colon (x,y) \in X \times Y\} is an orthonormal basis of \mathrm{Hom}(V,W) with respect to the Frobenius scalar product.

Proof: For any x,x' \in X and y,y' \in Y, we have

\langle E_{xy},E_{x'y'}\rangle = \sum\limits_{z \in Y} \langle E_{xy}z,E_{x'y'}z\rangle = \sum\limits_{z \in Y}\langle x\langle y,z\rangle,x'\langle y',z\rangle\rangle= \langle x,x'\rangle\sum\limits_{z \in Y} \langle y,z\rangle \langle y',z\rangle =\langle x,x'\rangle \langle y,y'\rangle,

which establishes orthonormality of the elementary transformations in \mathrm{Hom}(V,W) with respect to the Frobenius scalar product. Verification of the fact that \{E_{xy} \colon (x,y) \in X \times Y\} spans \mathrm{Hom}(V,W) is left as an exercise.

-QED

Problem 18.2. Prove that for any A \in \mathrm{Hom}(V,W) and B \in \mathrm{Hom}(W,V) we have \mathrm{Tr}\, AB=\mathrm{Tr}\, BA. Note that this is more surprising than in the case of square matrices, since the operators AB \in \mathcal{L}(W) and BA \in \mathcal{L}(V) act in different spaces.

Now that \mathrm{Hom}(V,W) is a Hilbert space with an explicit orthonormal basis, we are in position to consider the algebra \mathcal{L}(\mathrm{Hom}(V,W)) of linear operators on this Hilbert space (i.e. linear operators on linear transformations), and to concretely calculate the matrices of such operators. In particular, we consider the map

\omega \colon \mathcal{L}(V) \times \mathcal{L}(W) \longrightarrow \mathcal{L}(\mathrm{Hom}(V,W))

defined by

\omega(A,B)T = BTA^*, \quad T \in \mathrm{Hom}(V,W).

First of all, \omega(A,B)T is a composition of linear transformations V \to V \to W \to W, so it is indeed a linear transformation V \to W, and the map \omega is itself sesquilinear.

Proposition 18.45 If A \in \mathcal{L}(V) and B \in \mathcal{L}(W) are unitary, then \omega(A,B) \in \mathcal{L}(\mathrm{Hom}(V,W)) is unitary.

Proof: For any S,T \in \mathrm{Hom}(V,W), we have

\langle \omega(A,B)S,\omega(A,B)T\rangle = \langle BSA^*,BTA^*\rangle=\mathrm{Tr}\, (BSA^*)*(BTA^*)=\mathrm{Tr}\, AS^*B^*BTA^*=\mathrm{Tr}\, AS^*TA^*=\mathrm{Tr}\,S^*TA^*A= \langle A,B\rangle.

-QED

You have in fact seen this construction before, in Math 202A, in the context of the Singular Value Decomposition of a linear transformation.

Proposition 18.6. For any A \in \mathcal{L}(V) and B \in \mathcal{L}(W), the matrix elements of the operator \omega(A,B) \in \mathcal{L}(\mathrm{Hom}(V,W)) relative to the orthonormal basis \{E_{xy} \colon X \times Y\} are

\langle E_{xy},\omega(A,B)E_{x'y'}\rangle = \overline{\langle y,Ay'\rangle}\langle x,Bx'\rangle.

Proof: We have

\langle E_{xy},\omega(A,B)E_{x'y'}\rangle = \mathrm{Tr} E_{xy}^*BE_{xy}A^*=\mathrm{Tr}BE_{x'y'}A^*E_{xy}^*.

The trace on the right hand side is

\mathrm{Tr}BE_{x'y'}A^*E_{xy}^*=\sum\limits_{w \in X} \langle w,BE_{x'y'}A^*E_{xy}^*w\rangle=\sum\limits_{w \in X} \langle w,BE_{x'y'}A^*y\rangle\langle x,w\rangle=\langle x,BE_{x'y'}A^*y\rangle.

The scalar product on the right is

\langle x,BE_{x'y'}A^*y\rangle=\langle x,Bx'\rangle \langle y',A^*y\rangle = \langle x,Bx'\rangle \langle Ay',y\rangle=\overline{\langle y,Ay'\rangle} \langle x,Bx'\rangle.

-QED

Math 202B: Lecture 17

Let \mathcal{A} = \mathcal{C}(G) be the convolution algebra of a finite group G, and let \rho \colon \mathcal{A} \to \mathcal{L}(V) be a linear map from \mathcal{A} into the algebra of linear operators on a Hilbert space V. We define an associated function on the underlying group G by

U^\rho(g) = \rho(E_g), \quad g \in G,

where E_g(h) = \delta_{gh} is the elementary function on G indexed by the group element g \in G. By definition, U^\rho is a function with domain G and codomain \mathcal{L}(V).

Theorem 17.1. The linear map \rho \colon \mathcal{C}(G) \to \mathcal{L}(V) is an algebra homomorphism if and only if U^\rho is a group homomorphism from G into U(\mathcal{L}(V)), the unitary group of the algebra \mathcal{L}(V).

Proof: Suppose first that \rho is an algebra homomorphism from \mathcal{C}(G) to \mathcal{L}(V). Then, for any g,h \in G we have

U^\rho(gh) = \rho(E_{gh}) = \rho(E_gE_h) = \rho(E_g)\rho(E_h) = U^\rho(g)U^\rho(h),

and moreover for e \in G the group unit

U^\rho(e) =\rho(E_e) = I,

where I \in \mathcal{L}(V) is the identity operator on V. Consequently, we have

U^\rho(g) U^\rho(g^{-1}) = U^\rho(gg^{-1}) = U^\rho(e)=I,

which shows that U^\rho(g) is invertible in \mathcal{L}(V) with inverse U^\rho(g)^{-1}=U^\rho(g^{-1}). Thus U^\rho is a group homomorphism from G into the group of invertible elements in \mathcal{L}(V). It remains to show that the codomain of U^\rho is in fact the smaller group U(\mathcal{L}(V)). This is verified by

U^\rho(g)^*=\rho(E_g)^*=\rho(E_g^*)=\rho(E_{g^{-1}}) = U^\rho(g^{-1})=U^\rho(g)^{-1}.

Conversely, suppose we start from the assumption that U^\rho is a group homomorphism from G into the unitary group U(\mathcal{L}(V)). Note that \rho is a linear map by hypothesis; what we have to check is that it respects multiplication and conjugation in the convolution algebra \mathcal{C}(G), and maps E_e to I. It is sufficient to check this for the elementary basis, and we have

\rho(E_gE_h) = \rho(E_{gh})=U^\rho(gh)=U^\rho(gh) = U^\rho(g)U^\rho(h)=\rho(E_g)\rho(E_h),

and also

\rho(E_e) = U^\rho(e)=I,

where in both calculations we are using the hypothesis that U^\rho is a group homomorphism. Finally,

\rho(E_g^*)=\rho(E_{g^{-1}}) = U^\rho(g^{-1})=U^\rho(g)^{-1}=U^\rho(g)^*=\rho(E_g)^*,

where we used our hypothesis that the codomain of the group homomorphism U^\rho is the unitary group U(\mathcal{L}(V)).

-QED

You will recognize that Theorem 17.1 was proved previously in the special case where V is a one-dimensional Hilbert space, so that \mathcal{L}(V) \simeq \mathbb{C} is the algebra of complex numbers and U(\mathcal{L}(V)) is the unit circle in \mathbb{C}. In this one-dimensional setting, we gathered enough information to construct the Fourier transform on \mathcal{C}(G) for G abelian. For G nonabelian, our path is the same except that we have to consider homomorphisms from G into the unitary group of higher-dimensional Hilbert spaces V.

Definition 17.1. A unitary representation of G is a pair (V,U) consisting of a Hilbert space V together with a group homomorphism

U\colon G \longrightarrow U(\mathcal{L}(V)),

where U(\mathcal{L}(V)) is the unitary group of the algebra of all linear operators on V.

By Theorem 17.1, there is a bijective correspondence between linear representations of the convolution algebra \mathcal{C}(G) and unitary representations of the underlying group G. Therefore, in the case of convolution algebras, we can and do choose to work with unitary representations of G. All the basic notions we need about these object essentially coincide with those already developed in the more general setting of linear representations of algebras.

Definition 17.2. If (V_1,U_1) and (V_2,U_2) are unitary representations of G. A homomorphism of unitary representations is a linear map T \colon V_1 \to V_2 such that

T \circ U_1(g) =U_2(g) \circ T, \quad \text{for all }g \in G.

The set of all such linear maps is denoted \mathrm{Hom}_G(V_1,V_2). If T as above is a vector space isomorphism, then we say that (V_1,U_1) and (V_2,U_2) are isomorphic unitary representations of the group G.

Theorem 17.2. Suppose that (V_1,U_1) and (V_2,U_2) are isomorphic unitary representations of G, i.e. there exists a linear isomorphism in \mathrm{Hom}_G(V_1,V_2). Then, a stronger statement holds: there exists a unitary isomorphism T \in \mathrm{Hom}_G(V_1,V_2), i.e. an isometric isomorphism. To be completely explicit, there is T \in \mathrm{Hom}_G(V_1,V_2) such that

\langle Tv_1,Tw_1\rangle = \langle v_1,w_1\rangle, \quad \text{for all } v_1,w_1 \in V_1.

Proof: The proof is an (interesting) exercise in polar decomposition, or singular value decomposition if you prefer. -QED

To further connect up with linear representations of algebras, let (V,\rho) be a linear representation of \mathcal{C}(G), and let (V,U^\rho) be the corresponding unitary representation of G, as in Theorem 17.1

Theorem 17.3. The linear representation (V,\rho) is irreducible if and only if the unitary representation (V,U^\rho) is irreducible.

Problem 17.1. Prove Theorem 17.3. (This is straightforward and is just to get you accustomed to the definitions in play).

Theorem 17.4 (Schur’s Lemma) Let (V_1,U_1) and (V_2,U_2) be unitary representations of G. If (V_1,U_1) is irreducible, every homomorphism T \in \mathrm{Hom}_G(V_1,V_2) is injective. If (V_2,U_2) is irreducible, every T \in \mathrm{Hom}_G(V_1,V_2) is surjective.

Proof: Special case of Schur’s lemma for linear representations of algebras. -QED

Theorem 17.5. (Maschke’s Theorem) Let (V,U) be a unitary representation of G. Then, there exists a linear partition

V = \bigoplus\limits_{W \in \mathsf{W}} W

of V whose such that, for every W \in \mathsf{W}, the unitary representation (W,U_W) obtained by restricting each U(g) to W is irreducible. That is, every unitary representation decomposes into a direct sum of irreducible unitary representations.

Proof: Special case of Maschke’s theorem for linear representations of algebras.

-QED

Our goal is to classify all unitary representations of a finite group G, up to isomorphism. By Maschke’s theorem, it suffices to classify the irreducible ones (“irreps”). If G is abelian, this is the classification of homomorphisms G \to U(\mathbb{C}), which we successfully achieved in our development of the Fourier transform; we called the parameterization set \Lambda, and it is an explicit set of tuples of roots of unity corresponding to the decomposition of G into a product of cyclic groups.

Now suppose that G is a nonabelian group, and let \Lambda be a set parameterizing isomorphism classes of irreducible unitary representations of \group{G}. We know absolutely nothing about \Lambda yet, but still we can get going. For each \lambda \in \Lambda, let (V^\lambda,U^\lambda) be a representative of the corresponding isomorphism class of irreducible unitary representations. Let X^\lambda \subset V^\lambda be an orthonormal basis, and for each x,y \in X^\lambda

\mu_{xy}^\lambda \colon G \longrightarrow \mathbb{C}

be the function on G defined by

\mu_{xy}^\lambda(g) = \langle x,U^\lambda(g)y\rangle.

One of the two main theorems next week is the following.

Theorem 17.6. The set \{\mu_{xy}^\lambda \colon \lambda \in \Lambda,\ x,y \in X^\lambda\} is an orthogonal basis of the convolution algebra \mathcal{C}(G).

This theorem generalizes the construction of the Fourier basis in \mathcal{C}(G) in the case where G is abelian. In fact, it is not too difficult to prove – the hardest part is to make sure you absorb and understand the definitions needed to state it. If you can do this, the logic of the argument is not difficult to follow.

Math 202B: Lecture 16

Let \mathcal{A} be an algebra.

Definition 16.1. A linear representation of \mathcal{A} is a pair (V,\rho) consisting of a Hilbert space V together with an algebra homomorphism \rho \colon \mathcal{A} \to \mathcal{L}(V).

Note that Definition 16.1 does not stipulate that \rho be an algebra isomorphism, so \mathcal{B}=\rho(\mathcal{A}) is just a homomorphic image of \mathcal{A} inside \mathcal{L}(V). In that sense, the term “representation” is a bit misleading. When \rho is an algebra isomorphism, the representation (V,\rho) is said to be faithful.

We saw a special case of Definition 16.1, where we considered the special case of a subalgebra \mathcal{A}_0 of \mathcal{L}(V_0) for some given Hilbert space V_0. Then, Burnside’s theorem says that either \mathcal{A}_0 = \mathcal{L}(V_0) or there is a proper non-trivial subspace V \leq V_0 invariant under all operators A \in \mathcal{A}_0. In this situation, (V,\rho) is a linear representation of \mathcal{A}_0, where \rho \colon \mathcal{A}_0 \to \mathcal{L}(V) is defined by declaring \rho(A) to be the restriction of A to the invariant subspace V \leq V_0. In the case where \mathcal{A}_0=\mathcal{L}(V_0), the pair (V_0,\rho_0) is a linear representation of \mathcal{A}_0 with \rho_0(A)=A the identity homomorphism; this is called the tautological representation or defining representation of \mathcal{A}_0.

Definition 16.2. If (V,\rho) and (W,\sigma) are representations of \mathcal{A}, a linear transformation T \colon V \to W is said to be a homomorphism of representations (or an intertwining transformation) if

T \circ \rho(A) = \sigma(A) \circ T, \quad\text{for all} A \in \mathcal{A}.

The set of all such linear transformations is denoted \mathrm{Hom}_\mathcal{A}(V,W).

Problem 16.1. Prove that \mathrm{Hom}_\mathcal{A}(V,W) is a vector subspace of the space \mathrm{Hom}(V,W) of all linear transformations V \to W.

Definition 16.3. A linear isomorphism T \in \mathrm{Hom}_\mathcal{A}(V,W) is said to be an isomorphism of representations. When such an intertwining linear isomorphism exists, we say that (V,\rho) and (W,\sigma) are isomorphic representation of \mathcal{A}.

We also encountered Definition 16.3 in a special case in Lecture 15, when we discussed the classification of subalgebras of \mathcal{L}(V).

Problem 16.2. Let (V,\rho) and (W,\sigma) be linear representations of \mathcal{A}. Prove that they are isomorphic representations if and only if there exists a basis \{v_1,\dots,v_n\} of V and a basis \{w_1,\dots,w_n\} of W such that the matrix of \rho(A) in the v-basis of V equals the matrix of \sigma(A) in the w-basis of W, for every algebra element A \in \mathcal{A}.

Let us fix a linear representation (V,\rho) of \mathcal{A} and consider its image \mathcal{B}:=\rho(\mathcal{A})=\{\rho(A) \colon A \in \mathcal{A}\}, which is a subalgebra of \mathcal{L}(V). Recall from Week One (or maybe Week Two) that the centralizer \mathcal{C}=Z(\mathcal{B},\mathcal{L}(V)) is the subalgebra consisting of all operators C \in \mathcal{L}(V) that commute with every operator B \in \mathcal{B}.

Proposition 16.1. We have \mathcal{C} = \mathrm{Hom}_\mathcal{A}(V,V).

Proof: We have C \in \mathcal{C} if and only if C\rho(A)=\rho(A)C, which is exactly the condition for C to be an element of \mathrm{Hom}_\mathcal{A}(V,V).

-QED

Definition 16.4. A linear representation (V,\rho) of \mathcal{A} is said to be irreducible if the only \mathcal{B}-invariant subspaces of V are the zero space and the total space (where once again \mathcal{B} is the image of \mathcal{A} in \mathcal{L}(V) under \rho).

Problem 16.3. Prove that (V,\rho) is irreducible if and only if there does not exist a basis \{v_1,\dots,v_n\} of V such that the matrix of every \rho(A) in the v-basis has block diagonal form, with blocks of positive size.

Proposition 16.2. If (V,\rho) is an irreducible representation of \mathcal{A}, then \mathcal{B}=\mathcal{L}(V).

Proof: This is Burnside’s theorem from Lecture 15: the only subalgebra of \mathcal{L}(V) with exactly two invariant subspaces is \mathcal{B}=\mathcal{L}(V).

-QED

Continuing on with (V,\rho) an irreducible representation of \mathcal{A}, and maintaining the notation \mathcal{C}=Z(\mathcal{B},\mathcal{L}(V))=\mathrm{Hom}_\mathcal{A}(V,V), we have the following.

Corollary 16.3. \mathcal{C}=\mathbb{C}I.

Proof: Since \mathcal{C} is the centralizer of \mathcal{B}=\mathcal{L}(V), the statement is equivalent to the fact that the center of \mathcal{L}(V) consists of scalar multiples of the identity, which we proved in Lecture 15.

-QED

We can refine the above corollary to make the following statement about intertwining maps between two possibly distince representations (V,\rho) and (W,\sigma) of \mathcal{A}, at least one of which is irreducible.

Theorem 16.4. (Schur’s Lemma) If (V,\rho) is irreducible, then every nonzero homomorphism T \in \mathrm{Hom}_\mathcal{A}(V,W) is injective. If (W,\sigma) is irreducible, then every nonzero homomorphism T \in \mathrm{Hom}_\mathcal{A}(V,W) is surjective. If both (V,\rho) and (W,\sigma) are irreducible, then every nonzero T \in \mathrm{Hom}_\mathcal{A}(V,W) is an isomorphism of representations.

Proof: Suppose (V,\rho) is irreducible, and consider the kernel of any T \in \mathrm{Hom}_\mathcal{A}(V,W). We claim that \mathrm{Ker}(T) is a \rho(\mathcal{A})-invariant subspace of V. Indeed, if v \in \mathrm{Ker}(T), then for any A \in \mathcal{A} we have

T\rho(A)v=\sigma(A)Tv = \sigma(A)0_W = 0_W,

which shows that \rho(A)v is again in \mathrm{Ker}(T). Since (V,\rho) is irreducible we have either \mathrm{Ker}(T)=\{0_V\} or \mathrm{Ker}(T)=V, and since T is not the zero map we must have T injective.

Now suppose (W,\sigma) is irreducible. Then, we claim that the image of V under T\in \mathrm{Hom}_\mathcal{A}(V,W) is a \sigma(\mathcal{A})-invariant subspace of W. Indeed, suppose w \in \mathrm{Im}(T), i.e. w=Tv for v \in V. Then, for any A \in \mathcal{A} we have that

\sigma(A)w=\sigma(A)Tv=T(\rho(A)v),

showing that \sigma(A)w remains in the image of T. Since T is nonzero and (W,\sigma) is irreducible, we have \mathrm{Im}(T)=W as claimed.

The final statement follows from the two arguments above: any nonzero homomorphism of irreducible representations must be an isomorphism. Equivalently, there does not exist a nonzero homomorphism between two non-isomorphic irreducible representations.

-QED

Math 202B: Lecture 15

Let V be a Hilbert space of dimension 1 < V < \infty and \mathcal{L}(V) be the algebra of all linear operators A \colon V \to V. As we saw in Lecture 14, the trace on \mathcal{L}(V) is, up to scaling, the unique linear functional \tau \colon \mathcal{L}(V) \to \mathbb{C} such that \tau(AB)=\tau(BA). Thus, the only linear functional satisfying \tau(I)=1 and \tau(AB)=\tau(BA) is

\tau(A) = \frac{1}{\dim V}\mathrm{Tr}(A), \quad A \in \mathcal{L}(V).

This fact allows us to prove a fact claimed early on in the course.

Problem 15.1. Prove that there does not exist an algebra homomorphism \chi \colon \mathcal{L}(V) \to \mathbb{C}.

The fact that \mathcal{L}(V) admits not even a single homomorphism to \mathbb{C} indicates that this is an algebra with the highest degree of noncommutativity. According to our definitions in Week One, the degree of noncommutativity of an algebra is formalized as the dimension of its center. We now prove that the center of \mathcal{L}(V) is indeed one-dimensional. The key notion is the following.

Definition 15.1. Given a subalgebra \mathcal{A} of \mathcal{L}(V), a subspace W of V is said to be \mathcal{A}–invariant if for every A \in \mathcal{A} and every w \in W we have Aw \in W.

Every subalgebra \mathcal{A} of \mathcal{L}(V) has at least two invariant subspaces, namely the zero space \{0_V\} and the total space V. An important theorem of Burnside characterizes subalgebras of \mathcal{L}(V) for which this minimum is achieved.

Theorem 15.1. (Burnside) A subalgebra \mathcal{A} of \mathcal{L}(V) has exactly two invariant subspaces if and only if \mathcal{A}=\mathcal{L}(V).

Proof: An elementary proof of Burnside’s theorem by induction in the dimension of V can be found here.

-QED

Corollary 15.2. A proper subalgebra \mathcal{A} of \mathcal{L}(V) has at least four invariant subspaces.

Proof: The zero space and the total space are \mathcal{A}-invariant, and since \mathcal{A} \neq \mathcal{L}(V) there is a third \mathcal{A}-invariant subspace \{0_V\} < W < \mathcal{L}(V), by Burnside’s theorem. The orthogonal complement W^\perp of W is again a nonzero proper subspace of V, and we claim it is \mathcal{A}-invariant. Indeed, let x \in W^\perp and A \in \mathcal{A}. Then, for any w \in W we have

\langle w,Ax \rangle = \langle A^*w,x \rangle = 0,

where we used the fact that W is \mathcal{A}-invariant. This shows Ax \in W^\perp, i.e. W^\perp is \mathcal{A}-invariant.

– QED

Using Burnside’s theorem we can characterize the center of \mathcal{L}(V).

Theorem 15.2. The center of \mathcal{L}(V) is the set \mathbb{C}I of scalar operators.

Proof: Let Z \in \mathcal{L}(V) be a central element, and let \zeta \in \mathbb{C} be an eigenvalue of Z. Then, the \zeta-eigenspace of Z is an \mathcal{L}(V)-invariant subspace, since if v \in V is an eigenvector of Z corresponding to \zeta and A \in \mathcal{L}(V) is any operator we have

ZAv = AZv =\lambda Av,

which shows that Av is again in the \zeta-eigenspace of Z. Since this eigenspace is not the zero space, Burnside’s theorem forces it to be all of V, i.e. Z=\zeta I.

-QED

We have succeeded in describing the center of \mathcal{L}(V) explicitly, and ideally we would like to have an explicit description of all subalgebras of \mathcal{L}(V). We were able to do this for function algebras \mathcal{F}(X), where subalgebras are indexed by partitions \mathsf{P} of the finite set X. The classification is very simple: the subalgebra \mathcal{A}(\mathsf{P}) of a given partition \mathsf{P} of X is the set of all functions on X which are constant on the blocks of \mathsf{P}. A nice feature of this classification is that it shows every subalgebra of \mathcal{F}(X) is itself isomorphic to a function algebra, because \mathcal{A}(\mathsf{P}) is isomorphic to \mathcal{F}(\mathsf{P}).

We can attempt to classify subalgebras of \mathcal{L}(V) by adapting our approach to the classification of subalgebras of \mathcal{F}(X). The first step is to formulate the vector space version of partitions of a set.

Definition 15.2. A linear partition of V is a set \mathsf{W} of nonzero subspaces of V whose union spans V. We refer to the subspaces W \in \mathsf{W} as the blocks of \mathsf{W}.

Given a linear partition \mathsf{W} of V, it is not useful to consider the set of linear operators in \mathcal{L}(V) constant on the blocks of \mathsf{W}. Indeed, there is only one such operator, namely the zero operator. Rather, we define \mathcal{A}(\mathsf{W}) to be the set of all operators A \in \mathcal{L}(V) such that every block W \in \mathsf{W} is A-invariant.

Problem 15.2. Prove that A(\mathsf{W}) is a subalgebra of \mathcal{L}(V).

Here our construction starts to diverge from what we saw in the setting of function algebras: unless \mathsf{W}=\{V\} is the linear partition of V consisting of a single block, in which case \mathcal{A}(\mathsf{W})=\mathcal{L}(V), the subalgebra \mathcal{A}(\mathsf{W}) is definitely not isomorphic to the algebra of all linear operators on a Hilbert space. Indeed, consider the case where \mathsf{W}=\{W_1,W_2\} is a linear partition of V with two blocks. Let e_1,\dots,e_m be an ordered basis of W_1, and let f_1,\dots,f_n be an ordered basis of W_2. Then e_1,\dots,e_m,f_1,\dots,f_n is an ordered basis of V, and \mathcal{A}(\mathsf{W}) consists of all operators in A\in \mathcal{L}(V) whose matrix relative to this ordered basis has the form

[A]_{(e_1,\dots,e_m,f_1,\dots,f_n)}=\begin{bmatrix} M_1 & {} \\ {} & M_2 \end{bmatrix},

with M_1 \in \mathbb{C}^{m \times m} and M_2 \in \mathbb{C}^{n\times n}. Thus, \mathcal{A}(\{W_1,W_2\}) is not isomorphic to a linear algebra, but rather to a direct sum of linear algebras,

\mathcal{A}(\{W_1,W_2\}) = \mathcal{L}(W_1) \oplus \mathcal{L}(W_2).

The question now is whether the above can reversed: given a subalgebra \mathcal{A} of \mathcal{L}(V), does there exist a linear partition \mathsf{W} such that \mathcal{A}=\mathcal{A}(\mathsf{W})? If \mathcal{A}=\mathcal{L}(V), the answer is clearly “yes” since we have \mathcal{L}(V) = \mathcal{A}(\{V\}). Thus we consider the case where \mathcal{A} is a proper subalgebra of \mathcal{L}(V). Our starting point is Burnside’s theorem, which guarantees the existence of an \mathcal{A}-invariant subspace \{0_V\} < W < V. Together with the corollary to Burnside’s theorem, we thus have a linear partition \mathsf{W} = \{W,W^\perp\} of V whose blocks are \mathcal{A}-invariant. We now ask whether W has a proper non-trivial subspace invariant under \mathcal{A}. If it does we can split it into the orthogonal direct sum of two smaller \mathcal{A}-invariant subspaces; if not W is said to be \mathcal{A}-irreducible. Continuing this process, for both W and W^\perp, we get the following.

Theorem 15.4. (Maschke’s theorem) There exists a linear partition \mathsf{W} of V whose blocks are \mathcal{A}-invariant and \mathcal{A}-irreducible subspaces.

Now consider the linear partition \mathsf{W} of V into \mathcal{A}-invariant, \mathcal{A}-irreducible subspaces we have constructed. It is tempting to hope that \mathcal{A} = \mathcal{A}(\mathsf{W}). This is almost correct. Let us say that two blocks W,W' \in \mathsf{W} are \mathcal{A}–equivalent if it is possible to choose a basis in W and a basis in W' such that the matrix of every A \in \mathcal{A}, viewed as an operator on W, coincides with the matrix of A viewed as an operator on W'. This is an equivalence relation on the blocks of \mathsf{W}. Let \Lambda be a set parameterizing the corresponding equivalence classes of blocks, and for each \lambda \in \Lambda choose a representative V^\lambda of the corresponding equivalence class.

Theorem 15.5. (Classification Theorem for subalgebras of linear algebras) The subalgebra \mathcal{A} is isomorphic to \bigoplus_{\lambda \in \Lambda} \mathcal{L}(V^\lambda).

We are not going to prove the above classification in full right now – in the coming lectures we will prove a special case of it for homomorphic images of convolution algebras of finite groups. If time permits, we will prove the full theorem later.

Math 202B: Lecture 14

When things get confusing, it helps to come back to basics. Let V be a Hilbert space of dimension 1< n <\infty, and let X \subset V be an orthonormal basis.

Definition 14.1. The matrix elements relative to X are the linear functionals

M_{xy} \colon \mathcal{L}(V) \longrightarrow \mathbb{C}, \quad (x,y) \in X \times X

defined by

M_{xy}(A) =\langle x, Ay \rangle, \quad A \in \mathcal{L}(V).

Proposition 14.1. Two operators A,B \in \mathcal{L}(V) are equal if and only if M_{xy}(A) = M_{xy}(B) for all x,y \in X.

Problem 14.1. Prove Proposition 14.1.

Definition 14.2. The trace on \mathcal{L}(V) relative to X is the linear functional defined by

\mathrm{Tr}_X = \sum\limits_{x \in X} M_{xx}.

Thus,

\mathrm{Tr}_X(A) = \sum\limits_{x \in X} \langle x,Ax\rangle, \quad A \in \mathcal{L}(V).

Proposition 14.2. We have

  • \mathrm{Tr}_X(I) = \dim V;
  • \mathrm{Tr}_X(AB)=\mathrm{Tr}_X(BA);
  • \mathrm{Tr}_X(A^*A) \geq 0 with equality if and only if A=0.

Proof: First, we have

\mathrm{Tr}_X(I) = \sum\limits_{x \in X} \langle x,Ix\rangle = \sum\limits_{x\in X}1 = |X|=\dim V.

Second, we have

\mathrm{Tr}_X(AB)=\sum\limits_{x\in X} \langle x,ABx\rangle=\sum\limits_{x \in X} \langle x,A\sum\limits_{y \in X} \langle y,Bx\rangle y\rangle=\sum\limits_{x \in X}\sum\limits_{y\in X} \langle x,Ay\rangle \langle y,Bx \rangle,

and changing order of summation this is

\sum\limits_{y\in X}\sum\limits_{x \in X}\langle y,Bx \rangle\langle x,Ay\rangle=\sum\limits_{y \in X} \langle y,B\sum\limits_{x \in X} \langle x,Ay\rangle x\rangle=\sum\limits_{y \in X} \langle y,BAy\rangle=\mathrm{Tr}_X(BA).

Third, we have

\mathrm{Tr}_X(A^*A) = \sum\limits_{x \in X} \langle x,A^*Ax\rangle = \sum\limits_{x \in X} \langle Ax,Ax\rangle = \sum\limits_{x\in X} \|Ax\|^2,

which vanishes if and only if every term of the sum vanishes, and this occurs if and only if A maps every vector in the basis X to the zero vector.

-QED

Note that the definition of \mathrm{Tr}_X makes sense even if the basis X is not orthonormal. We claim that Proposition 14.2 remains valid in this case. Proof: define a new scalar product on V in which X is orthonormal.

Proposition 14.3. Let X and Y be two orthonormal bases of V. We have \mathrm{Tr}_X=\mathrm{Tr}_Y.

Proof: For any enumeration x_1,\dots,x_n of X, and any enumeration y_1,\dots,y_n of Y, the linear operator U \colon V \to V defined by Ux_i=y_i is unitary. We have

\mathrm{Tr}_Y(A) = \sum\limits_{i=1}^n \langle y_i,Ay_i\rangle=\sum\limits_{I=1}^n \langle Ux_i,AUx_i\rangle = \mathrm{Tr}_X(U^*AU).

By Proposition 14.2, we have \mathrm{Tr}_X(U^*AU)=\mathrm{Tr}_X(AUU^*)=\mathrm{Tr}_X(A).

-QED

Proposition 14.3 again holds without assuming orthonormality of X,Y, for the same reason as above. We have thus shown that Definition 14.3 gives the same linear functional on \mathcal{L}(V) no matter what basis X is used to define it, and we call this functional the trace on V.

Definition 14.3. The spectrum of A \in \mathcal{L}(V) is the subset of \mathbb{C} defined by

\sigma(A) = \{\alpha \in \mathbb{C} \colon \alpha I - A \text{ not invertible}\}.

The numbers in \sigma(A) are called the eigenvalues of A.

Theorem 14.4. (Fundamental Theorem of Algebra) Every A \in \mathcal{L}(V) has nonempty spectrum \sigma(A). That is, every operator has an eigenvalue.

Problem 14.2. Prove that, for any A \in \mathcal{L}(V), we have

\mathrm{Tr} A = \sum\limits_{\alpha \in A} \alpha.

Note Definition 14.3 makes sense in any algebra \mathcal{A}, though the term “eigenvalue” is only used in the case \mathcal{A}=\mathcal{L}(V).

Problem 14.3. What is the spectrum of a function A \in \mathcal{F}(X)?

One of the most important aspects of the trace on \mathcal{L}(V) is that it gives us a scalar product on \mathcal{L}(V) which interfaces well with its algebra structure.

Definition 14.4. The Frobenius scalar product on \mathcal{L}(V) is defined by \langle A,B \rangle = \mathrm{Tr} A^*B.

The definition of the Frobenius scalar product should be familiar from Math 202A. The associated norm is called the Frobenius norm, \|A\|=\sqrt{\mathrm{Tr A^*A}}.

Problem 14.4. Show that the Frobenius scalar product is indeed a scalar product, and moreover that it interacts with multiplication and conjugation in \mathcal{L}(V) according to \langle AB,C\rangle =\langle B,A^*C\rangle, for all A,B,C \in \mathcal{L}(V).

Definition 14.5. The elementary operators relative to X are the operators E_{xy} \in \mathcal{L}(V) defined by

E_{xy}z = x \langle y,z\rangle, \quad x,y,z \in X.

From the definition, we have

M_{wz}(E_{xy})=\langle w,E_{xy}z\rangle =\langle w,x\langle y,z\rangle\rangle=\langle w,x \rangle \langle y,z \rangle, \quad x,y,z,w \in X.

Problem 14.4. Show that the trace of an elementary operator is \mathrm{Tr} E_{xy}=\langle x,y \rangle.

Proposition 14.4. The elementary operators satisfy

E_{xy}^*=E_{yx} \quad\text{and}\quad E_{wx}E_{yz}=\langle x,y\rangle E_{wz}, \quad w,x,y,z \in X.

Proof: For any w,x,y,z \in X, we have

\langle w,E_{xy}z\rangle = \langle w,x\langle y,z\rangle\rangle=\langle w,x\rangle \langle y,z\rangle

and

\langle E_{yx}w,z\rangle = \langle y\langle x,w\rangle,z\rangle = \langle x,w\rangle \langle y,z \rangle,

so indeed E_{xy}^*=E_{yx}.

Now consider the v,a \in X and consider the matrix element

M_{va}(E_{wx}E_{yz})=\langle v,E_{wx}E_{yz}a\rangle =

-QED

Theorem 14.5. The elementary operators \{E_{xy}\colon (x,y) \in X \times X\} form an orthonormal basis of \mathcal{L}(V) relative to the Frobenius scalar product.

Proof: First we check that \{E_{xy}\colon (x,y) \in X \times X\} is an orthonormal set with respect to the Frobenius scalar product. For any $w,x,y,z \in X$ we have

\langle E_{wx},E_{yz}\rangle = \mathrm{Tr}(E_{wx}^*E_{yz})=\mathrm{Tr}(E_{xw}E_{yz}) = \langle w,y\rangle\mathrm{Tr}(E_{xz})=\langle w,y\rangle \langle x,z\rangle.

Now we check that \{E_{xy}\colon (x,y) \in X \times X\} spans \mathcal{L}(V). Let A \in \mathcal{L}(V) be an arbitrary operator, and write \alpha_{xy} = \langle x,Ay\rangle for its matrix elements relative X. We claim that

A = \sum\limits_{(x,y) \in X\times X} \alpha_{xy}E_{xy}.

Let R be the operator on the right hand side of the above equation. Then, the matrix elements of R are

\langle w,Rz\rangle = \sum\limits_{(x,y) \in X \times X} \alpha_{xy}\langle w,E_{xy}z\rangle \sum\limits_{(x,y) \in X\times X} \alpha_{xy}\langle w,x\rangle \langle y,z\rangle=\alpha_{wz},

which shows that all matrix elements of A and R are equal, so that these operators are the same by Proposition 14.1.

-QED

Unlike the elementary basis \{E_x \colon x \in X\} of the function algebra \mathcal{F}(X), the elementary basis \{E_{xy} \colon x,y \in X\} of the linear algebra \mathcal{L}(V) is not a basis of orthogonal projections — indeed, \mathcal{L}(V) is not a commutative algebra, so it does not admit a basis of orthogonal projections. However, the diagonal elementary operators \{E_{xx} \colon x\in X\} are orthogonal projections spanning a commutative subalgebra of \mathcal{L}(V) isomorphic to \mathcal{F}(X). Indeed, the isomorphism is simply

E_{xx} \mapsto E_x, \quad x \in X.

The trace on \mathcal{L}(V) is unique in the following sense.

Theorem 14.5. If \tau \colon \mathcal{L}(V) \to \mathbb{C} is a linear functional which satisfies \tau(AB)=\tau(BA) for all A,B \in \mathcal{L}(V), then \tau is a scalar multiple of the trace.

Proof: Since \tau is linear, it is uniquely determined by its values on the elementary basis E_{xy} of \mathcal{L}(V).

For distinct x,y \in X, we have

\tau(E_{xy})=\tau(E_{xy}E_{yy}) = \tau(E_{yy}E_{xy})=\tau(0)=0.

For the diagonal elementary operators,

\tau(E_{xx})=\tau(E_{xy}E_{yx})=\tau(E_{yx}E_{xy})=\tau(E_{yy}).

Thus, \tau is constant on the diagonal elementary operators, i.e. there is a constant \gamma such that \tau(E_{xx})=\gamma for all x \in X. Thus for any

A = \sum\limits_{x,y \in X} \alpha_{xy} E_{xy},

we have

\tau(A) = \sum\limits_{x \in X} \tau(E_{xx}) = \gamma \dim V = \gamma \mathrm{Tr} A.

-QED

Math 202B: Lecture 13

Let G be a finite abelian group.

Theorem 13.1. For every A \in \mathcal{C}(G), the Fourier basis of \mathcal{C}(G) is an eigenbasis for the image of A in the left regular representation, and the Fourier transform of A is its spectrum: we have

\mathsf{L}(A)F^\lambda = \widehat{A}(\lambda)F^\lambda, \quad \lambda \in \Lambda.

This result has many applications because many interesting matrices can be realized in the left regular representation of an abelian group, and in any such situation Theorem 13.1 explicitly gives us the eigenvalues and eigenvectors of the matrix in question. Last lecture we considered the case of cyclic group, whose elements become circulant matrices in the regular representation.

Let us give another example, this time taking

\Lambda = \{\alpha=(\alpha_1,\dots,\alpha_r) \colon \alpha_i \in \{0,1\}\}

to be the group of vertices of the r-dimensional unit cube. In this case, the convolution algebra \mathcal{C}(\Lambda) is 2^r-dimensional and the group basis is

E_\alpha=E_{(\alpha_1,\dots,\alpha_r)}

with (\alpha_1,\dots,\alpha_r) ranging over all r-dimensional vectors with entries in \{0,1\}. The convolution product of two basis vectors is

E_\alpha E_\beta = E_{(\alpha_1+\beta_1,\dots,\alpha_r+\beta_r)}

with coordinate addition performed mod 2. In addition to being a group, it is natural to view \Lambda as a graph which reflects the geometry of the r-dimensional cube: we consider two vertices \alpha,\beta of the r-dimensional cube adjacent if they differ in a single coordinate. Equivalently, we have \alpha+\varepsilon_i=\beta for some 1 \leq i \leq r, where \varepsilon_1,\dots,\varepsilon_r \in \Lambda are the vectors

\varepsilon_1=(1,0,0\dots,0),\ \varepsilon_2=(0,1,0,\dots,0),\ \dots,\ \varepsilon_r=(0,0,\dots,0,1).

In more general terminology, we are identifying \Lambda with its Cayley graph corresponding to the generators \{\varepsilon_1,\dots,\varepsilon_r\}.

Adjacency in \Lambda can also be expressed in \mathcal{C}(\Lambda), since \alpha,\beta \in \Lambda are adjacent if and only if

E_\alpha E_{\varepsilon_i}=E_\beta

for some 1 \leq i \leq r. This means that the Wedderburn transform of

A = E_{\varepsilon_1} +E_{\varepsilon_2}+\dots + E_{\varepsilon_r}

is the adjacency operator of the graph \Lambda, and hence that we can calculate the eigenvalues and eigenvectors of \rho(A) explicitly using Theorem 13.1.

For the Boolean cube \Lambda \simeq C_2 \times \dots \times C_2, the character table is

\chi^\lambda_\alpha = \chi^{(\lambda_1,\dots,\lambda_r)}_{(\alpha_1,\dots,\alpha_r)} = (-1)^{\alpha_1\lambda_1} \dots (-1)^{\alpha_r\lambda_r} =(-1)^{\alpha \cdot \lambda},

where \alpha \cdot \lambda = \alpha_1\lambda_1+\dots+\alpha_r\lambda_r is the usual dot product. Thus, the Fourier basis of \mathcal{C}(\Lambda) is

F^\lambda = \sum\limits_{\alpha \in \Lambda} (-1)^{\alpha \cdot \lambda}E_\alpha, \quad \lambda \in \Lambda,

and this is an eigenbasis of \rho(A). Now by direct computation

\rho(A)F^\lambda =AF^\lambda= \left( \sum\limits_{i=1}^n E_{\varepsilon_i}\right)\left(\sum\limits_{\alpha \in \Lambda} (-1)^{\alpha \cdot \lambda} E_\alpha\right)=\sum\limits_{\alpha \in \Lambda}\sum\limits_{i=1}^n (-1)^{\alpha \cdot \lambda}E_{\alpha+\varepsilon_i}=\left(\sum\limits_{i=1}^n(-1)^{\varepsilon_i\cdot\lambda}\right)F^\lambda.

The sum

\sum\limits_{i=1}^r(-1)^{\varepsilon_i\cdot\lambda}

giving the eigenvalue of \rho(A) on F^\lambda can be simplified: every coordinate of \lambda equal to zero contributes +1 and every coordinate of \lambda equal to 1 contributes -1. Thus writing w(\lambda) for the number of entries equal to one in \lambda (this is called the Hamming weight of \lambda in coding theory), we have r-w(\lambda) terms equal to one and w(\lambda) terms equal to minus one, giving

\sum\limits_{i=1}^n(-1)^{\varepsilon_i\cdot\lambda}=r-2w(\lambda).

From this we can conclude that the eigenvalues of the r-dimensional hypercube are the integers

r-0,\ r-2,\ r-4,\ \dots, r-2r,

and that the multiplicity of the eigenvalue r-2i is equal to the number of \{0,1\}-vectors with Hamming weight i, which is {r \choose i}.

Math 202B: Lecture 12

Let \mathcal{C}(G) be the convolution algebra of a finite group G. Then, the left regular representation

\mathsf{L} \colon \mathcal{C}(G) \longrightarrow \mathcal{L}(V)

is an injective algebra homomorphism from the convolution algebra \mathcal{C}(G) into the the linear algebra of \mathcal{C}(G) viewed as a Hilbert space. That is, the regular representation converts convolution of functions into matrix multiplication.

If G is abelian, then the Fourier transform

\widehat{\cdot} \colon \mathcal{C}(G) \longrightarrow \mathcal{F}(\Lambda)

is an algebra isomorphism between the convolution algebra \mathcal{C}(G) and the function algebra of the dual group \Lambda of G. That is, the Fourier transform converts convolution into pointwise multiplication.

Let us reconcile these two points of view on the convolution algebra of a finite abelian group. Let A \in mathcal{C}(G) be any function on G. Let

A = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)F^\lambda

be the Fourier expansion of A. Taking the left regular representation, we have

\mathsf{L}(A) = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)P^\lambda,

where P^\lambda=\mathsf{L}(F^\lambda) is the linear operator on the Hilbert space V=\mathcal{C}(G) given by the image of the Fourier basis vector F^\lambda under the injective algebra homomorphism \mathsf{L}. In other words,

P^\lambda, \quad \lambda \in \Lambda,

is a basis of orthogonal projections for the image of \mathcal{C}(G) in \mathcal{L}(V) under \mathsf{L}, and

\mathsf{L}(A) = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)P^\lambda

is the spectral decomposition of the operator \mathsf{L}(A). This says that the operators \mathsf{L}(A) making up the image of the commutative algebra \mathcal{C}(G) in \mathcal{L}(V) are simultaneously diagonalizable, which we already knew from our characterization of commutative subalgebras of \mathcal{L}(V), and moreover that calculating the eigenvalues of the operator \mathsf{L}(A) \in \mathcal{L}(V) is the same thing as calculating the Fourier transform of the function A \in \mathcal{C}(G).

Problem 12.3. A circulant matrix is a complex square matrix of the form

C=\begin{bmatrix} c_0 & c_1 & \dots & c_{n-1} \\ c_{n-1} & c_0 & \dots & c_{n-2} \\ \vdots & \vdots & {} & \vdots \\ c_1 & c_2 & \dots & c_0 \end{bmatrix}.

Determine the eigenvalues and eigenvectors of C. Hint: show that every n \times n circulant matrix is the image of a function on the cyclic group of order n in the regular representation.

As an extension to the above problem, you may consider how the uncertainty principle for the Fourier transform relates the sparsity of a circulant matrix to the dimension of its kernel. (Note to self: perhaps do this in Lecture in the next iteration of Math 202B).

Math 202B: Lecture 10

Let

G = C_{n_1} \times \dots \times C_{n_r}

be an abelian group, and let

\Lambda = \mathbb{Z}_{n_1} \times \dots \times \mathbb{Z}_{n_r}

be its dual group. In Lecture 8 we showed that the convolution algebra \mathcal{C}(G) admits a basis \{F^\lambda \colon \lambda \in \Lambda\} consisting of orthogonal projections, called its Fourier basis. Thus any function A \in \mathcal{C}(G) can be expanded in the group basis,

A = \sum\limits_{g \in G} A(g) E_g,

and in the Fourier basis

A = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda) F^\lambda.

The coefficients of the Fourier expansion of A \in \mathcal{C}(G) give a function \widehat{A} \in \mathcal{F}(\Lambda) defined by \lambda \mapsto \widehat{A}(\lambda).

Definition 9.1. The Fourier transform on \mathcal{C}(G) is the map \mathcal{C}(G) \to \mathcal{F}(\Lambda) defined by A \mapsto \widehat{A}.

In Lecture 3, we proved that any algebra which admits a basis of orthogonal projections is isomorphic to a function algebra, and the proof of this general fact given there applies here in particular, allowing us to conclude that the Fourier transform A \mapsto \widehat{A} is an algebra isomorphism \mathcal{C}(G) \to \mathcal{F}(G). Thus, we have

\widehat{A*B} = \widehat{A} \widehat{B},

where on the left hand side

[A*B](g) = \sum\limits_{h \in G} A(gh^{-1})B(h), \quad g \in G,

is the convolution product of two functions in \mathcal{C}(G), and on the right hand side

[\widehat{A}\widehat{B}](\lambda) = \widehat{A}(\lambda)\widehat{B}(\lambda), \quad \lambda \in \Lambda

is the pointwise product of two functions in \mathcal{F}(\Lambda).

By definition, the elements of the Fourier basis of \mathcal{C}(G) are

F^\lambda = \frac{1}{|G|} \sum\limits_{g \in G} \chi^\lambda(g) E_g,

where \{\chi^\lambda \colon \lambda \in \Lambda\} is the character basis of \mathcal{C}(G), which consists of all group homomorphisms G \to U(\mathbb{C}).

Theorem 9.1. For any A \in \mathcal{C}(G), we have

\widehat{A}(\lambda) = \frac{1}{|G|}\langle \chi^\lambda,A\rangle,

where the right hand side is an \ell^2-scalar product in \mathcal{C}(G).

Proof: Taking a slightly different normalization of the Fourier basis, we get an orthonormal basis of \mathcal{C}(G) consisting of the functions

E^\lambda = \sqrt{|G|}F^\lambda = \frac{1}{\sqrt{|G|}}\chi^\lambda, \quad \lambda \in \Lambda.

Because this basis is orthonormal, we have

A = \sum\limits_{\lambda \in \Lambda} \langle E^\lambda,A\rangle E^\lambda,

for any function A \in \mathcal{C}(G), and comparing with the Fourier expansion of A this gives

\widehat{A}(\lambda) = \frac{1}{\sqrt{|G|}}\langle E^\lambda,A\rangle E^\lambda=\frac{1}{|G|}\langle \chi^\lambda,A\rangle.

-QED

Theorem 9.1 is called the Fourier inversion formula, because

\widehat{A}(\lambda) = \frac{1}{|G|}\langle \chi^\lambda,A\rangle = \frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)}A(g)

recovers the values of the transformed function \widehat{A} \in \mathcal{F}(\Lambda) in terms of the values of the original function A \in \mathcal{C}(G). As an example, let us calculate the Fourier transform of the elementary function E_h \in \mathcal{C}(G) corresponding to a given group element h \in G. From the inversion formula, the Fourier transform of the indicator function of h is

\widehat{E_h}(\lambda) = \frac{1}{|G|}\langle \chi^\lambda,E_h\rangle = \frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)}E_h(g) = \frac{1}{|G|} \overline{\chi^\lambda(h)}.

Equivalently, we have

\widehat{E_h} = \frac{1}{|G|}\sum\limits_{\lambda \in \Lambda} \overline{\chi^\lambda(g)}E_\lambda,

where \{E_\lambda \colon \lambda \in \Lambda\} is the elementary basis of \mathcal{F}(\Lambda) consisting of indicator functions E_\lambda(\mu) = \delta_{\lambda\mu}.

Theorem 9.2. (Plancherel and Parseval formulas) For any A,B \in \mathcal{C}(G), we have

\langle \widehat{A},\widehat{B}\rangle=|G|\langle A,B\rangle

and

\|\widehat{A}\|_2 = \sqrt{|G|}\|A\|_2.

Proof: From the definition of the Fourier transform, we have

\langle A,B \rangle = \left\langle \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)F^\lambda,\sum\limits_{\mu \in \Lambda} \widehat{B}(\mu)F^\mu\right\rangle=\sum\limits_{\lambda \in \Lambda}\sum\limits_{\mu \in \Lambda}\overline{\widehat{A}(\lambda)}\widehat{B}(\mu)\langle F^\lambda,F^\mu\rangle=\sum\limits_{\lambda \in \Lambda}\overline{\widehat{A}(\lambda)}\widehat{B}(\lambda)\frac{1}{|G|}.

This is the Plancherel formula, and the Parseval formula is the special case where A=B, together with the definition of the norm associated to a scalar product.

-QED

The \ell^2-norm on functions A \colon G\to \mathbb{C} the p=2 case of the \ell^p-norm, which for any p>0 is defined by

\|A\|_p =\left(\sum\limits_{g\in G}|A(g)|^p\right)^{\frac{1}{p}}.

Also familiar from analysis is the case p=\infty, which is defined by

\|A\|_\infty = \max\limits_{g \in g} |A(g)|.

The intuition behind this definition is that, as p \to \infty the sum \|A\|_p^p is approximately equal to its largest term, and this concentration becomes exact in the p=\infty limit. This intuition leads to the following inequality.

Proposition 11.3. For any p>0 and A \colon G\to \mathbb{C}, we have

\|A\|_p^p \leq \|A\|^p_\infty |\mathrm{supp}(A)|,

where by definition the support of A is the cardinality of the complement of its vanishing set,

\mathrm{supp}(A) = \{g\in G\colon A(g) \neq 0\}.

Problem 11.2. Prove Proposition 11.3.

We now prove a discrete version of the famous uncertainty principle, whose origins are in quantum mechanics; in Fourier analysis, this principle effectively says that if the support of a function is small than the support of its Fourier transform is large.

Theorem 11.4. (Uncertainty principle) For any nonzero function A \in \mathcal{C}(G), we have

|\mathrm{supp}(A)||\mathrm{supp}(\widehat{A}| \geq |\Lambda|.

Proof: Start from the Fourier expansion of A, which in terms of the character basis \{\chi^\lambda \colon \lambda \in \Lambda\} of \mathcal{C}(G) is

A = \frac{1}{|G|} \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)\chi^\lambda.

For any g \in G, we have

|A(g)| =\left|\frac{1}{|G|} \sum\limits_{\lambda \in \Lambda} \widehat{A}\chi^\lambda(g)\right| \leq \frac{1}{|G|}\sum\limits_{\lambda \in \Lambda} |\widehat{A}(\lambda)| =\frac{1}{|G|}\sum\limits_{\lambda \in \mathrm{supp}(\widehat{A})} |\widehat{A}(\lambda)|,

where we used the triangle inequality and the fact that |\chi^\lambda(g)|=1. Since g\in G was arbitrary, the above shows that

\|A\|_\infty \leq \frac{1}{|G|}\sum\limits_{\lambda \in \mathrm{supp}(\widehat{A})} |\widehat{A}(\lambda)|.

Squaring both sides, we thus have

\|A\|_\infty^2 \leq \frac{1}{|G|^2}\left(\sum\limits_{\lambda \in \Lambda} \delta_{\lambda \in \mathrm{supp}(A)} |\widehat{A}(\lambda)|\right)^2\leq \frac{1}{|G|^2} |\mathrm{supp}(\widehat{A})| \|\widehat{A}\|_2^2=\frac{1}{|G|} |\mathrm{supp}(\widehat{A})| \|A\|_2^2\leq \frac{1}{|G|} |\mathrm{supp}(\widehat{A})| \|A\|_\infty^2|\mathrm{supp}(A)|,

where we applied the Cauchy-Schwarz inequality, the Parseval formula, and Proposition 11.3. Since \|A\|_\infty^2 \neq 0, it can be cancelled from both sides of the above inequality, leaving

1 \leq \frac{1}{|\Lambda|} |\mathrm{supp}(\widehat{A})||\mathrm{supp}(A)|.

-QED

Problem 11.3. Prove that the uncertainty principle is sharp by exhibiting a function A \colon G\to \mathbb{C} for which equality holds in Theorem 11.4.

Math 202B: Lecture 9

The convolution and class algebras \mathcal{C}(G) and \mathcal{Z}(G) of a group G are apparently quite different from the other types of algebras we have seen, namely function algebras \mathcal{F}(X) and linear algebras \mathcal{L}(V). However, they are in fact closely related.

Theorem 9.1. The convolution algebra \mathcal{C}(G) is isomorphic to a subalgebra of a linear algebra.

Proof: Since \mathcal{C}(G) has a natural scalar product, the \ell^2-scalar product

\langle A,B \rangle = \sum\limits_{g \in G} \overline{A(g)}B(g),

it is a Hilbert space as well as an algebra. For each A \in \mathcal{C}(G), consider the function \mathsf{L}(A) \colon \mathcal{C}(G) \to \mathcal{C}(G) defined by

\mathsf{L}(A)B = AB, \quad B \in \mathcal{C}(G).

Observe that \mathsf{L}(A) is a linear operator on \mathcal{C}(G), i.e.

\mathsf{L}(A)(\beta B + \gamma C) = A(\beta B + \gamma C) = \beta AB + \gamma AC = \beta \mathsf{L}(A)B + \gamma \mathsf{L}(A)C.

Thus, \mathsf{L} is a function from the convolution algebra \mathcal{C}(G) to the linear algebra \mathcal{L}(\mathcal{C}(G)). Moreover, the map \mathsf{L} itself is a linear transformation, i.e.

\mathsf{L}(\alpha A + \beta B) C= (\alpha A +\beta B)C=\alpha AB +\beta AC = \alpha \mathsf{L}(A)B + \gamma \mathsf{L}(B)C.

We claim that \mathsf{L} is an injective linear transformation. Since \mathsf{L} is a linear transformation of \mathcal{C}(G), it is completely determined by its action on the group basis \{E_g \colon g \in G\} of \mathcal{C}(G). Thus, it suffices to show that the linear operators \mathsf{L}(E_g) and \mathsf{L}(E_h) on \mathcal{C}(G) are distinct unless g=h. Since \mathsf{L}(E_g) and \mathsf{L}(E_h) are linear operators on \mathcal{C}(G), they are uniquely determined by their actions on the group basis \{E_g \colon g \in G\} of \mathcal{C}(G). We have

\mathsf{L}(E_g)E_k = E_gE_k = E_{gk} \quad\text{and}\quad \mathsf{L}(E_h)E_k = E_hE_k = E_{hk},

hence

\mathsf{L}(E_g)E_k=\mathsf{L}(E_h)E_k \iff gk = hk \iff g=h.

From this we can conclude that the image of \mathcal{C}(G) in \mathcal{L}(\mathcal{C}(G)) under \mathsf{L} is a subspace isomorphic to \mathcal{C}(G). It remains to show that \mathsf{L} is not just a linear transformation, but an algebra homomorphism, and we leave this as an exercise (that you really should do).

-QED

The proof of Theorem 8.1 is the linear version of Cayley’s theorem from group theory: instead of representing G as a subgroup of the group of permutations of G, it represents \mathcal{C}(G) as a subalgebra of the algebra of linear operators on \mathcal{C}(G). This is called the left regular representation of \mathcal{C}(G).

Corollary 9.2. The class algebra \mathcal{Z}(G) is isomorphic to a function algebra.

Proof: Since \mathcal{Z}(G) is a subalgebra of \mathcal{C}(G), and since \mathcal{C}(G) is isomorphic to a subalgebra of a linear algebra, \mathcal{Z}(G) is isomorphic to a subalgebra of a linear algebra. Thus, \mathcal{Z}(G) is isomorphic to a commutative subgalgebra of a linear algebra, and we have previously shown that all commutative subalgebras of linear algebras are isomorphic to function algebras.

-QED

For an example illustrating how the proof of Theorem 8.1 works, let us take G=\{g^0,g^1,g^2,g^3\} to be a cyclic group of order four with generator g, so that g^0=g^4=\dots=e is the group unit. The group basis of \mathcal{C}(G) is thus E_{g^0},E_{g^1},E_{g^2},E_{g^3}, and we denote these vectors by E_0,E_1,E_2,E_3 for brevity. Let A \in \mathcal{C}(G) be any function on G, and let

A = \alpha_0 E_0 + \alpha_1 E_1 + \alpha_2 E_2 + \alpha_3 E_3

be its expansion in the group basis, so that

\alpha_0 = A(g^0),\ \alpha_1=A(g),\ \alpha_2=A(g^2),\ \alpha_3=A(g^3).

Problem 9.1. Show by direct calculation that the matrix of \mathsf{L}(A) \in \mathcal{L}(\mathcal{C}(G)) with respect to the ordered basis E_0,E_1,E_2,E_3 of \mathcal{C}(G) is

\mathsf{L}(A) = \begin{bmatrix} \alpha_0 & \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_3 & \alpha_0 & \alpha_1 & \alpha_2 \\ \alpha_2 & \alpha_3 & \alpha_0 & \alpha_1 \\ \alpha_1 & \alpha_2 & \alpha_3 & \alpha_0 \end{bmatrix}.

In the case where G is an abelian group, as in the example above, we have that \mathcal{C}(G) = \mathcal{Z}(G). Moreover, it is possible to establish Corollary 8.1 directly, without appealing to Theorem 8.1. This is done constructively, by finding an explicit basis of orthogonal projections in the commutative convolution algebra \mathcal{C}(G) called its Fourier basis. The advantage of this direct approach is that it also gives us an explicit description of the spectrum of all matrices in the left regular representation of \mathcal{C}(G). This is very useful in applications – in particular, matrices in the left regular representation of a cyclical group are called circulant matrices and they are important in engineering.

To see begin to see where a projection basis of \mathcal{C}(G) might come from, recall that we previously showed the non-existence of an algebra homomorphism \mathsf{T} \colon \mathcal{L}(V) \to \mathbb{C} for V a Hilbert space of dimension at least two. This reflects the fact that linear algebras \mathcal{L}(V) are maximally noncommuative. But we have also seen that for G a group with at least two elements, the dimension of \mathcal{Z}(G) is at least two, so convolution algebras are always at least one degree more commutative than linear algebras, and therefore might admit homomorphisms to the complex numbers.

Theorem 9.3. A linear transformation \mathsf{T} \colon \mathcal{C}(G) \to \mathbb{C} is an algebra homomorphism if and only if the function \chi \in \mathcal{C}(G) defined by

\chi(g) = \mathsf{T}(E_g), \quad g \in G,

is a group homomorphism taking values in U(\mathbb{C}), the unitary group of the complex numbers.

Problem 9.2. Prove Theorem 8.3.

Definition 9.1. A group homomorphism \chi \colon G \to U(\mathbb{C}) is called a character of G. The set of all characters of G is denoted \widehat{G}.

Observe first of all that \widehat{G} is a set of functions on G, is a subset of \mathcal{C}(G), the space of all \mathbb{C}-valued functions on G. It is this special subset of functions tethered to the group law in G that we look to for orthogonal projections. No matter what the group G is, the set \widehat{G} is nonempty: it contains at least the trivial character defined by \chi(g)=1, g \in G. However, for highly noncommutative groups there may not be many nontrivial characters.

Problem 9.3. Determine the Fourier dual of the symmetric group S_n.

For abelian groups the situation is much better and in fact we always have |G|=|\widehat{G}|. To begin, recall that every finite abelian group is isomorphic to a product of cyclic groups. We thus fix a positive integer r \in \mathbb{N}, and r positive integers n_1,\dots,n_r \in \mathbb{N}, and consider the group

G = G_1 \times \dots \times G_r,

where G_i is a cyclic group of order n_i with generator g_i. Define the dual group of G to be

\Lambda = \{\alpha=(\alpha_1,\dots,\alpha_r) \colon \alpha_k \in \{0,1,\dots,n_k-1\},\ 1 \leq k \leq r\}.

That is,

\Lambda = \mathbb{Z}_{n_1} \times \dots \times \mathbb{Z}_{n_r},

where \mathbb{Z}_n= is the additive group of integers modulo n. We can parameterize G by the points of \Lambda, writing

g_\alpha=(g_1^{\alpha_1},\dots,g_r^{\alpha_r}), \quad \alpha \in \Lambda.

Indeed, the parameterization \alpha \mapsto g_\alpha is a group isomorphism \Lambda \to G (Exercise: prove this, noting that because |\Lambda|=|G| it is sufficient to show the parameterization is an injective group homomorphism).

Theorem 9.4. For every \lambda \in \Lambda, the function \chi^\lambda \colon G \to U(\mathbb{C}) defined by

\chi^\lambda(g_\alpha) = \omega_1^{\alpha_1\lambda_1} \dots \omega_r^{\alpha_r\lambda^r},

where \omega_k=\exp\left(\frac{2\pi i}{n_k}\right) is a principal kth root of unity, is a character of G, and every character of G is of this form.

Proof: For any \lambda \in \Lambda, it is clear that \chi^\lambda(e)=1, because the identity element e \in G has parameters \alpha=(0,0,\dots,0). Moreover, for any \alpha,\beta \in \Lambda we have

\chi^\lambda(g_\alpha g_\beta) = \chi^\lambda(g_{\alpha+\beta})=(\omega_1)^{(\alpha_1+\beta_1)\lambda_1} \dots (\omega_r)^{(\alpha_r+\beta_r)\lambda_r)} = (\omega_1)^{\alpha_1\lambda_1} \dots (\omega_r)^{\alpha_r\lambda_r}(\omega_1)^{\beta_1\lambda_1} \dots (\omega_r)^{\beta_r\lambda_r}=\chi^\lambda(g_\alpha)\chi^\lambda(g_\beta),

so \chi^\lambda is indeed a group homomorphism G \to U(\mathbb{C}). The fact that every homomorphism \chi \colon G \to U(\mathbb{C}) is \chi^\lambda for some \lambda \in \Lambda is left as an exercise.

-QED

We now have a special subset \widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\} of the convolution algebra \mathcal{C}(G) of the finite abelian group G, namely the set \widehat{G} of all homomorphisms to the unitary group U(\mathbb{C}). We now claim that the characters form a basis of \mathcal{C}(G). Since the number of characters is |\widehat{G}|=|\Lambda|=|G|, which is the dimension of \mathcal{C}(G), it is sufficient to show that \widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\} is a linearly independent set in \mathcal{C}(G).

Theorem 9.5. The set \{\chi^\lambda \colon \lambda \in \Lambda\} is orthogonal with respect to the \ell^2-scalar product on G – we have

\langle \chi^\lambda,\chi^\mu\rangle = \sum\limits_{g \in G} \overline{\chi^\lambda(g)} \chi^\mu(g) = \delta_{\lambda\mu}|G|.

Proof: For any \lambda,\mu \in \Lambda, we have

\langle \chi^\lambda,\chi^\mu \rangle = \sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda(g_\alpha)}\chi^\mu(g_\alpha) = \sum\limits_{\alpha \in \Lambda}(\omega_1)^{\alpha_1(\mu_1-\lambda_1)} \dots (\omega_r)^{\alpha_r(\mu_r-\lambda_r)}=\left(\sum\limits_{\alpha_1=0}^{n_1-1} \zeta_1^{\alpha_1}\right) \dots \left(\sum\limits_{\alpha_r=0}^{n_r-1} \zeta_r^{\alpha_r}\right),

where

\zeta_1 = \omega_1^{\mu_1-\lambda_1},\ \dots,\ \zeta_r=\omega_r^{\mu_r-\lambda_r}.

Thus if \lambda=\mu we have

\langle \chi^\lambda,\chi^\mu \rangle = n_1 \dots n_r = |\Lambda|=|G|,

and if \lambda \neq \mu we have

\langle \chi^\lambda,\chi^\mu \rangle =\frac{1-\zeta_1^{n_1}}{1-\zeta_1} \dots \frac{1-\zeta_r^{n_r}}{1-\zeta_r},

where the denominator of each fraction is nonzero and the numerator is zero, because \zeta_k=\omega_k^{\mu_k-\lambda_k} is an n_kth root of unity.

-QED

The orthogonal basis \widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\} of \mathcal{C}(G) is called its character basis. It is convenient to write \chi^\lambda_\alpha := \chi^\lambda(g_\alpha), since this highlights the symmetry \chi^\lambda_\alpha = \chi^\alpha_\lambda. The |G| \times |G| symmetric matrix X=[\chi^\lambda_\alpha] is called the character table of G, and Theorem 8.3 says that \frac{1}{\sqrt{|G|}} is a symmetric unitary matrix. Another way to say the same thing is that the rescaled character basis

E^\lambda = \frac{1}{\sqrt{|G|}} \chi^\lambda, \quad \lambda \in \Lambda,

is an orthonormal basis of the convolution algebra \mathcal{C}(G). In fact, the further scaling

F^\lambda = \frac{1}{|G|}F^\lambda, \quad \lambda \in \Lambda,

is even better, for the following reason.

Theorem 9.6. The elements of the basis \{F^\lambda \colon \lambda \in \Lambda\} are orthogonal projections in \mathcal{C}(G).

Proof: For any \lambda \in \Lambda, we have

(F^\lambda)^* = \left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g) E_g\right)^*=\frac{1}{|G|}\sum\limits_{g \in G} \overline{\chi^\lambda(g)} E_g^* =\frac{1}{|G|} \sum\limits_{g \in G} \chi^\lambda(g^{-1}) E_{g^{-1}} = F^\lambda.

For any \lambda,\mu \in \Lambda, we have

F^\lambda F^\mu = \left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g) E_g\right)\left( \frac{1}{|G|}\sum\limits_{g \in G} \chi^\mu(g) E_g\right)=\frac{1}{|G|^2}\sum\limits_{g \in G} \left( \sum\limits_{h \in G}\chi^\lambda(gh^{-1})\chi^\mu(h)\right)E_g = \frac{1}{|G|^2}\sum\limits_{g \in G} \chi^\lambda(g)\left( \sum\limits_{h \in G}\chi^\lambda(h^{-1})\chi^\mu(h)\right)E_g.

The internal sum is

\sum\limits_{h \in G}\chi^\lambda({h^{-1}})\chi^\mu(h)=\sum\limits_{h \in G}\overline{\chi^\lambda(h)}\chi^\mu(h)=\delta_{\lambda\mu}|G|,

where the final equality is Theorem 8.3. Thus

F^\lambda F^\mu = \frac{1}{|G|^2}\delta_{\lambda\mu}|G|\sum\limits_{g \in G} \chi^\lambda(g)E_g=\delta_{\lambda\mu}F^\lambda.

-QED

Since we know that any algebra with a basis of orthogonal projections is isomorphic to the function algebra (Lecture 3), Theorem 8.4 gives the promised second proof of the fact that the convolution algebra \mathcal{C}(G) of an abelian group G is isomorphic to a function algebra. In this particular case, the basis \{F^\lambda \colon \lambda \in \Lambda\} is known as the Fourier basis of \mathcal{C}(G).

Math 202B: Lecture 8

The convolution algebra \mathcal{C}(G) of a group G is commutative if and only if G is abelian. As in Lecture 2, we will refine this dichotomy by giving a quantitative measurement of how (non)commutative \mathcal{C}(G) is. In Lecture 2 we defined the commutativity index of an algebra to be the dimension of its center, so our goal now is to determine the dimension of the center of a convolution algebra.

The degree of non commutativity of \mathcal{C}(G) is determined by that of G, and in group theory we understand this using the action of G on itself defined by

g.h = ghg^{-1}.

The orbits of this action are the conjugacy classes of G, and the number of conjugacy classes in G is called its class number. The higher the class number, the more commutative the group – for abelian groups, conjugacy classes are singleton sets.

Problem 8.1. Prove that G is an abelian group if and only if its class number is equal to its cardinality, and more generally that the number of commuting pairs of elements in G is equal to the cardinality of G times its class number (hint: the orbit-stabilizer theorem may be helpful). Show also that every group except the trivial group contains at least two conjugacy classes.

For the function algebra \mathcal{F}(G), every partition of G gives rise to a subalgebra of \mathcal{F}(G), namely the set of functions constant on the blocks of the partition. As we discussed in Lecture 4, there is no reason that the convolution of two such functions will still be constant on the blocks of the given partition. On the other hand, the partition of G into conjugacy classes is determined by the group structure of G, and so functions on G which are constant on the blocks of this particular group-theoretic partition of G are relevant from the point of view of \mathcal{C}(G).

Definition 8.2. A function A \colon G \to \mathbb{C} is called a class function if it is constant on conjugacy classes, meaning that A(g) = A(hgh^{-1}) for all g,h \in G.

The following gives an alternative way to think about class functions as functions which are insensitive to any noncommutativity present in G.

Theorem 8.1. A function A \colon G \to \mathbb{C} is a class function if and only if A(gh)=A(hg) for all g,h \in G.

Proof: Suppose first that A is a class function on G. Then,

A(gh) = A(hghh^{-1}) = A(hg).

Conversely, suppose A is insensitive to noncommutativity. Then,

A(hgh^{-1}) = A(h^{-1}hg) = A(g).

-QED

We can now characterize the center of \mathcal{C}(G).

Theorem 8.2. A function A \in \mathcal{C}(G) belongs to Z(\mathcal{C}(G)) if and only if it is constant on conjugacy classes.

Proof: Suppose first that A \in \mathcal{C}(G) is a class function; we will prove that it commutes with every B \in \mathcal{C}(G). For any g \in G, we have

[AB](g) = \sum\limits_{h \in G} A(gh^{-1})B(h) = \sum\limits_{h \in G} A(g(gh)^{-1}) B(gh) = \sum\limits_{h \in G} A(gh^{-1}g^{-1})B(gh),

where the second inequality follows from the fact that the substitution h \rightsquigarrow gh simply permutes the terms of the sum. Continuing the calculation, we have

[AB](g) = \sum\limits_{h \in G} A(gh^{-1}g^{-1})B(gh) =\sum\limits_{h \in G} A(h^{-1})B(gh) = \sum\limits_{h \in G} B(gh)A(h^{-1}),

where the second inequality follows from the fact that A is a class function. We now conclude

[AB](g) = \sum\limits_{h \in G} B(gh^{-1})A(h) = [BA](g),

as required.

Now suppose that Z \in Z(\mathcal{C}(g)) is a central function; we will prove it is constant on conjugacy classes. Since Z commutes with all functions in \mathcal{C}(g), it commutes with every elementary function E_g. Since

[ZE_g](h) = \sum\limits_{k \in G}Z(hk^{-1})E_g(k)=Z(hg^{-1}),

and

[E_gZ](h) = \sum\limits_{k\in G}E_g(hk^{-1})Z(k)=Z(g^{-1}h),

the centrality of Z implies that Z(gh)=Z(hg) for all g,h \in G. Thus Z is a class function, by Theorem 5.1.

-QED

Thus, the set of all functions constant on conjugacy classes of G is a subalgebra of both \mathcal{F}(G) and \mathcal{C}(G). The subalgebra of class functions has no special significance in \mathcal{F}(G), not being any more or less special than the subalgebra associated with any other partition of G. But in \mathcal{C}(G), the subalgebra of class functions is the center of \mathcal{C}(G). The center of \mathcal{C}(G) is often called the class algebra of G and denoted \mathcal{Z}(G) rather than Z(\mathcal{C}(G)) to emphasize that it is worth thinking about as a standalone object, i.e. as a commutative algebra naturally associated to the finite group G.

Just as the convolution algebra \mathcal{C}(G) has a natural basis given by indicator functions of elements of G, the class algebra \mathcal{Z}(G) has a natural basis given by indicator functions of conjugacy classes in G. Let \Lambda be a set parameterizing the conjugacy classes of G, so the collection of these is

\{C_\alpha \colon \alpha \in \Lambda\},

and this is a subset of the power set of G. For each \alpha \in \Lambda, let K_\alpha \colon G \to \mathbb{C} be the indicator function of C_\alpha, so

K_\alpha(g) = \begin{cases} 1, \text{ if }g \in C_\alpha \\ 0, \text{ if }g \not\in C_\alpha \end{cases}.

Equivalently, in terms of the group basis \{E_g \colon g \in G\} of \mathcal{C}(G) we have

K_\alpha = \sum\limits_{g \in C_\alpha} E_g, \quad \alpha \in \Lambda.

Then, the functions \{K_\alpha \colon \alpha \in \Lambda\} span the class algebra of G, since any function Z \colon G \to \mathbb{C} constant on conjugacy classes can be written

Z = \sum\limits_{\alpha \in \Lambda} Z(\alpha) K_\alpha,

where Z(\alpha) denotes the value Z(g) for any g \in \mathcal{C}_\alpha. We will show that \{K_\alpha \colon \alpha \in \Lambda\} is a basis of \mathcal{Z}(G) using the \ell^2-scalar product on \mathcal{C}(G), which is

\langle A,B \rangle = \sum\limits_{g \in G} \overline{A(g)}B(g).

It is clear that \{E_g \colon g \in G\} is an orthonormal basis of \mathcal{C}(G) with respect to the \ell^2-scalar product, and we therefore have

\langle K_\alpha,K_\beta \rangle = \left\langle \sum\limits_{g \in C_\alpha} E_g , \sum\limits_{h \in C_\beta} E_h \right\rangle = \sum\limits_{g \in C_\alpha,\ h \in C_\beta} \langle E_g,E_h\rangle = |C_\alpha \cap C_\beta|.

Since \{C_\alpha \colon \alpha \in \Lambda\} is a partition of G, the scalar product is

\langle K_\alpha,K_\beta \rangle = \delta_{\alpha\beta}|C_\alpha|,

which shows that \{K_\gamma \colon \gamma \in \Gamma\} is an orthogonal (but not orthonormal) set of functions in \mathcal{Z}(G), hence linearly independent.

In conclusion, if G is any finite group, the dimension of the convolution algebra \mathcal{C}(G) is the cardinality of G, and the dimension of the class algebra \mathcal{Z}(G) is the class number of G.

Problem 8.3. Consider the multiplication tensor [c_{\alpha\beta\gamma}] of \mathcal{Z}(G), i.e. K_\alpha K_\beta = \sum_\gamma c_{\alpha\beta\gamma}K_\gamma. For any \alpha,\beta,\gamma \in \Gamma and any g \in C_\gamma, show that

c_{\alpha\beta\gamma} = |\{(x,y) \in C_\alpha \times C_\beta \colon xy=g\}|.

Thus, the multiplication tensor of \mathcal{Z}(G) is quite an interesting object: c_{\alpha\beta\gamma} counts solutions to the equation xy=g in G, where g is any particular point of the conjugacy class C_\gamma, and x,y \in G are required to belong to C_\alpha and C_\beta respectively.