202B – Jonathan Novak

Math 202B: Revenge of the Centralizer

Yesterday during office hours I had the unsettling experience of being shown that something I assumed to be true is, in fact, subtly incorrect. That’s what happens when you take things for granite. My sloppy thinking was pointed out to me by Amelia and Eliza in the context of the following problem which appeared on the Math 202B Exam: prove that if $\mathcal{B}$ is a maximal abelian subalgebra of an algebra $\mathcal{A}$ , then the centralizer $Z(\mathcal{B},\mathcal{A})$ of $\mathcal{B}$ in $\mathcal{A}$ equals $\mathcal{B}.$

This was supposed to be easy. Since $\mathcal{B}$ is abelian, it is certainly contained in its centralizer $Z(\mathcal{B},\mathcal{A}).$ Now, if the containment is proper, then there exists an element $C \in Z(\mathcal{B},\mathcal{A})$ which does not lie in $\mathcal{B}.$ Then,

$\mathcal{C}=\mathrm{CAlg}(\mathcal{B} \sqcup \{C\}),$

the commutative algebra generated by $\mathcal{B}$ with $C$ adjoined, is a commutative algebra properly containing $\mathcal{B}$ , contradicting the fact that $\mathcal{B}$ is a MASA.

The subtle issue here is that $\mathcal{C}$ is by definition the intersection of all commutative subalgebras of $\mathcal{A}$ containing the set $\mathcal{B} \sqcup \{C\},$ and this could conceivably be empty. To be concrete, imagine a single non-normal matrix $C$ . There is no commutative subalgebra of the full matrix algebra containing $C$ , since our definition of “algebra” requires $*$ -closure, and by hypothesis $C$ does not commute with its adjoint.

Luckily, our attempted solution can be salvaged: as Evan immediately recognized, if we can show that $Z(\mathcal{B},\mathcal{A})$ contains a *selfadjoint* element $D$ not contained in $\mathcal{B}$ this issue goes away. This is the case, since we can take $D$ to be either the real or imaginary part of $C$ — at least one of these does not lie in $\mathcal{B}$ , since if they both did so would $C.$

Hopefully I’m now getting this right, and my apologies for being such a bad professor. Try reading some of my papers, I swear they’re really good.

In any case, we will have to incorporate this correction into our census of reality. In particular, if this problem were to appear on the qual, a proposed solution which did not address this issue would no longer pass muster. Live and learn.

Math 202B: Lecture 26

We begin by recalling our general algebraic perspective on the discrete Fourier transform (DFT). A Fourier basis of an algebra $\mathcal{A}$ is a basis $\{F^\lambda \colon \Lambda\}$ of pairwise orthogonal selfadjoint idempotents. If such a basis exists, then $\mathcal{A}$ is necessarily commutative. Note also that since we have not mentioned a scalar product on the algebra $\mathcal{A},$ the orthogonality of the Fourier basis vectors can only refer to the fact that the vector product of two distinct Fourier basis vectors is the zero vector.

There is no guarantee that a given algebra $\mathcal{A}$ admits a Fourier basis, but if it does then every element $A \in \mathcal{A}$ can be expanded on the Fourier basis,

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

and the coefficients in this expansion define a function $\widehat{A} \in \mathcal{F}(\Lambda).$ One of our most basic but important theorems from early on in the course is that the map $A \mapsto \widehat{A}$ is an algebra isomorphism $\mathcal{A} \to \mathcal{F}(\Lambda).$ This algebra isomorphism is called the Fourier transform on $\mathcal{A}.$

The above construction cannot work for noncommutative algebras, in the sense that a noncommutative algebra cannot be isomorphic to a function algebra. We do however have an alternative path to pursue in this case. Instead of looking for a Fourier basis in $\mathcal{A}$ , we look for a Frobenius scalar product on $\mathcal{A}$ . When such a scalar product exists, then $\mathcal{A}$ becomes a Hilbert space in a way which is compatible with its algebra structure. In this case, we have shown that $\mathcal{A}$ is isomorphic not to an algebra of functions on a finite set, but rather to an algebra of operators on a finite-dimensional Hilbert space. More precisely, we have an algebra homomorphism $\Phi \colon \mathcal{A} \to \mathrm{End} \mathcal{A}$ which sends each $A \in \mathcal{A}$ to the corresponding left multiplication operator $\Phi(A) \in \mathrm{End}\mathcal{A}$ . A good name for the homomorphism $\Phi$ would be the Wedderburn transform; the pair $(\mathcal{A},\Phi)$ together is the left-regular representation of $\mathcal{A}.$ The Wedderburn transform is an injective algebra homomorphism, and it replaces the Fourier transform in the sense that it sets up an isomorphism between the abstract algebra $\mathcal{A}$ and the subalgebra of $\mathrm{End}\mathcal{A}$ consisting of left multiplication operators $\Phi(A)$ , $A \in \mathcal{A}.$

Although the above describes the Wedderburn transform as a replacement for the Fourier transform in the noncommutative setting, there is no reason that it cannot be applied in the commutative setting. Namely, the best case scenario is that we have a commutative algebra $\mathcal{A}$ in which we have successfully identified a Fourier basis $\{F^\lambda \colon \lambda \in \Lambda\}$ , and we also are able to identify a Frobenius scalar product on the commutative algebra $\mathcal{A}$ . In this case, both the Fourier transform $\mathcal{A} \to \mathcal{F}(\Lambda)$ and the Wedderburn transform $\mathcal{A} \to \mathrm{End}\mathcal{A}$ are in play. They are moreover related in a very useful way: the Fourier basis $\{F^\lambda \colon \lambda \in \Lambda\}$ is an eigenbasis for the Wedderburn transform $\Phi(A)$ of every $A \in \mathcal{A}$ , and the corresponding eigenvalues are given by the Fourier transform of $A,$

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

We have concretely implemented this compelling abstract picture in the case where $\mathcal{A}=\mathcal{C}(G)$ is the convolution algebra of a finite abelian group, $G.$ In order to construct a Fourier basis of $\mathcal{C}(G),$ we started from the observation that the set $\{\chi^\lambda \colon \lambda \in \Lambda\}$ of multiplicative characters of $G$ , which is parameterized by an explicit set $\Lambda$ corresponding to the decomposition of $G$ as a product of cyclical groups, forms an orthogonal basis of $\mathcal{C}(G).$ This allows us to define a Fourier basis in $\mathcal{C}(G)$ consisting of the normalized multiplicative characters

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

In order to see how the Fourier transform and the Wedderburn transform work together in the setting of convolution algebras of finite abelian groups, I asked you to work out the case where $G$ is a cyclic group in detail, where the synchrony between the two transforms gives a method to diagonalize circulant matrices. I hope this was instructive, and it is definitely something worth reviewing more than once.

Now suppose $G$ is a nonabelian group. Then, $\mathcal{C}(G)$ is a noncommutative algebra which cannot be isomorphic to a function algebra. However, the center $\mathcal{K}(G)$ of $\mathcal{C}(G)$ , which consists of functions on $G$ which are constant on conjugacy classes of $G,$ is a commutative algebra associated to $G$ . In this lecture we will see that the Fourier transform can indeed be concretely implemented for the class algebra of any finite group, which generalizes our prior implementation for convolution algebras of abelian groups. To achieve this extension, we will rely on our extension of multiplicative characters of finite abelian groups to irreducible characters of arbitrary finite groups. I hope that at this point you are close to internalizing the fact that, for abelian groups, multiplicative characters and irreducible characters are the same thing.

Let $\Lambda$ be a set parameterizing irreducible unitary representations of $G$ (or equivalently irreducible linear representations of $\mathcal{C}(G)$ ) up to isomorphism. For each $\lambda \in \Lambda,$ let $(V^\lambda,\varphi^\lambda)$ be a representative of the corresponding isomorphism class, and let

$\chi^\lambda(g) = \mathrm{Tr} \varphi^\lambda(g), \quad g \in G.$

We know that the set $\{\chi^\lambda \colon \lambda \in \Lambda\}$ forms an orthogonal basis of $\mathcal{K}(G).$ Let $\{C_\alpha \colon \alpha \in \Lambda\}$ be the set of conjugacy classes in $G.$ For each $\alpha \in \Lambda,$ let $K_\alpha$ be the indicator function of $C_\alpha.$ Then, the set $\{K_\alpha \colon \alpha \in \Lambda\}$ is an orthogonal basis of $\mathcal{K}(G),$

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

The connection coefficients of the class basis,

$K_\alpha K_\beta = \sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma}K_\gamma,$

count solutions to certain equations in the underlying group, $G$ . Namely, $c_{\alpha\beta\gamma}$ is the number of pairs $(h,k) \in C_\alpha \times C_\beta$ such that $g=hk,$ where $g \in C_\gamma.$ Let $\chi^\lambda_\alpha$ denote $\chi^\lambda(g)$ for any $g \in C_\alpha.$ In Lecture 25, we posed the following problem which expresses the connection coefficients of $\mathcal{K}(G)$ in terms of the irreducible characters of $G.$ You should reflect on what this says in the case where $G$ is abelian.

Theorem 26.1. For any $\alpha,\beta,\gamma \in \Lambda,$ we have

$c_{\alpha\beta\gamma} = \frac{|C_\alpha| |C_\beta|}{|G|}\sum\limits_{\lambda \in \Lambda}\frac{\chi^\lambda_\alpha \chi^\lambda_\beta \overline{\chi^\lambda_\gamma}}{\dim V^\lambda}.$

Using this result, we construct a Fourier basis of $\mathcal{K}(G).$ Here is the basic claim: the central functions

$F^\lambda = \frac{\dim V^\lambda}{|G|}\sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}K_\alpha, \quad \lambda \in \Lambda$

form a Fourier basis of $\mathcal{K}(G).$ Let us first check that these candidate functions are nonzero.

Problem 26.1. Compute the $L^2$ -norm of $F^\lambda,$ and show that $\sum_{\lambda \in \Lambda} \|F^\lambda\|^2=1.$

Now, on to the main event.

Theorem 26.1. The $|\Lambda|$ nonzero functions

$F^\lambda = \frac{\dim V^\lambda}{|G|}\sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}K_\alpha, \quad \lambda \in \Lambda$

are pairwise orthogonal selfadjoint idempotents in $\mathcal{K}(G)$ .

Proof: We have

$F^\lambda F^\mu = \frac{\dim V^\lambda}{|G|}\frac{\dim V^\mu}{|G|}\sum\limits_{\alpha \in \Lambda}\sum\limits_{\beta \in \Lambda} \overline{\chi^\lambda_\alpha} \overline{\chi^\mu_\beta} K_\alpha K_\beta=\frac{\dim V^\lambda}{|G|}\frac{\dim V^\mu}{|G|}\sum\limits_{\alpha \in \Lambda}\sum\limits_{\beta \in \Lambda} \sum\limits_{\gamma \in \Lambda}\overline{\chi^\lambda_\alpha} \overline{\chi^\mu_\beta} c_{\alpha\beta\gamma}K_\gamma.$

By Theorem 26.1, the right hand side is

$\frac{\dim V^\lambda}{|G|}\frac{\dim V^\mu}{|G|}\sum\limits_{\alpha \in \Lambda}\sum\limits_{\beta \in \Lambda} \sum\limits_{\gamma \in \Lambda}\overline{\chi^\lambda_\alpha} \overline{\chi^\mu_\beta} K_\gamma \frac{|C_\alpha||C_\beta|}{|G|}\sum\limits_{\nu \in \Lambda} \frac{\chi^\nu_\alpha\chi^\nu_\beta\overline{\chi^\nu_\gamma}}{\dim V^\nu}.$

This quadruple sum over $\Lambda$ can be reorganized as

$\frac{\dim V^\lambda \dim V^\mu}{|G|^3}\sum\limits_{\nu \in \Lambda}\frac{1}{\dim V^\nu} \left(\sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\nu_\alpha|C_\alpha| \right)\left(\sum\limits_{\beta \in \Lambda} \overline{\chi^\mu_\beta}\chi^\nu_\beta|C_\beta|\right)\sum\limits_{\gamma \in \Lambda}\overline{\chi^\nu_\gamma}K_\gamma.$

Character orthogonality (original, not dual) collapses the sums over $\alpha$ and $\beta,$ leaving

$\frac{\dim V^\lambda \dim V^\mu}{|G|}\sum\limits_{\nu \in \Lambda} \frac{1}{\dim V^\nu}\delta_{\lambda\nu}\delta_{\mu\nu}\sum\limits_{\gamma \in \Lambda}\overline{\chi^\nu_\gamma}K_\gamma =\delta_{\lambda\mu} \frac{\dim V^\lambda}{|G|}\sum\limits_{\gamma \in \Lambda}\overline{\chi^\lambda_\gamma}K_\gamma.$

This completes the orthogonal idempotent part of the proof. It remains to check that the proposed Fourier basis consists of selfadjoint elements, and we leave this as an exercise. $\square$

Math 202B: Lecture 25

Let $\Lambda$ be a set parameterizing irreducible unitary representations of $G$ up to isomorphism, and for each $\lambda \in \Lambda$ let $(V^\lambda,\varphi^\lambda)$ be a representative of the corresponding isomorphism class. Equivalently, $\Lambda$ parameterizes irreducible linear representations of the convolution algebra $\mathcal{C}(G)$ up to isomorphism, and for each $\lambda \in \Lambda$ we take the representative $(V^\lambda,\Phi^\lambda)$ determined by

$\Phi^\lambda(E_g) = \varphi^\lambda(g), \quad g \in G.$

Thus, for any function

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

in $\mathcal{C}(G)$ we have

$\Phi^\lambda(A) = \sum\limits_{g \in G} A(g)\phi^\lambda(g),$

and

$\chi^\lambda(A) = \mathrm{Tr}\ \Phi^\lambda(A) = \sum\limits_{g \in G} A(g) \mathrm{Tr}\, \varphi^\lambda(g) = \sum\limits_{g \in G} A(g) \chi^\lambda(g).$

Recall that the class algebra $\mathcal{K}(G)$ consists of those functions $A \in \mathcal{C}(G)$ which are constant on the conjugacy classes of $G,$ and that this is the center of $\mathcal{C}(G).$ By Schur’s Lemma, if $A \in \mathcal{K}(G)$ then the operator $\Phi^\lambda(A)$ is a scalar multiple of the the identity operator,

$\Phi^\lambda(A) = \omega^\lambda(A)I_{V^\lambda}.$

This determines a map

$\omega^\lambda \colon \mathcal{K}(G) \longrightarrow \mathbb{C}$

called the central character of the irrep $(V^\lambda,\Phi^\lambda).$

Problem 25.1. Prove that, for any $\lambda \in \Lambda,$ the central character $\omega^\lambda \colon \mathcal{K}(G) \to \mathbb{C}$ is an algebra homomorphism, given in terms of the character of $(V^\lambda,\Phi^\lambda)$ by

$\omega^\lambda(A) = \frac{1}{\dim V^\lambda} \chi^\lambda(A), \quad A \in \mathcal{K}(G).$

We are now ready to prove that the irreducible characters of $G$ form an orthogonal basis of the class algebra $\mathcal{K}(G).$ We

Theorem 25.2. The set $\{\chi^\lambda \colon \lambda \in \Lambda\}$ spans $\mathcal{K}(G).$

Proof: Let $W$ be the subspace of $\mathcal{K}(G)$ spanned by $\{\chi^\lambda \colon \lambda \in \Lambda\}.$ We will show that $W^\perp$ consists solely of the zero function. That is, if $A \in \mathcal{K}(G)$ is a central function orthogonal to every irreducible character,

$\langle A, \chi^\lambda \rangle = \sum\limits_{g \in G} \overline{A(g)}\chi^\lambda(g) =0, \quad \lambda \in \Lambda,$

then $A$ is the zero function on $G.$ Let

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

which is the conjugate of $A$ viewed as an element of the function algebra $\mathcal{F}(G)$ rather than the convolution algebra $\mathcal{C}(G)$ . Then,

$\omega^\lambda(\overline{A})= \frac{1}{\dim V^\lambda} \mathrm{Tr} \Phi^\lambda(\overline{A}) = \frac{1}{\dim V^\lambda} \sum\limits_{g \in G} \overline{A(g)} \chi^\lambda(g) \frac{1}{\dim V^\lambda} \langle A,\chi^\lambda\rangle= 0, \quad \lambda \in \Lambda.$

This says that the image of $\overline{A}$ in every irreducible representation $(V^\lambda,\Phi^\lambda)$ of $\mathcal{C}(G)$ is the zero operator. Since every linear representation of $\mathcal{C}(G)$ is a direct sum of irreducible linear representations, this means that $\overline{A}$ is the zero operator in every linear representation of $\mathcal{C}(G)$ . This includes the regular representation, which is faithful. We conclude that $\overline{A}$ and hence $A$ is the zero function on $G,$ as required. $\square$

We now know that the set $\{\chi^\lambda \colon \lambda \in \Lambda\}$ of irreducible characters of $G$ is an orthogonal basis of $\mathcal{K}(G)$ . In particular, $|\Lambda|=\mathrm{dim}\mathcal{K}(G),$ which says the following.

Corollary 25.3. The number of isomorphism classes of irreducible unitary representations of $G$ is equal to the number of conjugacy classes in $G.$

Let us consider character orthogonal more carefully. Explicitly, we have

$\langle \chi^\lambda,\chi^\mu\rangle= \sum\limits_{g \in G} \overline{\chi^\lambda(g)}\chi^\mu(g) = \sum\limits_{g \in G}\chi^\lambda(g^{-1})\chi^\mu(g)=\delta_{\lambda\mu}|G|.$

Now, let $\{C_\alpha \colon \alpha \in \Lambda\}$ be an enumeration of the conjugacy classes in $G$ , and let us write $\chi^\lambda_\alpha$ for $\chi^\lambda(g)$ with $g \in C_\alpha.$ Then, character orthogonality takes the form

$\langle \chi^\lambda,\chi^\mu\rangle= \sum\limits_{g \in G} \overline{\chi^\lambda(g)}\chi^\mu(g) = \sum\limits_{\alpha \in \Lambda} |C_\alpha|\overline{\chi^\lambda_\alpha} \chi^\mu_\alpha=\delta_{\lambda\mu}|G|.$

Definition 25.4. The character table of $G$ is the square matrix $\mathbf{T}$ with rows and columns indexed by $\Lambda$ and entries

$\mathbf{T}_{\alpha\lambda} = \chi^\lambda_\alpha.$

The modified character table of $G$ is the square matrix $\mathbf{M}$ with rows and columns indexed by $\Lambda$ and entries

$\mathbf{M}_{\alpha\lambda} = \sqrt{\frac{|C_\alpha|}{|G|}}\chi^\lambda_\alpha.$

Character orthogonality says that the modified character table of $G$ is a unitary matrix, i.e. that columns of $\mathbf{M}$ are pairwise orthogonal unit vectors,

$\sum\limits_{\alpha \in \Lambda} \overline{\mathbf{M}}_{\lambda\alpha} \mathbf{M}_{\mu\alpha}=\sum\limits_{\alpha \in \lambda} \sqrt{\frac{|C_\alpha|}{|G|}}\overline{\chi^\lambda_\alpha}\sqrt{\frac{|C_\alpha|}{|G|}}\chi^\mu_\alpha=\delta_{\lambda\mu}.$

Since the rows of a unitary matrix are also pairwise orthogonal unit vectors, this gives us a second orthogonality relation, namely

$\sum\limits_{\lambda \in \Lambda} \overline{\mathbf{M}}_{\lambda\alpha} \mathbf{M}_{\lambda\beta}=\sum\limits_{\lambda \in \lambda} \sqrt{\frac{|C_\alpha|}{|G|}}\overline{\chi^\lambda_\alpha}\sqrt{\frac{|C_\beta|}{|G|}}\chi^\lambda_\beta=\delta_{\alpha\beta}.$

This is called dual character orthogonality.

Theorem 25.5. For any $\alpha,\beta \in \Lambda,$ we have

$\sum\limits_{\lambda \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\lambda_\beta=\delta_{\alpha\beta}\frac{|G|}{|C_\alpha|}.$

As an application of Dual Character Orthogonality, we can obtain a formula for the connection coefficients of the class algebra $\mathcal{K}(G)$ in terms of the irreducible characters of $G.$ Recall that the class basis

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

of $\mathcal{K}(G)$ has the property that its connection coefficients

$K_\alpha K_\beta = \sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma}K_\gamma, \quad \alpha,\beta \in \Lambda$

solve a factorization problem in the group $G$ .

Problem 25.2. Prove that

$c_{\alpha\beta\gamma} = \frac{|C_\alpha||C_\beta|}{|G|}\sum\limits_{\lambda \in \lambda} \frac{\chi^\lambda_\alpha \chi^\lambda_\beta \overline{\chi^\lambda_\gamma}}{\dim V^\lambda}.$

Math 202B: Lecture 24

*** Problems in this lecture Due March 15 ***

*** Please complete a course evaluation ***

Let $G$ be a finite group, and let $\Lambda$ be a set parameterizing irreducible unitary representations of $G$ up to isomorphism. We know that $\Lambda$ is a finite set whose cardinality is bounded by the number of conjugacy classes in $G.$ For each $\lambda \in \Lambda,$ let $(V^\lambda,\varphi^\lambda)$ be an irreducible unitary representation of $G$ , and denote by

$\chi^\lambda(g) = \mathrm{Tr}\ \varphi^\lambda(g), \quad g \in G,$

the corresponding character. Then, $\{\chi^\lambda \colon \lambda \in \Lambda\}$ form an orthogonal set of functions in the class algebra $\mathcal{K}(G)$ ,

$\langle \chi^\lambda,\chi^\mu\rangle= \delta_{\lambda\mu}|G|.$

We do not yet know that $\{\chi^\lambda \colon \lambda \in \Lambda\}$ spans $\mathcal{K}(G),$ but we will get there soon.

Today we will discuss the representation-theoretic analogue of prime factorization. In fact, decomposing representations is easier than decomposing numbers because our set of irreducible objects $\{V^\lambda \colon \lambda \in \Lambda\}$ is finite, and there is a formula for the multiplicity of a given irreducible in a given representation.

Let $(V,\varphi)$ be any finite-dimensional unitary representation of $G.$ Either this representation is irreducible, or not. If it is, then $(V,\varphi)$ is isomorphic to $(V^\lambda,\varphi^\lambda)$ for some $\lambda \in \Lambda).$ If not, then it contains a proper subrepresentation, i.e. a nonzero subspace $V_1<V$ which is $G$ -stable.

Theorem 24.1 (Maschke’s Theorem). The orthogonal complement $V_1^\perp$ is also $G$ -stable.

Proof: For any $v_0 \in V_1^\perp$ and $v_1 \in V_1$ we have

$\langle \varphi^\lambda(g)v_0,v_1\rangle = \langle v_0,\varphi^\lambda(g^{-1})v_1\rangle = 0$

for all $g \in G.$ $\square$

We now have a decomposition $V=V_1 \oplus V_2$ of $V$ into two proper subrepresentations. Repeat this process on each of the “factors” $V_1$ and $V_2$ , and after finitely many steps you will produce a binary tree whose leaves are irreducible subrepresentations of $V$ . Each of these leaves is necessarily isomorphic to one of the “primes” $(V^\lambda,\varphi^\lambda).$ We thus have a “prime factorization”

$V \simeq \bigoplus\limits_{\lambda \in \Lambda} \mathrm{mult}(V^\lambda,V)V^\lambda,$

where $\mathrm{mult}(V^\lambda,V)$ is the number of leaves in the tree isomorphic to $V^\lambda.$ This is called the isotypic decomposition of $V$ . It remains to verify that however we decompose $(V,\varphi)$ into irreducibles, the end result will be the same.

Problem 24.1. State and prove a uniqueness theorem for “the” isotypic decomposition of a representation (hint: character orthogonality).

Theorem 24.2. We have

$\mathrm{mult}(V^\lambda,V)=\frac{1}{|G|}\langle \chi^\lambda,\chi^V\rangle.$

Proof: Taking traces on each side of the isotypic decomposition yields

$\chi^V = \sum\limits_{\lambda \in \Lambda} \mathrm{mult}(V^\lambda,V)\chi^\lambda,$

now use character orthogonality. $\square$

Problem 24.2. Prove that $\mathrm{mult}(V^\lambda,V) = \dim\mathrm{Hom}_G(V^\lambda,V).$

Now we are in position to prove that the character of a representation really does characterize it, up to isomorphism.

Theorem 24.3. Two unitary representation $(V,\varphi)$ and $(W,\psi)$ of $G$ are isomorphic if and only if $\chi^V=\chi^W.$

Proof: It is clear that isomorphic representations have the same character. For the converse, the fact that $\chi^V=\chi^W$ implies that $\langle \chi^\lambda,\chi^V\rangle = \langle \chi^\lambda,\chi^W\rangle$ for all $\lambda \in \Lambda,$ so that $V$ and $W$ have the same isotypic decomposition. $\square$

Theorem 24.4. A unitary representation $(V,\varphi)$ of $G$ is irreducible if and only if $\langle \chi^V,\chi^V \rangle=|G|.$

Proof: Let

$\chi^V = \sum\limits_{\lambda \in \Lambda} \mathrm{mult}(V^\lambda,V)\chi^\lambda$

denote the isotypic decomposition of $G$ in character form. We then have

$\langle \chi^V,\chi^V\rangle = |G|\sum\limits_{\lambda \in \Lambda} \mathrm{mult}(V^\lambda,V)^2,$

and you can take it from here. $\square$

Problem 24.3. Compute the isotypic decomposition of the regular representation.

Math 202B: Lecture 23

*** Problems in this lecture due March 15 ***

*** Please complete a course evaluation ***

Let $\mathbf{Rep}(G)$ be the category of finite-dimensional unitary representations $(V,\varphi)$ of a finite group $G$ .

Problem 23.1. One-dimensional unitary representations of $G$ are the same thing as multiplicative characters of $G.$ Explain this sentence carefully.

We also discussed a way to associate a scalar-valued function on $G$ to every unitary representation $(V,\varphi)$ of $G$ . Namely, the character of $(V,\varphi)$ is the function on $G$ defined by

$\chi^V(g)= \mathrm{Tr}\ \varphi(g), \quad g \in G.$

After objects come morphisms. We have been working on our understanding of the hom-spaces in $\mathbf{Rep}(G),$ which are vector spaces of linear transformations. Namely, for any two unitary representations $(V,\varphi)$ and $(W,\psi)$ of $G$ , the corresponding set of morphisms $\mathrm{Hom}_G(V,W)$ is the subspace of $\mathrm{Hom}(V,W)$ consisting of those linear transformations which satisfy

$T\circ \varphi(g) = \psi(g) \circ T, \quad g \in G.$

Transformations $T \in \mathrm{Hom}_G(V,W)$ are said to intertwine the actions $\varphi$ and $\psi.$ Equivalently, $\mathrm{Hom}_G(V,W)$ is the space of $G$ -equivariant linear maps from $V$ to $W.$ The first thing we did was prove a dimension formula, which we restate now for ease of reference.

Theorem 23.1. We have

$\dim \mathrm{Hom}_G(V,W) = \frac{1}{|G|} \langle \chi^V,\chi^W \rangle.$

This is at least reminiscent of our main result about multiplicative characters of a finite abelian group: they form an orthogonal basis of $\mathcal{C}(G).$ So, we want to see some sort of orthogonality emerging for characters of unitary representations which, once again, are higher-dimensional generalizations of multiplicative characters. We saw that the more general version of character orthogonality we are trying to develop is related to the concept of irreducibility.

Problem 23.2. Prove that if $G$ is abelian, all its irreducible representations are one-dimensional.

Last lecture, we proved the following.

Theorem 23.2 (Schur’s Lemma). If $(V,\varphi)$ and $(W,\psi)$ are irreducible unitary representations of $G,$ then every nonzero map $T \in \mathrm{Hom}_G(V,W)$ is an isomorphism.

As a consequence of Theorems 23.1 and 23.2 we have the following: if $(V,\varphi)$ and $(W,\psi)$ are irreducible and non-isomorphic, their characters are orthogonal: we have

$\chi^V,\chi^W \rangle = |G| \dim \mathrm{Hom}_G(V,W) =0.$

In particular, this generalizes character orthogonality as we know it for finite abelian groups. This leads to the following.

Conjecture 23.3. If $(V,\varphi)$ is an irreducible unitary representation of $G,$ then

$\langle \chi^V,\chi^V\rangle = |G|.$

The goal of this lecture is to prove Conjecture 23.3. First of all, because of Theorem 23.1, for any unitary representation $(V,\varphi)$ of $G$ we have

$\langle \chi^V,\chi^V\rangle = |G|\dim \mathrm{Hom}_G(V,V).$

So Conjecture 23.3 is equivalent to the following.

Conjecture 23.4. If $(V,\varphi)$ is an irreducible unitary representation of $G$ , then $\mathrm{End}_G(V) = \mathrm{Hom}_G(V,V)$ is one-dimensional.

First of all, let us clarify that $\mathrm{End}_G(V)$ is the space of all linear operators $T \in \mathrm{End}(V)$ such that

$T\rho(g)=\rho(g)T, \quad g \in G.$

As we have seen, this is in fact a subalgebra of $\mathrm{End}(V)$ , and e know that $\mathrm{End}_G(V)$ and every algebra has a unique one-dimensional subalgebra. So, Conjecture 23.4 is equivalent to the following.

Theorem 23.5. If $(V,\varphi)$ is an irreducible unitary representation of $G,$ then $\mathrm{End}_G(V) = \mathbb{C}I$ .

Proof: Let $T \in \mathrm{End}_G(V)$ and let $\alpha$ be an eigenvalue of $T.$ Then, $T - \alpha I$ is again in the algebra $\mathrm{End}_G(V),$ but at the same time $\mathrm{Ker}(T-\alpha I)$ is a nontrivial $G$ -stable subspace of $V$ . Since $(V,\varphi)$ is irreducible, $\mathrm{Ker}(T-\alpha I)=V$ , which is the same as $T=\alpha I.$ $\square$

The above argument relies on something we all take for granted. For any algebra $\mathcal{A}$ and any element $A \in \mathcal{A}$ , the spectrum of $A$ is the set

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

Theorem 23.6 (Fundamental Theorem of Algebra). Every $T \in \mathrm{End}(V)$ has nonempty spectrum $\mathrm{Spec}(T) \subset \mathbb{C}.$

Problem 23.3. Assuming the above, prove that the spectrum of any element of any finite-dimensional von Neumann algebra is nonempty.

The following generalizes Theorem 23.6.

Theorem 23.7 (Burnside’s Theorem). The only subalgebra $\mathcal{A} \leq \mathrm{End}(V)$ which admits no proper non-trivial stable subspace $W \leq V$ is $\mathcal{A}=\mathrm{End}(V).$

Let $(V,\varphi)$ be any unitary representation of $G,$ and consider the corresponding linear representation of $\mathcal{C}(G)$ defined by

$\Phi(E_g) = \varphi(g), \quad g \in G.$

Problem 23.4. Prove that $(V,\varphi)$ is irreducible if and only if $(V,\Phi)$ is irreducible.

Combining this problem with Burnside’s theorem, we have that if $(V,\varphi)$ and hence $(V,\Phi)$ is irreducible, then the image of $\mathcal{C}(G)$ in $\mathrm{End}(V)$ is $\mathrm{End}(V).$ Consequently, $\mathrm{End}_G(V)$ is the center of $\mathrm{End}(V)$ and a second proof of Theorem 23.5 follows if we prove directly that the center of an endomorphism algebra is minimal.

Theorem 23.8. The center of $\mathrm{End}(V)$ is $\mathbb{C}I.$

Proof: Let $X \subset V$ be an orthonormal basis and let $\{E_{yx} \colon x,y \in X\}$ be the corresponding elementary basis of $\mathrm{End}(V).$ Suppose that $A \in \mathrm{End}(V)$ commutes with all elementary operators. Then, for any $p,q \in X$ we have

$E_{pp}AE_{qq} = \sum\limits_{x,y \in X} \langle y,Ax\rangle E_{pp}E_{yx}E_{qq} = \sum\limits_{y \in X} \langle p,Ax\rangle E_{px}E_{qq} = \langle p,Aq \rangle E_{pq},$

and likewise

$E_{qq}AE_{pp} = \langle q,Ap \rangle E_{qp},$

so that

$p \neq q \implies \langle p,Aq\rangle = \langle q,Ap \rangle =0.$

Furthermore, we have

$E_{pq}AE_{qp} = \sum\limits_{x,y \in X} \langle y,Ax\rangle E_{pq}E_{yx}E_{qp} =\sum\limits_{x \in X} \langle q,Ax\rangle E_{px}E_{qp} = \langle q,Aq\rangle E_{pp},$

and also

$E_{pq}E_{qp}A = E_{pp}A = \sum\limits_{x,y \in X} \langle y,Ax \rangle E_{pp}E_{yx} = \sum\limits_{x \in X} \langle p,Ax\rangle E_{px} = \langle p,Ap\rangle E_{pp},$

which forces $\langle p,Ap\rangle = \langle q,Aq\rangle$ . $\square$

Math 202B: Lecture 22

Let $\mathbf{Rep}(G)$ be the category of finite-dimensional unitary representations of a finite group $G.$ As we discussed last week, $\mathbf{Rep}(G)$ is both “bigger” and “smaller” than the category $\mathbf{FHil}$ of finite-dimensional Hilbert spaces. What this means is that the same Hilbert space $V$ may appear in $\mathbf{Rep}(G)$ as two distinct representations $(V,\varphi)$ and $(V,\varphi')$ , but at the same time hom-spaces in $\mathbf{Rep}(G)$ may be smaller because, unlike $\mathrm{Hom}(V,W)$ , linear maps in $\mathrm{Hom}_G(V,W)$ must intertwine the actions $\varphi$ and $\psi$ of two unitary representations $(V,\varphi)$ and $(W,\psi)$ . In fact, we have already made this “smaller” feature quantitative by deriving the dimension formula

$\dim \mathrm{Hom}_G(V,W) = \frac{1}{|G|} \langle \chi^V,\chi^W\rangle.$

In this lecture we will compare $\mathrm{Hom}(V,W)$ and $\mathrm{Hom}_G(V,W)$ structurally as opposed to numerically.

One of the simplest things we can say about a linear map $T \in \mathrm{Hom}(V,W)$ is that its kernel and image are subspaces of $V$ and $W$ , respectively. It is therefore natural to hope that a $G$ -equivariant linear map $T \in \mathrm{Hom}_G(V,W)$ would have kernel and image that are “subrepresentations” of $V$ and $W$ , respectively. To make sense of this we first need to define subobjects in $\mathbf{Rep}(G).$

Definition 22.1. Let $(V,\varphi)$ be a unitary representation of $G,$ and let $W$ be a vector subspace of $V.$ We say that $W$ is $G$ -stable if $\varphi(g)w \in W$ for all $g \in G$ and $w \in W.$

Problem 22.1. Prove that if $T \in \mathrm{Hom}_G(V,W)$ then $\mathrm{Ker}(T)$ is a $G$ -stable subspace of $V$ and $\mathrm{Im}(T)$ is a $G$ -stable subspace of $W.$

The only reason that the above is not our definition of “subrepresentation” is that we require representations to be nonzero.

Definition 22.2. A subrepresentation of $(V,\varphi)$ is a nonzero $G$ -stable subspace $W \leq V.$ In particular, $(W,\varphi)$ is a unitary representation of $G$ in its own right provided we restrict each $\varphi(g)$ to $W.$

Now let us recall the main structure theorem we have for linear maps between finite-dimensional vector spaces, which was the main result of Math 202A.

Theorem (SVD): For any linear map $T \in \mathrm{Hom}(V,W)$ between finite-dimensional Hilbert spaces, there exist distinct positive numbers $\sigma_1> \dots > \sigma_r > 0$ and nonzero pairwise orthogonal subspaces $V_1,\dots,V_r$ and $W_1,\dots,W_r$ of $V$ and $W$ such that

$V = V_1 \oplus \dots \oplus V_r \oplus \mathrm{Ker}(T) \quad\text{and}\quad W=W_1 \oplus \dots \oplus W_r \oplus \mathrm{Im}(T)^\perp$

and the restriction of $T$ to $V_i$ has the form $T=\sigma_iU_i$ with $U_i \in \mathrm{Hom}(V_i,W_i)$ an isometric isomorphism.

It is an under-appreciated fact that the SVD holds in $\mathbf{Rep}(G)$ in a very natural way.

Theorem 22.3 ( $G$ -equivariant SVD). Let $(V,\varphi)$ and $(W,\psi)$ be unitary representations of $G$ and let $T \in \mathrm{Hom}_G(V,W)$ be a $G$ -equivariant linear map. Then, the left singular spaces $V_1,\dots,V_r$ of $T$ are subrepresentations of $V$ , the right singular spaces $W_1,\dots,W_r$ are subrepresentations of $W,$ and the restriction of $T$ to $V_i$ is an isomorphism in $\mathrm{Hom}_G(V_i,W_i).$ Moreover, the map $U_i = \sigma_iT_i|_{V_i}$ is an isometric isomorphism in $\mathrm{Hom}(V_i,W_i).$

In $\mathbf{FHil},$ a space $V$ whose only subspace is $V$ is necessarily one-dimensional. In $\mathbf{Rep}(G),$ the situation is not as simple.

Definition 22.1. A unitary representation $(V,\varphi)$ of $G$ is said to be irreducible if the only subrepresentation it contains is $(V,\varphi)$ itself.

Let $\mathbf{Irr}(G)$ be the full subcategory of $\mathbf{Rep}(G)$ whose objects are irreducible representations. The hom-spaces in this category are highly restricted.

Theorem 22.4 (Schur’s Lemma). If $(V,\varphi)$ and $(W,\psi)$ are irreducible unitary representations of $G$ , then any nonzero map $T \in \mathrm{Hom}_G(V,W)$ is an isomorphism.

Problem 22.2. Prove Schur’s Lemma (it follows directly from the $latex $G-equivariant SVD).$

Another way to state Schur’s Lemma is that if $(V,\varphi)$ and $(W,\psi)$ are irreducible representations which are not isomorphic, then

$\mathrm{Hom}_G(V,W) = \{0_{\mathrm{Hom}(V,W)}\}.$

Together with our dimension formula for $\mathrm{Hom}_G(V,W),$ this yields the following.

Corollary 22.5. The characters of two non-isomorphic irreducible unitary representations $(V,\varphi)$ and $(W,\psi)$ are orthogonal: we have $\langle \chi^V,\chi^W\rangle=0.$

Problem 22.3. Prove that the character of any unitary representation $(V,\varphi)$ of $G$ is a class function on $G.$ Prove that the characters of any two isomorphic representations are equal. Combining these facts, deduce an upper bound on the number of isomorphism classes of irreducible unitary representations of $G.$

Math 202B: Lecture 21

***Problems in this lecture due March 8***

Let $G$ be a finite group. The notion of adjoint representation plays a key role in understanding the unitary representation category $\mathbf{Rep}(G)$ because it leads to a dimension formula for its hom-spaces, $\mathrm{Hom}_G(V,W).$ We now explain this.

Definition 21.1. For any unitary representation $(V,\varphi)$ of $G$ , the space of $G$ -invariant vectors in $V$ is

$V^G = \{ v \in V \colon \varphi(g)v=v\text{ for all }g \in G\}.$

In other words, $V^G$ is the set of all vectors $v\in V$ which are fixed points of each unitary operator $\varphi(g)$ , $g \in G.$

Proposition 21.2. The space $V^G$ is a vector subspace of $V$ , which is moreover mapped into itself by each of the operators $\varphi(g)$ , $g \in G.$

Problem 21.1. Prove Proposition 21.2.

We now come to the key point, which is that the space $\mathrm{Hom}_G(V,W)$ of intertwining maps between two unitary representations $(V,\varphi)$ and $(W,\psi)$ of $G$ is the same thing as the space of $G$ -invariant vectors in the corresponding adjoint representation $(\mathrm{Hom}(V,W),\omega).$

Proposition 21.3. $\mathrm{Hom}(V,W)^G = \mathrm{Hom}_G(V,W).$

Proof: We have

$\mathrm{Hom}(V,W)^G = \{T \in \mathrm{Hom}(V,W) \colon \omega(g)T =T \text{ for all }g \in G\} \\ = \{T \in \mathrm{Hom}(V,W) \colon \psi(g)T\varphi(g)^* =T \text{ for all }g \in G\} \\ = \{T \in \mathrm{Hom}(V,W) \colon \psi(g)T =T\varphi(g) \text{ for all }g \in G\}.$

-QED

Proposition 21.3 reduces the problem of describing $\mathrm{Hom}_G(V,W)$ to the problem of describing $\mathrm{Hom}(V,W)^G.$ This is progress, because the problem of describing $V^G$ can be solved for any unitary representation $(V,\varphi)$ of $G.$

Theorem 21.4. (First Projection Formula) For any unitary representation $(V,\varphi)$ of $G,$ the operator $P \in \mathrm{End}(V)$ defined by averaging all operators in the representation,

$P = \frac{1}{|G|} \sum\limits_{g \in G} \varphi(g),$

is a selfadjoint projections whose image is $V^G.$

Problem 21.2. Prove Theorem 21.4, which is our third application of averaging over a group.

We will apply the First Projection Formula to the adjoint representation $(\mathrm{Hom}(V,W),\omega)$ corresponding to two given unitary representations $(V,\varphi)$ and $(V,\psi)$ of $G.$ On one hand, we have

$\dim \mathrm{Hom}(V,W)^G = \dim \mathrm{Hom}_G(V,W)$

as a direct consequence of Proposition 18.2. On the other hand, by the First Projection Formula the operator

$P = \frac{1}{|G|} \sum\limits_{g \in G} \omega(g)$

is the orthogonal projection of $\mathrm{Hom}(V,W)$ onto the subspace $\mathrm{Hom}(V,W)^G = \mathrm{Hom}_G(V,W),$ so that in particular

$\mathrm{Tr}\, P = \dim \mathrm{Hom}(V,W)^G = \dim \mathrm{Hom}_G(V,W).$

This is because, as you know from Math 202A, the trace of a projection is the dimension of its image. We want to compute this trace. Since

$\mathrm{Tr}P = \frac{1}{|G|} \sum\limits_{g \in G} \mathrm{Tr} \omega(g)=\frac{1}{|G|} \sum\limits_{g \in G}\chi^{\mathrm{Hom}(V,W)}(g),$

and since we showed in Lecture 20 that

$\chi^{\mathrm{Hom}(V,W)}(g)=\overline{\chi^V(g)}\chi^W(g),$

we have established the following.

Theorem 21.5. For any two unitary representations $(V,\varphi)$ and $(W,\psi)$ of $G,$ we have

$\dim \mathrm{Hom}_G(V,W) = \frac{1}{|G|}\langle \chi^V,\chi^W\rangle,$

where $\langle \cdot,\cdot \rangle$ is the $L^2$ -scalar product on $\mathcal{C}(G)$ .

Theorem 21.5 opens up a path to establishing a version of character orthogonality in $\mathcal{C}(G).$ Namely, we want to build a special subcategory of $\mathrm{Rep}(G)$ such that if $(V,\rho)$ and $(W,\psi)$ are distinct objects in this subcategory then the space $\mathrm{Hom}_G(V,W)$ of morphisms between them is the zero space. We will see how to do this next week using a structural (as opposed to numerical) description of $\mathrm{Hom}_G(V,W)$ based on the notion of irreducible representations – those which cannot be be decomposed into smaller ones.

Math 202B: Lecture 20

Let $G$ be a finite group. recall that a unitary representation of $G$ is a pair $(V,\rho)$ consisting of a nonzero Hilbert space $V$ together with a group homomorphism $\varphi \colon G \to U(V),$ where $U(V)$ is the group of unitary operators on $V.$

We are going to spend the rest of the course studying the category $\mathbf{Rep}(G)$ whose objects are finite-dimensional unitary representations of $G.$ Morphisms in $\mathbf{Rep}(G)$ are defined as follows.

Definition 20.1. Let $(V,\rho)$ and $(W,\psi)$ be unitary representations of $G.$ A linear transformation $T \in \mathrm{Hom}(V,W)$ is said to intertwine the actions $\varphi$ and $\psi$ if

$T \circ \varphi(g) = \psi(g) \circ T, \quad g \in G.$

Declare the set of $\mathrm{Hom}_G(V,W)$ of morphisms from $(V,\rho)$ to $(W,\psi)$ to be the set of intertwining maps as above.

We have now defined the category $\mathbf{Rep}(G)$ of unitary representations of $G.$ This category encodes all the ways in which $G$ can act on finite-dimensional Hilbert spaces. It has more objects than the category $\mathbf{FHil}$ of finite-dimensional Hilbert spaces, but its Hom-spaces are smaller.

Problem 20.1. Prove that $\mathrm{Hom}_G(V,W)$ is a subspace of $\mathrm{Hom}(V,W).$ Moreover, prove that if $T \in \mathrm{Hom}_G(V,W)$ then $T^* \in \mathrm{Hom}_G(W,V).$

Now let us discuss a few ways in which the definition of $\mathrm{Hom}_G(V,W)$ is well-chosen.

First consider $\mathrm{End}_G(V)=\mathrm{Hom}_G(V,V),$ the space of endomorphisms of a unitary representation $(V,\varphi)$ of $G.$ This is the subspace of $\mathrm{End}(V)=\mathrm{Hom}(V,V)$ consisting of linear operators $A$ on $V$ which commute with each of the unitary operators $\varphi(g),$

$T \varphi(g) = \varphi(g)T, \quad g \in G.$

Equivalently, if we let $(V,\Phi)$ be the linear representation of $\mathcal{C}(G)$ corresponding to the unitary representation $(V,\phi)$ via $\Phi(E_g)=\varphi(g)$ , then $\mathrm{End}_G(V)$ is the centralizer of the image of $\mathcal{C}(G)$ in $\mathrm{End}(V)$ under $\Phi.$ Thus, since the centralizer of any subalgebra is again a subalgebra, we have that $\mathrm{End}_G(V)$ is a subalgebra of $\mathrm{End}(V).$

Isomorphisms in $\mathbf{FHIl}$ are easy to understand: $\mathrm{Iso}(V,W)$ is nonempty if and only if $\dim V=\dim W,$ and it consists of all linear bijections $T \colon V \to W.$ We know from Math 202A that if $\mathrm{Iso}(V,W)$ is nonempty then there exists an isometry $U \in \mathrm{Iso}(V,W).$ The set $\mathrm{Iso}_G(V,W)$ of isomorphisms between unitary representations $(V,\varphi)$ and $(W,\psi)$ is more constrained, since it consists of linear bijections $T \colon V \to W$ which intertwine the actions $\varphi$ and $\psi,$ and it may be empty even if a linear bijection $V \to W$ exists.

Problem 20.2. Prove that $\mathrm{Iso}_G(V,W)$ is nonempty if and only if it contains an isometry $U \colon V \to W.$ (This can be done using either singular value decomposition or polar decomposition).

Problem 20.2 gives a matrix interpretation of what it means for two unitary representations $(V,\varphi)$ and $(W,\psi)$ to be isomorphic. Namely, let $U \in \mathrm{Iso}_G(V,W)$ be an isometry. Choose any orthonormal basis $X \subset V,$ and let $Y = \{Ux \colon x \in V\}$ be the corresponding orthonormal basis of $W.$ Then, we have

$\langle y',\psi(g)y\rangle_W = \langle Ux',\psi(g)Ux\rangle_W = \langle Ux',U\varphi(g)x\rangle_W=\langle x',\varphi(g)x\rangle_V$ .

In other words, for any ordering $x_1,\dots,x_n$ , taking the corresponding ordering $Ux_1,\dots,Ux_n$ of $Y$ we have that the $n \times n$ unitary matrices $[\varphi(g)]_X$ and $[\psi(g)]_Y$ coincide for all $g \in G.$ So, isomorphic representations of $G$ represent the elements of $G$ as exactly the same unitary matrices. An obvious but important consequence of this is the following.

Theorem 20.1. Isomorphic objects in $\mathbf{Rep}(G)$ have the same character.

One of the many miracles of group representation theory is that the converse of Theorem 20.1 holds: the character of a unitary representation characterizes it up to isomorphism (hence the name). As a first step towards establishing this result, we consider another important aspect of $\mathbf{Rep}(G)$ , namely that given any two objects we can build a third as follow.

Definition 20.2. Let $(V,\rho)$ and $(W,\psi)$ be unitary representations of $G.$ The corresponding adjoint representation $(\mathrm{Hom}(V,W),\omega)$ is defined by

$\omega(g)T=\psi(g)T\varphi(g^{-1}), \quad g \in G.$

Let us place this construction in a more general framework. Given two finite-dimensional Hilbert spaces $V$ and $W,$ equip $\mathrm{Hom}(V,W)$ with the Hilbert-Schmidt scalar product.

Definition 20.3. The adjoint action of $U(V) \times U(W)$ on $\mathrm{Hom}(V,W)$ is the group homomorphism

$\omega \colon U(V) \times U(W) \longrightarrow U(\mathrm{Hom}(V,W))$

defined by

$\omega(A,B)T=BTA^*.$

Problem 20.3. Check that $\omega(A,B)$ is indeed a unitary operator on $\mathrm{Hom}(V,W)$ , and that $\omega$ is indeed a group homomorphism.

Now let us calculate the matrix elements of $\omega(A,B) \in U(\mathrm{Hom}(V,W))$ in terms of the matrix elements of $A \in U(V)$ and $B\in U(W).$ To do so, let $X \subset V$ and $Y \subset W$ be orthonormal bases, and let $\{E_{yx} \colon x \in X, y \in Y\}$ be the corresponding orthonormal basis of $\mathrm{Hom}(V,W).$

Proposition 20.3. We have

$\langle E_{y'x'},\omega(A,B)E_{yx}\rangle = \overline{\langle x',Ax\rangle}{\langle y',By\rangle}.$

Proof: We have

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

which is the trace of an operator in $\mathrm{End}(W).$ We compute this trace as

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

Now expand

$A^*x' = \sum\limits_{x'' \in X} \langle x'',A^*x'\rangle x''$

to get

$\langle y',BE_{yx}A^*x'\rangle=\sum\limits_{x'' \in X} \langle y',BE_{yx}x''\rangle \langle x'',A^*x'\rangle=\langle y',By\rangle \langle x,A^*x'\rangle = \overline{\langle x',Ax \rangle}\langle y',By\rangle,$

which completes the proof. $\square$

Corollary 20.4. We have

$\mathrm{Tr}\omega(A,B) = \overline{\mathrm{Tr} A}\ \mathrm{Tr}B.$

Proof: Proposition 20.3 gives

$\mathrm{Tr}\omega(A,B) = \sum\limits_{x \in X}\sum\limits_{y \in Y}\langle E_{yx},\omega(A,B)E_{yx}\rangle =\sum\limits_{x \in X}\sum\limits_{y \in Y} \overline{\langle x,Ax \rangle}\langle y,By\rangle=\overline{\mathrm{Tr} A}\ \mathrm{Tr}B,$

as claimed. $\square$

From the above, we get the following relationship between the characters of two given unitary representations $(V,\varphi)$ and $(W,\psi),$ and the character of the corresponding adjoint representation $(\mathrm{Hom}(V,W),\omega).$

Corollary 20.5. For all $g \in G,$ we have

$\chi^{\mathrm{Hom}(V,W)}(g) = \overline{\chi^V(g)}\chi^W(g).$

Math 202B: Lecture 19

We now understand the convolution algebra $\mathcal{C}(G)$ very well for

$G=G_1 \times \dots \times G_r$

be a product group with $G_i$ a cyclic group of order $n_i.$ The basic mechanism is that we can parameterize both the group $G=\{g_\alpha \colon \mu \in \Lambda\}$ and the set $\widehat{G}=\{\chi^\lambda \colon \lambda \in \Lambda\}$ of multiplicative characters of $G$ by

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

in a very explicit and useful way. Namely, for every $\alpha,\lambda \in \Lambda$ we have

$\chi^\lambda(g_\alpha) = \omega_1^{\alpha_1\lambda_1} \dots \omega_r^{\alpha_r\lambda_r}, \quad \omega_k=\exp\left(\frac{2\pi i}{n_k}\right),$

and this is the complete list of homomorphisms $G \to U(1)$ from $G$ into the unit circle $U(1)$ in $\mathbb{C}.$ Using this description, we were able to show that $\widehat{G}=\{\chi^\lambda \in \lambda \in \Lambda\}$ is an orthogonal basis of $\mathcal{C}(G)$ ,

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

Moreover, the normalized characters

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

form a Fourier basis $\{F^\lambda \colon \lambda \in \Lambda\}$ of $\mathcal{C}(G)$ . Thus we get an algebra isomorphism $\mathcal{C}(G) \to \mathcal{F}(\Lambda),$ the Fourier transform.

Now let us compare the Fourier transform on $\mathcal{C}(G)$ with another kind of “transform” we have seen already, the regular representation of an algebra $\mathcal{A}$ equipped with a Frobenius scalar product $\langle \cdot,\cdot \rangle$ , i.e. a von Neumann algebra). In this setting we looked at the algebra homomorphism

$\Phi \colon \mathcal{A} \longrightarrow \mathrm{End}\mathcal{A}$

which sends each algebra element $A \in \mathcal{A}$ to the left multiplication operator

$\Phi(A)B=AB, \quad B \in \mathcal{A}.$

Specializing this construction to the case where $\mathcal{A}=\mathcal{C}(G)$ is the convolution algebra of a finite abelian group $G,$ we obtain the following spectral interpretation of the Fourier transform.

Theorem 19.1. For every $A \in \mathcal{C}(G),$ the Fourier basis $\{F^\lambda \colon \lambda \in \Lambda\}$ is an eigenbasis of $\Phi(A)$ . More precisely, we have

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

Proof: We have

$\Phi(A)F^\lambda = \left(\sum\limits_{\mu \in \Lambda} \widehat{A}(\mu)F^\mu\right)F^\lambda = \sum\limits_{\mu \in \Lambda} \widehat{A}(\mu)F^\mu F^\lambda = \widehat{A}(\lambda)F^\lambda,$

where we used the fact that $F^\lambda F^\mu = \delta_{\lambda\mu}F^\lambda.$ $\square$

Problem 19.1. Explicitly calculate the eigenvalues and eigenvectors of circulant matrices. Use the uncertainty principle to relate the sparsity (number of nonzero entries) and nullity (number of zero eigenvalues) of circulant matrices.

Since every finite abelian group is isomorphic to a product of cyclic groups, we understand $\mathcal{C}(G)$ very well for arbitrary finite abelian groups $G.$ On the other hand, we have seen that there are noncommutative groups which only admit a single “trivial” character sending every element to $1.$ Thus, we cannot expect the multiplicative characters of $G$ to provide us with such a complete theory of $\mathcal{C}(G)$ for nonabelian groups $G.$

We now take a leap of faith and generalize the character concept by looking at homomorphisms from $G$ into “noncommutative circles.” You might wonder how someone would make such an inspired guess, and the answer is that there was no particular “someone” who did. Rather, over time the work of many people gradually made it clear that this would be a good approach to develop. You can read more about this here. The resulting theory has now been gone over many times by successive generalizations of algebraists (any analysts, and physicists), resulting in a very elegant and streamlined flow of ideas which we will follow in our remaining lectures.

Definition 19.2. A unitary representation of $G$ is a pair $(V,\varphi)$ consisting of a Hilbert space $V$ together with a group homomorphism $\varphi \colon G \to U(V),$ where $U(V)$ is the unitary group of $V.$

If we take $V=\mathbb{C},$ a group homomorphism $\varphi \colon G \to \mathbb{U}(V)$ is the same thing as a multiplicative character of $G,$ so unitary representations are indeed generalizations of multiplicative characters. Note that Definition 19.1 makes sense with $G$ being any group, whether finite or infinite, and $V$ being any Hilbert space, whether finite or infinite dimensional. However, in Math 202B we will work exclusively with finite-dimensional unitary representations of finite groups.

The basic difference between multiplicative characters $\chi \colon G \to U(1)$ and unitary representations $\varphi \colon G \to U(V)$ is that $\chi$ is a scalar-valued function on $G$ whereas $\varphi$ is an operator-valued function on $G$ . Thus, $\chi \in \mathcal{C}(G)$ but $\varphi \not\in \mathcal{C}(G).$ However, the operator $\varphi(g)$ can be described by a matrix and as $g$ varies over $G$ this gives us a collection of functions in $\mathcal{C}(G).$

Definition 19.3. Let $(V,\varphi)$ be a unitary representation of $G$ and let $X \subset V$ be an orthonormal basis of $V$ . We then have a corresponding set of functions $\{\mu_{yx}^V \colon (x,y) \in X \times X\}$ in $\mathcal{C}(G)$ defined by

$\mu_{yx}^V(g) = \langle y,\varphi(g)x\rangle, \quad g \in G.$

The functions $\mu_{xy}^V \in \mathcal{C}(G)$ are called the matrix elements of the $(V,\varphi)$ relative to the basis $X.$

The matrix elements of $(V,\varphi)$ depend on the chosen basis $X \subset V,$ but as we know a particular combination of them does not.

Definition 19.3. The character of a unitary representation $(V,\varphi)$ of $G$ is the function $\chi^V \in \mathcal{C}(G)$ defined by

$\chi^V(g) =\mathrm{Tr}\ \varphi(g), \quad g \in G.$

Here is an example of a unitary representation $(V,\varphi)$ of a group $G.$ We take $V=\mathcal{C}(G)=\mathcal{F}(G)$ to be the space of complex-valued functions on $G$ with its $L^2$ -scalar product. We define a group homomorphism $\varphi \colon G \to U(V)$ by

$\varphi(g)A = E_gA, \quad A \in \mathcal{C}(G),$

where $E_g$ is the elementary function corresponding to $g \in G$ . Then, $\varphi(g)$ is indeed a linear operator on $V,$ since

$\varphi(g)(\alpha A +\beta B) = \alpha E_gA +\beta E_gB=\alpha \varphi(g)A + \beta \varphi(g)B.$

Moreover, the linear operator $\varphi(g) \in \mathrm{End}\mathcal{C}(G)$ is unitary because it permutes the elements of the orthonormal basis $\{E_g \colon g \in G\},$

$\varphi(g)E_h = E_gE_g = E_{gh}, \quad h \in G.$

Thus, the matrix of $\varphi(g)$ relative to the elementary basis $\{E_g \colon g \in G\}$ is not just a unitary matrix, it is a permutation matrix. This unitary representation is called the regular representation of $G,$ and it is “the same thing” as the regular representation of $\mathcal{C}(G)$ discussed above in the following sense: we have

$\Phi(E_g) = \varphi(g), \quad g \in G.$

Problem 19.2. Calculate the character of the regular representation of $G.$

For an arbitrary algebra $\mathcal{A}$ , a linear representation of $\mathcal{A}$ is a pair $(V,\varphi)$ consisting of a Hilbert space $V$ together with an algebra homomorphism

$\Phi \colon \mathcal{A} \longrightarrow \mathrm{End}(V).$

In the special case where $\mathcal{A}=\mathcal{C}(G)$ , every linear representation $(V,\Phi)$ of $\mathcal{C}(G)$ gives a unitary representation $(V,\varphi)$ of $G$ defined by

$\varphi(g)=\Phi(E_g),\quad g \in G.$

Conversely, every unitary representation $(V,\varphi)$ of $G$ gives a linear representation $(V,\Phi)$ of $\mathcal{C}(G)$ defined by

$\Phi(E_g) = \varphi(g), \quad g \in G.$

So, for convolution algebras $\mathcal{C}(G)$ , the study of linear representations of $\mathcal{C}(G)$ is equivalent to the study of unitary representations of $G.$

One could drop the “unitary” clause and consider a more general construction in which we define a representation of $G$ to be a pair $(V,\rho)$ consisting of a Hilbert space $V$ together with a group homomorphism

$\varphi \colon G \longrightarrow GL(V)$

from $G$ into the group of all invertible linear transformations $V \to V.$ Thus, we do not require $\varphi(g) \in GL(V)$ to preserve the scalar product $\langle \cdot,\cdot \rangle$ on $V.$ We will only consider unitary representations, and it turns out that there is in fact no loss of generality here.

Problem 19.3. Given a representation $(V,\varphi)$ of $G,$ define a new scalar product on $V$ by

$\langle v,w \rangle_G = \frac{1}{|G|}\sum\limits_{g \in G} \langle \varphi(g)v,\varphi(g)w\rangle.$

Show that each of the operators $\varphi(g)$ , $g \in G$ , is unitary with respect to this new scalar product, i.e.

$\langle \varphi(g)v,\varphi(g)w\rangle_G = \langle v,w\rangle_G, \quad v,w \in V.$

Problem 19.3 is our second use of the averaging trick, and we will see a third application of this technique very soon. In problem 19.3, it is not important that $V$ is finite-dimensional, but it is significant that $G$ is finite. This result does hold for certain infinite groups called compact groups, but for noncompact groups unitary representations actually are a special subclass of representations.

Math 202B: Lecture 18

*** Problems in this lecture due March 1 ***

Any algebra $\mathcal{A}$ which admits a basis $\{F^\lambda \colon \lambda \in \Lambda\}$ of orthogonal projections is isomorphic to a function algebra, namely $\mathcal{F}(\Lambda).$ The isomorphism $\mathcal{A} \to \mathcal{F}(\Lambda)$ is called the Fourier transform on $\mathcal{A}$ , and it sends $A \in \mathcal{A}$ to the function $\widehat{A} \in \mathcal{F}(\Lambda)$ uniquely determined by

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

We have implemented this abstract isomorphism concretely for the convolution algebra $\mathcal{C}(G)$ of an abelian group $G,$ which we take to be a product

$G = G_1 \times \dots \times G_r$

of cyclic groups. More precisely, $G_i$ is a cyclic group of order $n_i$ with generator $g_i.$ The dual group is then

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

and we label elements of $G$ with elements of $\Lambda$ using the isomorphism $\Lambda \to G$ which sends $\alpha=(\alpha_1,\dots,\alpha_r)$ to

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

Last lecture we showed that the functions

$\chi^\lambda(g_\alpha) = \omega_1^{\alpha_1\lambda_1} \dots \omega_r^{\alpha_r\lambda_r}, \quad \omega_k = \exp\left(\frac{2\pi i}{n_k}\right)$

are precisely the multiplicative characters of $G,$ and that $\{\chi^\lambda \colon \lambda \in \Lambda\}$ is an orthogonal basis of $\mathcal{C}(G),$

$\langle \chi^\lambda,\chi^\mu\rangle = \sum\limits_{g \in G} \overline{\chi^\lambda(g)}\chi^\mu(g)=\sum\limits_{g \in G} \chi^\lambda(g^{-1})\chi^\mu(g)=\delta_{\lambda\mu}|G|.$

The normalized characters

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

form a Fourier basis $\{F^\lambda \colon \lambda \in \Lambda\}$ . Thus for any function $A \in \mathcal{C}(G)$ we have a corresponding Fourier expansion,

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

and this in turn defines the Fourier transform $\mathcal{C}(G) \to \mathcal{F}(\Lambda),$ an algebra isomorphism.

Because we are working in a concrete setting, it is natural to ask for an explicit formula for the Fourier transform $\widehat{A} \in \mathcal{F}(\Lambda)$ of a given function $A \in \mathcal{C}(G).$

Theorem 18.1. For any $A \in \mathcal{C}(G),$ we have

$\widehat{A}(\lambda) = \sum\limits_{g \in G} \overline{\chi^\lambda(g)}A(g), \quad \lambda \in \Lambda.$

Problem 18.1. Prove Theorem 18.1.

For example, let us compute the Fourier transform of an elementary function $E_g \in \mathcal{C}(G).$ Theorem 18.1 gives

$\widehat{E}_g(\lambda) = \frac{1}{|G|}\sum\limits_{h \in G} \overline{\chi^\lambda(h)}E_g(h) = \frac{1}{|G|}\overline{\chi^\lambda(h)}, \quad \lambda \in \Lambda.$

Notice that the function $E_g$ is highly concentrated: it is nonzero at a single point, namely $g \in G$ , where it has value $1.$ However, its Fourier transform $\widehat{E}_g$ is very spread out: all of its values are complex numbers of modulus one, so it never equals zero. We will see below that this is a general phenomenon.

Because the Fourier transform is an isomorphism $\mathcal{C}(G) \to \mathcal{F}(\Lambda),$ we can also ask for a formula which gives $A$ in terms of $\widehat{A}.$

Theorem 18.2 (Fourier inversion). For any $A \in \mathcal{C}(G)$ , we have

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

Proof: The elementary expansion of $A$ is

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

while its Fourier expansion is

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

Comparing the two yields the stated formula. $\square$

The Fourier inversion formula gives a useful extremal result: the maximum modulus of a function $A \in \mathcal{C}(G)$ is bounded by the average modulus of its Fourier transform $\widehat{A} \in \mathcal{C}(G).$

Corollary 18.3 (Fourier bound). For all $g \in G,$ we have

$|A(g)|\leq \frac{1}{|\Lambda|} \sum\limits_{\lambda \in \Lambda} |\widehat{A}(\lambda)|.$

The above can also be written as the inequality

$\|A\|_\infty \leq \frac{1}{|G|}\|A\|_1.$

More generally, for any $p >0$ we have the corresponding $p$ -norm on $\mathcal{C}(G)$ defined by

$\|A\|_p = \left( \sum_{g \in G} |A(g)|^p\right)^{1/p}.$

The sup norm of a function $A \in \mathcal{C}(G)$ is defined by

$\|A\|_\infty = \max\limits_{g \in G} |A(g)|,$

and heuristically it is the case $p=\infty$ of the $p$ -norm.

Theorem 18.4. (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

Proposition 18.5. 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 18.2. Prove Proposition 18.5.

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 18.6. (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