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.

Math 202B: Lecture 3

Let \mathcal{A} be an algebra.

Definition 3.1. A basis of orthogonal projections in \mathcal{A} is said to be a Fourier basis of \mathcal{A}.

Let X be a finite nonempty set.

Definition 3.2. The function algebra of X is the vector space

\mathcal{F}(X) = \{A \colon X \to \mathbb{C}\}

of \mathbb{C}-valued functions on X with multiplication defined by

[AB](x) = A(x)B(x), \quad x \in X,

and conjugation defined by

A^*(x) = \overline{A(x)}, \quad x \in X.

One says that the operations in \mathcal{F}(X) are defined pointwise.

Problem 3.1. Prove that \mathcal{F}(X) is indeed an algebra.

There is a natural Fourier basis in \mathcal{F}(X) indexed by the points of X. For each x \in X, define the corresponding elementary function E_x \in \mathcal{F}(X) by

E_x(y) = \begin{cases} 1, \text{ if }x=y \\ 0,\text{ if }x \neq y\end{cases}.

That is, E_x(y) = \delta_{xy} where \delta_{xy} is the Kronecker delta. The elementary functions are selfadjoint elements of \mathcal{F}(X) because they are real-valued, and they are orthogonal projections because

[E_xE_y](z)=E_x(z)E_y(z)=\delta_{xz}\delta_{yz}=\delta_{xy}E_x(z).

Thus, \{E_x \colon x \in X\} is a linearly independent set in \mathcal{F}(X) by Theorem 1.1 in Lecture 1. To see that \{E_x \colon x \in X\} spans \mathcal{F}(X), observe that any function A \in \mathcal{F}(X) can be written as

A = \sum\limits_{x \in X} A(x)E_x.

Using the elementary basis \{E_x \colon x \in X\} of \mathcal{F}(X), we can forget that the elements of this algebra are functions on X and view them as linear combinations

A = \sum\limits_{x\in X} \alpha_xE_x, \quad \alpha_x \in \mathbb{C}.

Combining the fact that the elementary functions are orthogonal projections with the axioms of bilinearity and antilinearity, we recover multiplication and conjugation in \mathcal{F}(X) in the form

AB = \left(\sum\limits_{x \in X} \alpha_xE_x\right)\left(\sum\limits_{x \in X} \beta_xE_x\right)=\sum\limits_{x,y\in X}\alpha_x\beta_yE_xE_y = \sum\limits_{x \in X} \alpha_x\beta_xE_x

and

A^*= \left(\sum\limits_{x \in X} \alpha_xE_x\right)^*=\sum\limits_{x \in X} \bar{\alpha}_xE_x^* = \sum\limits_{x \in X} \bar{\alpha}_xE_x.

Two sets X and Y are said to be isomorphic if there exists a bijection between them.

Problem 3.1. Prove that two finite nonempty sets X and Y are isomorphic if and only if their function algebras \mathcal{F}(X) and \mathcal{F}(Y) are isomorphic.

Theorem 3.1. An algebra \mathcal{A} admits a Fourier basis if and only if it is isomorphic to a function algebra.

Proof: We have already shown that a function algebra has a Fourier basis. Conversely, let \mathcal{A} be an algebra, and let \{F^\lambda \colon \lambda \in \Lambda\} be a vector space basis of \mathcal{A} indexed by the points of a finite set \Lambda whose cardinality is the dimension of \mathcal{A}. Let \mathcal{F}(\Lambda) be the function algebra of this set, and define a linear transformation

\mathsf{T} \colon \mathcal{A} \longrightarrow \mathcal{F}(\Lambda)

by \mathsf{T}(F^\lambda)=E_\lambda, where E_\lambda \in \mathcal{F}(\Lambda) is the elementary function corresponding to \lambda \in \Lambda. This is a vector space isomorphism, since it maps a basis of \mathcal{A} onto a basis of \mathcal{F}(\Lambda). If \{F^\lambda \colon \lambda \in \Lambda\} is a basis of orthogonal projections in \mathcal{A}, then \mathsf{T} is also an algebra homomorphism. Indeed, by Theorem 1.2 and linearity of \mathsf{T} we have

\mathsf{T}(I_\mathcal{A}) = \sum\limits_{\lambda \in \Lambda} \mathsf{T}(F^\lambda) = \sum\limits_{\lambda \in \Lambda} E_\lambda = I_{\mathcal{F}(\Lambda)},

so \mathsf{T} maps the multiplicative unit of \mathcal{A} to that of \mathcal{F}(\Lambda). Next,

\mathsf{T}(F^\lambda F^\mu) = \mathsf{T}(\delta_{\lambda\mu}F^\lambda)=\delta_{\lambda\mu}E_\lambda = E_\lambda E_\mu = \mathsf{T}(F^\lambda)\mathsf{T}(F^\mu),

so \mathsf{T} respects multiplication. Finally,

\mathsf{T}((F^\lambda)^*)=\mathsf{T}(F^\lambda)=E_\lambda=E_\lambda^*=\mathsf{T}(F^\lambda)^*.

-QED

Theorem 3.1 is simple but important and you should go over its proof carefully. The argument shows that if \{F^\lambda \colon \lambda \in \Lambda\} is a Fourier basis of \mathcal{A}, then the linear transformation

\mathsf{T} \colon \mathcal{A} \longrightarrow \mathcal{F}(\Lambda)

defined by

\mathsf{T}(F^\lambda) = E_\lambda, \quad \lambda \in \Lambda,

is an algebra isomorphism. This isomorphism is called the Fourier transform on \mathcal{A}, and it is generally denoted as A \mapsto \widehat{A} rather than A \mapsto \mathsf{T}(A). Thus,

\widehat{F^\lambda} = E_\lambda,\quad \lambda \in \Lambda,

and the argument above shows that the linear isomorphism \mathcal{A} \to \mathcal{F}(\Lambda) so defined is an algebra isomorphism. Note that the Fourier transform on \mathcal{A} is not canonical – it is defined in terms of a specified basis \{F^\lambda \colon \lambda \in \Lambda\} of orthogonal projections in \mathcal{A}. For any element A \in \mathcal{A}, its expansion in the Fourier basis is written

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

and the coefficients in this expansion are called the Fourier coefficients of A. We thus have

\widehat{A} = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)E_\lambda,

which is the elementary expansion of a function \widehat{A} \in \mathcal{F}(\Lambda). The function \widehat{A} \in \mathcal{F}(\Lambda) is called the Fourier transform of the algebra element A \in \mathcal{A}. The practical value of the Fourier transform lies in the fact that

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

where A*B is the product of A,B \in \mathcal{A}, which might be convoluted and difficult to calculate. On the other hand, the pointwise product of the corresponding functions \widehat{A},\widehat{B}\in \mathcal{F}(\Lambda) is very simple, both conceptually and computationally.

Math 202B: Lecture 1

This course is the second quarter of Math 202, a three-quarter graduate course sequence in applied algebra at UCSD. Briefly, the 202 sequence is arranged as follows.

Math 202A (Fall): vectors and transformations.

Math 202B (Winter): algebras and representations.

Math 202C (Spring): tensors and invariants.

In Math 202B, the term “vector space” will always mean a finite-dimensional complex vector space.

Definition 1.1. An algebra is a vector space \mathcal{A} of positive dimension equipped with an associative, bilinear, unital multiplication and an antilinear, antimultiplicative, involutive conjugation.

The prototypical example of an algebra is \mathcal{A}=\mathbb{C}, the complex number system, elements of which are called scalars and denoted by lower-case Greek letters,

\alpha,\beta,\gamma, \dots, \omega.

There are some exceptions: integers like zero and one are denoted 0 and 1 as usual, and we write i for the imaginary unit. Elements of a general algebra \mathcal{A} are denoted by upper-case Roman letters,

A,B,C,\dots,Z.

Multiplication in \mathcal{A} is a function \mathcal{A} \times \mathcal{A} \to \mathcal{A} whose values are denoted by concatenating its arguments: (A,B) \mapsto AB. Associativity means that the symbol ABC is unambiguous because its two possible meanings coincide:

(AB)C = A(BC).

Bilinearity means that multiplication in \mathcal{A} interacts with its vector space structure according to the rule

(\alpha_1A_1+\alpha_2A_2)(\beta_1B_1+\beta_2B_2) = \alpha_1\beta_1A_1B_1+\alpha_1\beta_2A_1B_2 + \alpha_2\beta_1A_2B_1+\alpha_2\beta_2A_2B_2.

We do not assume multiplication is commutative.

Problem 1.1. Let 0_\mathcal{A} denote the zero vector in \mathcal{A}. Prove that A0_\mathcal{A}=0_\mathcal{A}A=0_\mathcal{A} for all A \in \mathcal{A}.

Later, when we are more familiar with algebras and there is less chance of confusion, we will sometimes omit the subscript and write 0 for the zero vector in a general algebra \mathcal{A}, as it will generally be clear from context whether this symbol represents a scalar or a vector.

Unital means that there exists a vector I \in \mathcal{A} such that

IA=AI=A

for all A \in \mathcal{A}. Any such vector is called a multiplicative unit. Note that because the dimension of \mathcal{A} is positive, any multiplicative unit I is distinct from the additive unit 0_\mathcal{A}. In fact, there is only one multiplicative unit.

Problem 1.2. Let I,J be multiplicative units in \mathcal{A}. Prove that I=J.

Henceforth we write I_\mathcal{A} for the unique multiplicative unit. Later on, we may omit the subscript and simply write I for the multiplicative unit if it causes no confusion to do so. An element A \in \mathcal{A} is said to be invertible if there exists B \in \mathcal{A} such that AB=BA=I_\mathcal{A}.

Problem 1.3. Suppose A,B,C \in \mathcal{A} are such that AB=BA=I_\mathcal{A} and AC=CA=I_\mathcal{A}. Prove that B=C.

When AB=BA=I_\mathcal{A} we say that B is the inverse of A, and that A is the inverse of B. This is written B=A^{-1} and A=B^{-1}.

Multiplication in an algebra can be described numerically as follows. Let \{E_x \colon x \in X\} be a vector space basis of \mathcal{A} indexed by the points of some finite nonempty set X. Then, A,B \in \mathcal{A} can be represented as a linear combinations

A = \sum\limits_{x \in X} \alpha_x E_x\quad\text{and}\quad B = \sum\limits_{x \in X}\beta_x E_x.

According to bilinearity we have

AB = \sum\limits_{x,y \in X} \alpha_x\beta_y E_xE_y.

Each product of basis vectors can also be resolved into a linear combination of basis vectors,

E_xE_y = \sum\limits_{z \in X} \gamma_{xyz} E_z.

As the indices x,y,z range over X we get a three dimensional array [\gamma_{xyz}] of complex numbers called the multiplication tensor of \mathcal{A} relative to the basis E_x, x \in X. The elements of this three-tensor are called the connection coefficients of \mathcal{A} relative to the basis \{E_x \colon x \in X\}. This set of (\dim V)^3 numbers completely determines multiplication in \mathcal{A}, since

AB = \sum\limits_{x,y,z \in X} \alpha_x\beta_y\gamma_{xyz}E_z.

From a practical perspective, one would like to find a vector space basis of \mathcal{A} such that the corresponding multiplication tensor is sparse, i.e. many connection coefficients are zero, so that the computational cost of performing multiplication is minimized – this is the basic idea behind Strassen’s algorithm for matrix multiplication.

Problem 1.4. Prove that a two-dimensional algebra must be commutative.

Conjugation is a function \mathcal{A} \to \mathcal{A} whose values are denoted by a superscript asterisk: A \mapsto A^*. Antilinearity means that conjugation interacts with the vector space operations according to the rule

(\alpha A +\beta B)^* = \overline{\alpha}A^* + \overline{\beta}B^*.

Antimultiplicativity means that conjugation interacts with multiplication according to the rule

(AB)^*=B^*A^*.

Involutive means that conjugation is 2-periodic,

(A^*)^*=A.

Just like multiplication, conjugation in \mathcal{A} can be described with respect to a linear basis \{E_x \colon x \in X\}. Indeed, for each basis vector we can write its conjugate as

E_x^* = \sum\limits_{y \in X}\eta_{xy}E_y.

This gives a two-dimensional array which completely describes conjugation in \mathcal{A}, the conjugation tensor [\eta_{xy}] relative to the the basis \{E_x \colon x \in X\}. For any

A=\sum\limits_{x \in X} \alpha_x E_x,

then

A^*=\sum\limits_{x,y \in X}\overline{\alpha}_x \eta_{xy}E_y.

Problem 1.5. Prove that the set I(\mathcal{A}) of invertible elements in an algebra \mathcal{A} is a multiplicative group. Moreover, prove that I(\mathcal{A}) is closed under conjugation: A is invertible if and only if A^* is invertible, and in fact (A^*)^{-1} = (A^{-1})^*.

In any algebra \mathcal{A}, we define the following element classes:

  • Selfadjoint: X^*=X.
  • Idempotent: P^2=P.
  • Unitary: U^*U=UU^*=I.
  • Normal : A^*A=AA^*.

Problem 1.6. Prove the the set H(\mathcal{A}) of all selfadjoint elements in an algebra \mathcal{A} is an additive group, and in fact a real vector space. Show that every A \in \mathcal{A} can be written uniquely in the form A= X+iY with X,Y selfadjoint. We say that X is the real part of A, and that Y is its imaginary part.

Definition 1.2. A nonzero selfadjoint idempotent P \in \mathcal{A} is called a projection. Projections P,Q \in \mathcal{A} are said to be orthogonal if PQ=0_\mathcal{A}.

Sets of pairwise orthogonal projections play an important role in the study of algebras.

Theorem 1.1. Any set of pairwise orthogonal projections in an algebra \mathcal{A} is linearly independent.

Proof: Let \{E_x \colon x \in X\} be a set of pairwise orthogonal projections in \mathcal{A} indexed by the elements of some set X. Thus E_x \neq 0 are such that E_xE_y = \delta_{xy}E_x, where \delta_{xy} is the Kronecker delta. Let

A = \sum\limits_{x \in X} \alpha_x E_x

be a vector in the span of \{E_x \colon x \in X\}. Then, for any y \in X we have

AE_y = \sum\limits_{x \in X}\alpha_xE_xE_y =\alpha_yE_y.

Thus if A=0_\mathcal{A}, we must have \alpha_x = 0 for each x \in X.

-QED

According to Theorem 1.1, the maximum cardinality of a set of pairwise orthogonal projections in \mathcal{A} is \dim \mathcal{A}.

Definition 1.3. A basis of \mathcal{A} consisting of pairwise orthogonal projections is called a Fourier basis.

If \mathcal{A} admits a Fourier basis, it is a commutative algebra, and the corresponding conjugation and multiplication tensors are the two- and three-dimensional identity matrices. In this sense, algebras which admit a Fourier basis are the simplest algebras.

Theorem 1.2. Let \{E_x \colon x \in X\} be a Fourier basis of \mathcal{A}. Then,

I_\mathcal{A}= \sum\limits_{x \in X} E_x.

Proof: Take any A \in \mathcal{A} and let

A = \sum\limits_{x \in X} \alpha_x E_x

be its expansion in the given basis. Then, we have

\left(\sum_{x \in X} E_x\right)A = \sum\limits_{x,y \in X}\alpha_yE_xE_y = \sum\limits_{x \in X}\alpha_xE_x=A

and

A\left(\sum\limits_{y \in X} E_y\right) = \sum\limits_{x,y \in X} \alpha_xE_xE_y = \sum\limits_{x \in X} \alpha_x = A.

By uniqueness of the multiplicative unit in \mathcal{A}, we conclude that \sum\limits_{x \in X} E_x = I_\mathcal{A}.

-QED

Just as selfadjoint elements in \mathcal{A} are analogous to real numbers, unitary elements in \mathcal{A} are analogous to complex numbers of modulus one.

Problem 1.8. Prove that the set U(\mathcal{A}) of all unitary elements in \mathcal{A} is a subgroup of I(\mathcal{A}). We call U(\mathcal{A}) the unitary group of \mathcal{A}.

As for normal elements, these are in bijection with pairs of commuting selfadjoint elements.

Theorem 1.3. Given A \in \mathcal{A}, let A=X+iY be its decomposition into real and imaginary parts. Then A is normal if and only if X and Y commute.

Proof: Suppose first that X and Y are commuting selfadjoint elements. We will prove that A=X+iY is normal. We have

A^*A = (X+iY)^*(X+iY) = (X-iY)(X+iY) = XX +iXY-iYX+YY

and

AA^*= (X+iY)(X+iY)^* = (X+iY)(X-iY) = XX-iXY+iYX+YY,

so

A^*A-AA^*=i(XY-YX)-i(YX-XY)=0.

Now suppose that A=X+iY is a normal element. We have

XY = \frac{A+A^*}{2}\frac{A-A^*}{2i} = \frac{AA-AA^*+A^*A-A^*A^*}{4i} = \frac{AA-A^*A^*}{4i}

and

YX = \frac{A-A^*}{2i}\frac{A+A^*}{2}=\frac{AA+AA^*-A^*A-A^*A^*}{4i} = \frac{AA-A^*A^*}{4i}.

The two expressions agree: XY=YX.

-QED

Commutativity of real and imaginary parts characterizes normalcy at the level of elements. Normalcy itself characterizes commutativity at the level of algebras.

Theorem 1.4. An algebra is commutative if and only if all its elements are normal.

Proof: One direction is obvious: if \mathcal{A} is a commutative algebra, then certainly every element commutes with its conjugate.

Conversely, suppose that every element of \mathcal{A} is normal. Let X,Y \in \mathcal{A} be any two selfadjoint elements, and set A=X+iY. Then, since A is normal, we have

A^*A-AA^* =2i(XY-YX)=0,

which shows that XY=YX. Since X,Y were arbitrary selfadjoint elements of \mathcal{A}, we have shown that any two selfadjoint elements of \mathcal{A} commute. It remains to show that A_1,A_2 \in \mathcal{A} commute even if they are not selfadjoint. Then we can write A_1=X_1+iY_1 and A_2=X_2+iY_2 where X_1,Y_1,X_2,Y_2 are selfadjoint and thus commute with one another. Thus,

A_1A_2=(X_1+iY_1)(X_2+iY_2) = (X_1X_2-Y_1Y_2)+i(X_1Y_2+Y_1X_2)

and

A_2A_1=(X_2+iY_2)(X_1+iY_1)=(X_2X_1-Y_2Y_1)+i(X_2Y_1+Y_2X_1)

are equal.

-QED

Now let us consider functions between possibly different algebras \mathcal{A} and \mathcal{B}.

Definition 1.4. A linear transformation \mathsf{T} \colon \mathcal{A} \to \mathcal{B} is said to be an algebra homomorphism if

\mathsf{T}(I_\mathcal{A}) = I_\mathcal{B}

and

\mathsf{T}(A_1A_2)=\mathsf{T}(A_1)\mathsf{T}(A_2), \quad \text{for all }A_1,A_2 \in \mathcal{A},

and moreover

\mathsf{T}(A^*)=\mathsf{T}(A)^*, \quad \text{for all }A \in \mathcal{A}.

We say that \mathcal{A} and \mathcal{B} are isomorphic if there is \mathsf{T} \colon \mathcal{A} \to \mathcal{B} which is both a vector space isomorphism and an algebra homomorphism; such a map is called an algebra isomorphism.

The word “isomorphic” means “same shape” in Greek. Two objects which have the same shape need not be the same in all ways, and similarly saying that two algebras are isomorphic should not be taken to mean that they are the same set. To emphasize this distinction, one writes \mathcal{A} \simeq \mathcal{B} to indicate that \mathcal{A} and \mathcal{B} are isomorphic algebras.

Problem 1.9. Prove that every one-dimensional algebra \mathcal{A} is isomorphic to the complex number system \mathbb{C}.

As stipulated above, all vector spaces (and hence all algebras) in Math 202B are defined over \mathbb{C}. You may wonder about algebras with real scalars, and as we now explain these can be naturally included in our framework. Let \mathcal{B} be a real algebra, i.e. a finite-dimensional vector space over \mathbb{R} together with an associative, bilinear, unital multiplication and a linear, involutive conjugation.

Definition 1.5. The complexification of \mathcal{B} is the algebra \mathcal{A} whose elements A are ordered pairs of elements X,Y \in \mathcal{B}. We write A=(X,Y) as A=X+iY and define algebraic operations in \mathcal{A} from those in \mathcal{B} as follows: for \alpha,\beta \in \mathbb{R} and X_1,X_2,Y_1,Y_2 \in \mathcal{B} we declare

(X_1+iY_1)+(X_2+iY_2) = (X_1+X_2) + i(Y_1+Y_2),

(\alpha + i\beta)(X+iY) = (\alpha X-\beta Y)+i(\beta X + \alpha Y),

(X_1+iY_1)(X_2+iY_2) = (X_1X_2-Y_1Y_2) + i(X_1Y_2+Y_1X_2)

(X+iY)^*=X^*-iY^*.

Problem 1.10. Prove that Definition 1.5 does indeed define an algebra in the sense of Definition 1.1.

We say that an element of the complexificiation \mathcal{A} of a real algebra \mathcal{B} is real if it has the form A = X +i0_\mathcal{B} for some X \in \mathcal{B}.

Theorem 1.5. The complexification \mathcal{A} of a real algebra \mathcal{B} is commutative if every real element of \mathcal{A} is selfadjoint.

Proof: Let A_1=X_1+i0_\mathcal{B} and A_2=X_2+i0_\mathcal{B} be real elements of \mathcal{A}. Then, the product A_1A_2 = X_1X_2+i0_\mathcal{B} is also a real element of \mathcal{A}. By hypothesis, A_1,A_2, and A_1A_2 are selfadjoint elements of \mathcal{A}, and therefore

A_1A_2=(A_1A_2)* =A_2^*A_1^*=A_2A_1.

Now writing A=A_1+iA_2, we have that the real part of A \in \mathcal{A} is A_1 and its imaginary part is A_2. Since A_1,A_2 commute, A is normal, by Theorem 1.3. Thus every element of \mathcal{A} is normal, hence \mathcal{A} is commutative by Theorem 1.4.

-QED

Math 202B: Lecture 2

In this lecture we move beyond the commutative vs noncommutative dichotomy and quantify the degree of (non)commutativity of a given algebra \mathcal{A}. This is done using the notion of a subalgebra.

Definition 2.1. A subspace \mathcal{B} of \mathcal{A} is called a subalgebra if it contains I_\mathcal{A} and is closed under multiplication and conjugation.

Being a subalgebra is a stronger condition than being a subspace. In particular, the zero subspace \{0_\mathcal{A}\} of \mathcal{A} is not a subalgebra because it does not contain I_\mathcal{A}. Indeed, every subalgebra of \mathcal{A} must contain the one-dimensional subspace

\mathbb{C}I_\mathcal{A} = \{\alpha I_\mathcal{A} \colon \alpha \in \mathbb{C}\}

of scalar multiples of I_\mathcal{A}, which is a commutative subalgebra of \mathcal{A} isomorphic to \mathbb{C}. In this sense, \mathbb{C}I_\mathcal{A} is the smallest subalgebra of \mathcal{A}, and not only is it commutative but its elements commute with all elements in \mathcal{A}. We can consider the set of all elements which have the “commute with everything” property.

Definition 2.2. The center of \mathcal{A} is the set

Z(\mathcal{A})=\{Z \in \mathcal{A} \colon ZA=AZ \text{ for all }A \in \mathcal{A}\}.

Elements of Z(\mathcal{A}) are called central elements.

The use of the letter “Z” here comes from the German word for “center,” which is “zentrum.” Don’t forget to take your zentrum.

Theorem 2.1. The center Z(\mathcal{A}) is a subalgebra of \mathcal{A}.

Proof: We have to check that the conditions stipulated by Definition 2.1 hold for Z(\mathcal{A}) – it would be bad if the center did not hold. The argument is straightforward but also worth spelling out in detail, since all the defining features of an algebra (Definition 1.1) are used.

First we need to verify that Z(\mathcal{A}) is a vector subspace of \mathcal{A}, i.e. that it is closed under linear combinations. Let Z_1,Z_2 \in Z(\mathcal{A}) be any two central elements and A \in \mathcal{A} be any element. For any two scalars \alpha_1,\alpha_2 \in \mathbb{C}, we have

(\alpha_1 Z_1 + \alpha_2 Z_2)A = \alpha_1 Z_1A + \alpha_2 Z_2A = \alpha_1 AZ_1 + \alpha_2 AZ_2 = A(\alpha_1 Z_1 + \alpha_2 Z_2).

Now we check that Z(\mathcal{A}) is closed under multiplication:

Z_1Z_2A = Z_1AZ_2=AZ_1Z_2.

Now we check that Z(\mathcal{A}) is closed under conjugation:

Z^*A = Z^*(A^*)^*=(A^*Z)^*=(ZA^*)^*=AZ^*.

Finally, it is clear that I_\mathcal{A} \in Z(\mathcal{A}).

-QED

We can now quantify commutativity.

Definition 2.3. The commutativity index of \mathcal{A} is the dimension of Z(\mathcal{A}).

The commutativity index of \mathcal{A} is a positive integer between 1 (minimally commutative, maximally noncommutative) and \dim \mathcal{A} (maximally commutative, minimally noncommutative). Any algebra \mathcal{A} whose commutativity index is less than \dim \mathcal{A} is called noncommutative, which is a bit misleading because every algebra contains (uncountably) many elements which commute with one another. Algebras with one-dimensional center Z(\mathcal{A})=\mathbb{C}I have the lowest commutativity index; such maximally noncommutative algebras are called central-simple algebras.

We now generalize the center of an algebra as follows.

Definition 2.4. Given a subalgebra \mathcal{B} of \mathcal{A}, its centralizer Z(\mathcal{B},\mathcal{A}) is the set of all elements in \mathcal{A} which commute with every element of \mathcal{B}:

Z(\mathcal{B},\mathcal{A}) = \{A \in \mathcal{A} \colon AB=BA \text{ for all }B \in \mathcal{B}\}.

The symbol Z(\mathcal{B},\mathcal{A}) is read as “the centralizer of \mathcal{B} in \mathcal{A}.’’ In particular, the centralizer of \mathcal{A} in \mathcal{A} is Z(\mathcal{A},\mathcal{A})=Z(\mathcal{A}).

Problem 2.1. Prove that \mathcal{Z}(\mathcal{B},\mathcal{A}) is indeed a subalgebra of \mathcal{A}. Moreover, show that if \mathcal{B},\mathcal{C} are subalgebras of \mathcal{A} with \mathcal{B} \subseteq \mathcal{C}, then Z(\mathcal{B},\mathcal{A}) \supseteq Z(\mathcal{C},\mathcal{A}).

Problem 2.1 suggests a way to organize the set of all subalgebras of a given algebra \mathcal{A}. We know that this is a nonempty set, whose “smallest” element is \mathbb{C}I and whose “largest” element is \mathcal{A} itself — we want to order everything in between.

Defintion 2.3. Given a set \Omega, a relation on \Omega is a subset \mathrm{R} of \Omega \times \Omega.

In the context of relations, instead of ordered pairs of elements it is standard to write X \text{ symbol }Y to denote the membership of (X,Y) in \mathrm{R}. For example, you are familiar with the notion of an equivalence relation on a set \Omega, which is a relation written X \sim Y with the following properties :

  • Reflexivity: X \sim X for all X \in \Omega;
  • Symmetry: X\sim Y iff Y \sim X;
  • Transitivity: if X \sim Y and Y \sim Z then X \sim Z.

There is a different relation which formalizes order in the same way as the above formalizes equivalence.

Defintion 2.3. A partial order on \Omega is a relation with the following properties:

  • Reflexivity: X \leq X for all X \in \Omega;
  • Antisymmetry: if X\leq Y and Y \leq X then X=Y;
  • Transitivity: if X \leq Y and Y \leq Z then X \leq Z.

A pair (\Omega,\leq) consisting of a set together with a partial order on it is called a partially ordered set, or “poset.” The symbol X \geq Y by definition means Y\leq X. Importantly, note the use of the adjective “partial” – Definition 2.3 allows for the possibility that there are pairs of elements X,Y \in \Omega for which neither X \leq Y nor X \geq Y holds true, i.e. X,Y are incomparable. The axiomatic study of partially ordered sets is a branch of algebra called order theory.

A very basic example of a partial order is obtained by taking any set A and defining a partial order on the power set \Omega = \{X\subseteq A\} by inclusion:

X \leq Y \iff X \subseteq Y.

For example, if A=\{x,y,z\}, then \Omega consists of the 2^3=8 subsets of X ordered as in the following Hasse diagram:

The poset \Omega of subsets X of an arbitrary set A has an additional attribute: every pair of subsets X,Y \in \Omega has a minimum and a maximum. More precisely, if we define

\min(X,Y) := X \cap Y \quad\text{ and }\quad \max(X,Y) := X \cup Y,

then \min(X,Y) is the largest subset of A contained in both X and Y,

Z \leq X \text{ and }Z \leq Y \implies Z \leq \min(X,Y),

and \max(X,Y) is the smallest subset of A containing both X and Y,

Z \geq X \text{ and }Z \geq Y \implies Z \geq \max(X,Y).

A partially ordered set in which we have such a notion of greatest lower bound (min) and least upper bound (max) is called a lattice, and the power set \Omega of subsets of an arbitrary set A partially ordered by inclusion is such an object. Moreover, this lattice has a unique smallest element: the empty X=\emptyset set is the only subset of A satisfying X \leq Y for all Y \in \Omega. Similarly, the whole set X=A is the only set satisfying X \geq Y for all other Y, making it the unique largest element of \Omega.

We now want to use the partial order concept to organize the subalgebgras of a given algebra \mathcal{A}. Starting very simply, we could forget the structure of \mathcal{A} entirely and just view it as a set, which would result in the construction above with \Omega the power set of \mathcal{A}. Of course, since \mathcal{A} is a not a finite set \Omega is uncountably infinite.

Now let us remember the vector space structure on \mathcal{A} and accordingly take \Omega' \subseteq \Omega to be the set of all vector subspaces of \mathcal{A}, partially ordered by inclusion, so \Omega' is an induced subposet of \Omega. If we want to make \Omega' into a lattice we can still use the set-theoretic definition of \min as intersection, because the intersection of two subspaces is again a subspace. However, the union of two subspaces is not, and we have to modify the definition to

\max(V,W) = \mathrm{span}(V \cup W).

This makes \Omega' into a lattice with largest element \mathcal{A} and smallest element the zero subspace \{0\}, as the empty subset of \mathcal{A} is not a vector space.

Now let us remember the algebra structure on \mathcal{A} and declare \Omega'' \subseteq \Omega' to be the set of all subalgebras of \mathcal{A} partially ordered by inclusion. Then \Omega" is an induced subposet of \Omega', and once again taking \min to be intersection gives us a greatest lower bound operation.

Problem 2.2. Show that the intersection of any nonempty set \mathfrak{F} of subalgebras of \mathcal{A} is a subalgebra. Show moreover that if all members of the family \mathfrak{F} are commutative, so is their intersection.

However, our notion of least upper bound must be modified in order to make \Omega'' into a lattice, because the span of the union \mathcal{B} \cup \mathcal{C} of two subalgebras of \mathcal{A} is a subspace but not necessarily a subalgebra.

Definition 2.4. For any subset X \subseteq \mathcal{A}, define \mathrm{alg}(X) to be the intersection of all subalgebras of \mathcal{A} containing X. This is called the subalgebra generated by X. It contains all scalar products, sums, products, and conjugates of elements of X, and is the smallest subset of \mathcal{A} containing all of these elements.

Note that the set of subalgebras containing X is always nonempty, since it contains the whole algebra \mathcal{A}, and consequently Problem 2.1 legitimizes Definition 2.4 and allows us to define the least upper bound of a pair of subalgebras by

\max(\mathcal{B},\mathcal{C}) = \mathrm{alg}(\mathcal{B} \cup \mathcal{C}).

This makes \Omega'' into a lattice, the lattice of subalgebras of \mathcal{A}. The maximal element of this lattice is still \mathcal{A}, and its minimal element is \mathbb{C}I, since the zero space is not a subalgebra.

Finally, let \Omega''' be the set of all commutative subalgebras of \mathcal{A}, partially ordered by inclusion, an induced subposet of the lattice \Omega'' of algebras of \mathcal{A}. To latticize this we can keep the same greatest lower bound operation, but must modify least upper bound to

\max(\mathcal{B},\mathcal{C}) = \mathrm{calg}(\mathcal{B} \cup \mathcal{C}),

where the right hand should be the intersection of all commutative subalgebras of the ambient algebra \mathcal{A} which contain the set \mathcal{B} \cup \mathcal{C}. This definition will fail if one of B or C is not contained in any larger commutative subalgebra of \mathcal{A}.

Definition 2.5. A maximal abelian subalgebra (MASA) of \mathcal{A} is a commutative subalgebra \mathcal{B} \subseteq \mathcal{A} with the following property: if \mathcal{C} is a commutative subalgebra with \mathcal{B} \leq \mathcal{C}, then \mathcal{B}=\mathcal{C}.

Note that “abelian” is a synonym for “commutative” and the two are used interchangeably.

Problem 2.3. Prove that any two distinct MASAs are incomparable.

MASAs can be characterized using centralizers.

Theorem 2.2. An abelian subalgebra \mathcal{B} of an algebra \mathcal{A} is a MASA if and only if it is its own centralizer: Z(\mathcal{B},\mathcal{A})=\mathcal{B}.

Proof: Suppose first that \mathcal{B} is a MASA. Since \mathcal{B} is abelian we have \mathcal{B} \subseteq Z(\mathcal{A},\mathcal{B}). If \mathcal{B} is a proper subset of its centralizer, then there exists C \in Z(\mathcal{A},\mathcal{B}) such that C \not\in \mathcal{B}. In this case, the algebra generated by \{C\} \cup \mathcal{B} is a commutative subalgebra of \mathcal{A} properly containing \mathcal{B}, and this contradicts the maximality of \mathcal{B}.

Conversely, suppose \mathcal{B} is an abelian subalgebra of \mathcal{A} such that \mathcal{B} = Z(\mathcal{A},\mathcal{B}). Then for a commutative subalgebra \mathcal{C} \geq \mathcal{B} we have

\mathcal{C} \leq Z(\mathcal{A},\mathcal{C}) \leq Z(\mathcal{A},\mathcal{B}) = \mathcal{B},

so we also have \mathcal{C} \leq \mathcal{B} (note that the conclusion of Problem 2.1 was used here) .

-QED

Math 220A: Lecture 0

Schedule

LectureDateTopicModality
0S 23Course descriptionIn person
1S 26Numbers: complexIn person
2S 28Numbers: hypercomplexIn person
3S 30CodaOnline
4O 03Formal power series: algebra structureIn person
5O 05Formal power series: group structureIn person
6O 07CodaOnline
7O 10Convergent power series: algebra structureIn person
8O 12Convergent power series: group structureIn person
9O 14CodaOnline
10O 17Analytic functions: local inverse function theoremIn person
11O 19Analytic functions: local maximum modulus principleIn person
12O 21CodaOnline
13O 24Convergent power series: partial sumsIn person
14O 26Convergent power series: Jentzsch’s theoremIn person
15O 28CodaOnline
16O 31Exponential function: partial sumsIn person
17N 02Exponential function: Szego’s theoremIn person
18N 04CodaOnline
19N 07Analytic continuation: logarithmIn person
20N 09Analytic continuation: Hadamard gap theoremIn person
N 11Veteran’s Day
21N 14
22N 16
23N 18
24N 21
25N 23
N 25Thanksgiving
26N 28
27N 30
28D 02
Schedule subject to change.

Books

Articles

  • Buckholtz, A characterization of the exponential function. American Mathematical Monthly 73 (1966), 121-123.
  • Gronwall, On the power series for \log(1+z). Annals of Mathematics 18 (1916), p. 70-73.
  • Palais, The classification of real division algebras, American Mathematical Monthly 75 (1968), 366-368.

Evaluation

  • Problem sets: 70%
  • Final exam: 30%

Instructors

  • Lecturer: Jonathan Novak, APM 7157. Office hours MW after lecture.
  • Teaching Assistant: Shubham Sinha.

Communication

  • Piazza. Public posts preferred, private messages when necessary.
  • No email.