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 202B: Lecture 2

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