Math 202B: Lecture 2

In Lecture 1, we classified commutative algebras in terms of normal elements. In this lecture, we move beyond the commutative/noncommutative dichotomy and introduce a way to quantify the degree of commutativity of a given algebra \mathcal A. This method is based on measuring the dimension of a certain subalgebra of \mathcal A, so we begin by defining subalgebras.

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

A subalgebra \mathcal B of \mathcal A is by definition a subspace, so it contains the additive identity 0_{\mathcal A} and is closed under taking linear combinations. However, being a subalgebra is a strictly stronger condition than being a subspace, just as being an algebra is stronger than being a vector space: Definition 2.1 is equivalent to saying that \mathcal B is a subspace of \mathcal A which is itself an algebra. In particular, the zero subspace {0} is not a subalgebra; the smallest subalgebra of \mathcal A is

\mathbb C I = \{\alpha I\colon \alpha \in \mathbb C\}.

Problem 2.1 Prove that if \mathcal B is a one-dimensional subalgebra of \mathcal A, then \mathcal B = \mathbb C I.

The minimal subalgebra \mathbb C I is commutative, and in fact each of its elements commutes with every element of \mathcal A. We can consider the set of all elements in \mathcal{A} which have the “commutes with everything” property.

Definition 2.2 The center of \mathcal A is

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

Clearly, \mathbb C I_{\mathcal A} \subseteq Z(\mathcal A).

Proposition 2.3 Z(\mathcal A) is a subalgebra of \mathcal A.

Definition 2.4 The commutativity index of an algebra is the dimension of its center.

A minimally commutative/maximally noncommutative algebra \mathcal{A} has \dim Z(\mathcal{A})=1, which forces Z(\mathcal{A})=\mathbb{C}I. At the other extreme, \dim Z(\mathcal{A})=\dim \mathcal{A} forces Z(\mathcal{A})=\mathcal{A} provided \mathcal{A} is finite-dimensional, so this measurement of commutativity is most useful in the category of finite-dimensional algebras. We have not yet restricted to finite-dimensional algebras, but we will soon do so.

Definition 2.5 The centralizer of \mathcal B in \mathcal A is

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

the set of all elements of \mathcal{A} which commute with every element of \mathcal{B}.

Problem 2.2 Prove that Z(\mathcal B,\mathcal A) is a subalgebra of \mathcal A. Show that if \mathcal B \subseteq \mathcal C, then Z(\mathcal C,\mathcal A) \subseteq Z(\mathcal B,\mathcal A).

We now consider the set of all subalgebras of \mathcal A, ordered by inclusion:

\mathcal B \le \mathcal C \iff \mathcal B \subseteq \mathcal C.

This is a sub-poset of the lattice \mathcal L(\mathcal A) of all subspaces of \mathcal A, which only sees the vector-space structure. The poset of subalgebras of \mathcal{A} is strictly smaller than the lattice of subspaces, because we lose the zero-dimensional subspace and all one-dimensional subspaces except \mathbb{C}I. Furthermore, if we wish to promote the poset of subalgebras to a lattice we must work a bit harder.

Problem 2.3 Let \mathfrak F be a family of subalgebras of \mathcal A.

  1. Show that \bigcap_{\mathcal B \in \mathfrak F} \mathcal B is a subalgebra.
  2. Show that if all \mathcal B \in \mathfrak F are commutative, then so is the intersection.

For any subset X \subseteq \mathcal A, let

\mathcal F(X) = \{ \mathcal B \subseteq \mathcal A : X \subseteq \mathcal B \text{ and } \mathcal B \text{ is an algebra} \}.

Definition 2.6 The subalgebra generated by X is

\mathrm{alg}(X) = \bigcap\limits_{\mathcal B \in \mathcal F(X)} \mathcal B.

For subalgebras \mathcal B, \mathcal C \subseteq \mathcal A, define:

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

These operations make the poset of subalgebras into a lattice.

We can restrict even further to the poset of commutative subalgebras of \mathcal{A}, ordered by inclusion. If \mathcal{A} is itself commutative this restriction is vacuous, but if not then things change substantially. In particular, the poset of commutative subalgebras of \mathcal{A} no longer contains \mathcal{A} as a maximal element.

Definition 2.7 A maximal abelian subalgebra of \mathcal{A}, or MASA, is a commutative subalgebra \mathcal{B} which is not contained in any strictly larger commutative subalgebra: if \mathcal{C} is a commutative subalgebra with \mathcal{B} \leq \mathcal{C} then \mathcal{B}=\mathcal{C}.

In the context of commutative subalgebras we can modify Definition 2.6 as follows. Given any set X \subseteq \mathcal{A}, we consider the family

\mathcal F(X) = \{ \mathcal C \subseteq \mathcal A : X \subseteq \mathcal C \text{ and } \mathcal C \text{ is a commutative algebra} \}.

Definition 2.8 The commutative subalgebra of \mathcal{A} generated by X is

\mathrm{calg}(X) = \bigcap\limits_{\mathcal{C} \in \mathfrak{F}(X)} \mathcal{C}.

Returning to MASAs, the following characterization is very useful.

Problem 2.4. Prove that \mathcal{B} is a MASA in \mathcal{A} if and only if Z(\mathcal{B},\mathcal{A})=\mathcal{B}. Furthermore, show that any two MASAs are incomparable.

Published by Jonathan Novak

Mathematician at UC San Diego.

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