Topics in Operator Theory, Lecture 1: Introduction

by Orr Shalit

This is a summary of the first lecture, which was introductory in nature.

H will always denote a Hilbert space over \mathbb{C}. B(H) will always denote the algebra of bounded operators on H. I am interested in operators on Hilbert space; various subspaces and algebras of operators that come with various structures, as well as the relationship between these subspaces and structures; and connections and applications of the above to other areas, in particular complex function theory and matrix theory.

I expect students to know the spectral theorem for normal operators on Hilbert space (see here. A proof in the selfadjoint case that assumes very little from the reader can be found in my notes, see Section 3 and 4). I also will assume some familiarity with Banach algebras and commutative C*-algebras – the student should contact me for references.

We begin by surveying different kinds of structures of interest. 

1. Norm structure

B(H) is a Banach space when equipped with the norm

\|A\| := \sup_{\|h\|=1}\|Ah\|.

Definition 1: An operator space X is a linear subspace X \subseteq B(H).

We always consider an operator space as a normed space with the norm induced from B(H).

Remarks: 

  1. We usually assume without mention that our operator spaces are closed. For most parts of the theory it doesn’t make a difference.
  2. What we defined above is sometimes called a “concrete operator space”. We will encounter “abstract operator spaces” later on.

We can already say something quite interesting about the collection of operator spaces: every normed (Banach) space E is isometric to a (closed) operator space X \subset B(H) (so long as we allow \dim H to be big enough). 

(Quick proof: E \hookrightarrow C(K) \hookrightarrow B(\ell^2(K)), where K is the closed unit ball of E^* with the weak-* topology).

By isometric we really mean isometrically isomorphic, by which we mean the following: there exists a linear isomorphism T : E \to X \subseteq B(H) such that \|Tx\| = \|x\| for all x \in E.

For example \ell^2_2, \ell^\infty_2 are easily seen to be isometric to \begin{pmatrix} * & * \\ 0 & 0 \end{pmatrix} and \begin{pmatrix} * & 0 \\ 0 & * \end{pmatrix}, respectively.

Exercise A: What can you say about embedding \ell^1_2 in B(H) for H finite dimensional?

The above fact is in extreme contrast with the situation for Hilbert spaces: every closed subspace of a Hilbert space is a Hilbert space.

2. Algebraic structure

B(H) is also an algebra, with product

A,B \mapsto AB := A \circ B.

In fact, \|AB\| \leq \|A\|\|B\|, so B(H) is a Banach algebra.

Note: I expect students in this course to be familiar with Banach algebras. Those who are not should ask me for references (and I will provide time for catching up).

Definition 2: An operator algebra \mathcal{A} is a subalgebra \mathcal{A} \subseteq B(H).

By subalgebra we just mean A,B \in \mathcal{A} \Rightarrow AB \in \mathcal{A}. We always take operator algebras with the induced algebraic structure and norm.

Remark: The two remarks made after Definition 1 hold for operator algebras as well.

Examples: B(H), the upper triangular matrices, the operator algebra \overline{\textrm{alg}}(T) generated by an operator T.

Interesting fact: not every Banach algebra is an operator algebra. By this we mean that not every Banach algebra is isometrically isomorphic to an operator algebra. In the context of Banach algebras, and isometric isomorphism \phi is understood to be an isometric isomorphism in the sense of normed spaces, which also preserves the product \phi(ab) = \phi(a) \phi(b).

How can we see this? We shall prove later in this course von Neumann’s inequality, which says that

\|p(T)\| \leq \sup_{|z|\leq 1} |p(z)|

for every polynomial p and every contraction on a Hilbert space T (contraction means T \in B(H), \|T\|\leq 1). Conversely, if E is a Banach space such that von Neumann’s inequality holds for every contraction, then E must be a Hilbert space (a result of Foias ). So if E is not a Hilbert space, there exists T \in B(E) such that \textrm{alg}(T) is not an operator algebra.

Exercise B: Find explicitly such an operator T for which vN inequality fails.

So all Banach spaces are operator spaces, while not all Banach algebras are operator algebras. Structure matters. The more structure, the more “rigidity”.

3. Involutive structure

Definition 3: An involution on an algebra \mathcal{A} is a map a \mapsto a^* such that

  1. a^{**} := (a^*)^* = a
  2. (a+\lambda b)^* = a^* + \overline{\lambda} b^*
  3. (ab)^* = b^*a^*

for all a,b \in \mathcal{A} and \lambda \in \mathbb{C}. An algebra with involution is said to be a *-algebra.

Definition 4:Banach *-algebra is a *-algebra which is also a Banach algebra such that \|a^*\| = \|a\| for all a \in \mathcal{A}.

Definition 5: A C*-algebra is a Banach *-algebra such that

\|a^*a\| = \|a\|^2

holds for all a \in \mathcal{A}; this identity is called the C* identity.

Examples: B(H) has a natural involution, the adjoint T \mapsto T^*. So B(H) is a *-algebra, and so is every “*-subalgebra”. The norm and adjoint in B(H) satisfy the C* identity (prove this if you never did!), so B(H) is a C*-algebra, as are all its closed *-subalgebras.

Definition 6: concrete C*-algebra is a closed *-subalgebra of B(H).

Clearly concrete C*-algebras are C*-algebras, with norm and algebraic structure inherited from B(H). Note that we do require C*-algebras to be closed. The maps in the setting of C*-algebras are *-homomorphisms and *-isomorphims.

Theorem (Gelfand-Naimark): Every C*-algebra is (isometrically) *-isomorphic to a concrete C*-algebra. 

The word “isometrically” is in parenthesis, because it will turn out that every *-isomorphism is isometric. This is part of the beautiful rigidity properties that C*-algebras enjoy: the algebraic structure captures the metric structure.

As hinted above, there exist notions of “abstract operator space/algebras”, and there are theorems that these are isomorphic in the appropriate sense to concrete operator spaces/algebras. What could the abstract definition be based on?

Question: If abstract objects end up being concrete, why not just work in the concrete setting?

Answer: The abstract setting is more flexible, and allows for more constructions (but indeed sometimes it is indeed more convenient to work in the concrete setting). For example: quotient C*-algebras: easy to establish the axioms, not trivial to represent.

4. Order structure

Recall that A \in B(H) is said to be positive (or positive semidefinite) if

\langle A h , h \rangle \geq 0

for all h \in H. This is equivalent to the two following conditions holding (together)

  1. A = A^* (A is selfadjoint), and
  2. \sigma(A):= \{ \lambda \in \mathbb{C} : \lambda I - A \textrm{ is not invertible }\} \subset [0, \infty).

(Note that the above two conditions make sense in any Banach *-algebra, and so we have selfadjoint and positive elements in such algebras as well. Also the following definition of order makes sense there.)

We then say that A \geq B if A-B \geq 0. This gives a partial order on the selfadjoint elements.

Definition 7: An operator system is an operator space \mathcal{S} \subseteq B(H) such that

  1. 1 = I_H \in \mathcal{S}, and
  2. A \in \mathcal{S} \Rightarrow A^* \in \mathcal{S}.

Examples: Concrete C*-algebras are operator systems, as are spaces of the form \mathcal{M} + \mathcal{M}^* + \mathbb{C}I where \mathcal{M} is some operator space.

Non-example: The selfadjoint elements are not an operator system, they form a real subspace but not a subspace.

However, operator systems have “sufficiently many” selfadjoint and positive elements.

Facts:

  1. A = \frac{A+A^*}{2}+i \frac{A - A^*}{2i}, so an operator system \mathcal{S} is spanned by its selfadjoints.
  2. If \|A\|\leq 1 is selfadjoint then A = \frac{1+A}{2} - \frac{1-A}{2} so \mathcal{S} is actually spanned by its positive elements.

Notation: \mathcal{S}_+ := \{a \in \mathcal{S} : a \geq 0\}.

A linear map \phi : \mathcal{S} \to \mathcal{T} is positive if \phi(\mathcal{S}_+) \subseteq \mathcal{T}_+.

5. Matrix structure

For n \in \mathbb{N}, we write H^n = H \oplus \cdots \oplus H (direct sum n times), with inner product

\langle h,g \rangle = \sum \langle h_i, g_i \rangle_H

for h = (h_1, \ldots, h_n), g = (g_1, \ldots, g_n).

We have an identification B(H^n) = M_n(B(H)). Thus M_n(B(H)) has natural linear and algebraic structure, norm, * operation and order.

Key (and simple) fact: If \mathcal{M} is a concrete ZZZ in B(H), then M_n(\mathcal{M}) is a concrete ZZZ in M_n(B(H)) = B(H^n), where

ZZZ \in \{ op. space, op. alg., op. system, C*-subalgebra \}.

Let X,Y be operator spaces, and \phi: M \to N a linear map. We define \phi : M_n(X) \to M_n(Y) by

\phi([a_{ij}]) = [\phi(a_{ij})].

Definition 8:

  1. If \|\phi\|_{cb} := \sup_n \|\phi_n\| < \infty then \phi is said to be completely bounded  (and \|\phi\|_{cb} is referred to as the CB norm).
  2. If \|\phi\|_{cb}\leq 1 then \phi is completely contractive
  3. If \phi_n is isometric for all n then \phi is completely isometric. 
  4. If \phi_n is positive for all n then \phi is completely positive.

We will see that there is a big difference between being bounded/positive to being completely bounded/completely positive.