Reflexive operator algebra
|
|
In functional analysis, a reflexive operator algebra A is an operator algebra that has enough invariant subspaces to characterize it. Formally, A is reflexive if is equal to the algebra of bounded operators which leave invariant each subspace left invariant by every operator in A.
This should not be confused with a reflexive space.
| Contents |
Examples
Nest algebras are examples of reflexive operator algebras. In finite dimensions, these are simply algebras of all matrices of a given size whose nonzero entries lie in an upper-triangular pattern.
In fact if we fix any pattern of entries in an n by n matrix containing the diagonal, then the set of all n by n matrices whose nonzero entries lie in this pattern forms a reflexive algebra.
An example of an algebra which is not reflexive is the set of 2 by 2 matrices
- <math>\left\{
\begin{pmatrix} a&b\\ 0 & a \end{pmatrix} \ :\ a,b\in\mathbb{C}\right\}.<math>
This algebra is smaller than the nest algebra
- <math>\left\{
\begin{pmatrix} a&b\\ 0 & c \end{pmatrix} \ :\ a,b,c\in\mathbb{C}\right\}<math>
but has the same invariant subspaces, so it is not reflexive.
If T is a fixed n by n matrix then the set of all polynomials in T and the identity operator forms a unital operator algebra. A theorem of Deddens and Fillmore states that this algebra is reflexive if and only if the largest two blocks the Jordan normal form of T differ in size by at most one. For example, the algebra
- <math>\left\{
\begin{pmatrix} a & b & 0\\ 0 & a & 0\\ 0 & 0 & a \end{pmatrix} \ :\ a,b\in\mathbb{C}\right\}<math>
which is equal to the set of all polynomials in
- <math>
T=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} <math>
and the identity is reflexive.
Hyper-reflexivity
Let <math>\mathcal{A}<math> be an operator algebra contained in B(H), the set of all bounded operators on a Hilbert space H and for T any operator in B(H), let
- <math>\beta(T,\mathcal{A})=\sup \{ \| P^\perp TP \|\ :\ P\mbox{ is a projection and } P^\perp \mathcal{A} P = (0) \}<math>.
Observe that P is a projection involved in this supremum precisely if the range of P is an invariant subspace of <math>\mathcal{A}<math>.
The algebra <math>\mathcal{A}<math> is reflexive if and only if the following inequality is satisfied for every T in B(H):
- <math>\beta(T,\mathcal{A})\le \mbox{dist}(T,\mathcal{A})<math>.
Here <math>\mbox{dist}(T,\mathcal{A})<math> is the distance of T from the algebra, namely the smallest norm of an operator T-A where A runs over the algebra. We call <math>\mathcal{A}<math> hyperreflexive if there is a constant K such that for every operator T in B(H),
- <math>\mbox{dist}(T,\mathcal{A})\le K \beta(T,\mathcal{A})<math>.
The smallest such K is called the distance constant for <math>\mathcal{A}<math>. A hyper-reflexive operator algebra is automatically reflexive.
In the case of a reflexive algebra of matrices with nonzero entries specified by a given pattern, the problem of finding the distance constant can be rephrased as a matrix-filling problem: if we fill the entries in the complement of the pattern with arbitrary entries, what choice of entries in the pattern gives the smallest operator norm?
Examples
- Every finite-dimensional algebra is hyper-reflexive. However, there are examples of infinite-dimensional reflexive operator algebras which are not hyper-reflexive.
- The distance constant for a one dimensional algebra is 1.
- Nest algebras are hyper-reflexive with distance constant 1.
- Any von Neumann algebra is hyper-reflexive with distance constant at most 4.
- A type I von Neumann algebra is hyper-reflexive with distance constant at most 2.