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Reflexive operator algebra - Wikipedia, the free encyclopedia

Reflexive operator algebra

From Wikipedia, the free encyclopedia

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 it 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

[edit] 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

\left\{ 
\begin{pmatrix}
a&b\\ 0 & a
\end{pmatrix}
\ :\  a,b\in\mathbb{C}\right\}.

This algebra is smaller than the Nest algebra

\left\{ 
\begin{pmatrix}
a&b\\ 0 & c
\end{pmatrix}
\ :\  a,b,c\in\mathbb{C}\right\}

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 in the Jordan normal form of T differ in size by at most one. For example, the algebra

\left\{ 
\begin{pmatrix}
a & b & 0\\ 0 & a & 0\\ 0 & 0 & a
\end{pmatrix}
\ :\  a,b\in\mathbb{C}\right\}

which is equal to the set of all polynomials in


T=\begin{pmatrix}
0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0
\end{pmatrix}

and the identity is reflexive.

[edit] Hyper-reflexivity

Let \mathcal{A} be a weak*-closed 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

\beta(T,\mathcal{A})=\sup \{ \| P^\perp TP \|\ :\ P\mbox{ is a projection and } P^\perp \mathcal{A} P = (0) \}.

Observe that P is a projection involved in this supremum precisely if the range of P is an invariant subspace of \mathcal{A}.

The algebra \mathcal{A} is reflexive if and only if for every T in B(H):

\beta(T,\mathcal{A})=0 \mbox{ implies that } T \mbox{ is in } \mathcal{A}.

We note that for any T in B(H) the following inequality is satisfied:

\beta(T,\mathcal{A})\le \mbox{dist}(T,\mathcal{A}).

Here \mbox{dist}(T,\mathcal{A}) 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 \mathcal{A} hyperreflexive if there is a constant K such that for every operator T in B(H),

\mbox{dist}(T,\mathcal{A})\le K \beta(T,\mathcal{A}).

The smallest such K is called the distance constant for \mathcal{A}. 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?

[edit] 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.

[edit] See also

[edit] References


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