Subnormal operator explained

In mathematics, especially operator theory, subnormal operators are bounded operators on a Hilbert space defined by weakening the requirements for normal operators. Some examples of subnormal operators are isometries and Toeplitz operators with analytic symbols.

Definition

Let H be a Hilbert space. A bounded operator A on H is said to be subnormal if A has a normal extension. In other words, A is subnormal if there exists a Hilbert space K such that H can be embedded in K and there exists a normal operator N of the form

N = \begin{bmatrix} A & B\ 0 & C\end{bmatrix}

for some bounded operators

B : H ^{\perp} → H, and C : H ^{\perp} → H ^{\perp} .

Normality, quasinormality, and subnormality

Normal operators

Every normal operator is subnormal by definition, but the converse is not true in general. A simple class of examples can be obtained by weakening the properties of unitary operators. A unitary operator is an isometry with dense range. Consider now an isometry A whose range is not necessarily dense. A concrete example of such is the unilateral shift, which is not normal. But A is subnormal and this can be shown explicitly. Define an operator U on

H ⊕ H

by

U = \begin{bmatrix} A & I - AA ^{*} \ 0 & - A ^{*} \end{bmatrix}.

Direct calculation shows that U is unitary, therefore a normal extension of A. The operator U is called the unitary dilation of the isometry A.

Quasinormal operators

An operator A is said to be quasinormal if A commutes with A*A. A normal operator is thus quasinormal; the converse is not true. A counter example is given, as above, by the unilateral shift. Therefore, the family of normal operators is a proper subset of both quasinormal and subnormal operators. A natural question is how are the quasinormal and subnormal operators related.

We will show that a quasinormal operator is necessarily subnormal but not vice versa. Thus the normal operators is a proper subfamily of quasinormal operators, which in turn are contained by the subnormal operators. To argue the claim that a quasinormal operator is subnormal, recall the following property of quasinormal operators:

Fact: A bounded operator A is quasinormal if and only if in its polar decomposition A = UP, the partial isometry U and positive operator P commute.

Given a quasinormal A, the idea is to construct dilations for U and P in a sufficiently nice way so everything commutes. Suppose for the moment that U is an isometry. Let V be the unitary dilation of U,

V = \begin{bmatrix} U & I - UU ^{*} \ 0 & - U ^{*} \end{bmatrix} = \begin{bmatrix} U & D _{U ^{*}} \ 0 & - U ^{*} \end{bmatrix} .

Define

Q = \begin{bmatrix} P & 0 \ 0 & P \end{bmatrix}.

The operator N = VQ is clearly an extension of A. We show it is a normal extension via direct calculation. Unitarity of V means

N ^{*N} = QV ^{*VQ} = Q ^{2} = \begin{bmatrix} P ^{2} & 0 \ 0 & P ^{2} \end{bmatrix}.

On the other hand,

N N ^{*} = \begin{bmatrix} UP ^{2U} ^{*} + D _{U ^{*}} P ^{2} D _{U ^{*}} & - D _{U ^{*}} P ^{2} U \ - U ^{*} P ^{2} D _{U ^{*}} & U ^{*} P ^{2} U \end{bmatrix}.

Because UP = PU and P is self adjoint, we have U*P = PU* and D_U*P = D_U*P. Comparing entries then shows N is normal. This proves quasinormality implies subnormality.

For a counter example that shows the converse is not true, consider again the unilateral shift A. The operator B = A + s for some scalar s remains subnormal. But if B is quasinormal, a straightforward calculation shows that A*A = AA*, which is a contradiction.

Minimal normal extension

Non-uniqueness of normal extensions

Given a subnormal operator A, its normal extension B is not unique. For example, let A be the unilateral shift, on l²(N). One normal extension is the bilateral shift B on l²(Z) defined by

B (\ldots, a _{-1} , {\hat a _{0},} a _{1,} \ldots) = (\ldots, {\hat a _{-1} }, a_0, a_1, \ldots),

where ˆ denotes the zero-th position. B can be expressed in terms of the operator matrix

B = \begin{bmatrix} A & I - AA ^{*} \ 0 & A ^{*} \end{bmatrix}.

Another normal extension is given by the unitary dilation B' of A defined above:

B' = \begin{bmatrix} A & I - AA ^{*} \ 0 & - A ^{*} \end{bmatrix}

whose action is described by

B' (\ldots, a _{-2} , a _{-1} , {\hat a _{0},} a _{1,} a _{2,} \ldots) = (\ldots, - a _{-2} , {\hat a _{-1} }, a_0, a_1, a_2, \ldots).

Minimality

Thus one is interested in the normal extension that is, in some sense, smallest. More precisely, a normal operator B acting on a Hilbert space K is said to be a minimal extension of a subnormal A if K' ⊂ K is a reducing subspace of B and H ⊂ K' , then K' = K. (A subspace is a reducing subspace of B if it is invariant under both B and B*.)

One can show that if two operators B₁ and B₂ are minimal extensions on K₁ and K₂, respectively, then there exists a unitary operator

U: K _{1} → K _{2.}

Also, the following intertwining relationship holds:

U B _{1} = B _{2} U.

This can be shown constructively. Consider the set S consisting of vectors of the following form:

\sum ^{n} _{i =0} (B ^{*)} _{1} ^{i} h _{i} = h _{0+} B ^{*} _{1} h _{1} + (B ^{*)} _{1} ^{2} h _{2} + … + (B ^{*)} _{1} ^{n} h _{n} where h _{i} \in H.

Let K' ⊂ K₁ be the subspace that is the closure of the linear span of S. By definition, K' is invariant under B₁* and contains H. The normality of B₁ and the assumption that H is invariant under B₁ imply K' is invariant under B₁. Therefore, K' = K₁. The Hilbert space K₂ can be identified in exactly the same way. Now we define the operator U as follows:

U \sum ^{n} _{i =0} (B ^{*)} _{1} ^{i} h _{i} = \sum ^{n} _{i =0} (B ^{*)} _{2} ^{i} h _{i}

Because

\left\langle \sum ^{n} _{i =0} (B ^{*)} _{1} ^{i} h _{i,} \sum ^{n} _{j =0} (B ^{*)} _{1} ^{j} h _{j\right\rangle =} \sum _{i} \langle h _{i,} (B ^{i} _{1 )} (B ^{*)} _{1} ^{j} h _{j\rangle =} \sum _{i} \langle (B ^{j} _{2 )} h _{i,} (B ^{i} _{2 )} h _{j\rangle =} \left\langle \sum ^{n} _{i =0} (B ^{*)} _{2} ^{i} h _{i,} \sum ^{n} _{j =0} (B ^{*)} _{2} ^{j} h _{j\right\rangle} ,

, the operator U is unitary. Direct computation also shows (the assumption that both B₁ and B₂ are extensions of A are needed here)

if g = \sum ^{n} _{i =0} (B ^{*)} _{1} ^{i} h _{i} , then U B _{1} g = B _{2} U g = \sum ^{n} _{i =0} (B ^{*)} _{2} ^{i} A h _{i.}

When B₁ and B₂ are not assumed to be minimal, the same calculation shows that above claim holds verbatim with U being a partial isometry.