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

for some bounded operators

B : H ⊥ → H , and C : H ⊥ → H ⊥ . {\displaystyle B:H^{\perp }\rightarrow H,\quad {\mbox{and}}\quad C:H^{\perp }\rightarrow 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 {\displaystyle H\oplus H}

by

U = [ A I − A A ∗ 0 − A ∗ ] . {\displaystyle 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.² 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.³

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 = [ U I − U U ∗ 0 − U ∗ ] = [ U D U ∗ 0 − U ∗ ] . {\displaystyle V={\begin{bmatrix}U&I-UU^{*}\\0&-U^{*}\end{bmatrix}}={\begin{bmatrix}U&D_{U^{*}}\\0&-U^{*}\end{bmatrix}}.}

Define

Q = [ P 0 0 P ] . {\displaystyle 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 = Q V ∗ V Q = Q 2 = [ P 2 0 0 P 2 ] . {\displaystyle N^{*}N=QV^{*}VQ=Q^{2}={\begin{bmatrix}P^{2}&0\\0&P^{2}\end{bmatrix}}.}

On the other hand,

N N ∗ = [ U P 2 U ∗ + D U ∗ P 2 D U ∗ − D U ∗ P 2 U − U ∗ P 2 D U ∗ U ∗ P 2 U ] . {\displaystyle NN^{*}={\begin{bmatrix}UP^{2}U^{*}+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 DU*P = DU*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 ( … , a − 1 , a ^ 0 , a 1 , … ) = ( … , a ^ − 1 , a 0 , a 1 , … ) , {\displaystyle 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 = [ A I − A A ∗ 0 A ∗ ] . {\displaystyle 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 ′ = [ A I − A A ∗ 0 − A ∗ ] {\displaystyle B'={\begin{bmatrix}A&I-AA^{*}\\0&-A^{*}\end{bmatrix}}}

whose action is described by

B ′ ( … , a − 2 , a − 1 , a ^ 0 , a 1 , a 2 , … ) = ( … , − a − 2 , a ^ − 1 , a 0 , a 1 , a 2 , … ) . {\displaystyle 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*.)⁴

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 . {\displaystyle U:K_{1}\rightarrow K_{2}.}

Also, the following intertwining relationship holds:

U B 1 = B 2 U . {\displaystyle UB_{1}=B_{2}U.\,}

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

∑ i = 0 n ( B 1 ∗ ) i h i = h 0 + B 1 ∗ h 1 + ( B 1 ∗ ) 2 h 2 + ⋯ + ( B 1 ∗ ) n h n where h i ∈ H . {\displaystyle \sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i}=h_{0}+B_{1}^{*}h_{1}+(B_{1}^{*})^{2}h_{2}+\cdots +(B_{1}^{*})^{n}h_{n}\quad {\text{where}}\quad 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 ∑ i = 0 n ( B 1 ∗ ) i h i = ∑ i = 0 n ( B 2 ∗ ) i h i {\displaystyle U\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i}=\sum _{i=0}^{n}(B_{2}^{*})^{i}h_{i}}

Because

⟨ ∑ i = 0 n ( B 1 ∗ ) i h i , ∑ j = 0 n ( B 1 ∗ ) j h j ⟩ = ∑ i j ⟨ h i , ( B 1 ) i ( B 1 ∗ ) j h j ⟩ = ∑ i j ⟨ ( B 2 ) j h i , ( B 2 ) i h j ⟩ = ⟨ ∑ i = 0 n ( B 2 ∗ ) i h i , ∑ j = 0 n ( B 2 ∗ ) j h j ⟩ , {\displaystyle \left\langle \sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i},\sum _{j=0}^{n}(B_{1}^{*})^{j}h_{j}\right\rangle =\sum _{ij}\langle h_{i},(B_{1})^{i}(B_{1}^{*})^{j}h_{j}\rangle =\sum _{ij}\langle (B_{2})^{j}h_{i},(B_{2})^{i}h_{j}\rangle =\left\langle \sum _{i=0}^{n}(B_{2}^{*})^{i}h_{i},\sum _{j=0}^{n}(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 = ∑ i = 0 n ( B 1 ∗ ) i h i , {\displaystyle {\text{if }}g=\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i},} then  U B 1 g = B 2 U g = ∑ i = 0 n ( B 2 ∗ ) i A h i . {\displaystyle {\text{then }}UB_{1}g=B_{2}Ug=\sum _{i=0}^{n}(B_{2}^{*})^{i}Ah_{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.

References

1. John B. Conway (1991), "11", The Theory of Subnormal Operators, American Mathematical Soc., p. 27, ISBN 978-0-8218-1536-6, retrieved 15 June 2017 978-0-8218-1536-6 ↩

2. John B. Conway (1991), "11", The Theory of Subnormal Operators, American Mathematical Soc., p. 29, ISBN 978-0-8218-1536-6, retrieved 15 June 2017 978-0-8218-1536-6 ↩

3. John B. Conway; Robert F. Olin (1977), A Functional Calculus for Subnormal Operators II, American Mathematical Soc., p. 51, ISBN 978-0-8218-2184-8, retrieved 15 June 2017 978-0-8218-2184-8 ↩

4. John B. Conway (1991), The Theory of Subnormal Operators, American Mathematical Soc., pp. 38–, ISBN 978-0-8218-1536-6, retrieved 15 June 2017 978-0-8218-1536-6 ↩