Subnormal operator: Difference between revisions

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In [[mathematics]], especially [[operator theory]], '''subnormal operators''' are [[bounded operatorsoperator]]s on a [[Hilbert space]] defined by weakening the requirements for [[normal operator]]s. <ref name="Conway1991">{{citation|author=John B. Conway|title=The Theory of Subnormal Operators|url=https://books.google.com/books?id=Ho7yBwAAQBAJ|accessdate=15 June 2017|year=1991|publisher=American Mathematical Soc.|isbn=978-0-8218-1536-6|page=27|chapter=11}}</ref> Some examples of subnormal operators are [[isometry|isometries]] and [[Toeplitz operator]]s 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

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:<math>B : H^{\perp} \rightarrow H, \quad \mbox{and} \quad C : H^{\perp} \rightarrow H^{\perp}.</math>

== Normality, quasinormality, and subnormality ==

=== Normal operators ===

Every normal operator is subnormal by definition, but the converse is not true in general. AnA simple class of examples can be obtained by weakening the properties of [[unitary operator]]s. A unitary operator is an isometry with [[dense set|dense]] [[range of a function|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

:<math>H \oplus H</math>

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:<math> U = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}.</math>

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 operator|quasinormal]]''' if ''A'' commutes with ''A*A''.<ref>{{citation|author=John B. Conway|title=The Theory of Subnormal Operators|url=https://books.google.com/books?id=Ho7yBwAAQBAJ|accessdate=15 June 2017|year=1991|publisher=American Mathematical Soc.|isbn=978-0-8218-1536-6|page=29|chapter=11}}</ref> 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.<ref name="ConwayOlin1977">{{citation|author1=John B. Conway|author2=Robert F. Olin|title=A Functional Calculus for Subnormal Operators II|url=https://books.google.com/books?id=yQXUCQAAQBAJ|accessdate=15 June 2017|year=1977|publisher=American Mathematical Soc.|isbn=978-0-8218-2184-8|page=51}}</ref>

Given a quasinormal ''A'', the idealidea 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'',

:<math> V = \begin{bmatrix} U & I - UU^* \\ 0 & - U^* \end{bmatrix}

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:<math> Q = \begin{bmatrix} P & 0 \\ 0 & P \end{bmatrix}.</math>

The operator ''N'' = ''VQ'' is clearly an extension of ''A'';. directWe calculation shows thatshow it is a normal extension via direct calculation. Unitarity of ''V'' means

:<math>N^*N = QV^*VQ = Q^2 = \begin{bmatrix} P^2 & 0 \\ 0 & P^2 \end{bmatrix}.</math>

On the other hand,

:<math>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}.</math>

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 ''l''²('''N'''). One normal extension is the bilateral shift ''B'' on ''l''²('''Z''') defined by

:<math>B (\ldots, a_{-1}, {\hat a_0}, a_1, \ldots) = (\ldots, {\hat a_{-1}}, a_0, a_1, \ldots),</math>

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

:<math> B = \begin{bmatrix} A & I - AA^* \\ 0 & A^* \end{bmatrix}.</math>

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

:<math> B' = \begin{bmatrix} A & I - AA^* \\ 0 & - A^* \end{bmatrix}</math>

whose action is described by

:<math>

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).

</math>

===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*''.)<ref>{{citation|author=John B. Conway|title=The Theory of Subnormal Operators|url=https://books.google.com/books?id=Ho7yBwAAQBAJ&pg=PA38|accessdate=15 June 2017|year=1991|publisher=American Mathematical Soc.|isbn=978-0-8218-1536-6|pages=38–}}</ref>

One can show that if two operators ''B''₁ and ''B''₂ are minimal extensions on ''K''₁ and ''K''₂, respectively, then there exists a unitary operator

:<math>U: K_1 \rightarrow K_2.</math>

Also, the following intertwining relationship holds:

:<math>U B_1 = B_2 U. \,</math>

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

:<math>

\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.

</math>

Let ''K' '' ⊂ ''K''₁ be the subspace that is the closure of the linear span of ''S''. By definition, ''K' '' is invariant under ''B''₁* and contains ''H''. The normality of ''B''₁ and the assumption that ''H'' is invariant under ''B''₁ imply ''K' '' is invariant under ''B''₁. Therefore, ''K' '' = ''K''₁. The Hilbert space ''K''₂ can be identified in exactly the same way. Now we define the operator ''U'' as follows:

:<math>

U \sum_{i=0}^n (B_1^*)^i h_i = \sum_{i=0}^n (B_2^*)^i h_i

</math>

Because

:<math>

\left\langle \sum_{i=0}^n (B_1^*)^i h_i, \sum_{j=0}^n (B_1^*)^j h_j\right\rangle

= \sum_{i j} \langle h_i, (B_1)^i (B_1^*)^j h_j\rangle

= \sum_{i j} \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 ,

</math>

, the operator ''U'' is unitary. Direct computation also shows (the assumption that both ''B''₁ and ''B''₂ are extensions of ''A'' are needed here)

:<math>\text{if } g = \sum_{i=0}^n (B_1^*)^i h_i ,</math>

:<math>\text{then } U B_1 g = B_2 U g = \sum_{i=0}^n (B_2^*)^i A h_i.</math>

When ''B''₁ and ''B''₂ are not assumed to be minimal, the same calculation shows that above claim holds verbatim with ''U'' being a [[partial isometry]].

==References==

{{reflist}}

{{DEFAULTSORT:Subnormal Operator}}

[[Category:Operator theory]]

[[Category:Linear operators]]