Content deleted Content added
→Minimal normal extension: +unitary freedom of minimal extensions |
→Minimal normal extension: +argument for claim |
||
Line 55:
== 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''<sup>2</sup>('''N'''). One normal extension is the bilateral shift ''B'' on ''l''<sup>2</sup>('''Z''') defined by
Line 73 ⟶ 75:
B' (\cdots, a_{-2}, a_{-1}, {\hat a_0}, a_1, a_2, \cdots) = (\cdots, - a_{-2}, {\hat a_{-1}}, a_0, a_1, a_2, \cdots).
</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*''.)
Line 82 ⟶ 86:
Also, the following interwining 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 \mbox{where} \quad h_i \in H.
</math>
Let ''K' '' ⊂ ''K''<sub>1</sub> be the subspace that is the closure of the linear span of ''S''. By definition, ''K' '' is invariant under ''B''<sub>1</sub>* and contains ''H''. The normality of ''B''<sub>1</sub> and the assumption that ''H'' is invariant under ''B''<sub>1</sub> imply ''K' '' is invariant under ''B''<sub>1</sub>. Therefore ''K' '' = ''K''<sub>1</sub>. The Hilbert space ''K''<sub>2</sub> 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>
\langle \sum_{i=0}^n (B_1^*)^i h_i, \sum_{j=0}^n (B_1^*)^j h_j\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
= \langle \sum_{i=0}^n (B_2^*)^i h_i, \sum_{j=0}^n (B_2^*)^j h_j\rangle ,
</math>
, the operator ''U'' is unitary. Direct computation also shows (the assumption that both ''B''<sub>1</sub> and ''B''<sub>2</sub> are extensions of ''A'' are needed here)
:<math>\mbox{if} \quad g = \sum_{i=0}^n (B_1^*)^i h_i ,</math>
:<math>\mbox{then} \quad U B_1 g = B_2 U g = \sum_{i=0}^n (B_2^*)^i A h_i.</math>
[[Category:Operator theory]]
| |||