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A measurable two-dimensional operator field

Example

Assume AC. On the preceding multiplicity-two field over $[0, 1]$ , with $H_{t} = C^{2}$ , Lebesgue measure $λ$ , and diagonal algebra $D = {M_{f} \otimes I_{2} : f \in L^{\infty} ([0, 1], λ)}$ from Multiplicity-two diagonal representation, define $T_{t} = (\begin{array}{cc} 0 & t \\ 1 - t & 0 \end{array}) (0 \leq t \leq 1) .$ This is a weakly measurable, essentially bounded operator field. Its induced operator $T = \int_{[0, 1]}^{\oplus} T_{t} d λ (t)$ has norm $1$ , adjoint field $T_{t}^{*} = (\begin{array}{cc} 0 & 1 - t \\ t & 0 \end{array}),$ and square field $T_{t}^{2} = (\begin{array}{cc} t (1 - t) & 0 \\ 0 & t (1 - t) \end{array}) .$ The operator $T$ commutes with every element of $D$ , but $T$ is not itself in $D$ .

Facts & Assumptions

Given: AC and the multiplicity-two constant field, measure, Hilbert space, and diagonal algebra of the preceding example.

[F1]

The preceding example has base $[0, 1]$ with Borel Lebesgue measure, fibre $C^{2}$ , direct integral $H = L^{2} (λ) \oplus L^{2} (λ)$ , and diagonal algebra $D$ acting by $f (t) I_{2}$ (Multiplicity-two diagonal representation).

[F2]

Weak measurability is tested by the fundamental matrix coefficients, and essential boundedness means the measurable pointwise operator norm has finite essential supremum (Measurable and decomposable operator fields).

[F3]

Under AC, every weakly measurable essentially bounded field induces a bounded direct-integral operator with norm equal to the essential supremum; adjoint and product fields induce the operator adjoint and product (Measurable essentially bounded operator fields act decomposably).

[F4]

The operator norm is the supremum of $∥ T x ∥$ over the unit ball (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F5]

The essential supremum is the least almost-everywhere bound in $[0, + \infty]$ (The essential supremum of a measurable function with respect to a measure).

[F6]

Under Countable Choice, every one-dimensional box with any choice of faces is Borel and has measure equal to its length; in particular $λ ([0, δ)) = δ$ for $0 < δ \leq 1$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F7]

AC is assumed here. It implies DC and Countable Choice, which supplies the hypothesis of [F6] (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), AC implies DC implies countable choice).

[F8]

The action theorem gives the exact norm of the induced operator as the essential supremum of the fibre norms (Measurable essentially bounded operator fields act decomposably).

[F9]

The action theorem identifies the induced adjoint and product fields with the operator adjoint and product (Measurable essentially bounded operator fields act decomposably).

[F10]

The preceding example defines the diagonal algebra as scalar multiplication by $f (t) I_{2}$ (Multiplicity-two diagonal representation).

Verification

Proof technique: compute the coefficient functions and pointwise norms, then apply the direct-integral action theorem and exhibit a vector separating $T$ from every scalar diagonal operator.

Given: the preceding multiplicity-two field and $T_{t}$ as in the Example.

1.1F1F2algebra

In the constant standard basis $e_{1}, e_{2}$ , the four fundamental matrix coefficients of $T_{t}$ are the Borel functions $0, t, 1 - t, 0$ , so $T_{t}$ is weakly measurable.

1.2

The operator-norm and essential-supremum calculations use [F1, F4, F5, F6, F7, algebra]. For $z = (z_{1}, z_{2}) \in C^{2}$ , $∥ T_{t} z ∥^{2} = t^{2} ∣ z_{2} ∣^{2} + (1 - t)^{2} ∣ z_{1} ∣^{2} .$ The operator norm definition [F4], with the two standard unit vectors as witnesses for the larger coefficient, gives $∥ T_{t} ∥ = \max {t, 1 - t} .$ This is a Borel function bounded by $1$ . For every $M \in [0, 1)$ , $∥ T_{t} ∥ > M$ on $[0, 1 - M)$ , whose measure is $1 - M > 0$ by [F6] and [F7]. So no $M < 1$ is an almost-everywhere bound, while $1$ is a pointwise bound; therefore ${ess sup}_{t \in [0, 1]} ∥ T_{t} ∥ = 1$ by [F5]. The field is essentially bounded. At $t = 0$ and $t = 1$ its rank is one, while for $0 < t < 1$ its rank is two; the norm formula holds in all cases.

2.1

The action theorem induces $T$ with norm one and identifies its adjoint and square fields. [F3, F7, F8, F9, step 1.1, step 1.2, algebra] It gives $T = \int^{\oplus} T_{t} d λ (t)$ and $∥ T ∥ = 1$ . Direct matrix multiplication yields $T_{t}^{*} = (\begin{array}{cc} 0 & 1 - t \\ t & 0 \end{array}), T_{t}^{2} = (\begin{array}{cc} t (1 - t) & 0 \\ 0 & t (1 - t) \end{array}) .$

3.1

Pointwise commutation and the action theorem put $T$ in $D^{'}$ . [F3, F9, F10, step 1.1, step 1.2, step 2.1, algebra] For $f \in L^{\infty} ([0, 1], λ)$ , the scalar field $f (t) I_{2}$ induces the corresponding $M_{f} \otimes I_{2}$ by [F10] and [F3]. At every $t$ , $T_{t} (f (t) I_{2}) = f (t) T_{t}$ . The product clause of [F3] therefore shows that $T (M_{f} \otimes I_{2}) = (M_{f} \otimes I_{2}) T$ . Hence $T \in D^{'}$ .

3.2

A vector witness separates $T$ from every $R \in D$ , proving $T \notin D$ . [F1, step 2.1, algebra] Let $η = (1, 0) \in L^{2} (λ) \oplus L^{2} (λ)$ , so $∥ η ∥ = 1$ by [F1]. Then $T η = (0, 1 - t)$ . For any $R = M_{f} \otimes I_{2} \in D$ , $R η = (f, 0)$ , and $∥ T η - R η ∥^{2} = ∥ f ∥_{2}^{2} + \int_{0}^{1} (1 - t)^{2} d λ (t) \geq \frac{1}{3} > 0.$ Thus $T \neq R$ for every $R \in D$ .

4.1step 1.1step 1.2step 2.1step 3.1step 3.2

Steps 1.1–1.2 prove weak measurability and the exact essential norm; step 2.1 gives the induced operator, its adjoint and square; steps 3.1 and 3.2 prove that $T \in D^{'}$ and $T \notin D$ . $□$

Source qualifications

Bekka–de la Harpe, Chapter 1 §1.H, state the action and essential-supremum norm formula for measurable essentially bounded fields on a constant Hilbert space and prove the constant-field commutant characterization in Theorem 1.H.4, with Corollary 1.H.5 identifying the nonabelian commutant for fibres of dimension greater than one. The present calculation checks all hypotheses for the continuous $C^{2}$ field explicitly. Bruhat, Part III Chapter 10 §§1.7–1.8, states the corresponding matrix-coefficient, action and norm results in a locally compact/Lusin field convention; the polynomial matrix coefficients here are continuous and the local action theorem supplies the standard-Borel operator conventions used in this item.

Boundary cases

Empty: Not applicable because the base is the fixed nonempty interval $[0, 1]$ and $λ ([0, 1]) = 1$ .
Zero: Not applicable because every fibre is $C^{2}$ and the base has measure $1$ , so $H \neq {0}$ .
One: Not applicable because the example fixes two-dimensional fibres throughout and makes no one-dimensional-fibre claim.
Degenerate: Checked at $t = 0, 1$ , where $T_{t}$ has rank one, and on $0 < t < 1$ , where it has rank two; the norm, adjoint and square formulas hold on all of $[0, 1]$ .
Endpoints: Checked explicitly: $∥ T_{0} ∥ = ∥ T_{1} ∥ = 1$ , and for every $M < 1$ the set $[0, 1 - M)$ where $∥ T_{t} ∥ > M$ has positive measure.
Nonempty choice: AC is stated; it supplies the action theorem hypothesis and, via DC and Countable Choice, the box-measure input. The matrix field and vector witness are explicit, with no further choice.
Iff directions: Not applicable because the example gives one explicit commuting operator outside $D$ and asserts no equivalence.

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Sources

B. Bekka and P. de la Harpe, Unitary Representations of Groups, Duals, and Characters (standard reference, not scraped)
F. Bruhat, Lectures on Lie Groups and Representations of Locally Compact Groups, Ch. 10 (standard reference, not scraped)