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Unitary intertwiners preserve fibre multiplicity over a standard Borel base

Statement

Assume AC. Let $X$ be a standard Borel space, $μ$ a nonzero finite Borel measure on $X$ , and let $m, m^{'} : X \to {1, 2, \dots} \cup {\infty}$ be Borel multiplicity functions. If there is a unitary $U : L^{2} (X, μ; m) ⟶ L^{2} (X, μ; m^{'})$ with $U M_{f} = M_{f} U$ for every bounded Borel $f : X \to C$ , then $m = m^{'}$ $μ$ -almost everywhere. Consequently a multiplicity model of a projection-valued measure over a fixed base is unique in multiplicity, and a unitary intertwiner of two such models is a decomposable operator whose fibres are unitary almost everywhere.

Facts & Assumptions

Given: AC, a standard Borel space $X$ with a nonzero finite Borel measure $μ$ , Borel multiplicity functions $m, m^{'}$ , and a unitary $U : L^{2} (X, μ; m) \to L^{2} (X, μ; m^{'})$ with $U M_{f} = M_{f} U$ for all bounded Borel $f$ .

[F1]

For a Borel function $k : X \to N \cup {\infty}$ the field with fibre $C^{k (x)}$ and fundamental family $e_{j} (x) =$ the $j$ -th coordinate vector for $j \leq k (x)$ and $0$ otherwise has Borel Gram coefficients $x \mapsto δ_{i j} 1_{j \leq k (x)}$ and spans a dense subspace of each fibre; its direct integral $L^{2} (X, μ; k)$ is a Hilbert space of measurable square-integrable sections, and in the constant case $k \equiv r$ the fibre family $e_{1}, \dots, e_{r}$ is orthonormal and complete, so Parseval in each fibre makes $[ξ] \mapsto (⟨ ξ, e_{j} ⟩)_{j \leq r}$ an isometry onto the vector-valued $L^{2}$ -space $⨁_{j \leq r} L^{2} (X, μ)$ (Measurable Hilbert field from a countable fundamental family, Direct integral of a measurable Hilbert field, Measurable sections have measurable pointwise inner products, Parseval equivalences for an orthonormal family).

[F2]

For $f \in L^{\infty} (X, μ)$ the multiplication $M_{f}$ is a bounded operator on $L^{2} (X, μ; k)$ , $∥ M_{f} ∥ \leq ∥ f ∥_{\infty}$ , and for a Borel $B$ the operator $M_{1_{B}}$ is the orthogonal projection onto the closed subspace of sections supported in $B$ (Direct integral of a measurable Hilbert field, Hilbert space).

[F3]

On a $σ$ -finite standard Borel base with a measurable Hilbert field, the commutant of the diagonal multiplications $D = {M_{f} : f \in L^{\infty}}$ is exactly the set of decomposable operators; an operator commuting with $D$ is induced by a weakly measurable, essentially bounded field of fibre operators, and that field is unique up to a null set (Decomposable operators are the commutant of diagonal multiplication, Measurable and decomposable operator fields, Measurable essentially bounded operator fields act decomposably).

[F4]

A unitary between complex inner product spaces is a bijective linear isometry, so it exists only between fibres of equal dimension $k = k^{'}$ in $N \cup {\infty}$ : a finite-dimensional $C^{k}$ cannot be linearly isomorphic to $C^{\infty}$ , and $C^{k} ≇ C^{k^{'}}$ for distinct finite $k, k^{'}$ (Hilbert space, Separability: the existence of an at most countable dense subset, Orthonormal families, complete orthonormal systems and Hilbert bases).

[F5]

The sets ${x : k (x) > j}$ are Borel for a Borel $k$ , and $X$ is the countable disjoint union of the Borel sets $B_{k, k^{'}} = {m = k} \cap {m^{'} = k^{'}}$ , so measures on $X$ are countably additive over this partition; the standard Borel base is $σ$ -finite for the finite measure $μ$ (Standard Borel spaces, Monotone convergence for the integral, A sigma-finite signed measure that is absolutely continuous with respect to a sigma-finite positive measure has a unique almost-everywhere density).

Proof

technique · direct

Given: AC, the data and the unitary $U$ of the statement.

1.1F2

For each Borel set $B \subseteq X$ , the identity $U M_{1_{B}} = M_{1_{B}}^{'} U$ and unitarity of $U$ give $M_{1_{B}}^{'} = U M_{1_{B}} U^{- 1}$ , so $U$ carries the range of the projection $M_{1_{B}}$ onto the range of $M_{1_{B}}^{'}$ ; restricting to these closed subspaces yields a unitary $U_{B}$ from the sections of $L^{2} (X, μ; m)$ supported in $B$ to the sections of $L^{2} (X, μ; m^{'})$ supported in $B$ .

2.1step 1.1F1F2

Fix $k, k^{'} \in N \cup {\infty}$ and put $B = B_{k, k^{'}}$ , a Borel set by [F5]. The supported subspace of $L^{2} (X, μ; m)$ over $B$ is, by [F1], the direct integral over $(B, μ ∣_{B})$ of the constant field with fibre $C^{k}$ ; similarly the target over $B$ is the constant field with fibre $C^{k^{'}}$ . The unitary $U_{B}$ of [step 1.1] intertwines the multiplication operators for all bounded Borel $f$ on $B$ , because $U_{B}$ is the restriction of $U$ and $M_{f}$ preserves the supported subspaces.

3.1step 2.1F3

Assume $μ (B) > 0$ . Apply [F3] to the sum field $C^{k} \oplus C^{k'}$ and its block operator whose only nonzero block is $U_{B}$ from the first summand to the second. This operator commutes with all diagonal multiplications, so its off-diagonal block is decomposable: there is a weakly measurable, essentially bounded operator field $x \mapsto T_{x} \in B (C^{k}, C^{k^{'}})$ with $U_{B}$ acting by $T_{x}$ fibrewise; the same applies to $U_{B}^{*} = U_{B}^{- 1}$ , and since $U_{B}^{*} U_{B} = I$ and $U_{B} U_{B}^{*} = I$ , the uniqueness of decomposable fields in [F3] gives $T_{x}^{*} T_{x} = I_{k}$ and $T_{x} T_{x}^{*} = I_{k^{'}}$ for $μ$ -almost every $x \in B$ . Thus for almost every $x$ the fibre map $T_{x}$ is a unitary between $C^{k}$ and $C^{k^{'}}$ .

4.1step 3.1F4

Hence $k = k^{'}$ whenever $μ (B_{k, k^{'}}) > 0$ : by [step 3.1] a unitary $C^{k} \to C^{k^{'}}$ exists for some $x$ , and [F4] says this forces $k = k^{'}$ .

5.1step 4.1F5

Therefore ${m \neq m^{'}} = ⋃_{k \neq k^{'}} B_{k, k^{'}}$ is a countable union of sets of $μ$ -measure zero, hence $μ$ -null by countable additivity; that is, $m = m^{'}$ $μ$ -almost everywhere.

6.1step 5.1F3∎

The intertwiner $U$ itself is decomposable: its block operator on the direct sum of the two fields commutes with all diagonal multiplications, so [F3] represents its off-diagonal block by a weakly measurable essentially bounded field. Applying the same to $U^{- 1} = U^{*}$ , whose field is the fibrewise adjoint up to a null set by the uniqueness clause of [F3], and using $U^{*} U = U U^{*} = I$ as in [step 3.1], its fibres are unitary almost everywhere. Thus every unitary intertwiner of two multiplicity models over the fixed base $(X, μ)$ has unitary fibres a.e., and the multiplicity is unique, which is exactly the rigidity statement a multiplicity model of a projection-valued measure over a fixed base invokes.

Depends on

Used by

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Sources

J. B. Conway, A Course in Functional Analysis, 2nd ed., Chapter IX §10, Propositions 10.17-10.19 (multiplicity rigidity) (standard reference, not scraped)
G. Misra, E. K. Narayanan and C. Varughese, Mackey Imprimitivity and commuting tuples of homogeneous normal operators, arXiv:2402.15737 (standard reference, not scraped)