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A commutator vanishing on a dense invariant domain does not imply strong spectral commutation

Statement refuted

Let $H \neq {0}$ be a complex Hilbert space, let $A, B$ be self-adjoint operators on $H$ , and let $D \subseteq D (A) \cap D (B)$ be a dense linear subspace with $A D \subseteq D (B)$ and $B D \subseteq D (A)$ . If $A B u = B A u$ for every $u \in D$ , then the spectral projection-valued measures of $A$ and $B$ strongly commute in the sense of Strong spectral commutation of self-adjoint operators clause (b).

Facts & Assumptions

Given: The Axiom of Choice and Countable Choice; the interval $(0, 1)$ ; the twisted-periodic momentum operators $P_{0}$ and $P_{π}$ on $L^{2} (0, 1; C)$ ; the space $D : = C_{c}^{\infty} (0, 1)$ ; the exponential functions $φ_{0} = 1$ and $ψ_{0} (x) = e^{i π x}$ .

[F1]

Twisted-periodic momentum. For $0 < L < \infty$ and $θ \in [0, 2 π)$ put $D_{θ} : = {u \in H^{1} (0, L) : u^{*} (L) = e^{i θ} u^{*} (0)}$ and $P_{θ} u : = - i D u$ . Then $D_{θ}$ is dense, $P_{θ}$ is self-adjoint, the functions $φ_{n} (x) = L^{- 1 / 2} e^{i (2 π n + θ) x / L}$ form a complete orthonormal family, $P_{θ} φ_{n} = κ_{n} φ_{n}$ with $κ_{n} = (2 π n + θ) / L$ , the eigenvalues are simple, and $σ (P_{θ}) = {κ_{n} : n \in Z}$ (The twisted-periodic momentum operator on an interval clauses (a)-(c)).

[F2]

The trace and integration by parts. $H^{1} (0, L)$ consists of the classes with an $L^{2}$ weak derivative, each class has a continuous absolutely continuous representative $u^{*}$ with $\int_{0}^{L} u \overline{v^{'}} = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{L} u^{'} \overline{v}$ for $u, v \in H^{1} (0, L)$ ; a compactly supported smooth function vanishes near the endpoints, so $C_{c}^{\infty} (0, L) \subseteq D_{θ}$ for every $θ$ (One-dimensional Sobolev spaces on intervals, Integration by parts and the endpoint-vanishing space on intervals clauses (a),(b)).

[F3]

Density. $C_{c}^{\infty} (0, 1)$ is dense in $L^{2} (0, 1; C)$ (Smooth compactly supported functions of an open set are dense in $L^{2}$ ).

[F4]

Atoms are eigenspace projections. For a self-adjoint operator with spectral projection-valued measure $E$ and eigenvalue $λ$ , the projection $E ({λ})$ is the orthogonal projection onto the eigenspace $\ker (A - λ)$ (Atoms of the spectral measure and eigenvalues of a self-adjoint operator clause (a), Projection valued measure, Spectral theorem for unbounded self-adjoint operators (PVM form)).

[F5]

The complex exponential and its integral. The function $x \mapsto e^{i α x}$ is differentiable with derivative $i α e^{i α x}$ (The complex exponential by its power series, The complex exponential is entire and its complex derivative is itself), and for a continuous function on an interval with a continuous primitive $G$ one has $\int_{a}^{b} G^{'} = G (b) - G (a)$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , applied to real and imaginary parts).

Counterexample

technique · exhibit two distinct self-adjoint extensions of the same symmetric operator that agree on the dense invariant domain of test functions, so that the commutator vanishes there, and show by explicit eigenprojections that their spectral projection-valued measures fail to commute

Given: The data of the refuted statement and [F1]-[F5]; take $H = L^{2} (0, 1; C)$ , $A : = P_{0}$ , $B : = P_{π}$ , $D : = C_{c}^{\infty} (0, 1)$ .

1.1givenF2F3algebra

[The hypothesis of the refuted statement holds] By [F2], $D \subseteq D_{0} \cap D_{π} = D (A) \cap D (B)$ , and both operators act on $D$ by $u \mapsto - i D u$ , which maps $D$ into itself; hence for $u \in D$ the products $A B u = - D^{2} u$ and $B A u = - D^{2} u$ are defined and $[A, B] u = 0$ . By [F3] the space $D$ is dense in $H$ . So the hypotheses of the refuted statement are satisfied.

1.2givenF1

[Spectral data] Applying [F1] with $L = 1$ : $A = P_{0}$ has the complete orthonormal family $φ_{n} (x) = e^{2 π i n x}$ with simple eigenvalues $2 π n$ , and $B = P_{π}$ has the complete orthonormal family $ψ_{n} (x) = e^{i (2 n + 1) π x}$ with simple eigenvalues $(2 n + 1) π$ , for $n \in Z$ . In particular $φ_{0} = 1$ spans the eigenspace of $A$ at $0$ and $ψ_{0} (x) = e^{i π x}$ spans the eigenspace of $B$ at $π$ .

2.1step 1.2F1F4algebra

[The two spectral projections] By [F4], $E_{A} ({0})$ is the orthogonal projection onto $C φ_{0}$ , and $E_{B} ({π})$ is the orthogonal projection onto $C ψ_{0}$ . Since $φ_{0} = 1$ is a unit vector, $E_{A} ({0}) u = ⟨ u, φ_{0} ⟩ φ_{0}$ ; since $∥ ψ_{0} ∥^{2} = \int_{0}^{1} ∣ e^{i π x} ∣^{2} d x = 1$ , likewise $E_{B} ({π}) u = ⟨ u, ψ_{0} ⟩ ψ_{0}$ .

2.2step 1.2F5algebra

[The inner products] By [F5], $\int_{0}^{1} e^{- i π x} d x = [e^{- i π x} / (- i π)]_{0}^{1} = (e^{- i π} - 1) / (- i π) = 2 / (i π) = - 2 i / π$ , and by conjugation $\int_{0}^{1} e^{i π x} d x = 2 i / π$ . Hence $⟨ φ_{0}, ψ_{0} ⟩ = - 2 i / π$ and $⟨ ψ_{0}, φ_{0} ⟩ = 2 i / π$ .

3.1step 1.1step 2.1step 2.2algebra∎

[Noncommutation] With $u : = φ_{0} = 1$ and the projections of step 2.1, $E_{B} ({π}) u = ⟨ u, ψ_{0} ⟩ ψ_{0} = - \frac{2 i}{π} ψ_{0}, E_{A} ({0}) E_{B} ({π}) u = ⟨ u, ψ_{0} ⟩ ⟨ ψ_{0}, u ⟩ u = (\frac{2}{π})^{2} u,$ because $⟨ ψ_{0}, u ⟩ = \overline{⟨ u, ψ_{0} ⟩} = 2 i / π$ , so the product of the two coefficients is $(- 2 i / π) (2 i / π) = 4 / π^{2}$ . Thus $E_{A} ({0}) E_{B} ({π}) u = (4 / π^{2}) u$ while $E_{B} ({π}) E_{A} ({0}) u = E_{B} ({π}) u = - (2 i / π) ψ_{0}$ . The vectors $u = 1$ and $ψ_{0} = e^{i π x}$ are linearly independent in $L^{2} (0, 1; C)$ (the continuous function $e^{i π x}$ is not constant on $(0, 1)$ ), so the two results differ. Therefore $E_{A} ({0}) E_{B} ({π}) \neq E_{B} ({π}) E_{A} ({0})$ : the spectral projection-valued measures of $A$ and $B$ do not strongly commute, and the refuted statement fails for this pair.

Remarks

Where the domain identity loses information. Both operators are the differential expression $- i D$ on different self-adjoint domains; the dense invariant domain $C_{c}^{\infty} (0, 1)$ sees only the action, not the boundary conditions that determine the spectra. No graph-norm core property for either self-adjoint extension is asserted. The domain identity $[A, B] u = 0$ is therefore not a sufficient condition for strong spectral commutation. Strong spectral commutation still implies equality of the products on every common invariant domain where both compositions are defined.
Boundedness is the missing hypothesis. For bounded everywhere-defined self-adjoint operators, a commutator identity on a dense subspace extends by continuity to $A B = B A$ on all of $H$ , which is equivalent to strong spectral commutation. This counterexample refutes only the converse implication for unbounded operators; it does not refute the implication from strong spectral commutation to equality on a common invariant domain. This is why Strong spectral commutation of self-adjoint operators is defined at the level of projection-valued measures.

A commutator vanishing on a dense invariant domain does not imply strong spectral commutation

Statement refuted

Facts & Assumptions

Counterexample

Remarks

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