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The unilateral shift obstructs a cyclic linear trace extension

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $ℓ^{2} : = ℓ^{2} (N, F)$ with standard basis $(u_{n})_{n \in N}$ given by $u_{n} (m) = δ_{m n}$ , and let $S \in B (ℓ^{2})$ be the unilateral forward shift $S u_{n} : = u_{n + 1} (n \in N),$ extended linearly and by continuity, with $P_{0} : = ⟨ \cdot, u_{0} ⟩ u_{0}$ . Then $S^{*} S = I, S S^{*} = I - P_{0}, S^{*} S - S S^{*} = P_{0},$ neither $S^{*} S = I$ nor $S S^{*} = I - P_{0}$ is trace class (Trace class operator), while $P_{0}$ is rank one with $tr (P_{0}) = 1$ (Adjoint, norm and trace of an operator of rank at most one). Consequently there is no linear functional $τ$ on a linear subspace of $B (ℓ^{2})$ that contains the trace-class operators, $S^{*} S$ , and $S S^{*}$ , agrees with the usual trace on trace-class operators, and satisfies $τ (S^{*} S) = τ (S S^{*})$ . Thus the cyclicity identity $tr (S T) = tr (T S)$ of Cyclicity of the trace has no linear cyclic extension whose domain contains this pair of nonsummable products.

Facts & Assumptions

Given: Countable Choice, the space $ℓ^{2}$ with its standard basis $(u_{n})$ , the forward shift $S$ , and the projection $P_{0} = ⟨ \cdot, u_{0} ⟩ u_{0}$ .

[A1]

The standard basis and shifts. The vectors $u_{n} \in ℓ^{2}$ satisfy $⟨ u_{m}, u_{n} ⟩ = δ_{m n}$ and $∥ u_{n} ∥_{2} = 1$ , and if $a \in ℓ^{2}$ has $⟨ a, u_{n} ⟩ = 0$ for every $n$ then $a = 0$ , so the zero-complement characterisation makes $(u_{n})$ a complete orthonormal family, a Hilbert basis of $ℓ^{2}$ ; hence every $x \in ℓ^{2}$ equals $\sum_{n} ⟨ x, u_{n} ⟩ u_{n}$ and two vectors with equal coefficients coincide (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , Parseval equivalences for an orthonormal family, Fourier expansion in a Hilbert space, Orthonormal families, complete orthonormal systems and Hilbert bases, Real and complex inner-product spaces and their induced length). The forward shift has $∥ S u_{n} ∥ = 1$ , is an isometry, and its adjoint satisfies $S^{*} u_{0} = 0$ , $S^{*} u_{n} = u_{n - 1}$ for $n \geq 1$ , the adjoint being characterised by $⟨ S x, y ⟩ = ⟨ x, S^{*} y ⟩$ (Hilbert-adjoint identities, The Hilbert-space adjoint of a bounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A2]

Trace-class diagonal test. If $T$ is trace class and $E$ is a Hilbert basis, then $\sum_{e \in E} ∣ ⟨ T e, e ⟩ ∣ \leq ∥ T ∥_{1} < + \infty$ ; hence an operator $T$ for which some Hilbert basis has infinitely many $e$ with $⟨ T e, e ⟩ = 1$ is not trace class (Trace of a trace class operator, Trace is absolutely convergent and basis independent, Trace class operator, Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[A3]

Rank-one operators. For $u, v \in H$ the operator $⟨ \cdot, v ⟩ u$ has adjoint $⟨ \cdot, u ⟩ v$ , norm $∥ u ∥ ∥ v ∥$ , and trace $⟨ u, v ⟩$ ; in particular $P_{0} = ⟨ \cdot, u_{0} ⟩ u_{0}$ has trace $⟨ u_{0}, u_{0} ⟩ = 1$ (Adjoint, norm and trace of an operator of rank at most one, Orthonormal families, complete orthonormal systems and Hilbert bases).

[A4]

Cyclicity theorem. If $T$ is trace class and $S$ is bounded then $S T$ and $T S$ are trace class and have equal traces (Cyclicity of the trace).

Counterexample

technique · direct

Given: Countable Choice, the shift $S$ , the projection $P_{0}$ , and the standard basis.

1.1

The products. Since $S$ is an isometry, $⟨ S^{*} S x, y ⟩ = ⟨ S x, S y ⟩ = ⟨ x, y ⟩$ for all $x, y$ by [A1], so $S^{*} S = I$ . Similarly $⟨ S S^{*} x, u_{n} ⟩ = ⟨ S^{*} x, S^{*} u_{n} ⟩$ for $n \geq 1$ equals $⟨ S^{*} x, u_{n - 1} ⟩ = ⟨ x, u_{n} ⟩$ , while $⟨ S S^{*} x, u_{0} ⟩ = ⟨ S^{*} x, S^{*} u_{0} ⟩ = 0$ ; hence $S S^{*}$ fixes each $u_{n}$ with $n \geq 1$ and annihilates $u_{0}$ , that is $S S^{*} x = x - ⟨ x, u_{0} ⟩ u_{0} = (I - P_{0}) x$ for all $x$ , and $S^{*} S - S S^{*} = P_{0}$ .

1.2

Neither product is trace class. For $I$ with the Hilbert basis $(u_{n})$ , every diagonal coefficient is $⟨ I u_{n}, u_{n} ⟩ = 1$ , so $\sum_{n} ∣ ⟨ I u_{n}, u_{n} ⟩ ∣ = + \infty$ and $I$ is not trace class by [A2]. For $I - P_{0}$ , the coefficients at $u_{n}$ with $n \geq 1$ are $⟨ (I - P_{0}) u_{n}, u_{n} ⟩ = 1$ , again infinitely many equal to $1$ , so $I - P_{0}$ is not trace class by [A2].

A1A2

1.3

The difference has trace one. $P_{0} = ⟨ \cdot, u_{0} ⟩ u_{0}$ is a rank-one operator whose trace is $⟨ u_{0}, u_{0} ⟩ = 1$ by [A3]; note that $u_{0} \neq 0$ because it is a unit vector.

2.1

Conclusion. Suppose that a linear functional $τ$ on a linear subspace containing the trace-class operators, $S^{*} S$ , and $S S^{*}$ agreed with the usual trace on trace-class operators and satisfied $τ (S^{*} S) = τ (S S^{*})$ . By linearity, [step 1.1], and [step 1.3], $0 = τ (S^{*} S) - τ (S S^{*}) = τ (P_{0}) = tr (P_{0}) = 1,$ a contradiction. Hence no such cyclic linear extension exists. The products themselves are not trace class by [step 1.2], so [A4] neither asserts nor assigns their individual traces.

step 1.1step 1.2step 1.3A4assume-contradischarge-contradiction∎

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Sources

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.6, the shift and the failure of cyclicity without trace class (standard reference, not scraped)