Consistency of a counterexample to Naimark's problem

Unique Extension of Pure States to Crossed Products

In this section we consider the problem: if f and g are inequivalent pure states on a C^*-algebra 𝒜, is it possible to find a C^*-algebra ℬ that contains 𝒜 such that (i) f and g have unique state extensions f′ and g′ to ℬ and (ii) f′ and g′ are equivalent? We find that the answer is yes if A is simple, separable, and unital, and moreover we can ensure that ℬ is also simple, separable, and unital, with the same unit as 𝒜.

A net (a _λ) of positive, norm-one elements of a C^*-algebra 𝒜 is said to excise a state f on 𝒜 if → for 0 all (ref. 18, p. 1239). By ref. 18, proposition 2.2, for each pure state f of a C ^*-algebra 𝒜, there is a decreasing net (a _λ) that excises f. As described in ref. 18, p. 1240 (see also ref. 6, p. 87), the support projection p of f is a rank 1 projection in . Further, as shown in the last 7 lines of the proof of ref. 18, proposition 2.3, a _λ → p in the strong operator topology on A ^**.

Lemma 1. Suppose f and g are inequivalent pure states on a C ^* -algebra 𝒜, (a _λ ) excises f, and (b _λ ) excises g. Also assume that (b _λ ) is decreasing. Then ∥a _λ xb _λ∥ → 0 for all .

Proof: Note that we are assuming (a _λ) and (b _λ) have the same index set. This is easy to arrange by replacing two different index sets with their product. Let q denote the rank one projection in that supports the pure state g. Suppose the lemma is false for some . Then, multiplying x by a large enough positive number and passing to a subnet, we can assume that ∥a _λ xb _λ∥ > 1 for all λ.

We first show that f(xb _κ x ^*) 0. In , xb _κ x ^* xqx ^*. Let y denote the central cover of q in . Since g is pure, is isomorphic to some B(H). Thus, f must be 0 on lest it be equivalent to g. Therefore, f(xb_κx^*)f(xqx^*)=f(xyqx^*)=f(y(xqx^*))=0. Fix κ₀ so that , and note that because (a _λ) excises f, there exists λ₀ such that λ ≥ λ₀ implies . Choose λ so that λ ≥ λ₀ and λ ≥ κ₀. Using b _λ ≤ b _κ0, we have a contradiction.

Let θ be an action of a discrete group G on a unital C^*-algebra 𝒜. The reduced crossed product can be defined as the C^*-algebra that acts on the right Hilbert module and is generated by (i) for each , the multiplication operator M_x defined by M_xφ(g) = θ_g^-1(x)φ(g) and (ii) for each h ∈ G, the translation operator T_h defined by T_h φ(g) = φ(h ^-1 g). Observe that the map isomorphically embeds 𝒜 in , so we can regard 𝒜 as a subalgebra of by identifying x with M_x. Denote the identity of G by e.

Theorem 2. Let 𝒜 be a unital C ^* -algebra, let f be a pure state on 𝒜, and let θ be an action of a discrete group G on 𝒜. Embed 𝒜 in in the manner just described. Then f has a unique state extension to if and only if f is inequivalent to for all g ≠ e.

Proof: (⇒) Suppose that for some g ∈ G not the identity, f is equivalent to ; we must show that f does not extend uniquely to the crossed product. Let u ∈ A be a unitary such that (ref. 6, proposition 3.13.4). Let be the GNS representation associated to f, and let ξ ∈ H_f be the image of the unit of 𝒜, so that for all . Let be any positive, norm-one element such that f(b) = 1; then π(b)(ξ) = ξ. Furthermore, θ_g(u ^* bu) is also a positive, norm-one element such that f(θ_g(u ^* bu)) = f(b) = 1, so the same is true of this element, and it follows that π(bθ_g(u ^* bu))(ξ) = ξ. Hence ∥bθ_g(u ^* bu)∥ = 1.

Now let x = θ_g(u ^*). Let be any positive, norm-one element such that f(b) = 1 and let . We will show that ∥y - b(xT_g)b∥ ≥ 1 (computing the norm in the crossed product), which implies nonunique extension by ref. 19, theorem 3.2. Observe first that where has norm 1. [It is the product of bθ_g(u ^* bu), which we showed above has norm 1, with the unitary θ_{g ^-1}(u).] Now to estimate the norm of y - b(xT_g)b = y - cT_g, consider the element that satisfies (the unit of 𝒜) and δ_e(h) = 0 for all h ≠ e. Since g ≠ e, we have (y - cT_g)(δ_e) = φ where φ(e) = y, φ(g)=-θ_g^-1(c), and φ(h) = 0 for all other h. The norm of δ_e in is and the norm of φ is since ∥θ_{g ^-1}(c)∥=∥c∥=1. Thus the norm of y - cT_g is at least 1, as we needed to show.

(⇐) Suppose f is not equivalent to for any g ≠ e. To verify that f has a unique extension to the crossed product, according to (ref. 19, theorem 3.2), we must, for every element z of the crossed product and every ε > 0, find and a positive norm-one element such that f(b) = 1 and ∥x - bzb∥ ≤ ε. It is sufficient to accomplish this only for a dense set of elements z, so let z = Σ_gx_gT_g where each x_g is an element of 𝒜 and x_g ≠ 0 for only finitely many g. (Such sums are clearly dense in .)

Let (a _λ) be a decreasing net in 𝒜 that excises f and satisfies f(a _λ) = 1 for all λ (ref. 18, proposition 2.2). We claim that , which will complete the proof. Observe first that since (a _λ) excises f. Since z is a finite sum and x_e = x_eT_e we now need only show that ∥a _λ(x_gT_g)a _λ∥ → 0 for each g ≠ e. But a _λ(x_gT_g)a _λ = (a _λ x_gb _λ)T_g where b _λ = θ_g(a _λ) and (b _λ) is a decreasing net that excises . By hypothesis, f and are inequivalent, so Lemma 1 now implies ∥a _λ x_gb _λ∥ → 0. Thus ∥(a _λ x_gb _λ)T_g∥ → 0, as desired.

Note that in the proof of the reverse direction of Theorem 2, the fact that z = Σx_gT_g can be “compressed” to x_e implies that the unique extension f′ of f satisfies f′(z) = f(x_e).

By ref. 20, the pure state space of any simple, separable C^*-algebra 𝒜 is homogeneous: given any two inequivalent pure states f and g, there is an automorphism ω of 𝒜 such that . In the following corollary we need an even more powerful “strong transitivity” result that was proven in ref. 21, theorem 7.5 for a certain class of simple, separable C^*-algebras. By combining the methods of the two papers, that result can be achieved for all simple, separable C^*-algebras (A. Kishimoto, personal communication).

The result we need states the following: Suppose 𝒜 is a simple, separable C ^* -algebra and (π_n) and (ρ_n) are sequences of irreducible representations such that the π_n are mutually inequivalent, as are the ρ_n. Then there is an automorphism ω of 𝒜 such that π_n is equivalent to for all n. It can be proven by first replacing every result in ref. 20 involving a single pure state f (or a pair of pure states f and g) with a corresponding result involving a finite set of mutually inequivalent pure states f_i (or a pair of such sets, with, for each i, f_i and g_i related in the same way that f and g were). The only difference in the proofs will be that every application of Kadison's transitivity theorem will now require the n-fold version for a finite family of mutually inequivalent pure states (ref. 7, theorem 1.21.16). This achieves a proof of ref. 21, theorem 7.3, for any simple separable C^*-algebra, which can be converted into a proof of the result we need in the same way that this is done in ref. 21, theorem 7.5.

Corollary 3. Let 𝒜 be a simple, separable, unital C ^* -algebra and let f and g be inequivalent pure states on 𝒜. Then there is a simple, separable, unital C ^* -algebra ℬ that unitally contains 𝒜 such that f and g have unique state extensions to ℬ and these extensions are equivalent.

Proof: Observe first that if f ₁ and f ₂ are equivalent pure states on 𝒜 and f ₁ has a unique state extension to ℬ, then so does f ₂. Indeed, if is a unitary such that f ₁ = u ^* f ₂ u then u also belongs to ℬ and conjugates the set of extensions of f ₁ with the set of extensions of f ₂.

It follows from the result quoted before this corollary that given any sequence (π_n) (n ∈ Z) of inequivalent irreps, there is an automorphism ω of 𝒜 such that is equivalent to π_{n +1} for all n. Now, since A is simple and separable and has inequivalent pure states it cannot be type I, and therefore it has uncountably many inequivalent irreps (ref. 6, corollary 6.8.5). Let (π_n) be any sequence of mutually inequivalent irreps such that π₁ and π₂ are the GNS irreps arising from f and g, and find an automorphism ω as above. Then g is equivalent to , and neither f nor g is equivalent to itself composed with any nonzero power of ω.

Define θ: Z → Aut(𝒜) by θ_n = ωⁿ and let ℬ be the crossed product of 𝒜 by this action of Z. By Theorem 2, f and g extend uniquely to ℬ, and therefore so do all pure states equivalent to g, in particular , by the comment at the start of this proof. Now if f′ and are the unique extensions of f and to ℬ, then for any finite sum Σx_nT_n using the comment following Theorem 2. This shows that , and hence f′ is equivalent to . Using the first paragraph again, we see that the unique extensions of f and g to ℬ are equivalent.

ℬ is clearly separable and unital, and it unitally contains 𝒜. It is simple by ref. 22, theorem 3.1.

We thank A. Kishimoto for a simplification in the above proof.

The Counterexample

A subset S of is said to be closed if for every countable S ₀ ⊂ S we have sup S ₀ ∈ S. It is unbounded if for every there exists β ∈ S such that β > α.

We call a nested transfinite sequence of C^*-algebras () continuous if for every limit ordinal α we have .

Lemma 4. Let (), , be a continuous nested transfinite sequence of separable C ^*-algebras and set . Then 𝒜 is a C ^*-algebra, and if f is a pure state on 𝒜 then {α: f restricts to a pure state on } is closed and unbounded.

Proof: We observe first that 𝒜 is automatically complete. For any sequence we can find indices α_n such that , and then (x_n) ⊂ A _α for α = supα_n. Thus if (x_n) is Cauchy, its limit belongs to and hence to 𝒜. This shows that 𝒜 is a C^*-algebra.

Let f be a pure state on 𝒜 and let S = {α: f restricts to a pure state on }. First we verify that S is closed. Suppose S ₀ ⊂ S is countable and let α = sup S ₀. If is not pure then we can write , where f ₁ and f ₂ are distinct states on . Now is dense in by continuity (unless α ∈ S ₀, when this is true vacuously). Thus there exists β ∈ S ₀ such that . But then contradicts the fact that is pure. We conclude that is pure and hence that S is closed.

Next observe that there is a sequence (x_n) in 𝒜 such that |f(x_n)|/∥x_n∥ → 1. Then for some , so that the restriction of f to has norm one, and hence is a state, for all β ≥ α. Thus, without loss of generality, in proving unboundedness we can assume the restriction of f to any is a state.

Let . We first claim that for any , for sufficiently large β ≥ α we have whenever f ₁ and f ₂ are states on . That is, for each there exists α′ ≥ α such that the above holds for all β ≥ α′. If the displayed condition holds for all states f ₁ and f ₂ on , then we say that is pure on x.

Suppose the claim fails for some . Then there exist ε > 0 and an unbounded set together with states and on for all β ∈ T, such that (If no such ε and T existed, then for each n ∈ N we could find such that for any β ≥ α_n, no states and on satisfy ^* with ε = 1/n. Then would be pure on x for all β ≥ sup α_n, contradicting our assumption that the claim fails for x.) Now let 𝒰 be an ultrafilter on T that contains the set {β ∈ T: β > β₀} for each , and define states g ₁ and g ₂ on 𝒜 by and , where the limits are taken pointwise on elements of 𝒜. Then f = (g ₁ + g ₂)/2 and |g ₁(x) - f(x)| ≥ ε hence f ≠ g ₁, contradicting purity of f. (This argument could also be carried out by using universal nets.) This establishes the claim.

Now for any , since is separable we can find a dense sequence , and the claim implies that for sufficiently large β, f|_Aβ is pure on every x_n. By density, is then pure on every . Let α^* be the least ordinal larger than α such that β ≥ α^* implies that is pure on every .

Finally, to see that S is unbounded, fix . Let α₁ = α^*, α₂ = α^**, etc., and let β = supα_n. Then is pure on every , and hence by continuity is a pure state on . This shows that β ∈ S, as desired.

A subset of is stationary if it intersects every closed unbounded subset of . We require the following version of the diamond principle (⋄): there exists a transfinite sequence of functions such that for any function the set {α: h|_α = h _α} is stationary (see ref. 13, 22.20, or ref. 14, exercise II.51).

Let denote the set of states on a C^*-algebra 𝒜.

Theorem 5. Assume ⋄. Then there is a counterexample to Naimark's problem that is generated by elements.

Proof: Let (h _α) be a transfinite sequence of functions which verifies ⋄ in the form given above. For we recursively construct a continuous nested transfinite sequence of simple separable unital C^*-algebras , all with the same unit; pure states f _α on with the property that α < β implies f _β is the unique state extension of f _α; and injective functions as follows. Let be any simple, separable, infinite dimensional, unital C^*-algebra and let f ₀ be any pure state on . Since ⋄ implies the continuum hypothesis (see ref. 13 or 14), has cardinality at most , so there exists an injective function from into ; let φ₀ be any such function.

To proceed from stage α of the construction to stage α + 1 when α is a limit ordinal, first check whether there is a pure state g _α on , not equivalent to f _α, such that for all β < α. (By injectivity of all φ_β, there is at most one such g _α.) If so, let be the C^*-algebra ℬ given by Corollary 3 with f = f _α and g = g _α. Let f _α+1 be the unique extension of f _α to , and as above let φ_α+1 be any injective function from into . If there is no such state g _α, and whenever α is a successor ordinal (or α = 0), let , and φ_α+1 = φ_α. At limit ordinals α, let (it is standard that will be simple, given that each is simple), define f _α by requiring for all β < α (f _α will be pure, by the argument in the second paragraph of the proof of Lemma 4), and as before let φ_α be any injective function from into .

Let and define by f|_Aα = f _α. Here f is well defined since f _β is an extension of f _α whenever α < β. Also, f is a pure state, again by the reasoning in the second paragraph of the proof of Lemma 4 coupled with the fact that each f _α is pure. We claim that every pure state g on 𝒜 is equivalent to f. To see this, it is enough to verify that is equivalent to f _α for some α; then since f _α extends uniquely to 𝒜 the same must be true of (see the first paragraph of the proof of Corollary 3), and the unitary in that implements the equivalence of f _α and must then also implement an equivalence between f and g. Now define by setting . Let S be the set of limit ordinals α such that is pure. According to Lemma 4, S is closed and unbounded. (The intersection of any two closed unbounded sets is always closed and unbounded.) Therefore, by ⋄ there exists a limit ordinal α such that is pure and h _α = h|_α, i.e., for all β < α. If is equivalent to f _α then we are done, and otherwise the construction at stage α guarantees that f _α+1 is equivalent to the unique extension of to , which must be . This completes the proof that f and g are equivalent.

Finally, 𝒜 is infinite dimensional and unital, so it cannot be isomorphic to any .

We conclude by observing that if the continuum hypothesis fails then there is no counterexample to Naimark's problem that is generated by elements.

Proposition 6. Let 𝒜 be a counterexample to Naimark's problem. Then 𝒜 cannot be generated by fewer than elements.

Proof: Since 𝒜 cannot be type I, it follows that there is a subalgebra ℬ of 𝒜 and a surjective ^*-homomorphism of ℬ onto the Fermion algebra (ref. 6, corollary 6.7.4). Let f ₁ and f ₂ be the pure states on M ₂(C) defined by Then for any sequence of indices i_n ∈ {1, 2} the product state is a pure state on ℱ. Moreover, ∥f - f′∥ = 2 for any two distinct states of this form. Lifting to ℬ and extending to 𝒜, we obtain a family of pure states on 𝒜, the distance between any two of which is 2. As 𝒜 is a counterexample to Naimark's problem, there exists a family of unitaries in the unitization 𝒜 of 𝒜 that conjugate this family of pure states to a single (arbitrary) pure state g on 𝒜. Uniform separation of the pure states implies uniform separation of the conjugating unitaries: if , then for all where , so . Thus , and hence 𝒜, cannot contain a dense set with fewer than elements, and consequently it cannot be generated by fewer than elements.

Corollary 7. If the continuum hypothesis fails, then no C^*-algebra generated by elements is a counterexample to Naimark's problem. The existence of a counterexample to Naimark's problem which is generated by elements is independent of ZFC.

The first assertion of Corollary 7 follows immediately from Proposition 6, and the second assertion follows from Theorem 5 coupled with the first assertion, together with the fact that both diamond and the negation of the continuum hypothesis are consistent with ZFC (assuming ZFC is consistent).