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Consistency of a counterexample to Naimark's problem

Edited by Vaughan F. Jones, University of California, Berkeley, CA (received for review March 2, 2004)
Abstract
We construct a C^{*}algebra that has only one irreducible representation up to unitary equivalence but is not isomorphic to the algebra of compact operators on any Hilbert space. This answers an old question of Naimark. Our construction uses a combinatorial statement called the diamond principle, which is known to be consistent with but not provable from the standard axioms of set theory (assuming that these axioms are consistent). We prove that the statement “there exists a counterexample to Naimark's problem which is generated by elements” is undecidable in standard set theory.
Let denote the C^{*}algebra of compact operators on a complex, not necessarily separable, Hilbert space H. In ref. 1 Naimark observed that every irreducible representation (irrep) of is unitarily equivalent to the identity representation, so that each of these algebras has only one irrep up to equivalence, and in ref. 2 he asked whether this property characterizes the algebras . In other words: if 𝒜 is a C^{*}algebra with only one irrep up to unitary equivalence, is 𝒜 isomorphic to some ? We call any algebra 𝒜 that satisfies the premise of this question but not its conclusion a counterexample to Naimark's problem.
The problem was quickly settled in the separable case. Building on results in ref. 2, Rosenberg (3) showed that there are no separable (indeed, no separably acting) counterexamples to Naimark's problem. Around the same time, Kaplansky (4) introduced the socalled “type I” (or “GCR”) C^{*}algebras and began developing their representation theory. (See refs. 57 for general background on type I C^{*}algebras.) This development was carried further by Fell and Dixmier, who showed in particular that any two irreps of a type I C^{*}algebra with the same kernel are unitarily equivalent (8) and conversely, any separable C^{*}algebra that is not type I has inequivalent irreps with the same kernel (9). As it is easy to see that no type I C^{*}algebra can be a counterexample to Naimark's problem, the latter result (partially) recovers Rosenberg's theorem, doing so, moreover, in the context of a general theory.
Next, in a celebrated paper Glimm (10) gave several characterizations of separable type I C^{*}algebras, showing in particular that every separable C^{*}algebra which is not type I has uncountably many inequivalent irreps. Since counterexamples to Naimark's problem cannot be type I, this reestablished the nonexistence of separable counterexamples in an especially dramatic way.
Some of Glimm's results were initially proven without assuming separability, and subsequent work by Sakai (11, 12) went further in this direction. However, separability was never removed from the reverse implication of the equivalence “type I ⇔ irreps with the same kernel are equivalent,” and it was recognized that Naimark's problem potentially represented a fundamental obstruction to a nonseparable generalization of this result. However, it was reasonable to expect that there were no counterexamples to Naimark's problem because it seemed likely that the separability assumption could be dropped in Glimm's theorem that nontype I C^{*}algebras have uncountably many inequivalent irreps, but this was never achieved.
That is the background for the present investigation. We construct a (necessarily nonseparable and not type I) counterexample to Naimark's problem. Our construction is not carried out in ZFC (ZermeloFrankel set theory with the axion of choice): it requires Jensen's “diamond” principle, which follows from Gödel's axiom of constructibility and hence is relatively consistent with ZFC, i.e., if ZFC is consistent then so is ZFC plus diamond (see refs. 13 and 14 for general background on set theory). The diamond principle has been used to prove a variety of consistency results in mainstream mathematics (see, e.g., refs. 1517).
Presumably, the existence of a counterexample to Naimark's problem is independent of ZFC, but we have not yet been able to show this. We can prove the relative consistency of the assertion “no C^{*}algebra generated by elements is a counterexample to Naimark's problem”; indeed, this follows easily from ref. 10. Since our counterexample is generated by elements, it follows that the existence of an generated counterexample is independent of ZFC (see Corollary 7).
The basic idea of our construction is to create a nested transfinite sequence of separable C^{*}algebras , for , each equipped with a distinguished pure state f _{α}, such that for any α < β, f _{β} is the unique extension of f _{α} to a state on . At the same time, for each α (or at least “enough” α) we want to select a pure state g _{α} on that is not equivalent to f _{α}, such that g _{α} has a unique state extension to , and is equivalent to f _{α+1}. Thus, we build up a continually expanding pool of equivalent pure states in an attempt to ensure that all pure states on will be equivalent.
There are two significant difficulties with this approach, which incidentally is essentially the only way an generated counterexample to Naimark's problem could be constructed. First, there is the technical problem of finding a suitable algebra that contains and such that f _{α} and g _{α} extend uniquely to and become equivalent. We accomplish this via a crossed product construction which takes advantage of a powerful recent result of Kishimoto, Ozawa, and Sakai that, together with earlier work of Futamura, Kataoka, and Kishimoto, ensures the existence of automorphisms on separable C^{*}algebras that relate inequivalent pure states in a certain manner. The second fundamental challenge is to choose the states g _{α} in such a way that all pure states are eventually made equivalent to one another. This is especially troublesome because pure states can be expected to proliferate exponentially as α increases (every pure state on extends to at least one, but probably many, pure states on ) and we can only make a single pair of states equivalent at each step. We handle this issue by using the diamond principle, one version of which states that it is possible to select a single vertex from each level of the standard tree of height and width , such that every path down the tree contains in some sense “many” selected vertices. In our application the vertices of the tree at level α model the states on and diamond informs us how to choose g _{α}. Then every pure state on 𝒜 induces a path down the model tree, and hence its restriction to some equals g _{α}, so that all pure states are indeed taken care of at some point in our construction.
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, normone 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_{κ}x^{*})f(xqx^{*})=f(xyqx^{*})=f(y(xqx^{*}))=0. Fix κ_{0} so that , and note that because (a _{λ}) excises f, there exists λ_{0} such that λ ≥ λ_{0} implies . Choose λ so that λ ≥ λ_{0} and λ ≥ κ_{0}. 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_{x} φ(g) = θ_{g1}(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, normone element such that f(b) = 1; then π(b)(ξ) = ξ. Furthermore, θ _{g} (u ^{*} bu) is also a positive, normone 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, normone 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_{g1}(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 normone 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 _{g}x_{g}T_{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_{e}T_{e} we now need only show that ∥a _{λ}(x_{g}T_{g} )a _{λ}∥ → 0 for each g ≠ e. But a _{λ}(x_{g}T_{g} )a _{λ} = (a _{λ} x_{g}b _{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_{g}b _{λ}∥ → 0. Thus ∥(a _{λ} x_{g}b _{λ})T_{g} ∥ → 0, as desired.
Note that in the proof of the reverse direction of Theorem 2, the fact that z = Σx_{g}T_{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 nfold 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 π_{1} and π_{2} 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} = ω ^{n} 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_{n}T_{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 > β_{0}} 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 α_{1} = α^{*}, α_{2} = 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 φ_{0} 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).
Acknowledgments
N.W. thanks Gert Pedersen for motivating his work on this problem. N.W. was partially supported by National Science Foundation Grant DMS0070634.
Footnotes

↵ § To whom correspondence should be addressed. Email: nweaver{at}math.wustl.edu.

This paper was submitted directly (Track II) to the PNAS office.

Abbreviations: irrep, irreducible representation; ZFC, ZermeloFrankel set theory with the axiom of choice.
 Copyright © 2004, The National Academy of Sciences
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