A ZFC counterexample to Naimark's problem

The explanation

Start with a grid of numbers two rows by two columns, a 2×2 matrix. Its normalized trace is the average of its diagonal entries. The corner matrix with a single 1 at top left has trace 1/2. Place that 2×2 corner inside a 4×4 matrix, and the corner's trace becomes 1/4. At size 8×8 it is 1/8, at 16×16 it is 1/16. The traces halve forever and shrink toward zero. Glue all these sizes into one infinite object. That object is where this paper begins.

Mathematicians call such objects C*-algebras (collections that can be added, multiplied and measured for size, like matrices). A representation lets an algebra act as operators on a space of vectors. It is irreducible when no smaller closed part of the space is left untouched. For matrices, every irreducible representation is the usual one, up to renaming the coordinates. The infinite version of matrix algebras is the compact operators. The same holds for them. Each irreducible representation is unitarily equivalent to the usual one, meaning the same after a change of coordinates.

Naimark's question dates to 1951. It asks whether this property singles out the compact operators. Suppose an algebra has, up to equivalence, only one irreducible representation. Must it be an algebra of compact operators? Part of the answer is yes. Suppose that one representation acts on a separable space (one with a countable dense set of vectors). Then the algebra is the compact operators.

The rest stayed open for decades. ZFC is the standard list of axioms behind nearly all of mathematics. In 2004, Charles Akemann and Nik Weaver built a counterexample using an extra axiom called Jensen's diamond principle. Diamond is added on top of ZFC. Daniel Calderón and Ilijas Farah later got a counterexample from different extra axioms. Andrea Vaccaro, still assuming diamond, controlled the traces of such examples. Each counterexample leaned on an assumption beyond ZFC.

Why was the extra axiom useful? The construction grows the algebra step by step. Every irreducible representation comes from a pure state (an extreme kind of averaging rule on the algebra). Each step glues one chosen pure state to a fixed one, so their representations become equivalent. In the end every pure state must be glued. But the algebra keeps growing, and new pure states keep appearing. Diamond acts like a prediction oracle. It lets the builder guess, in advance, the states the finished algebra will have.

This paper removes the need to guess. Its key step argues that each pure state of the final algebra is fixed by a countable pieceSource 1. That piece already exists at some earlier stage, so a schedule chosen in advance can reach itSource 2. The schedule is fixed before any algebra is built, using no states at allSource 3. Each gluing step adds one new element. A projection is an element like the corner matrix above: applying it twice changes nothing. The new element moves one shrinking chain of projections onto another, and both chains must have equal tracesSource 4. Here the opening example returns. Both chains have traces 1, 1/2, 1/4, 1/8 and so on, halving toward zero.

The finished algebra has a unit (an identity element) and infinite dimension. It is simple: it has no ideal except zero and itself. An ideal is a part that absorbs multiplication, so anything times a member stays a member. It carries a faithful tracial state (an averaging rule that ignores order and is positive on every nonzero element squared). And its irreducible representations are all equivalentSource 5. It cannot be the compact operators, because the compact operators have a unit only on finite-dimensional spacesSource 6. So the answer to Naimark's question is no, using ZFC alone.

This was not the first ZFC answer. Ryotaro Tanaka gave another ZFC construction with a faithful tracial state in September 2026. His algebra also has a faithful representation on a separable space. The new paper offers an alternative proof by a different routeSource 7. Its algebra is very large: the recursion runs for more steps than there are real numbers. By the earlier separable result, its single irreducible representation cannot act on a separable space.

How sure is this? A computer proof checker called Lean has a formal version of the main theorem. It states the counterexample in ZFC with no extra set-theory axiom. It is matched against a fixed reference statement by a tool called Comparator. The Lean scope note says nothing about the paper's other intermediate results. No specialist review of the paper has been published.

What it doesn't show. This does not give the first ZFC counterexample, and its algebra has no separable irreducible representation.