Dixmier’s problem about spectra of C*-algebras

Let $A$ and $B$ be simple separable $C^*$ -algebras, and let $\widehat{A}$ and $\widehat{B}$ denote their spectra, namely the spaces of unitary-equivalence classes of irreducible unitary representations equipped with their relevant Borel structures. Is there always a Borel-definable bijection between $\widehat{A}$ and $\widehat{B}$ , where a function between spectra is Borel-definable when it is induced by a Borel map between standard Borel spaces of unitary representations?

References

Primary source

arXiv

Additional references

Solution to Dixmier's Problem about spectra of C*-algebras — arXiv — Martino Lupini

descriptive-set-theory operator-algebras

Posted 2 weeks ago · edited 2 weeks ago

Progress summary

Refreshed 2 weeks ago

Claimed solved

A September 2026 unrefereed preprint claims to settle Dixmier’s long-standing spectrum problem, but the result has not yet been independently verified.

Dixmier posed the problem in 1967: whether the relevant Borel structure on spectra of simple separable $C^*$ -algebras is uniquely determined in the nonstandard case. It is also connected with a question of Simon Thomas.

Known results

A 2009 dichotomy shows that the pure-state equivalence relation is either smooth or contains a continuous reduction of the $$-equivalence relation, ruling out classification by countable structures in the nonsmooth case.
Elliott proved that the corresponding Borel structures of simple separable AF algebras are isomorphic.

September 22, 2026 claimed solution

Martino Lupini’s preprint claims nonexistence of a Borel-definable injection for a specific nuclear/non-nuclear pair, while proving contrasting positive results for separable nuclear non-type-I algebras. If correct, this resolves Dixmier’s 1967 problem and answers Thomas’s related question; the claim is presently unverified.

Current status (as of September 2026): A preprint claims a complete resolution, but independent verification is not recorded, so the problem is not established as solved.

Sources

No solutions have been posted yet.