Kitaev determinant trivialization conjecture

Let H\mathcal{H} be a complex Hilbert space and let U,VB(H)U,V\in\mathcal{B}(\mathcal{H}) be invertible. If (U1)(V1)(U-\mathbf{1})(V-\mathbf{1}) and (V1)(U1)(V-\mathbf{1})(U-\mathbf{1}) belong to the trace-class ideal L1(H)\mathcal{L}_1(\mathcal{H}), then UVU1V11L1(H)UVU^{-1}V^{-1}-\mathbf{1}\in\mathcal{L}_1(\mathcal{H}) and det ⁣(UVU1V1)=1\det\!\left(UVU^{-1}V^{-1}\right)=1, where the determinant is the Fredholm determinant.

Progress summary

Solved

A journal paper published in July 2026 reports a proof of the conjecture, and no subsequent objection or counterexample was found.

The conjecture asserts that invertible bounded operators satisfying the stated trace-class hypotheses have a commutator whose determinant is one. Guo Chuan Thiang’s paper claims the full adjoint-free statement.

Known results

Earlier work proved the result with additional hypotheses involving the adjoint, and established several special cases, including cases where one operator is normal. It explicitly described the adjoint-free conjecture as unresolved at that time.

July 2026 proof

The published paper claims that for invertible bounded operators U,VU,V with (U1)(V1),(V1)(U1)L1(U-1)(V-1),(V-1)(U-1)\in\mathcal{L}_1, one has det(UVU1V1)=1\det(UVU^{-1}V^{-1})=1. Its argument uses algebraic KK-theory, the Pincus–Helton–Howe formula, and Lidskii’s trace theorem.

Current status (as of August 2026): The conjecture is resolved by a journal-published proof; no counterexample, retraction, or standing gap was found.

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