For every K∈Q,R,C, every n,d≥ 1, and every symmetric tensor T∈operatornameSym^d(K^n), the ordinary tensor rank equals the symmetric tensor rank: operatornamerank_K(T)=operatornamesrank_K(T). Here operatornamerank_K(T) is the least r such that T is a sum of r arbitrary rank-one tensors, while operatornamesrank_K(T) is the least r such that T=∑_i=1^r lambda_i v_i^otimes d with lambda_i∈ K and v_i∈ K^n.

Comon’s conjecture

Comon’s conjecture asserts that symmetric tensors have equal ordinary and symmetric ranks. Earlier work had refuted the complex version, but the new claim is substantially smaller and also covers rational and real settings.

Known results

Shitov, 2017: a complex $800\times800\times800$ counterexample, with ordinary rank at most $903$ but symmetric rank greater than $903$ .
Chiantini, Ottaviani, and Vannieuwenhoven, 2018: equality for cubic surfaces, equivalently format $4\times4\times4$ , and whenever symmetric rank is at most $7$ .
The 2018 survey recorded border-rank equality as completely open.
Wang and Seigal, 2022: a real order-six counterexample with differing real ranks.

September 2026 smaller counterexample

Benjamin Lovitz’s new preprint claims an explicit $27\times27\times27$ tensor whose ordinary and symmetric ranks differ over $\mathbb{Q}$ , $\mathbb{R}$ , and $\mathbb{C}$ . This would refute Comon’s conjecture in all three settings, but the computational and algebraic claims are unrefereed and unverified.

Current status (as of September 2026): The conjecture is claimed refuted by a $27\times27\times27$ example over $\mathbb{Q}$ , $\mathbb{R}$ , and $\mathbb{C}$ , but that claim has not been independently verified.

Comon’s conjecture

Primary source

Additional references

Progress summary

Solutions 0