The transitive-array conjecture for solutions of the classical Yang–Baxter equation

Less than 1 year old · traced to 2026

Let $C$ be a set, let $\boldsymbol{a}$ be a transitive $n\times n$ matrix with entries in $C$ , and let $\mathbf r=\{r^{(c)}\}_{c\in C}\subset \mathfrak{g}\otimes\mathfrak{g}$ be a solution of the transitive classical Yang–Baxter equation. Write $\mathbf r^{(\boldsymbol{a})}$ for the corresponding element constructed from $\boldsymbol{a}$ and $\mathbf r$ . Transitive-array conjecture. For every such $\boldsymbol{a}$ and $\mathbf r$ , the element $\mathbf r^{(\boldsymbol{a})}$ solves the classical Yang–Baxter equation. This is a natural generalization of the preceding conjecture and is verified for $n\leq 4$ ; the assertion remains open in general.

References

Primary source

Arkady Berenstein, Jacob Greenstein and Jian-Rong Li, “Monomial bialgebras”, arXiv:2602.02342 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2026 paper leaves the conjecture open beyond four dimensions, while a posted argument claims a proof in every dimension but has not been independently checked.

Berenstein, Greenstein, and Li formulate the transitive-array conjecture in their 2026 paper: every transitive array and compatible transitive Yang–Baxter family should produce a classical Yang–Baxter solution.

Known results

Verified for $n\leq 4$ in Berenstein, Greenstein, and Li (2026).

Posted attempt

A posted argument claims a complete proof for every $n$ : it decomposes the classical Yang–Baxter expression into direct-summand components and applies array transitivity to each component. The attempt has not been independently verified.

Current status (as of August 2026): The conjecture is verified for $n\leq 4$ , but no published general proof or counterexample is recorded; a complete-proof claim remains unverified.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

Let g be any Lie algebra, C any index set, and {r^(c):c∈C}⊂g⊗g a family satisfying the source's transitive classical Yang–Baxter equation: [r^(c)_12,r^(c')_13]+[r^(c)_12,r^(c'')_23]+[r^(c')_13,r^(c'')_23]=0 whenever c'∈{c,c''}.

Let a=(a_ij) be any transitive n×n array, so a_ik∈{a_ij,a_jk} for every i,j,k. Writing ι_u:g→g^{⊕n} for inclusion into the uth direct summand, the source's constructed tensor is R=Σ_{u,v=1}^n (ι_u⊗ι_v)r^(a_{v,u}).

Distinct direct summands of g^{⊕n} commute. Therefore, for every triple (u,v,w), the component of CYB(R)=[R_12,R_13]+[R_12,R_23]+[R_13,R_23] in ι_u(g)⊗ι_v(g)⊗ι_w(g) is exactly [r^(a_{v,u})12,r^(a{w,u})13] +[r^(a{v,u})12,r^(a{w,v})23] +[r^(a{w,u})13,r^(a{w,v})23]. Taking (i,j,k)=(w,v,u) in array transitivity gives a{w,u}∈{a_{w,v},a_{v,u}}. Hence the displayed component vanishes by the transitive classical Yang–Baxter equation with c=a_{v,u}, c'=a_{w,u}, c''=a_{w,v}.

The argument applies without change when indices coincide. Since the tensor cube of g^{⊕n} is the direct sum of all such (u,v,w)-components, every component of CYB(R) vanishes. Consequently CYB(R)=0 for every n, every transitive array, every index set C, and every transitive Yang–Baxter family, proving the full conjecture.