Deligne–Drinfeld Freeness Conjecture for grt_1

Deligne–Drinfeld Freeness Conjecture for grt_1

OpenAI's claim

OpenAI presents the following claims relevant to this question, with manuscript numbers following paper order in Sources. In manuscript 1 (family 008; 2026-09-23), the asserted result is that the rational Lie algebra of Grothendieck--Teichmüller solutions has a free presentation with generators indexed by odd weights $\ge3$, and the corresponding completed presentation.

Mathematical statement of the claimed result

Manuscript 1.

Let $L=\operatorname{Lie}_{\mathbb Q}\langle x,y\rangle$, graded by word length. Let $W$ be its solutions of \[ \psi(x,y)+\psi(y,x)=0,\qquad \psi(x,y)+\psi(y,-x-y)+\psi(-x-y,x)=0, \] and the pentagon \[ \psi(t_{12},t_{23}+t_{24})+\psi(t_{13}+t_{23},t_{34}) =\psi(t_{23},t_{34})+\psi(t_{12}+t_{13},t_{24}+t_{34})+\psi(t_{12},t_{23}) \] in the infinitesimal pure braid Lie algebra, defined by $t_{ij}=t_{ji}$, $[t_{ij},t_{k\ell}]=0$ on distinct indices and $[t_{ij},t_{ik}+t_{jk}]=0$. Define $D_{\psi}(x)=0$, $D_{\psi}(y)=[y,\psi]$ and Ihara bracket $\{\psi,\phi\}=D_{\psi}\phi-D_{\phi}\psi+[\psi,\phi]$. The claim gives homogeneous $\sigma_{2k+1}\in W_{2k+1}$ for every $k\ge1$ with \[ \operatorname{Lie}_{\mathbb Q}\langle e_3,e_5,e_7,\ldots\rangle \xrightarrow{\sim}(W,\{\ ,\ \}),\qquad e_{2k+1}\mapsto\sigma_{2k+1}. \] It asserts generation as well as freeness, bracket closure and an induced continuous isomorphism of weight completions.

Outline of the argument

For manuscript 1 (family 008; 2026-09-23), a degeneration of the pentagon relation and deletion operators bound the leading-depth image. Regularized braid transport supplies nonzero odd-weight elements, and a weight-by-weight dimension comparison establishes both independence and spanning.

Formal verification

For manuscript 1 (family 008; 2026-09-23), OpenAI lists Lean statements for this assertion: the rational Lie algebra of Grothendieck--Teichmüller solutions has a free presentation with generators indexed by odd weights $\ge3$, and the corresponding completed presentation.

The central assertion is within the documented formal scope; the additional applications retain their own scope limits.

The linked family Comparator checks are DeligneDrinfeld.lean.

Their named declarations include OAI.DeligneDrinfeld.main.

This site has not executed the Lean code. Described statement coverage is not an independent proof check.

Status caveat

The manuscripts were produced by an internal OpenAI model. OpenAI cautions that results without formalization can contain mistakes. This site has not independently checked the proofs; the label records the scope of OpenAI's claim.

Sources

• OpenAI, The Deligne–Drinfeld conjecture, 2026-09-23; citation key OAI:The-Deligne-Drinfeld-conjecture-September-23-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main.
• Family 008: Lean scope.
• Comparator DeligneDrinfeld.json.
• Comparator DeligneDrinfeld.lean.
• OpenAI catalogue entry.

Deligne–Drinfeld Freeness Conjecture for grt_1

Definitions and mathematical statement

Let $\mathfrak{grt}_1$ denote the weight-graded Grothendieck--Teichmüller Lie algebra over ${\mathbb{Q}}$. It is defined by Lie series in two generators satisfying Drinfeld's infinitesimal antisymmetry, three-cycle, and pentagon identities, with its Ihara bracket; the defining identities and completion conventions are those of the references in the cited paper. Weight is total degree in the two generators. This is not the full Grothendieck--Teichmüller algebra with its additional degree-zero grading direction.

The Deligne--Drinfeld freeness assertion is \[ \mathfrak{grt}_1\cong \mathbb L_{{\mathbb{Q}}}(\sigma_3,\sigma_5,\sigma_7,\ldots), \qquad \operatorname{wt}(\sigma_{2j+1})=2j+1\quad(j\ge1), \] where $\mathbb L$ denotes the free graded Lie algebra: all relations follow from bilinearity, antisymmetry, and Jacobi. For completed Lie algebras, both sides are completed by weight. The source also discusses larger conjectural chains of isomorphisms; those extra isomorphisms are not added to this entry's freeness target.

Formulation and concept references: [S1].

Short English statement

Is the graded Grothendieck--Teichmüller Lie algebra freely generated by exactly one generator in each odd weight starting at three?

Research attempt

GPT-6 Astra Ultra; 27 September 2026. Status: UNRESOLVED.

I separate the existence of a free odd-generator subalgebra from the assertion that it exhausts $\mathfrak{grt}_1$. The known inclusion of such a free algebra, recalled with references in [S1, E1], supplies a lower bound. The missing assertion is an upper bound in every weight. This distinction permits an exact dimension calculation and a sharp test of what finite computations can establish, without treating the larger chain of conjectural isomorphisms as part of the present target.

The predicted dimensions from first principles.

Let $V$ have one basis vector in each odd weight at least three, and let $L=\mathbb L(V)$. Write $\ell_n=\dim L_n$. There are finitely many generators in each bounded weight, so all formal series below have well-defined coefficients. The generating series of $V$ is \[ v(t)=\frac{t^3}{1-t^2}. \] The enveloping algebra of a free Lie algebra is the tensor algebra on its generators, whose Hilbert series is $1/(1-v(t))$. On the other hand, the Poincare--Birkhoff--Witt basis gives its Hilbert series as a product over a homogeneous basis of $L$. Consequently \[ \prod_{n\ge1}(1-t^n)^{\ell_n} =1-v(t)=\frac{1-t^2-t^3}{1-t^2}. \tag{1} \] This identity describes the candidate free algebra; it is not an independent computation of the dimensions of $\mathfrak{grt}_1$.

Put $c_n=[t^n](-\log(1-v(t)))$. Taking logarithms in $1$ gives \[ nc_n=\sum_{d\mid n}d\ell_d,\qquad \ell_n=\frac1n\sum_{d\mid n}\mu(d)\frac nd c_{n/d}. \tag{2} \] Every coefficient can therefore be calculated exactly using rational arithmetic and divisors. This avoids numerical rank tolerances. In low weights the formula yields \[ \begin{array}{c|rrrrrrrrrrrrr} n&3&4&5&6&7&8&9&10&11&12&13&14&15\\\hline \ell_n&1&0&1&0&1&1&1&1&2&2&3&3&4 \end{array} \] There is no generator in weight eight, but the bracket $[\sigma_3,\sigma_5]$ contributes there. In weight eleven the generator $\sigma_{11}$ and $[\sigma_3,[\sigma_3,\sigma_5]]$ account for the two dimensions. Thus "odd generators" emphatically does not mean that all even weight spaces vanish. Antisymmetry also matters: the apparent weight-six bracket $[\sigma_3,\sigma_3]$ is zero.

An exact conditional completion criterion.

Fix the known graded injection $\iota:L\hookrightarrow\mathfrak{grt}_1$. If one proves \[ \dim\mathfrak{grt}_{1,n}\le\ell_n\quad\hbox{for every }n, \tag{3} \] then injectivity gives the reverse inequality and each component map is surjective. Their direct sum is the required graded Lie isomorphism. No additional bracket-compatibility argument is necessary because $\iota$ is already a Lie map. Conversely the conjectured isomorphism implies $3$. This reduces the remaining task, relative to the established injection, to a family of upper bounds rather than to a search for new generators satisfying the relations.

There is a useful filtered variant. Suppose that, in each weight $n$, the depth filtration is finite and separated, and there is an injection \[ \operatorname{gr}_{\rm depth}\mathfrak{grt}_{1,n} \hookrightarrow U_n \] into a finite-dimensional explicitly defined comparison space. Then \[ \dim\mathfrak{grt}_{1,n} =\dim\operatorname{gr}_{\rm depth}\mathfrak{grt}_{1,n} \le\dim U_n. \] An upper bound $\dim U_n\le\ell_n$ proves the desired equality in that weight. Summing all depth components is essential: an equality in depth one says nothing by itself about the brackets in higher depth or about additional independent elements there.

This is the sound dimension-squeezing mechanism used in [S1]. Its reported computations establish the relevant equalities through weight 29. The paper's claim is a bounded-weight verification, not an all-weight proof. I have inspected the stated inclusions and the dimension argument, but have not independently rerun that paper's large linear-algebra computations. Our small exact calculation checks the predicted series, a separate and much more limited task.

Why a long initial table does not finish the proof.

For any cutoff $N$, choose $m>N$ and form \[ L^{\rm extra}=L\oplus\mathbb Q z_m, \qquad [z_m,L^{\rm extra}]=0, \] with $z_m$ homogeneous of weight $m$. This is a graded Lie algebra with the same components and brackets as $L$ through weight $N$, and it contains $L$ injectively, but its weight-$m$ dimension is larger. The construction is not claimed to satisfy the Grothendieck--Teichmuller identities. Its purpose is exact: low-weight agreement plus a free subalgebra is logically insufficient to exclude a new higher-weight class. To use the pentagon against such a class requires an argument uniform in weight, absent from this construction.

There is also a different finite-table ambiguity if one does not know injectivity. A quotient by a homogeneous relation of weight larger than $N$ has unchanged components up to $N$ and may fail to be free later. The established injection rules out this second defect for the chosen odd generators in the actual $\mathfrak{grt}_1$. It does not rule out the first defect, namely extra elements outside their image. Confusing these two roles of freeness would reverse the direction of the remaining work.

Completion and a possible algebraic route.

Because all generators have positive weight, a componentwise isomorphism induces an isomorphism of weight completions $\prod_n L_n\cong\prod_n\mathfrak{grt}_{1,n}$, continuous for the weight topology. The completed statement must use completed objects on both sides. Adding an Euler or scaling derivation in degree zero changes the algebra and invalidates the Hilbert series used here.

One possible uniform route would construct, from the antisymmetry, three-cycle, and pentagon identities, a spanning set with at most $\ell_n$ elements in each weight. To be a proof it must give a terminating reduction procedure and verify that every defining relation used is valid with the Ihara bracket. Counting formally plausible bracket words gives a spanning set only after generation has already been proved; using that count as its proof is circular. The explicit criterion $3$ exposes this first gap.

Verification and outcome.

An exact rational-series check recomputes $1$--$2$ through weight 40 and verifies the displayed table and the PBW product to that order. It does not compute any new actual $\mathfrak{grt}_1$ ranks. The result of the attempt is a verified target series, a precise upper-bound criterion, and countertests for finite-table extrapolation. No uniform upper bound has been proved. The full freeness conjecture remains UNRESOLVED.

Sources

• S1 Florian Naef and Thomas Willwacher, Numerical computation of linearized KV and the Deligne-Drinfeld and Broadhurst-Kreimer conjectures (2025 preprint). https://arxiv.org/html/2508.08081. Formulation source. Consulted 24 September 2026: Graded freeness question and distinction from the larger conjectural chain.
• E1 Florian Naef and Thomas Willwacher, Numerical computation of linearized KV and the Deligne--Drinfeld and Broadhurst--Kreimer conjectures, arXiv:2508.08081v1, Introduction, Theorem 5 and Corollary 7. https://arxiv.org/html/2508.08081v1. Version-specific scope check of [S1]: the inclusions and the reported verification through weight 29 were inspected; the large computations were not independently reproduced. Consulted 27 September 2026.