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The Euclidean Steinitz-Bergstrom conjecture: is the Euclidean Steinitz constant O(sqrt d)?

Theorem 1.2: there is an absolute constant $C$ such that every zero-sum family of vectors in the Euclidean unit ball of $\mathbb R^d$ (repetitions allowed) has an ordering with all partial sums of norm at most $C\sqrt d$ ; with the simplex lower bound, $\tfrac12\sqrt d\le S_2(d)\le C\sqrt d$ . It follows from the prescribed-order signing theorem (separate entry). The constant is not made explicit, and the construction is existential: no online rule or efficient algorithm is given.

Exact FrontierDelta

Prior state unknown→proved

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Occurred: Sep 24, 2026

Delta type: SOURCE CLAIM

Assumptions: VibeMathed verification: lean-checked. Publication: announcement. AI contribution: ai-discovered. VibeMathed editorial classifications, scores, notes, relations, and dataset structure are CC BY 4.0. Source statements and linked content retain their own rights.

Canonical aliases: The Euclidean Steinitz-Bergstrom conjecture: is the Euclidean Steinitz constant O(sqrt d)? · Euclidean Steinitz-Bergstrom conjecture

Confidence: Not scored

Registry verification: lean checked · announcement · candidate

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VibeMathed record: The Euclidean Steinitz-Bergstrom conjecture: is the Euclidean Steinitz constant O(sqrt d)?

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No independent mathematician has checked this yet. Checked here: Theorems 1.1 and 1.2 were read against the conjecture as stated in the manuscript; the ordering bound follows from the signing bound by Chobanyan's transference, proved in the introduction. The proof was not refereed. lean/formalization.yaml lists a main result for this manuscript (comparator SteinitzBergstrom, declaration OAI.EuclideanSteinitzBergstrom.main). ComparatorChallenges/SteinitzBergstrom.lean was read here: it asserts one real $C$ such that for all $d,N\ge1$ and all $v:\mathrm{Fin}\,N\to\mathbb R^d$ with $\|v_i\|\le1$ there are signs $\pm1$ keeping every prefix within $C\sqrt d$ , and every zero-sum such family has a permutation keeping every prefix within $C\sqrt d$ . This states the headline claim. Not rebuilt here.