arXiv math.GT digest — 03 Jun 2026 (3 papers)

Three new papers appeared in the arXiv math.GT listing for Wednesday 3 June 2026. All three sit squarely in classical knot theory and its higher-dimensional generalisations: ribbon knots, knotted surfaces in

S^{4}

, and finiteness properties of link groups.

Paper 1 — The 2-twist spun trefoil has crossing number six

arXiv:2606.03799 · Sherry Gong, Samuel Lewis-Monkman, Jesse Osnes · 9 pp., 14 figures

arxiv.orghttps://arxiv.org/abs/2606.03799外部リンク

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Original abstract

We study the tri-plane crossing number, that is, the minimal number of crossings in a tri-plane diagram for a bridge trisection of a knotted sphere in $S^{4}$ . We show that every 2-knot in $S^{4}$ that admits a bridge trisection with at most five crossings is ribbon. As a consequence, we show that the 2-twist spin of the trefoil has crossing number 6. This is the first such computation for a non-trivial knotted surface.

Commentary

Key result. The paper computes the tri-plane crossing number of the 2-twist spun trefoil and shows it equals 6. This is the first exact crossing-number determination for a non-trivial knotted

S^{2}

S^{4}

; previously no analogous computation existed for a 2-knot that is not ribbon by construction.

Main techniques and tools. The authors work with bridge trisections and the associated tri-plane diagrams introduced by Meier–Zupan. The central engine is a complete ribbon enumeration: they classify all 2-knots in

S^{4}

whose bridge trisection admits a tri-plane diagram with at most five crossings, and show every such 2-knot is ribbon. The argument combines diagrammatic moves on tri-plane diagrams with Gauss-code style tabulation to rule out non-ribbon candidates at low crossing count.

Core proof idea. By exhaustively treating all tri-plane diagrams up to five crossings and verifying that each realises a ribbon knot, the authors show no non-ribbon 2-knot exists below crossing number 6. The 2-twist spun trefoil is known not to be ribbon (its knot group distinguishes it), so its crossing number is at least 6. An explicit tri-plane diagram with exactly 6 crossings provides the upper bound, completing the computation.

Paper 2 — On the BNSR invariants of link groups

arXiv:2606.02978 · Yuta Nozaki · 8 pp.

arxiv.orghttps://arxiv.org/abs/2606.02978外部リンク

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Original abstract

For a finitely generated group $G$, the Bieri-Neumann-Strebel-Renz (BNSR) invariants are subsets of the character sphere of $G$ that govern the finiteness properties of normal subgroups containing the commutator subgroup. We investigate the BNSR invariants of link groups and $2$-knot groups. In particular, for a link $L$ with at least two components, we prove that the commutator subgroup of the link group is finitely generated if and only if $L$ is a Hopf link. Moreover, we show that there exists a ribbon $2$-knot whose knot group has a non-symmetric BNS invariant.

Commentary

Key results. Two independent theorems are proved. First, for a multi-component link $L$, the commutator subgroup of

π_{1} (S^{3} ∖ L)

is finitely generated if and only if $L$ is a Hopf link — a complete characterisation in terms of the link's topology. Second, there exists a ribbon 2-knot whose knot group has a non-symmetric Bieri-Neumann-Strebel (BNS) invariant, meaning the invariant is not preserved under the antipodal map on the character sphere.

Main techniques and tools. The paper uses the BNSR invariant machinery (the

Σ^{n}

invariants of Bieri–Neumann–Strebel–Renz), which encode finiteness properties of kernels of characters from a group to

R

. For the commutator-subgroup criterion the key input is a careful analysis of the Alexander matrix and the structure of the longitude classes. The non-symmetry example exploits a specific ribbon presentation for a 2-knot constructed from a ribbon band move, and the BNS asymmetry is detected by a character whose kernel is finitely generated while the antipodal kernel is not.

Core proof idea. For the first theorem, finite generation of the commutator subgroup of a link group forces the Alexander polynomial to have a specific factored form; for links with at least two components this rigidity is strong enough to force the link to be the Hopf link. For the second theorem, the key is that ribbon 2-knots have knot groups presented via ribbon relators, and the resulting group admits a character whose positive and negative halves (in the sense of the BNS half-space structure) behave differently — one has a finitely presented kernel, the other does not.

Paper 3 — A ribbon knot which is not a symmetric union

arXiv:2606.02968 · Michel Boileau, Teruaki Kitano, Yuta Nozaki · 13 pp., 3 figures

arxiv.orghttps://arxiv.org/abs/2606.02968外部リンク

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Original abstract

A basic open question motivated by the study of ribbon knot diagrams asks whether every ribbon knot can be presented as a symmetric union. In this article, we give a negative answer to this question by exhibiting a ribbon Montesinos knot which does not admit a symmetric union presentation.

Commentary

Key result. The paper settles a long-standing open question by constructing an explicit ribbon knot — a Montesinos knot — that is ribbon but admits no symmetric union presentation. Symmetric unions are a diagrammatic class of ribbon knots introduced by Lamm; the question of whether every ribbon knot is a symmetric union had been open for over two decades.

Main techniques and tools. Symmetric unions come with a strong constraint: their Alexander polynomials must be squares (up to units) in

Z [t, t^{- 1}]

— a consequence of the folded structure of the diagram. The authors identify a ribbon Montesinos knot whose Alexander polynomial is not a perfect square in this sense, ruling out any symmetric union presentation. The ribbon structure of the Montesinos knot is established through an explicit ribbon disk construction. The argument for non-symmetry uses properties of the Levine–Tristram signatures and character-variety obstructions to further constrain the possible symmetric union forms.

Core proof idea. The obstruction to being a symmetric union is purely algebraic: if a knot $K$ is a symmetric union, its Alexander polynomial

Δ_{K} (t)

satisfies

Δ_{K} (t) = f (t) \cdot f (t^{- 1})

for some polynomial $f$. The chosen Montesinos knot has an Alexander polynomial that does not factorise this way, so no symmetric union diagram can represent it, regardless of the ribbon structure.

Note: Kitano and Nozaki also appear as authors on the previous session's paper 2606.01017 (real analytic foliations, Godbillon–Vey). Nozaki is sole author on paper 2606.02978 and co-author on 2606.02968 in the same session — an unusually productive single-day output.