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For $\ell >1$, we develop $L^{(2)}$-signature obstructions for $(4\ell-3)$-dimensional knots with metabelian knot groups to be doubly slice. For each $\ell>1$, we construct an infinite family of knots on which our obstructions are non-zero,…

Geometric Topology · Mathematics 2019-09-19 Patrick Orson , Mark Powell

Let K be a knot in the 3-sphere. A slope p/q is said to be characterising for K if whenever p/q surgery on K is homeomorphic, via an orientation-preserving homeomorphism, to p/q surgery on another knot K' in the 3-sphere, then K and K' are…

Geometric Topology · Mathematics 2018-08-08 Marc Lackenby

For any knot $K$ in $S^3$ and any positive rational $r$, we show that smooth $(-r)$-surgery on $K$ always admits a tight contact structure. More specifically, the tightness is detected by the non-vanishing Heegaard Floer contact invariant.

Geometric Topology · Mathematics 2025-10-09 Zhenkun Li , Shunyu Wan , Hugo Zhou

We show that if the connected sum of two knots with coprime Alexander polynomials is doubly slice, then the Ozsv\'ath-Szab\'o correction terms as smooth double sliceness obstructions vanish for both knots. Recently, Jeffrey Meier gave…

Geometric Topology · Mathematics 2016-11-24 Se-Goo Kim , Taehee Kim

Bing doubling is an operation which produces a 2-component boundary link B(K) from a knot K. If K is slice, then B(K) is easily seen to be boundary slice. In this paper, we investigate whether the converse holds. Our main result is that if…

Geometric Topology · Mathematics 2009-07-06 David Cimasoni

A construction of a spatial graph from a strongly invertible knot was developed by the second author, and a necessary and sufficient condition for the given spatial graph to be hyperbolic was provided as well. The condition is improved in…

General Topology · Mathematics 2007-11-05 Kazuhiro Ichihara , Akira Ushijima

Ribbon concordances between knots generalize the notion of ribbon knots. Agol, building on work of Gordon, proved ribbon concordance gives a partial order on knots in $S^3$. In previous work, the author and Greene conjectured that positive…

Geometric Topology · Mathematics 2025-04-09 Joe Boninger

We introduce and characterize extremal 2-cuts for good fractal necklaces. Using the characterization and the related topological properties of extremal 2-cuts, we prove that every good necklace has a unique necklace IFS in a certain sense.…

Dynamical Systems · Mathematics 2021-10-14 Fan Wen

A knot is said to be slice if it bounds a smooth disk in the 4-ball. For 50 years, it was unknown whether a certain 11 crossing knot, called the Conway knot, was slice or not, and until recently, this was the only one of the thousands of…

Geometric Topology · Mathematics 2021-07-21 Jennifer Hom

We prove that nonsingular retract rational algebraic varieties over any infinite field are uniformly retract rational. As a consequence, every rational, projective, nonsingular complex variety is algebraically elliptic.

Algebraic Geometry · Mathematics 2025-04-03 Juliusz Banecki

By using analytic method, we prove that there exist rational curves on compact Hermitian manifolds with positive holomorphic bisectional curvature. It confirms a question of S.-T. Yau. It is well-known that Mori proved in \cite{Mori79} that…

Differential Geometry · Mathematics 2014-10-07 Huitao Feng , Kefeng Liu , Xueyuan Wan , Xiaokui Yang

The stable Kauffman conjecture posits that a knot in $S^3$ is slice if and only if it admits a slice derivative. We prove a related statement: A knot is handle-ribbon (also called strongly homotopy-ribbon) in a homotopy 4-ball $B$ if and…

Geometric Topology · Mathematics 2020-05-25 Maggie Miller , Alexander Zupan

Akbulut and Kirby conjectured that two knots with the same $0$-surgery are concordant. In this paper, we prove that if the slice-ribbon conjecture is true, then the modified Akbulut-Kirby's conjecture is false. We also give a fibered…

Geometric Topology · Mathematics 2016-09-09 Tetsuya Abe , Keiji Tagami

A knot K in the 3-sphere is said to have Property nR if, whenever K is a component of an n-component link L and some integral surgery on L produces the connected sum of n copies of S^1 x S^2, there is a sequence of handle slides on L that…

Geometric Topology · Mathematics 2009-08-20 Robert E. Gompf , Martin Scharlemann

Given a simply-connected 4-manifold with boundary the 3-sphere, this paper establishes sufficient conditions for a knot in the boundary to be sliced by a locally flat disc in the 4-manifold, whose complement has finite cyclic fundamental…

Geometric Topology · Mathematics 2025-07-02 Anthony Conway , Patrick Orson , Mark Pencovitch

We consider the question of when the operation of contact surgery with positive surgery coefficient, along a knot $K$ in a contact 3-manifold $Y$, gives rise to a weakly fillable contact structure. We show that this happens if and only if…

Geometric Topology · Mathematics 2023-01-25 Thomas Mark , Bülent Tosun

In this paper we give necessary and sufficient conditions for a knot type to admit non-loose Legendrian and transverse representatives in some overtwisted contact structure, classify all non-loose rational unknots in lens spaces, and…

Geometric Topology · Mathematics 2023-10-10 Rima Chatterjee , John B. Etnyre , Hyunki Min , Anubhav Mukherjee

We construct infinitely many smoothly slice knots having topological slice discs that are non-approximable by smooth slice discs.

Geometric Topology · Mathematics 2025-07-08 Min Hoon Kim , Mark Powell

In his recent work \cite{Y1}, X. Yang proved a conjecture raised by Yau in 1982 (\cite{Yau82}), which states that any compact K\"{a}hler manifold with positive holomorphic sectional curvature must be projective. In this note, we prove that…

Differential Geometry · Mathematics 2019-06-18 Kai Tang

When the plane is pie-sliced in $n\leq 4$ parts (with nonempty interior and common vertex at the origin) our main result provides a sufficient condition for any map $L$, that is continuous and piecewise linear relatively to this slicing, to…

Classical Analysis and ODEs · Mathematics 2011-10-07 Laura Poggiolini , Marco Spadini
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