Celebratio Mathematica — Birman

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Works connected to Nancy Court Wrinkle

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J. S. Birman and N. C. Wrinkle: “Holonomic and Legendrian parametrizations of knots,” J. Knot Theor. Ramif. 9 : 3 (May 2000), pp. 293–309. MR 1753797 Zbl 1001.57015 article

Holonomic parametrizations of knots were introduced in 1997 by Vassiliev, who proved that every knot type can be given a holonomic parametrization. Our main result is that any two holonomic knots which represent the same knot type are isotopic in the space of holonomic knots. A second result emerges through the techniques used to prove the main result: strong and unexpected connections between the topology of knots and the algebraic solution to the conjugacy problem in the braid groups, via the work of Garside. We also discuss related parametrizations of Legendrian knots, and uncover connections between the concepts of holonomic and Legendrian parametrizations of knots.

J. S. Birman and N. C. Wrinkle: “On transversally simple knots,” J. Diff. Geom. 55 : 2 (2000), pp. 325–354. MR 1847313 Zbl 1026.57005 article

This paper studies knots that are transversal to the standard contact structure in \( \mathbb{R}^3 \) bringing techniques from topological knot theory to bear on their transversal classification. We say that a transversal knot type \( \mathcal{TK} \) is transversally simple if it is determined by its topological knot type \( \mathcal{K} \) and its Bennequin number. The main theorem asserts that any \( \mathcal{TK} \) whose associated \( \mathcal{K} \) satisfies a condition that we call exchange reducibility is transversally simple.

As a first application, we prove that the unlink is transversally simple, extending the main theorem in [Eliashberg 1991]. As a second application we use a new theorem of Menasco [2001] to extend a result of Etnyre [1999] to prove that all iterated torus knots are transversally simple. We also give a formula for their maximum Bennequin number. We show that the concept of exchange reducibility is the simplest of the constraints that one can place on \( \mathcal{K} \) in order to prove that any associated \( \mathcal{TK} \) is transversally simple. We also give examples of pairs of transversal knots that we conjecture are not transversally simple.