Questions tagged [legendrian-submanifolds]

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Reeb Chords as Posets

I have been thinking recently about Legendrian knots (specifically, the paper by Shende, Treumann, and Zaslow). Basically, a Legendrian knot $\Lambda$ is derivable by permuting the Reidemeister moves [...
Ryan Buchanan's user avatar
4 votes
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Maximal Thurston--Bennequin number of boundary knot classes in contact handlebodies

Let $H$ be a contact handlebody. In other words, $H$ is a small regular neighborhood of a Legendrian graph in a contact $3$-manifold (wlog $\mathbb R^3$). Equivalently, $H=(\Sigma\times[0,1],dt+\...
John Pardon's user avatar
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Confusion about the definition of a formal Legendrian isotopy

We say two Legendrian embeddings $f_0,f_1:L^n\rightarrow (Y^{2n+1},\xi)$ are formally isotopic if there is a smooth isotopy $f_t$ connecting $f_0$ and $f_1$ and a bundle monomorphism $F_t^s:TL\...
milou123's user avatar
6 votes
1 answer

stabilization of Legendrian knots

There are two ways to stabilize a Legendrian knot $k$ in standard contact sphere $(S^3,\xi_{st})$ i.e. adding right cusps or left cusps, let's call these two stabilized Legendrian knots $k_R$ and $k_{...
nikita's user avatar
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4 votes
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Legendrian knot in 3-sphere

We are given a Legendrian knot, fixed up to Legendrian isotopy, in $(S^3,\xi)$ ($\xi$ is the standard contact structure). Does it necessarily bound a symplectic surface in $(B^4,\omega)$ (again $\...
nikita's user avatar
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5 votes
1 answer

Is there a known Legendrian simple link?

Several knots like unknot, $4_1$, $3_1$ are known to be Legendrian simple, i.e., Thurston-Bennequin number and rotation number determine Legendrian type completely. How about the same notion for link ...
Seonhwa  Kim's user avatar
3 votes
2 answers

Legendrian Tubular Neighborhood Theorem

Given a Lagrangian submanifold $L\subset(M,\omega)$ of a symplectic manifold, we have Alan Weinstein's celebrated Lagrangian tubular neighborhood theorem. I now look for the analog on Legendrian ...
Chris Gerig's user avatar
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