Math-3193 · mathematics · Topology

Math-3193

Kirby Problem 1.66

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

Shared notation.
$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

Put $\alpha=dz-y\,dx$ on $\mathbb R^3$, $\xi_{\mathrm{std}}=\ker\alpha$, and equip the symplectization $\mathbb R_t\times\mathbb R^3$ with $\omega=d\lambda$, where $\lambda=e^t\alpha$. A Legendrian link is tangent to $\xi_{\mathrm{std}}$. An exact Lagrangian cobordism $F$ from $\Lambda_-$ to $\Lambda_+$ is a properly embedded surface with $\omega|_F=0$ and $\lambda|_F=df$, cylindrical over these links outside a compact region; there are constants $c_-,c_+\in\mathbb R$ such that the primitive equals $c_-$ on the entire negative cylindrical end and $c_+$ on the entire positive cylindrical end. Each constant is common to all link components at its corresponding end. A filling has $\Lambda_-=\varnothing$.

Statement recorded in the supplied source.
On Legendrian isotopy classes of links, set $[\Lambda_1]\preceq[\Lambda_2]$ when there exists an exact Lagrangian cobordism from $\Lambda_1$ to $\Lambda_2$. Is this relation a partial order? In particular, if exact cobordisms exist in both directions, must $\Lambda_1$ and $\Lambda_2$ be Legendrian isotopic? [S2] [S3]

Short English statement

Does existence of an exact Lagrangian cobordism define a partial order on Legendrian link types?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.