Math-3195 · mathematics · Topology

Math-3195

Kirby Problem 1.68

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 $\Lambda_-=\varnothing$. A smooth knot type is an ambient smooth isotopy class of embeddings $S^1\hookrightarrow S^3$ (equivalently into $\mathbb R^3$). Orientability below refers to the filling surface, not merely its boundary knot.

Statement recorded in the supplied source.
Determine exactly the smooth knot types $K$ for which some Legendrian representative $\Lambda$ of $K$ in $(\mathbb R^3,\xi_{\mathrm{std}})$ admits an orientable exact Lagrangian filling. [S2] [S3]

Short English statement

Which smooth knot types have at least one Legendrian representative with an orientable exact Lagrangian filling?

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.