A funny isotopy on the 3-sphere

In the papers on the conormal torus (I, II), we showed that if two knots $K, K’ \subset \mathbb{R}^3$ have Legendrian isotopic conormal tori $T, T’$ then the knots themselves are isotopic. Usually the difference between $\mathbb{R}^3$ and $S^3$ is not important in knot theory, but it’s not so clear if that’s the case here: …

Read more

Complete cluster structures from Legendrian knots

In some recent papers, we considered moduli spaces $M(\Sigma, \Lambda)$ (I) or more generally $M(\mathbb{L})$ (II) associated to certain configurations in 4d symplectic geometry.  Basically the point was that in this setting, finding an exact Lagrangian in the geometry gives rise to an algebraic torus chart, and disk surgery gives rise to cluster transformations of charts. Here …

Read more

Immersed Lagrangians from Skeleta

Some augmentations of the Legendrian DGA come from smooth fillings; in “augmentations are sheaves” we showed that, at least for Legendrian knots in the standard contact 3-space, all augmentations are geometric in the sense that the category of augmentations is equivalent to a geometrically defined category of sheaves. I will outline a strategy for making this into the perhaps more …

Read more

On Cornwell’s constructions

I want to record here some ideas on interpretations of the work of Chris Cornwell; these come from discussions with him and others.  Mostly I am concerned with the paper where he constructs representations of $\pi_1(\mathbb{R}^3 \setminus K)$ from augmentations of the Legendrian contact algebra $A(S^*\mathbb{R}^3; S^*_K \mathbb{R}^3)$ of the conormal torus.  This gives an inverse to a …

Read more

The support of an augmentation

Let $\Lambda \subset J^1(X)$ be Legendrian.  In an earlier post, I gave a conceptual argument for why there should be a fully faithful morphism from the category $Aug(\Lambda)$ defined as in [NRSSZ] to the category $Shv_\Lambda(X \times \mathbb{R})$ of sheaves on $X \times \mathbb{R}$ with microsupport at infinity contained in $\Lambda$. Here I will explain that …

Read more