Let’s assume for now that if $\Lambda$ is a Legendrian knot in the standard $\mathbb{R}^3$, then the groupoid cardinality of $Aug(\Lambda,\mathbb{F}_q)$ is given by $$|Aug(\Lambda,\mathbb{F}_q)| = R_\Lambda(q^{1/2}-q^{-1/2}) \cdot \frac{q^{(tb(\Lambda)+1)/2}}{q-1}.$$ My convention for the ruling polynomial $R_\Lambda(z)$ is that a ruling with s switches and c cusps contributes $z^{s-c+1}$, so for example the $tb=-1$ unknot $U$ has ruling polynomial 1. I’ll set $q=2$ from now on, and in what follows $\Lambda$ will be a Legendrian knot which admits concordances both from $U$ (i.e. $U \prec \Lambda$) and to $U$ ($\Lambda \prec U$).
Chris Cornwell, Lenny, and I [CNS] proved that $U \prec \Lambda \prec U$ implies $R_\Lambda(z) = 1$. The concordance $U \prec \Lambda$ provides a Lagrangian disk $L$ bounded by $\Lambda$, so there’s a geometric augmentation $\epsilon_L: \mathcal{A}(\Lambda) \to \mathbb{F}_2$, which according to [NRSSZ, Proposition 5.7] satisfies $H^*Hom(\epsilon_L,\epsilon_L) = \mathbb{F}_2$ supported in degree 0.
Since $H^0Hom(\epsilon_L,\epsilon_L)$ has one nonzero element, it’s the only automorphism of $\epsilon_L$; and the hom space has no cohomology in negative degrees, so the isomorphism class $[\epsilon_L]$ contributes exactly 1 to the groupoid cardinality of $Aug(\Lambda,\mathbb{F}_2)$. But since $tb(\Lambda)=-1$ and $R_\Lambda(z)=1$, that’s (conjecturally) all the groupoid cardinality there is. Recalling that equivalence in $H^*Aug$ is the same as DGA homotopy [NRSSZ, Proposition 5.17], we conclude:
Theorem 1. Every augmentation of $\Lambda$ is homotopic to the geometric augmentation $\epsilon_L$.
This already strengthens a result of Chantraine et al. [CDRGG, Corollary 1.3] in dimension 3: if $U \prec \Lambda \prec U$, then for every pair of augmentations $\epsilon,\epsilon’$ of $\Lambda$ we must have $H^*Hom(\epsilon,\epsilon’) = H^*(L) = \mathbb{F}_2$ in degree 0. In particular we don’t need the augmentations to be geometric, because they’re both isomorphic to $\epsilon_L$ anyway. In fact, there’s more:
Theorem 2. The DGA morphism $\mathcal{A}(\Lambda) \to \mathcal{A}(U)$ induces an equivalence of categories $Aug(U, \mathbb{F}_2) \xrightarrow{\sim} Aug(\Lambda, \mathbb{F}_2)$.
The DGA morphism is the one defined by Ekholm-Honda-Kálmán [EHK], strengthened slightly to include $t^{\pm 1}$ generators which don’t commute with base points (one should decorate the Lagrangian with a path connecting the base points, and keep track of where boundaries of holomorphic disks hit this path); and then the functor it induces comes from [NRSSZ, Proposition 3.29].
To prove the theorem, we can use a result of Eliashberg-Polterovich, as originally applied by Chantraine [C]: the concordance $U \prec \Lambda \prec U$ is Hamiltonian isotopic to the identity, so by [EHK] the DGA morphism $\mathcal{A}(U) \to \mathcal{A}(\Lambda) \to \mathcal{A}(U)$ is homotopic to the identity. In particular, the induced map $Hom_U(\epsilon,\epsilon) \to Hom_\Lambda(\epsilon_L,\epsilon_L) \to Hom_U(\epsilon,\epsilon)$ is a quasi-isomorphism, hence so is the first half of it (recall that $H^*Hom_\Lambda(\epsilon_L,\epsilon_L) \cong H^*Hom_U(\epsilon,\epsilon) = \mathbb{F}_2$). This means that the functor $Aug(U,\mathbb{F}_2) \to Aug(\Lambda,\mathbb{F}_2)$, which had to be essentially surjective because each category has only one isomorphism class of objects, is also fully faithful.
Hi Steven.
I didn’t realize it was known, that there is a single Hamiltonian isotopy class of (exact?) self-concordances of the Legendrian U. Is that the “result of Eliashberg-Polterovich” you mean? Which EP paper?
I haven’t really appreciated how this could be true, but it is still unknown whether $U < \Lambda < U$ implies $\Lambda = U$. If that's not true, what could the image of $\Lambda$ under the Hamiltonian isotopy from $U < \Lambda < U$ to the trivial concordance $U = U$ look like?
I think I was being a little sloppy when I made that claim. The E-P result is in “Local Lagrangian 2-knots are trivial”; they prove that a Lagrangian disk which fills the unknot is isotopic to the standard one by a compactly supported Hamiltonian isotopy, and in fact that the space of such disks is contractible. More recently, CDRGG proved in section 4 of the paper I cited above that every Lagrangian concordance from U to itself is the trace of a Legendrian isotopy, so there are actually a Z family of concordances but that’s it. (The result I was thinking of in [C] is that they all induce the same map on LCH, because they’re equivalent to the product cylinder by a symplectomorphism.)
I don’t know about the image of a nontrivial knot $\Lambda$ in the middle of such a concordance, and I’m inclined to believe they don’t exist, but this is possible for smooth knots: they’re called “doubly slice” and defined as knots K for which there is an unknotted 2-sphere in $S^4$ whose intersection with the equator $S^3$ is K. The first example is $9_{46}$, none of whose Legendrian representatives are what one might call doubly Lagrangian slice.
Good answer, I didn’t know about doubly-slice knots.
If $\Lambda$ is a doubly-slice Legendrian knot, you conclude that it has a unique $\mathbf{F}_2$-valued augmentation. What about $\mathbf{F}_q$-valued augmentations?
The answer should be the same, but this is a little harder because a lot of the relevant results are proved over $\mathbb{F}_2$ (even though as far as I know they should work with signs). If you don’t want to think about orientation issues to lift [EHK] to $\mathbb{Z}$ coefficients, you can try to argue as follows: Leverson [L] showed that the $\mathbb{F}_2$-augmentation I used above can be lifted to $\mathbb{Z}$, so by reducing mod q we get an $\mathbb{F}_q$-valued augmentation $\epsilon$. Presumably we still have $H^*Hom(\epsilon,\epsilon) = H^*(L) = \mathbb{F}_q$, though something needs to be done to prove this (could there be torsion in the homology over $\mathbb{Z}$?); and then this augmentation has automorphism group $\mathbb{F}_q^\times$, so it contributes $\frac{1}{q-1}$ to the cardinality and again that’s all there is.
Can a quasi-positive knot be doubly slice?
Peter Feller pointed out to me that $8_{20}$ is quasipositive and slice (but not doubly slice).
$9_{46}$ is quasipositive, as is anything else which admits a Lagrangian filling.
I’d like to get to know $9_{46}$! It is filled by an embedded Lagrangian disk?
In Lenny’s atlas, $9_{46}$ and its mirror each have a preferred Legendrian representative. Is there a binary presentation?
It’s actually filled by two different Lagrangian disks: if you look at figure 3 of [CNS], on page 7, you can see two different normal rulings which correspond to different disks. The one on the left (with the switches in blue) has a disk which you can build by first inserting a Lagrangian saddle, pinching off the bottom two strands in the middle and thus replacing the strands $=$ with a pair of cusps $\succ\prec$; and then what remains is a 2-component Legendrian unlink, and you cap off each component with a disk. The reason the blue ruling corresponds to this disk is that the 2-component unlink has a unique normal ruling, and when you reverse the pinching move you just glue the corresponding pairs of strands together. For the ruling on the right, with the red switches, you just flip everything on the left upside-down, so that you pinch off the top two strands and then cap off the remaining unlink.
This knot is actually $m(9_{46})$ in the atlas (and in [CNS]) — I think some other sources disagree on whether it’s the mirror or not. I don’t immediately know whether either the knot or its mirror have 0-1 Maslov potentials (is that what you mean by binary presentation?), but none of the obstructions I mentioned in my answer to a question Vivek asked on MathOverflow seem to rule it out.
It looks like both $9_{46}$ and its mirror have Legendrian representatives with 0-1 Maslov potentials! I used Gridlink‘s “simplify” move repeatedly until it gave me the following picture for $m(9_{46})$ (the one with $tb=-1$ and a Lagrangian disk filling):
https://math.berkeley.edu/wp/microlocal/wp-content/uploads/sites/7/2015/04/m9_46.png
(This can be viewed as a front projection by smoothing all the NE/SW corners and turning your head 45 degrees clockwise.) The augmentation category over $\mathbb{F}_2$ has two isomorphism classes of objects $[\epsilon]$ and $[\epsilon’]$, one for each disk; we have $H^*Hom(\epsilon, \epsilon) = H^*Hom(\epsilon’,\epsilon’) = H^*(D^2)$, while for $H^*Hom(\epsilon,\epsilon’)$ the Poincaré polynomial is $1+t+t^2$.
And for $9_{46}$, which has $tb=-7$ and no fillings:
https://math.berkeley.edu/wp/microlocal/wp-content/uploads/sites/7/2015/04/9_46.png
This has one isomorphism class of objects, with $H^*Hom(\epsilon,\epsilon)$ having Poincaré polynomial $4+3t^2$ and an automorphism group of order 8.
That’s wonderful. Here‘s another view of m9_46. It looks like it will be fun to give a lin alg description of its $\mathcal{M}_1$.
Of course, it’s programmable, but there may be some shortcuts. There are two three-dimensional regions, but they share a 2d region, so we should be able to sit the thing in a 4d vector space with the two 3d regions being the xyz and xyw axes.
Nice picture! How did you produce it?
I wrote a sage routine that turns a pair of permutations into a picture like that. I read the two permutations off of your picture:
24061735 and 67534201.
Can you post this sage routine on the blog?