In Conjecture 7.5 of [STZ], we hypothesized that, for $\Lambda$ a Legendrian knot in the standard 3-space, the following were equal over any finite field:
(1) The groupoid cardinality of the augmentation category
(2) The groupoid cardinality of the category of rank 1 sheaves on $\Lambda$
(3) The raw number of augmentations, times some fudge factor $q^{?} (q-1)^{-c}$
The equality of (1) and (2) in general follows from the main result of [NRSSZ], where we showed that the augmentation and sheaf categories are equivalent. In [STZ], we showed that in fact (2) and (3) are equal for the “rainbow closure” of a positive braid, by computing this ourselves for the sheaf category and comparing it with computations of Brad Henry and Dan Rutherford [HR]. Specifically, Henry and Rutherford had shown that the ruling polynomial counts augmentations, and we showed it counted objects in the sheaf category.
This conjecture was particularly interesting because Dan had previously shown that $\mathbb{Z}/2\mathbb{Z}$-graded rulings are counted by a certain term in the HOMFLY polynomial. Thus it would follow from the conjecture that this piece of HOMFLY is counting objects in some natural category associated to the knot!
However, recent calculations of Steven Sivek have shown that this conjecture is false, or more precisely, my naive interpretation of the word groupoid cardinality was false. The correct interpretation was pointed out to me by some helpful mathoverflowers.
Since $\mathcal{A} := Aug(\Lambda)$ is a higher category, we should take its higher groupoid cardinality. That is, compute
$$\sum_{[o] \in \pi_0(\mathcal{A})} \frac{|\pi_2(o,\mathcal{A})|\cdot |\pi_4(o, \mathcal{A})| \cdots}{|\pi_1(o,\mathcal{A}) |\cdot |\pi_3(o, \mathcal{A})| \cdots}$$
What Steven had computed, and found not to match with (3), was the above formula without any $\pi_{> 1}$.
I am told that (at least in the connective case, which we are not in…) computing the above thing amounts to computing
$$ \sum_{[o] \in \pi_0(\mathcal{A})} \frac{1}{|Aut(o)|} \cdot \frac{|H^{-1} End(o, o)|\cdot |H^{-3} End(o, o)| \cdots}{|H^{-2} End(o,o)|\cdot |H^{-4}End(o,o)| \cdots} = \sum_{[o] \in \pi_0(\mathcal{A})} \frac{q^{\chi_-(o)}}{|Aut(o)|}$$
where $\chi_-(o) = \sum_{i < 0} (-1)^{i+1} \dim H^i End(o, o)$.
At least for now, I think Steven’s calculations do not show that the conjecture in the above corrected form is false.
I’ve now computed the higher version of groupoid cardinality over Z/2 for every knot in the Melvin-Shrestha table (see arXiv:0411206), which contains one tb-maximizing Legendrian of each knot type through 9 crossings, and also for 23 knots which bound Lagrangian disks (arXiv:1411.1364). I can confirm that they all satisfy the corrected form of the conjecture.
Wow! How come this works at all? We have Exts in positive degrees.
The cardinality is only supposed to count objects up to the right notion of equivalence, so it’s ignoring a lot of structure in the category: all that matters are the isomorphisms and the homotopies between them and so on, and none of these live in positive degree.
Question for Vivek, or anyone else who might know: what’s the right notion of groupoid cardinality for the category of 2-graded augmentations / 2-periodic sheaves? I realize the answer should be “whatever gives us the appropriate piece of the HOMFLY polynomial”, but since Hom spaces are now Z/2-graded I don’t know how we’re supposed to interpret the terms $\pi_2$, $\pi_3$, and so on.
I don’t know — also Brad and Dan run into this problem in their paper, in some mysterious form: see the top of page 8 of [HR].
Then I guess the right thing to do is experiment until that piece of the HOMFLY polynomial shows up.
Just to record the thing we’re obliged to believe in the form which I think it should take for $\Lambda$ connected: if $R_\Lambda(z)$ is the ruling polynomial, then the groupoid cardinality of $Aug(\Lambda, \mathbb{F}_q)$ should probably be $R_\Lambda(q^{1/2}-q^{-1/2}) q^{(tb(\Lambda)+1)/2} / (q-1)$. Hopefully your huge amount of cancellation arrives at the same answer.
For the unknot, I think the groupoid cardinality is $1/(q-1)$, but what you’ve written gives I think just $1$. (I think you left off the scalar multiplication from your endomorphisms…)
I think it’s right, unless we have different normalization conventions for the ruling polynomial or something: I’ve weighted each ruling with s switches and c cusps by $z^{s-c+1}$ (some people use $z^{s-c}$), so that $R(z)=1$, and then the above formula gives $1 \cdot q^0 / (q-1) = 1/(q-1)$ as expected.
Oh, yes, the problem was just that I was using $z^{s-c}$.