Finally, let’s prove this counting formula.

Theorem.

$$\# \underline{Aug}(\mathbb{F}_q) = q^{(tb-\chi^*)/2 } \cdot (q – 1)^{-c}  \cdot \# Aug(\mathbb{F}_q)$$

Proof.

Note that the Thurston-Bennequin number is also $tb = \sum_{i} (-1)^{i} a_i$, so that

$$(tb-\chi^*)/2 = \sum_{i < 0} (-1)^{i} a_i$$

Recall that the complex we use to compute $Aug_+$ hom spaces has generators in degrees shifted up one from the DGA generators.  We write $C^i$ for the degree $i$ piece of the chain complex that computes $Aug_+$ hom, so that $a_i = \dim C^i$.  So,

$$(tb-\chi^*)/2 = c + \sum_{i \le 0} (-1)^{i-1} \dim C^i$$

Expanding out the conjecture:

$$\sum_{[o] \in \pi_0(\underline{Aug(\mathbb{F}_q)})} \frac{1}{q^{\sum_{i < 0} (-1)^{i}\dim H^i Hom(o, o)}  |Aut(o)|} = \frac{\# Aug(\mathbb{F}_q)}{q^{- c + \sum_{i \le 0} (-1)^{i} \dim C^i}(q-1)^c}$$

or, rearranging,

$$\# Aug(\mathbb{F}_q) =  \sum_{[o] \in \pi_0(\underline{Aug(\mathbb{F}_q)})}\frac{q^{ \sum_{i < 0} (-1)^{i} \dim C^i}}{q^{\sum_{i < 0} (-1)^{i}\dim H^i Hom(o, o)}} \cdot \frac{q^{- c + \dim C^0} (q-1)^c}{|Aut(o)|}$$

We now perform the ritual of homological algebra.

$$ \sum_{i < 0} (-1)^{i} \dim C^i = -\dim B^0 + \sum_{i < 0} (-1)^{i}\dim H^i Hom(o, o)$$

Here, $B^0$ is the 0-boundaries, or in other words, the homotopies amongst morphisms (!).  Thus the conjecture simplifies to

$$\# Aug(\mathbb{F}_q) =  \sum_{[o] \in \pi_0(\underline{Aug(\mathbb{F}_q)})} q^{-\dim B^0} \cdot \frac{q^{- c + \dim C^0} (q-1)^c}{|Aut(o)|}$$

Thus to prove the theorem, it will suffice to prove the following stronger result:

Proposition.  The number of augmentations isomorphic to a given augmentation $o$ is $q^{\dim C^0 -\dim B^0 – c} \cdot \frac{(q-1)^c}{|Aut(o)|}$.

Proof.  Let me first explain it for the lines category.  Here are we are saying the following thing.  An isomorphism two elements in the Morse complex category is a degree zero morphism $\phi: [d] \to [d’]$ such that $\phi d \pm d’ \phi = 0$.  But given any degree zero invertible linear map $\phi$, we can take $d’ = \phi d \phi^{-1}$.  The stabilizer of this action is precisely the $\phi$ annihilated by the differential on $C^0$.  So I want to compute linear isomorphisms modulo their intersection with $Z^0$.  That’s presumably what the above formula says, and if it doesn’t, that’s presumably because I miscalculated somewhere in the last 2 pieces of this note.

The argument in general is essentially similar, although just at the moment I cannot tell whether or not I need to use that isomorphism = dga homotopy…