Let's follow the reference of EES (Ekholm-Etnyre-Sabloff). The standard contact form on $\R^3$ is $dz - ydx$.

Our basic symplectic example is $\C$, where we have $$ \omega = dx \wedge dy, \; J(\d_x) = \d_y \; g(v,v) = \omega(Jv,v) \geq 0. $$

The boundary of $\C$ (at infinity) is $S^1$, where we want the Reeb flow to be $\d_\theta$. That requires $\alpha = d\theta$.

If we want to use function $H=r^2/2$ to generate this flow, then $dH = r dr$, and $X_H = \d_\theta$, $\omega =r dr \wedge d\theta$, hence, we better require $$ \iota_{X_H} \omega = - dH. $$

We want to choose the Liouville form $\lambda$ on the Weinstein domain $W$, so that $\lambda |_{\d W} = \alpha$. Here $W = \{ |z| \leq 1\}$.

So, we can use $\lambda = -y dx + x dy$ (which restricts to $d\theta$).

Given the above sign convention, we want $$ \lambda = -y dx, \quad \omega = dx \wedge dy $$ and Reeb flow is the negative geodesic flow on the boundary.

Let $(W, \lambda)$ be a Weinstein manifold. Let $(M = W \times \R_z, \alpha = dz \pm \lambda)$ be the contact form. Here, if we use $\lambda = -y dx$, then it make sense to use $dz + \lambda = dz - y dx$ as the contact form.

If $(M, \alpha)$ is a contact manifold, then we can do symplectization, by using $W = (M \times \R, \lambda = e^t \alpha)$. We then have $$ d\lambda = e^t (d\alpha + dt \wedge \alpha) $$

If we have an exact immersed Lagrangian $L \In W$, we may lift it to $L^+ \In W\times \R$, by choosing the $z$ coordinate, such that $dz + \lambda = 0$. Such choice is of course unique only up to translation.

Suppose we have an immersed curve, goes like $\theta \mapsto (e^{i\theta} - 1)^2$, so image is a figure 8. Then, we have one double point. There are two disks ending on this point, so possibly contribute to $m_0(1) = 2$.

In general, suppose we have a disk that ends on one double point, then $$ 0 < \int_D \omega = \int_{\d D} \lambda = \int_{\d D} -dz $$ So, as we go around $\d D$ in the positive direction, we are going downward in the $z$ direction. Thus, to come back up, we need to use the upward pointing Reeb chord.

The differential of this upward Reeb chord would be $2$. (since we have two such disks, and there is no further output).

Another example is a bigon in $T^*\R$, top one is $L_0$, bottom one is $L_1$, left intersection is $p$ in (degree 1), and right intersection point is $q$ (as output). Regardless of choices, we have $\deg(q) - \deg(p) = 1$, and $\d p = q$.

We are considering the special case where $M = W \times \R$, contactization of a Weinstein manifold.

Given an immersed Lagrangian $L$ in $W$ (no wrapping turned on) equipped with a potential, we can do the lift $L_+$ to $M$. Then, we have a free algebra generated by the double point in $L$, or Reeb chord ending in $L_+$. It is hugely non-commutative. The only thing makes geometric sense is the differential, whose input is a positive chord, and outputs are many negative chord.

So CDGG (Definition 3.4)'s sign convention for the punctures is correct after all.

Fix a comm ring $F$, and DGA $A$ over $F$. Suppose we have augmentation from a DGA $\epsilon: A \to F$.

Consider two Legendrians $L_0^+, L_1^+$, with DGA $A_0, A_1$. Consider the set of self-Reeb chords $D_1, D_2$, and $C$ the Reeb chords from $L_1^+$ to $L_0^+$. We build a total immersed Lagrangian $L = L_0 \cup L_1$.

Then, given the augmentation data, we have $LCC(L_0^+, L_1^+)$ is an $F$-module freely generated by the wrong way Reeb chord from $L_1$ to $L_0$. Then differential counting 'jagged bigon', with one edge going around $L_0$, and the other going around $L_1$, and whenever we meet a double point along the boundary