cheatsheets:symplectic-contact-manifold
Differences
This shows you the differences between two versions of the page.
Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
cheatsheets:symplectic-contact-manifold [2022/12/12 02:23] – pzhou | cheatsheets:symplectic-contact-manifold [2023/06/25 15:53] (current) – external edit 127.0.0.1 | ||
---|---|---|---|
Line 1: | Line 1: | ||
====== Symplectic and Contact manifolds ====== | ====== Symplectic and Contact manifolds ====== | ||
+ | Let's follow the reference of [[https:// | ||
===== Example of $\C$ ===== | ===== Example of $\C$ ===== | ||
Line 20: | Line 21: | ||
$$ \lambda = -y dx, \quad \omega = dx \wedge dy $$ | $$ \lambda = -y dx, \quad \omega = dx \wedge dy $$ | ||
and Reeb flow is the negative geodesic flow on the boundary. | and Reeb flow is the negative geodesic flow on the boundary. | ||
+ | |||
+ | ===== Weinstein, Contact and Back ===== | ||
+ | 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, | ||
+ | $$ d\lambda = e^t (d\alpha + dt \wedge \alpha) $$ | ||
+ | |||
+ | |||
+ | |||
+ | ===== Exact Immersed Lagrangian ===== | ||
+ | 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, | ||
+ | |||
+ | |||
+ | 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 $\d p = q$. | ||
+ | |||
+ | |||
+ | ===== Legendrian Contact Homology ===== | ||
+ | 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, | ||
+ | |||
+ | So [[https:// | ||
+ | |||
+ | ==== Linearized version ==== | ||
+ | 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 ' | ||
+ | |||
+ | |||
+ | |||
cheatsheets/symplectic-contact-manifold.1670811787.txt.gz · Last modified: 2023/06/25 15:53 (external edit)