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cheatsheets:symplectic-contact-manifold [2022/12/12 02:23] pzhoucheatsheets:symplectic-contact-manifold [2023/06/25 15:53] (current) – external edit 127.0.0.1
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 ====== Symplectic and Contact manifolds ====== ====== Symplectic and Contact manifolds ======
  
 +Let's follow the reference of [[https://arxiv.org/pdf/0803.2455.pdf | EES]] (Ekholm-Etnyre-Sabloff). The standard contact form on $\R^3$ is $dz - ydx$.
  
 ===== Example of $\C$ ===== ===== Example of $\C$ =====
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 $$ \lambda = -y dx, \quad \omega = dx \wedge dy $$ $$ \lambda = -y dx, \quad \omega = dx \wedge dy $$
 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, by using $W = (M \times \R, \lambda = e^t \alpha)$. We then have
 +$$ d\lambda = e^t (d\alpha + dt \wedge \alpha) $$
 +
 +
 +
 +===== 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,  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 $\d p = q$. 
 +
 +
 +===== 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, whose input is a positive chord, and outputs are many negative chord. 
 +
 +So [[https://hal.archives-ouvertes.fr/hal-01873728/file/generation-v2.pdf | CDGG ]] (Definition 3.4)'s sign convention for the punctures is correct after all. 
 +
 +==== 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 '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
 +
 +
 +
  
  
cheatsheets/symplectic-contact-manifold.1670811787.txt.gz · Last modified: 2023/06/25 15:53 (external edit)