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Chapter 22: Problem 22

Suppose \((M, H)\) is a contact manifold and \(X\) is a smooth vector field on \(M\). Prove that \(X\) is the Reeb field of some contact form for \(H\) if and only if it is a contact vector field that takes no values in \(H\).

Short Answer

Expert verified
A vector field is a Reeb field iff it is a contact vector field not lying in the contact distribution.

Step by step solution

01

Understand the Definitions

A contact manifold \(M, H\) is a manifold equipped with a hyperplane distribution \(H\) that locally can be described as the kernel of a contact form. A Reeb field on \(M\) is a vector field \(X\) satisfying \(\alpha(X) = 1\) and \(L_X \alpha = 0\), where \(\alpha\) is a contact form for \(H\). A contact vector field is one that preserves the contact structure, i.e., \(L_X \alpha = f \alpha\) for some function \(f\). The condition that \(X\) takes no values in \(H\) implies that \(\alpha(X) eq 0\).
02

Show X as a Reeb Field Implies X is a Contact Vector Field with No Values in H

Assume \(X\) is a Reeb field for some contact form \(\alpha\) of \(H\). This means \(\alpha(X) = 1\) and \(L_X \alpha = 0\). Since \(\alpha(X) = 1\), \(X\) takes no values in \(H = \ker(\alpha)\). Therefore, \(X\) is trivially a contact vector field as \(L_X \alpha = 0\) implies \(f = 0\). Thus, \(X\) preserves the contact form.
03

Show X as a Contact Vector Field with No Values in H Implies X is a Reeb Field

Assume \(X\) is a contact vector field that takes no values in \(H\). Then \(\alpha(X) eq 0\) since \(X\) does not lie in \(\ker(\alpha)\). Set \(\beta = \frac{\alpha}{\alpha(X)}\). Then \(\beta(X) = 1\) and \(L_X \beta = \frac{1}{\alpha(X)} L_X \alpha - \frac{\beta}{\alpha(X)} d\alpha(X) = 0\) because \(L_X \alpha = f \alpha\). Thus, \(X\) is the Reeb field for the contact form \(\beta\).
04

Conclude the Proof

We have shown that if \(X\) is a Reeb field, then it must be a contact vector field not taking values in \(H\). Conversely, if \(X\) is a contact vector field not taking values in \(H\), then \(X\) is a Reeb field for some contact form. Therefore, \(X\) is a Reeb field of some contact form for \(H\) if and only if it is a contact vector field that takes no values in \(H\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reeb vector field
A Reeb vector field is a special type of vector field associated with a contact manifold. On a manifold with a contact structure, the Reeb field
  • satisfies two crucial conditions: the contact form evaluated on the Reeb vector equals one, i.e., \( \alpha(X) = 1 \),
  • and the Lie derivative of the contact form in the direction of the Reeb vector vanishes, i.e., \( L_X \alpha = 0 \).
These conditions ensure that the Reeb vector remains outside the kernel defined by the contact form, playing a pivotal role in the dynamics on contact manifolds. Because it doesn't change the contact form, it helps preserve the contact structure's properties.
contact form
A contact form is a differential 1-form on a manifold, represented as \( \alpha \), integral to defining a contact structure. It's not just any 1-form; a contact form makes a manifold more than just a regular differentiable structure. It characterizes the directionality of the contact manifold by defining a distribution that assigns a hyperplane tangent to the manifold at each point.
The hyperplane distribution is the kernel of the contact form, expressed as \( H = \ker(\alpha) \). This means every vector field that lies in \( H \) results in the contact form yielding zero when evaluated on those vectors. To keep things dynamic and interesting, the contact form isn't non-degenerate across the whole manifold鈥攊t's contact.
vector field
Vector fields are everywhere in the world of manifolds. They are essentially an assignment of a vector to each point in a manifold. Imagine a gentle wind blowing over a landscape, assigning a wind direction and strength to every spot鈥攖hat's akin to what a vector field does. In mathematical terms, each vector field is a smooth section from the tangent bundle of a manifold.
  • They help describe dynamics and movement across manifolds,
  • and can interact with various structures, such as the contact form,
  • by preservation or transformation of structures like hyperplanes.
Some vector fields, such as the Reeb vector fields, have special rules they follow within the context of contact manifolds.
contact structure
Contact structures paint the geometric picture of a contact manifold, adding a layer above the typical smooth structure found on manifolds. They are characterized by a non-integrable hyperplane distribution, often defined as the kernel of a contact form \( \alpha \). This hyperplane distribution is called the contact distribution.

In contrast to integrable distributions, which lead to foliations, a contact structure resists integration鈥攊llustrating the nontrivial nature of these hyperplanes. In 3 dimensions, for example, the contact structure embodies this non-integrability, leading to twisting and complex interactions.
  • They appear in mechanical systems as constraints, where movement is restricted,
  • but provide freedom for dynamics in the manifold,
  • and fundamentally connect with Reeb vector fields to showcase elegant geometric dynamics.
Together, the contact form and contact structure define the unique properties of contact manifolds, crucial in areas like Hamiltonian dynamics.

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Most popular questions from this chapter

Let \((V, \omega)\) be a symplectic vector space of dimension \(2 n\). Show that for each symplectic, isotropic, coisotropic, or Lagrangian subspace \(S \subseteq V\), there exists a symplectic basis \(\left(A_{i}, B_{i}\right)\) for \(V\) with the following property: (a) If \(S\) is symplectic, \(S=\operatorname{span}\left(A_{1}, B_{1}, \ldots, A_{k}, B_{k}\right)\) for some \(k\). (b) If \(S\) is isotropic, \(S=\operatorname{span}\left(A_{1}, \ldots, A_{k}\right)\) for some \(k\). (c) If \(S\) is coisotropic, \(S=\operatorname{span}\left(A_{1}, \ldots, A_{n}, B_{1}, \ldots, B_{k}\right)\) for some \(k\). (d) If \(S\) is Lagrangian, \(S=\operatorname{span}\left(A_{1}, \ldots, A_{n}\right)\).

Let \((M, \omega)\) be a symplectic manifold, and suppose \(F: N \rightarrow M\) is a smooth map such that \(F^{*} \omega\) is symplectic. Show that \(F\) is a smooth immersion.

Suppose \((M, H)\) is a contact manifold of dimension \(2 n+1\). Show that if \(n\) is odd, then \(M\) is orientable, while if \(n\) is even, then \(M\) is orientable if and only if there exists a global contact form for \(H\).

Let \((M, \omega)\) be a symplectic manifold. Let \(S(M) \subseteq \mathscr{E}(M)\) denote the set of symplectic vector fields on \(M\), and \(\mathscr{H}(M) \subseteq \mathscr{E}(M)\) the set of Hamiltonian vector fields. (a) Show that \(S(M)\) is a Lie subalgebra of \(\mathscr{X}(M)\), and \(\mathscr{H}(M)\) is a Lie subalgebra of \(S(M)\). (b) Show that the map from \(S(M)\) to \(\Omega^{1}(M)\) given by \(\left.X \mapsto X\right\lrcorner \omega\) descends to a vector space isomorphism between \(S(M) / \mathscr{H}(M)\) and \(H_{\mathrm{dR}}^{1}(M)\).

Let \(Q\) be a smooth manifold, and let \(S\) be an embedded Lagrangian submanifold of the total space of \(T^{*} Q\). Prove the following statements. (a) If \(S\) is transverse to the fiber of \(T^{*} Q\) at a point \(q \in T^{*} Q\), then there exist a neighborhood \(V\) of \(q\) in \(S\) and a neighborhood \(U\) of \(\pi(q)\) in \(Q\) such that \(V\) is the image of a smooth closed 1-form defined on \(U\). (b) \(S\) is the image of a globally defined smooth closed 1-form on \(Q\) if and only if \(S\) intersects each fiber transversely in exactly one point.
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