Homework 3 Problem 4

Problem

Show that there can exist no constant nonzero Hamiltonian vector field on $\mathbb{T}^2$ with the standard symplectic form $\Omega = d \theta \wedge d \phi$. What if “Hamiltonian” were replaced with “locally Hamiltonian”?

Solution

Proof by Contradiction.

Assume there exists a constant nonzero Hamiltonian vector field $X_H$ on $P = \mathbb{T}^2$. This means,

(1)
\begin{align} \Omega(X_H , v) = dH(x) . v \\ \forall x \in P, v \in T_xP \end{align}

$\mathbb{T}^2$ is a compact manifold without boundary, and H is a smooth function on $\mathbb{T}^2$. The image of a continuous function on a compact set is compact. This means H takes on a maximum and a minimum on $\mathbb{T}^2$. Let $y \in \mathbb{T}^2$ be such a point where H takes on its extreme value. Such a point is a stationary point of H, and hence,

(2)
\begin{align} dH(y) . v = 0 \\ \forall v \in T_yP \end{align}

This means,

(3)
\begin{align} \Omega(X_H , v) = dH(y) . v = 0\\ \forall v \in T_yP \end{align}

If $\Omega(X_H , v) = 0$ $\forall v \in T_yP$ , then since $X_H$ is non-zero, the matrix determined by $\Omega$ is singular at that point. That means $\Omega$ is not a non-degenerate 2-form on $\mathbb{T}^2$. This contradicts the fact that $\Omega$ is non-degenerate! Hence there can be no non-zero constant vector field on $\mathbb{T}^2$.

We can have a constant non-zero locally hamiltonian vector field on $\mathbb{T}^2$. Let $X_H = a\frac{\partial}{\partial \theta} + b\frac{\partial}{\partial \phi}$ for $a,b \in $\mathbb{R}$ and at least one of $a$ or $b$ not zero.

We compute $X_H$ "hook" $\Omega$.

(4)
\begin{align} i_{X_H}\Omega = ad\phi - bd\theta \end{align}

$X_H$ is locally hamiltonian if $d (i_{X_H}\Omega$) = 0$. This is obviously true. So $X_H$ is locally hamiltonian.

Explanations

Here are some detailed calculations, and informal (and possibly inaccurate) descriptions of concepts used above. Not part of the formal solution.

1. Compactness.

A set E is compact, if given any infinite collection of open sets which cover E, there is a finite sub-collection of the original open sets which cover E. Whats an example of a non-compact set? Consider the real line , $\mathbb{R}$ . A covering for the real line using open sets is given by the open intervals (-n,n) for $n \in \mathbb{N}$. It is obvious any finite subcollection of this covering will not cover the whole real line, and the real line is not compact.

So what sets are compact? For finite dimensional spaces, a set is compact if and only if it is closed and bounded . Thus $\mathbb{T}^2$ is compact. (Note: For infinite dimensional space, a compact set is closed and bounded, but the converse may not be true).

2. The continuous image of a compact set is compact.

The proof of this goes something like this. Say the continuous function f maps the compact set to f[A]. Given any open covering of the image, we pull back the open sets to the compact set, and these sets cover A. Using compactness of A, we get a finite sub-cover. Pushing forward this finite subcover via f, gives a finite covering of f[A] by open sets, and hence making that compact.

3. A continuous function , say f, takes it maximum or minimum value on any compact set A.

We know f[A] is compact, and hence it is closed and bounded. This means the infimum and supremum of f[A] is in f[A].

Consider a counter example, by taking a non-compact set, say (0,1). Then 1 is the supremum of this set (its bigger than any other element in this set, and between any other number of the set and 1 there exists another number between them also in the set), but it does not belong to (0,1). )

4. Stationary point for a function $f$ means $df.(v) = 0$ $\forall v$.

This is equivalent to the condition that the gradient of the function is zero.
A function takes on its extreme values either at its stationary points, or on the boundary. Since $\mathbb{T}^2$ has no boundary, we considered only the stationary points.

5. If $\Omega(X_H , v) = 0$ $\forall v \in T_yP$, it is degenerate.

Let $A$ denote the matrix for $\Omega$, and $x$ denote the vector for $X_H$. The above statement is equivalent to saying:

(5)
\begin{align} v^TAx = 0 \ \forall v\\ \Rightarrow Ax = 0 \ (\ where \ 0 \ is \ thought \ of \ as \ a \ vector \ here) \\ \Rightarrow \ A \ is \ singular \ since \ x \neq 0 \end{align}
6. $i_{X_H}\Omega = ad\phi - bd\theta$

Let $X_H = a\frac{\partial}{\partial \theta} + b\frac{\partial}{\partial \phi}$ and $v = c\frac{\partial}{\partial \theta} + d\frac{\partial}{\partial \phi}$.

(6)
\begin{align} \Omega(X_H,v) = d\theta\wedge d\phi(X_H,v ) \\ = d\theta(X_H)d\phi(v) - d\theta(v)d\phi(X_H) \\ = ad - bc \end{align}

The conclusion then follows.

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