Integral Curves

Let $\alpha, \beta : I \longrightarrow V$ be integral curves in a manifold $M$ then $A = \{ t : \alpha (t) = \beta (t) \}$ is closed.

Proof:

Let $\{t_n\}$ be have a limit $t$ in $I$. Let $\xi : U \longrightarrow \mathbb{R}^m$ be a chart then $\xi (\alpha (t_n))$ converges to $\xi (\alpha (t))$ by continuity. But $\xi (\beta (t_n))$ also converges to $\xi (\alpha (t))$. Hence $\xi (\alpha (t)) = \xi (\beta (t))$ and so $t \in A$. Thus $A$ is closed.

6 thoughts on “Integral Curves”

1. John $f(t)=f(s)$ does not mean $s=t$.
For example, $-2\times -2 = 2\times 2$; does $\xi$ have the right properties?

2. idontgetoutmuch

Yes because $\xi$ is a chart hence one-to-one.

3. john $\alpha (t)$ and $\beta (t)$ both cts $\implies f(t) = \alpha (t) - \beta (t)$ is cts.

Let $A=\{t | f(t)=0\} = \{t | \alpha (t) = \beta (t) \}$ and $B=\{t | f(t) \neq 0 \}$ $A \cap B =\emptyset$ and $A \cup B = I$

Claim $B$ is open. $\forall t \in B f(t) = X$ where $|X| = |2\epsilon| > 0$ $f$ is continuous $\implies$ given $\epsilon > 0 \exists \delta$ s.t. $\forall t\prime$ $|t - t\prime| < \delta \implies |f(t) - f(t\prime)| < \epsilon$ $\implies |f(t)| - |f(t\prime)| < \epsilon$ $\implies 2\epsilon - |f(t\prime)| < \epsilon$ $\implies \epsilon < |f(t\prime)|$ $\implies f(t\prime) \neq = 0$ $\implies t\prime \in B$ $\forall t \in B \exists \delta$ s.t. $(t-\delta,t+\delta) \in B$ $\implies B$ is open $\implies B$ complement closed $\implies A$ closed.

4. idontgetoutmuch

1. The minus function ( $-$) doesn’t necessarily exist in the manifold so you can’t even define $f$.

2. Allowing for this, why should $f(t) = X$ and this always be $2\epsilon$?

5. john

1 you are right need to apply $\xi$ to $\alpha$ and $\beta$ then define $f$ as the difference of these two images now in $\mathbb{R}^m$. This $f$ is still cts and we now have a metric.

2 2 is my nonsense, now improved by pre-applying $\xi$; $f(t)$ is now a non zero element of $\mathbb{R}^m$ so has a non-zero length, if we call its length $X$ then halve this length to be our $\epsilon$ to be used in the $\epsilon, \delta$ argument.

intuitively any non zero $f(t)$ has a cloud of non-zero $f(t')$ around it and the $t'$ pre-images belong to an open ball in $I$

6. idontgetoutmuch

I see what you are saying. Wouldn’t it be better to argue that if $x \in B$ then there is an open interval around $x$ such that this interval is in $B$. Take the union of these open intervals and since the arbitrary unions of open sets is open then $B$ must be open. Hence $A$ must be closed.