Integral Curves

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


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. 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?

  2. 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

  3. 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.

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