I think I understand why a vector field must have a unique set of integral curves, but I don't see why they must define a one-parameter group of diffeomorphisms.(adsbygoogle = window.adsbygoogle || []).push({});

Let X be a vector field on a manifold M, and p a point in M. A smooth curve C through p is said to be an integral curve of X if C(0)=p and

[tex]X_{C(t)}=\dot C_{C(t)}[/tex]

for all t in some interval (-a,a). (I hope it's obvious that what I mean by "dot C" is the tangent vector of C). I'm confused about how this defines a one-parameter group of diffeomorphisms. Supposedly, if we write the integral curve at q as C^{q}, we can define [itex]\phi_t(q)=C^q(t)[/itex] for all t and q, and now this [itex]\phi[/itex] is a one-parameter group of diffeomorphisms. In particular, this would imply that

[tex]\phi_t(\phi_s(p))=\phi_{t+s}(p)[/tex]

which is equivalent to

[tex]C^{C^p(s)}(t)=C^p(t+s)[/tex]

Let's simplify the notation by calling the integral curve at p C, and the integral curve at C(s) B. The equation becomes

[tex]B(t)=C(t+s)[/tex]

How do we prove that this equation holds?

Also, why does the definition of an integral curve require that [tex]X=\dot C[/tex] holds over anintervalrather than just at a single point? I suspect that the answer is that as long as the curve is smooth, it doesn't matter.

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# Integral curves and one-parameter groups of diffeomorphisms

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