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Local flow as a diffeomorphism.

  1. Jan 25, 2014 #1
    Given a smooth vector field ##V## on a smooth manifold ##M## the uniqueness of differential equations assures
    that there exists a unique integral curve ##\phi^{(p)}: J \to M## for some open interval ##J \subseteq \mathbb{R}## for which ##0 \in J## and ##\dot \phi^{(p)} (0) = V_{\phi^{(p)} (0)}##. We can now define a map ##\phi_t: \phi^{(p)}(J) \to M## such that ##\phi_t(p) = \phi^{(p)}(t)##. From the fact that we can reparametrize the above solution it follows that we have the properties ##\phi_t \circ \phi_s = \phi_{t+s}##. As an application of the chain rule it's easy to show that this map satisfies ##\phi_t \circ \phi_s = \phi_{t+s}## and it obviously satisfies ##\phi_0(p) = p##. However it's also often stated that it is a diffeomorphism. How does one go about proving that? Sure the map is differentiable, but how does one prove it has an differentiable inverse? Furthermore exactly what subsets of ##M## are the domain and codomains of the diffeomorphism?
     
  2. jcsd
  3. Jan 25, 2014 #2

    jgens

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    Gold Member

    Notice that φtφ-t = φ0 = φ-tφt which shows the diffeomorphism property. In general, the second question is complicated (and depends heavily on your choice of manifold), but in many special cases it turns out to be the whole manifold.
     
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