- #1
center o bass
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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?
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?