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A curve $\gamma : (a,b)\rightarrow M$ is an integral curve of X if and only if \begin{equation} d\gamma\bigg(\frac{d}{dr}\bigg|_{t}\bigg)= X(\gamma(t)) \end{equation} $t \in (a,b)$ Suppose we require an integral curve through $m\in M$ with vector field X such that $\gamma(0)=m$ Choosing a co-ordinate system around $m, (U,\phi)$ with co-ordinate function $x_1,x_2,...x_d$. Then \begin{equation} X|_{U} = \sum_{i=1}^{d}f_i\frac{\partial}{\partial x_i} \end{equation}

Also, \begin{equation} d\gamma\bigg(\frac{d}{dr}\bigg|_{t}\bigg) = \sum_{i=1}^{d}\frac{d(x_i\circ\gamma)}{dr}\bigg|_{t}\frac{\partial}{\partial x_i}\bigg|_{\gamma(t)} \end{equation}

From above equations $\gamma$ is integral curve if \begin{equation} \sum_{i=1}^{d}\frac{d(x_i\circ\gamma)}{dr}\bigg|_{t}\frac{\partial}{\partial x_i}\bigg|_{\gamma(t)} = \sum_{i=1}^{d}f_i(\gamma(t))\frac{\partial}{\partial x_i}\bigg|_{\gamma(t)} \end{equation} Thus $\gamma$ is an integral curve of X on $\gamma^{-1}(U)$ if and only if \begin{equation} \frac{d\gamma_i}{dr}\bigg|_t = f_{i}\circ \phi^{-1}(\gamma_1(t),\gamma_2(t),...,\gamma_d(t)) \end{equation} where $\gamma_i=x_i\circ\gamma$. Such a unique solution exists(from theory of differential equations)

Let $\Gamma(t) := (\gamma_1(t),\gamma_2(t),...,\gamma_d(t))$ Then $\gamma(t) = \phi^{-1} \circ \Gamma(t)$ and the above equation is \begin{equation} \frac{d\Gamma}{dt}= X\circ\phi^{-1}(\Gamma(t)) \end{equation} Question: I wish to prove that $\frac{d}{dt}\gamma(t) = X (\gamma(t))$. Applying the chain rule i get: \begin{equation} d(\phi^{-1}\circ\Gamma)(t)(1) = d\phi^{-1}(\Gamma(t)) (d\Gamma(t)(1)) = d\phi^{-1}(\Gamma(t))(X\circ\phi^{-1}(\Gamma(t))) = d\phi^{-1}(\Gamma(t))(X(\gamma(t))) \end{equation} I am not able to carry this forward.

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  • $\begingroup$ $d\phi$ is an identity function (in the sense it carries basis to basis) and so $d\phi^{-1}$ should be an identity function, $\phi$ being a diffeomorphism. i think that should solve it ? $\endgroup$ – Himanshu Oct 4 '15 at 14:50

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