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Lets say I have two (compact) manifolds $U$,$V$ and a diffeomorphism $\psi:U\rightarrow V $.

The shortest way between two points $a$ , $b \in V$ is given by a parametrisation $\gamma :W \rightarrow V $ which I found using Euler-Langrange-Equation. Now I want to have a parametrisation $\eta: W \rightarrow U$ with the same property, that means $\psi(\eta(t))=\gamma(t)$.

Moreover, I need more theory according to lengths on manifolds, I minimized the length using $L=\int \left \|\gamma '(t)) \right \|dt$ but I don't know if this is an elegent way of doing it. The problem which I want to solve is: what is the Length of to points $a,b\in U$ regarding to $V$, without transforming to $V$, finding a parametrisation and integrating over it.

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The Euler-Lagrange equation for geodesics is easier to work with if you minimize the energy $\int \|\gamma'\|^2$ instead of the length $\int \|\gamma'\|$. The energy is better behaved, and its minimizers are geodesics parametrized with constant speed. –  user53153 Feb 11 '13 at 0:20
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When you are talking about lengths of curves at least $V$ is a Riemannian manifold to begin with: For any system of local coordinates $(v_i)_{1\leq i\leq n}$ on $V$ a metric tensor $v\mapsto\bigl(g_{ij}(v)\bigr)$ is given such that the length of a curve $$\gamma:\quad t\mapsto v(t)\in V\qquad(a\leq t\leq b)$$ computes to $$L(\gamma)=\int_a^b\sqrt{\sum_{i,j} g_{ij}\bigl(v(t)\bigr)\dot v_i(t)\dot v_j(t)}\ dt\ .$$ If a diffeomorphism $\psi:\ U\to V$ is given then any curve $$\gamma:\quad t\mapsto u(t)\in U\qquad(a\leq t\leq b)$$ has an image curve $$\gamma':=\psi(\gamma):\quad t\mapsto \psi\bigl(u(t)\bigr)\in V\qquad(a\leq t\leq b)\ .$$ You then wish to define the length of $\gamma$ as $\tilde L(\gamma):=L\bigl(\psi(\gamma)\bigr)$. It turns out that this idea can be implemented without reference to a particular $\gamma$. What is at stake here is the so-called pullback of the given Riemannian metric $(g_{ij})$ on $V$ to a Riemannian metric $u\mapsto \bigl(\tilde g_{kl}(u)\bigr)$ on $U$.

The $\tilde g_{kl}(u)$ can be computed from the $g_{ij}(v)$ and the Jacobian $\left[{\partial\psi_i\over\partial u_k}\right]_{ik}$ without eigenvalues or inverting a matrix. One just has to plug in $v(t)=\psi\bigl(u(t)\bigr)$ into $$\sum_{i,j} g_{ij}\bigl(v(t)\bigr)\dot v_i(t)\dot v_j(t)$$ and to apply the chain rule. It then becomes a matter of linear algebra.

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