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Let $\gamma:[0,1] \longrightarrow M$ a geodesic in M, and V a Jacobi field. Let $(E_1(t),...,E_n(t))$ an orthonormal base of $T_{\gamma(t)}M$ with $E_1(0)=\frac{\gamma'(0)}{||\gamma'(0)||}$. Define functions on $[0,1]$ by $R(E_j(t),E_h(t))E_k(t)=R^i_{j,h,k}(t)E_i(t)$. We also have $J(t)=J_i(t)E_i(t)$ (the Einstein summation convention is used). So the Jacobi equation becomes $$ J_i''(t)E_i(t)=R(\gamma_j'E_j(t),J_hE_h(t))(\gamma_k'E_k(t))=\gamma_j'\gamma_k'J_hR^i_{j,h,k}E_i $$. Why does it become in the end $$ J_i''+||\gamma'(0)||^2R^i_{j,1,1}J_j $$ ? It is probably very simple, but I don't get it.. Thanks for any help

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Could you provide some reference of where you take this formulae from? –  matgaio Jun 8 '12 at 0:06

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You'll need to require that $E_1(t) = \frac{\gamma'(t)}{||\gamma'(t)||}$ for all $t$ (which is certainly not a problem locally). If you make this additional assumption, then there are only two small steps that are needed to get from your penultimate equation to the final form. When $E_1(t) = \frac{\gamma'(t)}{||\gamma'(t)||}$, then $$ \gamma_j'E_j(t) = ||\gamma'(t)|| E_1(t). $$ Then the Jacobi equation can be put into your final form: $$ J_i'' E_i = ||\gamma'(t)||J_j\ R(E_1, E_j)E_1 = -||\gamma'(t)||J_j R_{j11}^i E_i, $$ where I have used the anti-symmetry $$ R_{ijk}^l = -R_{jik}^l $$ coming from the definition of $R(X,Y)Z$ via $$ R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla X Z - \nabla_{[X,Y]} Z. $$ Of course, since $\gamma$ is a geodesic, it has constant speed, so that $||\gamma'(t)|| = ||\gamma'(0)||$ for all $t$.

If we don't require that $E_1$ point along $\gamma'$ for all $t$, then the equation $$ J_i''+||\gamma'(0)||^2R^i_{j,1,1}J_j = 0 $$ will be false. The general form of the Jacobi equation is $$ J''(t) + R(J, \gamma')\gamma' = 0, $$ and if the frame $\{E_i(t)\}$ is picked in such a way that $\gamma'(t)$ does not lie in the $E_1$ direction for all time, then the above equation does not simplify to your desired form.

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