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Let $M$ be a Riemannian manifold equipped with the Levi-Civita connection $\nabla$, $\gamma : I \rightarrow M$ a geodesic of $M$ such that $0 \in I$. Let $Y,Z$ two Jacobi fields along $\gamma$ vanishing at $0$, that is $$ \left\{ \begin{array}{ll} \nabla^2_{\dot{\gamma}} Y + R(\dot{\gamma}, Y)\dot{\gamma} = 0,\quad Y(0) = 0 \\ \nabla^2_{\dot{\gamma}} Z + R(\dot{\gamma}, Z)\dot{\gamma} = 0, \quad Z(0) = 0 \end{array} \right. $$ where $R$ is the curvature tensor of $\nabla$.


Now I would like to show that $$\big\langle \nabla_{\dot{\gamma}}(R(\dot{\gamma}, Y)\dot{\gamma}), \nabla_{\dot{\gamma}} Z \big\rangle = \big\langle R(\dot{\gamma}, \nabla_{\dot{\gamma}}Y)\dot{\gamma}, \nabla_{\dot{\gamma}} Z \big\rangle$$ I tried to go back to the definition of $R$, or used the fact that $\partial_t\langle Y,Z \rangle = \langle \nabla_{\dot{\gamma}}Y,Z \rangle + \langle Y,\nabla_{\dot{\gamma}}Z \rangle$ for every vector fields along $\gamma$, but I didn't succeed. Do you have any idea how to proceed ?

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Note that this equation holds at $t=0$. For any vector $V$, we have at $t=0$ $$ \begin{align} \langle \nabla_{\dot\gamma} (R(\dot\gamma, Y)\dot\gamma), V\rangle &= \frac{d}{dt}\langle R(\dot\gamma, Y)\dot\gamma, V\rangle - \langle R(\dot\gamma, Y)\dot\gamma, \nabla_{\dot\gamma}V\rangle \tag{1}\\ &=\frac{d}{dt}\langle R(\dot\gamma, V)\dot\gamma, Y\rangle \tag{2}\\ &=\langle \nabla_{\dot\gamma} (R(\dot\gamma, V)\dot\gamma), Y\rangle+ \langle R(\dot\gamma, V)\dot\gamma, \nabla_{\dot\gamma}Y\rangle \tag{3}\\ &= \langle R(\dot\gamma, \nabla_{\dot\gamma}Y)\dot\gamma, V\rangle .\tag{4} \end{align} $$ In $(1)$ we used $\langle R(\dot\gamma, Y)\dot\gamma, \nabla_{\dot\gamma}V\rangle = 0$ since $Y(0)=0$. For the same reason, we know that $\langle \nabla_{\dot\gamma} (R(\dot\gamma, V)\dot\gamma), Y\rangle$ vanishes in $(3)$. In equations $(2)$ and $(4)$ we used the symmetry property $$\langle R(X,Y)Z,W\rangle = \langle R(Z,W)X,Y\rangle.$$ Setting $V = \nabla_{\dot\gamma}Z$ gives the equation.

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  • $\begingroup$ You are absolutely right, thank you ! $\endgroup$ Sep 18, 2019 at 3:08

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