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In a two dimensional riemannian manifold, we take a normal ball $B_\delta(p)=\text{exp}_p(B(0_{T_pM},\delta))$. We consider $v(\theta)$ a parametrization of the unit circle in $T_pM$, and the define the map $$f(\rho,\theta)=\text{exp}_p(\rho v(\theta))=c_{v(\theta)}^p(\rho).$$ for $0<\rho<\delta$ and $-\pi<\theta<\pi$, these are coordinates for $B_\delta(p)$, except along the geodesic $c_{-v(0)}^p(\rho)$. Fixed a $\theta_0$, the field $$\frac{\partial f}{\partial \rho}|_{(\rho,\theta_0)}$$ is precisely the velocity of the geodesic $\gamma(\rho)=C_{v(\theta_0)}^p(\rho)$. Apparently, the other derivative $$\frac{\partial f}{\partial\theta}|_{(\rho,\theta_0)}$$ is a Jacobi field along the same geodesic, but I am unable to show this. To simplify, we can briefly denote the respective fields $W_\rho$ and $W_\theta$. In order to be a Jacobi field, using that $\gamma'=W_\rho$, $W_\theta$ should satisfy $$\frac{D^2}{d\rho^2}W_\theta+R(W_\theta,W_\rho)W_\rho=0,$$ which in turn is equivalent to $$\nabla_{W_\rho}\nabla_{W_\rho}W_\theta+\nabla_{W_\theta}\nabla_{W_\rho}W_\rho+\nabla_{W_\rho}\nabla_{W_\theta}W_\rho=\nabla_{[W_\theta,W_\rho]}W_\rho.$$ From here I am clueless on how to proceed. I suppose my notation is not the best, for the nature of both fields must be relevant, but I don't know how to take advantage of it.

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Your last equation is off by a sign, then it will hold. You need to note the following equations.

  1. $[W_\theta, W_\rho]=0$ since they are pushforwards of the coordinate vector fields $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \rho}$ by $f$.

  2. Therefore, since the Levi-Civita connection is torsion-free, we have $$ \nabla_{W_\theta}W_\rho = \nabla_{W_\rho}W_\theta + [W_\theta, W_\rho]=\nabla_{W_\rho}W_\theta. $$

  3. $W_\rho = \gamma'$ and $\gamma$ is a geodesic, so $ \nabla_{W_\rho}W_\rho = 0. $

Therefore, you last equation with a fixed sign is \begin{align*} & \nabla_{W_\rho}\nabla_{W_\rho}W_\theta + \nabla_{W_\theta}\nabla_{W_\rho}W_\rho - \nabla_{W_\rho}\nabla_{W_\theta}W_\rho\\ =& \nabla_{W\rho}\big(\nabla_{W_\rho}W_\theta-\nabla_{W_\theta}W_\rho) + 0 =0=\nabla_{[W_\theta, W_\rho]} W_\rho. \end{align*}

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  • $\begingroup$ Thank you very much. I just don't see why the Lie bracket should be 0. Could you develop a bit more on that? $\endgroup$ Commented Jan 4 at 18:01
  • $\begingroup$ This is because pushforward preserves Lie bracket, so $[W_\theta, W_\rho]=[f_*\frac{\partial}{\partial \theta}, f_* \frac{\partial}{\partial \rho}] = f_*[\frac{\partial}{\partial \theta}, \frac{\partial}{\partial \rho}] = f_*0 = 0$. See math.stackexchange.com/questions/365535/…. Also note $[\frac{\partial}{\partial \theta}, \frac{\partial}{\partial \rho}]=0$ since mixed partials are equal. $\endgroup$ Commented Jan 4 at 18:08
  • $\begingroup$ I'm sorry, I'm not familiar with this concept of the push-forward of a vector field... In particular, I don't see how it would act on the Lie bracket. $\endgroup$ Commented Jan 4 at 18:21
  • $\begingroup$ Fair enough. It takes some getting used to. There are a lot of chain rule and Jacobians. (I don't even think the above linked question is posed or answered in the best way.) $\endgroup$ Commented Jan 4 at 18:38

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