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Let $M={\mathbb R}^{3} $ with the usual metric $g=ds^{2} =dx^{2} +dy^{2} +dz^{2} $. Let $\gamma :I\to M$ be a unit speed curve. How can I prove that $\nabla _{\gamma '} \gamma '=\gamma ''$ , where $\nabla $ is the Levi-Civita connection and $\gamma '=\gamma '(s)=\frac{d}{ds} \gamma (s)$ with the arc-length parameter s?

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    $\begingroup$ What did you try to solve this problem yourself? Did you try to write the equation for the covariant derivative using Christoffel symbols? $\endgroup$ Feb 21 '14 at 21:17
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First we have to notice that $\nabla_{\frac{\partial}{\partial x_i}}\frac{\partial}{\partial x_j}=0$ for all $i,j=1,2,3$. Second, let $\gamma(s)=(\gamma_1,\gamma_2,\gamma_3)$ and $\gamma'(s)=(\gamma'_1,\gamma'_2,\gamma'_3)$, then $$\nabla_{\gamma'}\gamma'=\nabla_{\gamma'(s)}(\gamma'_1,\gamma'_2,\gamma'_3) \\=\nabla_{\gamma'(s)}(\gamma'_1\partial_1+\gamma'_2\partial_2+\gamma'_3\partial_3) \\=\nabla_{\gamma'(s)}\gamma'_1\partial_1+\nabla_{\gamma'(s)}\gamma'_2\partial_2+\nabla_{\gamma'(s)}\gamma'_3\partial_3 \\=\gamma''_1\partial_1+\gamma''_2\partial_2+\gamma''_3\partial_3 \\=\gamma''(s)$$You have to know that $$\nabla_{\gamma'(s)}\gamma'_1\partial_1=\frac{d}{ds}(\gamma'_1(s))\partial_1+\gamma'_1(s)\nabla_{\gamma'}\partial_1 \\=\gamma''_1(s))\partial_1+0$$

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  • $\begingroup$ Thank you for your valueable answer. $\endgroup$
    – shabo
    Feb 22 '14 at 20:25

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