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Let $M$ a smooth manifold and $\nabla $ a connexion. Let $\gamma :[a,b]\longrightarrow M$ a $\mathcal C^\infty $ curvature. I recall that if $X,Y\in \Gamma(M)$, and $f,g\in \mathcal C^\infty (M)$, then $$\nabla_{fX}(gY)= f X(g)Y+fg\nabla _XY.$$ So $\nabla _{\dot \gamma (t)}$ is the covariante derivative (the derivate along $\gamma $). We denote $Y$ the vector field along $\gamma $, i.e. $Y_t\in T_{\gamma (t)}M$ for all $t$. Let $x^1,...,x^n$ a coordinate system around $p=\gamma (t)$. So I want to understand why

$$\nabla _{\dot \gamma (t)}Y_t=\dot x^i \frac{\mathrm d a^j(t)}{\mathrm d t}\partial _j+\dot x^i a^j\nabla _{\partial _i}\partial _j$$ (using Einstein sommation convention and the fact that $\partial _i=\frac{\partial }{\partial x_i}$).

My Idea

In the coordinate $x^1,...,x^n$, $$\gamma (t)=(x^1(t),...,x^n(t)),$$ $$\dot\gamma (t)=x^{i}(t)\partial_i$$ and $$Y_t=a^j(t)\partial _j|_{\gamma (t)}.$$

Then $$\nabla _{\dot \gamma (t)}Y_t=\nabla _{x^i(t)\partial _i}(a^j\partial _j)=x^{i}(t)\partial _i (a^j(t))\partial _j+x^i(t)a^j\nabla _{\partial _i}\partial _j,$$

Q1) I don't understand why $\partial _i(a^j(t))=\frac{\mathrm d a^j(t)}{\mathrm d t}$

Q2) In my formula, does $x^i(t)$ is a scalar or a function ? I have the impression that with my notations, thing are a little bit ambiguous, isn't it ?

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  • $\begingroup$ yes the problem with your notations is that it seems at first you fixed $t$, but you need $t$ to be variable for computing the differentials $\frac{ \ }{dt}$ $\endgroup$ – reuns Jun 8 '16 at 8:45
  • $\begingroup$ also you should $\dot x^i(t)\partial_i$ in $\dot\gamma(t)$ $\endgroup$ – Spotty Jun 8 '16 at 12:38
  • $\begingroup$ $\dot r(t)=\dot x^i(t)\partial_i$ $\endgroup$ – lanse7pty Jun 15 '16 at 14:16
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Your notation are a bit confused. But I think there is a mistakes, it should be $$\nabla _{\dot \gamma (t)}Y_t=\left(\frac{\mathrm d a(\gamma (t))}{\mathrm d t}+\dot\gamma ^ia^j\Gamma_{ij}^k\right)\partial _k$$

Indeed, let $\gamma(t)= (\gamma ^1(t),...,\gamma ^n(t))$, and $X:=\dot\gamma =\dot\gamma ^i\partial _i$ and $Y=a^j\partial _j$. Now, \begin{align*} \nabla _{\dot\gamma }Y_t&=\nabla _{X}Y\\ &=\nabla _{\dot \gamma ^i\partial _i}a^j\partial _j\\ &=\dot \gamma ^i\partial_i(a^j)\partial _j+\dot\gamma ^ia^j\nabla _{\partial _i}\partial _j\\ &=\Big(\dot\gamma ^i\partial _i(a^k)+\dot\gamma ^ia^j\Gamma_{ij}^k\Big)\partial _k \end{align*}

Now , $$\dot\gamma^i\partial _i(a^k)=X(a^k)=\frac{\mathrm d a^k}{\mathrm d t}(\gamma (t)),$$

it prove the claim.

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