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I know that $\gamma $ is a geodesic if and only if $$\nabla _{\dot \gamma}\dot\gamma =0.$$ Using this, I'm trying to re find the equation $$\ddot x^k +\Gamma_{i\ell}^k \dot x^i\dot x^\ell=0,$$ but I don't get it.

Let $x^1,...,x^n$ local coordinate. Then $\gamma (t)=(x^1(t),...,x^n(t))$ and $\dot\gamma (t)=\dot x^i(t)\frac{\partial }{\partial x^i}$ (using Einstein convention). Then, $$0=\nabla _{\dot \gamma (t)}\dot \gamma (t)=\nabla _{\dot x^i \frac{\partial }{\partial x^i}}\dot x^\ell \frac{\partial }{\partial x^\ell}=\dot x^i\frac{\partial \dot x^\ell}{\partial x^i}\frac{\partial }{\partial x^\ell}+\dot x^i\dot x^\ell \underbrace{\nabla _{\frac{\partial }{\partial x^i}}\frac{\partial }{\partial x^\ell}}_{=\Gamma_{i\ell}^m\frac{\partial }{\partial x^m}}$$

but I don't get the right equation. Any idea ?

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    $\begingroup$ You lost a dot. Try looking under the couch. $\endgroup$ – amcalde Jun 9 '16 at 11:04
  • $\begingroup$ @amcalde: I corrected it, thanks. $\endgroup$ – MSE Jun 9 '16 at 12:08
  • $\begingroup$ You actually are very close. Just remember that you're on the curve $\gamma(t)$, so you need to interpret $\dfrac{\partial \dot x^\ell}{\partial x^i}$ by using the chain rule, and you'll get $\dfrac{\ddot x^\ell}{\dot x^i}$. Then just fiddle with indices and you have it. It's probably better to write everything explicitly with a composition and use the chain rule from the start. $\endgroup$ – Ted Shifrin Jun 9 '16 at 16:02
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First of all, remember that the arguments of $\nabla$ must be tangent fields on the manifold, which $\dot \gamma$ is not, therefore one has to first give a rigorous meaning to the notation $\nabla _{\dot \gamma} \dot \gamma$. Once we've clarified this, the rest will come naturally, without any effort.

Let $\gamma : [a,b] \to M$ and let $t_0 \in [a,b]$. Let $p = \gamma (t_0)$. Let $U$ be some small neighbourhood of $p$ such that the portion of $\gamma$ that stays inside $U$ has no self-intersection, and let $X \in \mathcal X (U)$ be a local field that extends $\dot \gamma$, i.e. $X _{\gamma (t)} = \dot \gamma (t)$. Then $\left( \nabla _{\dot \gamma} \dot \gamma \right) (t_0)$ means $(\nabla _X X) _p$.

With this, things become easy because the condition $\nabla _{\dot \gamma} \dot \gamma = 0$ expands as:

$$0 = \nabla _X X = \nabla _{X^i \partial _i} (X^j \partial _j) = X^i \nabla _{\partial _i} (X^j \partial _j) = X^i \Big( (\nabla _{\partial _i} X^j) \partial _j + X^j (\nabla _{\partial _i} \partial _j) \Big) = \\ X^i \Big( (\partial _i X^j) \partial _j + X^j \Gamma _{ij} ^k \partial _k \Big) = (X^i \partial _i) (X^j) \partial _j + \Gamma _{ij} ^k X^i X^j \partial _k = \Big( X (X^k) + \Gamma _{ij} ^k X^i X^j \Big) \partial _k ,$$

which implies that

$$\tag {#} X (X^k) + \Gamma _{ij} ^k X^i X^j = 0 \; \forall k .$$

Notice now that for any smooth $f$,

$$X(f) (p) = \textrm d f (X) (p) = \textrm d _p f (X_p) = \textrm d _{\gamma (t_0)} f (\dot \gamma (t_0)) = \frac {\textrm d} {\textrm d t} \Bigg| _{t = t_0} f \circ \gamma ,$$

so

$$X(X^k) (p) = \frac {\textrm d} {\textrm d t} \Bigg| _{t = t_0} X^k \circ \gamma = \frac {\textrm d} {\textrm d t} \Bigg| _{t = t_0} \dot \gamma ^k = \ddot \gamma ^k (t_0) ,$$

hence, upon evaluation in $p$, $(\#)$ becomes

$$\ddot \gamma ^k (t_0) + \Gamma _{ij} ^k (\gamma (t_0)) \ \dot \gamma ^i (t_0) \dot \gamma ^j (t_0) = 0 ,$$

which upon removal of the arguments becomes the desired

$$\ddot \gamma ^k + (\Gamma _{ij} ^k \circ \gamma) \ \dot \gamma ^i \dot \gamma ^j = 0 .$$

All this pain that we have gone through was required by the fact that $\nabla _{\dot \gamma} {\dot \gamma}$ does not obey the usual algebraic manipulation rules as $\nabla _X Y$ does. (For instance, what would you make of $\nabla _{\partial _i} \dot \gamma ^j$, given that $\dfrac {\partial \dot \gamma ^j} {\partial x_i}$ has no meaning - how would you derive a thing that depends on $t$ with respect to the variables $x_1, \dots, x_n$?)

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  • $\begingroup$ Gorgeous ! Thank you for the time you spent to write it. I'll stud this and make you a feedback if necessary. Thank you for everything :) $\endgroup$ – MSE Jun 9 '16 at 13:54
  • $\begingroup$ Actually, it's perfectly clear ! Nothing to say except : thank you very very much. $\endgroup$ – MSE Jun 9 '16 at 14:11
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Reconsider your line $$ 0=\nabla _{\dot \gamma (t)}\dot \gamma (t)=\nabla _{\dot x^i \frac{\partial }{\partial x^i}}\dot x^\ell \frac{\partial }{\partial x^\ell}=\dot x^i\underbrace{\frac{\partial x^\ell}{\partial x^i}}_{=\delta_{i\ell}}\frac{\partial }{\partial x^\ell}+\dot x^i\dot x^\ell \underbrace{\nabla _{\frac{\partial }{\partial x^i}}\frac{\partial }{\partial x^\ell}}_{=\Gamma_{i\ell}^m\frac{\partial }{\partial x^m}}. $$ In particular I think you have lost a 'dot' in $$ \dot x^i\underbrace{\frac{\partial x^\ell}{\partial x^i}}_{=\delta_{i\ell}}\frac{\partial }{\partial x^\ell}, $$ which should be $$ \dot x^i\underbrace{\frac{\partial \dot x^\ell}{\partial x^i}}_{}\frac{\partial }{\partial x^\ell}. $$

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  • $\begingroup$ You right, I corrected it. But I still cannot conclude :-) $\endgroup$ – MSE Jun 9 '16 at 11:46
  • $\begingroup$ Unfortunately, the formula $\nabla _X Y = \nabla _{X^i \partial _i} {Y^j \partial _j}$ is valid only for fields defined on some neighbourhood of $\gamma ([0,1])$, which $\dot \gamma$ is not. Things are more complicated than what you write, even though the basic idea is correct. The missing steps, though, are significant and should not be omitted. $\endgroup$ – Alex M. Jun 9 '16 at 12:23
  • $\begingroup$ @AlexM.: Could you please post a complete answer (if you can), because I really have problem with this exercise, and I have problem to see every identification and step. I think, to have one detailed example would help me a lot. Thank you. $\endgroup$ – MSE Jun 9 '16 at 13:37
  • $\begingroup$ @MSE: Complete answer posted. :) $\endgroup$ – Alex M. Jun 9 '16 at 13:53

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