Show that geodesic equation is given by $\ddot x^k +\Gamma_{ij}^k \dot x^i\dot x^j=0$ 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 ?
 A: 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 $x_1, \dots, x_n$?)
A: 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
$$
\dot x^i\underbrace{\frac{\partial \dot x^\ell}{\partial x^i}}_{}\frac{\partial }{\partial x^\ell}.
$$
