Lagrange's Equation on a Manifold I know that, if $L: \mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}$, then the Euler-Lagrange equation is:
$$ \nabla_x L - (\nabla_{\dot{x}}L)' \equiv 0$$
In trying to generalize this for a given riemannian manifold. My idea is to consider:
$$L: TM \times \mathbb{R} \rightarrow \mathbb{R}$$
And then split $TTM$ in its horizontal and vertical components $T_hTM $ and $T_vTM$ by the riemannian connection. Therefore:
$TL: TTM \times \mathbb{R} \rightarrow \mathbb{R}$
$TL: T_hTM \times T_vTM \times \mathbb{R} \rightarrow \mathbb{R}$
Now, given a path $\gamma:(0,1) \rightarrow M$ and using the canonical metric of $T_hTM$ and $T_vTM$ due to identifications with $T_xM$, I would get two differentiable sections:
$\nabla_xL: (0,1) \rightarrow \gamma ^*T_hTM$ that maps $t \mapsto (t,\nabla_1 L(\gamma(t)))$, where $\nabla_1 L(x)$ is the vector that determines the linear functional $TL |_{T_hT_xM}$ 
Analogously for $\nabla_{\dot{x}}L$ (with respect to $TL |_{T_vT_xM}$).
And now, the euler-lagrange would takes its normal form:
$$ \nabla_x L - (\nabla_{\dot{x}}L)' \equiv 0$$
Now, my question is: Is my idea correct? Is this the way to generalize it? I think there may be a simpler way to do so. Not only that, I'm not sure that the splitting is the right idea to consider.
 A: The key point, I think, is that the Euler-Lagrange equations are not geometric objects.
You can see this by deriving the equations from the least action principle: for boundary conditions $p_0, p_1$ you can write down the variational problem
$$\mathop{\textrm{ext}}_{\substack{\gamma:[0,1]\to M\\ {\gamma(0)=p_0},\ \gamma(1)=p_1}} \int_0^1 L(\gamma'(t),t)\,dt$$
and notice that the variational problem requires nothing of $M$ beyond differential structure. 
If you have a metric you can use it to solve the variational problem using the calculus of variations in a coordinate-free way, e.g. $\gamma$ is a solution to the variational problem if, for every vector field $\delta(t)$ along $\gamma$ with $\delta(0)=\delta(1)=0$,
$$0=\frac{d}{d\epsilon}\int_0^1 L(\exp[\gamma(t),\epsilon\delta(t)]'(t),t)\,dt\Bigg\vert_{\epsilon\to 0},$$
and you can get E-L equations in terms of Jacobi fields on $TTM$ etc.
But if you don't, you can instead work in coordinates on charts, and the Euler-Lagrange equations in local coordinates must be the familiar ones from Euclidean space.
