Second-order Euler-Lagrange equations on homogeneous spaces

Consider a Lie Group $$G$$ with a Lie subgroup $$H$$, and the resulting Homogeneous space $$M := G/H$$ with canonical projection map $$\pi$$. Suppose we have a Lagrangian $$L: TM \to \mathbb{R}$$ and consider the Lagrangian $$\tilde{L}: TG \to \mathbb{R}$$ defined by $$\tilde{L} = L \circ \pi_\ast$$.

Now let $$x$$ be a solution to the Euler-Lagrange equations for $$L$$ and $$\tilde{x}$$ be a lift of $$x$$. If $$\tilde{x}_\epsilon$$ is a smooth deformation of $$\tilde{x}$$ with fixed endpoints, then $$x_\epsilon := \pi \circ \tilde{x}_\epsilon$$ is clearly a smooth deformation of $$x$$ with fixed endpoints. Hence,

$$\frac{\partial}{\partial \epsilon}\Big{\vert}_{\epsilon = 0}\int \tilde{L}(\tilde{x}_\epsilon, \dot{\tilde{x}}_\epsilon)dt = \frac{\partial}{\partial \epsilon}\Big{\vert}_{\epsilon = 0} \int L(x_\epsilon, \dot{x}_\epsilon)dt = 0$$

from which it follows thar $$\tilde{x}$$ solves the Euler-Lagrange equations for $$\tilde{L}$$.

However, I was recently reading about a problem in which $$G$$ has a Riemannian metric $$\left< \cdot, \cdot \right>_G$$ with corresponding Levi-Civita connection $$\tilde{\nabla}$$ and $$M$$ is a Riemannian homogeneous space (that is, there is a metric $$\left< \cdot, \cdot \right>_M$$ on $$M$$ which makes $$\pi_\ast$$ a linear isometry between the horizontal subspace $$\ker(\pi_\ast \vert_g)^\perp$$ and $$T_{\pi(g)}M$$ for all $$g \in G$$, and corresponding Levi-Civita connection $$\nabla$$). They instead consider the problem of critical points of $$\int \left_M dt$$, called Riemannian cubics.

It can be shown that $$\tilde{D}_t \dot{\tilde{x}} = \widetilde{D_t \dot{x}}$$ for any curve $$x$$ on $$M$$, where $$\tilde{x}$$ and $$\widetilde{D_t \dot{x}}$$ are the horizontal lifts of $$x$$ and $$D_t \dot{x}$$, respectively. Using the same argument for the deformations as above, I would expect that if $$x$$ is a Riemannian cubic on $$(M, \left<\cdot, \cdot\right>_M)$$, then $$\tilde{x}$$ would be a Riemannain cubic on $$(G, \left<\cdot, \cdot\right>_G)$$, because:

$$\frac{\partial}{\partial \epsilon}\Big{\vert}_{\epsilon = 0}\int \left<\tilde{D}_t \dot{\tilde{x}}, \tilde{D}_t \dot{\tilde{x}}\right>_G dt = \frac{\partial}{\partial \epsilon}\Big{\vert}_{\epsilon = 0} \int \left_Hdt = 0.$$

However this is not the case, in general $$\tilde{x}$$ is not a Riemannian cubic. Why does the previous argument for Lagrangians fail here?

Your argument for Riemannian cubics only applies to horizintal paths in $$G$$. This is inadequate in a number of ways:
• Even if a variation $$x_\epsilon$$ satisfies the boundary conditions on $$M$$ (fixed values and derivatives on endpoints), It may not be possible to choose horizontal lifts $$\tilde{x}_\epsilon$$ which satisfy the boundary conditions on $$G$$.
• A horizontal Riemannian cubic on $$G$$ must be a critical point w.r.t. all variations which satisfy the boundary conditions, most of which are not horizontal.
Thus, you have successfully shown that $$\tilde{x}$$ is a critical point of the Lagrangian on $$G$$ only with respect to a small number of variations, where you need to show that it is critical with respect of all of them.
If you want to understand the relationship between cubics on $$G$$ and $$M$$. It might be better to look at the Euler-Lagrange equations directly. It seems that the curvature term will lead to some subtleties, since the curvatures of $$M$$ and $$G$$ are related in a somewhat complicated way.