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 $\int \left<D_t \dot{x}, D_t \dot{x}\right>_M dt$, 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:

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

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


1 Answer 1


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.

  • $\begingroup$ Yes, you're right. I originally expected that we could simply use horizontal variations, but I now see that that's not the case. I found another way to solve the problem by carefully using horizontal projections (so arbitrary variations may be considered, but they become horizontal in the Lagrangian). Thanks! $\endgroup$ Apr 8, 2022 at 9:07

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