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Consider $M \cong G/K$ ($G$ a lie group with a transitive action on $M$ and $K$ a subgroup) and consider a Lagrangian $\mathcal{L}: TM \rightarrow \ \mathbb{R}$ (no time dependence). Consider also that $\mathcal{L}$ is also homogeneous order one in tangent vectors i.e. $\mathcal{L}(p, \lambda x ) = \lambda \mathcal{L}(p, x)$ so that actions assigned to curves are independent of parametrization.

When exactly are the stationary curves $\gamma(t)$ of $\mathcal{L}$ on $M$ of the form $e^{tA} \circ \gamma_0$ where $\gamma_0$ is one of the end points of the curve and $\circ$ is the action of $G$ on $M$.

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    $\begingroup$ Are you assuming any invariance properties of the Lagrangian, e.g., that $\mathcal{L}$ is invariant under the induced action of $G$ on $TM$? Also, would you rather have $\mathcal{L}(\lambda v) = |\lambda|\mathcal{L}(v)$ for all $\lambda\in\mathbb{R}$ and $v\in TM$? (Without the absolute value sign, the value of the action will depend on the orientation of the curve.) $\endgroup$
    – Robert Bryant
    Commented Jan 25, 2014 at 17:02
  • $\begingroup$ Let's say initially that absolute homogeneity is not guaranteed. I guess I'm asking what the needed invariance properties are. $\endgroup$
    – Benjamin
    Commented Jan 25, 2014 at 17:53
  • $\begingroup$ I should have said $\lambda > 0$. $\endgroup$
    – Benjamin
    Commented Jan 25, 2014 at 18:09
  • $\begingroup$ Are you at least going to assume that $\mathcal{L}^2:TM\to\mathbb{R}$ is nondegenerate? Also, I suppose that you mean only to ask that the stationary curves will be of the desired form up to reparametrization, since, otherwise, your desired form could never hold for all stationary curves. $\endgroup$
    – Robert Bryant
    Commented Jan 25, 2014 at 20:52
  • $\begingroup$ Yes, I mean only the images of the curves coincide $\endgroup$
    – Benjamin
    Commented Jan 25, 2014 at 22:03

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This is not really an answer as much as it is a caution that the problem, even with the extra assumptions I was able to solicit from the OP, is not going to have a very nice answer unless one adds some further hypotheses.

To see why, just look at the case that the dimension of the homogeneous space $M$ is equal to $2$. Even for $M=\mathbb{R}^2$ and $G$ is the group of Euclidean motions of the plane, there exists a huge family of Lagrangians on $TM$ such that the geodesics (up to parametrization) are the straight lines (basically two arbitrary functions of two variables worth of such Lagrangians, according to Darboux). Since, for this $M$ and $G$, each straight line is of the form desired by the OP, that means that there are many, many solutions to the problem, even in this simple case. (Note that most of these Lagrangians will not be invariant under $G$, even though the resulting geodesic path geometry is invariant under $G$.)

More generally, if $M=G/K$ is a surface endowed with just about any path geometry such that the (unparametrized) paths are all of the form $exp(tx)\cdot p$ for some $x\in \frak{g}$ and $p\in M$ and such that the set of paths forms a smooth surface of its own (and it is easy to construct many such examples), there will be a large family of nondegenerate Lagrangians on $TM$ whose geodesics are the curves in the path geometry. This family will depend on two arbitrary functions of two variables, though it will not be easy to write them down explicitly.

Thus, in my opinion, the question has too many solutions, even in the $2$-dimensional case, for there to be a good answer to the OP's question.

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    $\begingroup$ If you take a 2-torus and your Lagrangians are smooth Finsler metrics you will only have flat metrics. For $C^1$ metrics you have a lot more, but a lot depends on the gobal character of the path geometry. $\endgroup$
    – alvarezpaiva
    Commented Jan 27, 2014 at 15:33
  • $\begingroup$ But even the flat smooth Finsler metrics on the $2$-torus depend on an arbitrary function because the unit circle at one point can be any smooth closed curve that is strongly convex towards the origin. Also, the path geometry of the flat $2$-torus can't be written as $G/K$ where all the (unparametrized) geodesics are orbits of $1$-parameter subgroups of $G$, nor is the space of geodesics a smooth $2$-manifold. $\endgroup$
    – Robert Bryant
    Commented Jan 27, 2014 at 15:44
  • $\begingroup$ Perhaps it's a matter of the topology of G and M? The plane is non compact but the torus is compact and the answers seems vacuous for the plane but interesting for the torus. I'm specifically interested in the case $SU(N+1)/U(N) \cong \mathbb{C}P^N$ all of which are compact. $\endgroup$
    – Benjamin
    Commented Jan 27, 2014 at 15:44
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    $\begingroup$ @Benjamin: Even in that case, say $N=1$, there are many different Finsler metrics on the $2$-sphere ($=\mathbb{CP}^1$) whose geodesics are the (oriented) great circles, and hence all of these are examples of the kind of Lagrangian you are hoping to classify, with $G = \mathrm{SU}(2)$ and $K= \mathrm{U}(1) = S^1$. Even in this simple case, writing down all of these Lagrangians explicitly is not trivial. $\endgroup$
    – Robert Bryant
    Commented Jan 27, 2014 at 16:05
  • $\begingroup$ Thanks for your helpful comments, I'll have to refine my hypotheses then! $\endgroup$
    – Benjamin
    Commented Jan 27, 2014 at 16:28

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