A famous open problem in Geometric Control Theory and in the study of sub-Riemannian manifolds is whether constant-speed length minimizers in a sub-Riemannian manifold are always smooth (see also [this question](https://mathoverflow.net/questions/156678/why-is-proving-c-infty-regularity-of-sub-riemannian-geodesics-so-hard)). This is open also in the special case of Carnot groups.


When $G$ is a Carnot group, one can try at least to prove that, given a geodesic $\gamma:[0,T]\to G$ with $\gamma(0)=e$, the family of geodesics
$$ \gamma\_r:[0,r^{-1}T]\to G,\quad\gamma\_r(t):=\delta\_{r^{-1}}(\gamma(rt)) $$
tend locally uniformly to a straight line, as $r\to 0$. (It is well-known that Carnot groups come equipped with a one-parameter family of automorphisms $(\delta\_r)\_{r\in\mathbb{R}^+}$ satisfying $d(\delta\_r(x),\delta\_r(y))=rd(x,y)$. Here $d$ is the Carnot-Carathéodory distance.)


It is easy to see that any limit curve $\gamma\_\infty:[0,\infty)\to G$ along a subsequence of $r$'s is a *ray*, i.e. a length minimizer between any two of its points.


Hence a very natural question is the following:



> 
> Given any unit-speed ray $\alpha:[0,\infty)\to G$ (i.e. a continuous curve
>  such that $d(\alpha(s),\alpha(t))=t-s$), is it true that $\alpha$ belongs to a one-parameter subgroup of $G$? In other words, is it true that $\alpha(t)=\exp(tX)$, for a suitable $X$ in the Lie algebra?
> 
> 
> 



Are rays in Carnot groups straight?
No, infinite rays are not necessarily one-parameter subgroups.


An example can be found in the paper ["Cut time in sub-riemannian problem on engel group"](https://arxiv.org/abs/1408.6651) by Ardentov and Sachkov.


In the paper the geodesics of the Engel group (a rank 2, step 3 Carnot group) are separated into 7 different classes $C\_1,...C\_7$, of which $C\_3, C\_4, C\_5$ and $C\_7$ give infinite geodesics. The geodesics of class $C\_3$ give examples of infinite rays that are not one-parameter subgroups.


