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YCor
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Are rays in Carnot groups straight?

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). 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?

Mizar
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