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?

  • $\begingroup$ No: if $V\subset\mathfrak{g}$ is the subspace defining the CC-metric, then the only geodesics that would have the form $\exp(tX)$ should satisfy $X\in V$, and hence can only reach points of $\exp(V)$. $\endgroup$ – YCor Apr 30 '16 at 17:04
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    $\begingroup$ Ok, but how does this answer the question? $\endgroup$ – Mizar Apr 30 '16 at 17:37
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    $\begingroup$ @YCor: Yeah, what you say is true, but I don't think it resolves the question. For instance, in the Heisenberg group, I think it's simply true that all rays lie in $\exp(V)$. You can draw a length-minimizing geodesic from the identity to a point outside of $\exp(V)$ (its projection in the $xy$-plane looks like an arc) but if you try to extend it indefinitely, it ceases to be globally length-minimizing (specifically, after the arc has completed a full circle and started wrapping around again). So the extension won't be a ray. $\endgroup$ – Nate Eldredge Apr 30 '16 at 18:04
  • $\begingroup$ OK, thanks, I was mislead by another (wrong) interpretation of the question $\endgroup$ – YCor Apr 30 '16 at 19:44
  • $\begingroup$ I guess that $X$ is not only in the Lie algebra, but must be horizontal (i.e. $X$ belongs to the first layer of the stratified Lie algebra). $\endgroup$ – Raziel May 15 '16 at 8:09

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" 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.

  • $\begingroup$ The Engel group is indeed one of the few other Carnot groups (but the one mentioned in my comment) of which the cut times are explicitly known (are there others?). And there it is the counterexample to the original question! $\endgroup$ – Raziel May 18 '16 at 18:40

This is far from being a complete answer, but it's too long for a comment.

Your statement is true, by direct inspection, for corank 1 and 2 Carnot groups, where the cut time of normal geodesics (or, equivalently, the cotangent injectivity domain for the normal exponential map) is known explicitly.

I am not sure whether the claim is true in general. However, if a counter-example exists, one should prove global optimality for a horizontal curve $\gamma : [0,+\infty)$ that is not a one-parameter subgroup. Even assuming that $\gamma$ is a normal geodesic (and thus solves some Hamiltonian equations) to simplify the task, this would be extremely hard to do (except for one-parameter subgroups, that coincide with straight lines in the first layer $\mathfrak{g}_1$ of the Carnot group $G \simeq \mathfrak{g}$).


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