Carnot rectifiability of sub-Riemannian manifolds with constant tangent

Abstract

We consider 2-step free-Carnot groups equipped with sub-Finsler distances. We prove that the metric spheres are codimension-one rectifiable from the Euclidean viewpoint. The result is obtained by studying how the Lipschitz constant for the distance function behaves near abnormal geodesics.

Characterising rectifiable metric spaces using tangent spaces

“…Let (Z , ζ ) be the metric space obtained from Lemma 2.22 item 2. Observe that for any x ∈ B(x, 1/ ), ζ (x , f (x )) ≤ , proving and the first containment in (20). If y ∈ Y ∩ B(y, 1/ − ) then there exists x ∈ X ∩ B(x, 1/ ) with ρ(y , f (x )) ≤ , so that ζ (x , y ) ≤ 2 , proving the second containment in (20), for replaced by 2 .…”

“…Relationships between tangent spaces and parametrising maps have been explored in specific non-Euclidean metric spaces such as the Heisenberg group (or other Carnot groups) [19,20]. In this case, the additional structure of a particular ambient metric space such as the Heisenberg group enables one to define a much stronger notion of a tangent than a Gromov-Hausdorff tangent.…”

Carnot rectifiability and Alberti representations

“…We stress that, as shown by the example constructed in [53], see Remark 7.7, in the item (i) below it is not possible to substitute biLipschitz G-rectifiable with G-Lipschitz rectifiable. We notice that a distinguished class of G-biLipschitz rectifiable spaces is that of subRiemannian manifolds with constant nilpotentization G, see [56].…”

“…It would be interesting to understand whether the metric version of Federer's projection theorem in [18] can be extended to our setting, and this seems to be an important step to prove the previous conjecture. The implication 1⇒2 in Conjecture 7.1 is known to be true for equiregular sub-Riemannian manifolds, as a consequence of the results in [56]. We stress that, since there exists a nilpotent Lie group equipped with a left-invariant sub-Riemannian metric that is not locally quasiconformally equivalent to its tangent cone, see [55, Theorem 1.1], the implication 1⇒2 in Conjecture 7.1 might be false if we require the biLipschtiz charts to be defined on open sets.…”

“…Hence item (a') is verified with k + 1 instead of k, recalling that we will construct E k+1 in such a way that E k+1 ⊂ E k . Finally, let us check item (b') with k + 1 instead of k. Thanks to the inductive step in item (b'), and (56) we know that…”