In this Lecture we introduce a canonical distance on a Carnot group. This distance is naturally associated to the sub-Riemannian structure which is carried by a Carnot group. It plays a fundamental role in the rough paths topology. Let be the free Carnot group over . Remember that if , then we denote by the lift of in . The first important concept is the notion of horizontal curve.
Definition: A curve is said to be horizontal if there exists such that .
It is remarkable that any two points of can be connected by a horizontal curve.
Proposition: Given two points and , there is at least one such that .
Proof: Let us denote by the subgroup of diffeomorphisms generated by the one-parameter subgroups corresponding to . The Lie algebra of can be identified with the Lie algebra generated by , i.e. . We deduce that can be identified with itself, so that it acts transitively on . It means that for every , the map , is surjective. Thus, every two points in can be joined by a piecewise smooth horizontal curve where each piece is a segment of an integral curve of one of the vector fields
In the above proof, the horizontal curve constructed to join the two points is not smooth. Nevertheless, it can be shown that it is always possible to connect two points with a smooth horizontal curve.
Let us also remark that this theorem is a actually a very special case of the so-called Chow-Rashevski theorem which is one of the cornerstones of sub-Riemannian geometry. We now are ready for the definition of the Carnot-Carathéodory distance.
Definition For , we define
is called the Carnot-Carathéodory distance between and .
The first thing to prove is that is indeed a distance.
Lemma: The Carnot-Carathéodory distance is indeed a distance.
Proof: The symmetry and the triangle inequality are easy to check and we let the reader find the arguments. The last thing to prove is that implies . From the definition of it clear that where is the Riemmanian measure on . It follows that implies
We then observe the following properties of :
- For ,
- Let be the one parameter family of dilations on . For , and , .
Proof: The first part of the proposition stems from the fact that for every , , so that is equivalent to which also equivalent to . For the second part, we observe that for ,
The Carnot-Carathéodory distance is pretty difficult to explicitly compute in general. It is often much more convenient to estimate using a so-called homogeneous norm.
Definition: A homogeneous norm on is a continuous function , such that:
- , , ;
- , ;
- if and only if .
It turns out that the Carnot-Carathéodory distance is equivalent to any homogeneous norm in the following sense:
Theorem: Let be a homogeneous norm on . There exist two positive constants and such that for every ,
By using the left invariance of , it is of course enough to prove that for every ,
We first prove that the function is bounded on compact sets (of the Riemannian topology of the Lie group ). As we have seen before, every can be written as a product:
From the very definition of the distance, we have then
It is not difficult to see that can uniformly be bounded on compact sets, therefore is bounded on compact sets. Consider now the compact set
Since is bounded on , we deduce that there exists a constant such that for every ,
Since , where is the Riemannian distance, we deduce that there exists a constant such that for every ,
Now, for every , , we deduce that
This yields the expected result
Let us give an example of a homogeneous norm which is particularly adapted to rough paths theory. Write the stratification of as:
and denote by the projection onto . Let us denote by the norm on that comes from the norm on formal series. Then, it is easily checked that
is an homogeneous norm on . This homogeneous norm is particulary adapted to the study of paths because if , then one has:
We finally quote the following result, not difficult to prove which is often referred to as the ball-box estimate.
Proposition: There exists a constant such that for every ,
In particular, for every compact set , there is a constant such that for every ,
Proof: See the book by Friz-Victoir, page 152