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Any Sub-Riemannian Metric has Points of A. A. Agrachev
 

Summary: Any Sub-Riemannian Metric has Points of
Smoothness
A. A. Agrachev
Abstract
We prove the result stated in the title; it is equivalent to the exis-
tence of a regular point of the sub-Riemannian exponential mapping.
We also prove that the metric is analytic on an open everywhere dense
subset in the case of a complete real-analytic sub-Riemannian mani-
fold.
1. Preliminaries
Let M be a smooth (i. e. C
) Riemannian manifold and TM a smooth
vector distribution on M (a vector subbundle of TM). We denote by
the space of smooth sections of that is a subspace of the space VecM of
smooth vector fields on M. The Lie bracket of vector fields X, Y is denoted
by [X, Y ]. We assume that is bracket generating; in other words, q M,
span{[X1, [ , [Xm-1, Xm] ](q) : Xi , i = 1, . . . m, m N} = TqM.
Given q0, q1 M, we define the space of starting from q0 admissible paths:
q0 = { H1
([0, 1], M) : (0) = q0, (t) (t) for almost all t}

  

Source: Agrachev, Andrei - Functional Analysis Sector, Scuola Internazionale Superiore di Studi Avanzati (SISSA)

 

Collections: Engineering; Mathematics