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%%%%%%%%%%%% PAPER DATA %%%%%%%%%%%%%%%%%%%
\author{Jos\'e Ricardo Arteaga, Mikhail Malakhaltsev}

\title{Lectures on non-holonomic geometry} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{document}
\section{ Distributions on manifolds}
Let $M$ be a smooth manifold.  Assume that for each $p \in M$ a subspace $\Delta(p) \subset T_p M$ 
is assigned such that 1) $\dim \Delta(p) = k$ is constant, 2) for each $q \in M$ there exists a neigbourhood
$U$ with vector fields $E_1, E_2, \ldots, E_k \in \gX(U)$ such that,  
for each $q´ \in U$, $\Delta(q´) = span\{E_1(q´), E_2(q´), \ldots, E_k(q´)\}$. 
We say that the dimension of the distribution $\Delta$ equals $k$.

A distribution $\Delta$ on a manifold $M$ is called \emph{totally integrable} if
for each $p \in M$ there exists a submanifold $\Sigma$ passing through $p$ such that 
$T_q \Sigma = \Delta(q)$, for any $q \in \Sigma$. This submanifold is called an integral 
submanifold of $\Delta$.
A totally integrable distibution  is called  a foliation, 
and the integral submanifolds are called the leaves.

Locally, a distribution $\Delta$ can be given either by vector fields $E_a$, $a=1, \ldots, k$,
spanning the subspaces $\Delta(p) \subset T_p M$, or by differential forms 
$\omega^1, \omega^2, \ldots, \omega^{n-k}$ such that, 
for any $V \in T_p M$, $V \in \Delta(p)$ if and only if $\omega^\alpha(V)=0$, 
$\alpha=1,\ldots,n-k$ 

The following theorem gives necessary and sufficient conditions for a distribution 
be completely integrable.

\begin{thm}[Frobenius]
Let $\Delta$ be a distribution on a manifold $M$ locally 
given by a frame field $E_a$, $a=1, \ldots, k$, or by the fields of 
1-forms $\omega^\alpha$.
The following conditions are equivalent:

a) The distribution $\Delta$ is completely integrable;

b) $[E_a, E_b] = Q^c_{ab} E_c$, where $Q^c_{ab}$ are functions; 

c) $d\omega^\gamma =   Q^\gamma_\beta \wedge\omega^\beta$, where  $Q^\gamma_\beta$
are 1-forms. 
\end{thm}

\begin{ex}
Let a distribution $\Delta$ on $\R^3$ be given by the 1-form $\omega = dz - xdx - ydy$.
Then $d\omega = 0$, and $\Delta$ is completely integrable because this distribution 
consists of pĺanes tangent to hyperbolic paraboloids of the family
$z = \frac{1}{2}(x^2 + y^2)$. 
\end{ex}
\begin{ex}
Let a distribution $\Delta$ on $\R^3$ be given by the 1-form $\omega = dy - zdx$.
Then one can take the local frame spanning $\Delta$ as follows:
$E_1 = \frac{\pd}{\pd x}+z\frac{\pd}{\pd y}$, $E_2 = \frac{\pd}{\pd z}$.
Then $[E_1,E_2] = -\frac{\pd}{\pd y}$, and this vector field does not lie in $\Delta$, because
$\omega(\frac{\pd}{\pd y}) = -1 \ne 0$. Thus this distribution is not completely integrable.
This means that one cannot find a surface tangent to $\Delta$.
\end{ex}

\section{Non-holonomic surface in $E^3$}

Let $E_3$ be a three-dimensional Euclidean space, and $\Delta$ be a distribution on $E_3$.
We denote by $(\cdot, \cdot)$ the scalar product in $E_3$. 
This distribution can be given by a unit vector field $\vec n$ normal to $\Delta$.
The integrability condition for $\Delta$ can be formulated in terms of $\vec n$.
\begin{thm}
$\Delta$ is integrable if and only if $(\vec n, \rot\vec n) = 0$
\end{thm}

\end{document}

