\documentclass[12pt]{article}
\usepackage{amssymb,amsfonts}
\newcommand{\ind}{\mathop{\rm ind}\nolimits}
\newcommand{\rank}{\mathop{\rm rank}\nolimits}
\newcommand{\sign}{\mathop{\rm sign}\nolimits}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{remark}{Remark}[section]
\title{Jacobi Curves for Singular Extremals}
\author{A.~A.~Agrachev\thanks{Steklov Mathematical Institute,
Moscow \& Laboratoire de Topologie, Universit\'e de Bourgogne, Dijon.\quad
Email: agrachev@mi.ras.ru}}
\date{}
\begin{document}
\maketitle
\subsection*{Introduction}
This paper is in a sense a continuation of \cite{ag97}.
Jacobi curves were defined, computed,
and studied in \cite{ag97} for regular extremals of smooth control systems.
Here we do the same for singular extremals.

The points of the Jacobi curves are the ${\cal L}$-derivatives of
endpoint mappings. The notion of ${\cal L}$-derivative
was introduced in \cite{ag97}; in section~1 of the present paper we prove
a general existence theorem for the ${\cal L}$-derivatives of smooth
mappings and indicate a way to compute them. In section~2 we actually
compute the ${\cal L}$-derivatives of the endpoint mappings and the
Jacobi curves for a wide class of control systems and their extremals.
In section~3 we apply the theory to a low dimensional example: we
define the curvature for 3-dimensional control systems
of the form $\dot x=g^0(x)+ug^1(x)$ and give a local classification
of such systems under some nondegeneracy conditions.

The Jacobi curves studied here and in \cite{ag97} are determined by
the ${\cal L}$-derivatives and are automatically feedback--invariant.
To prevent a possible misunderstanding I have to mention that in all
earlier papers we used the term "Jacobi curve" for similar but not
the same (and not feedback--invariant) objects. We, however, extensively
apply results and techniques of that earlier papers for the computing
and studying this new feedback--invariant Jacobi curves.


\section{The Existence of the ${\cal L}$-derivative}
We use definitions and notations from \cite[Sec.~1]{ag97}.
Let
$(u,\lambda)$
be a Lagrangian point of the mapping $f:U\to M$
and let $N$
be a germ of a submanifold  in $U$ at $u$,
hence
$(u,\lambda)$
is a Lagrangian point of
$f\bigr|_N$.
If
$N$
is finite--dimensional, then
${\cal L}_{(u,\lambda)}(f\bigr|_N)$
is a Lagrangian subspace. Denote by
${\cal N}$
the set of all such germs partially ordered by inclusion. Then
$\left\{{\cal L}_{(u,\lambda)}(f\bigr|_N)\right\}_{N\in{\cal N}}$
is a generalized sequence of points of the Lagrangian Grassmannian
$L(T_\lambda(T^*M))$.

\begin{theorem}
The limit \
${\cal N}$-$lim\,{\cal L}_{(u,\lambda)}(f\bigr|_N)$
exists if and only if the negative or the positive inertia index of
$Hess_{(u,\lambda)}f$
is finite.
\end{theorem}
{\bf Proof.} We fix local coordinates in $U$, $M$ and hence in $T^*M$.
In these coordinates, $f(u)$ is the origin 0 in $\mathbb R^n$,
$\lambda=(p,0)$, where $p\in\mathbb R^{n*}$. Let
$\Pi=\{(\xi,0):\xi\in\mathbb R^{n*}\}$ be the fiber of $T^*M$ at $f(u)$
identified with the tangent space to the fiber.
Set $Qv=pf''_uv,$ $\bar Q(v)=\langle Qv,v\rangle$, $v\in T_uU$.
 We may deal only with the negative inertia
index and may assume that $\ker Q=0$ without lack of generality.
\begin{lemma}
Let $N_1\subset N_2$ be germs at $u$ of finite--dimensional submanifolds
in $U$, $V_i=T_uN_i$, $V^0_i=V_i\cap\ker f'_u$ $i=1,2$.
If $\rank f'_u\bigr|_{V_1}=\rank f'_u\bigr|_{V_2}$ and
$\ker\bar Q\bigr|_{V^0_1}=\ker\bar Q\bigr|_{V^0_2}$, $i=1,2$,
then $$
\ind_-\bar Q\bigr|_{V^0_2}\ge\ind_-\bar Q\bigr|_{V^0_1}+
\ind_\Pi ({\cal L}(f\bigr|_{N_1}),{\cal L}(f\bigr|_{N_2})).
$$
\end{lemma}
{\bf Proof.} The assumption on kernels of $\bar Q\bigr|_{V^0_i}$ implies
$$
\ind_- \bar Q\bigr|_{V^0_2}=\ind_- \bar Q\bigr|_{V^0_1}+
\ind_- \bar Q\bigr|_{V_2\cap W_1},
$$
where $W_1=\{w:\langle Qw+\xi f'_u,V_1\rangle =0,\ f'_uw=0,\mbox{for some}
\ \xi\in R^{n*}\}.$

On the other hand, $$
\ind_\Pi ({\cal L}(f\bigr|_{N_1}),{\cal L}(f\bigr|_{N_2}))= \ind_- S+
$$
$$
+\frac 12\left(\dim({\cal L}(f\bigr|_{N_1})\cap\Pi)+
\dim({\cal L}(f\bigr|_{N_2})\cap\Pi)\right)
-\dim\left(\bigcap\limits_{i=1}^2{\cal L}(f\bigr|_{N_i})\cap\Pi\right),
$$
where S is a standard quadratic form defined on
$\left(({\cal L}(f\bigr|_{N_1}+{\cal L}(f\bigr|_{N_2})\right)\cap\Pi.$
(see Append.). Recall that
$$
{\cal L}(f\bigr|_{N_i})=\{(\xi_i,f'_uv_i):
\langle Qv_i+\xi_i f'_u,V_i\rangle =0,\ v_i\in V_i\}.
$$
The assumption on kernels of $\bar Q\bigr|_{V^0_i}$ implies that
${\cal L}(f\bigr|_{N_1})\cap\Pi={\cal L}(f\bigr|_{N_2})\cap\Pi.$
Hence
$\ind_\Pi ({\cal L}(f\bigr|_{N_1}),{\cal L}(f\bigr|_{N_2}))= \ind S.$
The quadratic form $S$ is computed in terms of the symplectic structure
$\sigma$. Let $(\xi_i, f'_uv_i)\in{\cal L}(f\bigr|_{N_i})$, $f'_u(v_1+v_2)=0,$
$\xi=\xi_1+\xi_2$. Then
$$
S(\xi)=\sigma((\xi_1, f'_uv_1),(\xi_2, f'_uv_2))=
\xi_1f'_uv_2-\xi_2f'_uv_1=-\xi f'_uv_1.
$$
We set $w=v_1+v_2$, then $w\in V_2\cap W_1$ and we obtain
$$
\bar Q(w)=\langle Qw,w\rangle=\langle Q(v_1+v_2),v_1+v_2\rangle=
\langle Qv_1,v_1+v_2\rangle=
$$
$$
=\langle Q(v_1+v_2),v_1\rangle=
\langle Qw,v_1\rangle=-\xi f'_uv_1=S(\xi).
$$
Hence $S$ is equivalent to the restriction of $\bar Q$ to the subspace in
$V_2\cap W_1$ and
$\ind_- S\le \ind_- \bar Q\bigr|_{V_2\cap W_1}$.\quad $\square$

\begin{lemma} If $\ind_- Q\bigr|_{\ker f'_u}<\infty$, then there exists a
finite--dimensional subspaces $V_o\subset T_uU$, $V_o^0=V_o\cap\ker f'_0$
such that $\rank f'_u=\rank f'_u\bigr|_{V_o}$,
$\ker\bar Q\bigr|_W=\ker\bar Q\bigr|_{V^0_o}$, and
$\ind_-\bar Q\bigr|_W=\ind_-\bar Q\bigr|_{V^0_o}$ for any $W$ satisfying
the inclusions $V^0_o\subset W\subset\ker f'_u$.
\end{lemma}
The proof is left to the reader.\quad $\square$

Let $V=T_uU$, then ${\cal L}(f\bigr|_N)$ depends only on $V$. We
introduce a shorter notation $\Lambda_V={\cal L}(f\bigr|_N)$.

For any $V\supset V_o$, we have $\Pi\cap\Lambda_V=\Pi\cap\Lambda_{V_o}$.
We introduce a reduced symplectic space
$\Sigma=(\Pi\cap\Lambda_{V_o})^\angle /(\Pi\cap\Lambda_{V_o})$,
$\dim\Sigma =2m\le 2n$. Then $\forall\, V_2\supset V_1\supset V_o$,
$\Lambda_{V_i}$ belongs to the Lagrange Grassmannian $L(\Sigma)$,
$\ind_\Pi(\Lambda_{V_1},\Lambda_{V_2})=0$, and $\Lambda_{V_i}$
are transversal to $\Pi$ in $\Sigma$.

Take $\Delta\in L(\Sigma)$ such that $\ind_\Pi(\Lambda_{V_o},\Delta)=m$.
The triangle inequality
$$
\ind_\Pi(\Lambda_{V_o},\Delta)\le \ind_\Pi(\Lambda_{V_o},\Lambda_V)
+\ind_\Pi(\Lambda_V,\Delta)
$$
implies that $\ind_\Pi(\Lambda_V,\Delta)=m$ $\forall\, V\supset V_o$.
In particular, $\Lambda_V$ is transversal to $\Delta$ in $L(\Sigma)$.

Identifying the chart $\Delta^\pitchfork$ in $L(\Sigma)$ with the affine
space over the linear space of quadratic forms on
$\Delta^*\cong\sigma /\Delta$ we obtain
$\ind_\Pi(\Lambda_V,\Delta)=\ind(\Pi-\Lambda_V)$ (the inertia index of the
quadratic form in the right--hand side). In particular, $\Pi-\Lambda_V$
is identified with a negative definite quadratic form. Finally, we make
the affine space $\Lambda^\pitchfork$ linear fixing $\Pi$ as the origin.
Then $\Lambda_V$ is identified with a positive definite quadratic form.

The relation $\ind_\Pi(\Lambda_{V_1},\Lambda_{V_2})=0$ implies that
there exists a monotonically decreasing curve in $\Pi^\pitchfork$
connecting $\Lambda_{V_1}$ with $\Lambda_{V_2}$. Hence
$\Lambda_{V_2}\le\Lambda_{V_1}$ in the space of quadratic forms.

\begin{remark} {\rm The signs of quadratic forms under consideration and
the monotonicity types of curves in the Lagrange Grassmannian depend
on the general sign agreement which varies from paper to paper. We
follow the agreement of the paper \cite{ag97} and also identify the
coset $(z+\Delta)\in\Sigma /\Delta$ with the linear functional
$\sigma(z,\cdot)$ on $\Delta$.}
\end{remark}
So $\Lambda_V,\ V\supset V_o$, form a monotonically decreasing
generalized sequence of positive definite quadratic forms. The sequence
must have the limit which is a nonnegative quadratic form or, in other
words, an element of $\Delta^\pitchfork\subset L(\Sigma)$.

We didn't prove yet the "only if" part of the theorem. This easier and
less important part is left to the reader.\quad $\square$

\bigskip
Let $a$ be a smooth function on $M$, $da\subset T^*M$ be the differential
of $a$, which is a Lagrangian submanifold of $T^*M$.  Let
$\lambda\in da$, we set $\Pi_\lambda(a)=T_\lambda(da)$. Then $u$ is a
critical point of the scalar function $a\circ f$ and the Hessian
$Hess_u(a\circ f)$ is a well--defined quadratic form on $T_uU$,
$Hess_u(a\circ f)(v)=\langle(a\circ f)''_uv,v\rangle$. One can prove the
following modification of Lemma 1 in the same way as that lemma.
\begin{lemma}
Let $N_1\subset N_2$ be germs at $u$ of finite--dimensional submanifolds
in $U$, $V_i=T_uN_i$, $V^0_i=V_i\cap\ker f'_u$ $i=1,2$.
If $\rank f'_u\bigr|_{V_1}=\rank f'_u\bigr|_{V_2}$ and
$\ker(Hess_u(a\circ f)\bigr|_{V^0_1})=
\ker(Hess_u(a\circ f)\bigr|_{V^0_2})$, $i=1,2$,
then $$
\ind_-Hess_u(a\circ f)\bigr|_{V^0_2}\ge\ind_-Hess_u(a\circ f)\mid_{V^0_1}+
\ind_{\Pi_\lambda(a)} ({\cal L}(f\bigr|_{N_1}),{\cal L}(f\bigr|_{N_2})).
$$
$\square$
\end{lemma}


The limit from Theorem 1.1 is the precise definition of the
$\cal L$-derivative of $f$ at the Lagrangian point
$(u,\lambda)$.
But it is not enough to prove the existence of the limit, we must compute it.
Introducing local coordinates, we may assume that
$U$
is a Banach space and
$u$
its origin.
Let
$U_0\subset U$
be an arbitrary linear subspace which is dense in
$U$,
and
${\cal N}_0$
be the set of all finite--dimensional subspaces
$N_0\subset U_0$,
partially ordered by inclusion.  Thus,
${\cal N}_0\subset{\cal N}$.
The following assertion is an essential addition to Theorem 1.1, making
possible to
explicitly compute the  limit indicated in the theorem.

\begin{prop}
 Under the hypothesis of Theorem 1.1 the following equality holds:
$$
{\cal N}_0\mbox{-}\lim{\cal L}_{(u,\lambda)}(f\bigr|_{N_0})=
{\cal N}\mbox{-}\lim{\cal L}_{(u,\lambda)}(f\bigr|_N).
$$
\end{prop}
The proof is based on Lemma 1.3. We didn't use the completeness of $T_uU$
in the proof of Theorem 1.1, hence the limit in the left--hand side
does exist. Suppose
$$
\Lambda_0={\cal N}_0\mbox{-}\lim{\cal L}_{(u,\lambda)}(f\bigr|_{N_0}),\quad
\Lambda={\cal N}\mbox{-}\lim{\cal L}_{(u,\lambda)}(f\bigr|_N).
$$
Take an arbitrary $\Delta\in\Pi^\pitchfork$
and the function $a$ such that $T_\lambda(da)=\Delta$.
Using the fact that
$\ind_-(Hess_u(a\circ f)\bigr|_{U_0})=\ind_-Hess_u(a\circ f)$
and applying Lemma 1.3 one can prove that
$\ind_\Delta(\Lambda_0,\Lambda)=0.$
This is possible for all $\Delta$ only if $\Lambda_0=\Lambda$
(in fact, it would be enough
to consider any everywhere dense subset of "points" $\Delta$ in the
Lagrange Grassmannian).\quad $\square$
\begin{remark}{\rm The ${\cal L}$-derivative $\Lambda$ contains the
subspace $L=\{(\xi,f'_uv):Qv+\xi f'_u=0,\ v\in U\}$ and coincides
with $L$ if $Q(U)$ is closed in $U$. In particular,
$\dim L\le n$. It is often possible to endow the space
$U$ with a natural weaker topology preserving the continuity of $Q$
and $f'_u$ and such that $Q(\bar U)$ is closed in $\bar U$, where
$\bar U$ is the completion of $U$ in the new topology. Then
$\Lambda$ coincides with $\bar L=\{(\xi,f'_uv):Qv+\xi f'_u=0,\
v\in\bar U\}$. It follows from
Proposition 1.1 applied to the space $\bar U$ with the everywhere dense
subspace $U$ and the mapping $(f'_u,\bar Q)$ from $\bar U$ into
$\mathbb R^n=\{p\}^\bot\times\mathbb R$. The equality  $\Lambda=\bar L$
still holds if
$Q(\bar U)$ is not closed but $\bar L$ is $n$-dimensional.}
\end{remark}

\section{Computing the Jacobi Curves}
We use definitions and notations from \cite[Sec.~2]{ag97} and we deal with
a given extremal $\tau\mapsto (\xi(\tau),\lambda_\tau)$, where the
"control" $\xi$ is presented in the form
$\xi(\tau)=(\tilde u(\tau),\tilde x(\tau))$,
$\frac{d}{d\tau}\tilde x=
f_\tau(\tilde x(\tau),\tilde u(\tau))$, $0\le\tau\le t$.
Moreover, we assume that isomorphisms
$T_{\tilde u(\tau)}U\approx {\mathbb R^r}$ are fixed; it makes
$\frac{\partial f_\tau}{\partial u}(\tilde x(\tau),\tilde u(\tau))$ a linear
mapping from ${\mathbb R^r}$ into $T_{\tilde x(\tau)}M$. We also assume
that $\frac{\partial f_\tau}{\partial u}(\tilde x(\tau),\tilde u(\tau))$
is Lipschitzian with respect to $\tau$.

Recall that the extremal is regular if and only if
$\frac{\partial^2(\lambda_\tau f_\tau)}{\partial u^2}
(\tilde x(\tau),\tilde u(\tau))$ \linebreak
is a nondegenerate
quadratic form for all $\tau\in [0,t]$. We assume that \linebreak
$\frac{\partial^2(\lambda_\tau f_\tau)}{\partial u^2}
(\tilde x(\tau),\tilde u(\tau))=0$
$\forall\tau\in[0,t]$, so that $\tau\mapsto(\xi(\tau),\lambda_\tau)$
is a totally singular extremal. We set
$$
g_\tau(\lambda,u)=\lambda(\overrightarrow\exp\int\limits_t^\tau
ad\,\tilde f_\theta\,d\theta(f_\tau(\cdot,u)-\tilde f_\tau)),
$$
the same notation as in \cite[Sec.~3]{ag97}.
In particular,
$$
g_\tau(\lambda_t,u)=
\lambda_\tau(f_\tau(\tilde x(\tau),u)-f_\tau(\tilde x(\tau),\tilde u(\tau)).
$$
We also denote
$X_\tau=\frac{\partial\vec g_\tau}{\partial u}(\lambda_t,\tilde u(\tau))$;
then $X_\tau :\mathbb R^r\to T_{\tilde x(\tau)}M$ is a
Lipschitzian with respect to $\tau\in[0,t]$ family of linear mappings.

Let $F_{0,t}=F_t\bigr|_{\Omega_{x_0}}^t:\xi\mapsto pr\circ\xi(t)=x(t)$
be the endpoint mapping, which is defined on the space of admissible
controls subject to the condition $x(0)=pr\circ\xi(0)=x_0$. It follows
from proposition 3.1 and equality (3.2) of \cite{ag97} that
$$
{\cal L}_{(\xi,\lambda_t)}(F_{0,t})=
\{\eta_t\in T_{\lambda_t}(T^*M):
\exists\eta_\tau\in T_{\lambda_t}(T^*M), v(\tau)\in{\mathbb R}^r\
(0\le\tau\le t),
$$
$$
\mbox{ s.~t. }\dot\eta=X_\tau v(\tau),
\sigma(X_\tau\cdot,\eta_\tau)=0, \eta_0\in
T_{\lambda_t}(T^*_{\tilde x(\tau)}M)\},
$$
where $\tau\mapsto\eta_\tau$ is Lipschitzian and $\tau\mapsto v_\tau$
is bounded measurable.
${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$ is an isotropic subspace of
$T_{\lambda_t}(T^*M)$; recall that ${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$
is contained in the ${\cal L}$-derivative of $F_{0,t}$ at $(\xi,\lambda_t$,
if the ${\cal L}$-derivative does exist. It is worth to note that,
contrary to the regular case, the dimension of
${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$ is usually stricly less than $n$,
so that ${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$ is not a Lagrangian
subspace and hence it is not the ${\cal L}$-derivative. Indeed, the
vectors $X_\tau v$ are skew-orthogonal to
${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$ though these vectors do not
belong to ${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$ except of very
degenerate cases.
\begin{prop}[\cite{a88,ag86,as96}]
If the negative inertia index  of the
quadratic form \linebreak
$Hess_{(\xi,\lambda_t)}F_{0,t}$ is finite, i.e.
$\ind_-Hess_{(\xi,\lambda_t)}F_{0,t}<\infty$, then
$$
\sigma(X_\tau v_1,X_\tau v_2)=0,\quad \sigma(\dot X_\tau v_1,X_\tau v_1)\ge 0,
\quad \forall v_i\in {\mathbb R^r}, \tau\in [0,t].
$$
If, additionally, $\sigma(\dot X_\tau v,X_\tau v)\ge \alpha |v|^2$
for some $\alpha >0$, any $v\in\mathbb R^r$,
and almost all $\tau\in [0,t]$, then indeed
$\ind_-Hess_{(\xi,\lambda_t)}F_{0,t}<\infty$.
\end{prop}
The identity from the above proposition is called the Goh condition while
the inequality is called the (first) generalized Legendre condition.

The finiteness of $\ind_-Hess_{(\xi,\lambda_t)}F_{0,t}$ provides the
existence of the ${\cal L}$-deriva\-tive of $F_{0,t}$ at $(\xi,\lambda_t)$.
Our nearest goal is to compute the ${\cal L}$-deriva\-tive under
conditions
$$
\sigma(X_\tau v_1,X_\tau v_2)=0,\quad
\sigma(\dot X_\tau v,X_\tau v)\ge \alpha |v|^2, \eqno (2.1)
$$
i.e. under the Goh condition and the strengthened (first) generalized
Legendre condition. Differentiating the identity
$\sigma(X_\tau v_1,X_\tau v_2)=0$ with respect to $\tau$ we obtain
that $\sigma(\dot X_\tau v_1,X_\tau v_2)$ is a symmetric bilinear form
of $v_1,v_2$. We define the linear mapping
$b_\tau :{\mathbb R}^r\to {\mathbb R}^{r*}$ by the identity
$\langle b_\tau v_1,v_2\rangle=\sigma(\dot X_\tau v_1,X_\tau v_2)$.
Then $b_\tau$ is selfadjoint and invertible. In particular,
$b_\tau^{-1} :{\mathbb R}^{r*}\to {\mathbb R}^r$ is a selfadjoint
mapping and it may be identified with a quadratic form on
${\mathbb R}^{r*}$; we denote this quadratic form by $b_\tau^{-1}(\cdot)$
so that $b_\tau^{-1}(w)\stackrel{def}{=}\langle w,b_\tau^{-1}w\rangle$,
$w\in {\mathbb R}^{r*}$.

\begin{prop} Under conditions (2.1), the ${\cal L}$-derivative of
$F_{0,t}$ at $(\xi,\lambda_t)$ is a subspace
$\Lambda_t\subset T_{\lambda_t}(T^*M)$ which is the sum of the subspace
$\{X_tv : v\in {\mathbb R}^r\}$ and the subspace consisting of the
values at $t$ of the solutions $\tau\mapsto\eta(\tau)$ of the linear
Hamiltonian system on $T_{\lambda_t}(T^*M)$ associated to the
nonstationary quadratic Hamiltonian
$$
q_\tau(\eta)=-\frac 12b_\tau^{-1}(\sigma(\dot X_\tau\cdot,\eta))
$$
with the initial conditions $\eta(0)\in T_{\lambda_t}(T^*_{\tilde x(t)}M)$,
$\sigma(\eta(0),X_0v)=0 \ \forall v\in \mathbb R^r$.
\end{prop}
{\bf Proof.} Let us first show that $\Lambda_t$ is a Lagrangian subspace.
We have
$\vec q_\tau(\eta)=\dot X_\tau b_\tau^{-1}\sigma(\dot X_\tau\cdot,\eta)$.
In particular,
$$
\vec q_\tau(X_\tau v)=\dot X_\tau
b_\tau^{-1}\sigma(\dot X_\tau\cdot ,X_\tau v)=\dot X_\tau
b_\tau^{-1}b_\tau v=\dot X_\tau v.
$$
Hence $\tau\mapsto X_\tau v$ is a solution of the Hamiltonian system
$\dot\eta =\vec q_\tau(\eta)$. Let $\Phi_\tau\in Sp(T_{\lambda_t}(T^*M))$,
$\tau\in [0,t]$, be a linear symplectic flow generated by this system,
$\Phi_0=id$. We also denote
$\Gamma_\tau =\{X_\tau v:v\in\mathbb R^r\}$, $\Pi_t=
T_{\lambda_t}(T_{\tilde x(t)}^*M)$. Then $\Gamma_\tau$ is a family of
isotropic subspaces and
$$
\Lambda_t=\Phi(\Pi_t\cap\Gamma_0^\angle)+\Gamma_t=\Phi_t(\Pi_t)\cap
\Gamma_t^\angle+\Gamma_t=\Phi(\Pi_t\cap\Gamma_0^\angle+\Gamma_0).
$$
In other words, $\Lambda_t=\Phi(\Pi_t^{\Gamma_0}=\Phi(\Pi_t)^{\Gamma_t}$
(see Appendix for the notation).

So $\Lambda_t$ is a Lagrangian subspace. In order to prove that
$\Lambda_t$ is the ${\cal L}$-derivative we'll use the method described in
Remark 1.2. Let us recall the construction of
${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$. This subspace consists of the
vectors $(\eta_0+\int\limits_0^tX_\tau v(\tau)\,d\tau)$, where
$\eta_0\in\Pi_t$, the vector function $v(\cdot)$ belongs to
$L_\infty^r[0,t]$ and satisfies the equation
$$
\sigma(\eta_0+\int\limits_0^\tau X_\theta v(\theta)\,d\theta,X_\tau\cdot)=0,
\quad 0\le\tau\le t. \eqno (2.2)
$$
The operator $v(\cdot)\mapsto
\sigma((\eta_0+\int\limits_0^\cdot X_\theta v(\theta)\,d\theta,X_\cdot)$
is continuous in a much weaker topology than the topology of
$L_\infty^r[0,t]$; namely, this operator is continuous in the topology
of the Sobolev space $H_{-1}^r[0,t]$. More precisely, the norm
$\|\cdot\|_{-1}$ of $H_{-1}^r[0,t]$ has a form
$\|v\|_{-1}=\left(|w_t|^2+\int\limits_0^tw_\tau^2\,d\tau\right)^{\frac 12}$,
where $w_\tau =\int\limits_0^tv(\tau)\,d\tau$.

The completion of $L_\infty^r[0,t]$ in the norm $\|\cdot\|_{-1}$ consists
of the all pairs $(w_t,w(\cdot))$, $w_t\in\mathbb R^r$,
$w(\cdot)\in L_2^r[0,t]$. Integrating by part and applying
the Goh condition we obtain that the extension
of equation (2.2) to the space $H_{-1}^r[0,t]$ has a form
$$
\sigma(\eta_0-\int\limits_0^\tau \dot X_\theta w(\theta)\,
d\theta,X_\tau\cdot)=0,\quad 0\le\tau\le t,
$$
$$
\eta_0\in\Pi_t,\quad,w_t\in\mathbb R^r,\quad w(\cdot)\in L_2^r[0,t];
\eqno (2.3)
$$
the corresponding extension of ${\cal L}_{(\xi,\lambda_t)}(F_{0,t})$
consists of the vectors
$\eta_0+X_tw_t-\int\limits_0^t \dot X_\tau w(\tau)\,d\tau$,
where $\eta_0,w_t,w(\cdot)$ satisfy equation (2.3). Set
$$
\eta_\tau=\eta_0+X_\tau w_t-
\int\limits_0^\tau \dot X_\theta w(\theta)\,d\theta.
$$
If the data satisfy (2.3), then
$\eta(0)\in(\Pi\cap\Gamma_0^\angle+\Gamma_0)=\Pi_t^{\Gamma_0}$.
We have to show that
$$
\Lambda_t=\{\eta(t) :
\sigma(\eta(\tau),X_\tau\cdot)\equiv 0,
\dot \eta(\tau)=X_\tau w(\tau), w\in L_2^r[0,t],
\eta(0)\in\Pi_t^{\Gamma_0}\}.
$$
Differentiating the equality $\sigma(\eta(\tau),X_\tau\cdot)=0$
gives the identity
$$
-\sigma(\dot X_\tau w(\tau),X_\tau\cdot)+
\sigma(\eta(\tau),\dot X_\tau\cdot)=0,
$$
which can be rewritten in the form
$b_\tau w(\tau)=\sigma(\eta(\tau),\dot X_\tau\cdot)$, or
$$
w(t)=b_\tau^{-1}\sigma(\eta(\tau),\dot X_\tau\cdot).
$$
Thus $$
\dot\eta=-\dot X_\tau b_\tau^{-1}\sigma(\eta(\tau),\dot X_\tau\cdot)
=\vec q_\tau(\eta)
$$
and we are done.\quad $\square$

\medskip
We stay with the notations of the paper \cite{ag97} as long as it is
possible. In particular,
$F_{\tau,t}=F_t\bigr|_{\Omega^{t-\tau}_{\tilde x(\tau)}}$
(see \cite[Sec.~4]{ag97}). The difference with the mentioned paper is
the singularity of the extremal $(\xi,\lambda_t)$. Because of this,
the definition of the quadratic Hamiltonian $q_\tau$ must be modified.
The definition of the Jacobi curve must be modified as well. Suppose
that $(\xi,\lambda_t)$ satisfies relations (2.1), and let
$\Lambda_{\tau,t}$ be the $\cal L$-derivative of $F_{\tau,t}$ at
the point $(\xi\bigr|_{[\tau,t]},\lambda_t)$. We may apply
Proposition 2.2 to construct $\Lambda_{\tau,t}$ since
$(\xi\bigr|_{[\tau,t]},\lambda_t)$ satisfies the Goh condition and the
strengthened generalized Legendre condition.

Let $\Phi_{\tau,s}\in Sp(T_{\lambda_t}(T^*M))$ be the family of linear
symplectic transformations defined by the relations
$$
\frac\partial{\partial s}\Phi_{\tau,s}=\vec q_s\Phi_{\tau,s}, \quad
\Phi_{\tau,\tau}=id.
$$
In other words, $\Phi_{\tau,s}$ is the fundamental matrix of the
Hamiltonian system associated to the nonstationary quadratic Hamiltonian
$q_s$. We also set
$\Gamma_\tau =\{X_\tau v:v\in\mathbb R^r\}$, $\Pi_t=
T_{\lambda_t}(T_{\tilde x(t)}^*M)$. Then
$$
\Lambda_{\tau,t}=\Phi_{\tau,t}(\Pi_t^{\Gamma_\tau})=
\Phi_{\tau,t}(\Pi_t\cap\Gamma_\tau^\angle)+\Gamma_t. \eqno (2.4)
$$
In particular, all the spaces $\Lambda_{\tau,t}$, $\tau\in[0,t]$
contain the subspace $\Gamma_t$; it is also easy to see that the
subspaces $\Lambda_{\tau,t}$ contain the line
$\mathbb R\lambda_t\subset\Pi_t$, this line is in the kernel of
the quadratic form $q_s$, $\forall s$. Hence
$\Lambda_{\tau,t}\subset(\Gamma_t+\mathbb R\lambda_t)^\angle
\subset(T_{\lambda_t}(T^*M))$.

We set
$$
\Sigma_{(\xi,\lambda_t)}=(\Gamma_t+\mathbb R\lambda_t)^\angle /
(\Gamma_t+\mathbb R\lambda_t);  \eqno (2.5)
$$
then $\Sigma_{(\xi,\lambda_t)}$ is a $2(n-1-r)$-dimensional symplectic
space and $\Lambda_{\tau,t}$ are Lagrangian subspaces of
$\Sigma_{(\xi,\lambda_t)}$. The curve
$$
J_{(\xi,\lambda)}:\tau\mapsto\Lambda_{\tau,t},\quad \tau\in [0,t],
$$
in the Lagrange Grassmannian $L(\Sigma_{(\xi,\lambda_t)}$ will be
called the {\it Jacobi curve} associated with the extremal $(\xi,\lambda)$.

Originally defined on $[0,t)$ the curve $J_{(\xi,\lambda)}$ has the
smooth extension to $[0,t]$ such that $J_{(\xi,\lambda)}(t)=
\Pi_t^{(\Gamma_t+\mathbb R\lambda_t)}$.

\begin{remark} The same construction of the ${\cal L}$-derivative and of
the Jacobi curve works if one replace conditions (2.1) by their
opposite-sign version
$$
\sigma(X_\tau v_1,X_\tau v_2)=0,\quad
\sigma(\dot X_\tau v_1,X_\tau v_1)\le -\alpha |v_1|^2. \eqno (2.1-)
$$
Under conditions (2.1--), the positive inertia index of the
$Hess_{(\xi,\lambda_t)}F_{0,t}$ is finite. The difference of these two cases
disappears if one defines the Lagrange multiplier $\lambda_t$ up to
a scalar nonzero factor, i.e. as a point in the projectivization of
$T^*M$ (see Remark in \cite[Sec.~1]{ag97}).
\end{remark}
\bigskip
Suppose that we have an affine in control system:
$$
f_\tau (x,u)=g^0(x)+G(x)u,
$$
where $G(x):\mathbb R^r\to T_xM$ is a smooth with respect to $x$ family
of linear mappings. Then
$\frac{\partial f_\tau}{\partial u}(x,u)=G(x)$ and any extremal is singular.
The Goh condition has the following form:
$$
\lambda_\tau [Gv_1,Gv_2](\tilde x(\tau))=0,\quad
\forall v_1,v_2\in\mathbb R^r, \tau\in [0,t].   \eqno (2.6)
$$
Besides that, $X_tv=\overrightarrow{\lambda Gv}\bigr|_{\lambda=\lambda_t}$.
If $\tilde u(\tau)\equiv 0$, then the (first) generalized Legendre
condition takes the form
$$
\lambda_\tau [[g^0,Gv],Gv](\tilde x(\tau))\ge 0,\quad
\forall v\in\mathbb R^r, \tau\in [0,t].
$$
To obtain the strenghtened generalized Legendre condition one has to
replace $0$ by $\alpha|v|^2$ in the right--hand side of the last
inequality, for some $\alpha>0$.

\bigskip
Now consider a linear in control system: $f_\tau(x,u)=G(x)u$.
This is, of course, a special case of the affine in control systems,
but there is an essential pecularity in the linear case. Namely,
if an extremal satisfy the Goh condition, then it may also satisfy
the generalized Legendre condition but never the strengthened one.
The reason is simple. Any reparametrization of an admissible trajectory
of the linear in control system is again an admissible trajectory.
At the same time, the reparametrizations do not change the endpoint
mapping! This gives a big kernel of the Hessian of the endpoint mapping
and implies the degeneration of the generalized Legendre condition.
Nevertheless, the described above method for computing the
${\cal L}$-derivative works perfectly after a simple modification of the
original system.

Let $\tilde u(\tau)\equiv u_0$, $\xi=(u_0,\tilde x)$, and let $W$ be
a transversal to $u_0$ linear hyperplane in $\mathbb R^r$,
i.e. $\mathbb R^r=\mathbb Ru_0\oplus W$. Then
$((0,\tilde x),\lambda)$ is an extremal of the affine in control
system
$$
\dot x=G(x)u_0+G(x)\bigr|_Wu, \quad u\in W. \eqno (2.7)
$$
The Goh condition and the (first) generalized Legendre condition
for the extremal $(\xi,\lambda)$ of the original system and the
extremal $((0,\tilde x),\lambda)$ of the system (2.7) are equivalent.
The strengthened generalized Legendre condition for the given extremal
of system (2.7) is equivalent to the inequality
$$
\lambda_\tau [[Gu_0,Gv],Gv](\tilde x(\tau))\ge\alpha|v\wedge u_0|^2 ,\quad
\forall v\in\mathbb R^r, \tau\in [0,t],  \eqno (2.8)
$$
for some $\alpha>0$.
\begin{prop} Let $f_\tau(x,u)=G(x)u$, $\tilde u(\tau)\equiv u_0$.
Under conditions (2.6), (2.8), if $\Lambda\subset T_{\lambda_t}(T^*M)$
is the ${\cal L}$-derivative of the endpoint mapping of the reduced
system (2.7) at $((0,\tilde x),\lambda_t)$, then
$\Lambda^{\mathbb R\dot\lambda_t}$ is the ${\cal L}$-derivative of
$F_{0,t}$ at $(\xi,\lambda_t)$. The same is true if not $\lambda_\cdot$
but $(-\lambda_\cdot)$ satisfy relations (2.6), (2.8). $\quad\square$
\end{prop}
Let $\Lambda_{\tau,t}$ be the ${\cal L}$-derivative of
$F_{\tau,t}$ at $(\xi\bigr|_{[\tau,t]},\lambda_t)$ and
$\Gamma_\tau=\{X_\tau v:v\in\mathbb R^r\}$. Then
$\Lambda_{\tau,t}\supset (\Gamma_t+\mathbb R\lambda_t)$ so that
$$
\Lambda_{\tau,t}\subset\Sigma_{(\xi,\lambda_t)}\stackrel{def}{=}
(\Gamma_t+\mathbb R\lambda_t)^\angle /(\Gamma_t+\mathbb R\lambda_t).
$$
The curve $J_{(\xi,\lambda)}:\tau\mapsto\Lambda_{\tau,t}$ is the
{\it Jacobi curve} associated with the extremal $(\xi,\lambda)$.
Note that the isotropic subspace $\Gamma_t$ is defined for the original
system, not for the reduced system (2.7) and is, in general,
$r$-dimensional.

\bigskip
Let us consider the case of rank 2 distributions on $M$, which are linear
in control systems with $r=2$. Any extremal automatically satisfies
the Goh condition in this case and there is a nice intrinsic description
of the Jacobi curves. So let $\Delta_x\subset T_xM$, $x\in M$, be a
rank 2 distribution (a smooth two-dimensional linear subbundle of $TM$).
By $\Delta^1$ we denote the space of all smooth sections of the
distribution, $\Delta^1\subset Vec\,M$. Set
$$
\Delta^2=span\{[g_1,g_2]:g_i\in\Delta^1, i=1,2\},
$$
$$
\Delta^3=span\{[[g_1,g_2],g_3] : g_i\in\Delta^1, i=1,2,3\},
$$
$$
\Delta^{i\perp}=\{\lambda\in T^*M:\lambda g(\pi(\lambda))=0,\
\forall g\in\Delta^i\},\quad i=1,2,3,
$$
where $\pi:T^*M\to M$ is the canonical projection. It is easy to check
that $\Delta_x^1\subset\Delta_x^2\subset\Delta_x^3$ $\forall x\in M$,
where $\Delta_x^i=\{g(x):g\in\Delta^i\}$. Hence
$\Delta^{3\perp}\subset\Delta^{2\perp}\subset\Delta^{1\perp}$.
We also introduce a linear subbundle $\vec\Delta_{\lambda_0}$,
$\lambda_0\in T^*M$, in $T(T^*M)$ as follows:
$$
\vec\Delta_{\lambda_0}=\{\overrightarrow{\lambda g}\bigr|_{\lambda=\lambda_0}
: g\in\Delta^1\}.
$$
Then $T_\lambda(T^*_xM)\subset\vec\Delta_{\lambda}$,
$\pi_*\vec\Delta_{\lambda}=\Delta_x$ for any
$\lambda\in T^*_xM\setminus\{0\}, x\in M$.

Let $\bar\sigma$ be the standard symplectic structure of the symplectic
manifold $T^*M$, then $\sigma=\bar\sigma_{\lambda_t}$. To say that
$(\xi,\lambda)$ is an extremal is the same as to say that
$\tau\mapsto\lambda_\tau$ is a characteristic curve of
$\bar\sigma\bigr|_{\Delta^{1\perp}}$, i.e
$\bar\sigma(\dot\lambda_\tau,\cdot)\bigr|_{\Delta^{1\perp}}=0$,
and that $\xi$ is the 1-jet of $\pi(\lambda_\cdot)$.
The form $\bar\sigma$ is nondegenerate on
$(\Delta^{1\perp}\setminus\Delta^{2\perp})$, hence all extremals live in
$\Delta^{2\perp}$ and thus satisfy the Goh condition. The extremal
$\lambda_\cdot$ satisfies the strengthened generalized Legendre
condition (or the opposite sign version of this condition) if and only
if $\lambda_\tau\in(\Delta^{2\perp}\setminus\Delta^{3\perp})$,
$\forall\tau\in[0,t]$.

We assume that $\Delta^2\ne\Delta^3$; then
$(\Delta^{2\perp}\setminus\Delta^{3\perp})$ is a $(2n-3)$-dimensional
submanifold in $T^*M$ and
$$
\rank\bar\sigma_\lambda\bigr|_{\Delta^{2\perp}}=(2n-4) \quad
\forall\lambda\in(\Delta^{1\perp}\setminus\Delta^{2\perp}).
$$
Hence $(\ker\bar\sigma_\lambda)\cap T_\lambda\Delta^{2\perp}$
form a line distribution on
$(\Delta^{2\perp}\setminus\Delta^{3\perp})$ and characteristics of
$\bar\sigma_\lambda\bigr|_{\Delta^{2\perp}}$ form a smooth
{\it characteristic} 1-foliation of
$(\Delta^{1\perp}\setminus\Delta^{2\perp})$.

Suppose that given extremal $(\xi,\lambda)$ satisfies the strengthened
generalized Legendre condition. Then $\tau\mapsto\lambda_\tau$,
$\tau\in[0,t]$, is a segment of a leaf of the characteristic foliation.
We assume that this segment has no double points. Let
${\cal O}_\lambda$ be  a tubular neighborhood of
$\{\lambda_\tau :0\le\tau\le t\}$ in $\Delta^{2\perp}$ and let
${\cal D}$ be the restriction of the characteristic foliation to
${\cal O}_\lambda$ such that the quotient space
${\cal O}_\lambda /{\cal D}$ is well--defined and is a smooth
$2(n-2)$-dimensional manifold. Let
$\nu :{\cal O}_\lambda\to{\cal O}_\lambda /{\cal D}$ be the canonical
projection. The form
$\bar\sigma_\lambda\bigr|_{\Delta^{2\perp}}$ induces a canonical
symplecic structure
$\hat\sigma$ on the quotient space ${\cal O}_\lambda /{\cal D}$
by the relation
$\nu^*\hat\sigma=\bar\sigma\bigr|_{\cal O}$. Then
$T_{\lambda_\cdot}({\cal O}_\lambda /{\cal D})$ is a
$2(n-2)$-dimensional symplectic space. It is easy to see that the line
$\nu_*(\mathbb R\lambda_\tau)\subset
T_{\lambda_\cdot}({\cal O}_\lambda /{\cal D})$ does not depend on
$\tau\in[0,t]$. We set $\mathbb R\hat\lambda=\nu_*(\mathbb R\lambda_\tau)$
and $\Sigma_\lambda=(\mathbb R\hat\lambda)^\angle/\mathbb R\hat\lambda$.
The symplectic space $\Sigma_\lambda$ is naturally isomorphic to the
space $\Sigma_{(\xi,\lambda_t)}$ (see (2.5)) for the control problem
under consideration. Finally, set
$$
J_\lambda(\tau)=
\nu_*(\vec\Delta_{\lambda_\tau}\cap T_{\lambda_\tau}\Delta^{2\perp});
$$
then $J_\lambda(\tau)$ is a Lagrangian subspace of $\Sigma_\lambda$
and the curve $\tau\mapsto J_\lambda(\tau)$ is the Jacobi curve of the
extremal $(\xi,\lambda)$ up to the isomorphism
$\Sigma_\lambda\cong\Sigma_{(\xi,\lambda_t)}$ .

\bigskip
A similar intrinsic description of Jacobi curves is valid for rank 1
affine distributions on $M$, i.e. for affine in control systems
with scalar controls. Let $\ell_x\subset T_xM\setminus\{0\},$
$x\in M$, be a rank 1 affine distribution (a smooth one-dimensional
affine subbundle of $TM$). We'll consider only normal extremals of
this control system, i.e. such that they are not extremals for the
rank 2 distribution $span\,\ell_x$, $x\in M$. For the normal extremal
$(\xi,\lambda)$ we have $\lambda_\tau\dot{\tilde x}(\tau)\ne 0$. Since
$\lambda_\tau\dot{\tilde x}(\tau)$ does not depend on $\tau$ and
$\lambda_\cdot$ is defined up to a scalar multiplier, we may (and will)
suppose that $\lambda_\tau\dot{\tilde x}(\tau)=1$.

By $\ell^1$ we denote the space of all smooth sections of the affine
subbundle, $\ell^1\subset Vec\,M$. Set
$$
\ell^0=\{g_1-g_2 : g_i\in\ell^1, i=1,2\},
$$
$$
\ell^2=span\{[g_1,g_2]:g_i\in\ell^1, i=1,2\},
$$
$$
\ell^3=span\{[[g_1,g_2],g_3] : g_i\in\ell^1, i=1,2,3\}.
$$
Then $\ell^0$ is the space of smooth sections of the one-dimensional
vector subbundle associated with given affine subbundle and
$\ell^0\subset\ell^2\subset\ell^3$. We also introduce dual objects
similar to the linear case:
$$
\ell^{1\circ}=\{\lambda\in T^*M:\lambda g(\pi(\lambda))=1,\
\forall g\in\ell^1\},
$$
$$
\ell^{i\circ}=\{\lambda\in\ell^{1\circ} :\lambda g(\pi(\lambda))=0,\
\forall g\in\ell^i\},\quad i=0,2,3,
$$
$$
\vec{\ell^0_{\lambda_0}}=\{\overrightarrow{\lambda g}\bigr|_{\lambda=\lambda_0}
: g\in\ell^0\},\quad\lambda_0\in T_x^*M\setminus\{0\}, \ x\in M.
$$
Normal extremals that satisfy the strengthened generalized Legendre
condition (or the opposite--sign version of this condition) are just
properly paramet\-rized characteristic curves of
$\bar\sigma_\lambda\bigr|_{(\ell^{2\circ}\setminus\ell^{3\circ})}$.
If $\ell^{2\circ}\ne\ell^{3\circ}$, then
$\ell^{2\circ}\setminus\ell^{3\circ}$ is a $(2n-3)$-dimensional
submanifold and characteristic curves of
$\bar\sigma_\lambda\bigr|_{(\ell^{2\circ}\setminus\ell^{3\circ})}$
form a smooth 1-foliation of this submanifold. Let
$\tau\mapsto\lambda_\tau$,
$\tau\in[0,t]$, be a segment of a leaf of this foliation,
${\cal O}_\lambda$ be  a tubular neighborhood of
$\{\lambda_\tau :0\le\tau\le t\}$ in $\ell^{2\circ}$, and
${\cal D}$ be the restriction of the foliation to
${\cal O}_\lambda$ such that the quotient space
${\cal O}_\lambda /{\cal D}$ is well--defined and is a smooth
$2(n-2)$-dimensional manifold.
The quotient space ${\cal O}_\lambda /{\cal D}$ is endowed with
a symplectic structure induced by
$\bar\sigma_\lambda\bigr|_{\ell^{2\circ}}$. The symplectic space
$T_{\lambda_\cdot}({\cal O}_\lambda /{\cal D})$ is naturally
isomorphic to $\Sigma_{(\xi,\lambda_t)}$ (the space
$T_{\lambda_\cdot}({\cal O}_\lambda /{\cal D})$ does not contain
$\mathbb R\lambda_\tau$ in contrast to the case of linear
distributions).
Let
$\nu :{\cal O}_\lambda\to{\cal O}_\lambda /{\cal D}$ be the canonical
projection. The curve
$$
J_\lambda :\tau\mapsto
\nu_*(\vec{\ell^0_{\lambda_\tau}}\cap T_{\lambda_\tau}\ell^{2\circ})
$$
in the Lagrange Grassmannian of
$T_{\lambda_\cdot}({\cal O}_\lambda /{\cal D})$ is the Jacobi curve
of the extremal $(\xi,\lambda)$ up to the isomorphism
$T_{\lambda_\cdot}({\cal O}_\lambda /{\cal D})\cong\Sigma_{(\xi,\lambda_t)}$.

\bigskip
We now go back to the general nonlinear system and finish this Section
with an expression for the ${\cal L}$-derivative which unifies the
regular case, the case of Proposition 2.2, and also works in much more
degenerate cases. What follows is, in fact, a corollary of results
proved in \cite{a88}.

So we cancel the above assumption on the total singularity of the
extremal $(\xi,\lambda)$ in order to include the regular case.
We assume, however, that $X_\tau$ is not just Lipschitzian but
piecewise smooth with respect to $\tau$; we need this assumption
to handle highly degenerate cases. We agree to take all piecewise
smooth functions to be left--smooth and to have limits from the right
of all derivatives under consideration; the expression $X_\tau^{(k)}v$ below
denotes the derivative of $X_\tau v$ of order $k\ge 0$ with respect
to $\tau$. Let $m=\dim span\{\pi_*X_\tau v:v\in\mathbb R^r,\tau\in(0,t]\}$,
then $m$ equals the rank of the differential of the mapping $F_{0,t}$
at $\xi$.
With each $\tau\in(0,t]$ we
associate an integer $k_\tau\ge 0$ and a quadratic form $b_\tau$
on $\mathbb R^r$ as follows: if the form
$\frac{\partial^2(\lambda_\theta f_\theta)}{\partial u^2}
(\tilde x(\theta),\tilde u(\theta))$ is not identical zero on any interval
$\bar\tau <\theta <\tau$, then let $k_\tau =0$ and
$b_\tau=\frac{\partial^2(\lambda_\tau f_\tau)}{\partial u^2}
(\tilde x(\tau),\tilde u(\tau))$; otherwise, let $k_\tau$ be the maximal
number $k$ such that $\sigma(X_\theta^{(i)}v_1,X_\theta^{(j)}v_2)\equiv 0$
for $i+j<2(k-1)$ and $v_1,v_2\in \mathbb R^r$ on some interval
$\bar\tau <\theta <\tau$, and let
$b_\tau(v)=\sigma(X_\theta^{(k_\tau)}v,X_\theta^{(k_\tau -1)}v)$,
$v\in\mathbb R^r$ (if the maximal $k$ exists, then it does not exceed $m$;
if it does not exist, then we set $k_\tau =m+1$ and $b_\tau =0$).

\begin{prop}[\cite{a88,ag86}]
If
$\ \ind_-Hess_{(\xi,\lambda_t)}F_{0,t}<\infty$, then
$$
\sigma(X_\tau^{(k_\tau-1)} v_1,X_\tau^{(k_\tau-1)} v_2)=0,\
 \forall v_1,v_2\in\mathbb R^r, \quad b_\tau\ge 0, \ \tau\in (0,t].
$$
If, additionally, $b_\tau(v)\ge \alpha |v|^2$
for some $\alpha >0$ and all $v\in\mathbb R^r, \tau\in (0,t]$, then indeed
$\ \ind_-Hess_{(\xi,\lambda_t)}F_{0,t}<\infty$.
\end{prop}
In what follows we assume the sufficient condition in Proposition 2.4
for $\ind_-Hess_{(\xi,\lambda_t)}F_{0,t}<\infty$ to be finite. Let
$$
\Gamma_\tau=span\{X_\tau^{(i)}v :0\le i<k_\tau, \ v\in \mathbb R^r\},
$$
an $rk_\tau$-dimensional isotrpic subspace of $T_{\lambda_t}(T^*M)$.

The quadratic form $b_\tau$ is identified with an invertible selfadjoint
mapping from $\mathbb R^r$ in $\mathbb R^{r*}$, and the inverse mapping
determines a quadratic form $b_\tau^{-1}$ on $\mathbb R^{r*}$.
The invertibility of $b_\tau$ implies that $\Gamma_\tau$ and $b_\tau$
are smooth with respect to $\tau$ on any interval where $X_\tau$ is
smooth. In particular, $\Gamma_\tau$ and $b_\tau^{-1}$ are piecewise
smooth on $[0,t]$.

Set $q_\tau(\eta)=-\frac 12b_\tau^{-1}(\sigma(X_\tau^{(k_\tau)}\cdot,\eta))$,
a nonstationary quadratic Hamiltonian on $T_{\lambda_t}(T^*M)$.
Hamiltonian $q_\tau$ defines a linear Hamiltonian system on
$T_{\lambda_t}(T^*M)$ and, hence, a nonautonomous differential equation
on the Lagrange Grassmannian $L(T_{\lambda_t}(T^*M))$. We need the last
one, which becomes the matrix Riccati equation in "affine" local
coordinates on $L(T_{\lambda_t}(T^*M))$. We prefer coordinate free
notations and denote $q_\tau(\Lambda)$ the value at
$\Lambda\in L(T_{\lambda_t}(T^*M))$ of the vector field on
$L(T_{\lambda_t}(T^*M))$ induced by the Hamiltonian field $\vec q_\tau$
on $T_{\lambda_t}(T^*M)$. This notation is coordinated with the
canonical identification of the tangent vectors to $L(T_{\lambda_t}(T^*M))$
and quadratic forms (see Append.).

So we have a differential equation
$$
\dot\Lambda =q_\tau(\Lambda),\quad \Lambda\in L(T_{\lambda_t}(T^*M));
\eqno (2.9)
$$
what is important and is not quite standard, we allow piecewise smooth
(maybe discontinuous in some points) solutions of this equations.
Thus to determine a solution of (2.9) we need not only the initial
condition but also the relations for "jumps" at the points of
discontinuity. In fact, the solution may have a "jump" even at the
initial moment since piecewise continuous functions are supposed to be
left--smooth but not right--smooth.
\begin{prop} Differential equation (2.9) and the relations
$$
\Lambda(0)=T_{\lambda_t}(T_{\tilde x(t)}^*M),\quad
\Lambda(\tau+0)=\Lambda(\tau)^{\Gamma_{\tau+0}},\ \tau\in [0,t),
$$
imply the inclusions $\Gamma_\tau\subset\Lambda(\tau)$,
$\tau\in (0,t]$, and uniquely
determine a piecewise smooth curve $\Lambda(\tau), \tau\in [0,t]$,
in  $L(T_{\lambda_t}(T^*M))$. Then $\Lambda(t)$ is the
${\cal L}$-derivative of $F_{0,t}$ at $(\xi,\lambda_t)$. $\quad \square$
\end{prop}
\begin{theorem} Under conditions of Proposition 2.5, let
$\tau_1,\ldots,\tau_l$ be all points of discontinuity of the curve
$\Lambda(\cdot)$. Suppose $\bar\Lambda(\cdot)$ is a continuous closed
curve in the Lagrange Grassmannian obtained from the curve
$\tau\mapsto\Lambda(t-\tau)$ by the connecting of $\Lambda(\tau_i+0)$
with $\Lambda(\tau_i)$, $i=1,\ldots,l$, and $\Lambda(0)$ with
$\Lambda(t)$ by simple nondecreasing curves (see~Append.). Then
$$
\ind_-Hess_{(\xi,\lambda_t)}F_{0,t}=\ind\bar\Lambda(\cdot)-m,
$$
where $\ind\bar\Lambda(\cdot)$ is the Maslov index of $\bar\Lambda(\cdot)$.
$\quad \square$
\end{theorem}



\section{Three-dimensional systems}
We are going to study generic germs of affine in control systems
admitting extremals, and we are going to do it in a lowest possible
dimension. Generic germs do not admit extremals if $n=2$ and if
$n=3,r=2$, therefore we focus on the case $n=3,r=1$. So we deal with
the germ at $x_0\in M$ of a system
$$
\dot x=g^0(x)+ug^1(x),\ u\in\mathbb R,\ x\in M,\quad \dim M=3. \eqno (3.1)
$$
Generic assumptions are as follows:
$$
\begin{array}{c}
g^0(x)\wedge g^1(x)\wedge [g^0,g^1](x)\ne 0,\\
g^1(x)\wedge [g^1,g^0](x)\wedge [g^1,[g^1,g^0]](x)\ne 0. \\
\end{array}  \eqno (3.2)
$$
We'll describe a normal form of the germ and a complete system of
invariants under feedback and state transformations.

First let us roughly
evaluate the "number of parameters". Affine lines in
$\mathbb R^3$ form a 4-dimensional manifold, hence the space of
the equivalence classes of systems of form (3.1) with respect to purely
feedback (not the state!) transformations can be locally parametrized
by 4 germs of real smooth functions of 3 variables. The group of
state transformations is parametrized by 3 smooth functions of 3 variables.
We thus may expect $4-3=1$ arbitrary germ of a smooth function of 3
variables as a principal invariant and a number of germs of functions of
2, 1, and 0 variables as some additional invariants.

We have $\ell_x=g^0(x)+\mathbb Rg^1(x)$,
$$
\ell^{2\circ}=\{\lambda\in T^*M :\lambda g^0=1, \
\lambda g^1=\lambda [g^0, g^1]=0\},\quad \ell^{3\circ}=\emptyset .
$$
Hence $\ell^{2\circ}=\ell^{2\circ}\setminus\ell^{3\circ}$ is just a
section of the bundle $T^*M$ and $\ell^{2\circ}$ is foliated by the
extremals. In other words, extremal trajectories form a local flow in M;
there exists a unique smooth function $v$ on $M$ such that the extremal
trajectories are exactly trajectories of the differential equation
$\dot x=g^0(x)+v(x)g^1(x)$. Replacing $g^0$ by $g^0+vg^1$, if necessary,
we may assume from the very beginning that the trajectories of the
system $\dot x=g^0(x)$ are the extremal trajectories. The last property
is equivalent to the identity
$$
g^0(x)\wedge [g^0,g^1](x)\wedge [g^0,[g^0,g^1]](x)=0,
\quad x\in M.  \eqno (3.3)
$$
The only feedback transformations preserving identity (3.3) are the
multiplications of $g^1$ by a nonvanishing function.

The extremals do not admit reparametrizations and a Jacobi curve is
associated to every extremal. The Jacobi curves are curves in
the Lagrange Grassmannian of the symplectic space of dimension
$2(3-2)=2$. In other words, these are curves in the projective line.
The curvature tensors associated to germs of regular curves in the
Lagrange Grassmannians are just real numbers in this low dimensional
case (see \cite[Sec.~4]{ag97}). Through every $x\in M$ passes exactly
one extremal trajectory corresponding to a uniquely determined extremal
in $\ell^{2\circ}$. Take the germ of the extremal with the endpoint
$\lambda_t=\ell^{2\circ}\cap T_xM$. Assume that $R(x)$ is the curvature
of the germ at $t$ of the Jacobi curve associated to the extremal.
We'll see that $R(x),\ x\in M$, is a smooth function of $x$ and the
function $R$ turns into the required principal functional invariant as
soon as one fix some canonical coordinates in the neighborhood of $x_0$.

Let us denote $g^2=[g^0,g^1]$; then $g^0(x),g^1(x),g^2(x)$ form a
basis of $T_x^*M$ and we have
$$
[g^0,g^1]=g^2,\quad [g^2,g^0]=c_{02}^1g^1+c_{02}^2g^2,\quad
[g^2,g^1]=\sum\limits_{k=0}^2c_{12}^kg^k,
$$
where the "structural constants" $c_{ij}^k$ are actually smooth
functions on $M$ and $c_{12}^0\ne 0$.
\begin{prop} $R=c_{02}^1-\frac 14(c_{02}^2)^2-\frac 12g^0c_{02}^2$.
\end{prop}
{\bf Proof.} Let $\omega_0,\omega_1,\omega_2$ be the frame in $T^*M$
dual to the frame $g^0,g^1,g^2$ in $TM$, i.e.
$\langle\omega_i,g^j\rangle=\delta_{ij}$, $i,j=0,1,2$. Then
$\ell^{2\circ}\subset T^*M$ is the graph of $\omega_0$,
$$
\bar\sigma\bigr|_{\ell^{2\circ}}=\left(\pi\bigr|_{\ell^{2\circ}}\right)^*
(c^0_{12}\omega_1\wedge\omega_2), \quad
\vec{\ell^0}\cap T\ell^{2\circ}=
\left(\pi\bigr|_{\ell^{2\circ}}\right)_*^{-1}g^1.
$$
The diffeomorphism $\pi\bigr|_{\ell^{2\circ}}$ provides the identification
of $\ell^{2\circ}$ with $M$ and we make the remaining calculations in $M$
instead of $\ell^{2\circ}$. Let $\tau\mapsto J_x(\tau)$ be the Jacobi
curve associated to the extremal trajectory $\tau\mapsto xe^{(\tau-t)g^0}$.
Then $J_x(\tau)$ is a line in the plane $T_xM/(\mathbb Rg^0(x))$ endowed
with the symplectic structure $(c^0_{12}\omega_1\wedge\omega_2)_x$.
More precisely,
$J_x(\tau)=\mathbb R\left(e_*^{(t-\tau)g^0}g^1\right)(x)+
\mathbb Rg^0(x)$.  The vectors $g^1(x),\ \frac 1{c^0_{12}(x)}g^2(x)$
form a canonical basis of the symplectic plane; what we need are
the coordinates of the vector $\left(e_*^{(t-\tau)g^0}g^1\right)(x)=
\left(e^{(\tau-t)ad\, g^0}g^1\right)(x)$ in this basis. We have
$e^{(\tau-t)ad\, g^0}g^1=a^1_\tau g^1+a^2_\tau g^2$, where
$a^1_t=\dot a^2_t=1$, $\dot a^1_t=a^2_t=0$,
$$
\ddot a^i_\tau =-\left(e^{(\tau-t)g^0}c^1_{02}\right)a^i_\tau -
\left(e^{(\tau-t)g^0}c^2_{02}\right)\dot a^i_\tau,\quad i=1,2; \eqno (3.4)
$$
here $\dot{\rm\ }$ is the differentiation with respect to $\tau$.

According to \cite[Sec. 4]{ag97},
the curvature $R(x)$ is the Schwartzian derivative at $t$ of the function
$S_\tau=c^0_{12}(x)\frac{a^2_\tau(x)}{a^1_\tau(x)}$,
$$
R(x)=\frac 12\frac{\stackrel{\ldots}{S}_t}{\dot S_t}-
\frac 34\left(\frac{\ddot S_t}{\dot S_t}\right)^2. \eqno (3.5)
$$
The desired expression for $R$ is derived from (3.4), (3.5) by
a straightforward calculation. $\quad\square$

\medskip The curvature $R$ is feedback--invariant, while the "structural
constants" $c^k_{ij}$ are not. A feedback transformation
$g^1\mapsto bg^1$, where $b\in C^\infty(M)$, transforms
$c^0_{12}\mapsto b^2c^0_{12}$. Hence $\sign c^0_{12}$ is
feedback--invariant and the vector field
$\hat g=|c^0_{12}|^{-\frac 12}g^1$ may change only the sign under feedback
transformations.

We put $\nu=\sign c^0_{12}$ and define a feedback invariant germ
$N$ of a two-dimensional submanifold of $M$ as follows:
$$
N=\{x_0e^{y_2[g^0,\hat g]}\circ e^{y_1\hat g} : y_1,y_2
\mbox{ are close to }0\mbox{ real numbers}\}.
$$
\begin{lemma} There is a unique up to a sign feedback normalization
of the germ of $g^1$ at $x_0$ such that the following identities hold
for the normalized germ:
$$
c^2_{02}\equiv 0, \quad |c^0_{12}|\Bigr|_N\equiv 1 \eqno (3.6)
$$
\end{lemma}
{\bf Proof.} We set $g^1\bigr|_N=\hat g\bigr|_N$. Note that the field
$g^1$ is tangent to $N$. The field $g^0$ is transversal to $N$,
therefore the trajectories of $g^0$ define a tubular neighborhood
of $N$ in $M$. A renormalization $g^1\mapsto bg^1$ transforms
$c^2_{02}\mapsto c^2_{02}-\frac 2bg^0b$. The renormalized field
satisfies the required conditions if and only if $b\bigr|_N=\pm 1$,
$g^0b=\frac 12bc^2_{02}$. The second identity is equivalent to the
following one:
$$
e^{tg^0}b=be^{\frac 12\int\limits_0^te^{\tau g^0}c^2_{02}\,d\tau}.
$$
This means that $b$ is uniquely defined on a trajectory of the field
$g^0$ as soon as we fix $b$ at a point of the trajectory, and
this is all we need to complete the proof. $\quad\square$

\medskip Assume that $g^1$ is normalized and (3.6) is valid. We
introduce {\it normal coordinates} $(y_0,y_1,y_2)$ in a neighborhood
of $x_0$ by the formula
$$
\Phi(y_0,y_1,y_2)=x_0e^{y_2g^2}\circ e^{y_1g^1}\circ e^{y_0g^0}.
$$
In particular, $\Phi^{-1}(N)$ is the coordinate plane $y_0=0$.

Let $t\mapsto \rho_i(t,y_1,y_2),\ i=0,1$, be the solutions of the Hill
equation \linebreak
$\ddot\rho +R(t,y_1,y_2)\rho=0$ with the initial conditions
$$
\left(\begin{array}{cc}
\rho_0(0,y_1,y_2)&\dot\rho_0(0,y_1,y_2)\\
\rho_1(0,y_1,y_2)&\dot\rho_1(0,y_1,y_2)\\
\end{array}\right) =
\left(\begin{array}{cc}1&0\\0&1\end{array}\right).
$$
In the coordinates $y=(y_0,y_1,y_2)$, the fields $g^0, g^1$ have the form:
$$
g^0=\frac\partial{\partial y_0},\quad
g^1=\rho_0(y)\frac\partial{\partial y_1}+
\rho_1(y)\sum\limits_{i=0}^2\alpha_i(y_1,y_2)\frac\partial{\partial y_i},
\eqno (3.7)
$$
where $\alpha_1,\alpha_2$ are arbitrary germs of smooth functions meeting
the conditions
$$
\alpha_1(0,y_2)=0,\quad \alpha_2(0,y_2)=1, \eqno (3.8)
$$
and $\alpha_0(y_1,y_2)=-\nu
\int\limits_0^{y_1}\frac{\alpha_2(y_1,y_2)}{\alpha_2(t,y_2)}\,dt$.
The only ambiguity of the construction is the sign of $g^1$. Changing
the sign leads to the transformation
$$
\begin{array}{c}
R(y_1,y_2,y_3)\mapsto R(y_1,-y_2,-y_3),\\
(\alpha_1(y_1,y_2),\alpha_2(y_1,y_2))\mapsto
(\alpha_1(-y_1,-y_2),\alpha_2(-y_1,-y_2)).\\
\end{array}  \eqno (3.9)
$$
Summing up, we obtain the following
\begin{prop} Any system of the form (3.1) subject to conditions (3.2)
has a local normal form (3.7) under smooth feedback and state
transformations. In the normal form, $R$ is an arbitrary germ of smooth
function on $\mathbb R^3$, $\alpha_1,\alpha_2$ are arbitrary germs of
smooth function on $\mathbb R^2$ meeting conditions (3.8);
$\rho_0,\rho_1$ are determined by $R$, $\alpha_0$ is determined by
$\alpha_2$ and the discrete invariant $\nu\in\{-1,1\}$. The germs
$R,\alpha_1,\alpha_2$ are feedback and state invariants up to the
involution (3.9). $\quad\square$
\end{prop}

Recall that $R$ is the curvature, i.e. a well--defined function on $M$
with a clear geometric and variational meaning. The invariant $\nu$
has the following nice interpretation. Small enough pieces of the extremal
trajectories realize local extrema in the time optimal problem for system
(3.1); they are local time minimizers if $\nu=-1$ and local time
maximizers if $\nu=1$. Invariants $\alpha_1,\alpha_2$ are
well--defined germs of smooth functions on $N$; the geometric meaning
of these germs is not so clear yet.

In the "flat" case $R\equiv 0,\alpha_1\equiv 0,\alpha_2\equiv 1$
we obtain the relations
$$
[g^0,[g^0,g^1]]=0,\quad [g^1,[g^1,g^0]]=\nu g^0.
$$
The fields $g^0,g^1$ thus generate a solvable three-dimensional Lie
algebra: the Lie algebra of the group of isometries of the Euclidean plane
for $\nu=-1$ and of the pseudo-Euclidean plane for $\nu=1$. Then system
(3.1) is locally equivalent to a left--invariant system on the
corresponding group of isometries.
\begin{remark} Actually, the results of this section give more than
a normal form in a neighborhood of a point. It is easy to see that the
same normal form is valid in a neighborhood of an as long as we want
segment of the extremal trajectory, while the trajectory has no double
points and conditions (3.2) are satisfied.
\end{remark}


\section{Appendix. On Lagrange Grassmannians}

Here we introduce some notions of linear symplectic geometry
that we use in the paper (see \cite{ag,Ar1,GS,LV} for more details).
A {\em symplectic structure} in an even-dimensional linear
space $\Sigma$ is defined by a nondegenerate bilinear
skew-symmetric 2-form $\sigma(\cdot,\cdot).$ Two vectors
$\xi_{1},\xi_{2} \in \Sigma$ are called skew-orthogonal,
written $\xi_{1} \angle \xi_{2},$ if $\sigma(\xi_{1},\xi_{2})=0.$
If $N$ is a subspace of $\Sigma$, let us denote by
$N^{\angle}$ its skew-orthogonal complement:
$N^{\angle}=\{\xi \in \Sigma \mid \sigma (\xi,\nu)=0,
\forall \nu \in N\}.$ Evidently
$\dim N + \dim N^{\angle}=\dim \Sigma.$
A subspace $\Gamma \subseteq \Sigma$ is called
{\em isotropic}, if $\Gamma \subseteq \Gamma^{\angle},$
and {\em coisotropic}, if $\Gamma \supseteq \Gamma^{\angle}.$
A subspace $\Lambda \subset \Sigma$ is called
{\em Lagrangian}, if $\Lambda^{\angle}=\Lambda.$
Such subspaces have dimension $\frac{1}{2}\dim \Sigma$.
If $\Lambda$ is a Lagrangian subspace and $\Gamma$ is
isotropic, then it is easy to prove, that
$(\Lambda \cap \Gamma^{\angle})+\Gamma=
(\Lambda+\Gamma) \cap \Gamma^{\angle}$
is a Lagrangian subspace. We denote it by $\Lambda^{\Gamma}.$

The symplectic group $Sp(\Sigma)$ is the group of those
linear transformations of $\Sigma,$ which preserve the
symplectic form:
$$
Sp(\Sigma)=\{S \in GL (\Sigma) :
\sigma( S \xi_{1}, S \xi_{2})=
\sigma( \xi_{1}, \xi_{2}) \
\forall \xi_{1}, \xi_{2} \in \Sigma \}.
$$
The elements of this group are called
symplectic transformations of $\Sigma .$
The Lie algebra of the symplectic group is:
\[ sp(\Sigma)=\{A \in gl(\Sigma) :
\sigma( A \xi_{1}, \xi_{2})=
\sigma( A \xi_{2}, \xi_{1}) \
\forall \xi_{1}, \xi_{2} \in \Sigma \}.   \]

Let $H$ be a real quadratic form on
$\Sigma$ and $d_{\xi}H$ be the differential
of $H$ at a point $\xi \in \Sigma .$
Then
$d_{\xi}H$
is a linear form on $\Sigma$ 
which depends linearly on $\xi .$ 
For every $\xi \in \Sigma$ there exists 
a unique vector 
$\stackrel{\rightarrow}{H}(\xi) \in \Sigma$ 
which satisfies equality 
$\sigma(\cdot,\stackrel{\rightarrow}{H}(\xi))=
d_{\xi}H .$  
It is easy to show that the linear operator 
$\stackrel{\rightarrow}{H}
:\Sigma \rightarrow \Sigma $  
belongs to $sp(\Sigma),$ 
and the mapping $H \mapsto 
\stackrel{\rightarrow}{H}$ 
is an isomorphism of the space of 
quadratic forms onto 
$sp(\Sigma).$ The differential equation 
$\dot{\xi}=
\stackrel{\rightarrow}{H}(\xi)$ 
is called the linear Hamiltonian system 
corresponding to the quadratic Hamiltonian $H.$ 

Denote by 
$L(\Sigma)$ 
the Grassmanian of Lagrangian
subspaces of $\Sigma$. This is a smooth compact manifold of 
dimension $\frac{1}{8} \dim \Sigma(\dim \Sigma +2).$ 

Certainly symplectic transformations 
transform Lagrangian subspaces into Lagrangian ones, 
hence the symplectic group acts on  
$L(\Sigma).$ It is easy to show that it acts 
transitively. 

Let us consider a tangent space  
$T_{\Lambda}L(\Sigma), 
\ 
\Lambda \in L(\Sigma). $  
To every quadratic form $h$ on $\Sigma$ 
there corresponds a linear Hamiltonian vector field 
$\stackrel{\rightarrow}{h}$ and a one-parameter subgroup 
$t \mapsto e^{t \stackrel{\rightarrow}{h}}$ in 
$Sp(\Sigma).$ Let us consider the linear mapping 
$2h \mapsto\frac{d}{dt}e^{t \stackrel{\rightarrow}{h}}(\Lambda)\bigr|_{t=0}$ 
of the space of quadratic forms into 
$T_{\Lambda}L(\Sigma).$ 
This mapping is surjective and its kernel 
consists of all quadratic forms which vanish on 
$\Lambda .$ Thus two different quadratic forms 
correspond to the same vector from 
$T_{\Lambda}L(\Sigma)$ 
if and only if the restrictions of these forms 
on $\Lambda$ coincide. Hence we obtain a natural 
identification of the space 
$T_{\Lambda}L(\Sigma)$ 
with the space of quadratic forms on $\Lambda .$  

A tangent vector $\eta \in 
T_{\Lambda}L(\Sigma)$ 
is called nonnegative if the corresponding quadratic form 
is nonnegative on $\Lambda.$ An absolutely continuous curve 
$\Lambda_{\tau}
\ (\tau \in [0,t])$ in $L(\Sigma)$ 
is called nondecreasing if the velocities 
$\dot{\Lambda}_{\tau} \in T_{\Lambda_{\tau}}L(\Sigma)$       
are nonnegative for almost all $\tau \in [0,T].$ 

Treating the action of symplectic group $Sp(\Sigma)$ on $ 
L(\Sigma)$ one can easily 
verify, that pairs of Lagrangian subspaces $(\Lambda,\Lambda')$ 
have only one invariant w.r.t. this action: it is  
$\dim (\Lambda \cap \Lambda ').$ For triples of 
Lagrangian subspaces, there are more invariants. 

Let $\Lambda_{1}, \Lambda_{2}, \Lambda_{3}$ be Lagrangian 
subspaces. Let us present a vector 
$\lambda \in (\Lambda_{1}+ \Lambda_{3}) \cap \Lambda_{2}$ 
as a sum $\lambda= \lambda_{1}+ \lambda_{3}$ and consider 
a properly defined on $(\Lambda_{1}+ \Lambda_{3}) \cap \Lambda_{2}$   
quadratic form 
$S(\lambda)=\sigma (\lambda_{1}, \lambda_{3})$. 
{\em Maslov index} of the triple 
$(\Lambda_{1}, \Lambda_{2}, \Lambda_{3})$
is signature of $S(\lambda).$ It is an invariant 
of the action of symplectic group. 

In \cite{a88} a bit different invariant of a triple of 
Lagrangian planes $(\Lambda_{1}, \Lambda_{2}, \Lambda_{3})$ 
was exploited for 
computation of Morse index for singular extremals. 

\newtheorem{Maslov}{Definition}[section] 
\begin{Maslov}                              
\label{masin} 
Consider the quadratic form 
$S(\lambda)=\sigma (\lambda_{1}, \lambda_{3})$ 
with the domain  
$((\Lambda_{1}+ \Lambda_{3}) \cap \Lambda_{2})/ 
\bigcap^{3}_{i=1} \Lambda _{i}.$ 
A sum 
$\frac{1}{2} \dim \ker S +\ind_-S$, where 
$\ind_-S$ is the negative inertia index of $S ,$ 
is an invariant of the triple      
$(\Lambda_{1}, \Lambda_{2}, \Lambda_{3})$ 
of Lagrangian subspaces. It is denoted by 
$\ind_{\Lambda_{2}}(\Lambda_{1},\Lambda_{3})$ and 
is called Maslov-type index.
\end{Maslov}

Let us note, that 
$\ker S=((\Lambda_{1} \cap \Lambda_{2})+ 
(\Lambda_{2} \cap \Lambda_{3}))/\bigcap^{3}_{i=1}\Lambda_{i}.$ 
We refer to \cite{a88} for a simple formula connecting this 
Maslov-type index with Maslov index of the triple 
and for the proof of the following 'triangle inequality': 
\[ {\rm ind}_{\Lambda_{0}}(\Lambda_{1},\Lambda_{3}) \leq 
{\rm ind}_{\Lambda_{0}}(\Lambda_{1},\Lambda_{2})+
{\rm ind}_{\Lambda_{0}}(\Lambda_{2},\Lambda_{3}). \] 
It also follows directly from the definition, that 
$$ 
\ind_{\Lambda_{1}}(\Lambda_{1},\Lambda_{3}) 
=\frac{1}{2} \dim\ker S=\frac{1}{2}(\dim \Lambda_{1}-
\dim(\Lambda_{1} \cap \Lambda_{3})). 
$$
A continuous curve 
$\Lambda (\tau) \in L(\Sigma), \ 
0 \leq \tau \leq 1,$ is called 
{\em simple} if there exists 
$\Delta \in {\cal L}(\Sigma)$ such that 
$\Lambda (\tau) \cap \Delta ={0} \ 
\forall \tau \in [0,1].$ 
    

\begin{lemma}                              
\label{isim} 
If 
$\Lambda (\tau) \in L(\Sigma), \ 
0 \leq \tau \leq 1,$ 
is a simple nondecreasing curve in 
${\cal L}(\Sigma),$ 
and $\Pi \in {\cal L}(\Sigma),$ 
then 
\[ {\rm ind}_{\Pi}(\Lambda(0),\Lambda(1))=  
{\rm ind}_{\Pi}(\Lambda(0),\Lambda(\tau))+
{\rm ind}_{\Pi}(\Lambda(\tau),\Lambda(1)), \ 
\forall \tau \in [0,1].  \]  
\end{lemma} 

 
\begin{lemma} 
\label{sico}
Let $\Lambda^{0},\Lambda^{1} \in L(\Sigma).$   
There exist $\Delta \in L(\Sigma)$ 
and neighborhoods 
$V^{0}~\ni~\Lambda^{0},$        
$V^{1}~\ni~\Lambda^{1}$ 
in ${\cal L}(\Sigma)$ 
such that whenever $\Lambda \in V^{0}, \Lambda' \in V^{1}$      
and 
$\dim (\Lambda \cap \Lambda')=\dim(\Lambda^{0} \cap \Lambda^{1})$  
then there exists a simple nondecreasing curve 
$\Lambda(\tau), \ \tau \in [0,1]$ 
such that $\Lambda(0)=\Lambda, \Lambda(1)=\Lambda', \ 
\Lambda(\tau) \cap \Delta =0 \ \forall \tau \in [0,1]. \       
$ 
\end{lemma} 
Both Lemmas are proved in \cite{a88}. 

\newtheorem{Mascurve}[Maslov]{Definition} 
\begin{Mascurve} 
\label{mascu} 
Let $\Lambda(\tau), \ 0 \leq \tau \leq t,$ 
be a nondecreasing curve in   
$L(\Sigma)$ and 
$0=\tau_{0}<\tau_{1}< \cdots <\tau_{l}=t$ 
are such, that the curves 
$\Lambda(\cdot)\bigr|_{[\tau_{i},\tau_{i+1}]}, 
\ i=0, \ldots l-1,$ are simple and 
$\Pi \in L(\Sigma).$ 
The expression 
$$ 
{\rm ind}_{\Pi}\Lambda(\cdot)=
\sum_{i=0}^{l-1}{\rm ind}_{\Pi}
(\Lambda(\tau_{i}),\Lambda(\tau_{i+1}))  \eqno (4.1)
$$ 
is called Maslov index of the curve 
$\Lambda(\cdot)$ with respect to $\Pi. \ $ 
\end{Mascurve} 

It follows from Lemma~\ref{isim}  
that (4.1) does not depend on a choice of 
$\tau_{1}< \cdots <\tau_{l-1}.$ 
If $\Lambda(0)\cap\Pi=\Lambda(t)\cap\Pi=0$, then the intersection
number $\Lambda(\cdot)\cdot{\cal M}_\Pi$ of the curve $\Lambda(\cdot)$
with the hypersurface (train) ${\cal M}_\Pi=\{\Delta\in L(\Sigma) :
\Delta\cap\Pi\ne 0\}$ is well--defined and the identity
$\Lambda(\cdot)\cdot{\cal M}_\Pi={\rm ind}_{\Pi}\Lambda(\cdot)$
is valid.  
If the curve $\Lambda(\tau)$ is closed 
($\Lambda(0)=\Lambda(t)$), then 
${\rm ind}_{\Pi}\Lambda(\cdot)$ 
does not depend also on the choice of $\Pi $ and is denoted
$\ind\Lambda(\cdot)$ (cf. \cite{a88}).   



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\end{document}
