###### Proof.

###### Proof.

###### Functional determinants for the second variation

###### Abstract

We study the determinant of the second variation of an optimal control problem
for general boundary conditions. Generically, this operators are not trace
class and the determinant is defined as a principal value limit. We provide a
formula to compute this determinant in terms of the linearisation of the
extremal flow. We illustrate the procedure in some special cases, proving some
Hill-type formulas.

##  1 Introduction

The main focus of this paper is to study the spectrum of a particular class of
Fredholm operators that arise in the context of Optimal Control. Our main
result is a formula which relates the determinant of these operators to the
fundamental solutions of an ODE system in a finite dimensional space, much in
the spirit of Gelfand-Yaglom Theorem.

For operators of the form $1+K$, where $K$ is a compact operator, various ways
of defining a _determinant_ function can be found in the literature. Going all
the way back to Poincaré, Fredholm and Hilbert. If the operator $K$ one
considers is in the so called _trace class_ , i.e. the sequence of its
eigenvalues (with multiplicity) gives an absolutely convergent series, a
definition of determinant which involves the (infinite) product of its
eigenvalues is possible.

In our case, however, the classical approach is not immediately applicable
since, typically, the operators one encounters are not trace class. Under some
technical assumptions, the operators arising as second variation have a
symmetric spectrum, as shown in [3] and [9]. Generically, there is a non
negative real number $\xi(K)$, which we call _capacity_ , for which the
following asymptotic for the ordered sequence of the eigenvalues holds:

$\lambda_{n}(K)=\frac{\xi(K)}{n}+O(n^{5/3}),\quad n\in\mathbb{Z}.$ (1)

This symmetry allows us to talk about trace and determinant of the operators
$K$ and $1+K$ in the sense of _principal value_ limits. A similar approach has
been independently adopted in the works [19, 18] to study the spectrum of
Hamiltonian differential operators.

There are, of course, many other ways to define a _determinant function_ for
classes of Fredholm operators. For instance one could apply the theory of
regularized determinants, see [26], or rely on the so called
$\zeta-$regularization, see [14] for details. The literature concerning these
topics is vast. To mention a few works, one could refer to [24, 23] for some
results about Sturm-Liouville problems, to [16] for graphs and to [15] results
proved in the general framework of elliptic operators on section of vector
bundles. A relation between regularized determinants and
$\zeta$-regularization is given in [17].

Our approach, even if less general, provides an actual extension of the
definition of determinant given for trace class operators. It involves some
_principal value_ limit of the product of the eigenvalues. Thus, whenever the
compact part of the second variation is trace class, it gives exactly the
usual Fredholm determinant defined for trace class operators.

It is worth pointing out another feature of the construction. Our formula
relates the determinant of $1+K$ to the fundamental solution of a finite
dimensional system of (linear) ODEs. This provides a way to actually compute
the determinant and allows to recover some classical results such as Hill’s
formula for periodic trajectories. This kind of formulas have important
applications since allow to relate variational properties of an extremal (i.e.
the eigenvalues of the second variation), to dynamical properties such as
_stability_. These properties are usually expressed through the eigenvalues of
the linearisation of the Hamiltonian system of which the extremal we are
considering is solution. Take the case of a periodic, non degenerate
trajectory. On one hand, knowing the sign of the determinant amounts to know
the parity of the Morse index of the extremal. On the other hand this sign is
completely determined by the number of positive eigenvalues of the
linearisation grater than one. Applications of these kind of ideas go back to
Poincaré’s result about the instability of closed geodesics and can be found
for example in [12] or [20]. For several interesting examples of the interplay
between parity of Morse index and stability see for instance [10, 11] or [8].
Other related works in this direction are [25, 27] and [21] .

We stress that our results are formulated in a quite general framework which
encompasses Riemannian, sub-Riemannian, Finsler geometry and mechanical
systems on manifolds to name a few. Moreover the techniques can be applied
without virtually any modifications to treat constrained variational problems
on compact graphs as already done in [7] to compute the Morse index.

The structure of the paper is the following: in Section 1.1 we recall the
notation and the setting we will use through out the paper and give the full
statement of our results. Section 2 contains some information about the second
variation and the structure of the space of variation we will employ.

In Section 3 we deal with a couple of applications, such as Hill’s formula and
the eigenvalue problem for Schrödinger operators. The results deserve some
interest on their own, however, the main focus of the section is to provide a
worked out example of how to apply the formulas to concrete situations.

The last part of the paper is Sections 4.1 and 4.2. They are devoted to the
proof of 1. We first prove the result for boundary conditions of the type
$N_{0}\times N_{1}$ (Section 4.1) and than extend it to the general case
(Section 4.2). Finally we give formulas to compute the trace of the compact
part of the second variation $K$ (Lemmas 3 and 4) and some normalization
constant appearing in 1.

###  1.1 Problem statement and main results

We begin this section recalling briefly the setting and the notations that
will be used throughout the paper. The reader is referred to [6, 4] for more
information on optimal control and sub-Riemannian problems. By an optimal
control problem we mean the following data: a _configuration space_ , i.e.
some smooth manifold $M$ and a family of smooth (and complete if $M$ is non
compact) vector fields $f_{u}$. They depend on a parameter $u$ living in some
open set $U\subseteq\mathbb{R}^{k}$. We will always assume that the $f_{u}$
are smooth jointly in both variables, i.e. they belong to $C^{\infty}(U\times
M,TM)$. We can think of the parameter $u$ as our way of interacting with the
system and moving a particle from one state to another.

To any function $u(\cdot)\in L^{\infty}([0,1],U)$ we can associate a
trajectory in the configuration space considering the solution of:

$\dot{q}=f_{u(t)}(q),\quad q(0)=q_{0}\in M.$ (2)

We will usually call the function $u(\cdot)$ _control_.

We say that a control $u$ is admissible if the corresponding trajectory,
denoted by $\gamma_{u}$, is defined on the whole interval $[0,1]$. We can
impose further restrictions on the trajectory $\gamma_{u}$ specifying proper
boundary conditions. The most general situation that we are going to treat in
the paper, is the case in which the boundary conditions are given by a
submanifold $N\subseteq M\times M$. We say that (a Lipschitz continuous)
$\gamma$ is admissible if $\gamma=\gamma_{u}$ for some admissible control $u$
and if $(\gamma(0),\gamma(1))\in N$.

Given a smooth function $\varphi:U\times M\to\mathbb{R}$, we are interested in
the following minimization problem on the space of _admissible_ curves:

$\min_{\gamma_{u}\text{
admissible}}\mathcal{J}(\gamma_{u})=\min_{\gamma_{u}\text{
admissible}}\int_{0}^{1}\varphi(u(t),\gamma_{u}(t))dt$ (3)

It is customary to parametrized the space of admissible curves using the
_velocity_ , i.e. the control function $u(\cdot)$, and a finite dimensional
parameter space which takes into account initial data and boundary conditions.
We are going to follow this approach. However, this is just a technical point
which is independent of the main statements. Hence we postpone the discussion
of the structure of our space of variations to Section 2. Let us just mention
that, under some natural assumptions, the space of variations can be endowed,
locally, with a smooth Banach manifold structure. Thus, it makes sense to
consider the tangent space to the space of variations. It is a finite
codimension subspace $\mathcal{V}$ of
$L^{\infty}([0,1],\mathbb{R}^{k})\oplus\mathbb{R}^{\dim(N)}$. Suppose that $u$
is critical point of the functional $\mathcal{J}$ restricted to our space of
variations and consider the Hessian of $\mathcal{J}$. It is a quadratic form
on $\mathcal{V}$. We denote this quadratic form by
$Q(v)=d_{u}^{2}\mathcal{J}|_{\mathcal{V}}(v,v)$. Instead of working with
$L^{\infty}$ topology, we will work with the weaker $L^{2}$ one since
everything extends by continuity. For an appropriate choice of scalar product
on $L^{2}([0,1],\mathbb{R}^{k})$, it turns out that the quadratic form
$(Q-I)(v)=Q(v)-\langle v,v\rangle$ is compact, but in general not trace-class.
For a detailed account on the spectrum of the second variation the reader is
referred to [3, 9]. Given an eigenvalue of $Q$, $\lambda$, denote by
$m(\lambda)$ its multiplicity. We define the determinant of the second
variation as the following limit:

$\det(Q)=\lim_{\epsilon\to
0}\prod_{|\lambda-1|>\epsilon}\lambda^{m(\lambda)},\text{ where
}\lambda\in\text{Spec}(Q).$

As already stated in the introduction, the computation of this determinant for
general boundary conditions is the main contribution of this work. We provide
a formula for this determinant involving essentially two ingredients:

  * •

the fundamental solution of a linear (non autonomous) system of ODE’s which we
call _Jacobi_ equation;

  * •

the annihilator to the boundary condition manifold, a Lagrangian submanifold
of $T^{*}M$.

To state the main Theorem and define precisely the objects above, we need to
introduce a little bit of notation. We will just sketch here what is needed to
this purpose, further details are collected in A or given along the proofs.
From now on, assume that a strictly normal extremal $\lambda_{t}$ with optimal
control $\tilde{u}$ is fixed (see A).

The first tool we introduce is the following family of Hamiltonians. It is
strictly related to Legendre transform and quite useful when dealing with
problems in the cotangent bundle. Given our optimal control $\tilde{u}(\cdot)$
define:

$h^{t}_{\tilde{u}}(\lambda)=\langle\lambda,f_{\tilde{u}(t)}(q)\rangle-\varphi(\tilde{u}(t),q),\quad\dot{\lambda}=\vec{h}^{t}_{\tilde{u}}(\lambda).$

Denote its flow at time $t$ by $\tilde{\Phi}_{t}$ and its differential by
$(\tilde{\Phi}_{t})_{*}$. Pontryagin Maximum Principle (see A) tells us that
normal extremals $\lambda_{t}$ satisfy
$\lambda_{t}=\tilde{\Phi}_{t}(\lambda_{0})$. We will use the map
$\tilde{\Phi}_{t}$ to connect the tangent spaces to each point of
$\lambda_{t}$ to the starting one, $\lambda_{0}$. This flow, in some sense,
plays the role of the choice of a connection (or parallel transport as in
[12]).

The second object we are going to introduce, is a kind of quadratic
approximation of our starting system. It is given by a quadratic Hamiltonian
on $T_{\lambda_{0}}T^{*}M$ (see for detail [1] or [6][Chapters 20 and 21]). To
define it we need to introduce two matrix valued functions
$Z_{t}\in\mathrm{Mat}_{k\times 2\dim(M)}(\mathbb{R})$ and
$H_{t}\in\mathrm{Mat}_{k\times k}(\mathbb{R})$. The precise way to compute
them is given in (32). However, for the moment, a precise understanding of how
this matrices are obtained is not strictly needed. Heuristically, the matrix
$Z_{t}$ represents a linear approximation of the Endpoint map of the original
system whereas $H_{t}$ is a quadratic approximation of the Lagrangian
$\varphi$ along the extremal.

Let $\pi:T^{*}M\to M$ be the natural projection and set $\Pi:=\ker\pi_{*}$,
the _fibres_. Define $\delta^{s}$ as the dilation by $s\in\mathbb{R}$ of
$\Pi$. It is determined by the relations $\pi_{*}(\delta^{s}w)=\pi_{*}(w)$ for
all $w\in T_{\lambda_{0}}T^{*}M$ and $\delta^{s}v=sv$ for all
$v\in\ker(\pi_{*})$. Let $J$ be some coordinates representation of the
standard symplectic form on $T_{\lambda_{0}}T^{*}M$. Let us define the
following quadratic form:

$\beta_{s}(\lambda)=\frac{1}{2}\langle\lambda,J\delta^{s}Z_{t}H_{t}^{-1}(\delta^{s}Z_{t})^{*}J\lambda\rangle,\quad\lambda\in
T_{\lambda_{0}}T^{*}M.$

We will call _Jacobi (or Jacobi type) equation_ the following ODEs system on
$T_{\lambda_{0}}T^{*}M$:

$\dot{\lambda}=\vec{\beta}_{s}(\lambda),\quad\lambda\in
T_{\lambda_{0}}T^{*}M.$

and denote its fundamental solution at time $t$ as $\Phi_{t}^{s}$. Here, and
for the rest of the paper, we will call _fundamental solution_ any family
$\Theta_{t}$, $t\in\mathbb{R}$ of linear maps which satisfies a linear ODE and
have initial condition $\Theta_{0}=I$.

######  Remark 1.

Whenever PMP’s maximum condition determines a $C^{2}$ function $h$ on
$T^{*}M$, normal extremals satisfy a Hamiltonian ODE on the cotangent bundle
of the form $\dot{\lambda}=\vec{h}(\lambda)$. Jacobi equation for $s=1$ is
closely related to the linearisation of $\vec{h}$ along the extremal we are
fixing. Suppose local coordinates are fixed and let $d^{2}_{\lambda_{t}}h$ be
the Hessian of $h$ along the extremal. Let $\Psi_{t}$ be the fundamental
solution of:

$\dot{\Psi}_{t}=d^{2}_{\lambda_{t}}h\,\Psi_{t}.$

It can be shown (see for example [6][Proposition 21.3]) that:

$\Psi_{t}=(\tilde{\Phi}_{t})_{*}\Phi_{t}^{1}.$

The last maps we will need are a family of symplectomorphism of
$T_{\lambda_{0}}T^{*}M$ and $T_{\lambda_{1}}T^{*}M$. Their definition depends
on the choice of a scalar product on each tangent space. Let $g_{0}$ and
$g_{1}$ be two scalar products on these spaces. Assume that at each
$\lambda_{i}$, $\Pi_{i}:=\ker(\pi_{*})\subseteq T_{\lambda_{i}}T^{*}M$ has a
Lagrangian orthogonal complement with respect to $g_{i}$ which we denote by
$\Pi_{i}^{\perp}$. For a subspace $V$, denote by $pr_{V}$ the orthogonal
projection onto $V$. We set:

$\begin{split}A_{0}^{s}(\eta)&=\eta+(1-s)J_{0}^{-1}pr_{\Pi_{0}^{\perp}}\eta,\quad\eta\in
T_{\lambda_{0}}(T^{*}M),\\\
A_{1}^{s}(\eta)&=\eta+(1-s)(J_{1}^{-1}+\tilde{\Phi}_{*}\circ
pr_{\Pi_{0}}\circ\tilde{\Phi}_{*}^{-1})pr_{\Pi_{1}^{\perp}}\eta,\quad\eta\in
T_{\lambda_{1}}(T^{*}M).\end{split}$ (4)

The datum of the boundary condition is encoded in a Lagrangian submanifold of
$\big{(}T^{*}(M\times M),(-\sigma)\oplus\sigma\big{)}$, the annihilator of
$N$. It can be thought of as the symplectic version of the normal bundle in
Riemannian geometry and is define as follows. Take a sub-manifold $N\subseteq
M\times M$ and consider:

$Ann(N)=\bigcup_{q\in N}\\{(\lambda_{0},\lambda_{1})\in T_{q}^{*}(M\times
M):\langle\lambda_{0},X_{0}\rangle=\langle\lambda_{1},X_{1}\rangle,\forall(X_{0},X_{1})\in
T_{q}N\\}.$

In light of PMP (see A), critical points of $\mathcal{J}$ with boundary
conditions given by $N$, lift to the cotangent bundle to curves $\lambda_{t}$
such that $(\lambda_{0},\lambda_{1})\in Ann(N)$.

Fix now a complement to $T_{(\lambda_{0},\lambda_{1})}Ann(N)$, say $V_{N}$,
and denote by $\pi_{N}$ the projection on $V_{N}$ having
$T_{(\lambda_{0},\lambda_{1})}Ann(N)$ as kernel. We are ready now to define a
function that plays the role of the _characteristic polynomial_ of the Hessian
of $\mathcal{J}$. For a map $f$ denote by $\Gamma(f)$ its graph, set:

$\mathfrak{p}_{Q}(s)=\det(\pi_{N}|_{\Gamma(A_{1}^{s}\tilde{\Phi}_{*}\Phi^{s}_{1}A_{0}^{s})}).$

With this notation, our main result reads as follows:

######  Theorem 1.

Suppose that $\lambda_{t}$ is a strictly normal extremal for problem (3) and
$\tilde{u}$ is its optimal control. Moreover, suppose that $\lambda_{t}$
satisfies Legendre strong condition, i.e. that $\exists\alpha>0$ such that,
$\forall v\in\mathbb{R}^{k}$

$\langle-H_{t}v,v\rangle\geq\alpha\langle v,v\rangle$

and that at least one of the following holds:

  * •

the maps $t\mapsto Z_{t}$ and $t\mapsto H_{t}$ are piecewise analytic in $t$;

  * •

the dimension of the space of controls is $k\leq 2$;

  * •

the operator $I-Q$ is trace class;

Let $\lambda\in\mathrm{Spec}(Q)$ and $m(\lambda)$ be its multiplicity, the
limits:

$\det(Q)=\lim_{\epsilon\to
0}\prod_{|\lambda-1|>\epsilon}\lambda^{m(\lambda)},\quad\textrm{tr}(Q-I)=\lim_{\epsilon\to
0}\sum_{|\lambda-1|>\epsilon}m(\lambda)(\lambda-1).$

are well defined and finite. Moreover, for almost any choice of metrics
$g_{0}$ and $g_{1}$ we have that $\mathfrak{p}_{Q}(0)\neq 0$ and that:

$\det(I+s(Q-I))=\mathfrak{p}_{Q}(0)^{-1}e^{s(\textrm{tr}(Q-I)-\mathfrak{p}_{Q}^{\prime}(0))}\mathfrak{p}_{Q}(s).$

######  Remark 2.

The hypothesis about the regularity of $Z_{t}$ and $H_{t}$ are needed to
obtain the asymptotic for the spectrum of $Q-I$ that guarantees the existence
of the trace and of the determinant as limits. They can be weakened somehow by
requiring that the skew-symmetric $k\times k$ matrix $Z^{*}_{t}JZ_{t}$ is
continuously diagonalizable (see [3]).

######  Remark 3.

The constants $\mathfrak{p}_{Q}(0)$, $\mathfrak{p}_{Q}^{\prime}(0)$ and
$\textrm{tr}(Q-I)$ are completely explicit and are given in terms of iterated
integrals in Lemmas 4 and 3.

In particular we have the following corollary:

######  Corollary 1.

Under the assumption above, the determinant of the second variation $Q$
satisfies:

$\det(Q)=\mathfrak{p}_{Q}(0)^{-1}e^{\textrm{tr}(Q-I)-\mathfrak{p}_{Q}^{\prime}(0)}\det(\pi_{N}|_{\Gamma(\Psi_{t})}).$

Where $\Psi_{t}=\tilde{\Phi}_{*}\Phi_{1}$ and coincides with the fundamental
solution of the linearisation of the extremal flow, whenever the latter is
defined.

##  2 The Second Variation

The aim of this section is threefold: to define precisely what we mean by
$d^{2}\mathcal{J}|_{\mathcal{V}}$, to define precisely its domain and to
provide the integral representation of this quadratic form we will use
throughout the proof section of the paper.

Before going on, a little remark about topology is in order. Up until now we
have considered Lipschitz continuous curves and $L^{\infty}$ controls. Hence,
it would be natural to work on the Banach space
$L^{\infty}([0,1],\mathbb{R}^{k})\oplus\mathbb{R}^{\dim(N)}$. However it turns
out that, even if $d^{2}\mathcal{J}|_{\mathcal{V}}$ is defined on the latter
space, it extends to a continuous quadratic form on
$L^{2}([0,1],\mathbb{R}^{k})\oplus\mathbb{R}^{\dim(N)}$. Moreover, critical
points of $d^{2}\mathcal{J}|_{\mathcal{V}}$ in $L^{2}([0,1],\mathbb{R}^{k})$
are continuous and thus belong to $L^{\infty}([0,1],\mathbb{R}^{k})$. For this
reason (and Fredholm alternative), we will work with $L^{2}$ controls for the
rest of the paper.

Let $n_{0},n_{1}\in\mathbb{N}$ and consider the Hilbert space
$\mathcal{H}=\mathbb{R}^{n_{0}}\oplus
L^{2}([0,1],\mathbb{R}^{k})\oplus\mathbb{R}^{n_{1}}$ (its scalar product will
be defined in the next section). Let $(\Sigma,\sigma)$ be a symplectic space
and consider a linear map $Z:\mathcal{H}\to\Sigma$ defined as:

$Z(u)=Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1},\quad
u=(u_{0},u_{t},u_{1})\in\mathcal{H}.$

Suppose that $\Pi\subset\Sigma$ is a Lagrangian subspace transverse to the
image of the map $Z$ and define $\mathcal{V}=Z^{-1}(\Pi)$. For an appropriate
choice of $Z$ and $\Pi$ which depends on (2) and (3), the _second variation_
(at a strictly normal critical point) is the quadratic form given in the
following definition.

######  Definition 1 (Second Variation).

The second variation at $\tilde{u}$ is the quadratic form defined on
$\mathcal{V}\subseteq\mathcal{H}$:

$\displaystyle Q(u)=\int_{0}^{1}\left(-\langle
H_{t}u_{t},u_{t}\rangle+\sigma(Z_{t}u_{t},\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0})\right)dt+$
(5) $\displaystyle+\sigma(Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt,Z_{1}u_{1}).$

The definition of this quadratic form may be a bit strange at first glance.
Despite the appearances, the way one gets to such an expression is quite
natural. The construction is explained in detail in [7]. We will sketch here
just the main features, essentially to introduce the notation needed.

The idea is to reduce the problem with boundary conditions $N$ to a fixed
points (or Dirichlet) problem for an appropriate auxiliary system. We will
consider just the case of separated boundary conditions $N_{0}\times N_{1}$.
The general case reduces to this one using the procedure explained in Section
4.2.

The first step of the construction is to build the auxiliary system. We always
work with a fixed strictly normal extremal $\lambda_{t}$. Fix local foliations
in neighbourhoods of its initial and final points having a portion of $N_{0}$
and $N_{1}$ as leaves. This determines two integrable distributions in a
neighbourhood of those points. Suppose that said distributions are generated
by some fields $\\{X_{i}^{j}\\}_{i=1}^{\dim(N_{j})}$ and $j=0,1$. Consider the
extended system:

$f_{u}^{t}(q)=\begin{cases}\sum_{i=0}^{\dim(N_{0})}X_{i}^{0}(q)u_{0}^{i},&\text{
for }t\in[-1,0)\\\ f_{u}(q),&\text{ for }t\in[0,1]\\\
\sum_{i=0}^{\dim(N_{1})}X_{i}^{1}(q)u_{1}^{i},&\text{ for
}t\in(1,2].\end{cases}$

Denote the initial and final points of our original extremal curve by
$(q_{0},q_{1})$. We will use controls that are locally constant on
$[0,1]^{c}$, this will be enough to reach any neighbouring point of
$(q_{0},q_{1})$ in $N_{0}\times N_{1}$. Minimizing our original functional is
equivalent to minimize, with Dirichlet boundary conditions, the following one:

$\mathcal{J}(\gamma_{(u_{0},u,u_{1})})=\int_{0}^{1}\varphi(u(t),\gamma_{(u_{0},u,u_{1})}(t))dt.$

The second step is to differentiate the Endpoint map (see A) of the auxiliary
system. We employ the machinery of Chronological Calculus (see also
[6][Section 20.3]), which is standard for fixed endpoints. One of the main
steps of this differentiation, is to use a suitable family of
symplectomorphism to trivialize the cotangent bundle along the curve we are
fixing. This allows to write all the equations in the tangent space to the
initial point, $T_{\lambda_{0}}(T^{*}M)$. Let us consider the following
functions depending on the parameter $u$:

$h^{t}_{u}(\lambda)=\langle\lambda,f_{u}^{t}(q)\rangle-\varphi_{t}(u,q),\quad
q=\pi(\lambda).$

When an optimal control $\tilde{u}(t)$ is given, we consider
$h^{t}_{\tilde{u}(t)}(\lambda)$ and the Hamiltonian system:

$\dot{\tilde{\Phi}}_{t}=\vec{h}^{t}_{\tilde{u}(t)}(\tilde{\Phi}_{t}).$

We then define the following functions:

$\begin{split}Z_{t}&=\partial_{u}\vec{h}_{u}^{t}(\tilde{\Phi}_{t}(\lambda))|_{\lambda=\lambda_{0}},\quad
Z_{t}:\mathbb{R}^{k}\to T_{\lambda_{0}}(T^{*}M),\\\
H_{t}&=\partial_{u}^{2}h_{u}^{t}(\tilde{\Phi}_{t}(\lambda))|_{\lambda=\lambda_{0}},\quad
H_{t}:\mathbb{R}^{k}\to\mathbb{R}^{k}.\end{split}$

The asymptotic expansions of Chronological Calculus tell us that the second
variation at $\tilde{u}$ is the following quadratic form:

$Q(u)=\int_{-1}^{2}\langle-
H_{t}u,u\rangle-\int_{-1}^{2}\int_{-1}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}u_{t})d\tau
dt.$

It is defined on the tangent space $\mathcal{V}$ to the variations fixing the
endpoints of our curve. This space can be described explicitly as
$\mathcal{V}=\\{v:\int_{-1}^{2}Z_{t}v_{t}dt\in\Pi\\}$, where
$\Pi=\ker\pi_{*}$.

The third step is to specialize this representation to our auxiliary system.
Notice that an extremal of the original problem lifts naturally to an extremal
of the auxiliary one. If $\tilde{u}$ is the original optimal control,
extending it by zero on $[0,1]^{c}$ gives the optimal control for the
auxiliary problem. Applying the construction just sketched to the extended
system, we find that $Z_{t}$ and $H_{t}$ are locally constant on $[0,1]^{c}$.
We denote $Z_{0}$ to be its value on $[-1,0)$ and $Z_{1}$ the value on
$(1,2]$. $H_{t}$ is zero outside $[0,1]$. Thus, after substitution, we recover
precisely the operator given in (5).

######  Remark 4.

We always assume that our extremal is _strictly normal_ and satisfies
_Legendre strong condition_. In terms of the matrices $Z_{t}$ and $H_{t}$ this
means that for $t\in[0,1]$:

$X_{t}:=\pi_{*}Z_{t}\,\text{ satisfies
}\,\int_{0}^{1}X_{t}^{*}X_{t}>0,\quad\langle-H_{t}v,v\rangle\geq\alpha\langle
v,v\rangle>0,\,\forall v\in\mathbb{R}^{k}.$

As a last remark, notice that, by the first order optimality conditions, the
map $Z_{0}$ takes values in the space $T_{\lambda_{0}}Ann(N_{0})$ and the map
$Z_{1}$ in the space
$(\tilde{\Phi}^{-1}_{1})_{*}\left(T_{\lambda_{1}}Ann(N_{1})\right)$ ( see PMP
in A and [7]).

###  2.1 The scalar product on the space of variations

As already mentioned, we will assume through out this paper _Legendre strong
condition_. The matrix $-H_{t}$ is positive definite on $[0,1]$, with
uniformly bounded inverse. This allows to use $-H_{t}$ to define an Hilbert
structure on $L^{2}([0,1],\mathbb{R}^{k})$ equivalent to the standard one. We
have still to define the scalar product on a subspace transversal to
$\mathcal{V}_{0}=\\{u_{0}=u_{1}=0\\}$. A natural choice would be to introduce
two metrics on $T_{\lambda_{0}}T^{*}M$ and $T_{\lambda_{1}}T^{*}M$ and pull
them back to the space of controls using the maps $Z_{0}$ and
$\tilde{\Phi}_{*}Z_{1}:\mathbb{R}^{n}\to T_{\lambda_{i}}T^{*}M$. Let us call
any such metrics $g_{0}$ and $g_{1}$.

######  Definition 2.

For any $u,v\in\mathcal{H}$ define:

$\langle
u,v\rangle=-\int_{0}^{1}H_{t}(u_{t},v_{t})dt+g_{0}(Z_{0}u_{0},Z_{0}v_{0})+g_{1}(\tilde{\Phi}_{*}Z_{1}u_{1},\tilde{\Phi}_{*}Z_{1}v_{1})$

Since the symplectic form $\sigma$ is a skew-symmetric bilinear form, there
exists a $g_{i}-$skew-symmetric linear operator $J_{i}$ such that:

$g_{i}(J_{i}X_{1},X_{2})=\sigma(X_{1},X_{2}),\quad\forall X_{1},X_{2}\in
T_{\lambda_{i}}T^{*}M,\,i=0,1.$

In terms of the symplectic form the scalar product can be written as:

$\langle
u,v\rangle=-\int_{0}^{1}H_{t}(u_{t},v_{t})dt+\sigma(J_{0}^{-1}Z_{0}u_{0},Z_{0}u_{0})+\sigma(J_{1}^{-1}\tilde{\Phi}_{*}Z_{1}u_{1},\tilde{\Phi}_{*}Z_{1}u_{1})$

Now, we use the Hilbert structure just introduced to write the operator $K$
associated to the quadratic form $Q-I$, which is compact. To simplify
notation, we can perform the change of coordinates in $L^{2}$ sending
$v_{t}\mapsto(-H_{t})^{\frac{1}{2}}v_{t}$ and substitute $Z_{t}$ with
$Z_{t}(-H_{t})^{-\frac{1}{2}}$. In this way the Hilbert structure on the
interval becomes the standard one.

We introduce a further piece of notation, call $pr_{0}$ (respectively
$pr_{1}$) the orthogonal projection on $\textrm{Im}(Z_{0})$ (respectively
$\textrm{Im}(\tilde{\Phi}_{*}Z_{1})$) with respect to scalar product $g_{0}$
(respectively $g_{1}$). Let $L$ be a partial inverse to
$\tilde{\Phi}_{*}Z_{1}$ i.e. a map $L:T_{\lambda_{0}}T^{*}M\to\mathbb{R}^{n}$
defined by the relation $L\tilde{\Phi}_{*}Z_{1}v_{1}=v_{1}$. Set:

$\Lambda(u)=L\,pr_{1}\,J_{1}\tilde{\Phi}_{*}\Big{(}Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1}\Big{)}$
(6)

######  Lemma 1.

The second variation, as a bilinear form, can be expressed as: $Q(u,v)=\langle
u+Ku,v\rangle$ where $u,v\in\mathcal{V}$ and $K$ is the operator defined by:

$Ku=\begin{pmatrix}-u_{0}\\\
-Z_{t}^{*}J\Big{(}\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0}\Big{)}\\\
-u_{1}-\Lambda(u)\end{pmatrix}$ (7)

where $\Lambda(u)$ is given above, in eq. 6.

###### Proof.

A quick manipulation of the expression involving the symplectic form in 1,
yields the following:

$\displaystyle\int_{-1}^{2}\int_{-1}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}v_{t})d\tau
dt=\int_{-1}^{0}\int_{-1}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}v_{t})d\tau
dt+\int_{0}^{1}\int_{-1}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}v_{t})d\tau dt$
$\displaystyle+\int_{1}^{2}\int_{-1}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}v_{t})d\tau
dt$
$\displaystyle=\int_{0}^{1}\sigma\left(\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0},Z_{t}v_{t}\right)dt+\sigma\Big{(}Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1},Z_{1}v_{1}\Big{)}$
$\displaystyle=\int_{0}^{1}\sigma\left(\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0},Z_{t}v_{t}\right)dt+g_{1}\Big{(}J_{1}\tilde{\Phi}_{*}\Big{(}Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1}\Big{)},\tilde{\Phi}_{*}Z_{1}v_{1}\Big{)}$

Recall that $Z_{t}$ is constant on $[0,1]^{c}$. Moreover the images of the
maps $Z_{0}$ and $Z_{1}$ are isotropic subspaces. We used this fact to
simplify the expression in the first line. Now, it is clear that in the last
term:

$g_{1}\Big{(}J_{1}\tilde{\Phi}_{*}\Big{(}Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1}\Big{)},\tilde{\Phi}_{*}Z_{1}v_{1}\Big{)},$

only the projection onto the image of $\tilde{\Phi}_{*}Z_{1}$ plays a role. It
is straightforward to check that:

$g_{1}\Big{(}J_{1}\tilde{\Phi}_{*}\Big{(}Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1}\Big{)},\tilde{\Phi}_{*}Z_{1}v_{1}\Big{)}=g_{1}(\tilde{\Phi}_{*}Z_{1}\Lambda(u),\tilde{\Phi}_{*}Z_{1}v_{1}).$

Recall that we have normalized $H_{t}$ to $-1$, thus the first summand can be
rewritten as follows:

$\int_{0}^{1}\sigma\left(\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0},Z_{t}v_{t}\right)dt=\int_{0}^{1}\langle
Z_{t}^{*}J\Big{(}\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0}\Big{)},v_{t}\rangle
dt$

Adding and subtracting $g(Z_{0}u_{0},Z_{0}v_{0})$ and
$g(Z_{1}u_{1},Z_{1}v_{1})$ to single out the identity, we obtain the formula
in the statement. ∎

##  3 Hill-type formulas

Before going to the proof of 1 we present here some applications of the main
result. We deduce Hill’s formula for periodic trajectory and specify it to the
eigenvalue problem for Schrödinger operators. In the second sub-section we
present a variation of the classical Hill formula for systems with drift. We
will mainly deal with periodic and quasi-periodic boundary conditions. Namely,
we consider the case $N=\Gamma(f)$ for a diffeomorphism of the state space
$f:M\to M$.

The proofs of this section rely quite heavily on the machinery introduce in
Section 4.2, in particular in Lemmas 4 and 3. The statements, on the other
hand, do not and could shed some light on 1. Despite the appearance, proofs
are rather simple. They reduce to a (long) computation of the normalizing
factors appearing in the statement of 1 and can be skipped at first reading.

###  3.1 Driftless systems and classical Hill’s formula

In this section we consider driftless systems with periodic boundary
conditions on $\mathbb{R}^{n}$ and specify the formulas of 4 for this class of
problems.

First of all let us explain what we mean by _driftless_ systems. Let $t\mapsto
R_{t}$ a continuous family of symmetric matrices of size $n\times n$ and let
us denote by $u$ a function in $L^{\infty}([0,1],\mathbb{R}^{n})$.

Consider the following family of vector fields $f_{u}(q)$, their associated
trajectories $q_{u}(t)$ and the action functional $\mathcal{A}(u)$:

$\begin{split}f_{u}(q)=u(t),\quad\begin{cases}\dot{q}_{u}=f_{u}(q)=u(t),\\\
q(0)=q_{0}\in\mathbb{R}^{n}\end{cases}\\\
\mathcal{A}(u)=\frac{1}{2}\int_{0}^{1}|u|^{2}-\langle
R_{t}q_{u}(t),q_{u}(t)\rangle dt.\end{split}$ (8)

We impose periodic boundary conditions, i.e. we take
$N=\Delta=\\{(q,q)\in\mathbb{R}^{2n}:q\in\mathbb{R}^{n}\\}$. The Hamiltonian
coming from the Maximum Principle takes the form:

$H(p,q)=\max_{u\in\mathbb{R}^{k}}\langle
p,u\rangle-\frac{1}{2}(|u|^{2}-\langle R_{t}q,q\rangle)=\frac{1}{2}(\langle
p,p\rangle+\langle R_{t}q,q\rangle).$ (9)

Let us denote the flow generated by $H$ by $\Psi_{t}$. Fix a normal extremal
$\lambda_{t}$ for periodic boundary conditions and its optimal control
$\tilde{u}(t)$.

######  Theorem 2 (Hill’s formula).

Let $I+K$ be the second variation at $\tilde{u}$ and $\Psi$ the fundamental
solution of $\dot{\Psi}=\vec{H}\Psi$. The following equality holds:

$\det(I+K)=(-1)^{n}(2e)^{-n}\det(G^{-1})\det(1-\Psi)$

where $G$ is a scalar product on the tangent space to the initial point.

######  Remark 5.

If we are working on the interval $[0,T]$ instead of $[0,1]$ everything
remains essentially unchanged. The only difference is that extra factor
$T^{-n}$ appears in the right hand side. In the notation of the proof below
this corresponds to $\det(\Gamma)^{-1}$.

We can apply the previous result to study boundary value problems for Sturm-
Liouville operators. Let us illustrate the case of Schrödinger equation with
periodic boundary conditions. Fix, without loss of generality, the normal
extremal $(p(t),q(t))=(0,0)$ and with relative optimal control $\tilde{u}=0$.
Consider the cost $\tilde{R}_{t}=R_{t}+\lambda$, for $\lambda\in\mathbb{R}$.
Consider the second variation of the functional

$\mathcal{A}_{\lambda}(u)=\frac{1}{2}\int_{0}^{1}|u_{t}|^{2}+\langle(R_{t}+\lambda)q_{u}(t),q_{u}(t)\rangle
dt$

at the point $\tilde{u}=0$. It is given by the operator $1+K_{\lambda}$ where:

$\langle
K_{\lambda}(u),u\rangle=\lambda\left(\int_{0}^{1}\int_{0}^{t}\langle(\tau-t)u(\tau),u(t)\rangle
d\tau dt+\langle u_{0},u_{0}\rangle\right)+\langle K_{0}(u),u\rangle$

We have the following corollary:

######  Corollary 2.

Let $\lambda\in\mathbb{R}$, $\Psi_{\lambda}$ the fundamental matrix of the
lift to $\mathbb{R}^{2n}$ of the following $ODE$ on $\mathbb{R}^{2n}$:

$\ddot{q}(t)=(R_{t}+\lambda)q(t)$

The determinant of the operators $1+K_{\lambda}$ can be expressed as:

$\det(1+K_{\lambda})=(-1)^{n}(2e)^{-n}\det(G^{-1})\det(1-\Psi_{\lambda}),$

where $G$ as in the previous statement.

###### Proof of 2.

We are going now to describe explicitly all the objects involved in 1. Let us
start with the flow $\tilde{\Phi}$ we use to re-parametrize the space and its
differential. It is given by the Hamiltonian:

$\begin{split}&h_{\tilde{u}(t)}(p,q)=\langle
p,\tilde{u}(t)\rangle+\frac{1}{2}\langle
R_{t}q,q\rangle\quad\Rightarrow\quad\begin{cases}\dot{p}=-R_{t}q\\\
\dot{q}=\tilde{u}(t).\end{cases}\\\
&\tilde{\Phi}_{t}(p,q)=\begin{pmatrix}1&-\int_{0}^{t}R_{\tau}d\tau\\\
0&1\end{pmatrix}\begin{pmatrix}p\\\
q\end{pmatrix}+\begin{pmatrix}-\int_{0}^{t}\int_{0}^{\tau}R_{\tau}\tilde{u}(r)drd\tau\\\
\int_{0}^{t}\tilde{u}(\tau)d\tau\end{pmatrix}\\\
&(\tilde{\Phi}_{t})_{*}=\begin{pmatrix}1&-\int_{0}^{t}R_{\tau}d\tau\\\
0&1\end{pmatrix}.\end{split}$

The matrix $Z_{t}$ is the following:

$\quad
Z_{t}=(\tilde{\Phi}_{t}^{-1})_{*}\partial_{u}\vec{h}^{t}_{u}=\begin{pmatrix}\int_{0}^{t}R_{\tau}d\tau\\\
1\end{pmatrix}.$

To simplify notation, let us call $\hat{R}=-\int_{0}^{1}R_{\tau}d\tau$. The
annihilator of the diagonal is simply the graph of the identity. Hence the
following map, defined on $(T_{\lambda_{0}}T^{*}M)^{2}$, has the latter as
kernel:

$(\eta_{0},\eta_{1})\mapsto\eta_{1}-\eta_{0}.$

We will now define $Q^{s}$ as in eq. 28 (actually up to a scalar, but this is
irrelevant). For $\eta\in T_{\lambda_{0}}T^{*}M$ set:

$Q^{s}(\eta)=\begin{pmatrix}-1&1\end{pmatrix}\begin{pmatrix}\eta\\\
A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}\eta\end{pmatrix}=(A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}-1)\eta\in
T_{\lambda_{0}}T^{*}M.$

It is clear that the kernel of $Q^{s}$ is precisely the intersection of the
graph with the diagonal subspace. Since we are working on $\mathbb{R}^{2n}$ we
can define the determinant of this map as:

$\det(Q^{s})=\det(A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}-1)$

As already mentioned in Section 1.1 this function is a multiple of the
_characteristic polynomial_ of $K$. It satisfies (see sections 4.1 and 4.2):

$\det(Q^{s})=ae^{bs}\det(1+sK),\quad a\in\mathbb{C}^{*},b\in\mathbb{C}$

Let us compute the normalization factors. To do so, we have to evaluate
$\det(Q^{s})$ and its derivative at $s=0$. This will give us the relations:

$\det(Q^{0})=a,\quad a(b+\textrm{tr}(K))=\partial_{s}\det(Q^{s})|_{s=0}.$

We have to work a bit to write down precisely all the quantities appearing in
the formulas above. It is straightforward to compute the matrix
representations of the maps $A_{0}^{s}$ and $A_{1}^{s}$. In this setting the
projections onto $\Pi_{0}$ and $\Pi_{0}^{\perp}$ are given by:

$pr_{\Pi_{0}}=\begin{pmatrix}1&0\\\ 0&0\end{pmatrix},\quad
pr_{\Pi_{0}^{\perp}}=\begin{pmatrix}0&0\\\ 0&1\end{pmatrix}$

Recall that the definition of $A_{0}^{s}$ and $A_{1}^{s}$ given in (4) depends
on the choice of two scalar products $g_{0}$ and $g_{1}$. Denote by $G_{0}$
and $G_{1}$ their restriction to $\Pi_{0}^{\perp}$ and $\Pi_{1}^{\perp}$
respectively. We have that:

$\begin{split}A_{0}^{s}=\begin{pmatrix}1&(1-s)G_{0}\\\ 0&1\end{pmatrix},\quad
A_{1}^{s}=\begin{pmatrix}1&(1-s)(G_{1}-\hat{R})\\\ 0&1\end{pmatrix}\\\
A_{1}^{s}\tilde{\Phi}_{*}=\begin{pmatrix}1&s\hat{R}+(1-s)G_{1}\\\
0&1\end{pmatrix}\end{split}$

The value of $\Phi_{t}^{s}$ and its derivative at $s=0$ is given in Lemma 4.
Here, for the submatrices of $\Phi^{0}_{t}$ and
$\partial_{s}\Phi_{t}^{s}|_{s=0}$, we use the notation defined in the Lemma 4.
In this case, since $Y_{t}=\int_{0}^{t}R_{\tau}d\tau$ and $X_{t}=1$, we
obtain:

$\begin{split}\Phi_{t}^{0}&=\begin{pmatrix}1&0\\\
\Gamma&1\end{pmatrix}=\begin{pmatrix}1&0\\\ t&1\end{pmatrix}\\\
\partial_{s}\Phi_{t}^{s}|_{s=0}&=\begin{pmatrix}\Theta&0\\\
\Omega&-\Theta^{*}\end{pmatrix}=\begin{pmatrix}\int_{0}^{t}\int_{0}^{\tau}R_{\omega}d\omega&0\\\
\int_{0}^{t}\int_{0}^{\tau}\int_{r}^{\tau}R_{\omega}d\omega
drd\tau&-\int_{0}^{t}\int_{0}^{\tau}R_{\omega}d\omega\end{pmatrix}\end{split}$

Let us compute the value of $\det(Q^{s})$ in zero. Putting all together we
have:

$\det(Q^{s})|_{s=0}=\det\left(\begin{pmatrix}1&G_{1}\\\
0&1\end{pmatrix}\begin{pmatrix}1&0\\\
\Gamma&1\end{pmatrix}\begin{pmatrix}1&G_{0}\\\
0&1\end{pmatrix}-\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\right)$

After a little bit of computation we find that $Q_{s}|_{s=0}$ satisfies:

$\begin{split}Q^{s}|_{s=0}=\begin{pmatrix}G_{1}\Gamma&G_{0}+G_{1}+G_{1}\Gamma
G_{0}\\\ \Gamma&\Gamma
G_{0}\end{pmatrix},\quad\det(Q^{s}|_{s=0})=(-1)^{n}\det(\Gamma)\det(G_{1}+G_{0}),\\\
(Q^{s}|_{s=0})^{-1}=\begin{pmatrix}-G_{0}(G_{1}+G_{0})^{-1}&\Gamma^{-1}+G_{0}(G_{1}+G_{0})^{-1}G_{1}\\\
(G_{1}+G_{0})^{-1}&-(G_{1}+G_{0})^{-1}G_{1}\end{pmatrix}\end{split}$

We can compute the derivative $\det(Q^{s})$ at $s=0$, we find that:

$\begin{split}\partial_{s}Q^{s}|_{s=0}&=(\partial_{s}A_{1}^{s})\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}+A_{1}^{s}\tilde{\Phi}_{*}(\partial_{s}\Phi_{1}^{s})A_{0}^{s}+A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}(\partial_{s}A_{0}^{s})|_{s=0}\\\
&=\begin{pmatrix}(G_{1}-\hat{R})\Gamma&G_{1}-\hat{R}\\\
0&0\end{pmatrix}\begin{pmatrix}1&G_{0}\\\ 0&1\end{pmatrix}+\\\
&\quad+\begin{pmatrix}1&G_{1}\\\
0&1\end{pmatrix}\left(\begin{pmatrix}\Theta&0\\\
\Omega&-\Theta^{*}\end{pmatrix}\begin{pmatrix}1&G_{0}\\\
0&1\end{pmatrix}+\begin{pmatrix}0&G_{0}\\\ 0&\Gamma
G_{0}\end{pmatrix}\right)\end{split}$

We use now Jacobi formula for the derivative of the determinant of a family of
invertible matrices. It reads:

$\partial_{s}\det(M_{s})=\det(M_{s})\textrm{tr}(\partial_{s}M_{s}M_{s}^{-1}).$

Without going into the detail of the actual computation, which at this point
is just matrix multiplication, we have that:

$\begin{split}\partial_{s}\det(Q^{s})|_{s=0}&=\det(Q^{s})|_{s=0}\,\textrm{tr}(\partial_{s}(Q^{s})(Q^{s})^{-1})|_{s=0}\\\
&=\det(Q^{s})|_{s=0}\,\textrm{tr}((G_{1}+G_{0})^{-1}(G_{0}-G_{1}+\hat{R})+\Gamma^{-1}\Omega)\end{split}$

The last quantity we have to compute is $\textrm{tr}(K)$. To do so we use
Lemma 3. Mind that in the statement of the Lemma one works with twice the
variables, taking as state space $\mathbb{R}^{n}\times\mathbb{R}^{n}$ and
using the symplectic form $(-\sigma)\oplus\sigma$ on
$\mathbb{R}^{2n}\times\mathbb{R}^{2n}$. The quantities with $\tilde{\,}$ on
top always refer to the system in $\mathbb{R}^{4n}$, where we have a trivial
dynamic on the first factor and the boundary condition we impose are in this
case $\Delta\times\Delta$ (see the beginning of Section 4.2 for more details).
The formula given in the lemma reads:

$\begin{split}\textrm{tr}(K)=&-\dim(N)+\textrm{tr}[\pi_{*}^{1}\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\tilde{Z}_{0})]\\\
&+\textrm{tr}\left[\Gamma^{-1}\left(\Omega+(\pi^{2}_{*}-\pi_{*}^{1})\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J|_{\Pi}dt)\right)\right]\end{split}$
(10)

Let us explain all the objects appearing in the formula. $\tilde{\pi}_{*}^{i}$
denotes the differential of the natural projection on the $i-$th factor. The
matrix $\tilde{\Phi}_{*}$ is given here by:

$\tilde{\Phi}_{*}=\begin{pmatrix}\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}&0\\\
0&\begin{pmatrix}1&R\\\ 0&1\end{pmatrix}\end{pmatrix}$

Moreover the matrices $\tilde{Z}_{0}$, $\tilde{Z}_{t}$ and $\tilde{Z}_{1}$
are:

$\tilde{Z}_{0}=\begin{pmatrix}0\\\ 1\\\ 0\\\
1\end{pmatrix},\quad\tilde{Z}_{t}=\begin{pmatrix}0\\\ 0\\\
\int_{0}^{t}R_{\tau}d\tau\\\
1\end{pmatrix},\quad\tilde{Z}_{1}=\begin{pmatrix}0\\\ 1\\\ -R\\\
1\end{pmatrix},\quad\tilde{\Phi}_{*}\tilde{Z}_{1}=\begin{pmatrix}0\\\ 1\\\
0\\\ 1\end{pmatrix}$

The map $pr_{1}$ denotes the orthogonal projection onto the image of
$\tilde{Z}_{1}$. We are using the scalar product $g_{0}\oplus g_{1}$ on
$T_{\lambda_{0}}T^{*}M\times T_{\lambda_{1}}T^{*}M$ to define it. One can
check that the following map is the coordinate representation of $pr_{1}$:

$\displaystyle pr_{1}=(G_{0}+G_{1})^{-1}\begin{pmatrix}0&0&0&0\\\
0&G_{0}&0&G_{1}\\\ 0&0&0&0\\\ 0&G_{0}&0&G_{1}\\\ \end{pmatrix}$ $\displaystyle
pr_{1}\tilde{J}_{1}=(G_{0}+G_{1})^{-1}\begin{pmatrix}0&0&0&0\\\ -1&0&1&0\\\
0&0&0&0\\\ -1&0&1&0\\\ \end{pmatrix}$

Now everything reduces to some tedious matrix multiplications. The second term
in the right hand side of (10) simplifies as follows:

$\textrm{tr}[\pi_{*}^{1}\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\tilde{Z}_{0})]=\textrm{tr}(\hat{R}(G_{0}+G_{1})^{-1}).$

For the third term notice that
$(\pi_{*}^{2}-\pi_{*}^{1})(\tilde{\Phi}_{*})^{-1}pr_{1}$ is identically zero
since $\tilde{\Phi}_{*}$ does not change the projection on the horizontal part
and we are working with periodic boundary conditions. It follows we are left
with $\textrm{tr}(\Gamma^{-1}\Omega)$. Summing up we have that:

$\textrm{tr}(K)=\textrm{tr}(\Omega\Gamma^{-1}+\hat{R}(G_{0}+G_{1})^{-1}){-\dim(M).}$

It is natural to think of $g_{0}$ and $g_{1}$ as restrictions of globally
defined Riemannian metrics. Doing so, since we are working with periodic
boundary conditions, amounts to choose $G_{0}=G_{1}$. Hence the result. ∎

###  3.2 System with drift and Hill-type formulas

In this section we give a version of Hill’s formula for linear systems with
drift. They are again linear system with quadratic cost of the following form:

$\begin{split}f_{u}(q)&=A_{t}q+B_{t}u(t),\quad\begin{cases}\dot{q}_{u}=f_{u}(q)=A_{t}q_{u}+B_{t}u(t),\\\
q(0)=q_{0}\in\mathbb{R}^{n}\end{cases}\\\
\mathcal{A}(u)&=\frac{1}{2}\int_{0}^{1}|u|^{2}+\langle
R_{t}q_{u}(t),q_{u}(t)\rangle dt.\end{split}$ (11)

Here $A_{t}$ is $n\times n$ matrix and $B_{t}$ a $n\times k$ one, both with
possibly non-constant (but continuous) coefficients. The maximized Hamiltonian
takes the form:

$H(p,q)=\langle p,A_{t}q\rangle+\frac{1}{2}\left(\langle
B_{t}B^{*}_{t}p,p\rangle+\langle R_{t}q,q\rangle\right).$

Denote by $\hat{\Phi}_{t}$ the fundamental solution of $\dot{q}=A_{t}q$ at
time $t$. We can lift this map to a symplectomorphism of the cotangent bundle
which we denote by $\underline{\Phi}_{t}$. As boundary conditions we take the
following affine subspace of $\mathbb{R}^{4n}$:

$N=\Gamma(\hat{\Phi}+\hat{\Phi}\int_{0}^{1}\hat{\Phi}_{r}^{-1}B_{r}\tilde{u}(r)dr).$

Notice that, since only the tangent space matters in our formulas, the
translation is irrelevant and it would be the same as if we considered
$\Gamma(\hat{\Phi})$. An obvious choice of extremal is the point
$\lambda_{t}=(0,0)$, with control $\tilde{u}=0$.

######  Theorem 3 (Hill’s formula with drift).

Suppose that a critical point of the functional given in (11) is fixed and let
$\tilde{u}$ be its optimal control. Let $G_{0}$ and $G_{1}$ be our choices of
scalar product,
$\Gamma:=\int_{0}^{1}\hat{\Phi}_{\tau}B_{\tau}B_{\tau}^{*}\hat{\Phi}^{*}_{\tau}d\tau$
and $\Psi_{t}$ the fundamental solution of $\dot{\Psi}=\vec{H}\Psi$. Moreover,
define the following matrix:

$G=G_{0}+\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}.$

Let $I+K$ be the second variation at $\tilde{u}$. The following equality
holds:

$\det(I+K)=\frac{(-1)^{n}e^{-2\,\textrm{tr}(G^{-1}G_{0})}}{\det(G)\det(\Gamma)}\det(\Psi_{t}-\underline{\Phi}_{t}).$

###### Proof.

The proof is completely analogous to the one of 2. First of all the
Hamiltonian we use to re-parametrize is given as follows:

$h^{t}_{\tilde{u}(t)}=\langle
p,B_{t}\tilde{u}(t)+A_{t}q\rangle+\frac{1}{2}\langle R_{t}q,q\rangle.$

Hence the flow and its differential are given by:

$\begin{split}\tilde{\Phi}_{t}(p,q)=&\begin{pmatrix}(\hat{\Phi}_{t}^{*})^{-1}&-(\hat{\Phi}_{t}^{*})^{-1}\int_{0}^{t}\hat{\Phi}_{\tau}^{*}R_{\tau}\hat{\Phi}_{\tau}d\tau\\\
0&\hat{\Phi}_{t}\end{pmatrix}\begin{pmatrix}p\\\ q\end{pmatrix}\\\
&+\begin{pmatrix}-(\hat{\Phi}_{t}^{*})^{-1}\int_{0}^{t}\hat{\Phi}_{\tau}^{*}R_{\tau}\hat{\Phi}_{\tau}\int_{0}^{\tau}\hat{\Phi}_{r}^{-1}B_{r}\tilde{u}(r)drd\tau\\\
\hat{\Phi}_{t}\int_{0}^{t}\hat{\Phi}_{r}^{-1}B_{r}\tilde{u}(r)dr\end{pmatrix}\\\
(\tilde{\Phi}_{t})_{*}=&\begin{pmatrix}(\hat{\Phi}_{t}^{*})^{-1}_{*}&-(\hat{\Phi}_{t}^{*})^{-1}\int_{0}^{t}\hat{\Phi}_{\tau}^{*}R_{\tau}\hat{\Phi}_{\tau}d\tau\\\
0&\hat{\Phi}_{t}\end{pmatrix}\end{split}$

We have to define the map $Q^{s}$. Similarly as the previous case, we can
define a map having as kernel the annihilator to the boundary conditions as
follows:

$(\eta_{0},\eta_{1})\mapsto\begin{pmatrix}\hat{\Phi}^{*}_{t}&0\\\
0&(\hat{\Phi}_{t})^{-1}\end{pmatrix}\eta_{1}-\eta_{0}=:T_{1}\eta_{1}-\eta_{0}$

Hence:

$\det(Q_{s})=\det(T_{1}A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}-1).$

Set as before
$\hat{R}=-(\hat{\Phi}^{*})^{-1}\int_{0}^{1}\hat{\Phi}_{\tau}^{*}R_{\tau}\hat{\Phi}_{\tau}d\tau$,
the upper right minor of $\tilde{\Phi}_{*}$. A quick computation shows that:

$T_{1}A_{1}^{s}\tilde{\Phi}_{*}=\begin{pmatrix}1&(1-s)\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}+s\hat{\Phi}_{t}^{*}\hat{R}\\\
0&1\end{pmatrix}:=\hat{A}^{s}_{1}\tilde{\Phi}_{*}$

Hence, up to renaming $G_{1}$ and $\Gamma$ in the proof of 2, we have:

$\begin{split}\det(Q^{s})|_{s=0}&=(-1)^{n}\det(\Gamma)\det(\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}+G_{0})=(-1)^{n}\det(\Gamma)\det(G),\\\
\frac{\partial_{s}\det(Q^{s})}{\det(Q^{s})}|_{s=0}&=\textrm{tr}(((\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}+G_{0})^{-1}(G_{0}-\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}+\hat{\Phi}_{t}^{*}\hat{R})+\Gamma^{-1}\Omega)\end{split}$

Now we have to apply Lemma 3 to compute the trace of the compact part of the
second variation. Here $pr_{1}$ and $\tilde{Z}_{1}$ are different since we
have changed boundary conditions. However we have the same kind of
simplification as in the previous case. Let us write explicitly the new
objects:

$Z_{1}=\begin{pmatrix}0\\\ 1\\\ 0\\\ \hat{\Phi}_{t}\end{pmatrix},\quad
pr_{1}=\begin{pmatrix}0&0&0&0\\\ 0&LG_{0}&0&L\hat{\Phi}_{t}^{*}G_{1}\\\
0&0&0&0\\\
0&\hat{\Phi}_{t}LG_{0}&0&\hat{\Phi}_{t}L\hat{\Phi}_{t}^{*}G_{1}\end{pmatrix},\quad
L=(G_{0}+\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t})^{-1}$

In the end, the trace reads:

$\textrm{tr}(K)=\textrm{tr}(\Gamma^{-1}\Omega+\hat{\Phi}_{t}^{*}\hat{R}(\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}+G_{0})^{-1})-\dim(M).$

Contrary to the previous section the two term do not simplify. We are left
with the following equation for $b$:

$\displaystyle b$
$\displaystyle=\dim(M)+\textrm{tr}((\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}+G_{0})^{-1}(G_{0}-\hat{\Phi}_{t}^{*}G_{1}\hat{\Phi}_{t}))$
$\displaystyle=n+2\,\textrm{tr}(G^{-1}G_{0})-n=2\,\textrm{tr}(G^{-1}G_{0}).$

Hence, the statement follows evaluating $\det(Q^{s})$ at $s=1$:

$\begin{split}\det(Q^{s})&=\det(T_{1}\tilde{\Phi}_{*}\Phi_{1}^{1}-1)=\det(\underline{\Phi}_{t}^{-1}\tilde{\Phi}_{*}\Phi_{1}^{1}-1)\\\
&=\det(\Psi_{t}-\underline{\Phi}_{t}).\end{split}$

∎

##  4 Proof of the main Theorem

In this section we provide a proof of 1. First we work with separated boundary
conditions and then reduce the general case to the former. The proof is a bit
long so we try to give here a concise outline. The idea is to construct an
analytic function $f$ which vanishes precisely on the set
$\\{-1/\lambda:\lambda\in\text{Spec}(K)\\}\subseteq\mathbb{R}$. Particular
care is needed to show that the multiplicity of the zeros of this function
equals the multiplicity of the eigenvalues of $K$. We do this in 1 and 2
respectively. We show that this function decays exponentially and use a
classical factorization Theorem by Hadamard to represent it as

$f(s)=as^{k}e^{bs}\prod_{\lambda\in\text{Spec}(K)}(1+\lambda
s)^{m(\lambda)},\quad a,b\in\mathbb{C},a\neq 0,k\in\mathbb{N}.$

To prove the general case, we double the variables and consider general
boundary conditions as separated ones. In this framework we compute the value
of the parameters $a,b$ and $k$ appearing in the factorization.

###  4.1 Separated boundary conditions

We briefly recall the notation. We are working with an extremal $\lambda_{t}$
with initial and final point $(\lambda_{0},\lambda_{1})\in Ann(N)$, where
$N=N_{0}\times N_{1}$ are product boundary conditions. We are assuming that
$\lambda_{t}$ is strictly normal and satisfies Legendre strong conditions. We
work in a fixed tangent space, namely $T_{\lambda_{0}}T^{*}M$, to do so we
backtrack our curve to its starting point $\lambda_{0}$ using the flow
generated by the time dependent Hamiltonian:

$h^{t}_{\tilde{u}}(\lambda)=\langle\lambda,f_{\tilde{u}(t)}(q)\rangle-\varphi_{t}(q,\tilde{u}(t)).$

We denote the differential of said flow by $\tilde{\Phi}_{*}$. We have a
scalar product $g_{i}$ on $T_{\lambda_{i}}T^{*}M$, for each $i=0,1$. We assume
that the orthogonal complement to the fibre at $\lambda_{i}$,
$\Pi_{i}=T_{\lambda_{i}}T^{*}_{\pi(\lambda_{i})}M$ is a Lagrangian subspace
and that the range of $Z_{0}$ (and $\tilde{\Phi}_{*}Z_{1}$ respectively) is
contained in $\Pi_{0}^{\perp}$ (resp. $\Pi_{1}^{\perp}$).

######  Remark 6.

If we fix Darboux (i.e. _canonical_) coordinates coming from the splitting
$\Pi_{i}\oplus\Pi_{i}^{\perp}$ it is straightforward to check that $g_{i}$
takes a block diagonal form with symmetric $n\times n$ matrix $G^{j}_{i}$ on
the main diagonal. Similarly we can write down the coordinate representation
of the matrix $J_{i}$ and find:

$g_{i}(X,Y)=\left\langle\begin{pmatrix}G^{1}_{i}&0\\\
0&G^{2}_{i}\end{pmatrix}\begin{pmatrix}X_{1}\\\
X_{2}\end{pmatrix},\begin{pmatrix}Y_{1}\\\
Y_{2}\end{pmatrix}\right\rangle,\quad
J_{i}=\begin{pmatrix}0&-(G^{1}_{i})^{-1}\\\ (G^{2}_{i})^{-1}&0\end{pmatrix}.$

For $s\in\mathbb{R}$ (or $\mathbb{C}$) we introduce the following symplectic
maps:

$\begin{split}A_{0}^{s}(\eta)&=\eta+(1-s)J_{0}^{-1}pr_{\Pi_{0}^{\perp}}\eta,\quad\eta\in
T_{\lambda_{0}}(T^{*}M),\\\
A_{1}^{s}(\eta)&=\eta+(1-s)(J_{1}^{-1}+\tilde{\Phi}_{*}\circ
pr_{\Pi_{0}}\circ\tilde{\Phi}_{*}^{-1})pr_{\Pi_{1}^{\perp}}\eta,\quad\eta\in
T_{\lambda_{1}}(T^{*}M).\end{split}$ (12)

Notice that the transformation $A_{i}^{s}$ are indeed symplectomorphisms. In
(canonical) coordinates given by $\Pi_{i}$ and $\Pi_{i}^{\perp}$ they have the
following matrix representation:

$A_{i}^{s}=\begin{pmatrix}1&&(1-s)S_{i}\\\ 0&&1\end{pmatrix},\quad
S_{i}^{*}=S_{i}.$

The last map we are going introduce is two families of dilation in
$T_{\lambda_{0}}(T^{*}M)$, one of the vertical subspace and one of its
orthogonal complement. Let $s\in\mathbb{R}$ (or $\mathbb{C}$) and let us
define the following maps:

$\begin{split}\delta^{s}:T_{\lambda_{0}}T^{*}M\to
T_{\lambda_{0}}T^{*}M,\,\quad\delta^{s}\nu=s\,pr_{\Pi_{0}}\nu+pr_{\Pi_{0}^{\perp}}\nu\\\
\delta_{s}:T_{\lambda_{0}}T^{*}M\to
T_{\lambda_{0}}T^{*}M,\,\quad\delta_{s}\nu=pr_{\Pi_{0}}\nu+s\,pr_{\Pi_{0}^{\perp}}\nu\\\
\end{split}$ (13)

######  Proposition 1.

Let $A^{s}_{i}$ be the maps given in eq. 12 and let $\Phi_{1}^{s}$ be the
fundamental solution of the system:

$\dot{\eta}=Z_{t}^{s}(Z_{t}^{s})^{*}J\eta,\quad Z_{t}^{s}=\delta^{s}Z_{t},$

The operator $I+sK$ restricted to $\mathcal{V}$ has non trivial kernel if and
only if there exists a non zero $(\eta_{0},\eta_{1})\in
T_{(\lambda_{0},\lambda_{1})}(Ann(N))$ such that

$A_{1}^{s}\circ\tilde{\Phi}_{*}\circ\Phi_{1}^{s}\circ
A_{0}^{s}\,\eta_{0}=\eta_{1}$

In particular, the geometric multiplicity of the kernel of $I+sK$ equals the
number of linearly independent solutions of the above equation.

###### Proof.

$I+sK$ has a non trivial kernel if and only if

$\langle u,v\rangle+\langle sKu,v\rangle=0,\quad\forall u,v\in\mathcal{V}.$

Equivalently if and only if $\exists u$ such that
$u+sKu\in\mathcal{V}^{\perp}$ (see Lemma 2 below for a description of
$\mathcal{V}^{\perp}$). This is in turn equivalent to the following system

$\begin{cases}(1-s)u_{0}=v_{0}\\\
u_{t}=sZ_{t}^{*}J\Big{(}\int_{0}^{t}Z_{\tau}u_{\tau}d\tau+Z_{0}u_{0}\Big{)}+Z_{t}^{*}J\nu\\\
(1-s)u_{1}=s\Lambda(u)+v_{1}\end{cases}$ (14)

Let us substitute $Z_{t}$ with $Z_{t}^{s}=\delta^{s}Z_{t}$. It is
straightforward to check, using the definition in (13), that:

$sZ_{t}^{*}J\int_{0}^{t}Z_{\tau}u_{\tau}d\tau=(Z_{t}^{s})^{*}J\int_{0}^{t}Z^{s}_{\tau}u_{\tau}d\tau.$

Moreover $Z_{t}^{*}J\nu=(Z_{t}^{s})^{*}J\nu$ for any $\nu\in\Pi$ and
$sZ_{t}^{*}JZ_{0}=(Z_{t}^{s})^{*}JZ_{0}^{s}$ since $\delta^{s}Z_{0}=Z_{0}$.

All the computations we will do from here on are aimed at rewriting (14) as a
boundary value problem in $T_{\lambda_{0}}T^{*}M\times T_{\lambda_{0}}T^{*}M$.
Let us start with the second equality in (14). Define:

$\eta(t)=\int_{0}^{t}Z^{s}_{\tau}u_{\tau}d\tau+Z^{s}_{0}u_{0}+\nu,\quad\eta(0)=Z^{s}_{0}u_{0}+\nu.$

The linear constraint defining $\mathcal{V}$ implies:

$\begin{split}Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1}\in\Pi&\iff
Z_{0}u_{0}+\int_{0}^{1}Z^{s}_{t}u_{t}dt+Z^{s}_{1}u_{1}\in\Pi\\\
&\iff\eta(1)+Z^{s}_{1}u_{1}\in\Pi_{0}\\\
&\iff\tilde{\Phi}_{*}(\eta(1)+Z^{s}_{1}u_{1})\in\Pi_{1}\end{split}$

This imposes a condition on the initial and final value of $\eta(t)$. The
initial one must lie in $\pi_{*}^{-1}TN_{0}$ and final one in
$\pi_{*}^{-1}TN_{1}$. If we multiply by $Z_{t}^{s}$ the second equation in
(14), we are brought to consider the following problem:

$\begin{cases}\dot{\eta}(t)=Z_{t}^{s}(Z_{t}^{s})^{*}J\eta(t)\\\
(\pi_{*}\eta(0),\pi_{*}\eta(1))\in T(N_{0}\times N_{1})\\\ \end{cases}$ (15)

Now we use the remaining equations in (14) to reduce the space
$\pi_{*}^{-1}(T(N_{0}\times N_{1}))$ to a Lagrangian one. Let us reformulate
the first and third line in (14) as equations in $T_{\lambda_{0}}T^{*}M$ and
$T_{\lambda_{1}}T^{*}M$. Using the maps $Z_{0}$ and $\tilde{\Phi}_{*}Z_{1}$ we
obtain:

$\begin{cases}(1-s)Z_{0}u_{0}=Z_{0}v_{0}=pr_{0}J_{0}\nu\\\
(1-s)\tilde{\Phi}_{*}Z_{1}u_{1}=\tilde{\Phi}_{*}Z_{1}v_{1}+s\tilde{\Phi}_{*}Z_{1}\Lambda(u)=pr_{1}J_{1}\tilde{\Phi}_{*}\nu+s\tilde{\Phi}_{*}Z_{1}\Lambda(u)\end{cases}$

Where we used the fact that $(v_{0},v_{t},v_{1})$ is a vector in
$\mathcal{V}^{\perp}$. Notice that, for $u\in\mathcal{V}$ we have that:

$s(Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1})=Z_{0}u_{0}+\int_{0}^{1}Z^{s}_{t}u_{t}dt+Z^{s}_{1}u_{1}$

This implies that the term $s\tilde{\Phi}_{*}Z_{1}\Lambda(u)$ can be rewritten
as:

$\displaystyle
s\tilde{\Phi}_{*}Z_{1}\Lambda(u)=s\,pr_{1}J_{1}\tilde{\Phi}_{*}(Z_{0}u_{0}+\int_{0}^{1}Z_{t}u_{t}dt+Z_{1}u_{1})=pr_{1}J_{1}\tilde{\Phi}_{*}(\eta(1)+Z_{1}^{s}u_{1}-\nu)$

If substitute $Z_{0}u_{0}$ with $pr_{\Pi_{0}^{\perp}}\eta(0)$ we end up with
the equations:

$\begin{cases}(1-s)pr_{\Pi_{0}^{\perp}}\eta(0)=pr_{0}J_{0}\nu=pr_{0}J_{0}\eta(0)\\\
(1-s)\tilde{\Phi}_{*}Z_{1}u_{1}=pr_{1}J_{1}\tilde{\Phi}_{*}(\eta(1)+Z_{1}^{s}u_{1})\end{cases}$

Now we do the same kind of substitution for the term $Z_{1}^{s}u_{1}$. Using
the projections on $\Pi_{0}$ and $\Pi_{0}^{\perp}$ and recalling that
$\tilde{\Phi}_{*}$ sends $\Pi_{0}$ to $\Pi_{1}$, we have:

$\begin{split}Z_{1}^{s}u_{1}&=s\,pr_{\Pi_{0}}Z_{1}u_{1}+pr_{\Pi_{0}^{\perp}}Z_{1}u_{1}=(s-1)\,pr_{\Pi_{0}}Z_{1}u_{1}+Z_{1}u_{1}\\\
pr_{1}J_{1}\tilde{\Phi}_{*}Z_{1}^{s}u_{1}&=(s-1)\,pr_{1}J_{1}\tilde{\Phi}_{*}pr_{\Pi_{0}}Z_{1}u_{1}+pr_{1}J_{1}\tilde{\Phi}_{*}Z_{1}u_{1}\\\
&=(s-1)\,pr_{1}J_{1}\tilde{\Phi}_{*}pr_{\Pi_{0}}Z_{1}u_{1}\end{split}$

Last equality being due to the fact that the image of $\tilde{\Phi}_{*}Z_{1}$
is isotropic and thus
$J_{1}\textrm{Im}(\tilde{\Phi}_{*}Z_{1})\subset\textrm{Im}(\tilde{\Phi}_{*}Z_{1})^{\perp}$.
Moreover $\tilde{\Phi}_{*}Z_{1}u_{1}$ coincides with the projection of
$-\tilde{\Phi}_{*}\eta(1)$ on $\Pi_{1}^{\perp}$. Thus we are left with:

$\begin{cases}(1-s)pr_{\Pi_{0}^{\perp}}\eta(0)=pr_{0}J_{0}\nu=pr_{0}J_{0}\eta(0)\\\
(1-s)(-pr_{\Pi_{1}^{\perp}}\tilde{\Phi}_{*}\eta(1)+pr_{1}J_{1}\tilde{\Phi}_{*}pr_{\Pi_{0}}Z_{1}u_{1})=pr_{1}J_{1}\tilde{\Phi}_{*}\eta(1)\end{cases}$
(16)

It is straightforward to check that
$pr_{1}J_{1}\tilde{\Phi}_{*}pr_{\Pi_{0}}Z_{1}u_{1}$ depends only on the
projection of $Z_{1}u_{1}$ on $\Pi_{0}^{\perp}$. Moreover, expanding
$1=\tilde{\Phi}_{*}\circ\tilde{\Phi}_{*}^{-1}$ and using the relation
$\tilde{\Phi}_{*}Z_{1}u_{1}=-pr_{\Pi_{1}^{\perp}}\tilde{\Phi}_{*}\eta(1)$, the
second equality in (16) can be rewritten as:

$(s-1)pr_{\Pi_{1}^{\perp}}\tilde{\Phi}_{*}\eta(1)=pr_{1}J_{1}\tilde{\Phi}_{*}\eta(1)+(1-s)pr_{1}J_{1}\tilde{\Phi}_{*}pr_{\Pi_{0}}\tilde{\Phi}_{*}^{-1}pr_{\Pi_{1}^{\perp}}\tilde{\Phi}_{*}\eta(1)$
(17)

If $s=1$, the equations reduce to $pr_{0}(J_{0}\eta(0))=0$ and
$pr_{1}J_{1}\tilde{\Phi}_{*}\eta(1)=0$. Consider the first case, the relation
is equivalent to:

$\sigma(\eta(0),Z_{0}w_{0})=g_{0}(J_{0}\eta(0),Z_{0}w_{0})=g_{0}(pr_{0}J_{0}\eta(0),Z_{0}w_{0})=0,\,\forall
w_{0}\in\mathbb{R}^{\dim(N_{0})}.$

Thus we are looking for solution starting from $T_{\lambda_{0}}Ann(N_{0})$.
Similarly setting $s=1$ in the second equality we find that $\eta(1)$ must lie
inside $T_{\lambda_{1}}Ann(N_{1})$.

Now, for $s\neq 1$, we want to interpret the boundary conditions as an
analytic family of Lagrangian subspaces depending on $s$. To do so we employ
the following linear map defined in (12):

$A_{0}^{s}(\eta)=\eta+J_{0}^{-1}(1-s)pr_{\Pi_{0}^{\perp}}\eta$

If $\eta\in T_{\lambda_{0}}Ann(N_{0})$ we have that
$pr_{\Pi_{0}^{\perp}}\eta=pr_{0}\eta$ and $pr_{0}J_{0}\eta=0$ and thus:

$\displaystyle pr_{\Pi_{0}^{\perp}}(A_{0}^{s}(\eta))$
$\displaystyle=pr_{\Pi_{0}^{\perp}}(\eta)\in\textrm{Im}(Z_{0})$ $\displaystyle
pr_{0}J_{0}(A_{0}^{s}(\eta))$
$\displaystyle=pr_{0}(J_{0}\eta+(1-s)pr_{\Pi_{0}^{\perp}}\eta)$
$\displaystyle=pr_{0}J_{0}\eta+(1-s)pr_{\Pi_{0}^{\perp}}\eta$
$\displaystyle=(1-s)pr_{\Pi_{0}^{\perp}}(A_{0}^{s}(\eta))$

So we have shown that $A_{0}^{s}(T_{\lambda_{0}}Ann(N_{0}))$ is precisely the
space satisfying the first set of equations. A similar argument works for the
final point. Let us recall the definition of $A_{1}^{s}$ given in (12):

$A_{1}^{s}(\eta)=\eta+(1-s)(J_{1}^{-1}+\tilde{\Phi}_{*}pr_{\Pi_{0}}\tilde{\Phi}_{*}^{-1})pr_{\Pi_{1}^{\perp}}\eta.$

Now, we check that the boundary condition for the final point are satisfied if
and only if $A_{1}^{s}\circ\tilde{\Phi}_{*}\,\eta(1)\in
T_{\lambda_{1}}Ann(N_{1})$. In fact, take any $\eta$ in
$T_{\lambda_{1}}Ann(N_{1})$, it holds:

$\begin{split}pr_{\Pi_{1}^{\perp}}(A_{1}^{s})^{-1}\eta&=pr_{\Pi_{1}^{\perp}}\eta,\\\
pr_{\Pi_{1}}(A_{1}^{s})^{-1}\eta&=pr_{\Pi_{1}}\eta+(s-1)(J^{-1}_{1}+\tilde{\Phi}_{*}pr_{\Pi_{0}}\tilde{\Phi}_{*}^{-1})pr_{\Pi_{1}^{\perp}}\eta,\\\
pr_{1}J_{1}(A_{1}^{s})^{-1}\eta&=(s-1)pr_{1}(1+J_{1}\tilde{\Phi}_{*}pr_{\Pi_{0}}\tilde{\Phi}_{*}^{-1})pr_{\Pi_{1}^{\perp}}\eta.\end{split}$

It is straightforward to substitute the last equality in (17) and check that
indeed $(A^{s}_{1})^{-1}(T_{\lambda_{1}}Ann(N_{1}))$ is the right space.

Let us call $\Phi_{1}^{s}$ the fundamental solution of (15) at time $1$ and
denote by $\Gamma(\Phi_{1}^{s})$ its graph. It follows that
$s\in\mathbb{R}\setminus\\{0\\}$ is in the kernel of $1+sK$ if and only if:

$\Gamma(\tilde{\Phi}_{*}\circ\Phi_{1}^{s})\cap
A_{0}^{s}(T_{\lambda_{0}}Ann(N_{0}))\times(A_{1}^{s})^{-1}(T_{\lambda_{1}}Ann(N_{1}))\neq(0)$

which is equivalent to the condition:

$\Gamma(A_{1}^{s}\circ\tilde{\Phi}_{*}\circ\Phi_{1}^{s}\circ A_{0}^{s})\cap
T_{\lambda_{0}}Ann(N_{0})\times T_{\lambda_{1}}Ann(N_{1})\neq(0)$ (18)

Now we prove the part about the multiplicity. Suppose that two different
controls $u$ and $v$ give the same trajectory $\eta_{t}$ solving (15). Since
the maps $Z_{0}$ and $Z_{1}$ are injective it must hold that $v_{0}=u_{0}$ and
$v_{1}=u_{1}$. Moreover
$\int_{0}^{t}Z_{\tau}u_{\tau}d\tau=\int_{0}^{t}Z_{\tau}v_{\tau}d\tau$
$\forall\,t\in[0,1]$ and thus:

$Ku=Z_{t}^{*}J\int_{0}^{t}Z_{\tau}u_{\tau}d\tau=Z_{t}^{*}J\int_{0}^{t}Z_{\tau}v_{\tau}d\tau=Kv.$

However, Volterra operator are always injective and thus $u=v$.

Vice-versa, consider $u=0$ and see whether you get solutions of the system
above that do not correspond to any variation. Since $u_{0}$ and $u_{1}$ are
both zero we are considering solution starting from the fibre and reaching the
fibre. Plugging in $u_{t}=0$ we obtain:

$0=\dot{\eta}=Z_{t}^{s}(Z_{t}^{s})^{*}J\eta=Z_{t}^{s}(Z_{t}^{s})^{*}J\nu$

However $pr_{\Pi^{\perp}}Z_{t}^{s}(Z_{t}^{s})J\nu=X_{t}X_{t}^{*}\nu$ and by
assumption the matrix $\int_{0}^{1}X_{t}X_{t}^{*}dt$ is invertible. Thus we
get a contradiction. ∎

The following Lemma was used in the proof of 1. Gives the orthogonal
complement to $\mathcal{V}$ inside $\mathcal{H}$, using the Hilbert structure
introduced in 2. We will denote by the symbol $\perp_{i}$ the orthogonal
complement, in $T_{\lambda_{i}}T^{*}M$, with respect to the scalar product
$g_{i}$.

######  Lemma 2.

With this choice of scalar product the orthogonal complement to $\mathcal{V}$
is given by:

$\mathcal{V}^{\perp}=\\{(v_{0},Z_{t}^{*}J\nu,v_{1}):\nu\in\Pi\\}$

where $v_{0}$ and $v_{1}$ are determined by the following conditions:

$Z_{0}v_{0}-J_{0}\nu\in\textrm{Im}Z_{0}^{\perp_{0}},\quad\tilde{\Phi}_{*}Z_{1}v_{1}-J_{1}\tilde{\Phi}_{*}\nu\in\textrm{Im}\tilde{\Phi}_{*}Z_{1}^{\perp_{1}}.$

###### Proof.

Suppose that $u\in\mathcal{V}^{\perp}$. Let us test it against infinitesimal
variations that fix the endpoints, i.e. such that $u_{i}=0$ and
$\int_{0}^{1}Z_{t}u_{t}dt\in\Pi$. Recall that $\Pi$ is Lagrangian, thus the
condition $\int_{0}^{1}Z_{t}u_{t}dt\in\Pi$ can be equivalently formulated as
$\sigma(\int_{0}^{1}Z_{t}u_{t}dt,\nu)=0$ for all $\nu\in\Pi$. Hence, the
subspace $\\{u:u_{i}=0,\int_{0}^{1}Z_{t}u_{t}dt\in\Pi\\}$ is the intersection
of the kernels of $u_{t}\mapsto\int_{0}^{1}\sigma(Z_{t}u_{t},\nu)dt$. It
follows that:

$\langle v,u\rangle=-\int_{0}^{1}\langle v_{t},u_{t}\rangle dt=0,\,\forall
u\in V\iff v_{t}=\sigma(\nu,Z_{t}\cdot),\quad\nu\in\Pi.$

Hence $v_{t}=Z_{t}^{*}J\nu$. Now, take $u\in\mathcal{V}$ and compute:

$\langle
v,u\rangle=\sigma(\nu,\int_{0}^{1}Z_{t}u_{t}dt)+\sigma(J_{0}^{-1}Z_{0}v_{0},Z_{0}u_{0})+\sigma(J_{1}^{-1}\tilde{\Phi}_{*}Z_{1}v_{1},\tilde{\Phi}_{*}Z_{1}u_{1}).$

Since $u\in\mathcal{V}$, we have:

$\sigma\left(\nu,\int_{0}^{1}Z_{t}u_{t}dt\right)=-\sigma(\nu,Z_{0}u_{0}+Z_{1}u_{1}).$

It follows that:

$\begin{split}\langle
v,u\rangle&=-\sigma(\nu,Z_{0}u_{0}+Z_{1}u_{1})+\sigma(J_{0}^{-1}Z_{0}v_{0},Z_{0}u_{0})+\sigma(J_{1}^{-1}\tilde{\Phi}_{*}Z_{1}v_{1},\tilde{\Phi}_{*}Z_{1}u_{1})\\\
&=\sigma(J_{0}^{-1}Z_{0}v_{0}-\nu,Z_{0}u_{0})+\sigma(J_{1}^{-1}\tilde{\Phi}_{*}Z_{1}v_{1}-\tilde{\Phi}_{*}\nu,\tilde{\Phi}_{*}Z_{1}u_{1}).\end{split}$

Since we are assuming that $\int_{0}^{1}X_{t}X_{t}^{*}dt>0$, the image of the
map $u_{t}\mapsto\pi_{*}\int_{0}^{1}Z_{t}u_{t}dt$ is the whole $T_{q_{0}}M$.
In particular, for any $u_{0}$, is possible to find variations of the form
$(u_{0},u_{t},0)\in\mathcal{V}$. An analogous statements holds for variations
of the form $(0,u_{t},u_{1}).$

Hence, if $\langle u,v\rangle=0$ $\forall u\in\mathcal{V}$, then both
$\sigma(J_{0}^{-1}Z_{0}v_{0}-\nu,Z_{0}u_{0})$ and
$\sigma(J_{1}^{-1}\tilde{\Phi}_{*}Z_{1}v_{1}-\tilde{\Phi}_{*}\nu,\tilde{\Phi}_{*}Z_{1}u_{1})$
must be zero at the same time.

Moreover, this implies that $v_{0}$ and $v_{1}$ are completely determined by
the value of $\nu$. Finally notice that:

$\begin{split}\sigma(J_{0}^{-1}Z_{0}v_{0}-\nu,Z_{0}u_{0})=0\,\forall\,u_{0}&\iff
Z_{0}v_{0}-J_{0}\nu\in\textrm{Im}(Z_{0})^{\perp_{0}}\\\
\sigma(J_{1}^{-1}\tilde{\Phi}_{*}Z_{1}v_{1}-\tilde{\Phi}_{*}\nu,Z_{1}u_{1})=0\,\forall\,u_{1}&\iff\tilde{\Phi}_{*}Z_{1}v_{1}-J_{1}\tilde{\Phi}_{*}\nu\in\textrm{Im}(\tilde{\Phi}_{*}Z_{1})^{\perp_{1}}\end{split}$

∎

######  Remark 7.

If we complexify all the subspaces involved in the proof of 1, i.e. tensor
with $\mathbb{C}$ we can take also $s\in\mathbb{C}$.

We can reformulate the intersection problem in the statement of 1 as follows.
Let $\pi_{N_{1}}$ the orthogonal projection, with respect to $g_{1}$, onto the
subspace $T_{\lambda_{1}}Ann(N_{1})^{\perp}$ and define a map $Q^{s}$ as:

$Q^{s}:T_{\lambda_{0}}Ann(N_{0})\to T_{\lambda_{1}}Ann(N_{1})^{\perp},\quad
Q^{s}(\eta)=\pi_{N_{1}}(A_{1}^{s}\Phi_{1}^{s}A_{0}^{s})(\eta).$ (19)

Let us fix now two bases, one of $T_{\lambda_{0}}Ann(N_{0})$ and one of
$T_{\lambda_{1}}Ann(N_{1})$. Construct two $2n\times n$ matrices using the
elements of the chosen basis. Let us call the resulting objects $T_{0}$ and
$T_{1}$ respectively. It follows that $J_{1}T_{1}$ is a basis of
$T_{\lambda_{1}}Ann(N_{1})^{\perp}$. Define the function $\det(Q^{s})$ as the
determinant of the $n\times n$ matrix
$T_{1}^{*}J_{1}A_{1}^{s}\Phi_{1}^{s}A_{0}^{s}T_{0}$. Clearly, different
choices of basis give simply a scalar multiple of $\det(Q^{s})$ and thus is
well defined:

$\det(Q^{s})=\frac{\det(T_{1}^{*}J_{1}A_{1}^{s}\Phi_{1}^{s}A_{0}^{s}T_{0})}{\det(T_{0}^{*}T_{0})^{1/2}\det(T_{1}^{*}J_{1}J_{1}^{*}T_{1})^{1/2}}$

Moreover $\det(Q^{s})|_{s=s_{0}}=0$ if and only if there exists at least a
solution to our boundary problem. Notice that map $s\mapsto\det(Q^{s})$ is
analytic in $s$ since the fundamental matrix is an entire map in $s$ (see
[3][Proposition 4]). The following Proposition shows that the multiplicity of
any root $s_{0}\neq 0$ is equal to the number of independent solutions to the
boundary value problem.

######  Proposition 2.

The multiplicity of any root $s_{0}\neq 0$ of $\det Q^{s}$ equals the
dimension of the kernel of $Q^{s}$.

###### Proof.

The proof is done in two steps. First we show that the equation
$\det(Q^{s})=0$ is equivalent to $\det(R^{s})=0$ where $R^{s}$ is a symmetric
matrix, analytic in $s$. Once one knows this, it suffices to compute
$\partial_{s}R^{s}$ and show that it is non degenerate to prove that the
multiplicity of the equation is the same as the dimension of the kernel.

_Step 1: Replace $Q^{s}$ with a symmetric matrix_

Let us restrict to the case $s\in\mathbb{R}$, since all the roots are real. As
shown in (18) and remarked above, the determinant of the matrix $Q^{s}$ is
zero whenever the graph of $A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}$
intersect the subspace $L_{0}=T_{(\lambda_{0},\lambda_{1})}(Ann(N_{0}\times
N_{1}))$. Suppose that $s_{0}$ is a time of intersection and choose as
coordinates in the Lagrange Grassmannian $L_{0}$ and another subspace $L_{1}$
transversal to both $L_{0}$ and
$\Lambda_{s}:=\Gamma(A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s})$. This
means that, if
$(T_{\lambda_{0}}T^{*}M)^{2}\approx\\{(p,q)|p,q\in\mathbb{R}^{2n}\\}$, we
identify $L_{0}\approx\\{q=0\\}$ and $L_{1}\approx\\{p=0\\}$.

In this coordinates $\Lambda_{s}$ is given by the graph of a symmetric matrix,
i.e. is the following subspace $\Lambda_{s}=\\{(p,R^{s}p)\\}$ where again
$R^{s}$ is analytic in $s$.

The quadratic form associated to the derivative $\partial_{s}R(s)$ can be
interpreted as the velocity of the curve $s\mapsto\Lambda_{s}$ inside the
Grassmannian, it is possible to compute it choosing an arbitrary base of
$\Lambda_{s}$ and an arbitrary set of coordinates. Invariants such as
signature and nullity do not change (see for example [5, symplecticMethods] or
[2]). Take a curve $\lambda_{s}=(p_{s},R^{s}p_{s})$ inside $\Lambda_{s}$ then
one has:

$S(\lambda_{s})=\sigma(\lambda_{s},\dot{\lambda}_{s})=\langle
p_{s},\partial_{s}R^{s}p_{s}\rangle$

Recall that we will be using the symplectic form given by
$(-\sigma_{\lambda_{0}})\oplus\sigma_{\lambda_{1}}$, in order to have that
graph of a symplectic map is a Lagrangian subspace.

_Step 2: Replace $\Lambda_{s}$ with a positive curve_

We slightly modify our curve to exploit an hidden positivity of the Jacobi
equation. We substitute the fundamental solution $\Phi_{1}^{s}$ with the
following map:

$\Psi^{s}=\Psi^{s}_{1}=\delta_{s}\Phi^{s}_{1}\delta_{\frac{1}{s}}.$

It is straightforward to check that $\Psi^{s}$ is again a symplectomorphism
and that it is the fundamental solution of the following ODE system at time
$t=1$:

$\dot{\Psi}_{t}^{s}=sZ_{t}Z_{t}^{*}J\Psi_{t}^{s},\quad\Psi_{0}^{s}=Id.$ (20)

On one hand we are introducing a singularity at $s=0$ but on the other hand we
are going to show that the graph of $\Psi^{s}$ becomes a monotone curve and
its velocity is fairly easy to compute.

First of all, hoping that the slight abuse of notation will not create any
confusion, let us introduce a family of dilations similar to the
$\delta_{s}$,$\delta^{s}$ also in $T_{\lambda_{1}}T^{*}M$. The definition is
analogous to the one in (13) but with $\Pi_{1}$ and $\Pi_{1}^{\perp}$ instead
of $\Pi_{0}$ and $\Pi_{1}^{\perp}$. We will denote them with the same symbol.

Let us consider the following symplectomorphisms:

$\delta_{s}A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}\delta_{\frac{1}{s}}=\delta_{s}A_{1}^{s}\tilde{\Phi}_{*}\delta_{\frac{1}{s}}\Psi_{s}\delta_{s}A_{0}^{s}\delta_{\frac{1}{s}}$

Notice that the dilations $\delta_{s}$ preserve the subspaces
$T_{\lambda_{i}}Ann(N_{i})$ and thus the intersection points between the graph
of the above map and the subspace
$T_{(\lambda_{0},\lambda_{1})}Ann(N_{0}\times N_{1})$ are unchanged. Let us
rewrite the maps $\delta_{s}A_{0}^{s}\delta_{\frac{1}{s}}$ and
$\delta_{s}A_{1}^{s}\tilde{\Phi}_{*}\delta_{\frac{1}{s}}$. For the former:

$\delta_{s}A_{0}^{s}\delta_{\frac{1}{s}}=\delta_{s}(1+(1-s)J_{0}^{-1}pr_{\Pi_{0}^{\perp}})\delta_{\frac{1}{s}}=1+\frac{1-s}{s}J_{0}^{-1}pr_{\Pi_{0}^{\perp}}=B_{0}^{s}$

For the latter, a computation in local coordinates and the fact that the
dilations $\delta_{s}$ and $\tilde{\Phi}_{*}$ do not commute yield:

$\begin{split}\delta_{s}A_{1}^{s}\tilde{\Phi}_{*}\delta_{\frac{1}{s}}&=\delta_{s}(1+(1-s)(J_{1}^{-1}+\tilde{\Phi}_{*}pr_{\Pi_{0}}\tilde{\Phi}_{*}^{-1})pr_{\Pi_{1}^{\perp}})\tilde{\Phi}_{*}\delta_{\frac{1}{s}}\\\
&=(1+\frac{1-s}{s}J_{1}^{-1}pr_{\Pi_{1}^{\perp}})\tilde{\Phi}_{*}=B_{1}^{s}\tilde{\Phi}_{*}\end{split}$

Thus we take, for $s\neq 0$, as curve
$\Lambda_{s}:=\Gamma(B_{1}^{s}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s})$, the graph
of the symplectomorphism just introduced. Notice that $\Psi^{s}$ is actually
analytic, the singularity at $s=0$ comes only form the maps $B_{i}^{s}$.

_Step 3: Computation of the velocity_

Now we compute the velocity of the graph of
$B_{1}^{s}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}$. Take a curve
$\lambda_{s}=(\eta,B_{1}^{s}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta)$ inside the
$\Lambda_{s}$ and let us compute the quadratic form associated to the
velocity:

$\begin{split}S(\lambda_{s})&=-\sigma(\eta,\partial_{s}\eta)+\sigma(B_{1}^{s}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta,\partial_{s}(B_{1}^{s}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta))\\\
&=\sigma(B_{1}^{s}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta,(\partial_{s}B_{1}^{s})\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta)+\sigma(\Psi^{s}B_{0}^{s}\eta,(\partial_{s}\Psi^{s})B_{0}^{s}\eta)+\\\
&\qquad+\sigma(B_{0}^{s}\eta,\partial_{s}(B_{0}^{s})\eta)\end{split}$

Let us consider the terms of the type
$\sigma(B_{i}^{s}x,\partial_{s}B_{i}^{s}x)$. It immediate to compute the
derivative in this case, recall that
$B_{i}^{s}x=x+\frac{(1-s)}{s}J_{i}^{-1}pr_{\Pi_{i}^{\perp}}x$. It follows that
$\partial_{s}B_{i}^{s}=-\frac{1}{s^{2}}J_{i}^{-1}pr_{\Pi_{i}^{\perp}}$ thus
the first and last term read as:

$\begin{split}\sigma(B_{1}^{s}\xi,(\partial_{s}B_{1}^{s})\xi)&=-\frac{1}{s^{2}}\sigma(\xi,J_{1}^{-1}pr_{\Pi_{1}^{\perp}}\xi)\\\
&=\frac{1}{s^{2}}g_{1}(pr_{\Pi_{1}^{\perp}}\xi,pr_{\Pi_{1}^{\perp}}\xi),\quad\text{where
}\xi=\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta,\\\
\sigma(\partial_{s}(B_{0}^{s})\eta,B_{0}^{s}\eta)&=-\frac{1}{s^{2}}\sigma(B_{0}^{s}\eta,J_{0}^{-1}pr_{\Pi_{0}^{\perp}}\eta)=-\frac{1}{s^{2}}g_{0}(J_{0}\eta,J_{0}^{-1}pr_{\Pi_{0}^{\perp}}\eta)\\\
&=\frac{1}{s^{2}}g_{0}(pr_{\Pi_{0}^{\perp}}\eta,pr_{\Pi_{0}^{\perp}}\eta).\end{split}$

Notice we used the fact that $J_{i}$ (and thus $J_{i}^{-1}$) is $g_{i}-$skew
symmetric. Now we rewrite the middle term. We present it as the integral of
its derivative using the equation for $\Psi_{t}^{s}$. Let us use the shorthand
notation $x=B_{0}^{s}\eta$. We obtain:

$\begin{split}\frac{d}{dt}\left(\sigma(\Psi_{t}^{s}x,(\partial_{s}\Psi_{t}^{s})x)\right)=\sigma(\partial_{t}\Psi_{t}^{s}x,(\partial_{s}\Psi_{t}^{s})x)+\sigma(\Psi_{t}^{s}x,(\partial_{s}\partial_{t}\Psi_{t}^{s})x)\\\
=s\sigma(Z_{t}Z_{t}^{*}J\Psi_{t}^{s}x,\partial_{s}\Psi_{t}^{s}x)+s\sigma(\Psi_{t}^{s}x,Z_{t}Z_{t}^{*}J\partial_{s}\Psi_{t}^{s}x)\\\
+\sigma(\Psi_{t}^{s}x,\partial_{s}\big{(}sZ_{t}Z_{t}^{*}\big{)}J\Psi_{t}^{s}x).\end{split}$

The first and second term have opposite sign and thus cancel out. What remains
is:

$\frac{d}{dt}\left(\sigma((\partial_{s}\Psi_{t}^{s})x,\Psi_{t}^{s}x)\right)=\sigma(\Psi_{t}^{s}x,Z_{t}Z_{t}^{*}J\Psi_{t}^{s}x)=g(Z_{t}^{*}J\Psi_{t}^{s}x,Z_{t}^{*}J\Psi_{t}^{s}x)$

Integrating over $[0,1]$ and using the fact that
$\partial_{s}\Psi_{t}^{s}|_{(0,0)}=0$ we get that:

$\begin{split}\sigma(\partial_{s}\Psi^{s}x,\Psi^{s}x)&=\int_{0}^{1}g(Z^{*}_{t}J\Psi_{t}^{s}x,Z^{*}_{t}J\Psi_{t}^{s}x)dt.\end{split}$

Using the notation $||\cdot||$ to denote the norm with respect to the
corresponding metric and summing everything up we find the following
expression for the velocity of our curve:

$S_{s}(\lambda_{s})=\frac{1}{s^{2}}\left(\|pr_{\Pi_{0}^{\perp}}\eta\|^{2}+\|pr_{\Pi_{1}^{\perp}}\Psi^{s}B_{0}^{s}\eta\|^{2}\right)+\int_{0}^{1}\|Z_{t}^{*}J\Psi_{t}^{s}B_{0}^{s}\eta\|^{2}dt$

Since each term of the sum is non negative $S_{s}(\lambda_{s})$ is zero if and
only if each term is zero. From the first one we obtain that $\eta$ must be
contained in the fibre. Notice that $B^{s}_{0}$ acts as the identity on
$\Pi_{0}$ and thus in this case
$\|pr_{\Pi_{1}^{\perp}}\tilde{\Phi}_{*}\Psi^{s}B_{0}^{s}\eta\|^{2}=\|pr_{\Pi_{1}^{\perp}}\Psi^{s}\eta\|^{2}$.

It follows that $\Psi_{t}^{s}B^{s}\eta=\Psi_{t}^{s}\eta$ is a solution of the
Jacobi equation (15) starting and reaching the fibre (recall that
$\tilde{\Phi}_{*}(\Pi_{0})=\Pi_{1}$). Let us consider now the third piece,
since the integrand is positive it must hold that for almost any $t$,
$Z_{t}^{*}J\Psi_{t}^{s}\eta=0$. If we multiply this equation by $Z_{t}$ we
find that:

$Z_{t}Z_{t}^{*}J\Psi_{t}^{s}\eta=0=\dot{\Psi}_{t}^{s}\eta.$

It follow that we are dealing with a constant solution starting and reaching
the fibre. However this contradicts the assumption that the matrix
$\int_{0}^{1}X_{t}X_{t}^{*}dt$ is non degenerate. In fact, if we substitute a
non zero constant solutions starting from the fibre in (15), we find that
$pr_{\Pi_{0}^{\perp}}(\eta)=\int_{0}^{1}X_{t}X_{t}^{*}dt\,\eta\neq 0$ ∎

The following proposition is proved in [3].

######  Proposition 3.

There exists $c_{1},c_{2}>0$ such that:

$\|\Phi_{1}^{s}\|\leq c_{1}e^{c_{2}|s|}\quad\forall\,s\in\mathbb{C}$

Moreover $\Phi_{t}^{s}$ is analytic and the function $s\mapsto\det(Q^{s})$ is
entire and satisfy the same type of estimate.

This fact tells us that $\det Q^{s}$ is an entire function of order $\rho\leq
1$. We know its zeros, which are determined by the eigenvalues of $K$, and
thus we can apply Hadamard factorization theorem (see [13]) to present it as
an infinite product. It follows that we have the following identity:

$\det(Q^{s})=as^{k}e^{bs}\prod_{\lambda\in
Sp(K)}(1+s\lambda)^{m(\lambda)}\quad a,b\in\mathbb{C},a\neq 0,k\in\mathbb{N}$
(21)

where $m(\lambda)$ is the geometric multiplicity of the eigenvalue $\lambda$.
To determine the remaining parameters it sufficient to know the value of
$\det(Q^{s})$ and a certain number of its derivatives at $s=0$ (depending on
the value of $k$). Assume for now that $k=0$, a straightforward computation
shows that:

$\det(Q^{s})|_{s=0}=a,\quad\partial_{s}\det(Q^{s})|_{s=0}=a(b+\textrm{tr}(K)).$
(22)

We will compute these quantities in Lemmas 4 and 3 for general boundary
conditions. The proofs can be adapted to the case of separated conditions
easily.

###  4.2 General boundary condition

In this section we prove a determinant formula for general boundary conditions
$N\subseteq M\times M$. First, we reduce this case to the case of separate
boundary conditions. We have to slightly modify the proof of 1 since, after
this reduction, the Endpoint map will not be a submersion any more. Then, we
compute the normalization factors given in (22).

Let us consider $M\times M$ as state space, with the following dynamical
system:

$f_{u}(q^{\prime},q)=\begin{pmatrix}0\\\
f_{u}(q)\end{pmatrix},\quad(q^{\prime},q)\in M\times M.$ (23)

and boundary conditions $\Delta\times N$. With this definition, any extremal
between two points $q_{0}$ and $q_{1}$ lifts naturally to an extremal between
$(q_{0},q_{0})$ and $(q_{0},q_{1})$. However, the Endpoint map of the new
system is no longer surjective. In fact, any trajectory is confined to a
submanifold of the form $\\{\hat{q}\\}\times M$. Thus, even if we started with
a strictly normal extremal, we do not get a strictly normal extremal of the
new system. However, there is no real singularity of the Endpoint map here: we
have just introduced a certain number of conserved quantities. All the proofs
presented above work also in this case. We are going to discuss briefly how to
adapt them.

Let us start with Pontryagin maximum principle. It implies that the lift of
the extremal curve $\tilde{q}(t)=(q_{0},q(t))$ is the curve
$\tilde{\lambda}(t)=(-\lambda_{0},\lambda(t))$. This is because the initial
and final covector of the lift must annihilate the tangent space of the
boundary conditions manifold. In this case $N_{0}=\Delta$ and the annihilator
of the diagonal subspace is $\\{(\lambda,-\lambda):\lambda\in
T_{\lambda_{0}}T^{*}_{q_{0}}M\\}$. Moreover, by the orthogonality condition in
PMP (see A), we know that $(-\lambda(0),\lambda(1))$ annihilates the tangent
space of $N$.

Thus, if we want to work in one fixed tangent space, we have to multiply the
first covector by $-1$. This changes the sign of the symplectic form and we
are thus brought to work on $T_{\lambda_{0}}T^{*}M\times
T_{\lambda_{0}}T^{*}M$ with symplectic form $(-\sigma)\oplus\sigma$.

With this change of sign, the tangent space to the annihilator of the diagonal
gets mapped to the diagonal subspace of $T_{\lambda_{0}}T^{*}M\times
T_{\lambda_{0}}T^{*}M$ and the tangent space to the annihilator of the
boundary conditions $N$ is mapped to the tangent space of:

$A(N)=\\{(\mu_{0},\mu_{1}):\langle\mu_{0},X_{0}\rangle=\langle\mu_{1},X_{1}\rangle,\,\forall(X_{0},X_{1})\in
TN\\}.$

Let us make now some notational remarks. We will still denote by
$\tilde{\Phi}_{*}$ the map $1\times\tilde{\Phi}_{*}$, which correspond to the
new flow we are using to backtrack our trajectory to the starting point.

As explained in Section 2, to define the scalar product on our space of
variations, it is necessary to introduce two metrics on the tangent spaces to
the endpoints of our curve. We will choose them of the form
$\tilde{g}_{0}=g_{0}\oplus g_{0}$ and $\tilde{g}_{1}=g_{0}\oplus g_{1}$ where
$g_{0}$ and $g_{1}$ are two metrics on $T_{\lambda_{0}}T^{*}M$ and
$T_{\lambda_{1}}T^{*}M$ respectively.

Now we compute the second variation of the new system (23). As a general rule,
we will denote all the quantities relative to (23) on $M\times M$, putting a
$\tilde{}$ on top. We have:

$\tilde{Z}_{0}u_{0}=\begin{pmatrix}Z_{0}u_{0}\\\
Z_{0}u_{0}\end{pmatrix},\quad\tilde{Z}_{1}=\begin{pmatrix}Z_{1}^{0}u_{1}\\\
Z_{1}^{1}u_{1}\end{pmatrix},\quad\tilde{Z}_{t}=\begin{pmatrix}0\\\
Z_{t}\end{pmatrix}.$

Notice that $\tilde{\Phi}_{*}\tilde{Z}_{1}$ maps $\mathbb{R}^{\dim(N)}$ to the
tangent space to $A(N)$ and we can assume that its image is contained in
$\Pi_{0}^{\perp}\times\Pi_{1}^{\perp}$. We will denote by $\tilde{pr}_{1}$ the
orthogonal projection onto the image of $\tilde{Z}_{1}$.

The domain of the second variation is the subspace
$\mathcal{V}=\\{(u_{0},u_{t},u_{1}):\tilde{Z}_{0}u_{0}+\int_{0}^{1}\tilde{Z}_{t}u_{t}+\tilde{Z}_{1}u_{1}\in\Pi_{0}\times\Pi_{0}\\}$.
Clearly, this equation is equivalent to:

$Z_{0}u_{0}+Z_{1}^{0}u_{1}\in\Pi_{0}\quad\text{ and }\quad
Z_{0}u_{0}+Z^{1}_{1}u_{1}+\int_{0}^{1}Z_{t}u_{t}dt\in\Pi_{0}.$

It follows that the control $u_{0}$ is completely determined by $u_{1}$.
Moreover we can assume that $Z_{0}u_{0}=-Z_{1}^{0}u_{1}$ since we are free to
choose any system of coordinates and any trivialization of the tangent bundle
of the manifolds $\Delta$ and $N$.

Let $A_{i}^{s}$ be the maps given in (12). The following proposition is the
counterpart of 1 for general boundary conditions.

######  Proposition 4.

Let $\Phi_{1}^{s}$ be the fundamental solution of the Jacobi system:

$\dot{\eta}=Z_{t}^{s}(Z_{t}^{s})^{*}J\eta.$

The operator $1+sK$ restricted to $\mathcal{V}$ has non trivial kernel if and
only if there exists a non zero $(\eta_{0},\eta_{1})\in
T_{(\lambda_{0},\lambda_{1})}A(N)$ such that

$A_{1}^{s}\circ\tilde{\Phi}_{*}\circ\Phi_{1}^{s}\circ
A_{0}^{s}\,\eta_{0}=\eta_{1}$ (24)

The geometric multiplicity of the kernel equals the number of linearly
independent solutions of the above equation.

###### Proof.

The proof is completely analogous to the one of 1. However, some slight
modifications are in order since the Endpoint map is not surjective in this
case.

_Step 1: Characterize $\mathcal{V}^{\perp}$._

The orthogonal complement to $\mathcal{V}$ is given by:

$\mathcal{V}^{\perp}=\\{(v_{0},\tilde{Z}_{t}^{*}J\tilde{\nu},v_{1}):\tilde{pr}_{1}\left(\begin{pmatrix}2Z_{0}v_{0}\\\
0\end{pmatrix}+\tilde{J}_{1}\tilde{\nu}\right)=\tilde{\Phi}_{*}Z_{1}v_{1}\\}$

The proof is the same as the one of Lemma 2 and yields
$v_{t}=\tilde{Z}_{t}^{*}J\tilde{\nu}$. Here, however, we can not separate
$v_{0}$ from $v_{1}$. Take $u\in\mathcal{V}$ and $v\in V^{\perp}$:

$\displaystyle\langle
u,v\rangle=\sigma\left(\int_{0}^{1}Z_{t}u_{t}dt,\nu\right)+\tilde{g}_{0}(\tilde{Z}_{0}u_{0},\tilde{Z}_{0}v_{0})+\tilde{\Phi}^{*}\tilde{g}_{1}(\tilde{Z}_{1}u_{1},\tilde{Z}_{1}u_{1})$
$\displaystyle=\tilde{g}_{0}(\tilde{Z}_{0}u_{0},\tilde{Z}_{0}v_{0}-\tilde{J}_{0}\nu)+\tilde{\Phi}^{*}\tilde{g}_{1}(\tilde{Z}_{1}u_{1},\tilde{Z}_{1}v_{1}-\tilde{J}_{1}\nu)$
$\displaystyle=g_{0}(Z_{1}^{0}u_{1},J_{0}\nu-2Z_{0}v_{0}+Z_{1}^{0}v_{1})+\tilde{\Phi}^{*}g_{1}(Z_{1}^{1}u_{1},Z_{1}^{1}v_{1}-J_{1}\nu)=0.$

Hence:

$\tilde{pr}_{1}\begin{pmatrix}2Z_{0}v_{0}-J_{0}\nu\\\
J_{1}\nu\end{pmatrix}=\tilde{pr}_{1}\left(\begin{pmatrix}2Z_{0}v_{0}\\\
0\end{pmatrix}+\tilde{J}_{1}\tilde{\nu}\right)=\tilde{\Phi}_{*}\tilde{Z}_{1}v_{1}$
(25)

_Step 2: Derivation of Jacobi equation._

Now we can write down Jacobi equation in a fashion similar to the one of 1.
The system reads:

$\begin{cases}(1-s)\tilde{Z}_{0}u_{0}=\tilde{Z}_{0}v_{0}\\\
\dot{\tilde{\eta}}(t)=\tilde{Z}_{t}^{s}(\tilde{Z}_{t}^{s})J\tilde{\eta}(t)\\\
(1-s)\tilde{\Phi}_{*}\tilde{Z}_{1}u_{1}=\tilde{\Phi}_{*}\tilde{Z}_{1}v_{1}+s\,\tilde{pr}_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\tilde{Z}_{0}u_{0}+\int_{0}^{1}\tilde{Z}_{t}u_{t}dt+\tilde{Z}_{1}u_{1})\end{cases}$
(26)

Where
$\tilde{\eta}(t)=\int_{0}^{t}\tilde{Z}_{t}^{s}u_{t}dt+\tilde{Z}_{0}u_{0}+\tilde{\nu}$.
Arguing as in 1, we can rewrite the last term of the third equation. We have:

$\displaystyle{pr}_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\tilde{Z}_{0}^{s}u_{0}+\int_{0}^{1}\tilde{Z}^{s}_{t}u_{t}dt+\tilde{Z}^{s}_{1}u_{1})$
$\displaystyle={pr}_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\eta(1)-\tilde{\nu}+\tilde{Z}_{1}^{s}u_{1})$
$\displaystyle\tilde{\Phi}_{*}Z_{1}^{s}u_{1}=\tilde{\Phi}_{*}Z_{1}^{1}u_{1}+s\tilde{\Phi}_{*}\tilde{pr}_{\Pi}\tilde{Z}_{1}u_{1}$
$\displaystyle=\tilde{\Phi}_{*}Z_{1}^{1}u_{1}+s(\tilde{\Phi}_{*}\tilde{pr}_{\Pi}\tilde{\Phi}_{*}^{-1})\tilde{\Phi}_{*}\tilde{Z}_{1}u_{1}.$

By the equation defining $\mathcal{V}$ we have that:

$\tilde{pr}_{\Pi^{\perp}}\tilde{\Phi}_{*}\eta(1)=-\begin{pmatrix}Z_{1}^{0}u_{1}\\\
\tilde{\Phi}_{*}Z^{1}_{1}u_{1}\end{pmatrix}=-\tilde{\Phi}_{*}\tilde{Z}_{1}u_{1}.$

Lastly note the the first equation gives
$(1-s)\tilde{Z}_{0}u_{0}=(1-s)\tilde{pr}_{\Pi^{\perp}}\eta(0)=\tilde{Z}_{0}v_{0}$.
Let us now plug all this equations into (26) recalling that
$\tilde{pr}_{\Pi}+\tilde{pr}_{\Pi^{\perp}}=1$. The third line now reads:

$\displaystyle
pr_{1}\bigg{(}\begin{pmatrix}2(1-s)\tilde{pr}_{\Pi_{0}^{\perp}}\eta(0)\\\
0\end{pmatrix}+\tilde{J}_{1}\tilde{\Phi}_{*}(\tilde{pr}_{\Pi}\eta(1)-s\tilde{\Phi}_{*}\tilde{pr}_{\Pi}\tilde{\Phi}_{*}^{-1}\tilde{pr}_{\Pi^{\perp}}\tilde{\Phi}_{*}\eta(1))$
$\displaystyle+(1-s)\tilde{{pr}}_{\Pi^{\perp}}\tilde{\Phi}_{*}\eta(1)\bigg{)}=0.$

Writing the equation component-wise and using the fact that, since
$\tilde{\Phi}_{*}$ preserve the fibres, it holds:

$\tilde{\Phi}_{*}pr_{\Pi_{0}}=\tilde{\Phi}_{*}pr_{\Pi_{0}}\tilde{\Phi}_{*}^{-1}\tilde{p}r_{\Pi_{1}^{\perp}}\Phi_{*}+pr_{\Pi_{1}}\tilde{\Phi}_{*},$

we obtain a relation very similar to the one in (16). Namely:

$\tilde{pr}_{1}\begin{pmatrix}-J_{0}\left(3(s-1)J_{0}^{-1}pr_{\Pi_{0}^{\perp}}\eta(0)+\tilde{\Phi}_{*}pr_{\Pi_{0}}\eta(0)\right)\\\
J_{1}\left(pr_{\Pi_{1}}\tilde{\Phi}_{*}\eta(1)+(1-s)(\tilde{\Phi}_{*}pr_{\Pi_{1}}\tilde{\Phi}_{*}^{-1}+J_{1}^{-1}){pr}_{\Pi_{1}^{\perp}}\tilde{\Phi}_{*}\eta(1)\right)\end{pmatrix}=0$
(27)

At this point the argument is again the following. If $X$ belongs to
$A(N)+\Pi$ and $\tilde{pr_{1}}(\tilde{J}_{1}X)=0$, it follows that:

$\tilde{g}_{1}(\tilde{p}r_{1}\tilde{J}_{1}X,\tilde{Z}_{1}u_{1})=\sigma(X,\tilde{Z}_{1}u_{1})=0,\,\forall\,u_{1}\in\mathbb{R}^{\dim(N)}.$

Hence $X$ must lie in $\mathrm{Im}(\tilde{Z}_{1})^{\angle}\cap(A(N)+\Pi)$,
which is $A(N)$. Consider the maps $A^{s}_{i}$ defined in (12), they preserve
$A(N)+\Pi$. Thus we can rewrite (27) as:

$\begin{pmatrix}(A_{0}^{s})^{-1}\eta(0)\\\
A_{1}^{s}\tilde{\Phi}_{*}\eta(1)\end{pmatrix}\in
A(N)\iff\Gamma(A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s})\cap A(N).$

Notice the presence of the inverse of $A_{0}^{s}$ due to the sign of the
symplectic form.

_Step 3: Uniqueness_

Arguing again as in 1, to the trivial variation $(0,0,0)$ correspond constant
solutions of Jacobi equation starting from the fibre. However, since the
Endpoint map of the original system is regular, there are no such solutions.
Hence the correspondence between $\ker(1+sK)$ and
$\Gamma(A_{1}^{s}\tilde{\Phi}_{*}A_{0}^{s})\cap A(N)$ is one-to-one. ∎

Now we define an analogous map to the one in eq. 19. Let $\pi_{N}$ be the
orthogonal projection on the space $T_{(\lambda_{0},\lambda_{1})}A(N)^{\perp}$
and consider the map:

$Q^{s}:\Gamma(A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s})\to
T_{(\lambda_{0},\lambda_{1})}A(N)^{\perp},\quad Q^{s}(\eta)=\pi_{N}(\eta).$
(28)

Let $T=(T_{0},T_{1})$ be any linear invertible map from $\mathbb{R}^{2n}$ to
the tangent space $T_{(\lambda_{0},\lambda_{1})}A(N)$. We denote by $J$ the
map $(-J_{0})\oplus J_{1}$ representing the symplectic form
$(-\sigma_{\lambda_{0}})\oplus\sigma_{\lambda_{1}}$. As in the previous
section we define the following function:

$\det(Q^{s})=\frac{\det(T_{1}^{*}J_{1}A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}-T_{0}^{*}J_{0})}{\det(T_{0}^{*}J_{0}J_{0}^{*}T_{0}+T_{1}^{*}J_{1}J_{1}^{*}T_{1})^{1/2}}$

######  Remark 8.

One could also define (28) as a bilinear form, using just the symplectic
pairing. In fact, for $(\eta,(\xi_{0},\xi_{1}))$ in
$T_{\lambda_{0}}T^{*}M\times T_{(\lambda_{0},\lambda_{1})}A(N)$, define:

$\tilde{Q}^{s}(\eta,(\xi_{0},\xi_{1}))=\sigma(A_{1}^{s}\tilde{\Phi}_{*}\Phi_{1}^{s}A_{0}^{s}\eta,\xi_{1})-\sigma(\eta,\xi_{0}).$

This form is degenerate exactly when
$\Gamma(A_{1}^{s}\tilde{\Phi}_{*}A_{0}^{s})\cap A(N)\neq(0).$

######  Proposition 5.

The multiplicity of any roots $s_{0}\neq 0$ of the equation $\det(Q^{s})$ is
equal to the geometric multiplicity of the boundary value problem.

###### Proof.

The same proof of 2 works verbatim. Indeed, we are working with the same
curve, $\Gamma(A_{1}^{s}\tilde{\Phi}_{*}A_{0}^{s})$. ∎

In the remaining part of this section we carry out the computation of the
normalizing factors of the function $\det(Q^{s})$. As already mentioned at the
end of the previous section a classical factorization theorem by Hadamard (see
[13]) tells us that:

$\det(Q^{s})=as^{k}e^{bs}\prod_{\lambda\in
Sp(K)}(1+s\lambda)^{m(\lambda)}\quad a,b\in\mathbb{C},a\neq 0,k\in\mathbb{N}$

where $m(\lambda)$ is the geometric multiplicity of the eigenvalue. We are now
going to compute the values of $a,b\in\mathbb{C}$ and $k$.

######  Theorem 4.

For almost any choice of metrics $g_{0},g_{1}$ on $T_{\lambda_{i}}T^{*}M$,
$\det(Q^{s}|_{s=0})\neq 0$. Whenever this condition holds, the determinant of
the second variation is given by:

$\det(1+sK)=\det((Q^{s}|_{s=0})^{-1})e^{s(\textrm{tr}(K)-\textrm{tr}(\partial_{s}Q^{s}\,(Q^{s})^{-1}|_{s=0}))}\det(Q^{s})$
(29)

###### Proof.

We prove the first assertion: for almost any choice of scalar product, $k=0$
and thus $a=\det(Q_{1}^{s}|_{s=0})\neq 0$. This is equivalent to a
transversality condition between the graph of the symplectomorphism
$A_{1}^{s}\tilde{\Phi}_{*}\Phi^{s}A_{0}^{s}$ and the annihilator of the
boundary conditions $N$.

We can argue as follows: consider the following family of maps acting on the
Lagrange Grassmannian of $T_{\lambda_{0}}T^{*}M\times T_{\lambda_{1}}T^{*}M$
depending on the choice of scalar products $G_{0}$ and $G_{1}$:

$F_{G}=(A_{0}^{s})^{-1}\times A_{1}^{s}|_{s=0},\quad G=(G_{0},G_{1}),G_{i}>0.$

It is straightforward to see that they define a family of algebraic maps of
the Grassmannian to itself. For any chosen subspace $L_{0}$,
$F^{-1}_{G}(L_{0})$ is arbitrary close to $\Pi_{0}\times\Pi_{1}$, for $G_{i}$
large enough. Notice that
$\Gamma(A_{1}^{s}\tilde{\Phi}_{*}\Phi^{s}A_{0}^{s})\cap L_{0}\neq(0)$ if and
only if $\Gamma(\tilde{\Phi}_{*}\Phi^{s})\cap F^{-1}_{G}(L_{0})\neq(0)$. Using
the formula in Lemma 4 one has that $\Gamma(\tilde{\Phi}_{*}\Phi^{s})$ is
transversal to $\Pi_{0}\times\Pi_{1}$ and thus to $F^{-1}_{G}(L_{0})$ for any
fixed $L_{0}$ and $G_{i}$ sufficiently large. Now, since everything is
algebraic in $G$ and there is a Zariski open set in which the transversality
condition holds, the possible choices of $G_{i}$ for which $k>0$ are in
codimension 1.

Let us assume that $k=0$ and compute $b$. Differentiating the expression for
$\det(Q^{s})$ in eq. 21 at $s=0$ we find that:

$\partial_{s}\det(Q^{s})|_{s=0}=a(b+\textrm{tr}(K))$

An integral formula for the trace of $K$ is given in Lemma 3. The derivative
of $\det(Q^{s})$ can be computed using Jacobi formula:

$\partial_{s}\det(Q^{s})|_{s=0}=a\,\textrm{tr}(\partial_{s}Q^{s}\,(Q^{s})^{-1})|_{s=0}.$

An explicit expression of the derivatives of the map $Q^{s}$ can be computed
using Lemma 4. It follows that
$b=\textrm{tr}(\partial_{s}Q^{s}\,(Q^{s})^{-1})-\textrm{tr}(K)$ and we obtain
precisely the formula in the statement. ∎

Before giving the explicit formula for $\textrm{tr}(K)$ and the derivatives of
the fundamental solution to Jacobi equation at $s=0$ we need to make some
notational remark and write down a formula for the second variation in the
same spirit of Section 2 and eq. 5. We are working on the state space $M\times
M$ with twice the number of variables of the original system and trivial
dynamic on the first factor and separated boundary conditions. The left
boundary condition manifold is the diagonal of $M\times M$ and the right one
is our starting $N$. We apply the formula in eq. 5 to this particular system,
we denote by $\tilde{Z}_{t}$ and $\tilde{Z}_{i}$ the matrices for the
auxiliary problem, in general everything pertaining to it will be marked by a
tilde. Identifying $T_{(\lambda_{0},\lambda_{0})}T^{*}(M\times M)$ with
$T_{\lambda_{0}}T^{*}M\times T_{\lambda_{0}}T^{*}M$ we have that:

$\tilde{Z}_{0}u_{0}=\begin{pmatrix}Z_{0}u_{0}\\\
Z_{0}u_{0}\end{pmatrix},\quad\tilde{Z}_{1}=\begin{pmatrix}Z_{1}^{0}u_{1}\\\
Z_{1}^{1}u_{1}\end{pmatrix},\quad\tilde{Z}_{t}=\begin{pmatrix}0\\\
Z_{t}\end{pmatrix}.$

We still work on the subspace
$\mathcal{V}=\\{(u_{0},u_{t},u_{1}):\tilde{Z}_{0}u_{0}+\int_{0}^{1}\tilde{Z}_{t}u_{t}+\tilde{Z}_{1}u_{1}\in\Pi\\}$.
However it is clear that the this equation implies that:

$Z_{0}u_{0}+Z_{1}^{0}u_{1}\in\Pi_{0}\quad\text{ and }\quad
Z_{0}u_{0}+Z^{1}_{1}u_{1}+\int_{0}^{1}Z_{t}u_{t}dt\in\Pi_{0}$

It follows that control $u_{0}$ is completely determined by $u_{1}$. Moreover
we can assume that $Z_{0}u_{0}=-Z_{1}^{0}u_{1}$ since we are free to choose
any system of coordinates and any trivialization of the tangent bundle of the
manifolds $\Delta$ and $N$. Technically we are working with different scalar
products on each of the copies of $T_{\lambda_{0}}T^{*}M$. However it is easy
to see that on the space $\mathcal{V}$ only the sum of this metrics plays a
role. We will denote it $g_{0}$. Now we are ready to state the following:

######  Lemma 3.

The second variation of the extended system, as a quadratic form, can be
written as $\langle(I+K)u,u\rangle$ where $K$ is the symmetric (on
$\mathcal{V}$) compact operator given by:

$\begin{split}-\langle
Ku,u\rangle=\int_{0}^{1}\int_{0}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}u_{t})d\tau
dt-\sigma\left(Z_{1}^{0}u_{1},\int_{0}^{1}Z_{t}u_{t}dt\right)\\\
+\sigma\left(\int_{0}^{1}Z_{t}u_{t}dt-
Z_{1}^{0}u_{1},Z_{1}^{1}u_{1}\right)+g_{0}(Z_{1}^{0}u_{1},Z_{1}^{0}u_{1})\\\
+(\tilde{\Phi}^{*}g_{1})(Z_{1}^{1}u_{1},Z_{1}^{1}u_{1}).\end{split}$

Moreover, define the following matrices:

$\Gamma=\int_{0}^{1}X_{t}X_{t}^{*}dt,\quad\Omega=\int_{0}^{1}\int_{0}^{t}X_{t}Z_{t}^{*}JZ_{\tau}X^{*}_{\tau}d\tau
dt.$

Denote by $pr_{1}$ the projection onto $T_{(\lambda_{0},\lambda_{1})}A(N)$ and
by $\pi^{i}_{*}$ the differential of the natural projections
$\pi^{i}:T^{*}M\to M$ relative to the $i-$th component, $i=1,2$. The trace of
$K$ has the following expression:

$\begin{split}\textrm{tr}(K)=&-\dim(N)+\textrm{tr}[\pi_{*}^{1}\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\tilde{Z}_{0})]\\\
&+\textrm{tr}\left[\Gamma^{-1}\left(\Omega+(\pi^{2}_{*}-\pi_{*}^{1})\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J|_{\Pi}dt)\right)\right]\end{split}$

###### Proof.

The first part is a straightforward computation combining the expression
obtained in Lemma 1 for the second variation with the observation concerning
the structure of the maps $Z_{0}$ and $Z_{1}$ made before the statement and
the choice of the Riemannian metrics.

Now, notice that the codimension of the space giving fixed endpoints
variations $\mathcal{V}$ for the extended system in $\mathcal{H}$ is
$2\dim(M)$. Moreover, it is defined as the kernel of the linear functional:

$\rho:(u_{0},u_{t},u_{1})\mapsto\tilde{\pi}_{*}(\tilde{Z}_{0}u_{0}+\int_{0}^{1}\tilde{Z}_{t}u_{t}dt+\tilde{Z}_{1}u_{1})\in
T_{\pi(\lambda_{0})}M\times T_{\pi(\lambda_{0})}M.$

It is straightforward to check that the following subspace is
$2\dim(M)-$dimensional and transversal to $\ker\rho$:

$\mathcal{V}^{\prime}=\\{(u_{0},Z_{t}^{*}J\nu,0):\nu\in\Pi,u_{0}\in\mathbb{R}^{\dim
M}\\}$

The trace of $K$ on the whole space splits as a sum of two pieces, the trace
of $K|_{\mathcal{V}}$ and the trace of $K|_{\mathcal{V}^{\prime}}$. We can
then further simplify and compute separately the trace on
$\mathcal{V}^{\prime}\cap\\{u_{0}=0\\}$ and its complement
$\mathcal{V}^{\prime}\cap\\{\nu=0\\}$. We are going to compute the trace of
$K$ on the whole space and then the trace of $K|_{\mathcal{V}^{\prime}}$,
determining in this way the value of $K|_{\mathcal{V}}$.

Consider $\mathcal{H}=\mathcal{H}_{1}\oplus\mathcal{H}_{2}$. Where

$\mathcal{H}_{1}=\\{u:u=(0,u_{t},0)\\},\quad\mathcal{H}_{2}=\\{u:u=(u_{0},0,u_{1})\\}.$

It is straightforward to check that $\mathcal{H}_{1}^{\perp}=\mathcal{H}_{2}$
for our class of metrics and that $\mathcal{H}_{1}\equiv
L^{2}([0,1],\mathbb{R}^{k})$. Using eq. 5, the restrictions of the quadratic
form $\hat{K}(u)=\langle u,Ku\rangle$ to each one of the former subspaces
read:

$\hat{K}|_{\mathcal{H}_{1}}(u)=\int_{0}^{1}\int_{0}^{t}\sigma(Z_{t}u_{t},Z_{\tau}u_{\tau})dtd\tau,\quad\hat{K}|_{\mathcal{H}_{2}}(u)=\sigma(\tilde{Z}_{0}u_{0},\tilde{Z}_{1}u_{1})-\|u\|_{\mathcal{H}}^{2}.$

The trace of the first quadratic form is zero (see [3][Theorem 2]) whereas the
trace of the second part is just $-\dim(N)$. Thus we have that:

$\textrm{tr}(K|_{\mathcal{V}})=-\dim(N)-\textrm{tr}(K|_{\mathcal{V}^{\prime}})$

To compute the last piece we apply $K$ to a control
$u\in\mathcal{V}^{\prime}\cap\\{u_{0}=0\\}$ using the explicit expression of
the operator given in Lemma 1. Recall that $pr_{1}$ is projection onto the
image of $\tilde{\Phi}_{*}\tilde{Z}_{1}$. It follows that:

$\begin{split}K(u)&=\left(0,-Z_{t}^{*}J\int_{0}^{t}Z_{\tau}u_{\tau}d\tau,-L\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}\left(\int_{0}^{1}\tilde{Z}_{t}u_{t}dt\right)\right)\\\
&=\left(0,-Z_{t}^{*}J\int_{0}^{t}Z_{\tau}u_{\tau}d\tau,-\Lambda(u)\right).\end{split}$

Now we write $K(u)$ in coordinates given by the splitting
$\mathcal{V}\oplus\mathcal{V}^{\prime}$. To do so we have first to consider
$\rho\circ K(u)$. It is given by:

$\begin{split}-\rho\circ
K(u)&=\tilde{\pi}_{*}\left(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J\int_{0}^{t}Z_{\tau}Z_{\tau}^{*}J\nu+\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J\nu
dt)\right)\\\ &=\begin{pmatrix}0\\\
\Omega\nu\end{pmatrix}+\tilde{\pi}_{*}\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J\nu
dt).\end{split}$ (30)

Set $\Gamma=\int_{0}^{1}X_{t}^{*}X_{t}dt$, it easy to check that, if
$u\in\mathcal{V}^{\prime}$, then

$\rho(u)=\begin{pmatrix}X_{0}u_{0}\\\ \Gamma\nu+X_{0}u_{0}\end{pmatrix}$

for an invertible matrix $X_{0}$ which, without loss of generality, can be
taken to be identity. It follows that the projection on the first component of
$\rho\circ K(u)$ is completely determined by the second term in eq. 30. Let us
call $\pi_{*}^{i}$ for $i=1,2$ the projection on the $i-$th component. It
follows that an element
$(Z_{0}\hat{u}_{0},Z_{t}^{*}J\hat{\nu},0)=\hat{u}\in\mathcal{V}^{\prime}$ has
the same projection as $K(u)$ if and only if:

$\rho\circ(K(u)-\hat{u})=0\iff\begin{cases}\hat{u}_{0}&=X_{0}^{-1}\pi^{1}_{*}\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J\nu
dt)\\\
\hat{\nu}&=\Gamma^{-1}\left(\Omega\nu+(\pi^{2}_{*}-\pi^{1}_{*})\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}J\nu
dt)\right)\end{cases}$

In particular the restriction to $\mathcal{V}^{\prime}\cap\\{u_{0}=0\\}$ is
given by:

$\nu\mapsto\Gamma^{-1}\left(\Omega+(\pi^{2}_{*}-\pi^{1}_{*})\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}(\int_{0}^{1}\tilde{Z}_{t}Z_{t}^{*}Jdt)\right)\nu.$

A similar strategy applied to $\mathcal{V}^{\prime}\cap\\{\nu=0\\}$ tells us
that the last contribution for the trace is given by the following map:

$u_{0}\mapsto
X_{0}^{-1}\pi_{*}^{1}\tilde{\Phi}_{*}^{-1}\,pr_{1}\tilde{J}_{1}\tilde{\Phi}_{*}\tilde{Z}_{0}u_{0}.$

It is worth pointing out that indeed the trace does not depend on $X_{0}$ and
that the vector $\int_{0}^{1}\tilde{Z}_{t}Z_{t}J\nu dt$ is the following:

$\int_{0}^{1}\tilde{Z}_{t}Z_{t}J\nu dt=\begin{pmatrix}\begin{pmatrix}0\\\
0\end{pmatrix}\\\ \begin{pmatrix}\Theta\nu\\\
\Gamma\nu\end{pmatrix}\end{pmatrix}$

In particular if the boundary conditions are separated (i.e. $N=N_{0}\times
N_{1}$) the part of the trace coming from
$\mathcal{V}^{\prime}\cap\\{u_{0}=0\\}$ depends only on the projection onto
$T_{\lambda_{1}}N_{1}$. ∎

######  Lemma 4.

The flow $\Phi_{t}^{s}|_{s=0}$ and its derivative
$\partial_{s}\Phi_{t}^{s}|_{s=0}$ are given by:

$\Phi_{t}^{s}|_{s=0}=\begin{pmatrix}1&0\\\
\int_{0}^{t}X_{\tau}X_{\tau}^{*}d\tau&1\end{pmatrix},\quad\partial_{s}\Phi_{t}^{s}|_{s=0}=\begin{pmatrix}\int_{0}^{t}Y_{\tau}X_{\tau}^{*}d\tau&0\\\
\int_{0}^{t}\int_{0}^{\tau}X_{\tau}Z_{\tau}^{*}JZ_{r}X_{r}^{*}drd\tau&-\int_{0}^{t}X_{\tau}Y_{\tau}^{*}d\tau\end{pmatrix}$

###### Proof.

It is straightforward to check that $\Phi_{1}^{s}|_{s=0}$ and
$\partial_{s}\Phi_{1}^{s}|_{s=0}$ solve the following Cauchy problems:

$\begin{cases}\dot{\Phi}_{t}^{0}=\begin{pmatrix}0&0\\\
X_{t}X_{t}^{*}&0\end{pmatrix}\Phi_{t}^{0}\\\
\Phi_{0}^{0}=Id\end{cases}\quad\begin{cases}\partial_{s}\dot{\Phi}_{t}^{s}|_{s=0}=\begin{pmatrix}0&0\\\
X_{t}X_{t}^{*}&0\end{pmatrix}\partial_{s}\Phi_{t}^{s}|_{s=0}+\begin{pmatrix}Y_{t}X_{t}^{*}&0\\\
0&-X_{t}Y_{t}^{*}\end{pmatrix}\Phi_{t}^{0}\\\
\partial_{s}\Phi_{0}^{s}|_{s=0}=0.\end{cases}$

Solving the ODE one obtains the formula in the statement. ∎

## Appendix A

In this appendix we collect some information concerning Pontryagin Maximum
Principle (PMP) and the differentiation of the endpoint map used and mentioned
throughout the text. Everything is fairly standard material in geometric
control theory, the reader is referred to [6, 22, 4] for further details.

###  A.1 Pontryagin Maximum Principle

Let us introduce a useful family of Hamiltonian functions on $T^{*}M$. They
generate a family of Hamiltonian flows which we use to backtrack admissible
trajectories $\gamma$ to their initial point. Moreover, they appear in the
formulation of PMP and extend the flow of the fields $f_{u(t)}$ to the
cotangent bundle. Set:

$h^{t}_{u}:T^{*}M\to\mathbb{R},\quad
h^{t}_{u}(\lambda)=\langle\lambda,f_{u}\rangle+\nu\varphi(u,\pi(\lambda)),\quad\nu\leq
0.$

In particular, if $\tilde{\gamma}$ is and admissible curve, we can build a
lift, i.e. a curve $\tilde{\lambda}$ in $T^{*}M$ such that
$\pi(\tilde{\lambda})=\tilde{\gamma}$, solving
$\dot{\lambda}=\vec{h}_{u}(\lambda)$. The following wellknown theorem,
Pontryagin Maximum Principle, gives a characterization of critical points of
$\mathcal{J}$ (as defined in (3)), for any set of boundary conditions $N$.

######  Theorem 1 (PMP).

If a control $\tilde{u}\in L^{\infty}([0,1],U)$ is a local minimizer for the
functional in eq. 3 there exists a curve $\lambda:[0,1]\to T^{*}M$,
$\nu\in\mathbb{R}$ and an admissible curve $q:[0,1]\to M$ such that for almost
all $t\in[0,1]$

  1. 1.

$\lambda(t)$ is a lift of $q(t)$:

$q(t)=\pi(\lambda(t));$

  2. 2.

$\lambda(t)$ satisfies the following Hamiltonian system:

$\frac{d\lambda}{dt}=\vec{h}_{\tilde{u}(t)}(\lambda);$

  3. 3.

the control $\tilde{u}$ is determined by the maximum condition:

$h_{\tilde{u}(t)}(\lambda(t))=\max_{u\in U}h_{u}(\lambda(t)),\quad\nu\leq 0;$

  4. 4.

the non-triviality condition holds: $(\lambda(t),\nu)\neq(0,0)$;

  5. 5.

transversality condition holds:

$(-\lambda(0),\lambda(1))\in Ann(N).$

We call $q(t)$ an extremal curve (or trajectory) and $\lambda(t)$ an extremal.

There are essentially two possibility for the parameter $\nu$, it can be
either $0$ or, after appropriate normalization of $\lambda_{t}$, $-1$. The
extremals belonging to the first family are called _abnormal_ whereas the ones
belonging to second _normal_.

###  A.2 The Endpoint map and its differentiation

In this subsection we write down the integral expression for the first and
second derivative of the endpoint map. Further details can be found in
[6][Section 20.3]. Denote by $\mathcal{U}_{q_{0}}\subset L^{\infty}([0,1],U)$
be the space of admissible controls at point $q_{0}$ and define the following
map:

$E^{t}:\mathcal{U}_{q_{0}}\to M,\quad u\mapsto\gamma_{u}(t)$

It takes the control $u$ and gives the position at time $t$ of the solution
starting from $q_{0}$ of:

$\dot{q}=f_{u(\tau)}(q).$

We call this map _Endpoint map_. It turns out that $E^{t}$ is smooth, provided
that the fields $f_{u}(q)$ are smooth too.

For a fixed control $\tilde{u}$ consider the function
$h_{\tilde{u}}(\lambda):=h_{\tilde{u}(t)}(\lambda)$ and define the following
non autonomous flow which plays the role of parallel transport in this
context:

$\frac{d}{dt}\tilde{\Phi}_{t}=\vec{h}_{\tilde{u}}(\tilde{\Phi}_{t})\qquad\tilde{\Phi}_{0}=Id$
(31)

It has the following properties:

  * •

It extends to the cotangent bundle the flow which solves
$\dot{q}=f^{t}_{\tilde{u}}(q)$ on the base. In particular if $\lambda_{t}$ is
an extremal with initial condition $\lambda_{0}$,
$\pi(\tilde{\Phi}_{t}(\lambda_{0}))=q_{\tilde{u}}(t)$ where $q_{\tilde{u}}$ is
an extremal trajectory.

  * •

$\tilde{\Phi}_{t}$ preserves the fibre over each $q\in M$. The restriction
$\tilde{\Phi}_{t}:\,T^{*}_{q}M\to T^{*}_{\tilde{\Phi}_{t}(q)}M$ is an affine
transformation.

We suppose now that $\lambda(t)$ is an extremal and $\tilde{u}$ a critical
point of the functional $\mathcal{J}$. We use the symplectomorphism
$\tilde{\Phi}_{t}$ to pull back the whole curve $\lambda(t)$ to the starting
point $\lambda_{0}$. We can express all the first and second order information
about the extremal using the following map and its derivatives:

$b_{u}^{t}(\lambda)=(h_{u}^{t}-h_{\tilde{u}}^{t})\circ\tilde{\Phi}_{t}(\lambda)$

Notice that:

  * •

$b_{u}^{t}(\lambda_{0})|_{u=\tilde{u}(t)}=0=d_{\lambda_{0}}\,b_{u}^{t}|_{u=\tilde{u}(t)}$
by definition.

  * •

$\partial_{u}b_{u}^{t}|_{u=\tilde{u}(t)}=\partial_{u}(h_{u}^{t}\circ\tilde{\Phi}_{t})|_{u=\tilde{u}(t)}=0$
since $\lambda(t)$ is an extremal and $\tilde{u}$ the relative control.

Thus the first non zero derivatives are the order two ones. We define the
following maps:

$\begin{split}Z_{t}=\partial_{u}\vec{b}_{u}^{t}(\lambda_{0})|_{u=\tilde{u}(t)}:\mathbb{R}^{k}=T_{\tilde{u}(t)}U\to
T_{\lambda_{0}}(T^{*}M)\\\
H_{t}=\partial_{u}^{2}b_{t}(\lambda_{0})|_{u=\tilde{u}(t)}:\mathbb{R}^{k}=T_{\tilde{u}(t)}U\to
T^{*}_{\tilde{u}(t)}U=\mathbb{R}^{k}\end{split}$ (32)

We denote by $\Pi=\ker\pi_{*}$ the kernel of the differential of the natural
projection $\pi:T^{*}M\to M$.

######  Proposition 6 (Differential of the endpoint map).

Consider the endpoint map $E^{t}:\mathcal{U}_{q_{0}}\to M$. Fix a point
$\tilde{u}$ and consider the symplectomorphism $\tilde{\Phi}_{t}$ and the map
$Z_{t}$ defined above. The differential is the following map:

$d_{\tilde{u}}E(v_{t})=d_{\lambda(t)}\pi\circ
d_{\lambda_{0}}\tilde{\Phi}_{t}(\int_{0}^{t}Z_{\tau}v_{\tau}d\tau)\in
T_{q_{t}}M$

In particular, if we identify $T_{\lambda_{0}}(T^{*}M)$ with $\mathbb{R}^{2m}$
and write $Z_{t}=\begin{pmatrix}Y_{t}\\\ X_{t}\end{pmatrix}$, $\tilde{u}$ is a
regular point if and only if $v_{t}\mapsto\int_{0}^{t}X_{\tau}v_{\tau}d\tau$
is surjective. Equivalently if the following matrix is invertible:

$\Gamma_{t}=\int_{0}^{t}X_{\tau}X^{*}_{\tau}d\tau\in Mat_{n\times
n}(\mathbb{R}),\quad\det(\Gamma_{t})\neq 0$

If $d_{\tilde{u}}E^{t}$ is surjective then $(E^{t})^{-1}(q_{t})$ is smooth in
a neighbourhood of $\tilde{u}$ and is tangent space is given by:

$\begin{split}T_{\tilde{u}}(E^{t})^{-1}(q_{t})=\\{v\in
L^{\infty}([0,1],\mathbb{R}^{k}):\,\int_{0}^{t}X_{\tau}v_{\tau}d\tau=0\\}\\\
=\\{v\in
L^{\infty}([0,1],\mathbb{R}^{k}):\,\int_{0}^{t}Z_{\tau}v_{\tau}d\tau\in\Pi\\}\end{split}$

If the differential of the endpoint map is surjective, the set of admissible
control becomes smooth (at least locally) and our minimization problem can be
interpreted as a constrained optimization problem. We are looking for critical
points of $\mathcal{J}$ on the submanifold
$\\{u\in\mathcal{U}:E^{t}(u)=q_{1}\\}$.

######  Definition 3.

We say that a normal extremal $\lambda(t)$ with associated control
$\tilde{u}(t)$ is strictly normal if the differential of the endpoint map at
$\tilde{u}$ is surjective.

It makes sense to go on and consider higher order optimality conditions. At
critical points is well defined (i.e. independent of coordinates) the Hessian
of $\mathcal{J}$ (or the _second variation_). Using chronological calculus
(see again [6] or [1]) it is possible to write the second variation of
$\mathcal{J}$ on $\ker dE^{t}\subseteq L^{\infty}([0,1],\mathbb{R}^{k})$.

######  Proposition 7 (Second variation).

Suppose that $(\lambda(t),\tilde{u})$ is a strictly normal extremal, i.e. a
critical point of $\mathcal{J}$ for fixed initial and final point. For any
$u\in L^{\infty}([0,1],\mathbb{R}^{k})$ such that $\int_{0}^{1}X_{t}u_{t}dt=0$
the second variation of $\mathcal{J}$ has the following expression:

$d^{2}_{\tilde{u}}\mathcal{J}(u)=-\int_{0}^{1}\langle H_{t}u_{t},u_{t}\rangle
dt-\int_{0}^{1}\int_{0}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}u_{t})d\tau dt$

The associated bilinear form is symmetric provided that $u,v$ lie in a
subspace that projects to a Lagrangian one via the map
$u\mapsto\int_{0}^{1}Z_{t}u_{t}dt$.

$d^{2}_{\tilde{u}}\mathcal{J}(u,v)=-\int_{0}^{1}\langle
H_{t}u_{t},v_{t}\rangle
dt-\int_{0}^{1}\int_{0}^{t}\sigma(Z_{\tau}u_{\tau},Z_{t}v_{t})d\tau dt$

Through out the paper we make the assumption, which is customarily called
_strong Legendre condition_ , that the matrix $H_{t}$ is strictly negative
definite and has uniformly bounded inverse. This guarantees that the term:

$\int_{0}^{1}-\langle H_{t}u_{t},v_{t}\rangle dt$

is equivalent to the $L^{2}$ scalar product.

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## Acknowledgements

I would like to thank Prof. Andrei Agrachev for the stimulating discussions
during the preparation of this work and Ivan Beschastnyi for helpful comments
on preliminary versions of this article. I would also wish to warmly thank the
anonymous reviewers for the many helpful suggestions which improved greatly
the manuscript.

## Declarations of interest

None

## Funding

This research did not receive any specific grant from funding agencies in the
public, commercial, or not-for-profit sectors.