# On generalized Berwald manifolds of dimension three

Csaba Vincze  Institute of Mathematics, University of Debrecen, H-4002
Debrecen, P.O.Box 400, Hungary<EMAIL_ADDRESS>and  Márk Oláh
Institute of Mathematics, University of Debrecen, Doctoral School of
Mathematical and Computational Sciences, H-4002 Debrecen, P.O.Box 400, Hungary
<EMAIL_ADDRESS>
###### Abstract.

A linear connection on a Finsler manifold is called compatible to the Finsler
function if its parallel transports preserve the Finslerian length of tangent
vectors. Generalized Berwald manifolds are Finsler manifolds equipped with a
compatible linear connection. In the paper we present a general and intrinsic
method to characterize the compatible linear connections on a Finsler manifold
of dimension three. We prove that if a compatible linear connection is not
unique then the indicatrices must be Euclidean surfaces of revolution. The
surplus freedom of choosing compatible linear connections is related to
Euclidean symmetries. The unicity of the solution of the compatibility
equations can be provided by some additional requirements. Following the idea
in [11] we are also looking for the so-called extremal compatible linear
connection minimizing the norm of its torsion at each point of the manifold.

###### Key words and phrases:

Finsler spaces, Generalized Berwald spaces, Intrinsic Geometry, Extremal
compatible linear connections.

###### 1991 Mathematics Subject Classification:

53C60, 58B20

11footnotetext: Cs. Vincze is supported by the EFOP-3.6.1-16-2016-00022
project. The project is co-financed by the European Union and the European
Social Fund.22footnotetext: Cs. Vincze and M. Oláh are also supported by TKA-
DAAD 307818.

##  1\. Notation and terminology

Let $M$ be a smooth manifold with a local coordinate system
($u^{1},\ldots,u^{n}$). The induced local coordinate system on the tangent
manifold $TM$ consists of the functions $x^{1},\ldots,x^{n}$ and
$y^{1},\ldots,y^{n}$ given by $x^{i}(v):=u^{i}\circ\pi(v)=u^{i}(p)=:p^{i}$,
where $\pi\colon TM\to M$ is the canonical projection (the "foot map") and
$y^{i}(v)=v(u^{i})$, $i\in\\{1,\ldots,n\\}$.

######  Definition 1.

A Finsler function is a continuous function $F\colon TM\to\mathbb{R}$
satisfying the following conditions: $\displaystyle{F}$ is smooth on the
complement of the zero section _(_ regularity _)_ ,
$\displaystyle{F(tv)=tF(v)}$ for all $\displaystyle{t>0}$ _(_ positive
homogeneity _)_ and the Hessian

$(g_{ij})=\left(\frac{\partial^{2}E}{\partial y^{i}\partial y^{j}}\right)$

of the energy function $E=F^{2}/2$ is positive definite at all nonzero
elements $\displaystyle{v\in T_{p}M}$ _(_ strong convexity _)_. The pair
$(M,F)$ is called a Finsler manifold.

######  Definition 2.

Let $(M,F)$ be a Finsler manifold. A linear connection $\nabla$ on $M$ is
called _compatible_ to $F$ if the parallel transports with respect to $\nabla$
preserve the Finslerian length of tangent vectors. Then the triplet
$(M,F,\nabla)$ is called a generalized Berwald manifold.

It is well-known that any linear connection $\nabla$ determines a horizontal
distribution spanned by the _horizontal vector fields_

$X_{i}^{h}:=\frac{\partial}{\partial
x^{i}}-y^{j}{\Gamma}^{k}_{ij}\circ\pi\frac{\partial}{\partial y^{k}},\quad
i\in\\{1,\ldots,n\\}.$

The vanishing of the derivative of $F$ along the integral curves of horizontal
vector fields characterizes the compatibility of $\nabla$ to $F$ [11].
Therefore a linear connection $\nabla$ is compatible to a Finsler function if
and only if

(1) $\frac{\partial F}{\partial
x^{i}}-y^{j}{\Gamma}^{k}_{ij}\circ\pi\frac{\partial F}{\partial y^{k}}=0,\quad
i\in\\{1,\ldots,n\\}.$

System (1) is called _compatibility equations_.

######  Definition 3.

A Riemannian metric $\gamma$ on a Finsler manifold is _compatible_ to $F$ if
every linear connection compatible to $F$ is metrical with respect to
$\gamma$.

It is well-known that the so-called _averaged Riemannian metric_ (defined by
the integration of the metric components $g_{ij}$ over the indicatrices) is
compatible to the Finsler function, for details see [4] and [6]. Compatible
Riemannian metrics can be introduced in many different ways as well [2], [3].
They can be directly given as the Riemannian part of the Finsler function in
case of a Randers manifold [5]. In what follows, let $\gamma$ be a given
compatible Riemannian metric to $F$ with its uniquely determined Lévi-Civita
connection $\nabla^{*}$. The horizontal vector fields generated by
$\nabla^{*}$ are

$X_{i}^{h^{*}}:=\frac{\partial}{\partial
x^{i}}-{y^{j}\Gamma}^{*k}_{ij}\circ\pi\frac{\partial}{\partial y^{k}},\quad
i\in\\{1,\ldots,n\\}.$

##  2\. The compatibility equations in terms of the torsion components

Following the idea in [11], we formulate the compatibility equations (1) in
terms of the torsion components of the compatible linear connection $\nabla$
instead of the connection parameters. Using the Christoffel process, we have
the formula

(2)
$\Gamma_{ij}^{r}=\Gamma_{ij}^{*r}-\frac{1}{2}\left(T^{l}_{jk}\gamma^{kr}\gamma_{il}+T^{l}_{ik}\gamma^{kr}\gamma_{jl}-T_{ij}^{r}\right),$

and system (1) can be written as

$X_{i}^{h^{*}}F+\frac{1}{2}y^{j}\left(T^{l}_{jk}\gamma^{kr}\gamma_{il}+T^{l}_{ik}\gamma^{kr}\gamma_{jl}-T_{ij}^{r}\right)\circ\pi\frac{\partial
F}{\partial y^{r}}=0,\quad i\in\\{1,\ldots,n\\}.$

Evaluating at a nonzero tangent vector $v\in T_{p}M$, we have the
inhomogeneous system

$y^{j}(v)\left(T^{l}_{jk}\gamma^{kr}\gamma_{il}+T^{l}_{ik}\gamma^{kr}\gamma_{jl}-T_{ij}^{r}\right)(p)\frac{\partial
F}{\partial y^{r}}(v)=-2X_{i}^{h^{*}}F(v),\quad i\in\\{1,\ldots,n\\}$

of linear equations for the components $T_{ab}^{c}(p)$. The set $A_{p}(v)$ of
its solutions is an affine subspace of the vector space
$\wedge^{2}T_{p}^{*}M\otimes T_{p}M$ spanned by

$du^{i}_{p}\wedge du^{j}_{p}\otimes\left(\frac{\partial}{\partial
u^{k}}\right)(p),\quad 1\leq i<j\leq n,\quad k\in\\{1,\ldots,n\\}.$

It is of dimension $\displaystyle{\binom{n}{2}n}$. Running through all nonzero
tangent vectors $v\in T_{p}M$, the solution set

$A_{p}=\bigcap_{v\in T_{p}M\setminus\\{\bf{0}\\}}A_{p}(v)$

containing the restrictions of the torsion tensors of all compatible linear
connections to the Cartesian product $\displaystyle{T_{p}M\times T_{p}M}$ is
also an affine subspace of $\displaystyle{\wedge^{2}T_{p}^{*}M\otimes T_{p}M}$
(as an intersection of affine subspaces).

######  Theorem 1.

If $(M,F,\nabla)$ is a connected generalized Berwald manifold, then the
mapping $p\in M\mapsto A_{p}\subset\wedge^{2}T_{p}^{*}M\otimes T_{p}M$ is a
smooth affine distribution of constant rank on the torsion tensor bundle.

###### Proof.

Let $\displaystyle{H_{p}}\subset\wedge^{2}T_{p}^{*}M\otimes T_{p}M$ be the
directional space of the affine subspace $A_{p}$, and let $\nabla$ be a
compatible linear connection. We are going to prove that

(3) $T_{q}(v,w):=\varphi\circ T_{p}(\varphi^{-1}(v),\varphi^{-1}(w))$

belongs to the directional space $H_{q}$ for any $T_{p}\in H_{p}$, where
$\varphi$ is a linear isometry between the tangent spaces at the corresponding
points $p$ and $q$. Such an isometry is generated by parallel transports with
respect to $\nabla$. Let $(Q_{i}^{j})$ be the matrix representation of
$\varphi$, $(P_{j}^{i})=\left(Q_{i}^{j}\right)^{-1}$. The directional space
$H_{q}\subset\wedge^{2}T_{q}^{*}M\otimes T_{q}M$ can be given as

(4)
$\frac{1}{2}y^{j}\circ\varphi(v)\left(T^{l}_{jk}\gamma^{kr}\gamma_{il}+T^{l}_{ik}\gamma^{kr}\gamma_{jl}-T_{ij}^{r}\right)(q)\frac{\partial
F}{\partial y^{r}}\circ\varphi(v)=0,\quad i\in\\{1,\ldots,n\\},$

where $v$ runs through the nonzero elements of $T_{p}M$. So does $\varphi(v)$
in $T_{q}M$. Therefore (4) is equivalent to

$\frac{1}{2}y^{b}(v)\left(T^{l}_{jk}Q_{b}^{j}P_{r}^{c}\gamma^{kr}\gamma_{il}+T^{l}_{ik}Q_{b}^{j}P_{r}^{c}\gamma^{kr}\gamma_{jl}-T_{ij}^{r}Q_{b}^{j}P_{r}^{c}\right)(q)\frac{\partial
F}{\partial y^{c}}(v)=0,$

where $i\in\\{1,\ldots,n\\}$ and $v\in T_{p}M$. Indeed,
$\displaystyle{\varphi^{j}=y^{j}\circ\varphi=Q_{b}^{j}y^{b}}$ and the
invariance property $F\circ\varphi=F$ imply that

$\frac{\partial F}{\partial y^{c}}(v)=\frac{\partial F\circ\varphi}{\partial
y^{c}}(v)=Q_{c}^{r}\frac{\partial F}{\partial y^{r}}\circ\varphi(v)\
\Rightarrow\ P_{r}^{c}\frac{\partial F}{\partial y^{c}}(v)=\frac{\partial
F}{\partial y^{r}}\circ\varphi(v).$

Using the equalities

$Q_{a}^{i}Q_{b}^{j}\gamma_{ij}(q)=\gamma_{ab}(p),\
P_{i}^{a}P_{j}^{b}\gamma^{ij}(q)=\gamma^{ab}(p),\
Q_{b}^{j}\gamma_{jl}(q)=P_{l}^{j}\gamma_{jb}(p)\ \ \textrm{and}\ \
P_{j}^{b}\gamma^{jk}(q)=Q_{j}^{k}\gamma^{jb}(p),$

we have

$\frac{1}{2}y^{b}(v)\left(T^{l}_{jk}(q)Q_{b}^{j}Q_{r}^{k}\gamma^{rc}(p)\gamma_{il}(q)+T^{l}_{ik}(q)Q_{b}^{j}Q_{r}^{k}\gamma^{rc}(p)\gamma_{jl}(q)-T_{ij}^{r}(q)Q_{b}^{j}P_{r}^{c}\right)\frac{\partial
F}{\partial y^{c}}(v)=0.$

Taking the product by the matrix $(Q^{i}_{a})$, an equivalent system of
equations is

$\frac{1}{2}y^{b}(v)\left(T^{l}_{jk}(q)Q_{a}^{i}Q_{b}^{j}Q_{r}^{k}\gamma^{rc}(p)\gamma_{il}(q)+T^{l}_{ik}(q)Q_{a}^{i}Q_{b}^{j}Q_{r}^{k}\gamma^{rc}(p)\gamma_{jl}(q)-T_{ij}^{r}(q)Q_{a}^{i}Q_{b}^{j}P_{r}^{c}\right)\frac{\partial
F}{\partial y^{c}}(v)=0,$
$\frac{1}{2}y^{b}(v)\left(T^{l}_{jk}(q)P_{l}^{i}Q_{b}^{j}Q_{r}^{k}\gamma^{rc}(p)\gamma_{ia}(p)+T^{l}_{ik}(q)Q_{a}^{i}P_{l}^{j}Q_{r}^{k}\gamma^{rc}(p)\gamma_{jb}(p)-T_{ij}^{r}(q)Q_{a}^{i}Q_{b}^{j}P_{r}^{c}\right)\frac{\partial
F}{\partial y^{c}}(v)=0,$
$\frac{1}{2}y^{b}(v)\left(T_{br}^{i}\gamma^{rc}\gamma_{ia}+T_{ar}^{j}\gamma^{rc}\gamma_{jb}-T_{ab}^{c}\right)(p)\frac{\partial
F}{\partial y^{c}}(v)=0,\quad a\in\\{1,\ldots,n\\}$

for any nonzero $v\in T_{p}M$, where
$T_{ab}^{c}(p):=T_{ij}^{r}(q)Q_{a}^{i}Q_{b}^{j}P_{r}^{c}$, i.e.,
$T_{ij}^{r}(q)=P^{a}_{i}P_{j}^{b}T_{ab}^{c}(p)Q_{c}^{r}$ as in formula (3). ∎

Now we rewrite the compatibility equations in a more compact form. Due to the
skew-symmetry, it is enough to keep the components $T_{ab}^{c}$ with indices
$a<b$. Denoting by $\sigma_{ab;i}^{c}$ the coefficient of $T_{ab}^{c}$ ($a<b$)
in the $i$-th equation, we have

$\sigma_{ab;i}^{c}=\left(y^{a}\gamma^{br}-y^{b}\gamma^{ar}\right)\frac{\partial
F}{\partial
y^{r}}\gamma_{ic}+\left(\delta_{i}^{a}\gamma^{br}-\delta_{i}^{b}\gamma^{ar}\right)\frac{\partial
F}{\partial
y^{r}}y^{j}\gamma_{jc}-\left(\delta_{i}^{a}y^{b}-\delta_{i}^{b}y^{a}\right)\frac{\partial
F}{\partial y^{c}}.$

If $\displaystyle{(\partial/\partial u^{1},\ldots,\partial/\partial u^{n})}$
is an orthonormal basis at $p\in M$ with respect to the compatible Riemannian
metric $\gamma$, then

(5) $\sigma_{ab;i}^{c}=\delta_{i}^{c}\left(y^{a}\frac{\partial F}{\partial
y^{b}}-y^{b}\frac{\partial F}{\partial
y^{a}}\right)+\delta_{i}^{a}\left(y^{c}\frac{\partial F}{\partial
y^{b}}-y^{b}\frac{\partial F}{\partial
y^{c}}\right)-\delta_{i}^{b}\left(y^{c}\frac{\partial F}{\partial
y^{a}}-y^{a}\frac{\partial F}{\partial y^{c}}\right).$

Introducing the notation

(6) $f_{ab}=y^{a}\frac{\partial F}{\partial y^{b}}-y^{b}\frac{\partial
F}{\partial y^{a}},$

the coefficients are

(7)
$\sigma_{ab;i}^{c}=\delta_{i}^{c}f_{ab}+\delta_{i}^{a}f_{cb}+\delta_{i}^{b}f_{ac},$

and the compatibility equations take the form

(8) $\sum_{a<b,c}\sigma_{ab;i}^{c}T_{ab}^{c}=-2X_{i}^{h^{*}}F,\quad
i\in\\{1,\ldots,n\\},$

where the summation symbol means summing over the indices

$\\{(a,b,c)\in\mathbb{N}_{n}\times\mathbb{N}_{n}\times\mathbb{N}_{n}\,|\,a<b\\},\quad\mathbb{N}_{n}:=\\{1,\dots,n\\}.$

###  2.1. The case of Finsler surfaces

In case of dimension two, the compatibility equations are easy to solve
because we have only two variables $T_{12}^{1}$ and $T_{12}^{2}$ with
coefficients $\sigma_{12;i}^{1}=2\delta_{i}^{1}f_{12}$ and
$\sigma_{12;i}^{2}=2\delta_{i}^{2}f_{12}$. The compatibility equations are

$f_{12}T_{12}^{1}=-X_{1}^{h^{*}}F,\quad f_{12}T_{12}^{2}=-X_{2}^{h^{*}}F.$

The uniquely determined solutions can be expressed (for example) in terms of
quantities given by integration along the Euclidean unit circle with respect
to the compatible Riemannian metric in the tangent spaces:

$T_{12}^{1}=-\int_{c_{*}}f_{12}X_{1}^{h^{*}}F\bigg{/}\int_{c_{*}}f^{2}_{12},\quad
T_{12}^{2}=-\int_{c_{*}}f_{12}X_{2}^{h^{*}}F\bigg{/}\int_{c_{*}}f^{2}_{12}.$

If $f_{12}$ is identically zero in $T_{p}M$, then the Finslerian unit circle
is homothetic to the Riemannian unit circle because $F\circ c_{*}$ is
constant. In case of a generalized Berwald surface, this means that it is a
Riemannian surface due to the invariance of the indicatrices under the
parallel transports given by a linear connection. Another aspects and examples
in case of dimension 2 can be found in [8], [9] and [10].

###  2.2. The extremal compatible linear connection

Consider a generalized Berwald space. As we have seen above, the pointwise
solution sets $A_{p}$ of the compatibility equations are affine subspaces of
the vector spaces $\displaystyle{\wedge^{2}T_{p}^{*}M\otimes T_{p}M}$ ($p\in
M$). This means that the global solution is not necessarily unique even we
have a global solution. The unicity can be provided by some additional
requirements. Following the idea in [11] we are looking for the so-called
extremal compatible linear connection minimizing the norm of its torsion at
each point of the manifold in the following sense.

######  Definition 4.

Let $(M,F)$ be a Finsler manifold with a compatible Riemannian metric
$\gamma$, and suppose that the coordinate vector fields $\partial/\partial
u^{1},\ldots,\partial/\partial u^{n}$ form an orthonormal basis at a point
$p\in M$ with respect to $\gamma$. We introduce a Riemannian metric on
$\wedge^{2}T_{p}^{*}M\otimes T_{p}M$ in the following way: if
$\displaystyle{T=\sum_{i<j,k}T_{ij}^{k}du^{i}\wedge
du^{j}\otimes\frac{\partial}{\partial u^{k}}}$, then

(9) $\langle T_{p},S_{p}\rangle:=\sum_{i<j,k}T_{ij}^{k}(p)S_{ij}^{k}(p).$

The extremal compatible linear connection on a generalized Berwald space $M$
is the uniquely determined compatible linear connection whose torsion
minimizes the norm arising from _(_ 9 _)_.

Since the solution set $A_{p}\subset\wedge^{2}T_{p}^{*}M\otimes T_{p}M$ of the
compatibility equations at $p\in M$ is an affine (especially, convex) set, its
closest element to the origin is uniquely determined. In case of a generalized
Berwald manifold the pointwise extremal solutions form a continuous section of
the torsion bundle. Therefore the connection parameters of the corresponding
linear connection are obviously continuous. Conversely, the existence of a
continuous compatible linear connection implies that the manifold is
monochromatic and, consequently, it is a generalized Berwald manifold; for
details see [1] and [11].

##  3\. The compatibility equations in 3D

In what follows, we are going to solve the compatibility equations for
3-dimensional Finsler manifolds. All compatible linear connections will be
determined by their torsion components and we will find the extremal one among
them. The trick is to group the variables and work in a 3D space three times
instead of the $9$-dimensional fiber of the torsion tensor bundle.

###  3.1. Geometric structures on the tangent spaces

Let $p$ be a given point of $M$.

  1. (1)

The Finsler function $F$ restricted to $T_{p}M$ is a Minkowski norm, i.e., the
level surface $F^{-1}(1)\cap T_{p}M$ is the boundary of a (strictly) convex
body containing the origin in its interior. It is the Finslerian indicatrix
or, unit sphere, at the point $p\in M$.

  2. (2)

The compatible Riemannian metric $\gamma$ restricted to $T_{p}M$ is the
standard Euclidean metric after choosing the orthonormal basis
$\displaystyle{\partial/\partial u^{1},\partial/\partial
u^{2},\partial/\partial u^{3}}$ at the point $p\in M$. Its indicatrix is the
Euclidean unit sphere.

In what follows we simply write $\langle v,w\rangle$ instead of
$\gamma_{p}(v,w)$. Orthogonality, norms and normal vectors are taken in the
usual Euclidean sense. Since the Finslerian spheres are the level sets of $F$,
the Euclidean gradient vector field

$G:=\mathrm{grad}\ F=\dfrac{\partial F}{\partial
y^{1}}\dfrac{\partial}{\partial y^{1}}+\dfrac{\partial F}{\partial
y^{2}}\dfrac{\partial}{\partial y^{2}}+\dfrac{\partial F}{\partial
y^{3}}\dfrac{\partial}{\partial y^{3}}\sim\left(\dfrac{\partial F}{\partial
y^{1}},\dfrac{\partial F}{\partial y^{2}},\dfrac{\partial F}{\partial
y^{3}}\right)=\left(\partial_{1}F,\partial_{2}F,\partial_{3}F\right)$

gives the (outer) normals. The (outer) normal vectors of the tangent planes of
a Euclidean sphere are given by the radial or Liouville vector field

$C:=y^{1}\dfrac{\partial}{\partial y^{1}}+y^{2}\dfrac{\partial}{\partial
y^{2}}+y^{3}\dfrac{\partial}{\partial
y^{3}}\sim\left(y^{1},y^{2},y^{3}\right).$

In 3D two tangent planes are either parallel or intersect in a common line.

######  Definition 5.

_[11]_ We call a nonzero vector $v\in T_{p}M$

  * •

a _vertical contact point_ , if the Finslerian and Riemannian spheres have the
same tangent planes at $v$, i.e., the Euclidean normal vectors $G_{v}$ and
$C_{v}$ are parallel;

  * •

a _horizontal contact point_ , if $X_{i}^{h^{*}}F(v)=0$ for all
$i\in\\{1,\ldots,n\\}$.

The tangent space $T_{p}M$ is called vertical/horizontal contact if all of its
nonzero elements are vertical/horizontal contact.

We briefly summarize some basic facts about the vertical and horizontal
contact points, for details we refer to [11].

  * •

At vertical contact points the coefficients $\sigma_{ab;i}^{c}$ in the
compatibility equations are zero.

  * •

If the compatibility equations are solvable, i.e., we have a generalized
Berwald manifold, then all vertical contact points are horizontal contact.

  * •

If we have a _connected_ generalized Berwald manifold with a vertical contact
tangent space, then it is a Riemannian manifold and its extremal compatible
linear connection is the Lévi–Civita connection of the compatible Riemannian
metric.

  * •

If $T_{p}M$ is a horizontal contact tangent space, then the torsion of the
extremal compatible linear connection at $p$ is zero because the system of the
compatibility equations is homogeneous.

Suppose that the tangent space $T_{p}M$ contains at least one element $v$ that
is not vertical contact. Since $G_{v}$ and $C_{v}$ are linearly independent

  * •

we can use the Gram–Schmidt process to construct an orthogonal pair by
substituting $G$ with

$G^{\bot}_{v}:=G_{v}-\frac{\langle G_{v},C_{v}\rangle}{\langle
C_{v},C_{v}\rangle}C_{v}=G_{v}-\frac{C_{v}F}{\langle
v,v\rangle}C_{v}=G_{v}-\frac{F(v)}{\langle v,v\rangle}C_{v}$

because of the first order homogeneity of the Finsler function.

  * •

If $C\times G$ is the cross product of the vector fields $C$ and $G$, then

(10) $C\times G=\begin{vmatrix}e_{1}&e_{2}&e_{3}\\\ y^{1}&y^{2}&y^{3}\\\
\partial_{1}F&\partial_{2}F&\partial_{3}F\end{vmatrix}=\begin{pmatrix}y^{2}\partial_{3}F-y^{3}\partial_{2}F\\\
y^{3}\partial_{1}F-y^{1}\partial_{3}F\\\
y^{1}\partial_{2}F-y^{2}\partial_{1}F\end{pmatrix}=\begin{pmatrix}f_{23}\\\
f_{31}\\\ f_{12}\end{pmatrix}.$

It is clear that the cross product vanishes at the vertical contact elements
in the tangent spaces and vice versa. According to the orthogonality to both
$C_{v}$ and $G_{v}$, it lies in the tangent planes of both the Euclidean and
the Finslerian spheres passing through $v$. Finally, $(C,G^{\bot},C\times G)$
is an orthogonal frame on the complement of the vertical contact elements.

######  Remark 1.

_For any $p\in M$, the tangent space $T_{p}M$ has a vertical contact point.
Otherwise the vector field $C\times G$ would be a (continuous) non-vanishing
tangent vector field to the 3-dimensional Euclidean unit sphere contradicting
the hedgehog theorem. Another possible argument independently of the dimension
is based on the furthest point property and convexity: the tangent plane to
the Finslerian unit sphere at the furthest point $v$ from the origin with
respect to the Euclidean metric must be orthogonal to $v$._

######  Theorem 2.

In case of a three-dimensional Finsler manifold, the compatibility equations
can be written in the form

(11) $\left<C\times G,\begin{pmatrix}\ -T_{12}^{3}+T_{13}^{2}+T_{23}^{1}\\\
-2T_{13}^{1}\\\ \ \ 2T_{12}^{1}\end{pmatrix}\right>=-2X_{1}^{h^{*}}F,$ (12)
$\left<C\times G,\begin{pmatrix}2T_{23}^{2}\\\
-T_{12}^{3}-T_{13}^{2}-T_{23}^{1}\\\
2T_{12}^{2}\end{pmatrix}\right>=-2X_{2}^{h^{*}}F,$ (13) $\left<C\times
G,\begin{pmatrix}\ \ 2T_{23}^{3}\\\ -2T_{13}^{3}\\\
T_{12}^{3}+T_{13}^{2}-T_{23}^{1}\ \ \end{pmatrix}\right>=-2X_{3}^{h^{*}}F,$

where the new variables are groups of the torsion components and

$G:=\mathrm{grad}\ F\sim\left(\dfrac{\partial F}{\partial
y^{1}},\dfrac{\partial F}{\partial y^{2}},\dfrac{\partial F}{\partial
y^{3}}\right)=\left(\partial_{1}F,\partial_{2}F,\partial_{3}F\right).$

###### Proof.

Recall that the coefficient $\sigma_{ab;i}^{c}$ of the torsion component
$T_{ab}^{c}$ in the $i$-th compatibility equation is

(14)
$\sigma_{ab;i}^{c}=\delta_{i}^{c}f_{ab}+\delta_{i}^{a}f_{cb}+\delta_{i}^{b}f_{ac},\
\ \textrm{where}\ \ f_{ab}=y^{a}\frac{\partial F}{\partial
y^{b}}-y^{b}\frac{\partial F}{\partial y^{a}},\ \ \textrm{and so on}.$

Depending on the repetition among the indices $a<b$ and $c$, we have $8$ cases
as the following table shows. In the first row, they can take any value
different from $i$. Otherwise, equal indices are in the same cells, and
different cells contain different indices.

| indices | the coefficients
---|---|---
1 | $i$ | $a$ | $b$ | $c$ | $\sigma_{ab;i}^{c}=0$
2 | $i=a$ | $b$ | $c$ |  | $\sigma_{ib;i}^{c}=f_{cb}$
3 | $i=a$ | $b=c$ |  |  | $\sigma_{ib;i}^{b}=0$
4 | $i=b$ | $a$ | $c$ |  | $\sigma_{ai;i}^{c}=f_{ac}$
5 | $i=b$ | $a=c$ |  |  | $\sigma_{ai;i}^{a}=0$
6 | $i=c$ | $a$ | $b$ |  | $\sigma_{ab;i}^{i}=f_{ab}$
7 | $i=a=c$ | $b$ |  |  | $\sigma_{ib;i}^{i}=2f_{ib}$
8 | $i=b=c$ | $a$ |  |  | $\sigma_{ai;i}^{i}=2f_{ai}$

To sum up, the surviving coefficients in the $i$-th equation contain at least
one index equal to $i$ and, if exactly one index equals to $i$, then the
remaining two indices must be different. Thus, the compatibility equations in
3D (in matrix form) are

| $T_{12}^{1}$ | $T_{12}^{2}$ | $T_{12}^{3}$ | $T_{13}^{1}$ | $T_{13}^{2}$ | $T_{13}^{3}$ | $T_{23}^{1}$ | $T_{23}^{2}$ | $T_{23}^{3}$ | RHS
---|---|---|---|---|---|---|---|---|---|---
1 | $2f_{12}$ | $0$ | $f_{32}$ | $2f_{13}$ | $f_{23}$ | $0$ | $f_{23}$ | $0$ | $0$ | $-2X_{1}^{h^{*}}F$
2 | $0$ | $2f_{12}$ | $f_{13}$ | $0$ | $f_{13}$ | $0$ | $f_{13}$ | $2f_{23}$ | $0$ | $-2X_{2}^{h^{*}}F$
3 | $0$ | $0$ | $f_{12}$ | $0$ | $f_{12}$ | $2f_{13}$ | $f_{21}$ | $0$ | $2f_{23}$ | $-2X_{3}^{h^{*}}F$

,

and we can group the components in the following way:

$f_{23}(-T_{12}^{3}+T_{13}^{2}+T_{23}^{1})$ | $-$ | $2f_{31}T_{13}^{1}$ | $+$ | $2f_{12}T_{12}^{1}$ | $=$ |  $-2X_{1}^{h^{*}}F$,
---|---|---|---|---|---|---
$2f_{23}T_{23}^{2}$ | $+$ | $f_{31}(-T_{12}^{3}-T_{13}^{2}-T_{23}^{1})$ | $+$ | $2f_{12}T_{12}^{2}$ | $=$ |  $-2X_{2}^{h^{*}}F$,
$2f_{23}T_{23}^{3}$ | $-$ | $2f_{31}T_{13}^{3}$ | $+$ | $f_{12}(T_{12}^{3}+T_{13}^{2}-T_{23}^{1})$ | $=$ |  $-2X_{3}^{h^{*}}F$.

Since the coefficients are those of the cross product defined by formula (10),
we can indeed write the compatibility equations in the form (11), (12) and
(13). ∎

######  Lemma 1.

For any not vertical contact tangent space $T_{p}M$, the common directional
space of $(\ref{comp1})$, $(\ref{comp2})$ and $(\ref{comp3})$ is trivial or it
is a one-dimensional linear subspace in $T_{p}M$, which is the rotational axis
of the Finslerian indicatrix.

###### Proof.

The common directional space is given by

(15) $\left<C\times G,\begin{pmatrix}t^{1}\\\ t^{2}\\\
t^{3}\end{pmatrix}\right>=0.$

Taking a not vertical contact element $v\in T_{p}M$, (15) can be refined as
follows: since $C_{v}\times G_{v}\neq{\bf{0}}$ we can consider the non-trivial
integral curve $c$ of the cross product passing through $v$. Since
$c^{\prime}=(C\times G)\circ c$, it follows that $c^{\prime}(t)\ \bot\
c(t)\sim C_{c(t)}$ with respect to the Euclidean inner product, i.e.,

$\langle c,c\rangle=\textrm{const.}\ \ \Rightarrow\ \ \langle
c^{\prime},c\rangle(0)=0\ \ \Rightarrow\ \ \langle
c^{\prime\prime},c\rangle(0)=-\langle c^{\prime},c^{\prime}\rangle(0)=-\langle
C_{v}\times G_{v},C_{v}\times G_{v}\rangle\neq 0.$

This means that $c^{\prime}(0)$ and $c^{\prime\prime}(0)$ are linearly
independent. Therefore, evaluating (15) along $c$, a simple differentiation
gives

(16) $\left<c^{\prime}(0),\begin{pmatrix}t^{1}\\\ t^{2}\\\
t^{3}\end{pmatrix}\right>=0,\ \
\left<c^{\prime\prime}(0),\begin{pmatrix}t^{1}\\\ t^{2}\\\
t^{3}\end{pmatrix}\right>=0.$

Hence the directional space is at most one-dimensional. Suppose that
$\vec{t}\neq{\bf 0}$ is a solution of (15) for any not vertical contact $v\in
T_{p}M$. In case of vertical contact elements, (15) is automatically satisfied
because of the vanishing of the vector field $C\times G$. It follows that

$0=\left<C_{v}\times G_{v},\vec{t}\
\right>=-\left<C_{v}\times\vec{t},G_{v}\right>=C_{v}\times\vec{t}\ F\quad(v\in
T_{p}M).$

Since the integral curves of $C\times\vec{t}$ must satisfy the equation
$c(t)\times\vec{t}=c^{\prime}(t)$, we have $c(t)=e^{tA}c(0)$, where $A$ is the
(skew-symmetric) matrix of the linear transformation $v\mapsto
v\times\vec{t}$. This means that $(F\circ c)^{\prime}(t)=0$, i.e., $F$ is
constant along the orbits under the action of the one-parameter rotational
group generated by $A$. ∎

######  Corollary 1.

If $(M,F,\nabla)$ is a connected generalized Berwald manifold of dimension
three, then the mapping $p\in M\mapsto A_{p}\subset\wedge^{2}T_{p}^{*}M\otimes
T_{p}M$ is a smooth affine distribution on the torsion tensor bundle of
constant rank $0$ or $1$.

######  Theorem 3.

If $M$ is a connected non-Riemannian generalized Berwald manifold of dimension
three, then we have the following possible cases.

  * (UDC)

The Finslerian indicatrix is a Euclidean surface of revolution at some and
therefore all points of the manifold and the rotational axes are generated by
a globally well-defined nowhere vanishing covariant constant vector field
$D\in\mathfrak{X}(M)$ with respect to any compatible linear connection. One of
them is given by

(17) $\nabla_{X}Y=\nabla^{*}_{X}Y+\frac{\langle\nabla^{*}_{X}D,Y\rangle
D-\langle Y,D\rangle\nabla^{*}_{X}D}{K^{2}},$

where $K^{2}$ is the constant norm square of $D$.

  * (DC)

We have a uniquely determined flat compatible linear connection given by

(18) $\nabla_{X}Y=\nabla^{*}_{X}Y-\rho(X)\times Y,$

where $\rho$ is an endomorphism of $\mathfrak{X}(M)$ satisfying

$R^{*}(X,Y)Z=\bigg{(}(\nabla^{*}_{X}\rho)(Y)-(\nabla^{*}_{Y}\rho)(X)-\rho(X)\times\rho(Y)\bigg{)}\times
Z.$

###### Proof.

In the undetermined case (UDC) let $p$ be a point such that (15) has a non-
zero solution $\vec{t}$ in $T_{p}M$. By the previous lemma, the Finslerian
indicatrix at the point $p$ is a Euclidean surface of revolution. Using
parallel transports with respect to one of the compatible linear connections
from $p$ to an arbitrary point of the manifold, it follows that all Finslerian
indicatrices are Euclidean surfaces of revolution. Since we have a non-
Riemannian generalized Berwald manifold, the rotational axes must be uniquely
determined [7] (Corollary 8). Choosing a directional vector at a single point
$p\in M$, we can extend it to a globally well-defined nowhere vanishing
covariant constant vector field $D\in\mathfrak{X}(M)$ by parallel transports
with respect to a compatible linear connection. Especially, $\nabla D=0$,
i.e., the parallel transports with respect to $\nabla$ preserve $D$.
Conversely, since the Finslerian indicatrices are surfaces of revolution and
they are isometric to each other, it is enough to keep the rotational axes
invariant by the parallel transports of a metric linear connection in the
Riemannian sense to have a compatible linear connection to the Finsler
function. Therefore (17) is one of them. If we have (DC) (determined case),
then the solution space of (15) is trivial at any point of the manifold. To
avoid Euclidean surfaces of revolution and (UDC), the unit component of the
holonomy group of the compatible linear connection must be trivial. Therefore
the compatible linear connection is flat because $R_{p}(X,Y)\neq{\bf 0}$ would
generate a one-parameter subgroup in the holonomy group leaving the indicatrix
invariant. Since

$\nabla^{*}_{X}Y=\nabla_{X}Y+A(X,Y)$

for some tensor field of type ($1,2$), it follows that $\nabla$ is a metric
linear connection if and only if

$\langle A(X,Y),Z\rangle=-\langle A(X,Z),Y\rangle.$

The skew-symmetry implies that

$\nabla^{*}_{X}Y=\nabla_{X}Y+\rho(X)\times Y.$

A simple calculation shows

$\nabla_{X}\nabla_{Y}Z=\nabla^{*}_{X}\nabla_{Y}Z-\rho(X)\times\nabla_{Y}Z=$
$\nabla^{*}_{X}\bigg{(}\nabla^{*}_{Y}Z-\rho(Y)\times
Z\bigg{)}-\rho(X)\times\bigg{(}\nabla^{*}_{Y}Z-\rho(Y)\times Z\bigg{)}=$
$\nabla^{*}_{X}\nabla^{*}_{Y}Z-\bigg{(}(\nabla^{*}_{X}\rho)(Y)-\rho(\nabla^{*}_{X}Y)\bigg{)}\times
Z-\rho(Y)\times\nabla^{*}_{X}Z-\rho(X)\times\nabla^{*}_{Y}Z+$
$\rho(X)\times\bigg{(}\rho(Y)\times Z\bigg{)}.$

For the sake of simplicity suppose that $[X,Y]=0$ before changing the role of
$X$ and $Y$. We have that

$R(X,Y)Z=R^{*}(X,Y)Z-(\nabla^{*}_{X}\rho)(Y)\times
Z+(\nabla^{*}_{Y}\rho)(X)\times Z+\bigg{(}\rho(X)\times\rho(Y)\bigg{)}\times
Z$

because of $\nabla^{*}_{X}Y-\nabla^{*}_{Y}X=0$ and, using the Jacobi identity
for the cross product. Since $\nabla$ is flat, the formula for the curvature
of the Lévi-Civita connection follows immediately. ∎

###  3.2. General method for solution

Following the technique in the proof of Lemma 1, let us choose a not vertical
contact elemet $v\in T_{p}M$. The solution of (16) gives a one-dimensional
directional space. Using the right hand sides of (11), (12) and (13),
respectively, the solutions of the inhomogeneous version of (16) give one-
dimensional affine subspaces. Therefore we can speak about _solution lines_
belonging to a not vertical contact element $v\in T_{p}M$. The vertical
contact elements must be horizontal contact. The possible cases are the
following:

  * •

there is no solution of the compatibility equations,

  * •

(determined case) the solution is uniquely determined, i.e., the corresponding
solution lines intersect each other at the same point for all not vertical
contact elements and all vertical contact elements are horizontal contact;

  * •

(undetermined case) all vertical contact elements are horizontal contact and
the corresponding solution lines coincide for all not vertical contact element
in $T_{p}M$. The common directional space is generated by the rotational axis
of the Finslerian indicatrix.

Consider the equations

(19) $\left<C\times G,\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=b_{i},\quad i\in\\{1,2,3\\}$

in the tangent spaces. Especially,

$b_{i}=-2X_{i}^{h^{*}}F;$

cf. equations (11), (12) and (13).

######  Lemma 2.

Let $G=\mathrm{grad}\ F$ be the Euclidean gradient on the Finslerian spheres
with respect to the compatible Riemannian metric, and consider the vector
field $C\times G$. Taking the integral curve $c$ starting from a not vertical
contact element $v$ in the tangent space $T_{p}M$, the solutions of (19) at
the point $p$ are of the form

(20) $\begin{pmatrix}\vec{t}_{1}\\\ \vec{t}_{2}\\\
\vec{t}_{3}\end{pmatrix}=\begin{pmatrix}\omega_{11}&\omega_{12}&\omega_{13}\\\
\omega_{21}&\omega_{22}&\omega_{23}\\\
\omega_{31}&\omega_{32}&\omega_{33}\end{pmatrix}\begin{pmatrix}c^{\prime}(0)\\\
c^{\prime\prime}(0)\\\ c^{\prime}(0)\times c^{\prime\prime}(0)\end{pmatrix},$

where

$\omega_{i1}=\frac{b_{i}(v)}{\ |c^{\prime}(0)|^{2}}-\frac{\langle
c^{\prime}(0),c^{\prime\prime}(0)\rangle}{|c^{\prime}(0)|^{5}\kappa^{2}(0)}\left(\frac{b_{i}\circ
c}{|c^{\prime}|}\right)^{\prime}(0),\
\omega_{i2}=\frac{1}{|c^{\prime}(0)|^{3}\kappa^{2}(0)}\left(\frac{b_{i}\circ
c}{|c^{\prime}|}\right)^{\prime}(0)$

and $b_{i}=-2X_{i}^{h^{*}}F$, $i\in\\{1,2,3\\}$.

###### Proof.

Taking a not vertical contact element $v\in T_{p}M$, (19) can be refined as
follows: since $C_{v}\times G_{v}\neq{\bf{0}}$ we can consider the non-trivial
integral curve $c$ of the cross product starting at $v$. Since
$c^{\prime}=(C\times G)\circ c$, it follows that $c^{\prime}(t)\ \bot\
c(t)\sim C_{c(t)}$ with respect to the Euclidean inner product, i.e.,

$\langle c,c\rangle=\textrm{const.}\ \ \Rightarrow\ \ \langle
c^{\prime},c\rangle(0)=0\ \ \Rightarrow\ \ \langle
c^{\prime\prime},c\rangle(0)=-\langle c^{\prime},c^{\prime}\rangle(0)=-\langle
C_{v}\times G_{v},C_{v}\times G_{v}\rangle\neq 0.$

This means that $c^{\prime}(0)$ and $c^{\prime\prime}(0)$ are linearly
independent. Therefore, evaluating (19) along $c$, a simple differentiation
gives

$\left<c^{\prime}(0),\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=b_{i}(v),\
\left<c^{\prime\prime}(0),\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=(b_{i}\circ c)^{\prime}(0),\quad
i\in\\{1,2,3\\}.$

Finding the solution in the form

$\vec{t}_{i}=\omega_{i1}c^{\prime}(0)+\omega_{i2}c^{\prime\prime}(0)+\omega_{i3}c^{\prime}(0)\times
c^{\prime\prime}(0)$

we can express the coefficients $\omega_{i1}$ and $\omega_{i2}$. Since

$\omega_{i1}\left|c^{\prime}(0)\right|^{2}+\omega_{i2}\left<c^{\prime}(0),c^{\prime\prime}(0)\right>=b_{i}(v)\
\ \textrm{and}\ \
\omega_{i1}\left<c^{\prime\prime}(0),c^{\prime}(0)\right>+\omega_{i2}\left|c^{\prime\prime}(0)\right|^{2}=(b_{i}\circ
c)^{\prime}(0),$

it follows by Cramer’s rule that

$\omega_{i1}=\frac{b_{i}(v)\left|c^{\prime\prime}(0)\right|^{2}-\left<c^{\prime}(0),c^{\prime\prime}(0)\right>(b_{i}\circ
c)^{\prime}}{|c^{\prime}(0)\times c^{\prime\prime}(0)|^{2}}=\frac{b_{i}(v)}{\
|c^{\prime}(0)|^{2}}-\frac{\langle
c^{\prime}(0),c^{\prime\prime}(0)\rangle}{|c^{\prime}(0)|^{5}\kappa^{2}(0)}\left(\frac{b_{i}\circ
c}{|c^{\prime}|}\right)^{\prime}(0),$
$\omega_{i2}=\frac{\left|c^{\prime}(0)\right|^{2}(b_{i}\circ
c)^{\prime}-b_{i}(v)\left<c^{\prime\prime}(0),c^{\prime}(0)\right>}{|c^{\prime}(0)\times
c^{\prime\prime}(0)|^{2}}=\frac{1}{|c^{\prime}(0)|^{3}\kappa^{2}(0)}\left(\frac{b_{i}\circ
c}{|c^{\prime}|}\right)^{\prime}(0)$

because of

$|c^{\prime}(0)\times
c^{\prime\prime}(0)|^{2}=|c^{\prime}(0)|^{2}|c^{\prime\prime}(0)|^{2}-\langle
c^{\prime}(0),c^{\prime\prime}(0)\rangle^{2}$

and

$\left(\frac{b_{i}\circ
c}{|c^{\prime}|}\right)^{\prime}(0)=\frac{|c^{\prime}(0)|^{2}(b_{i}\circ
c)^{\prime}-b_{i}(v)\langle
c^{\prime}(0),c^{\prime\prime}(0)\rangle}{|c^{\prime}(0)|^{3}}.$

∎

According to the previous Lemma, the coefficients $\omega_{i1}$ and
$\omega_{i2}$ are uniquely determined. They are related to the curvature of
the integral curve of $C\times G$ (in a more general sense: the derivatives of
the integral curve up to order two). To provide the arclength parametrization
of the integral curve we are motivated to reformulate the argumentation by
considering the normalized vector field of $C\times G$, where
$G=\mathrm{grad}\ F$ is the Euclidean gradient on the indicatrix.

######  Corollary 2.

Let $G=\mathrm{grad}\ F$ be the Euclidean gradient on the Finslerian spheres
with respect to the compatible Riemannian metric, and consider the normalized
vector field of $C\times G$. Taking the integral curve $c_{0}$ starting from a
not vertical contact element $v$ in the tangent space $T_{p}M$, the solutions
of (19) at the point $p$ are of the form

(21) $\begin{pmatrix}\vec{t}_{1}\\\ \vec{t}_{2}\\\
\vec{t}_{3}\end{pmatrix}=\begin{pmatrix}\omega_{11}^{0}&\omega_{12}^{0}&\omega_{13}^{0}\\\
\omega_{21}^{0}&\omega_{22}^{0}&\omega_{23}^{0}\\\
\omega_{31}^{0}&\omega_{32}^{0}&\omega_{33}^{0}\end{pmatrix}\begin{pmatrix}\vec{T}(0)\\\
\vec{N}(0)\\\ \vec{B}(0)\end{pmatrix},$

where

$\omega_{i1}^{0}=b_{i}^{0}(v),\ \omega_{i2}^{0}=\frac{(b_{i}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)},\ b_{i}^{0}=\frac{b_{i}}{|C\times
G|}=-2\frac{X_{i}^{h^{*}}F}{|C\times G|},\quad i\in\\{1,2,3\\}$

and $(\vec{T},\vec{N},\vec{B})$ is the Frenet-Serret frame along $c_{0}$.

###### Proof.

Following the steps in the proof of the previous Lemma, the computations are
straightforward. ∎

Note that the determined part of (21) is lying in the osculating plane of the
curve $c_{0}$ (or the curve $c$) at the starting parameter $0$. To clarify the
contribution of the torsion of the integral curve to formula (21), we present
the following sufficient condition for the unicity of the solution.

######  Corollary 3.

Let $G=\mathrm{grad}\ F$ be the Euclidean gradient on the Finslerian spheres
with respect to the compatible Riemannian metric, and consider the normalized
vector field of $C\times G$. If the integral curve $c_{0}$ starting from a not
vertical contact element $v$ in the tangent space $T_{p}M$ has a non-vanishing
torsion at the starting parameter, then the solutions of (19) at the point $p$
are of the form

(22) $\begin{pmatrix}\vec{t}_{1}\\\ \vec{t}_{2}\\\
\vec{t}_{3}\end{pmatrix}=\begin{pmatrix}\omega_{11}^{0}&\omega_{12}^{0}&\omega_{13}^{0}\\\
\omega_{21}^{0}&\omega_{22}^{0}&\omega_{23}^{0}\\\
\omega_{31}^{0}&\omega_{32}^{0}&\omega_{33}^{0}\end{pmatrix}\begin{pmatrix}\vec{T}(0)\\\
\vec{N}(0)\\\ \vec{B}(0)\end{pmatrix},$

where

$\omega_{i1}^{0}=b_{i}^{0}(v),\ \omega_{i2}^{0}=\frac{(b_{i}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)},\ \omega_{i3}^{0}=\frac{(b_{i}^{0}\circ
c_{0})^{\prime\prime}(0)+\kappa^{2}(0)\omega_{i1}^{0}-\kappa^{\prime}(0)\omega_{i2}^{0}}{\kappa(0)\tau(0)},$
$b_{i}^{0}=\frac{b_{i}}{|C\times G|}=-2\frac{X_{i}^{h^{*}}F}{|C\times
G|},\quad i\in\\{1,2,3\\}.$

###### Proof.

Evaluating (19) along $c_{0}$, a simple differentiation gives

$\left<c_{0}^{\prime}(0),\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=b_{i}^{0}(v),\
\left<c_{0}^{\prime\prime}(0),\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=(b_{i}^{0}\circ c_{0})^{\prime}(0),\
\left<c_{0}^{\prime\prime\prime}(0),\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=(b_{i}^{0}\circ c_{0})^{\prime\prime}(0).$

The last equation allows us to express the missing parameters
$\omega_{13}^{0}$, $\omega_{23}^{0}$ and $\omega_{33}^{0}$ by substituting the
solution of the form (22). Note that
$\left<c_{0}^{\prime\prime}(0),c_{0}^{\prime}(0)\right>=0$ implies that

$\left<c_{0}^{\prime\prime\prime}(0),c_{0}^{\prime}(0)\right>=-\left<c_{0}^{\prime\prime}(0),c_{0}^{\prime\prime}(0)\right>=-\kappa^{2}(0)$

and
$\displaystyle{\left<c_{0}^{\prime\prime\prime}(0),c_{0}^{\prime\prime}(0)\right>=\frac{1}{2}\left<c_{0}^{\prime\prime},c_{0}^{\prime\prime}\right>^{\prime}(0)=\frac{1}{2}\left(\kappa^{2}\right)^{\prime}(0)=\kappa(0)\kappa^{\prime}(0)}$.
∎

######  Remark 2.

_If $v\in T_{p}M$ is not vertical contact, then the Euclidean and the
Finslerian indicatrices are transversally intersecting surfaces at $v$ and the
integral curve $c_{0}$ (or $c$) is a parametrization of the intersection
curve. The vanishing of the torsion means that the intersection is a plane
curve, i.e., it must be a part of a Euclidean circle. Otherwise, the torsion
of the intersection curve implies the possible values for the components of
the torsion of the uniquely determined compatible linear connection._

###  3.3. Compatible linear connections in the undetermined case (cf. Theorem
3)

Let $p\in M$ be a given point. If the solution sets of the compatibility
equations (11)–(13) are 1-dimensional, they are parallel lines in $T_{p}M$,
endowed with the inner product coming from the compatible Riemannian metric.
Taking a not vertical contact element $v\in T_{p}M$, formula (21) shows that
the general form of the solutions is

(23) $\begin{array}[]{rcl}\begin{pmatrix}-T_{12}^{3}+T_{13}^{2}+T_{23}^{1}\\\
-2T_{13}^{1}\\\ 2T_{12}^{1}\
\end{pmatrix}(p)&=&\omega_{11}^{0}\vec{T}(0)+\omega_{12}^{0}\vec{N}(0)+\omega_{13}^{0}\vec{B}(0),\\\
\\\ \begin{pmatrix}2T_{23}^{2}\\\ -T_{12}^{3}-T_{13}^{2}-T_{23}^{1}\\\
2T_{12}^{2}\end{pmatrix}(p)&=&\omega_{21}^{0}\vec{T}(0)+\omega_{22}^{0}\vec{N}(0)+\omega_{23}^{0}\vec{B}(0),\\\\[20.0pt]
\\\ \begin{pmatrix}\ \ 2T_{23}^{3}\\\ -2T_{13}^{3}\\\
T_{12}^{3}+T_{13}^{2}-T_{23}^{1}\ \
\end{pmatrix}(p)&=&\omega_{31}^{0}\vec{T}(0)+\omega_{32}^{0}\vec{N}(0)+\omega_{33}^{0}\vec{B}(0),\end{array}$

where

$\omega_{i1}^{0}=b_{i}^{0}(v),\ \omega_{i2}^{0}=\frac{(b_{i}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)},\ b_{i}^{0}=\frac{b_{i}}{|C\times
G|}=-2\frac{X_{i}^{h^{*}}F}{|C\times G|}$

and the parameters $\omega_{i3}^{0}$ can be arbitrarily chosen. The indicatrix
at the point $p$ is a Euclidean surface of revolution with respect to the axis
represented by the binormal vector $\vec{B}(0)$. For the torsion components
with different indices we have a linear system of equations with an invertible
coefficient matrix. Therefore formula (23) gives all the possible torsion
components at the point $p$. The corresponding compatible linear connections
are determined by formula (2).

###  3.4. The extremal compatible linear connection in the undetermined case

To find the extremal one among the compatible linear connections we need to
minimize the sum of the squares of the torsion components. Let us denote the
solutions as

(24) $\begin{array}[]{rcl}\begin{pmatrix}-T_{12}^{3}+T_{13}^{2}+T_{23}^{1}\\\
-2T_{13}^{1}\\\ 2T_{12}^{1}\ \end{pmatrix}(p)&=&2\begin{pmatrix}P_{1}\\\
P_{2}\\\ P_{3}\end{pmatrix}+2s\begin{pmatrix}D_{1}\\\ D_{2}\\\
D_{3}\end{pmatrix},\\\\[20.0pt] \\\ \begin{pmatrix}2T_{23}^{2}\\\
-T_{12}^{3}-T_{13}^{2}-T_{23}^{1}\\\
2T_{12}^{2}\end{pmatrix}(p)&=&2\begin{pmatrix}Q_{1}\\\ Q_{2}\\\
Q_{3}\end{pmatrix}+2t\begin{pmatrix}D_{1}\\\ D_{2}\\\
D_{3}\end{pmatrix},\\\\[20.0pt] \\\ \begin{pmatrix}\ \ 2T_{23}^{3}\\\
-2T_{13}^{3}\\\ T_{12}^{3}+T_{13}^{2}-T_{23}^{1}\ \
\end{pmatrix}(p)&=&2\begin{pmatrix}R_{1}\\\ R_{2}\\\
R_{3}\end{pmatrix}+2u\begin{pmatrix}D_{1}\\\ D_{2}\\\
D_{3}\end{pmatrix},\end{array}$

where

$\begin{pmatrix}D_{1}\\\ D_{2}\\\ D_{3}\end{pmatrix}=\vec{B}(0),\
\begin{pmatrix}P_{1}\\\ P_{2}\\\
P_{3}\end{pmatrix}=\frac{1}{2}\left(b_{1}^{0}(v)\vec{T}(0)+\frac{(b_{1}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)}\vec{N}(0)\right),$ $\begin{pmatrix}Q_{1}\\\
Q_{2}\\\
Q_{3}\end{pmatrix}=\frac{1}{2}\left(b_{2}^{0}(v)\vec{T}(0)+\frac{(b_{2}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)}\vec{N}(0)\right),\ \begin{pmatrix}R_{1}\\\
R_{2}\\\
R_{3}\end{pmatrix}=\frac{1}{2}\left(b_{3}^{0}(v)\vec{T}(0)+\frac{(b_{3}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)}\vec{N}(0)\right),$
$s=\frac{1}{2}\omega_{13}^{0},\ t=\frac{1}{2}\omega_{23}^{0},\
u=\frac{1}{2}\omega_{33}^{0}$

are the common directional vector, the determined parts in (23) and the free
parameters, respectively. Extra 2’s are inserted to make the forthcoming
formulas easy to review:

(25) $T_{13}^{1}(p)=-P_{2}-sD_{2}$, |  $T_{23}^{2}(p)=Q_{1}+tD_{1}$, |  $T_{23}^{3}(p)=R_{1}+uD_{1}$,
---|---|---
$T_{12}^{1}(p)=P_{3}+sD_{3}$, |  $T_{12}^{2}(p)=Q_{3}+tD_{3}$, |  $T_{13}^{3}(p)=-R_{2}-uD_{2}$.

For components with different indices, we have the linear system

$\begin{pmatrix}-1&\ \ 1&\ \ 1\\\ -1&-1&-1\\\ \ \ 1&\ \
1&-1\end{pmatrix}\begin{pmatrix}T_{12}^{3}\\\ T_{13}^{2}\\\
T_{23}^{1}\end{pmatrix}(p)=2\begin{pmatrix}P_{1}+sD_{1}\\\ Q_{2}+tD_{2}\\\
R_{3}+uD_{3}\end{pmatrix}.$

Here the coefficient matrix is invertible, and so the solution is

(26) $\begin{pmatrix}T_{12}^{3}\\\ T_{13}^{2}\\\
T_{23}^{1}\end{pmatrix}(p)=\dfrac{1}{2}\begin{pmatrix}-1&-1&\ \ 0\\\ \ \ 1&\ \
0&\ \ 1\\\ \ \ 0&-1&-1\end{pmatrix}2\begin{pmatrix}P_{1}+sD_{1}\\\
Q_{2}+tD_{2}\\\
R_{3}+uD_{3}\end{pmatrix}=\begin{pmatrix}-P_{1}-Q_{2}-sD_{1}-tD_{2}\\\ \ \
P_{1}+R_{3}+sD_{1}+uD_{3}\\\ -Q_{2}-R_{3}-tD_{2}-uD_{3}\end{pmatrix}.$

Using formulas (25)–(26):

$\begin{gathered}\left\lVert
T_{p}\right\rVert^{2}=(T_{12}^{1})^{2}(p)+(T_{12}^{2})^{2}(p)+(T_{12}^{3})^{2}(p)+(T_{13}^{1})^{2}(p)+(T_{13}^{2})^{2}(p)+(T_{13}^{3})^{2}(p)+(T_{23}^{1})^{2}(p)+\\\
(T_{23}^{2})^{2}(p)+(T_{23}^{3})^{2}(p)=as^{2}+bt^{2}+cu^{2}+ds+et+fu+gst+hsu+itu+j,\end{gathered}$

where the coefficients are

$a=2D_{1}^{2}+D_{2}^{2}+D_{3}^{2},\ b=D_{1}^{2}+2D_{2}^{2}+D_{3}^{2},\
c=D_{1}^{2}+D_{2}^{2}+2D_{3}^{2},$
$d=4D_{1}P_{1}+2D_{1}Q_{2}+2D_{1}R_{3}+2D_{2}P_{2}+2D_{3}P_{3},$
$e=2D_{1}Q_{1}+2D_{2}P_{1}+4D_{2}Q_{2}+2D_{2}R_{3}+2D_{3}Q_{3},$
$f=2D_{1}R_{1}+2D_{2}R_{2}+2D_{3}P_{1}+2D_{3}Q_{2}+4D_{3}R_{3},$
$g=2D_{1}D_{2},\ h=2D_{1}D_{3},\ i=2D_{2}D_{3},$
$j=2P_{1}^{2}+P_{2}^{2}+P_{3}^{2}+Q_{1}^{2}+2Q_{2}^{2}+Q_{3}^{2}+R_{1}^{2}+R_{2}^{2}+2R_{3}^{2}+2P_{1}Q_{2}+2P_{1}R_{3}+2Q_{2}R_{3}.$

######  Theorem 4.

Using the notation

$k=D_{1}^{2}+2D_{2}^{2}+2D_{3}^{2},\ l=2D_{1}^{2}+D_{2}^{2}+2D_{3}^{2},\
m=2D_{1}^{2}+2D_{2}^{2}+D_{3}^{2},\ \left\lVert
D\right\rVert^{2}=D_{1}^{2}+D_{2}^{2}+D_{3}^{2},$

the torsion components of the extremal compatible linear connection belong to
the parameters

$s^{0}=\dfrac{-2dk+eg+fh}{8\left\lVert D\right\rVert^{4}},\
t^{0}=\dfrac{dg-2el+fi}{8\left\lVert D\right\rVert^{4}},\
u^{0}=\dfrac{dh+ei-2fm}{8\left\lVert D\right\rVert^{4}}$

in formulas (25)–(26).

###### Proof.

Taking $\left\lVert T_{p}\right\rVert^{2}$ as a function of the variables
$s,t,u$, we have to find its global minimum point. At first, we are looking
for the critical points:

$\dfrac{\partial\left\lVert T_{p}\right\rVert^{2}}{\partial s}=2as+d+gt+hu=0,\
\ \dfrac{\partial\left\lVert T_{p}\right\rVert^{2}}{\partial
t}=2bt+e+gs+iu=0,$ $\dfrac{\partial\left\lVert T_{p}\right\rVert^{2}}{\partial
u}=2cu+f+hs+it=0.$

It is a linear system for the parameters $s,t,u$. Its matrix form is

(27) $\begin{pmatrix}2a&g&h\\\ g&2b&i\\\
h&i&2c\end{pmatrix}\begin{pmatrix}s\\\ t\\\
u\end{pmatrix}=\begin{pmatrix}-d\\\ -e\\\ -f\end{pmatrix}.$

In the second step we calculate the corner minors. To do this, we need the
coefficient matrix, as the Hessian. Using MAPLE program, we find

$\Delta_{1}=2a>0,\ \Delta_{2}=4m\left\lVert D\right\rVert^{2}>0,\
\Delta_{3}=16\left\lVert D\right\rVert^{6}>0.$

Thus the coefficient matrix is invertible, so we have a unique triplet of
stationary parameters, and the Hessian is positive definite. Therefore the
critical point is a global minimizer. Using the inverse of the coefficient
matrix of the linear system (27), we find

$\begin{pmatrix}s^{0}\\\ t^{0}\\\
u^{0}\end{pmatrix}=H^{-1}\begin{pmatrix}-d\\\ -e\\\ -f\end{pmatrix},\ \
\textrm{where}\ \ H^{-1}=\begin{pmatrix}2a&g&h\\\ g&2b&i\\\
h&i&2c\end{pmatrix}^{-1}=\dfrac{1}{8\left\lVert
D\right\rVert^{4}}\begin{pmatrix}\ 2k&-g&-h\\\ -g&\ 2l&-i\\\
-h&-i&2m\end{pmatrix}.$

∎

###  3.5. The determined case (cf. Theorem 3)

Let $G=\mathrm{grad}\ F$ be the Euclidean gradient on the Finslerian spheres
with respect to the compatible Riemannian metric, and consider the normalized
vector field of $C\times G$. If $c_{0}$ is its integral curve starting from a
not vertical contact element $v$ in the tangent space $T_{p}M$, then we can
write the solution in the form (21). To determine the missing parameters
$\omega_{i3}^{0}$ consider the action

$\Phi_{v}\colon\mathbb{R}\times T_{p}M\to\mathbb{R},\
(t,w)\to\Phi_{v}(t,w):=F(e^{At}w)$

of the one-parameter group of rotations around the binormal vector
$\vec{B}(0)$ on the Finslerian spheres, where $A$ denotes the skew-symmetric
matrix of the mapping $w\mapsto w\times\vec{B}(0)$.

######  Lemma 3.

If $\displaystyle{\dfrac{\partial\Phi_{v}}{\partial t}(0,w)=0}$ for any
nonzero element $w\in T_{p}M$, then the Finslerian spheres are Euclidean
surfaces of revolution about the axis generated by the binormal vector
$\vec{B}(0)$.

###### Proof.

Since $\Phi_{v}(t+s,w)=\Phi_{v}(t,e^{sA}w)$, it follows that

$\dfrac{\partial\Phi_{v}}{\partial t}(s,w)=\dfrac{\partial\Phi_{v}}{\partial
t}(0,e^{sA}w).$

Therefore the vanishing of the derivatives at $t=0$ implies that $\Phi_{v}$ is
independent of the first variable, i.e., the Finslerian spheres are Euclidean
surfaces of revolution about the axis generated by the binormal vector
$\vec{B}(0)$. ∎

Since the Euclidean surfaces of revolution belong to the undetermined case,
there is a non-zero element $w\in T_{p}M$ such that

$\dfrac{\partial\Phi_{v}}{\partial t}(0,w)\neq 0$

(see Lemma 1, Lemma 3 and Theorem 3), thus we have that

$\left<C_{w}\times
G_{w},\vec{B}(0)\right>=-\left<C_{w}\times\vec{B}(0),G_{w}\right>=\dfrac{\partial\Phi_{v}}{\partial
t}(0,w)\neq 0.$

To sum up, the uniquely determined solution must be of the form

(28) $\begin{pmatrix}\vec{t}_{1}\\\ \vec{t}_{2}\\\
\vec{t}_{3}\end{pmatrix}=\begin{pmatrix}\omega_{11}^{0}&\omega_{12}^{0}&\omega_{13}^{0}\\\
\omega_{21}^{0}&\omega_{22}^{0}&\omega_{23}^{0}\\\
\omega_{31}^{0}&\omega_{32}^{0}&\omega_{33}^{0}\end{pmatrix}\begin{pmatrix}\vec{T}(0)\\\
\vec{N}(0)\\\ \vec{B}(0)\end{pmatrix},$

where

$\omega_{i1}^{0}=b_{i}^{0}(v),\ \omega_{i2}^{0}=\frac{(b_{i}^{0}\circ
c_{0})^{\prime}(0)}{\kappa(0)},$ $\omega_{i3}^{0}=-\frac{1}{\left<C_{w}\times
G_{w},\vec{B}(0)\right>}\left(\omega_{i1}^{0}\left<C_{w}\times
G_{w},\vec{T}(0)\right>+\omega_{i2}^{0}\left<C_{w}\times
G_{w},\vec{N}(0)\right>+2X_{i}^{h^{*}}F(w)\right),$
$b_{i}^{0}=\frac{b_{i}}{|C\times G|}=-2\frac{X_{i}^{h^{*}}F}{|C\times
G|},\quad i\in\\{1,2,3\\}$

and $(\vec{T},\vec{N},\vec{B})$ is the Frenet-Serret frame along the integral
curve $c_{0}$ of the normalized vector field of $C\times G$ starting from a
not vertical contact element $v\in T_{p}M$. Instead of the specification of
the element $w$, we can integrate on the Euclidean unit sphere $\sigma_{*}$
with respect to the compatible Riemannian metric:

$\omega_{i3}^{0}\left<C_{w}\times G_{w},\vec{B}(0)\right>^{2}=$
$-\left<C_{w}\times
G_{w},\vec{B}(0)\right>\left(\omega_{i1}^{0}\left<C_{w}\times
G_{w},\vec{T}(0)\right>+\omega_{i2}^{0}\left<C_{w}\times
G_{w},\vec{N}(0)\right>+2X_{i}^{h^{*}}F(w)\right),$

i.e.,

$\omega_{i3}^{0}\int_{\sigma_{*}}\left<C\times G,\vec{B}(0)\right>^{2}=$
$-\int_{\sigma_{*}}\left<C\times
G,\vec{B}(0)\right>\left(\omega_{i1}^{0}\left<C\times
G,\vec{T}(0)\right>+\omega_{i2}^{0}\left<C\times
G,\vec{N}(0)\right>+2X_{i}^{h^{*}}F\right),$

where

$\int_{\sigma_{*}}\left<C\times G,\vec{B}(0)\right>^{2}\neq 0.$

##  4\. Examples: 3-dimensional Randers spaces

To illustrate the ideas and methods presented above, we consider a special
class of Finsler manifolds.

######  Definition 6.

A Finsler manifold is called a _Randers manifold_ if the Finsler function has
the form

$F(x,y)=\alpha(x,y)+\beta(x,y),$

where $\alpha$ is a norm coming from a Riemannian metric on $M$ given by
$\alpha(x,y)=\sqrt{\alpha_{ij}(x)y^{i}y^{j}}$ in a local basis
$\partial/\partial u^{1},\ldots,\partial/\partial u^{n}$ and $\beta$ comes
from a 1-form given by $\beta(x,y)=\beta_{j}(x)y^{j}$ such that
$\alpha^{ij}\beta_{i}\beta_{j}<1$.

Both the metric components of the Riemannian part and the components of the
perturbating term are considered on the tangent manifold as composite
functions $\alpha_{ij}(x)$ and $\beta_{k}(x)$, where $x=(x^{1},\ldots,x^{n})$.
It is well-known [5] that the Riemannian part is compatible to the Finsler
function of a Randers manifold, i.e., $\gamma:=\alpha$ is a convenient choice
for a compatible Riemannian metric. In what follows, we consider the Finsler
function $F=\alpha+\beta$ on a connected 3-dimensional manifold $M$. Let a
point $p\in M$ be given and suppose111If $\beta_{p}={\bf 0}$ then $T_{p}M$ is
a vertical contact tangent space and there is nothing to compute. Such a
Randers manifold is a generalized Berwald manifold if and only if it is a
Riemannian manifold, i.e., the perturbating term is zero at each point of the
manifold. Indeed, using the parallel transports induced by the compatible
linear connection, the quadratic Finslerian indicatrix at a single point
implies that the Finslerian indicatrix is quadratic at each point of the
manifold. that $\beta_{p}\neq{\bf 0}$. In order to make the computations
easier, we choose local coordinates around $p$ such that

  * •

the coordinate vector fields form an orthonormal basis at $p$ with respect to
the compatible Riemannian metric $\alpha$, i.e. $\alpha_{ij}(p)=\delta_{ij}$,

  * •

$\beta_{1}(p)=\beta_{2}(p)=0$ and $K:=\beta_{3}(p)\neq 0$, i.e., the
coordinate vector fields $\partial/\partial u^{1}$ and $\partial/\partial
u^{2}$ span the kernel of the linear functional $\beta$ at the point $p$.

Under these choices of the coordinate vector fields

$F(x,y)=\sqrt{\delta_{ij}y^{i}y^{j}}+\beta_{3}(x)y^{3}=\sqrt{(y^{1})^{2}+(y^{2})^{2}+(y^{3})^{2}}+Ky^{3}.$

The Finslerian spheres in $T_{p}M$ are given by the equations of type

$(y^{1})^{2}+(y^{2})^{2}+(1-K^{2})\left(y^{3}+\dfrac{Kc}{1-K^{2}}\right)^{2}=\dfrac{c^{2}}{1-K^{2}}.$

For any given $c>0$ it is an ellipsoid with center
$\left(0,0,-\dfrac{Kc}{1-K^{2}}\right)$ with axes belonging to the coordinate
vector fields at $p\in M$. The Finslerian spheres are rotationally symmetric
with respect to the axis of $y^{3}$. Therefore, by Theorem 3, we are in the
undetermined case, i.e., there must be infinitely many compatible linear
connections (if there are any), and the direction space of the solutions of
the compatibility equations in $T_{p}M$ is given by the vector $(0,0,1)$.

###  4.1. The vector field $C\times G$ and its integral curves

The partial derivatives of $F$ with respect to the vectorial directions are

$\dfrac{\partial F}{\partial
y^{k}}(x,y)=\dfrac{y^{k}}{\sqrt{(y^{1})^{2}+(y^{2})^{2}+(y^{3})^{2}}}+\delta^{3}_{k}K=\dfrac{y^{k}}{\alpha(x,y)}+\delta^{3}_{k}K$

and the Euclidean gradient vector field of $F$ is

$G(x,y)=\dfrac{C}{\alpha}(x,y)+\left(0,\,0,\,K\right)=\dfrac{1}{\alpha(x,y)}\left(y^{1},\,y^{2},\,y^{3}+K\alpha(x,y)\right),$

where $C$ is the radial vector field. The cross product is

$C\times G=\dfrac{1}{\alpha(x,y)}\begin{vmatrix}e_{1}&e_{2}&e_{3}\\\
y^{1}&y^{2}&y^{3}\\\
y^{1}&y^{2}&y^{3}+K\alpha(x,y)\end{vmatrix}=K\begin{pmatrix}\ \ y^{2}\\\
-y^{1}\\\ \ 0\end{pmatrix}.$

Therefore, an element $v\in T_{p}M$ is vertical contact if and only if
$v=(0,0,v^{3})$, i.e., the vertical contact points are the elements of the
$y^{3}$ coordinate axis. To simplify the formalism let us consider the vector
field

$\dfrac{C\times G}{K}=\begin{pmatrix}\ \ y^{2}\\\ -y^{1}\\\ \ 0\end{pmatrix}$

instead of the normalized version of $C\times G$. Taking a not vertical
contact element $v=(v^{1},v^{2},v^{3})\in T_{p}M$, it is easy to see that the
integral curve starting from $v$ and its derivatives can be given as

$\begin{array}[]{rclrccccc}c(t)&=&\begin{pmatrix}r\sin(t+t_{0})\\\
r\cos(t+t_{0})\\\ v^{3}\end{pmatrix},&&c(0)&=&\begin{pmatrix}r\sin t_{0}\\\
r\cos t_{0}\\\ v^{3}\end{pmatrix}&=&\begin{pmatrix}v^{1}\\\ v^{2}\\\
v^{3}\end{pmatrix},\\\\[25.0pt] c^{\prime}(t)&=&\begin{pmatrix}\ \
r\cos(t+t_{0})\\\ -r\sin(t+t_{0})\\\ \
0\end{pmatrix},&&c^{\prime}(0)&=&\begin{pmatrix}\ \ r\cos t_{0}\\\ -r\sin
t_{0}\\\ \ 0\end{pmatrix}&=&\begin{pmatrix}\ \ v^{2}\\\ -v^{1}\\\ \
0\end{pmatrix},\\\\[25.0pt]
c^{\prime\prime}(t)&=&\begin{pmatrix}-r\sin(t+t_{0})\\\ -r\cos(t+t_{0})\\\ \
0\end{pmatrix},&&c^{\prime\prime}(0)&=&\begin{pmatrix}-r\sin t_{0}\\\ -r\cos
t_{0}\\\ \ 0\end{pmatrix}&=&\begin{pmatrix}-v^{1}\\\ -v^{2}\\\ \
0\end{pmatrix}.\end{array}$

###  4.2. The solution lines

Let us write the compatibility equations in the form

(29) $\left<\dfrac{C\times G}{K},\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=-\dfrac{2}{K}X_{i}^{h^{*}}F,\quad
i\in\\{1,2,3\\},$

see (19). To determine the directional space of the solutions at a not
vertical contact $v$, we need to solve the system

$\left.\begin{array}[]{rcl}\langle c^{\prime}(0),\vec{t}\
\rangle&=&0\\\\[2.0pt] \langle c^{\prime\prime}(0),\vec{t}\ \rangle&=&0\\\
\end{array}\right\\}\Longleftrightarrow\left.\begin{array}[]{rcrcl}v^{2}t^{1}&-&v^{1}t^{2}&=&0\\\\[2.0pt]
-v^{1}t^{1}&-&v^{2}t^{2}&=&0\\\
\end{array}\right\\}\Longleftrightarrow\begin{pmatrix}\ \ v^{2}&-v^{1}\\\
-v^{1}&-v^{2}\end{pmatrix}\begin{pmatrix}t^{1}\\\
t^{2}\end{pmatrix}=\begin{pmatrix}0\\\ 0\end{pmatrix},$

see the proof of Lemma 1. It is clear that $t^{3}$ can be arbitrarily chosen
and the direction space is the line generated by $(0,0,1)$. Since $\nabla^{*}$
is the Lévi-Civita connection of $\alpha$, we have that

$X_{i}^{h^{*}}F=X_{i}^{h^{*}}\alpha+X_{i}^{h^{*}}\beta=X_{i}^{h^{*}}\beta.$

Furthermore,

$X_{i}^{h^{*}}\beta=\frac{\partial\beta_{s}y^{s}}{\partial
x^{i}}-y^{j}{\Gamma}^{k*}_{ij}\circ\pi\frac{\partial\beta_{s}y^{s}}{\partial
y^{k}}=\frac{\partial\beta_{s}}{\partial
x^{i}}y^{s}-y^{j}{\Gamma}^{k*}_{ij}\circ\pi\beta_{k}=y^{j}\left(\frac{\partial\beta_{j}}{\partial
x^{i}}-\Gamma^{3*}_{ij}\circ\pi\beta_{3}\right)$

and, consequently,

$-\frac{2}{K}X_{i}^{h^{*}}F=2\left({\Gamma}^{3*}_{ij}\circ\pi-\frac{1}{K}\frac{\partial\beta_{j}}{\partial
x^{i}}\right)y^{j}.$

Introducing the notation

(30)
$C_{j;i}:=\Gamma^{3*}_{ij}\circ\pi-\frac{1}{K}\frac{\partial\beta_{j}}{\partial
x^{i}},$

(29) takes the form

$\left<\begin{pmatrix}\ \ y^{2}\\\ -y^{1}\\\
0\end{pmatrix},\begin{pmatrix}t_{i}^{1}\\\ t_{i}^{2}\\\
t_{i}^{3}\end{pmatrix}\right>=2C_{j;i}y^{j},\ \ \textrm{i.e.},\ \
y^{2}t_{i}^{1}-y^{1}t_{i}^{2}=2\left(C_{1;i}y^{1}+C_{2;i}y^{2}+C_{3;i}y^{3}\right),\quad
i\in\\{1,2,3\\}.$

Differentiating these equations with respect to $y^{1}$, $y^{2}$ and $y^{3}$,
the comparison of the coefficients gives that the compatibility equations have
solutions if and only if $C_{3;i}=0$. Furthermore, the determined parts are
given by $t_{i}^{1}=2C_{2;i}$ and $t_{i}^{2}=-2C_{1;i}$, respectively.

######  Remark 3.

_The geometric meaning of conditions $C_{3;i}=0$ is that $\beta$ has a dual
vector field of constant length with respect to the compatible Riemannian
metric $\alpha$; for details see [5] and [12]._

###  4.3. The torsion components of the compatible linear connections

They can be given by substituting

$\begin{pmatrix}D_{1}\\\ D_{2}\\\ D_{3}\end{pmatrix}=\begin{pmatrix}0\\\ 0\\\
1\end{pmatrix},\ \begin{pmatrix}P_{1}\\\ P_{2}\\\
P_{3}\end{pmatrix}=\begin{pmatrix}\ \ C_{2;1}\\\ -C_{1;1}\\\ 0\end{pmatrix},\
\begin{pmatrix}Q_{1}\\\ Q_{2}\\\ Q_{3}\end{pmatrix}=\begin{pmatrix}\ \
C_{2;2}\\\ -C_{1;2}\\\ 0\end{pmatrix},\ \begin{pmatrix}R_{1}\\\ R_{2}\\\
R_{3}\end{pmatrix}=\begin{pmatrix}\ \ C_{2;3}\\\ -C_{1;3}\\\ 0\end{pmatrix}$

in formulas (25) and (26). The components

$T_{13}^{1}(p)=-P_{2}-sD_{2}=C_{1;1},\ T_{23}^{2}(p)=Q_{1}+tD_{1}=C_{2;2},\
T_{23}^{3}(p)=R_{1}+uD_{1}=C_{2;3},$ $T_{13}^{3}(p)=-R_{2}-uD_{2}=C_{1;3},\
T_{12}^{3}(p)=-P_{1}-Q_{2}-sD_{1}-tD_{2}=C_{1;2}-C_{2;1}$

are uniquely determined. Furthermore,

$T_{12}^{1}(p)=P_{3}+sD_{3}=s,\ T_{12}^{2}(p)=Q_{3}+tD_{3}=t,\
T_{13}^{2}(p)=P_{1}+R_{3}+sD_{1}+uD_{3}=C_{2;1}+u,$
$T_{23}^{1}(p)=-Q_{2}-R_{3}-tD_{2}-uD_{3}=C_{1;2}-u.$

Using that

$d=4D_{1}P_{1}+2D_{1}Q_{2}+2D_{1}R_{3}+2D_{2}P_{2}+2D_{3}P_{3}=0,$
$e=2D_{1}Q_{1}+2D_{2}P_{1}+4D_{2}Q_{2}+2D_{2}R_{3}+2D_{3}Q_{3}=0,$
$f=2D_{1}R_{1}+2D_{2}R_{2}+2D_{3}P_{1}+2D_{3}Q_{2}+4D_{3}R_{3}=2D_{3}(P_{1}+Q_{2})=2C_{2;1}-2C_{1;2},$
$g=2D_{1}D_{2}=0,\ h=2D_{1}D_{3}=0,\ i=2D_{2}D_{3}=0,\
k=D_{1}^{2}+2D_{2}^{2}+2D_{3}^{2}=2,$ $l=2D_{1}^{2}+D_{2}^{2}+2D_{3}^{2}=2,\
m=2D_{1}^{2}+2D_{2}^{2}+D_{3}^{2}=1,\ \left\lVert
D\right\rVert^{2}=D_{1}^{2}+D_{2}^{2}+D_{3}^{2}=1,$

Theorem 4 shows that the coefficients of the extremal compatible linear
connection belong to the parameters

$s^{0}=\dfrac{-2dk+eg+fh}{8\left\lVert D\right\rVert^{4}}=0,\
t^{0}=\dfrac{dg-2el+fi}{8\left\lVert D\right\rVert^{4}}=0,$
$u^{0}=\dfrac{dh+ei-2fm}{8\left\lVert
D\right\rVert^{4}}=\dfrac{4C_{2;1}-4C_{1;2}}{8}=\dfrac{C_{2;1}-C_{1;2}}{2}.$

##  5\. Acknowledgement

The authors would like to express their very great appreciation to Professor
József Szilasi for his valuable and constructive suggestions.

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