# Fractional-linear integrals of geodesic flows on surfaces and Nakai’s
geodesic 4-webs

Sergey I. Agafonov

Thaís G. P. Alves

Department of Mathematics,
São Paulo State University-UNESP,
São José do Rio Preto, Brazil
e-mail<EMAIL_ADDRESS>
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###### Abstract

We prove that if the geodesic flow on a surface has an integral, fractional-
linear in momenta, then the dimension of the space of such integrals is either
3 or 5, the latter case corresponding to constant gaussian curvature. We give
also a geometric criterion for existence of fractional-linear integrals: such
integral exists if and only if the surface carries a geodesic 4-web with
constant cross-ratio of the four directions tangent to the web leaves.

##  1 Introduction

Integrable geodesic flow on surfaces is a classical subject of differential
geometry and analytic mechanics since the 19th century works of Jacobi [J-84],
Darboux [D-91], Dini [D-69], Koenigs [K-96] and many others. Special attention
was payed to integrals polynomial in momenta. This problem is well motivated
even in the local setting due to the following observations: 1) if there is a
locally defined analytic integral then each homogeneous in momenta part of its
Taylor expansion is an integral polynomial in momenta, 2) the surface of zero
velocities in the tangent bundle of surface is the set of singular points of
the geodesic flow, the existence of regular integrals at singular point being
a nontrivial restriction.

Since the geodesic flow is Hamiltonian, it is enough to find one first
integral independent of the Hamiltonian function to integrate the flow due to
the Liouville Theorem. Polynomial integrals of fixed degree constitute a
vector space of finite dimension. Establishing this dimension turned out to be
a non-trivial task for degrees grater then two, in fact, the list of possible
dimensions is not known already for cubic integrals (see [K-08]). The case of
linear integrals is equivalent to listing the possible dimensions of isometry
group due to the Noether Theorem: the dimension is either 0, or 1, or 3. The
case of quadratic integrals was settled by Koenings: the possible dimensions
are 1, 2, 3, 4, and 6, the largest dimension in both cases corresponding to
constant gaussian curvature.

Modern interests to this topic is due to 1) a new discovered relation to
infinite-dimensional integrable systems of PDE and 2) a better understanding
of geometry behind the integrability of geodesic flows on surfaces.

For a fixed metric, the system of partial differential equations (PDE) for the
coefficients of the integral is over-determined but if one considers the
metric also as unknown then the system turns out to be of hydrodynamic type
with remarkable properties: it is diagonalizable, it possesses infinitely many
conservation laws, and it is linearizable by Tsarev’s generalized hodograph
method (see [K-82, T-97, BM-11, MP-17, T-85]).

On the other hand, the existence of quadratic and cubic integrals can be
described in geometric terms via geodesic webs. For example, consider a cubic
integral, which can be rewritten also as polynomial in velocity $v=(\xi,\eta)$
due to the canonical isomorphism between the tangent and cotangent bundles of
the surface with local coordinates $x,y$:

$I_{3}=a_{0}(x,y)\xi^{3}+a_{1}(x,y)\xi^{2}\eta+a_{2}(x,y)\xi\eta^{2}+a_{3}(x,y)\eta^{3}.$

Equation it to zero, one gets an implicit cubic ordinary differential equation
(ODE). If all its 3 roots $[\xi:\eta]$ are real, the integral curves of this
ODE form a hexagonal geodesic 3-web. And conversely, the existence of a
hexagonal geodesic 3-web implies existence of a cubic integral [Ah-21]. There
is also a geometric characterization of quadratic integrals [Aq-21]: existence
of quadratic integrals is equivalent to existence of geodesic net
$\mathcal{G}$ such that any 3-subweb of 4-web, formed by $\mathcal{G}$ and its
bisector net $\mathcal{N}$, is hexagonal.

Geodesic webs is a classical chapter of differential geometry, such webs were
a subject of intensive study, though without any relation to dynamics, see
[F-99, S-03, GS-24, S-26, V-29]. In his last book on the web theory [B-55]
Blaschke posed the problem of finding an intrinsic criterion for existence of
hexagonal geodesic 3-webs. Thus, the result of [Ah-21] provides such criterion
it dynamical terms and generalizes the Graf and Sauer Theorem [GS-24] for
metrics of non-constant gaussian curvature.

Another natural class of integrals of geodesic flows are integrals rational in
momenta, whose numerator and denominator are homogeneous polynomials in
momenta of the same degree. The study of such integrals was initiated by
Kozlov [K-14].

Agapov and Shubin investigated fractional-linear integrals [Ar-20, At-20,
AS-21]

$\dfrac{A(x,y)p+B(x,y)q}{C(x,y)p+E(x,y)q},$

in particular, they found explicit examples of metrics, whose geodesic flows
admit such integrals.

This paper contributes to both analytic and geometric description of
fractional-linear integrals. The Möbius group acts naturally on the space of
such integrals, thus giving a symmetry subgroup of the PDE system for the
integral. Splitting this system to automorphic and resolving parts (see [V-04,
O-82]), we manage to compute enough compatibility conditions to extract the
dimension of the space of fractional-linear integrals.

The first main result of the paper claims that if this space is nonempty then
this dimension is either 3 or 5, the maximal dimension corresponding to
constant gaussian curvature.

To give a geometric description, we construct a one-parametric family of
geodesic foliations, whose leaves are integral curves of the ODE

$\dfrac{\alpha(x,y)\xi+\beta(x,y)\eta}{\gamma(x,y)\xi+\delta(x,y)\eta}=\lambda,$

where on the left-hand side is a fractional-linear integral written on the
tangent bundle and $\lambda\in\mathbb{R}$ is a parameter. Taking four such
foliations one obtains a geodesic 4-web with constant cross-ratio of web
tangent directions.

Nakai studied 4-webs with constant cross-ratio of web tangent directions in
[N-98] and showed that the Blaschke curvatures of all its 3-subwebs are equal.
He excluded the case of parallelizable 4-web with hexagonal 3-subwebs as
trivial, though we will tolerate this degeneracy in our definition, as it
occurs for metrics with constant gaussian curvatures. Dufour e Lehmann proved
[DL-22] that Nakai’s webs can have rank 0 or 1 and gave examples of webs with
rank 1.

Our second main results claims that there is a fractional linear integral if
and only if the surface carries Nakai’s geodesic 4-web.

All the results of this paper are local, all functions, surfaces, fields, and
other objects are smooth.

##  2 Fractional-linear integrals of geodesic flow

Geodesic flow on a surface $S$ with local coordinates $(x,y)$ is Hamiltonian
with respect to canonical symplectic form $\Omega=dp\wedge dx+dq\wedge dy$ on
the cotangent bundle $T^{*}S$, where $f=(p,q)$ is the momentum. The
Hamiltonian $\frac{1}{2}g(v,v)$, where $v=(\xi,\eta)\in T_{s}S$ is the
velocity and $g$ metric, can be rewritten in momentum due to the canonical
isomorphism between $T_{s}S$ e $T_{s}^{*}S$ via $g(v,*)=\langle f,*\rangle$.
For metric $g=\Lambda(x,y)(dx^{2}+dy^{2})$ in conformal coordinates, we get
the Hamiltonian $H=\dfrac{p^{2}+q^{2}}{2\Lambda}$ and $p,q$ are conjugated to
the coordinates $(x,y)$.

###### Lemma 1

[K-14] Geodesic flow of the metric $g=\Lambda(x,y)(dx^{2}+dy^{2})$ admits a
fractional-linear integral

$I=\frac{A(x,y)p+B(x,y)q}{C(x,y)p+E(x,y)q}$ (1)

if and only if

$\begin{array}[]{l}2\Lambda(CA_{x}-AC_{x})-\Lambda_{y}(AE-BC)=0,\\\
2\Lambda(EA_{x}-AE_{x}+CA_{y}-AC_{y}+CB_{x}-BC_{x})+\Lambda_{x}(AE-BC)=0,\\\
2\Lambda(EA_{y}-AE_{y}+CB_{y}-BC_{y}+EB_{x}-BE_{x})-\Lambda_{y}(AE-BC)=0,\\\
2\Lambda(EB_{y}-BE_{y})+\Lambda_{x}(AE-BC)=0.\end{array}$ (2)

Proof: The condition $\\{I,H\\}=0$ writes as an equation, cubic and
homogeneous in $p,q$:

$\displaystyle(2\Lambda CA_{x}-2\Lambda
AC_{x}-\Lambda_{y}AE+\Lambda_{y}BC)p^{3}$ $\displaystyle+$
$\displaystyle(2\Lambda EA_{x}-2\Lambda AE_{x}+2\Lambda CA_{y}-2\Lambda
AC_{y}+2\Lambda CB_{x}-2\Lambda BC_{x}+\Lambda_{x}AE-\Lambda_{x}BC)p^{2}q$
$\displaystyle+$ $\displaystyle(2\Lambda EA_{y}-2\Lambda AE_{y}+2\Lambda
CB_{y}-2\Lambda BC_{y}+2\Lambda EB_{x}-2\Lambda
BE_{x}-\Lambda_{y}AE-\Lambda_{y}BC)pq^{2}$ $\displaystyle+$
$\displaystyle(2\Lambda EB_{y}-2\Lambda
BE_{y}+\Lambda_{x}AE-\Lambda_{x}BC)q^{3}=0.$

Since $A,B,C$ e $E$ are independent of $p,q$, the equation splits into four
given in (2). $\Box$

##  3 Differential invariants

Let us normalize a fractional linear integral (1) to
$A(x,y)E(x,y)-B(x,y)C(x,y)=1.$
Thus normalized integral defines the map

$\begin{array}[]{l}m:S\longrightarrow SL_{2}(\mathbb{R}),\\\ \\\ (x,y)\mapsto
m(x,y)=\left(\begin{array}[]{cc}A(x,y)&B(x,y)\\\
C(x,y)&E(x,y)\end{array}\right).\end{array}$ (3)

The matrix Lie group $SL_{2}(\mathbb{R})=\left\\{\left(\begin{array}[]{c
c}\alpha&\beta\\\ \gamma&\delta\end{array}\right):\
\alpha,\beta,\gamma,\delta\in\mathbb{R},\alpha\delta-\beta\gamma=1\right\\}$
naturally acts on the set of fractional-linear integrals:

$h\cdot I=\frac{\alpha I+\beta}{\gamma I+\delta}=\dfrac{(\alpha A+\beta
C)p+(\alpha B+\beta E)q}{(\gamma A+\delta C)p+(\gamma B+\delta E)q},$ (4)

where $h=\left(\begin{array}[]{c c}\alpha&\beta\\\
\gamma&\delta\end{array}\right)\in SL_{2}(\mathbb{R})$. Note that

$\tilde{I}=\dfrac{(\alpha A+\beta C)p+(\alpha B+\beta E)q}{(\gamma A+\delta
C)p+(\gamma B+\delta E)q}$

is also normalized

$(\alpha A+\beta C)(\gamma B+\delta E)-(\alpha B+\beta E)(\gamma A+\delta
C)=1.$

###### Definition 1

We say that two fractional-linear integrals $I$ e $\tilde{I}$ are equivalent
if there exists an element $h\in SL_{2}(\mathbb{R})$ such that $h\cdot
I=\tilde{I}$.

Thus the system (2) is invariant by the above defined action of
$SL_{2}(\mathbb{R})$ and can be rewritten in differential invariants. Since
the action is just multiplying $m$ by $h$ on the left, differential invariants
can be obtained via Darboux derivative of the map $m$ defined by (3). Recall
that the Darboux derivative of the map $m:S\rightarrow SL_{2}(\mathbb{R})$ is
the pull-back of the Maurer-Cartan form of $SL_{2}(\mathbb{R})$:
$\omega=m^{-1}\cdot dm$ (see [S-96]). In fact, it is invariant as

$(hm)^{-1}\cdot d(hm)=m^{-1}\cdot h^{-1}\cdot h\cdot d(m)=m^{-1}\cdot d(m).$

We have

$m^{-1}\cdot dm=\left(\begin{array}[]{cc}E&-B\\\
-C&A\end{array}\right)\left(\begin{array}[]{cc}dA&dB\\\
dC&dE\end{array}\right)=\left(\begin{array}[]{cc}EdA-BdC&EdB-BdE\\\
-CdA+AdC&-CdB+AdE\end{array}\right),$ (5)

where

$dA=A_{x}dx+A_{y}dy,\ \ \ dB=B_{x}dx+B_{y}dy,\ \ \ dC=C_{x}dx+C_{y}dy,\ \ \
dE=E_{x}dx+E_{y}dy.$

Substituting into (5) we obtain

$\displaystyle m^{-1}\cdot
dm=\left(\begin{array}[]{cc}EA_{x}-BC_{x}&EB_{x}-BE_{x}\\\
AC_{x}-CA_{x}&AE_{x}-CB_{x}\end{array}\right)dx+\left(\begin{array}[]{cc}EA_{y}-BC_{y}&EB_{y}-BE_{y}\\\
AC_{y}-CA_{y}&AE_{y}-CB_{y}\end{array}\right)dy.$

The group does not act on independent coordinates $x,y$, therefore all the
elements of the matrix coefficients of $dx$ and $dy$ are scalar diferential
invariants. Since the trace of $\omega$ vanishes we have 6 invariants

$\begin{array}[]{lll}P:=EA_{x}-BC_{x},&Q:=EB_{x}-BE_{x},&R:=AC_{x}-CA_{x},\\\
&&\\\ X:=EA_{y}-BC_{y},&Y:=EB_{y}-BE_{y},&Z:=AC_{y}-CA_{y}.\end{array}$

Using the normalization $det(m)=1$ we exclude $E=\frac{BC+1}{A}$ and its
derivatives

$E_{x}=\dfrac{BC_{x}+CB_{x}}{A}-\dfrac{(BC+1)A_{x}}{A^{2}},\ \ \ \
E_{y}=\dfrac{BC_{y}+CB_{y}}{A}-\dfrac{(BC+1)A_{y}}{A^{2}},$ (7)

and rewrite the invariants

$\begin{array}[]{lll}P=\frac{BC+1}{A}A_{x}-BC_{x},&Q=\dfrac{B^{2}C+B}{A^{2}}A_{x}+\dfrac{1}{A}B_{x}-\dfrac{B^{2}}{A}C_{x},&R=AC_{x}-CA_{x},\\\
&&\\\
X=\frac{BC+1}{A}A_{y}-BC_{y},&Y=\dfrac{B^{2}C+B}{A^{2}}A_{y}+\dfrac{1}{A}B_{y}-\dfrac{B^{2}}{A}C_{y},&Z=AC_{y}-CA_{y}.\end{array}$

We can express derivatives $A_{x},B_{x},C_{x},A_{y},B_{y},C_{y}$ in terms of
the invariants as follows:

$\begin{array}[]{lll}A_{x}=PA+RB,&B_{x}=QA-PB,&C_{x}=\dfrac{PAC+RBC+R}{A},\\\
&&\\\ A_{y}=XA+ZB,&B_{y}=YA-XB,&C_{y}=\dfrac{XAC+ZBC+Z}{A}.\\\ \end{array}$
(8)

Calculation of compatibility conditions is greatly simplified for the metric
form

$g=L(x,y)dxdy,$ (9)

related by formal complex substitution $x=u+iv$, $y=u-iv$ to the conformal
form:

$g=L(x,y)dxdy=\Lambda(u+iv,u-iv)(du^{2}+dv^{2}).$

Since all the expressions are rational, this substitution does not affect our
compatibility results obtained in a purely algebraic way (compare with
calculations of Lie in [L-82] and Koenigs in [K-96]).

###### Lemma 2

Geodesic flow on a surface admits a fractional-linear integral (1) if and only
if the invariants $P,Q,R,X,Y,Z$ are related to the metric (9) as follows:

$P=-\dfrac{Y}{2}-\dfrac{L_{x}}{2L},\,\ \ X=\dfrac{R}{2}+\dfrac{L_{y}}{2L},\ \
Q=Z=0.$ (10)

Proof: For the metric (9) the Hamiltonian assumes the form $H=\dfrac{2pq}{L},$
then the condition $\\{I,H\\}=0,$ writes as

$\begin{array}[]{l}L(AC_{y}-CA_{y})=0,\\\
L(CA_{x}-AE_{y}-AC_{x}+CB_{y}-BC_{y}+EA_{y})-L_{y}(AE-BC)=0,\\\
L(EA_{x}-BE_{y}-BC_{x}+EB_{y}-AE_{x}+CB_{x})+L_{x}(AE-BC)=0,\\\
L(BE_{x}-EB_{x})=0.\end{array}$

Using the normalization $AE-BC=1$, substituting $E,E_{x},E_{y}$ from (7) and
$A_{x},B_{x},C_{x},A_{y},B_{y},C_{y}$ from (8) we get

$LZ=LQ=0,\ \ 2LX-LR-L_{y}=0,\ \ 2LP+LY+L_{x}=0,\\\ $

hence (10). $\Box$

##  4 Dimension of the space of integrals

The gaussian curvature of metric (9) is

$K=-\dfrac{LL_{xy}-L_{x}L_{y}}{L^{3}},$ (11)

its partial derivatives are

$\displaystyle
K_{x}=-\dfrac{L_{xxy}}{L^{2}}+\dfrac{L_{xx}L_{y}+3L_{xy}L_{x}}{L^{3}}-\dfrac{3L_{x}^{2}L_{y}}{L^{4}},$
$\displaystyle
K_{y}=-\dfrac{L_{xyy}}{L^{2}}+\dfrac{L_{yy}L_{x}+3L_{xy}L_{y}}{L^{3}}-\dfrac{3L_{y}^{2}L_{x}}{L^{4}}.$

The Darboux derivative $\omega$ of $m$ satisfies the structure equation
$d\omega+\omega\wedge\omega=0$, which is the local condition for existence of
$m$ with the Darboux derivative $\omega$ (see [S-96]). The matrix structure
equation rewrites as three scalar equations

$\begin{array}[]{l}2RY+2LK-R_{x}-Y_{y}=0,\\\ \\\ LY^{2}+L_{x}Y-LY_{x}=0,\\\
\\\ LR^{2}+L_{y}R-LR_{y}=0.\\\ \end{array}$

Resolving for $Y_{x}$ e $R_{y}$ we obtain $Y_{x}=Y^{2}+\dfrac{L_{x}}{L}$ and
$R_{y}=R^{2}+\dfrac{L_{y}}{L}.$ Let us introduce $F$ by

$2F:=R_{x}-Y_{y},$ (12)

then $R_{x}=RY+LK+F,\ \ Y_{y}=RY+LK-F.$ Thinking of $F$ as an unknown function
of $x,y$, we get the following exterior differential system in
$\mathbb{R}^{8}$ with coordinates $A,B,C,R,Y,F,x,y$:

$\displaystyle\ dA$ $\displaystyle=$
$\displaystyle\left(RB-\dfrac{L_{x}}{2L}A-\dfrac{YA}{2}\right)dx+\left(\dfrac{L_{y}}{2L}A+\dfrac{RA}{2}\right)dy,$
(13) $\displaystyle dB$ $\displaystyle=$
$\displaystyle\left(\dfrac{L_{x}}{2L}B+\dfrac{YB}{2}\right)dx+\left(YA-\dfrac{L_{y}}{2L}B-\dfrac{RB}{2}\right)dy,$
(14) $\displaystyle dC$ $\displaystyle=$
$\displaystyle\left(\dfrac{BRC}{A}+\dfrac{R}{A}-\dfrac{L_{x}}{2L}C-\dfrac{YC}{2}\right)dx+\left(\dfrac{L_{y}}{2L}C+\dfrac{RC}{2}\right)dy,$
(15) $\displaystyle dR$ $\displaystyle=$
$\displaystyle(YR+LK+F)dx+\left(R^{2}+\dfrac{L_{y}}{L}R\right)dy,$ (16)
$\displaystyle dY$ $\displaystyle=$
$\displaystyle\left(Y^{2}+\dfrac{L_{x}}{L}Y\right)dx+(RY+LK-F)dy.$ (17)

###### Lemma 3

If the system (13-17) is compatible (i.e. has solutions) then

$\begin{array}[]{l}F_{x}=3FY+\dfrac{L_{x}}{L}F+LK_{x},\\\ \\\
F_{y}=3FR+\dfrac{L_{y}}{L}F-LK_{y},\\\ \\\
F^{2}=\dfrac{1}{3}LK_{xy}-\dfrac{1}{2}LK_{x}R-\dfrac{1}{2}LK_{y}Y.\end{array}$

Proof: The condition $d(dY)=0$ gives

$F_{x}-3FY+\left(\dfrac{LL_{xy}-L_{x}L_{y}}{L^{2}}\right)Y+LKY-\dfrac{L_{x}}{L}F-LK_{x}=0,$

and after using (11)

$F_{x}=3FY+\dfrac{L_{x}}{L}F+LK_{x}.$ (18)

Similarly from $d(dR)=0$ one gets

$F_{y}=3FR+\dfrac{L_{y}}{L}F-LK_{y}.$ (19)

Thus the forms for $dR$ and $dY$ are closed if and only if

$dF=\left(3FY+\dfrac{L_{x}}{L}F+LK_{x}\right)dx+\left(3FR+\dfrac{L_{y}}{L}F-LK_{y}\right)dy.$
(20)

Finally $d(dF)=0$ results in

$F^{2}=\dfrac{1}{3}LK_{xy}-\dfrac{1}{2}LK_{x}R-\dfrac{1}{2}LK_{y}Y.$ (21)

$\Box$
Now if we define $F$ in (16,17) by (21) we get a system of finite type: all
the derivatives of $R,Y$ are given.

###### Corollary 4

The system (16,17) with $F$ given by (21) is in involution if and only if
holds (18) and (19) due to (16,17,21)

If the system (16,17) with $F$ given by (21) is in involution then system
(13-17) with $F$ given by (21) is also in involution and, by the Frobenius
Theorem, the initial values $A_{0},B_{0},C_{0},R_{0},Y_{0}$ at point
$(x_{0},y_{0})$ define uniquely a (germ of) solution to the system (13-17).
Thus one can think of $A_{0},B_{0},C_{0},R_{0},Y_{0}$ as local coordinates on
the space of integrals, even though this space may be not well defined
globally.

Note that the subsystem (16,17) does not involve $A,B$ and $C$. Therefore any
solution to this subsystem fixes an orbit of $PSL_{2}(\mathbb{R})$-action,
since at a point $(x_{0},y_{0})$ any two triples of initial values for $A,B,C$
are Möbius equivalent.

Given a solution $R$ e $Y$ to (16,17), the system (13 \- 15) is automorphic:
any two solutions $A,B,C$ are Möbius equivalent and the system (16,17) is
resolving for (13-15), i.e. it describes the Möbius orbits on the space of
solutions to (13-15) (see [O-82, V-04] for more detail).

###### Lemma 5

Suppose that the geodesic flow on a surface with metric (9) admits a
fractional linear integral (1). Then the gaussian curvature is constant if and
only if

$F=\dfrac{R_{x}-Y_{y}}{2}=0.$

Proof: If the gaussian curvature $K$ is constant then $K_{xy}=K_{x}=K_{y}=0$
and therefore $F=0$ by (21).

On the other hand, if $F=0$, then also $F_{x}=F_{y}=0$ and equations (18) and
(19) imply $LK_{x}=0,$ $LK_{y}=0$, hence $K_{x}=K_{y}=0$ and $K$ is constant.
$\Box$

###### Lemma 6

If $K=const$ then the system (16,17) is in involution.

Proof: For $K=const$ we have $F=0$ and the claim follows by Corollary 4 .
$\Box$

For further analysis we need the following claim.

###### Proposition 7

Suppose that the geodesic flow on a surface admits a fractional linear
integral. Then if one partial derivative of the gaussian curvature vanishes
then the gaussian curvature is constant.

Proof: We can suppose that $K_{x}=0$ and $K_{y}\neq 0$. Then from equation
(21) we have $Y=-\dfrac{2F^{2}}{LK_{y}}$. Differentiating by $x$ we get

$Y_{x}=-\dfrac{12F^{2}}{LK_{y}}Y-\dfrac{2L_{x}F^{2}}{L^{2}K_{y}}.$

Comparing with (17) we obtain

$L^{2}K_{y}Y^{2}+(12LF^{2}+LL_{x}K_{y})Y+2F^{2}L_{x}=0.$

Substituting $Y=-\dfrac{2F^{2}}{LK_{y}}$ results in
$\dfrac{20F^{4}}{K_{y}}=0.$ Hence $F=0$ and by Lemma 5 the curvature $K$ is
constant. $\Box$

###### Theorem 8

Suppose that the geodesic flow on a surface admits a fractional linear
integral (1). Then the dimension of the space of such integrals is 3 if and
only if the gaussian curvature is not constant.

Proof: Suppose that the gaussian curvature $K$ is not constant. Then $F\neq 0$
and by Proposition 7 holds $K_{x}\neq 0$ and $K_{y}\neq 0$. Let us solve
equation (21) for $R$

$R=\dfrac{2K_{xy}}{3K_{x}}-\dfrac{K_{y}}{K_{x}}Y-\dfrac{2F^{2}}{LK_{x}}$ (22)

and consider the system of 2 equations (17,20) with $R$ given by (22). Then
this system for $F,Y$ is of finite type and its compatibility conditions reads
as (16).

Lets us show that $F$ and $Y$ are fixed.

Differentiating (22) we determine

$\begin{array}[]{l}R_{x}=\left(\dfrac{K_{xx}K_{y}}{K_{x}^{2}}-\dfrac{K_{xy}}{K_{x}}\right)Y-\dfrac{K_{y}}{K_{x}}Y_{x}+\left(\dfrac{2L_{x}}{L^{2}K_{x}}+\dfrac{2K_{xx}}{LK_{x}^{2}}\right)F^{2}-\dfrac{4F_{x}F}{LK_{x}}+\dfrac{2K_{xxy}}{3K_{x}}-\dfrac{2K_{xx}K_{xy}}{3K_{x}^{2}},\\\
\\\
R_{y}=\left(\dfrac{K_{xy}K_{y}}{K_{x}^{2}}-\dfrac{K_{yy}}{K_{x}}+\right)Y-\dfrac{K_{y}}{K_{x}}Y_{y}+\left(\dfrac{2L_{y}}{L^{2}K_{x}}+\dfrac{2K_{xy}}{LK_{x}^{2}}\right)F^{2}-\dfrac{4F_{y}F}{LK_{x}}+\dfrac{2K_{xyy}}{3K_{x}}-\dfrac{2K_{xy}^{2}}{3K_{x}^{2}}.\end{array}$

Comparing with (16) and substituting the above expression for $R$ , we get

$\displaystyle\left(\dfrac{2L_{x}}{L^{2}K_{x}}-\dfrac{2K_{xx}}{LK_{x}^{2}}\right)F^{2}+\left(\dfrac{10F^{2}}{LK_{x}}+a_{1}\right)Y+5F-a_{2}=0,$
(23)

where

$\displaystyle
a_{1}=-\dfrac{K_{xx}K_{y}}{K_{x}^{2}}+\dfrac{5K_{xy}}{3K_{x}}+\dfrac{L_{x}K_{y}}{LK_{x}};$
$\displaystyle
a_{2}=\dfrac{2K_{xxy}}{3K_{x}}-\dfrac{2K_{xx}K_{xy}}{3K_{x}^{2}}+\dfrac{L_{xy}}{L}-\dfrac{L_{x}L_{y}}{L^{2}}.$

and

$\displaystyle\frac{10K_{xy}}{3LK_{x}^{2}}F^{2}-\dfrac{20}{L^{2}K_{x}^{2}}F^{4}-\left(\dfrac{10K_{y}F^{2}}{LK_{x}^{2}}+b_{1}\right)Y-\dfrac{5K_{y}}{K_{x}}F+b_{2}=0,$
(24)

where

$\displaystyle
b_{1}=\dfrac{5K_{xy}K_{y}}{3K_{x}^{2}}-\dfrac{K_{yy}}{K_{x}}+\dfrac{L_{y}K_{y}}{LK_{x}},$
$\displaystyle
b_{2}=-\dfrac{2K_{xyy}}{3K_{x}}+\dfrac{10K_{xy}^{2}}{9K_{x}^{2}}+\dfrac{2L_{y}K_{xy}}{3LK_{x}}-\dfrac{L_{xy}K_{y}}{LK_{x}}+\dfrac{L_{x}L_{y}K_{y}}{L^{2}K_{x}}.$

If $\dfrac{10F^{2}}{LK_{x}}+a_{1}\neq 0$ then we can resolve equation (23) for
$Y$. As the equation (24) has a non-vanishing coefficient of $F^{4}$,
substituting the found expression for $Y$ into (24) we get a nontrivial
equation for $F$. Thus $F$ is fixed and consequently $Y$ is eventually fixed
by (23).

If $\dfrac{10F^{2}}{LK_{x}}+a_{1}=0$ then $F^{2}=-\dfrac{a_{1}LK_{x}}{10}$ and
$F$ is fixed as well as $F_{x}$. Finally, from equation (18) we see that $Y$
is fixed. Thus, locally there is only one orbit of the Möbius group and the
dimension is 3.

The converse claim follows from Lemma 6. $\Box$

###### Theorem 9

Suppose that the geodesic flow on a surface admits a fractional linear
integral (1). Then the dimension of the space of linear fractional integrals
is 5 if and only if the gaussian curvature is constant.

Proof: Follows from Lemma 6 and Theorem 8. $\Box$

##  5 Geodesic Nakai’s 4-webs

In this section we consider geometric questions and therefore return to
conformal coordinates for the metric
$g=\Lambda(x,y)\left(dx^{2}+dy^{2}\right)$.

Suppose there is a fractional-linear integral (1) of the geodesic flow. It can
be written as a function of local coordinates and the tangent vector
$v=(\xi,\eta)$. By the canonical isomorphism $\psi:TM\rightarrow T^{*}M$, the
momentum $(p,q)$ and the tangent vector $(\xi,\eta)$ are related as follows:

$(\xi,\eta)=\left(\dfrac{p}{\Lambda},\dfrac{q}{\Lambda}\right).$

So, in conformal coordinates, the integral is

$I=\dfrac{A\xi+B\eta}{C\xi+E\eta}.$

Given such integral $I$, we construct a one-parameter family of foliations as
follows: for any $\lambda\in\mathbb{R}$, the equation

$I=\dfrac{A\xi+B\eta}{C\xi+E\eta}=\lambda,$

gives an ODE

$(A-C\lambda)+(B-E\lambda)\dfrac{dy}{dx}=0$

on our surface after $\xi=dx$ e $\eta=dy$. Integral curves of this ODE form a
foliation on the surface.

###### Proposition 10

The constructed foliation is geodesic.

Proof: Follows from the uniqueness of an integral curve passing through
$(x_{0},y_{0})$: the function $I$ is constant and is equal to $\lambda$ along
the geodesic with the direction $(\xi,\eta)$ at $(x_{0},y_{0})$ fixed by
$I=\lambda$. $\Box$

###### Definition 2

Nakai’s 4-web is a planar 4-web such that the cross-ratio of the four
directions tangent to the web leaves is constant.

###### Theorem 11

There exists a fractional linear integral (1) of the geodesic flow on a
surface if and only if the surface carries a geodesic Nakai’s 4-web.

Proof: Suppose there is an integral. The equation defining the geodesic
foliations of Proposition 10 can be written as

$\dfrac{A+BP}{C+EP}=\lambda,$ (25)

where $P=\dfrac{dy}{dx}$ is the inclination of the foliation leaf. The
integral gives the map $m:S\to SL_{2}(\mathbb{R})$,

$m(x,y)=\left(\begin{array}[]{l r}A&B\\\ C&E\end{array}\right).$

Then $m\cdot P=\lambda$ and $P=m^{-1}\cdot\lambda$.

Consider a $4$-web, whose foliations are fixed by four values $\lambda_{i}$,
$i=1,2,3,4$. Then the cross-ratio of the inclinations $P_{i}$ is equal to the
cross-ratio of $\lambda$-s and therefore is constant.

Now suppose that there are 4 geodesic foliations tangent to 4 direction fields
$\partial_{x}+J\partial_{y}$, $\partial_{x}+M\partial_{y}$,
$\partial_{x}+N\partial_{y}$ e $\partial_{x}+T\partial_{y}$, such that

$\dfrac{J-M}{M-N}\cdot\dfrac{N-T}{T-J}=r$ (26)

with constant $r$. Resolving for $T$ we get

$T=\dfrac{rJM-(r-1)JN-MN}{J+(r-1)M-rN}$

and

$\displaystyle
T_{x}=\dfrac{r(r-1)(M-N)^{2}J_{x}+r(J-N)^{2}M_{x}-(r-1)(J-M)^{2}N_{x}}{(J+(r-1)M-rN)^{2}},$
$\displaystyle
T_{y}=\dfrac{r(r-1)(M-N)^{2}J_{y}+r(J-N)^{2}M_{y}-(r-1)(J-M)^{2}N_{y}}{(J+(r-1)M-rN)^{2}}.$

The total differentiation by $x$ along an integral curve $y=y(x)$ of the field
$\partial_{x}+P\partial_{y}$ is

$\dfrac{d^{2}y}{dx^{2}}=P_{x}+PP_{y}.$ (27)

If these curves are geodesics then $\partial_{x}+P\partial_{y}$ is the Jacobi
field. Using the equation

$\dfrac{d^{2}y}{dx^{2}}=-\Gamma_{11}^{2}+(\Gamma_{11}^{1}-2\Gamma_{12}^{2})\dfrac{dy}{dx}-(\Gamma_{22}^{2}-2\Gamma_{12}^{1})\left(\dfrac{dy}{dx}\right)^{2}+\Gamma_{22}^{1}\left(\dfrac{dy}{dx}\right)^{3},$

for unparametrized geodesics, we get

$P_{x}+PP_{y}=-\Gamma_{11}^{2}+(\Gamma_{11}^{1}-2\Gamma_{12}^{2})P-(\Gamma_{22}^{2}-2\Gamma_{12}^{1})P^{2}+\Gamma_{22}^{1}P^{3},$

where $\Gamma_{jk}^{i}$ are Christoffel symbols for the Levi-Civita
connection. Therefore for the fields $\partial_{x}+J\partial_{y}$,
$\partial_{x}+M\partial_{y}$, $\partial_{x}+N\partial_{y}$ e
$\partial_{x}+T\partial_{y}$ we have

$\displaystyle
J_{x}+JJ_{y}-\frac{\Lambda_{y}}{2\Lambda}+\frac{\Lambda_{x}}{2\Lambda}J-\frac{\Lambda_{y}}{2\Lambda}{J}^{2}+\frac{\Lambda_{x}}{2\Lambda}{J}^{3}=0,$
(28) $\displaystyle
M_{x}+MM_{y}-\frac{\Lambda_{y}}{2\Lambda}+\frac{\Lambda_{x}}{2\Lambda}M-\frac{\Lambda_{y}}{2\Lambda}{M}^{2}+\frac{\Lambda_{x}}{2\Lambda}{M}^{3}=0,$
(29) $\displaystyle
N_{x}+NN_{y}-\frac{\Lambda_{y}}{2\Lambda}+\frac{\Lambda_{x}}{2\Lambda}N-\frac{\Lambda_{y}}{2\Lambda}{N}^{2}+\frac{\Lambda_{x}}{2\Lambda}{N}^{3}=0,$
(30) $\displaystyle
T_{x}+TT_{y}-\frac{\Lambda_{y}}{2\Lambda}+\frac{\Lambda_{x}}{2\Lambda}T-\frac{\Lambda_{y}}{2\Lambda}{T}^{2}+\frac{\Lambda_{x}}{2\Lambda}{T}^{3}=0.$
(31)

Resolving (28,5,5) for $J_{x},M_{x},N_{x}$, we get

$\displaystyle
J_{x}=-\dfrac{{\Lambda_{x}J}^{3}-\Lambda_{y}{J}^{2}+(2\Lambda{J}_{y}+\Lambda_{x})J-\Lambda_{y}}{2\Lambda},$
(32) $\displaystyle
M_{x}=-\dfrac{\Lambda_{x}{M}^{3}-\Lambda_{y}{M}^{2}+(2\Lambda{M}_{y}+\Lambda_{x})M-\Lambda_{y}}{2\Lambda},$
(33) $\displaystyle
N_{x}=-\dfrac{\Lambda_{x}{N}^{3}-\Lambda_{y}{N}^{2}+(2\Lambda{N}_{y}+\Lambda_{x})N-\Lambda_{y}}{2\Lambda}.$
(34)

Substituting $J_{x},M_{x},N_{x},T,T_{x}$, and $T_{y}$ into (5), we obtain

$\displaystyle\dfrac{r(r-1)(J-M)(J-N)(M-N)u}{2\Lambda(J+(r-1)M-rN)^{3}}=0,$
(35)

where

$u=\Lambda_{x}(J-M)(J-N)(M-N)+2\Lambda((J-M)N_{y}-(J-N)M_{y}+(M-N)J_{y}).$

Each factor in the product $r(r-1)(J-M)(J-N)(M-N)$ is not zero hence $u=0$.
Note also that $J+(r-1)M-rN\neq 0$, since otherwise from (26) follows $J=N$.

Now we construct a map $m:S\rightarrow SL_{2}(\mathbb{R})$ ,
$m=\left(\begin{array}[]{cc}A&B\\\ C&E\end{array}\right),$ resolving

$\displaystyle\dfrac{A+BJ}{C+EJ}=0,\ \ \ \dfrac{A+BM}{C+EM}=1,\ \ \ C+EN=0,\ \
\ AE-BC=1,$ (36)

for $A,B,C,E$, and show that $I=\dfrac{Ap+Bq}{Cp+Eq}$ is a first integral. To
this end we verify that

$\begin{array}[]{l}2\Lambda(CA_{x}-AC_{x})-\Lambda_{y}=0,\\\ \\\
2\Lambda(EA_{x}-AE_{x}+CA_{y}-AC_{y}+CB_{x}-BC_{x})+\Lambda_{x}=0,\\\ \\\
2\Lambda(EA_{y}-AE_{y}+CB_{y}-BC_{y}+EB_{x}-BE_{x})-\Lambda_{y}=0,\\\ \\\
2\Lambda(EB_{y}-BE_{y})+\Lambda_{x}=0.\end{array}$ (37)

Then $I$ is a first integral by Lemma 1.

From (36) we obtain

$A=\dfrac{J(M-N)E}{J-M},\ \ \ B=-\dfrac{(M-N)E}{J-M},\ \ \ C=-NE.$

Differentiating, we have

$\begin{array}[]{l}A_{x}=-\dfrac{EM(M-N)J_{x}}{(J-M)^{2}}+\dfrac{EJ(J-N)M_{x}}{(J-M)^{2}}+\dfrac{E_{x}J(M-N)}{J-M}-\dfrac{EJN_{x}}{J-M},\\\
\\\
A_{y}=-\dfrac{EM(M-N)J_{y}}{(J-M)^{2}}+\dfrac{EJ(J-N)M_{y}}{(J-M)^{2}}+\dfrac{E_{y}J(M-N)}{J-M}-\dfrac{EJN_{y}}{J-M},\end{array}$

as well as

$\begin{array}[]{l}B_{x}=\dfrac{E(M-N)J_{x}}{(J-M)^{2}}-\dfrac{E(J-N)M_{x}}{(J-M)^{2}}-\dfrac{E_{x}(M-N)}{J-M}+\dfrac{EN_{x}}{J-M},\\\
\\\
B_{y}=\dfrac{E(M-N)J_{y}}{(J-M)^{2}}-\dfrac{E(J-N)M_{y}}{(J-M)^{2}}-\dfrac{E_{y}(M-N)}{J-M}+\dfrac{EN_{y}}{J-M},\end{array}$

and

$\begin{array}[]{l}C_{x}=-EN_{x}-E_{x}N,\\\ C_{y}=-EN_{y}-E_{y}N.\end{array}$

Now the expression $2\Lambda(CA_{x}-AC_{x})-\Lambda_{y}(AE-BC)$ writes as

$\displaystyle\dfrac{2\Lambda E^{2}MN(M-N)J_{x}}{(J-M)^{2}}-\dfrac{2\Lambda
E^{2}JN(J-N)M_{x}}{(J-M)^{2}}+\dfrac{2\Lambda
E^{2}JMN_{x}}{J-M}-\dfrac{\Lambda_{y}E^{2}(M-N)(J-N)}{J-M}.$

One checks that

$E^{2}=\dfrac{J-M}{(M-N)(J-N)},$

and has

$2\Lambda(CA_{x}-AC_{x})-\Lambda_{y}(AE-BC)=-\dfrac{uJMN}{(J-M)(M-N)(J-N)}.$

Similarly we get

$2\Lambda(EA_{x}-AE_{x}+CA_{y}-AC_{y}+CB_{x}-BC_{x})+\Lambda_{x}(AE-
BC)=\dfrac{u(JN+JM+MN)}{(J-M)(M-N)(J-N)}$

and

$2\Lambda(EA_{y}-AE_{y}+CB_{y}-BC_{y}+EB_{x}-BE_{x})-\Lambda_{y}(AE-
BC)=-\dfrac{u(J+M+N)}{(J-M)(M-N)(J-N)},$

and

$2\Lambda(EB_{y}-BE_{y})+\Lambda_{x}(AE-BC)=\dfrac{u}{(J-M)(M-N)(J-N)}.$

Since $u=0$ and $AE-BC=1$, the last 4 equations give (37). $\Box$

##  6 Concluding remarks

One easily checks that the explicit examples of metrics constructed by Agapov
and Shubin [AS-21] admit projective vector fields, whose local flows maps
geodesics into geodesics but do not have to respect metrics (see [L-82]). It
would be interesting to understand relation between existence conditions for
projective vector fields and for fractional-linear integrals.

## Acknowledgments

This research was supported by FAPESP grant # 2022/12813-5 (S.I.A) and CAPES
grant # 88882.434346/2019-01 (T.G.P.A).

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