Normal forms of two-dimensional metrics admitting exactly one
essential projective vector field
Gianni Manno∗, Andreas Vollmer†
arXiv:1705.06630v1 [math.DG] 18 May 2017
May 19, 2017
Abstract
We give a complete list of mutually non-isometric normal forms for the two-dimensional metrics that
admit an essential projective vector field. This extends the results of [6, 12], solving a problem posed by
Sophus Lie in 1882 [10].
1
Basic definitions and description of the main results
Let (M, g) be a smooth Riemannian or pseudo-Riemannian manifold of dimension 2. In the following, the
abbreviation “metric” is used for both Riemannian and pseudo-Riemannian metrics, unless otherwise specified.
Definition 1. A vector field on M is called projective if its (local) flow maps geodesics into geodesics (where
we view geodesics as unparametrized curves).
The set of projective vector fields of a metric g forms a Lie algebra [9], denoted by p(g) in the following.
Infinitesimal homotheties, i.e. vector fields w such that Lw g = λg for some λ ∈ R, are examples of projective
vector fields.
Definition 2. A projective vector field that is not an infinitesimal homothety is called essential.
Definition 3. Two metrics are called projectively equivalent if they have the same geodesics (as unparametrized
curves). The collection of all metrics projectively equivalent to a given metric g is called the projective class
of g. We denote it by P(g).
It turns out that metrics belonging to the same projective class admit the same algebra of projective vector
fields. A natural question is to characterize the projective classes of metrics with a prescribed Lie algebra
of projective symmetries. A more thorough investigation is to find a complete list of mutually non-isometric
normal forms of metrics within the same projective class.
In these directions, in his 1882 paper [10], Sophus Lie formulated the following problem: Find the metrics
that describe surfaces whose geodesic curves admit an infinitesimal transformation 1 (i.e. with p(g) ≥ 1).
Fubini referred to this problem as the “Lie problem”, see [1, 8]. An overview of the history of the problem
can be found, for instance, in [1, 2]. Important works in the field, with respect to the considerations that follow
below, are for instance [7, 11, 6, 12, 5].
In [6], the authors find mutually non-isometric normal forms of 2-dimensional metrics g with dim(p(g)) ≥ 2
around generic points. Metrics admitting an infinitesmal homothety w were described in [10]: locally around a
generic point there is a system of coordinates (x, y) such that w = ∂x and the metric is of the form
E(y) F (y)
λx
,
g=e
F (y) G(y)
where the matrix on the right hand side is a non-degenerate matrix.
On the other hand, metrics admitting one and only one essential projective vector field have been found in
[12], around generic points, via an explicit description of all projective classes. However, no normal forms of
mutually non-isometric metrics are provided in [12]. The main outcome of the present paper is to close this
∗ Dipartimento di Scienze Matematiche (DISMA), Politecnico di Torino, Corso Duca degli Abruzzi, 24, 10129 Torino,
[email protected]
† Istituto Nazionale di Alta Matematica – Dipartimento di Scienze Matematiche (DISMA), Politecnico di Torino, Corso Duca
degli Abruzzi, 24, 10129 Torino, [email protected]
1 German original [10]: “Es wird verlangt, die Form des Bogenelementes einer jeden Fläche zu bestimmen, deren geodätische
Curven eine infinitesimale Transformation gestatten.”
1
gap: Theorem 1 below provides a complete list of mutually non-isometric normal forms for the 2-dimensional
metrics admitting an essential projective vector field. Moreover, [12] discusses the projective classes of metrics
with one essential projective vector field, without specifying exactly which metrics admit an essential projective
vector field. In Proposition 4 it is clarified how to distinguish metrics with an essential projective vector field
from those with a homothetic one. For the formulation of the main result, we need the following proposition.
Proposition 1 ([3, 4, 7]). Let g and ǧ be projectively equivalent, non-proportional metrics. Then, in a neighborhood of almost every point, there are coordinates (x, y) such that the metrics assume one of the following
three normal forms:
g
ǧ
A
Liouville case
(X − Y )(dx2 ± dy 2 )
( X1 −
1
Y
2
)( dx
X ±
dy 2
Y )
B
complex Liouville case
(h(z) − h(z)) (dz 2 − dz 2 )
2
1
dz 2
1
dz
− h(z)
− h(z)
h(z)
h(z)
C
Jordan block case
(1 + xY ′ )dxdy
1+xY ′
(−2Y dxdy
Y4
+(1 + xY ′ )2 dy 2 )
Here, X = X(x) and Y = Y (y) are functions of one variable only, and in the complex Liouville case we use
coordinates z = x + iy, z = x − iy.
Now we can formulate our main result. The following theorem provides a list of normal forms of metrics
admitting exactly one essential projective vector field, indicating the type of the metric according to Proposition 1.
Notation. From now on we adopt, for brevity of exposition, the following convention: Whenever there is a
non-integer real exponent, the base expression is to be understood as the absolute value if it is real-valued,
unless explicitly stated otherwise. The same convention will apply if the exponent is real, but non-specified. For
2
instance, y ξ with ξ ∈ (0, 1) shall be understood as |y|ξ , for real y, and det(g) /3 is to be understood as identical
2/3
to | det(g)| . This convention shall not be applied if the base expression is complex-valued. For instance, for
z = ρeiθ ∈ C with ρ > 0 and θ ∈ [0, 2π), we have that z ξ = (ρeRiθ )ξ = ρξ eiθ ξ .
y
The indefinite integral of a function f (y) is denoted by
f (ξ) dξ. Note that in Theorem 1 below the
constant of integration is absorbed by a translation of the coordinate x.
Theorem 1. Let g be a Riemannian or pseudo-Riemannian metric on a 2-dimensional manifold M . Let us
assume that dim(p(g)) = 1 in any neighborhood of M and that p(g) is generated by an essential projective
vector field. Then, in a neighborhood of almost every point there exists a local coordinate system (x, y) such
that g assumes one of the following mutually non-isometric normal forms (organized according to their type,
see Proposition 1).
(A) Liouville metrics
λ−1
1. For k ∈ R \ {0}, ε2 ∈ {±1} and in the projective class determined by ξ = 2 2λ+1
∈ (0, 1) ∪ (1, 4], ε ∈ {±1}
and h ∈ R \ {0} (if ξ = 2: h 6= −ε1 )
(eξx − heξy ) e2x
(eξx − heξy ) e2y
2
2
g=k
.
dx + ε
dy
(1 + ε2 heξy )(1 + ε2 eξx )2
(1 + ε2 heξy )2 (1 + ε2 eξx )
Special cases: If h = −1, we require ε2 = 1. If h = +1 and ε = 1, the parameter k may only assume
positive values, k > 0.
2. For k ∈ R \ {0} and the projective class determined by h ∈ R \ {0}:
(y − x)e−3x 2 h(y − x)e−3y 2
g=k
dy .
dx +
x2 y
xy 2
In case h = 1, k > 0 is positive.
3. For θ ∈ [0, 2π), κ ∈ R \ {0} and the projective class determined by h ∈ R \ {0} and λ ∈ R (if λ = 0:
h 6= ±1):
For λ 6= 0:
for λ = 0:
−3λ x
e
h e−3λ y
sin(y − x)
2
2
,
dx +
dy
g=
sin(y + θ) sin(x + θ) sin(x + θ)
sin(y + θ)
sin(y − x)
dx2
h dy 2
.
g=κ
+
sin(y) sin(x) sin(x) sin(y)
2
(B) Complex–Liouville metrics
Below, we use the notation P1 = eiα where 0 ≤ α < π, in the complex plane C ∼ R2 .
λ−1
4. For κ ∈ R \ {0} and the projective class determined by h ∈ P1 , ξ = 2 2λ+1
∈ (0, 1) ∪ (1, 4] (if ξ = 2:
h 6= ±1):
hz ξ − hz ξ
hz ξ − hz ξ
2
2
g=κ
.
dz
+
dz
(1 + hz ξ )(1 + hz ξ )2
(1 + hz ξ )2 (1 + hz ξ )
5. For κ > 0 and the projective class determined by h ∈ P1 :
g = κ (z − z)
h e−3z 2 h e−3z 2
.
dz
dz
−
z2z
zz2
6. For θ ∈ [0, 2π), κ > 0 and the projective class determined by h ∈ P1 and λ ∈ R (if λ = 0: h 6= ±1):
h e−3λz
sin(z − z)
h e−3λz
dz 2 +
dz 2 ,
sin(z + θ) sin(z + θ) sin(z + θ)
sin(z + θ)
2 2
h dz
sin(z − z)
h dz
.
g=κ
+
sin(z) sin(z) sin(z) sin(z)
For λ 6= 0:
g=
for λ = 0:
(C) Jordan block metrics
λ−1
7. For ε ∈ {±1}, k ∈ R \ {0} and the projective class determined by ξ = 2 2λ+1
∈ (0, 12 ) ∪ ( 21 , 1) ∪ (1, 4]:

1
ξ
 (ε − y) + x
dxdy +
g = k −2
y3
8. For κ ∈ R \ {0}:
g = κ (Y (y) + x) dxdy
2
1
(ε − y) ξ + x
with Y (y) =
y4
Ry
3
e /2s
|s|3/2


dy 2 
ds
9. For the class determined by λ ∈ R:
g = κ (Y (λ, y) + x) dxdy
with κ ∈ R \ {0} if λ = p
0 and κ = eλα with α ∈ [0, 2π)
otherwise. The function Y (λ, y) is given by
p
Z y −3λ/2 arctan(s)
Z y
4
4
2+1
3
e
3
y
ξ2 + 1
√
e− 2 λ arctan(ξ)
dξ
=
ds .
+
Y (λ, y) = e− 2 λ arctan(y)
4 2
2
y − 3λ
(ξ − 3λ)
s +1
10. Superintegrable case: For κ ∈ R \ {0}, ε ∈ {±1} and c ∈ R, respectively:
y2 + x
(y 2 + x)2 2
(a) g = κ −2
dxdy +
dy
(y − ε)3
(y − ε)4
κ
9(y 2 + x)2 dx2 − 2(y 3 + 9xy − 2ε)(y 2 + x) dxdy + 12x(y 2 + x)2 dy 2
(b) g =
2
F (0, ε; x, y)
κ
9(y 2 + x)2 dx2 − 2(y 3 + 2ε y + 9xy − 2c)(y 2 + x) dxdy + 4(ε + 3x)(y 2 + x)2 dy 2
(c) g =
F (ε, c; x, y)2
with
F (ζ, c; x, y) = y 6 − 9xy 4 + 27x2 y 2 − 27x3 + 4c2 − (36xy + 4y 3 ) c + (18xy 2 − 5y 4 − 9x2 − 8cy) ζ + 4y 2 ζ 2 .
Remark 1. In Theorem 1, each item 1 − 10 corresponds to a different type of projective classes (see also
Proposition 3 below).
Structure of the paper. The paper is organized as follows: In Section 2, we discuss our approach and the
techniques that are made use of in the proof of Theorem 1. First, in Section 2.1, we present results that are
needed in the proof, then in Section 2.2 we discuss the general procedure followed in the proof of Theorem 1. This
proof is contained in Section 3, which is divided into three parts, according to the properties of Lie derivative
along the projective vector field, see the cases given in (7). Each part is subsequently divided further into three
or four subsections, which follow the cases of Proposition 1, and whose exact structure will be explained below
(see Proposition 3).
Acknowledgments. The authors wish to thank Vladimir Matveev for numerous suggestions, helpful discussions and comments on the manuscript. Andreas Vollmer is a research fellow of Istituto Nazionale di Alta
Matematica (INdAM). Both authors are members of GNSASA of INdAM and also gratefully acknowledge
support from the project FIR-2013 Geometria delle equazioni differenziali. Gianni Manno was also partially
supported by “Starting grant per giovani ricercatori” of Politecnico di Torino, code 53 RSG16MANGIO.
3
2
Methods
2.1
Main results used in the proof of Theorem 1
A metric g given in an explicit system of coordinates (x, y) gives rise, via its Levi-Civita connection, to a second
order ordinary differential equation (ODE)
y ′′ = −Γ211 + (Γ111 − 2Γ212 ) y ′ − (Γ222 − 2Γ112 ) y ′2 + Γ122 y ′3 ,
(1)
where y = y(x) and Γijk are the Christoffel symbols of g. The ODE (1) is called the projective connection
associated to g. The name is justified by the fact that, for a solution y(x) to (1), the curve
(x, y(x)) is a geodesic
of g up to reparametrization. Thus, (local) diffeomorphisms (x, y) → u(x, y), v(x, y) preserving (1) (finite
point symmetries) are projective transformations of g, i.e. they send geodesics into geodesics. Infinitesimal
point symmetries of (1) are projective vector fields of g and generate a 1-parametric family of projective
transformations. Let us consider a general (2-dimensional) projective connection,
y ′′ = f0 + f1 y ′ + f2 y ′2 + f3 y ′3 ,
fi = fi (x, y) .
(2)
A natural question is whether or not the projective connection (2) is metrizable, i.e. if there exists a 2-dimensional
metric g such that (2) is equal to (1).
The problem of metrisability is discussed from multiple perspectives in [5]. The metrisability condition can
be expressed in terms of a system of 4 partial differential equations (PDE) on the components of g. Its solution
is discussed in [12] and [6] for metrics with, respectively, one or more projective vector fields. These references
introduce a major simplification to the PDE problem: The system turns into a linear system of PDE when
replacing the unknowns, i.e. the components of the metric g, by the components of a weighted tensor section,
denoted by a = ψ −1 (g) and called the Liouville tensor associated to g in the following.2 Liouville tensors take
values in S 2 M ⊗ (vol M )−4/3 (S 2 M is the symmetric tensor product of the module of 1-forms on M , whereas
vol M is the one-dimensional bundle of volume forms on M ):
a = ψ −1 (g) =
g
,
| det(g)|2/3
g = ψ(a) =
a
.
det(a)2
(3)
More precisely we have the following proposition.
Proposition 2 ([6, 11]). The projective connection associated to the Levi-Civita connection of a metric g is (2)
g
if and only if the entries of the matrix aij describing a = ψ −1 (g) = | det(g)|
2/3 satisfy the linear system of PDEs

a11x − 32 f1 a11 + 2f0 a12 = 0



a11y + 2a12x − 34 f2 a11 + 32 f1 a12 + 2f0 a22 = 0
2a12y + a22x − 2f3 a11 − 32 f2 a12 + 34 f1 a22 = 0



a22y − 2f3 a12 + 32 f2 a22 = 0
(4)
where the subscripts x, y denote derivatives.
Of course, solutions to the linear system of PDE (4) span a linear space.
Definition 4. The space of solutions to the system (4) for a metric g is denoted by A(g): the dimension of
A(g) is called the degree of mobility of g. We usually abbreviate A = A(g) since there is no risk of confusion.
Remark 2. Let A be the space of solutions to the linear system of PDE (4), and let a ∈ A be a particular
solution. The mapping ψ defined in (3) is a bijection that identifies the solution a of (4) with the metric
g = ψ(a) that admits one projective vector field. In view of this correspondence, also the spaces A and P(g) are
identified, and in particular we have P(g) = ψ(A).
It is proven in [12] that, if we restrict our attention to metrics with an essential
projective vector field, dim(A)
Pm
is either 2 or 3. Therefore, if we denote by {ai } a basis of A, we have that a = i=1 Ki ai , where m = dim(A) ∈
{2, 3}. The corresponding metric, via Formula (3), is given by
Pm
gi
i=1 Ki det(gi )2/3
(5)
g = g[K1 , ..., Km ] = h
i2 , m = dim(A) ∈ {2, 3} .
Pm
gi
K
det
i det(gi )2/3
i=1
It can be easily seen that P(g1 ) = ... = P(gm ) = P(g[K1 , ..., Km ]), with (K1 , ..., Km ) ∈ Rm \ {0}, m =
dim(A) ∈ {2, 3}.
2 The
linearized system of PDE (4) has, to the best knowledge of the authors, first been discussed by Liouville in [11].
4
Notation. In what follows we sometimes alternatively denote the projective class P(g1 ) = ... = P(gm ), for
projectively equivalent metrics g1 , ..., gm , by P(g1 , ..., gm ), i.e.
P(g1 , ..., gm ) := P(g1 ) = ... = P(gm ).
In this way we underline that an arbitrary metric in the projective class can be obtained by g1 , ..., gm via
Formula (5). In the above notation we assume that the metrics g1 , . . . , gm form a basis of their projective class,
in the sense that the corresponding Liouville tensors ai = ψ −1 (gi ) form a basis of the linear space A.
The space A is invariant under the action of the Lie derivative Lw , where w is a projective vector field, the
action of Lw on the space of Liouville tensor being induced by the action of Lw on the metrics. More precisely
we have that
2
(6)
Lw a = det(g)−2/3 · Lw g − det(g)−2/3 trg (Lw g) · g
3
Consider the Lie derivative Lw |A : A → A. There exists a basis of A such that Lw |A is represented by a matrix
in Jordan normal form. Since dim(A) ∈ {2, 3}, there exists a two-dimensional, Lw -invariant subspace  ⊂ A
(for m = 2, Â = A). In a suitable basis, the (restricted) Lie derivative Lw |Â is represented by one of the
following three matrices (note that we can rescale the matrices by a constant factor since the projective vector
field is defined up to a constant factor only [12]):
λ 0
1 1
λ −1
(I)
with |λ| ≥ 1 ;
(II)
;
(III)
, with λ ∈ R.
(7)
0 1
0 1
1 λ
By using the above normal forms for each type of metrics described in Proposition 1, the author of [12] was able
to divide metrics that admit exactly one essential projective vector field into several non-equivalent projective
classes. The following proposition, which is one of the main tools we shall use, summarizes this result.
Proposition 3 ([12]). Let g be a 2-dimensional metric on M such that dim(p(g)) = 1. Moreover, let us assume
that g|U admits no homothetic vector field in any neighborhood U ⊂ M . Then, in a neighborhood of almost every
point there exists a coordinate system (x, y) such that g is of the form (5), where gi , i ∈ {1, ..., m}, m ∈ {2, 3},
are described below.
A) Liouville-type metrics
2
2
g1 = (X(x) − Y (y))(X1 (x) dx + Y1 (y) dy ) ,
g2 =
1
1
−
X
Y
X1 2 Y1 2
dx +
dy
X
Y
(8)
For h ∈ R \ {0}, ν ∈ (0, 1) ∪ (1, 4], ε ∈ {±1}, λ ∈ R:
(I) X(x) = eνx , Y (y) = heνy , X1 (x) = e2x , Y1 (y) = ε e2y (if ν = 2: h 6= −ε)
−3x
−3y
(II) X(x) = x1 , Y (y) = y1 , X1 (x) = e x , Y1 (y) = h e y
−3λx
e
e−3λy
(III) X(x) = tan(x), Y (y) = tan(y), X1 (x) = cos(x)
, Y1 (y) = h cos(y)
(if λ = 0: c 6= ±1)
B) complex Liouville metrics
2
2
g1 = (h(z) − h(z)) (h1 (z) dz + h1 (z) dz ) ,
g2 =
1
1
−
h(z) h(z)
!
h1 (z) 2 h1 (z) 2
dz +
dz
h(z)
h(z)
!
(9)
For C ∈ C, |C| = 1, ν ∈ (0, 1) ∪ (1, 4], λ ∈ R:
(I) h(z) = C eνz , h1 (z) = e2z (if ν = 2: C 6= ±1)
−3z
(II) h(z) = 1z , h1 (z) = C e z
e−3λz
(if λ = 0: C 6= ±1)
(III) h(z) = tan(z), h1 (z) = C cos(z)
C) Jordan-block metrics
g1 = (Y (y) + x) dxdy ,
g2 = −
2 (Y (y) + x)
(Y (y) + x)2 2
dxdy
+
dy
y3
y4
(10)
For λ ∈ R and η ∈ (0, 12 ) ∪ ( 12 , 1) ∪ (1, 4]:
(Ia) Y (y) = y 2 , and there is a third metric:
g3 =
y2 + x
9(y 2 + x) dx2 − 4y (9x + y 2 ) dxdy + 12x (y 2 + x) dy 2
2
6
(3x − y )
(Ib) Y (y) = y 1/η √
R y 3/2s √|s|
|y|
3
(II) Y (y) = e /2y y−3 +
e
(s−3)2 ds
√
4
R y − 3 λ arctan(s) √
2
4 2
3
y +1
s +1
e 2
(III) Y (y) = e− 2 λ arctan(y) y−3λ +
(s−3λ)2 ds
(11)
Remark 3. Note that two metrics belonging to different classes A(I)—C(III) of Proposition 3 are nonisometric, because isometries are projective transformations, and thus preserve the projective class.
5
2.2
General procedure of the proof of Theorem 1
We shall obtain the normal forms of Theorem 1 by thoroughly investigating possible isometries within the cases
A(I)—C(III) of the list of Proposition 3, as metrics belonging to different types among A(I)—C(III) cannot
be isometric since they are not even projectively equivalent. Of course, a (local) diffeomorphism that links
two metrics in a fixed projective class amongst A(I)—C(III) is a projective transformation, so the original
problem of characterizing metrics up to isometries (where the metrics admit exactly one essential projective
vector field) can be rephrased as follows: to characterize metrics with no homothetic vector field within the
classes A(I)—C(III) of Proposition 3, and then to determine their normal forms (up to isometries).
The current section is organized as follows: As a preliminary step, by Proposition 4, we characterize the
metrics in Proposition 3 that do not admit a homothetic vector field. The normal forms are then obtained using
two basic steps (see below) and a number of specific techniques that are explained when needed.
Let us begin with the characterization of metrics that admit a homothetic vector field.
Proposition 4. The projective vector field w of a metric g in a projective class of Proposition 3 is homothetic
if and only if the Liouville tensor a = ψ −1 (g) is a eigenvector of Lw |A .
In other words, the projective vector field w satisfies Lw g = η g for some η ∈ R if and only if there is an
eigenvalue µ such that Lw a = µ a where a = ψ −1 (g).
Note that here the restricted Lie derivative Lw |A is denoted simply by Lw , as we are going to do from now on
unless otherwise specified.
Proof. First, assume there is a homothetic vector field w. Then, since dim(p(g)) = 1, we have Lw g = η g.
Furthermore, according to Equation (6),
Lw a = det(g)−2/3 · Lw g −
4
η
2
det(g)−2/3 trg (Lw g) · g = η a − η · a = − a.
3
3
3
|{z}
=µ
On the other hand, if Lw a = µ a, then compute the components of Lw g, i.e.
Lw aij
ω∂k wk
aij
=
−
− 2 det(a)−3 wk ∂k (det(a)) aij
Lw gij = wk ∂k
det(a)2
det(a)2
det(a)2
=
ω∂k wk
wk trg (∂k a)
Lw aij
−
aij − 2
gij
2
2
det(a)
det(a)
det(a)2
Lw a=µ a
=
3µ gij + (4 + 3ω)∂k wk gij = −3µ gij
where ω = − 34 is the weight of the Liouville tensor a. This confirms Lw g = η g with η ∈ R, implying w is a
homothetic vector field.
Proposition 4 permits us to easily identify the metrics with one essential projective vector field. As announced
in the beginning of this section, we shall proceed further following two steps, in each of which a suitable invariant
is introduced. Following these steps, we shall obtain, in Section 3, the list of mutually non-isometric metrics of
Theorem 1.
Let us fix a system of coordinates (x, y) as in Proposition 3. Metrics of type A (resp. B, C not of type (Ia))
belong to P(g1 , g2 ) with g1 , g2 given by (8) (resp. (9), (10)). Metrics of type C(Ia) belong to P(g1 , g2 , g3 )
where gi (i ∈ {1, 2, 3}) are given by (10) and (11). We recall that a metric g belonging to one of the previous
projective classes P(g1 , . . . , gm ), m ∈ {2, 3}, is of the form (5). Moreover, since we are interested in metrics
admitting exactly one essential projective vector field, in view of Proposition 4 we can infer that, for the
purposes of Theorem 1, we have to consider exactly those combinations (5) of the metrics of Proposition 3 with
at least two non-zero parameters Ki , i = 1, ..., m (where m = 2, 3). Recall that, whereas Formula (5) is of
“geometric” character (it looks alike for any choice of coordinates), the coordinate expressions of the metrics gi
in Proposition 3 are not.
Step 1 (projective orbits). Let us consider a projective class among A(I)—C(III) of Proposition 3. Metrics
belonging to this class are of the form (5) and admit, up to a non-vanishing constant factor, one and only one
projective vector field: let us fix this vector field and denote it by w, and its local flow by φt .
Definition 5 (Projective orbits). Let w be a projective vector field and φt be its local flow. An orbit of φt in A
passing through a Liouville tensor a is called the projective orbit through a. In view of Remark 2, the term of
projective orbit will refer either to Liouville tensors or metrics, depending on the context.
6
The transformation φt sends a metric of the form (5) into a metric of the same form, so it induces a map
Rm \ {eigenspaces} → Rm \ {eigenspaces}, m = dim A ∈ {2, 3}. Here, we removed eigenspaces of Lw as we are
considering metrics admitting no homothetic vector fields (see discussions above). Below these eigenspaces are
axes of Rm . The first step therefore is to describe φt -orbits in terms of the parameters Ki .
We shall proceed as follows. Let g ∈PP(g1 , ..., gm ), m ∈ {2, 3}, and let a = ψ −1 (g) be its associated Liouville
tensor, see (3). It turns out that a = i Ki ai , for some Ki ∈ R, where ai are the associated Liouville tensors
of gi , i = 1, ..., m. We have already seen that the Lie derivative Lw acts on the space of Liouville tensors (see
Section 2.1) in such a way that the local flow φt acts on a as follows:
!
m
m
m
X
X
X
Ki′ ai .
(12)
Ki exp(tLw )ai =
K i ai =
φ∗t
i=1
i=1
i=1
From the last equality of (12), by eliminating the parameter t, one can derive relations on (Ki′ )i=1, ... ,m and
(Ki )i=1, ... ,m only, describing the φt -orbit of (isometric) metrics in P(g1 , ..., gm ) (such orbits do not depend
on the chosen projective vector field). This permits us to find a function f (K1 , ..., Km ) whose level sets are
orbits of the projective flow in A. In some cases it is helpful to replace the parameters Ki by another choice of
parameters, using f itself as a parameter. Note that this amounts to a change of coordinates in the space A.
Metrics on the same projective orbit are isometric, as they are linked by φ∗t . A complication that can occur is
that not all projective transformations are of type φt for some t, so that metrics belonging to different φt -orbits
could be isometric. In order to see if this happens, we need additional techniques. The most important one
to be used in the proof is the thorough investigation of the function describing the (square of the) length of
the projective vector field (Step 2 below). Other, more specific techniques, shall be introduced inside the proof
when needed.
Another complication appears if dim(A) = 3, because the resulting metrics become more cumbersome to
handle. In this case we use special elements of A, taken from eigenspaces of Lw , to obtain restrictions on possible
isometries.
Step 2 (isometries). Let τ be an isometry between g ′ and g = τ ∗ (g ′ ). Let us fix the projective vector field
w (we recall that it is defined up to a non-vanishing constant factor) by the condition (recall the action of Lw
on the space of Liouville tensors, see formula (6))
a
a
=M
,
(13)
Lw
ǎ
ǎ
where M is a 2×2-matrix as in (7). This condition depends only on the directions of a and ǎ in the 2-dimensional
invariant subspace  ⊂ A. The isometry τ preserves eigenspaces and thus condition (13) fixes w, in the sense
that τ ∗ (w) = w. Now, let φt be the local flow of w. We have that t → τ (φt (p0 )), with p0 being a point on the
manifold, is an integral curve of w starting at p′0 = τ (p0 ). Moreover, we have that
gφt (p0 ) (wφt (p0 ) , wφt (p0 ) ) = gτ′ (φt (p0 )) (wτ (φt (p0 )) , wτ (φt (p0 )) ) = gφ′ t (p′0 ) (wφt (p′0 ) , wφt (p′0 ) ) .
(14)
We are going to study the poles and zeros of each side of Equation (14), regarded as functions in t, which
coincide for isometric metrics. Thus we obtain extra conditions on Ki and Ki′ established in Step 1. We also
find conditions on p′0 and p0 , which provide information on the possible isometries between g ′ and g.
3
Proof of Theorem 1
In Sections 3.1, 3.2 and 3.3 we shall consider, respectively, the cases in which the Lie derivative Lw |Â restricted
to the 2-dimensional invariant subspace  ⊂ A is of type (I), (II) or (III) of (7); we recall that w is a projective
vector field as in Step 1 of Section 2.2 and A is the linear space of solutions of (4). For each of these three
cases, we proceed as follows: Using mainly the methods described in Section 2, we study the three types of
metrics from Proposition 1, thus obtaining the normal forms of Theorem 1. Specifically, the (real) Liouville-type
metrics, complex Liouville-type metrics and Jordan block-type metrics from Proposition 1 yield, respectively,
normal forms under (A), (B) and (C) in Theorem 1.
3.1
Case (I): Lw has 2 real eigenvalues
Here we are in the case when Lw is of the form (I) of (7). So, let (a1 , a2 ) be a basis of  ⊂ A in which Lw = Lw |Â
assumes such a form, i.e.
a1
λ 0
a1
=
,
(15)
Lw
0 1
a2
a2
7
K2
K2
K2
K1
K1
K2
K1
K2
K2
K1
K1
K1
Figure 1: Graph of the orbits of the projective action of φ∗t in the subspace A, for a metric g = ψ(K1 a1 + K2 a2 )
of type A(I) in Proposition 3, c.f. (5). The pictures show a = K1 a1 + K2 a2 as a point (K1 , K2 ), i.e. a1 lies on
the axis of abscissae, while a2 is on the axis of ordinates.
The graphs in the upper half represent the case λ > 0, whereas the graphs in the lower half are for λ < 0. The
plots to the left illustrate, for the generic situation, several orbits with different K1 and/or K2 (the thick curve
is one example of such an orbit). The two graphs in the middle and those to the right represent the orbits for
special cases, i.e. for projective classes that have discrete projective symmetries. In these plots, the thick solid
line is the orbit of the projective action of φ∗t , and the dashed line represents isometric metrics that are reached
due to a discrete projective symmetry.
where |λ| ≥ 1 and w is the projective vector field fixed by condition (15) (see Step 2 of Section 2.2 for more
details). Let us denote the flow of w by P
φt . Any metric g from the projective class P(g1 , g2 ), where gi = ψ(ai )
m
with ψ as in (3), is the ψ-image of a = 1 Ki ai , m ∈ {2, 3}. By Equation (15), taking into account (12), we
have
φ∗t (K1 a1 + K2 a2 ) = K1 eλt a1 + K2 et a2 =: K1′ a1 + K2′ a2 .
(16)
The following lemma, that can be proved by a direct computation, characterizes metrics on the same φt -orbit.
Lemma 1. The pairs (K1 , K2 ) and (K1′ , K2′ ) are related by (16) if and only if they satisfy one of the following
cases:
(a)
K1
>0
K1′
and
K2 = K2′ = 0
(b)
K2
>0
K2′
and
K1
K1′
=
.
λ
|K2 |
|K2′ |λ
Proof. Let us assume K1′ , K1 , K2′ , K2 6= 0 (otherwise the statement is trivial). Now, let us assume that (16)
K2
holds, i.e. K1′ = eλt K1 and K2′ = et K2 , that in particular implies K
′ > 0. Exponentiating the absolute values
2
in the second equation by λ 6= 0 and then eliminating t we obtain
K1′ |K2 |λ = K1 |K2′ |λ ,
i.e. the case (b) above.
Conversely, if we assume either (a) or (b), we arrive, by a straightforward computation, to the condition (16).
The graphs in Figure 1 illustrate the orbits of the projective flow inside Â.
3.1.1
Normal forms (A.1) of Theorem 1
In the case of Liouville metrics, A(I), we can construct any metric g in the projective class, according to
Proposition 3 and via Formula (5), from the following two metrics:
1
X1 2 Y1 2
1
2
2
g1 = (X − Y )(X1 dx + Y1 dy )
and
g2 =
(17)
−
dx + dy ,
X
Y
X
Y
where, after an obvious change of coordinates, X = xξ , Y = hy ξ , X1 = 1, Y1 = ε, and where
ξ=2
λ−1
,
2λ + 1
|λ| ≥ 1 and λ 6= 1 .
8
(18)
The projective vector field w is given by
w=−
and its flow reads
φt (x0 , y0 ) = (x0 e−
2λ+1
t
2
2λ + 1
(x∂x + y∂y )
2
, y0 e−
2λ+1
t
2
)
where
x0 > 0,
y0 > 0.
(19)
The projective connection (see (1)) of any metrics in P(g1 , g2 ), in the same coordinates (x, y), reads
y ′′ =
ξ xξ−1
hξ y ξ−1
ξ xξ−1
hξ y ξ−1
′
′2
+
y
+
y
+
y ′3 .
2ε (hy ξ − xξ ) 2 (hy ξ − xξ )
2ε (hy ξ − xξ )
2 (hy ξ − xξ )
It is different for different choices of (h, ε, ξ), i.e. these parameters characterize the projective class. Obviously,
if h = 0, the metric admits a Killing vector field. Thus, in the following, we require in particular that h 6= 0.
Any metric g belonging to the projective class P(g1 , g2 ) given by g1 and g2 from (17) reads, via Formula (5),
g = g[K1 , K2 ] =
xξ − hy ξ
xξ − hy ξ
2
dx
+
ε
dy 2 .
(K1 + K2 hy ξ )(K1 + K2 xξ )2
(K1 + K2 hy ξ )2 (K1 + K2 xξ )
(20)
In view of Step 2 of the general procedure outlined in Section 2.2, we from now on restrict to K1 , K2 6= 0,
since we are only interested in metrics with an essential projective vector field.
In this case, writing Equation (14) explicitly, taking into account (19), we obtain the requirement
xξ0 − hy0ξ
(K1 + K2 hy0ξ t)(K1 + K2 xξ0 t)2
=
x20 + ε
xξ0 − hy0ξ
(K1 + K2 hy0ξ t)2 (K1 + K2 xξ0 t)
′ξ
x′ξ
0 − hy0
2
(K1′ + K2′ hy0′ξ t)(K1′ + K2′ x′ξ
0 t)
x′2
0 +ε
y02
′ξ
x′ξ
0 − hy0
(K1′ + K2′ hy0′ξ t)2 (K1′ + K2′ x′ξ
0 t)
y0′2 . (21)
Let us study poles of the functions in t on the left and the right hand side of Equation (21). We find below
that this is enough for our purposes. Possible poles are the zeros of the denominators, provided xξ0 6= hy0ξ and
′ξ
x′ξ
0 6= hy0 , respectively. Determining these possible poles and equating them pairwisely, we find the following
cases:
(a)
y0′ξ
y0ξ
y0′ξ
xξ0
=
=
(b)
K1′ K2
K1 K2′
K1′ K2
K1 K2′ h
x′ξ
0
y0ξ
x′ξ
0
xξ0
=
=
(c)
y0′ξ
y0ξ
x′ξ
0
xξ0
K1′ K2 h
K1 K2′
K1′ K2
K1 K2′
=
K1′ K2
K1 K2′
=
K1′ K2
K1 K2′
(d)
x′ξ
0
y0ξ
′ξ
y0
xξ0
=
K1′ K2 h
K1 K2′
=
K1′ K2
K1 K2′ h
The first two cases (a) and (b) lead to xξ0 = hy0ξ , i.e. to zeros of the left hand side of (21), contrary to our
previous assumptions on x0 , y0 (for the metric under consideration such points lie on light lines). Since we want
to study only poles of (21), cases (a) and (b) can therefore be left out of further consideration.
′ 1/ξ
K K
In case (c), x′0 = Kx0 , y0′ = Ky0 where K = K11 K2′
. Isometries of the form (x, y) → (Kx, Ky), K > 0,
2
are embedded in the flow of the projective vector field. Substitute these expressions into (21) and obtain
λ
K λ−1
K1
= 1,
K1′
and thus
K1′
>0
K1
and
K2
> 0,
K2′
K′
which suffices to prove
|K1 |
|K1′ |
=
,
|K2 |λ
|K2′ |λ
K′
where the latter expression is identical to |KK21|λ = |K ′1|λ in view of K11 > 0, so the condition of Lemma 1 is
2
reobtained.
In case (d), supposing x0 , x′0 , y0 , y0′ > 0, we similarly find
1/ξ
1/ξ
′
′
K1 K2 h
K1 K2 h
′
′
y0 ,
y0 =
x0 .
(22)
x0 =
K1 K2′
K1 K2′
Such transformations are not embedded into the projective flow, cf. (19), since they include swapping x ↔ y.
One should therefore not expect the same results as in case (c).
Combining (22) and (21) we can deduce |h| = 1 as follows: First substitute (22) into (21), multiply by the
common denominator, and then collect terms w.r.t. t. The terms of degree zero and three of this polynomial in
t are enough to obtain |h| = 1. Finally, using again (21), we obtain
|K|
ξ+3
ξ
K13
= −ε sgn(K) ∈ {±1} ,
K1′3
9
where K =
K1′ K2
,
K1 K2′
(23)
implying
|K1 |
|K1′ |
,
=
λ
|K2 |
|K2′ |λ
so the constant
|K1 |
|K2 |λ
is an isometric invariant for the metrics g[K1 , K2 ]. Moreover, from (23) we can infer
sgn(K1 ) sgn(K1′ ) = −ε sgn(K)
and, using the definition of K,
sgn(K2′ ) = −ε sgn(K2 ).
|K1 |
Now consider a metric (20) with given k = |K
λ . After rescaling x and y by the same constant factor a
2|
with aξ = k |K2 |λ−1 , we have the metric in the form
ξ+3
k ξ
g = ε1 3
k
xξ − hy ξ
xξ − hy ξ
2
dx
+
ε
dy 2
(1 + ε2 hy ξ )(1 + ε2 xξ )2
(1 + ε2 hy ξ )2 (1 + ε2 xξ )
with ε, ε1 , ε2 ∈ {±1} where ε1 = sgn(K1 ) and ε2 = sgn(K2 ) ε1 . By renaming the multiplicative constant
ξ+3
κ=
|k| ξ
k3
, we arrive at the following metric
g = g[κ, ε2 ; h, ε] = κ
xξ − hy ξ
xξ − hy ξ
2
dx
+
ε
dy 2
(1 + ε2 hy ξ )(1 + ε2 xξ )2
(1 + ε2 hy ξ )2 (1 + ε2 xξ )
(24)
with κ ∈ R \ {0}, h ∈ R \ {0}, ε, ε2 ∈ {±1}. Note that we redefined ε2 → ε2 sgn(C1 ).
Notation. The notation g[κ, ε2 ; h, ε] in Equation (24) has the following meaning: the parameters after the
semicolon specify the projective class. Once the projective class is specified, the parameters before the semicolon
are those of the particular metric withing this class. This notation will be used several times in the remainder
of the paper.
Remark 4. In case h = ±1, we may apply another change of coordinates and swap the coordinates x ↔ y.
Thus, we achieve additional restrictions on the parameters. In case h = −1, we can achieve ε2 > 0. In case
h = 1, we need to distinguish between the case ε = 1 and the case ε = −1. Let (ε, h) = (1, 1). Then we can
achieve κ > 0. If, on the other hand, (ε, h) = (−1, 1), swapping x and y transforms the metric into itself.
The above reasoning ensures that, if g = g[κ, ε2 ; h, ε] and g ′ = g[κ′ , ε′2 ; h, ε] of the form (24) are isometric,
then |κ′ | = |κ|. Thus, it remains to prove that two metrics of the above type are isometric, (if and) only if
ε′2 = ε2 and sgn(κ′ ) = sgn(κ). Examine the poles of either side of (21). We have the four cases
(a)
ε′2 hy0′ξ
ε2 hy0ξ
ε′2 x′ξ
0
ε2 xξ0
=1
=1
(b)
(c)
ε′2 x′ξ
0
=1
ε2 y0ξ h
′ξ
ε′2 y0
h=1
ε2 xξ0
ε′2 hy0′ξ
ε2 hy0ξ
ε′2 y0′ξ
h
ε2 xξ0
(d)
=1
=1
ε′2 x′ξ
0
ε2 y0ξ h
ε′2 x′ξ
0
ε2 xξ0
=1
=1
In view of Remark 4, two different copies of (24) might be isometric. However, we now prove that Remark 4
provides all projective transformations that do not embed into the flow (19). This permits us to obtain the
following list of mutually non-isometric normal forms.
Proposition 5. A metric of the type A(I) in Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric below:
(i) For κ ∈ R \ {0}, ε, ε2 ∈ {±1} and h ∈ R \ {0, −1}, (h, ε) 6= (1, 1):
ε(xξ − hy ξ )
xξ − hy ξ
2
2
,
dx
+
dy
g=κ
(1 + ε2 hy ξ )(1 + ε2 xξ )2
(1 + ε2 hy ξ )2 (1 + ε2 xξ )
(ii) Special case h = −1 (we choose ε2 = 1, cf. Remark 4). For ε ∈ {±1} and κ ∈ R \ {0}, the normal forms
are:
ε(xξ + y ξ )
xξ + y ξ
2
2
,
dx
+
dy
g=κ
(1 − y ξ )(1 + xξ )2
(1 − y ξ )2 (1 + xξ )
10
(iii) Special case h = +1, ε = +1 (we choose κ > 0). For ε2 ∈ {±1}, the normal forms are:
(xξ − y ξ )
xξ − y ξ
2
2
,
dx
+
dy
g=κ
(1 + ε2 y ξ )(1 + ε2 xξ )2
(1 + ε2 y ξ )2 (1 + ε2 xξ )
The parameter ξ can assume values ξ ∈ (0, 1) or ξ ∈ (1, 4]. In case ξ = 2, we require h 6= −ε.
Proof. We need to show that two different copies of the above metrics are non-isometric. This can be achieved
by a thorough examination of all the cases in the table before Proposition 5. We need to confirm that in order
to have isometric metrics we need ε′2 = ε2 and κ′ = κ.
Analogously to Equation (21), we obtain the following requirement (we divide by k on both sides)
sgn(κ)
xξ0 − hy0ξ
(1 + ε2 hy0ξ sξ )(1 + ε2 xξ0 sξ )2
′
= sgn(κ )
x20 + ε
xξ0 − hy0ξ
(1 + ε2 hy0ξ sξ )2 (1 + ε2 xξ0 sξ )
′ξ
x′ξ
0 − hy0
(1 + ε′2 hy0′ξ sξ )(1 +
x′2
ξ )2 0
ε′2 x′ξ
s
0
+ε
y02
!
′ξ
x′ξ
0 − hy0
ξ
(1 + ε′2 hy0′ξ sξ )2 (1 + ε′2 x′ξ
0s )
y0′2
!
(25)
(1) Start with case (a). Recalling x0 > 0, y0 > 0 and x′0 > 0, y0′ > 0, we conclude from the second equation
′
that ε2/ε2 > 0 and thus ε2 = ε′2 . Thus, x0 = x′0 , again from the second equation. Now, infer hy0ξ = hy0′ξ and
thus y0 = y0′ from the first equation.
From (25) we then find κ = κ′ . To see this, replace unprimed objects by the relations we have just
ξ
derived. Comparing the left and the right hand side of (25), we have to equate xξ0 sgn(κκ′ ) = x′ξ
0 = x0 , proving
′
′
sgn(κ) = sgn(κ ) and therefore κ = κ.
′
(2) In case (b), we have ε2 xξ0 = ε′2 y0′ξ h and thus h ε2/ε2 > 0 from the second equation. This implies in
′
particular that ε2 = sgn(h)ε2 (and thus case (b) can only occur in case sgn(h) = 1).
Using the relations obtained so far, we can replace unprimed objects in (25). From the resulting expression
it follows that |h| = 1.
From the above we conclude h = 1 so that ε′2 = ε2 . Assuming ε = −1 and resubstituting this into (25)
confirms κ′ = κ. If, however, ε = 1, we see that case (b) cannot occur.
(3) Cases (c) and (d) can be discarded, as in these cases one can derive from the first and second equation
that xξ0 = hy0ξ . We may assume (x0 , y0 ) to be chosen anywhere except for light lines of the metric g. But,
xξ0 = hy0ξ would obviously put both sides of (25) to zero, contrary to the assumption.
3.1.2
Normal forms (B.4) of Theorem 1
In the complex Liouville case, B(I), we can construct any metric g in the projective class, via Formula (5), from
the following two metrics:
1
(Cz ξ − Cz ξ ) (dz 2 − dz 2 )
4
1
−1
−1
g2 = (C −1 z −ξ − C z −ξ ) C −1 z −ξ dz 2 − C z −ξ dz 2 ,
4
g1 =
λ−1
as in (18). Furthermore, we have
with C ∈ C, |C| = 1 and ξ = 2 2λ+1
2λ + 1
a
λ 0
a1
and
w=−
Lw 1 =
(z∂z + z∂z )
0 1
a2
a2
2
For convenience, we use real polar coordinates (ρ, θ) so that z = ρ cos(θ) + iρ sin(θ) . In these new coordinates,
the projective vector field w takes the concise form
w=
3
ρ∂ρ .
2(ξ − 1)
The flow of this projective vector field is
3
(ρ, θ) = (ρ0 e 2(ξ−1) t , θ0 ),
ρ0 > 0, θ0 ∈ [0, 2π).
For the projective connection (1) we find, with θ = θ(ρ),
θ′′ =
ξ
ξ ′
θ −
θ′2 ,
ρ
tan(ξθ + ϕ)
11
(26)
where ϕ is defined by C = eiϕ . Equation (26) shows that the projective connection is determined by the data
(ξ, tan ϕ), i.e. by (ξ, ϕ mod π). Let us now consider a general metric from the projective class P(g1 , g2 ),
g=
Cz ξ − C̄ z̄ ξ
Cz ξ − C̄ z̄ ξ
dz 2 −
dz̄ 2
ξ
ξ
2
(K1 + K2 C̄ z̄ )(K1 + K2 Cz )
(K1 + K2 C̄ z̄ ξ )2 (K1 + K2 Cz ξ )
(27)
|K1 |
From Lemma 1, we know that all metrics of the form (27) with the same value of |K
λ =: k are isometric if
2|
also the signs sgn(K1 ) and sgn(K2 ) are the same. Given k and εi = sgn(Ki ), i ∈ {1, 2}, we can therefore choose
a representative on each orbit, cf. Figure 1. For convenience let us choose the representative K1 = ±1, i.e. in
(27) we substitute K1 → sgn(K1 ) = ε1 . Thus, Metric (27) may in the following w.l.o.g. be assumed to be
g=
Cz ξ − C̄ z̄ ξ
Cz ξ − C̄ z̄ ξ
2
dz
−
dz̄ 2
(ε1 + K2 C̄ z̄ ξ )(ε1 + K2 Cz ξ )2
(ε1 + K2 C̄ z̄ ξ )2 (ε1 + K2 Cz ξ )
(28)
Now perform a change of coordinates z → ηz with η = |K2 |−1/ξ , so (28) is put into the form
eiϕ z ξ − e−iϕ z̄ ξ
eiϕ z ξ − e−iϕ z̄ ξ
2
2
.
dz
−
dz̄
g=K
(ε1 + ε2 e−iϕ z̄ ξ )(ε1 + ε2 eiϕ z ξ )2
(ε1 + ε2 e−iϕ z̄ ξ )2 (ε1 + ε2 eiϕ z ξ )
(29)
Here, we introduced K = |K2 |−1/ξ . Rewriting (29), we find
eiϕ z ξ − e−iϕ z̄ ξ
eiϕ z ξ − e−iϕ z̄ ξ
2
2
,
dz
−
dz̄
g = ε1 K
(1 + εe−iϕ z̄ ξ )(1 + εeiϕ z ξ )2
(1 + εe−iϕ z̄ ξ )2 (1 + εeiϕ z ξ )
where we introduced ε = ε1 ε2 .
Observation 1. We observe that under conjugation of the coordinates, z ↔ z, the metric g is transformed into
a metric of the same form, with the parameters being transformed according to (κ, ϕ) → (κ, −ϕ).
The parameter ϕ specifies the projective class (together with ξ), and w.l.o.g. we may assume ϕ ∈ (0, π] in
view of (26). Now suppose ε = −1 = e±iπ and introduce ϕ̃ such that
εeiϕ = eiϕ̃ .
where ϕ̃ ∈ (−π, 0]. By a conjugation z ↔ z we can subsequently achieve again ϕ̃ ∈ (0, π].
In view of Observation 1 we arrive at the normal forms
hz ξ − h̄z̄ ξ
hz ξ − h̄z̄ ξ
2
2
g = g[κ; h, ξ] = κ
dz −
dz̄
.
(1 + h̄z̄ ξ )(1 + hz ξ )2
(1 + h̄z̄ ξ )2 (1 + hz ξ )
(30)
where κ = ε2 K ∈ R \ {0} and h ∈ P1 .
Lemma 2. Two different copies g = g[κ; h, ξ] and g ′ = g[κ′ ; h, ξ] of (30) are non-isometric.
Proof. Since the metrics g and g ′ are inside the same projective class, and h ∈ P1 , the only parameters that can
distinguish non-isometric metrics are κ, κ′ . Consider Equation (14) for metrics of the form (30) with s = e(1−λ)t ,
κ
hz0ξ s − h̄z̄0ξ s
(1 + h̄z̄0ξ s)(1 +
z02
hz0ξ s)2
s−
hz ξ − h̄z̄0ξ s
(1 + h̄z̄0ξ s)2 (1 + hz0ξ s)
=κ
′
z̄02
s
!
hz0ξ s − h̄z̄0ξ s
(1 + h̄z̄0ξ s)(1 +
z02 s
ξ
2
hz0 s)
−
hz ξ − h̄z̄0ξ s
(1 + h̄z̄0ξ s)2 (1 + hz0ξ s)
z̄02 s
!
. (31)
Comparing the left and right hand side of (31), we infer κ′ = κ, which concludes the proof.
The reasoning made above assures that if the metrics g and g ′ are isometric then κ′ = κ (h and ξ define the
projective class).
Proposition 6. A metric of the type B(I) of Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric below:
hz ξ − hz ξ
hz ξ − hz ξ
2
2
dz +
dz ,
g=κ
(1 + hz ξ )(1 + hz ξ )2
(1 + hz ξ )2 (1 + hz ξ )
where κ ∈ R \ {0} and h ∈ P1 , ξ ∈ (0, 1) ∪ (1, 4]. If ξ = 2, h 6= ±1.
12
3.1.3
Normal forms (C.7) of Theorem 1
Consider metrics of case C(Ib) in Proposition 3. Such metrics are of Jordan block type and can be constructed,
via Formula (5), from the following two metrics:
g1 = 2(Y (y) + x)dxdy
(32a)
2
g2 = −2
Y (y) + x
(Y (y) + x)
dxdy +
dy 2
3
y
y4
(32b)
1
where y > 0 and Y (y) = Cy ξ with C ∈ R, ξ as in (18). The projective connection (1) is found to be, with
y = y(x),
3
y′
y 2(λ−1)
2λ + 1
′′
y = 2λ+1
−
y ′2 ,
2λ+1
y 2(λ−1) + x 2(λ − 1) y 2(λ−1) + x
i.e. the projective class is determined by λ such that |λ| ≥ 1, λ 6= 1.
Remark 5. If C = 0, the metric has constant curvature, then we can ignore this case. For C 6= 0, we can
w.l.o.g. restrict to C = 1, by grace of a coordinate transformation x → Cx and subsequent rescaling of the
metrics (32). In the following, however, we therefore continue only with C = 1.
λ−1
. The normal forms (C.7) of
We recall the definition (18) of ξ in terms of the eigenvalue λ: ξ = 2 2λ+1
Theorem 1 have degree of mobility 2 (see Proposition 3), i.e. dim(P(g1 , g2 )) = 2 according to Definition 4. This
requires
5
1
λ 6= , i.e. ξ 6= ,
2
2
for the present section, since for ξ = 21 the degree of mobility is 3 (this case is discussed in Section 3.1.4). The
projective vector field, fixed by (13) and case (I) of (7), reads, in the coordinates of (32),
1
w =− λ+
x∂x − (λ − 1)y∂y .
(33)
2
From (33) we infer the projective flow
1
φt (x0 , y0 ) = (x0 e−(λ+ 2 )t , y0 e−(λ−1)t )
(34)
A metric from the projective class P(g1 , g2 ) is of the form
2
1
yξ + x
1
ξ
g[K1 , K2 ] = 2
y +x
dxdy + K2
dy 2
(K1 − K2 y)3
(K1 − K2 y)4
(35)
Using this expression, we can write the square of the length of the projective vector field as
2
1
ξ + x
y
y +x
1
kwk2g (x, y) = 2
(λ − 1)xy + K2
λ+
(λ − 1)2 y 2 .
(K1 − K2 y)3
2
(K1 − K2 y)4
1
ξ
(36)
Restricting (36) to the flow of w, we obtain
1
(λ−1)
1
ξ −
t
ξ
1
+ x0 e−(λ+ 2 )t
1
y
0e
2
(λ − 1)x0 y0 e(−2λ+ 2 )t
λ+
kwkg (x0 , y0 ; t) = 2
2
(K1 − K2 y0 e−(λ−1)t )3
2
1
λ−1
1
y0ξ e− ξ t + x0 e−(λ+ 2 )t
(λ − 1)2 y02 e−2(λ−1)t
+ K2
(K1 − K2 y0 e−(λ−1)t )4
= e−3t 2
1
ξ
y0 + x0
(e(λ−1)t K1 − K2 y0 )3
1
2
ξ
y0 + x0
1
2 2
(λ − 1)x0 y0 + K2 (λ−1)t
λ+
(λ
−
1)
y
0
2
(e
K1 − K2 y0 )4
For the second identity, we have made use of Equation (18). Let g = g[K1 , K2 ] and g ′ = g[K1′ , K2′ ] be two
isometric copies of (35). Let us consider Equation (14) and introduce the new parameter s := e(λ−1)t , which
13
yields
1
2
y0ξ + x0
1
y0 + x0
x0 y0 + K2
λ+
(λ − 1)y02
2
(K1 s − K2 y0 )3
2
(K1 s − K2 y0 )4
1
ξ
1
y0′ ξ + x′0
=2
(K1′ s − K2′ y0′ )3
1
2
y0′ ξ + x′0
1
2
x′0 y0′ + K2′
λ+
(λ − 1)y0′
2
(K1′ s − K2′ y0′ )4
(37)
where ′ has the obvious meaning. If K1 = 0 then K1′ = 0, and analogously K2′ = 0 if K2 = 0. Therefore,
for metrics with an essential projective vector field, we have K1 6= 0, K2 6= 0, so that also K1′ 6= 0, K2′ 6= 0.
K2
y0 , and therefore also the right hand side has
Consider Equation (37). The left hand side has the pole s = K
1
K2
′ ′
2
to have a pole at s = K1 y0 . From this, we infer the condition K1′ K
K1 y0 − K2 y0 = 0, which implies
H :=
y0′
K ′ K2
= 1 ′ > 0.
y0
K1 K2
(38)
The constant H is greater than zero since, in view of Lemma 1, K1 and K1′ have the same sign, as well as K2
and K2′ . The same result follows from (34), from which we can deduce that the signs of y0 and y0′ (x0 and x′0 )
are the same. If we substitute y0′ = Hy0 into Equation (37), we obtain
1
y0ξ + x0
2
(K1 s − K2 y0 )3
1
(y0ξ + x0 )2
1
x0 y0 + K2
λ+
(λ − 1)y02
2
(K1 s − K2 y0 )4
1
1
1
1
K22 K12 (H ξ y0ξ + x′0 )2
1
K2 K12 H ξ y0ξ + x′0
′
x0 y0 + ′ ′ 2
λ+
(λ − 1)y02 .
= 2 ′ ′2
2
K2 K1 (K1 s − K2 y0 )3
K2 K1 (K1 s − K2 y0 )4
Multiplying the previous equation by (K1 s − K2 y0 )4 , we obtain a polynomial of first degree in s. If we put
equal to zero the coefficients of first degree and zero degree we obtain the following system

1
1
1
ξ
ξ
′
′

e
ξ

 (y0 + x0 )x0 = H(H y0 + x0 )x0
2
e := K′2 K′12
(39)
H
K2 K1

1

1
1

ξ
ξ
2
′
2
e
ξy
(y0 + x0 ) = H(H
0 + x0 )
that implies (by considering the square of the first equation and then considering the quotient of the two
equations)
e ′0 2 , H
e >0
x20 = Hx
(40)
e > 0 implies
e > 0 also follows from the second equation of system (39)). H
(note that H
K2
> 0 which in view of (38) implies
K2′
K1
>0
K1′
(41)
Substituting (40) into the first equation of (39), we obtain
e 1ξ x′ .
x0 = HH
0
(42)
2
e ξ . Now, substitute the explicit
By considering its square and taking into account (40), we find 1 = HH
e
expressions of H and H (see (38) and (39)) into this equation and obtain
1=
so
K2 K12
K2′ K1′ 2
K1′ K2
K1 K2′
ξ2
(41)
=
K2
K2′
ξ2 +1 K1′
K1
ξ2 −2
K2 λ K1′ ,
1= ′ K2
K1 (18)
=
K2
K2′
λ( ξ2 −2) K1′
K1
ξ2 −2
,
which together with (41) implies the second case of Lemma 1. So far we have essentially proven that different
|K1 |
λ
values of K := |K
into (35)
λ give non-isometric metrics of the type (35). Now, we substitute K1 = ε1 K|K2 |
2|
14
and obtain
1
ξ
g=2
2
1
yξ + x
y +x
dxdy + K2
dy 2
(ε1 K|K2 |λ − K2 y)3
(ε1 K|K2 |λ − K2 y)4
1
2
1
yξ + x
yξ + x
dxdy + 3
dy 2 . (43)
=2
|K2 |3 (ε1 K|K2 |λ−1 − sgn(K2 )y)3
K2 (ε1 K|K2 |λ−1 − sgn(K2 )y)4
By performing the change of coordinates
x → K|K2 |λ−1
1ξ
x,
y → K|K2 |λ−1 y ,
and taking into account (18), metrics (43) become


1
2
1
ξ + x
y
ξ
y +x


dxdy + ε2
dy 2  ,
g = k 2
(ε1 − ε2 y)3
(ε1 − ε2 y)4
εi ∈ {±1} ,
k>0
(44)
2
with k := K ξ −2 , ε1 = sgn(K1 ), ε2 = sgn(K2 ). Let us rename k → k/ε32 and define ε = ε1 ε32 , then we obtain


2
1
1
ξ + x
y

 yξ + x
g = k 2
dxdy +
dy 2  , ε ∈ {±1} , k ∈ R \ {0}
3
4
(ε − y)
(ε − y)
(note that k can now also be negative). Finally, we perform the coordinate change y → ǫ − y in order to put
the metric into the form,


2
1
1
ξ + x
(ε
−
y)
ξ

 (ε − y) + x
(45)
dxdy +
dy 2  , ε ∈ {±1} , k ∈ R \ {0}
g = g[k, ε; ξ] = k −2
y3
y4
Lemma 3. Two different copies g = g[k, ε; ξ] and g̃ = g[k̃, ε̃; ξ] of (45) are non-isometric.
Proof. We need to prove k̃ = k and ε̃ = ε. By construction, we already know that |k̃| = |k|, so it suffices
to prove that the sign of k and the sign ε remain unchanged under isometries. From (38), where we are only
interested in the signs, we have ε̃ ε = +1, and thus ε̃ = ε. Moreover, from (42), we similarly conclude ε̃ε22 = +1
(using the definition of ε2 from (44)), proving k̃ = k since sgn k = sgn(K 2/ξ−2 ε2 ) = ε2 .
Altogether, we have found:
Proposition 7. A metric of the type C(Ib) in Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric below:


2
1
1
ξ + x
(ε
−
y)
ξ
 (ε − y) + x

g = k −2
dxdy +
dy 2  ,
y3
y4
where ε ∈ {±1}, k ∈ R \ {0} and ξ ∈ (0, 21 ) ∪ ( 21 , 1) ∪ (1, 4].
3.1.4
Normal forms (C.10) of Theorem 1; degree of mobility 3
Amongst the projective classes of Proposition 3, almost all cases have a 2-dimensional space of solutions of
the linearized metrization equations (4), cf. Section 2.1. However, there is one case of dimension 3. This case
corresponds to a superintegrable system3 , see the Remark 6 below. In the superintegrable case, i.e. C(Ia) of
Proposition 3, we can construct any metric g in the projective class, via Formula (5), from the following three
metrics:
g1 = (x + y 2 ) dxdy
(x + y 2 )2 2
x + y2
dxdy
+
dy
y3
y4
y2 + x
g3 =
9 (y 2 + x) dx2 − 4y (9x + y 2 ) dxdy + 12x (y 2 + x) dy 2 ,
2
6
(3x − y )
g2 = −2
3 Actually,
there are several superintegrable systems but the solution spaces A are isomorphic.
15
and the projective vector field, fixed by (13) and case (I) of (7), has the form
w = 2x ∂x + y ∂y .
(47)
Remark 6 (Projective equivalence and superintegrable metrics). Given a metric g ∈ P(g1 , g2 , g3 ), the existence
of a metric ǧ projectively equivalent to g implies the existence of quadratic Killing tensors ι(g) for the geodesic
motion (i.e. integrals of motion that are homogeneous quadratic polynomials in the fiber coordinates of T ∗M ).
The correspondence is given by the mapping [6, 13]:
det(g) 2/3
ǧ.
ι(g) = det(ǧ) For the purposes of the present paper, a superintegrable system is a triple (g, L1 , L2 ) on a 2-dimensional
manifold with (pseudo-)Riemannian metric g and two quadratic Killing tensors L1 and L2 such that (H, I1 , I2 )
are functionally independent integrals of motion, where we define:
X
X ij
H=
g ij pi pj ,
and
Ik =
L k pi pj
for k ∈ {1, 2}.
ij
ij
Note that this definition of a superintegrable system is different from the definition used in [6]. In [6], a
(Darboux-)superintegrable system is defined by the requirement dim A = 4, i.e. the existence of four linearly
independent integrals of motion. The superintegrable metrics discussed in the present paper, in contrast, have
dim A = 3.
Before we proceed further, let us make the following observation:
Remark 7. For the metrics g1 , g2 and g3 , the projective vector field is homothetic. This follows from Proposition 4 and is easily verified by computing the Lie derivatives w.r.t. (47):
Lw g1 = 5g1 ,
Lw g2 = 2g2 ,
Lw g3 = −4g3 .
Any isometry is a projective transformation, and moreover any isometry preserves eigenspaces of Lw , as
eigenspaces are geometric objects. Therefore, for an isometry τ we have the identities
τ ∗ g 1 = µ1 g 1 ,
τ ∗ g 2 = µ2 g 2 ,
τ ∗ g 3 = µ3 g 3 ,
where µi ∈ R \ {0} , i ∈ {1, 2, 3} .
On the other hand, (46) suffices, via Formula (5), to describe any metric of the projective class P(g1 , g2 , g3 ),
using the same system of coordinates (x, y). In order to identify which metrics in P(g1 , g2 , g3 ) could be isometric,
we therefore need to understand those coordinate transformations that preserve the form of the metric.
Lemma 4. Any isometry of a metric in a projective class of the type C(Ia) that preserves the form of a metric
g ∈ P(g1 , g2 , g3 ) in the coordinates of (46), is given by
(xnew , ynew ) = (k 2 xold , kyold )
where k ∈ R \ {0}.
Proof. Consider the metrics g1 and g2 as in (46), and a generic coordinate transformation (x, y) = (x(u, v), y(u, v)).
For g1 obtain the equation
!
g1 = (x + y 2 ) (xu yu du2 + xv yv dv 2 + (xu yv + xv yu ) dudv) = µ1 (u + v 2 ) dudv
(48)
(by ! we mark the condition that eigenspaces are preserved). Thus, around almost every point, we have one of
the following possibilities:
(i) x = x(u),
y = y(v)
or
(ii) x = x(v),
y = y(u) ,
so that (48) gives
µ1 (u + v 2 ) = (x + y 2 ) (xu yv + xv yu ).
(49)
Now, do an analogous computation for g2 . Case (ii) turns out to be incompatible since it would replace the
dy 2 -term by a term in du2 instead of dv 2 . For alternative (i), we obtain
x + y2
(x + y 2 )2 2 2
x
y
dudv
+
yv dv
u
v
y3
y4
u + v2
u + v2
(u + v 2 )2 2
µ2 (u + v 2 )2 2 !
(49)
= −2µ1
dudv + 21
dv = −2µ2
dudv + µ2
dv .
3
4
3
y
xu
y
v
v4
g2 = −2
16
(50)
The coefficients in (50) of, respectively, dudv and dv 2 , yield the equations
µ2
µ1
= 3
y3
v
and
µ21
4
y x2u
µ2
,
v4
=
from which we can deduce
y=
r
3
µ1
v =: µ v
µ2
x2u =
and
µ21
µ2
r
3
4
µ2
2/3 √
= µ1 3 µ2 =: η ,
µ1
(51)
where the root takes account of the sign and where µ, η ∈ R\{0} are constants. In view of (50), we can integrate
the second equation of (51) and obtain x = η u. Note that µ and η are not independent, but, since their definition
is rather complicated, we shall continue with the investigation of the transformation (x, y) = (ηu, µv) without
prior study of the relation of µ and η. Apply this coordinate transformation to g3 of (46). The result again has
to be in the same eigenspace. Thus,
g3 =
µ2 v 2 + ηu
9 (µ2 v 2 + ηu) η 2 du2 − 4µ y (9ηu + µ2 v 2 ) ηµ dudv + 12ηu (µ2 v 2 + ηu) µ2 dv 2
(3ηu − µ2 v 2 )6
v2 + u
!
= µ3
9 (v 2 + u) du2 − 4y (9u + v 2 ) dudv + 12u (v 2 + u) dv 2 .
(3u − v 2 )6
The coefficients w.r.t. du2 and dv 2 of this equation give two equations from which we may infer
η2
2 2
2
(v 2 + u)2
(µ2 v 2 + ηu)2
2 (µ v + ηu)
=
µ
=
ηµ
.
3
(3ηu − µ2 v 2 )6
(3u − v 2 )6
(3ηu − µ2 v 2 )6
Therefore, we have verified η = µ2 and thus, in order to preserve the form of the metric, it is necessary that the
isometry takes the form (x, y) = (µ2 u, µv).
Observation 2. Consider an arbitrary metric g ∈ P(g1 , g2 , g3 ). By Formula (5), g is of the form
g=h
P3
i=1
det
Ki det(ggii )2/3
P3
gi
i=1 Ki det(gi )2/3
i2 .
Note that if we apply a change of coordinates (x, y) → (k 2 x, ky) then
P3
gi
−αi/3
i=1 Ki k
det(gi )2/3
g=h
i2 ,
P3
gi
αi/3
−
det
i=1 Ki k
det(gi )2/3
with (α1 , α2 , α3 ) = (5, 2, −12). Thus, by a suitable choice of k, we can normalize one of the coefficients K1 , K2
or K3 (up to its sign).
In view of Observation 2, we need to distinguish cases when one of the parameters Ki , i ∈ {1, 2, 3}, is zero,
since in such cases the respective parameters cannot be normalized. On the other hand, if one of the Ki is zero,
then the other two parameters definitely need to be non-zero since otherwise the metric is proportional to of
the metrics (46) and thus admits a homothety.
We therefore proceed in three steps: first, if K3 = 0, we normalize K2 (or K1 ). Second, if K2 = 0, we
normalize K1 . Finally, if both K2 and K3 are non-zero, we normalize K2 .
Vanishing coefficient K3 = 0. In this case we have to assume K1 , K2 6= 0 because otherwise the projective
vector field would not be essential. In view of Observation 2 we can put the metric into the form
(y 2 + x)2 2
y2 + x
,
(52)
dxdy
+
dy
g[ε, c] = ε −2
(y − c)3
(y − c)4
where c ∈ R \ {0} and ε ∈ {±1}. Alternatively, we could normalize the other coefficient and obtain
(y 2 + x)2 2
y2 + x
,
dxdy
+
dy
g[κ, ε] = κ −2
(y − ε)3
(y − ε)4
where κ ∈ R \ {0} and ε ∈ {±1}.
17
(53)
Lemma 5. Two different copies g[ε, c] and g[ε′ , c′ ] of (52) are non-isometric, as are two different copies
g = g[κ, ε] and g ′ = g[κ′ , ε′ ] of (53).
Proof. We prefer to use Metric (53) for the proof. Recall that the flow of the projective vector field (47) is
φs (x0 , y0 ) = (s2 x0 , sy0 ) with s = et > 0. Considering Equation (14), i.e.
!
2
2
y0′ + x′0 5 ′ ′
s6 (y02 + x0 )2 2
s6 (y0′ + x′0 )2 ′ 2
y02 + x0 5
′
,
(54)
s x0 y0 +
y = κ −2
s x0 y0 +
y
κ −2
(s y0 − ε)3
(s y0 − ε)4 0
(s y0′ − ε′ )3
(s y0′ − ε′ )4 0
and comparing poles of either side, we find
y0
y′
= 0′
ε
ε
Substituting this back into (54), an equation polynomial in the parameter s is obtained,
2 2
(κ − κ′ )y04 + (κ′ x′0 − κx20 ) y02 s6 + 2ε y0 (κx0 − κ′ x′0 )y02 + (κx20 − κ′ x′0 ) s5 = 0
(55)
Consider the system of polynomial equations obtained from the coefficients of (55) w.r.t. s. For generic values
of y0 , x0 , x′0 , it implies κ′ = κ and x′0 = x0 . Resubstituting these relations into (54), we have
−2
y02 + x0
s (y02 + x0 )2 2
ε′ ε y02 + x0
s (ε′ ε y02 + x′0 )2 ′ 2
x0 y0 +
y0 = −2
ε′ ε x0 y0 +
ε ε y0 ,
3
4
′
′
3
(s y0 − ε)
(s y0 − ε)
(s ε ε y0 − ε )
(s ε′ ε y0 − ε′ )4
from which we finally infer ε′ = ε.
Vanishing coefficient K2 = 0. To exclude metrics with homotheties we have to assume K1 , K3 6= 0. Via
Equation (5), we obtain
K1
K −3
2
2
2
3
(y 2 + x) dxdy + 12x)(y 2 + x)2 dy 2 ,
g= 3
2 9(y + x) dx − 2 y + 9xy − 2
K3
K1
; x, y
F K
3
with
F (c; x, y) = y 6 − 9xy 4 + 27x2 y 2 − 27x3 + 4c2 − (36xy + 4y 3 ) c .
In view of Observation 2, we may perform a change of coordinates (x, y) → (k 2 x, ky) so that the metric becomes
g[κ, ε] =
κ
9(y 2 + x)2 dx2 − 2(y 3 + 9xy − 2ε)(y 2 + x) dxdy + 12x(y 2 + x)2 dy 2 ,
2
F (ε; x, y)
(56)
where ε ∈ {±1} and κ ∈ R \ {0}.
Lemma 6. Two different copies g = g[κ, ε] and g ′ = g[κ′ , ε′ ] of metrics (56) are non-isometric.
Proof. If two metrics are isometric, they must, by virtue of Lemma 4, be related by a change of coordinates of
the form
(xnew , ynew ) = (k 2 xold , kyold )
with k ∈ R \ {0}. Plugging this relation into one of the metrics, we obtain an expression polynomial in
dx, dy, x, y. Its coefficients form a system of algebraic equations on κ, ε and κ′ , ε′ as well as on k (the primed
objects refer to the parameters of another copy of the metric (56)). Solving this system of equations, we find
κ′ = κ ,
ε′ = ε ,
k=1
as the only real solution with k 6= 0. This concludes the proof.
Both coefficients K2 , K3 6= 0. The constant K1 may or may not be zero. As before, let us first factor out K3
in (5), which gives a conformal factor K3−3 . Then, recalling Observation 2, we perform a change of coordinates
K2 3/14
with (x, y) → (k 2 x, ky) with k = | K
|
6= 0. Thus we arrive at the metric
3
g=
κ
9(y 2 + x)2 dx2 − 2(y 3 + 2ε y + 9xy − 2c)(y 2 + x) dxdy + 4(3x + ε)(y 2 + x)2 dy 2 ,
2
F (ε, c; x, y)
(57)
where ε = sgn(K2 ) ∈ {±1}, κ ∈ R \ {0}, c ∈ R, and where the function F is defined by
F (ζ, c; x, y) = y 6 − 9xy 4 + 27x2 y 2 − 27x3 + 4c2 − (36xy + 4y 3 ) c + (18xy 2 − 5y 4 − 9x2 − 8cy) ζ + 4y 2 ζ 2
18
Lemma 7. Two different copies g = g[κ, c, ε] and g ′ = g[κ′ , c′ , ε′ ] of (57) are non-isometric.
Proof. If two metrics are isometric, they must, by virtue of Lemma 4, be related by a change of coordinates of
the form
(xnew , ynew ) = (k 2 xold , kyold )
with k ∈ R \ {0}. Plugging this relation into one of the metrics, we obtain an expression polynomial in
dx, dy, x, y. Its coefficients form a system of algebraic equations on κ, c, ε and κ′ , c′ , ε′ as well as k (the primed
objects refer to the parameters of another copy of the metric (57)). Solving this system of equations, we find
κ′ = κ ,
c′ = c ,
ε′ = ε ,
k=1
as the only real solution with k 6= 0. This concludes the proof.
Finally, we have to make sure that two metrics of different form (53), (56) or (57) cannot be mutually isometric.
Lemma 8. Metrics of the form (53), (56) or (57) are pairwise non-isometric.
P
P
Proof. Let a = Ki ai and ǎ = Ki′ ai be the Liouville tensors of a pair of metrics g, ǧ, as in the statement of
Lemma 8. In view of Lemma 4 and Observation 2, isometries between g and ǧ rescale K1 , K2 , K3 ,
Ki′ = Ki k −
αi/3
,
i = 1, 2, 3 ,
with k ∈ R \ {0}, (α1 , α2 , α3 ) = (5, 2, −12). Therefore, if K2 = 0, then K2′ = 0, and if K3 = 0, then K3′ = 0. If
K2 , K3 6= 0, then K2′ , K3′ 6= 0. So, g and ǧ cannot be isometric.
Summarizing the statements of Lemmas 5, 6, 7 and 8, we have the following:
Proposition 8. A metric of the type C(Ia) in Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric below:
(y 2 +x)2
y 2 +x
2
,
(i) g = κ −2 (y−ε)
3 dxdy + (y−ε)4 dy
(ii) g =
κ
F (0,ε;x,y)2
(iii) g =
κ
F (ε,c;x,y)2
9(y 2 + x)2 dx2 − 2(y 3 + 9xy − 2ε)(y 2 + x) dxdy + 12x)(y 2 + x)2 dy 2 ,
9(y 2 + x)2 dx2 − 2(y 3 + 2ε y + 9xy − 2c)(y 2 + x) dxdy + 4(ε + 3x)(y 2 + x)2 dy 2 .
Here, κ ∈ R \ {0}, c ∈ R, ε ∈ {±1}, and
F (ζ, c; x, y) = y 6 − 9xy 4 + 27x2 y 2 − 27x3 + 4c2 − (36xy + 4y 3 ) c + (18xy 2 − 5y 4 − 9x2 − 8cy) ζ + 4y 2 ζ 2 .
3.2
Case (II): Lw has 1 real eigenvalue
Here we are in the case when Lw has the form (II) of (7). Let (a1 , a2 ) be a suitable basis of A, in which Lw
assumes such form. We have that
t
e tet
exp(tLw ) =
.
(58)
0 et
Let g1 = ψ(a1 ), g2 = ψ(a2 ) according to Equation (3). Furthermore let g = ψ(K1 a1 + K2 a2 ) be an arbitrarily
chosen metric in the projective class P(g1 , g2 ). By virtue of Equation (58), we have
φ∗t (K1 a1 + K2 a2 ) = K1 et a1 + K1 tet a2 + K2 et a2 =: K1′ a1 + K2′ a2 .
(59)
Lemma 9. The pairs (K1 , K2 ) and (K1′ , K2′ ) are related by (59) if and only if they satisfy one of the following
cases
′
K2
K2
K2′
1 K
K1′
1 K
′
1 =
(a) K1′ = K1 = 0 and
>0
(b)
> 0 and
e
e 1
′
K2
K1
K1
K1
Proof. We have K1′ = K1 et and K2′ = K1 tet + K2 et . Eliminating t we therefore have either K1′ = K1 = 0 and
K′
K′
then K22 > 0, or K11 > 0 and
K′
K′
K′
K2′ = K1 1 ln 1 + K2 1 .
K1 K1
K1
The converse direction is straightforward.
Figure 2 illustrates the orbits of a metric in the projective class P(g1 , g2 ) under the flow of φt .
19
K2
K2
K1
K1
Figure 2: Graph of the orbits under projective transformations for a metric g = ψ(K1 a1 + K2 a2 ) of type A(II)
of Proposition 3. The plot on the left depicts several generic orbits, the thick curve being one example. The
2
graph on the right illustrates the special case h = C
C1 = 1, when there are additional projective symmetries that
do not embed into the flow (solid line) of the projective vector field (entire isometric orbit: dashed and solid
line). This symmetry can be thought of as a reflection at the origin (dashed line).
3.2.1
Normal forms (A.2) of Theorem 1
In the case of Liouville metrics A(II), we can construct any metric g in the projective class, via Formula (5),
from the following two metrics, where h ∈ R \ {0}:
h 1
1
1
1 1
e−3x dx2 +
e−3y dy 2
−
−
g1 =
x x y
y x y
g2 = (y − x)e−3x dx2 + (y − x)he−3y dy 2 .
The metric g has the form
g=
(y − x)e−3x
h(y − x)e−3y
dy 2
dx2 +
2
(K1 x + K2 ) (K1 y + K2 )
(K1 x + K2 )(K1 y + K2 )2
(61)
Since we are only interested in metrics with an essential projective vector field, we assume K1 6= 0 (for K1 = 0
the metric lies in an eigenspace of Lw ). The projective vector field, using the same coordinates (x, y), reads
w = ∂x + ∂y . Therefore, the projective flow is given by
φt (x0 , y0 ) = (x0 + t, y0 + t).
The projective connection (1) is
y ′′ =
1 + 3(x − y) ′ 1 − 3(y − x) ′2 h e3(x−y) ′3
e3(y−x)
+
y +
y +
y ,
2h (y − x)
2 (y − x)
2 (y − x)
2 (y − x)
meaning that h ∈ R \ {0} determines the projective class. We find the following equation from (14).
h(y0 − x0 )e−3y0
(y0 − x0 )e−3x0
+
2
(K1 (t + x0 ) + K2 ) (K1 (t + y0 ) + K2 ) (K1 (t + x0 ) + K2 )(K1 (t + y0 ) + K2 )2
′
′
h(y0′ − x′0 )e−3y0
(y0′ − x′0 )e−3x0
+
=
′
′
′
′
′
′
′
′
2
(K1 (t + x0 ) + K2 ) (K1 (t + y0 ) + K2 ) (K1 (t + x0 ) + K2′ )(K1′ (t + y0′ ) + K2′ )2
Looking at the poles of this expression (respectively, those of the left and right side of this equation, which need
to be the same) we obtain the following cases:
(a)
x′0
x′0
− x0 +
− y0 +
K2′
K1′
K2′
K1′
(b)
−
−
K2
K1
K2
K1
=0
=0
y0′
y0′
− x0 +
− y0 +
K2′
K1′
K2′
K1′
(c)
−
−
K2
K1
K2
K1
x′0
y0′
=0
=0
− x0 +
− y0 +
K2′
K1′
K2′
K1′
(d)
−
−
K2
K1
K2
K1
=0
=0
y0′ − x0
x′0 − y0
+
+
K2′
K1′
K2′
K1′
−
−
K2
K1
K2
K1
=0
=0
The first two cases (a) and (b) imply x0 = y0 and are thus irrelevant.
For the case (c) we find y0 − x0 = y0′ − x′0 and thus the above equation reduces to
h e−3y0
e−3x0
+
2
(K1 (t + x0 ) + K2 ) (K1 (t + y0 ) + K2 ) (K1 (t + x0 ) + K2 )(K1 (t + y0 ) + K2 )2
′
=
′
e−3x0
h e−3y0
. (62)
+
′
′
′
′
′
′
′
′
2
(K1 (t + x0 ) + K2 ) (K1 (t + y0 ) + K2 ) (K1 (t + x0 ) + K2′ )(K1′ (t + y0′ ) + K2′ )2
20
By denoting K =
simplifies to
K2
K1
−
K2′
K1′ ,
we have that x′0 = x0 + K, y0′ = y0 + K, which is in the flow of φt . Thus, (62)
K1′3 −3x0
K ′3
(K1 (t + y0 ) + K2 ) + h 13 e−3y0 (K1 (t + x0 ) + K2 )
e
3
K1
K1
= e−3x0 e−3K (K1 (t + y0 ) + K2 ) + h e−3y0 e−3K (K1 (t + x0 ) + K2 ),
which is a polynomial in t. By considering its first-degree component, we obtain
K′
K′
e−K = 1
thus 1 > 0 ,
K1
K1
which, in view of the definition of K, is the relation from Lemma 9.
Finally, in case (d) we have
x′0 = y0 + K,
y0′ = x0 + K,
with
K=
K2
K′
− 2′
K1
K1
(note that this transformation is not inside the flow of φt ), which implies y0 − x0 = x′0 − y0′ . Analogously to case
(c) we find
K1′
K′
⇒
< 0.
e−K = − 1
K1
K1
Thus, in view of Lemma 9, the pairs (K1 , K2 ) and (K1′ , K2′ ) are not related by (59). However, this case is
structurally similar to case (c). Consider the exchange of coordinates x ↔ y, so Metric (61) becomes
g=
(x − y) e−3y
h (x − y) e−3x
2
dx
+
dy 2
(K1 y + K2 )(K1 x + K2 )2
(K1 y + K2 )2 (K1 x + K2 )
Note that if we set (K1′ , K2′ ) = (−K1 , −K2 ), we obtain
g=
h (y − x) e−3x
(y − x) e−3y
2
dx
+
dy 2
(K1′ y + K2′ )(K1′ x + K2′ )2
(K1′ y + K2′ )2 (K1′ x + K2′ )
and this is identical to (61) if h = 1.
K2
Let us now look for the normal forms for metrics (61). Our discussions imply that different values of K =
give non-isometric metrics. By substituting K2 = K1 ln(KK1 ) into (61), we hence obtain
g=
K13 (x
h (y − x)e−3y
(y − x)e−3x
dy 2
dx2 + 3
2
+ ln(KK1 )) (y + ln(KK1 ))
K1 (x + ln(KK1 ))(y + K1 ln(K))2
e K1
K1
(63)
We have the following proposition:
Proposition 9. A metric of the type A(II) in Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric below:
(y − x)e−3x 2 h(y − x)e−3y 2
(64)
dy
dx
+
g=k
x2 y
xy 2
with k ∈ R \ {0}. If h = 1, we require k > 0.
Proof. In (63), let us perform the change of coordinates
x → x − ln(KK1 ),
y → y − ln(KK1 ) ,
and let us introduce k = K 3 .
For h 6= 1, the statement is clear by construction of the metric. For the case h = 1, recall our freedom
to swap coordinates, x ↔ y, which produces the conformal factor −1. Therefore, we obtain the additional
requirement k > 0.
Remark 8. Note that the normal forms (64) are constant multiples of metric g1 of (60). In Figure 2, these
metrics lie on the axis of abscissa. In the special case h = 1, the normal forms (64) lie in the right half of the
horizontal axis only.
21
3.2.2
Normal forms (B.5) of Theorem 1
In the case of complex Liouville metrics B(II), we can construct any metric g in the projective class, via
Formula (5), from the following two metrics:
e−3z 2
1 1
e−3z 2
C
−
dz − C
dz ,
g1 =
z z
z
z
g2 = (z − z) C e−3z dz 2 − C e−3z dz 2 .
A metric from the projective class P(g1 , g2 ) is given by
g = (z − z)
C e−3z
C e−3z
dz 2 −
dz 2
2
(K2 − K1 z)(K2 − K1 z)
(K2 − K1 z)2 (K2 − K1 z)
,
(65)
where we require K1 6= 0 because we only consider metrics with one essential vector field (metrics with K1 = 0
admit a homothetic vector field). The projective vector field is w = ∂z + ∂z , and thus the projective flow reads
φt (z0 , z 0 ) = (z0 + t, z 0 + t).
Remark 9. Let C = eiϕ . The projective connection (1) is specified completely by the value of ϕ mod π. Instead
of writing down (1) explicitly, which can be done but is rather lengthy in the present case, we give the following
qualitative argument: Since multiplication by a constant conformal factor does not affect the Christoffel symbols
of a metric, we may divide g2 by cos(ϕ). In view of Euler’s formula, eiϕ = cos(ϕ) + i sin(ϕ), one therefore finds
that
g2
= (z − z) ((1 + i tan(ϕ)) e−3z dz 2 − (1 − i tan(ϕ)) e−3z dz 2 )
gnew =
cos(ϕ)
lies in the projective class P(g1 , g2 ). Thus the projective connection (1) depends on tan(ϕ) only, and therefore
it depends on ϕ mod π only, which is exactly the result of the full computation.
The square of the length of the projective vector field w = ∂z + ∂z w.r.t. the metric (65) is
2
kwk (z0 , z 0 , t) = (z 0 − z0 )
C e−3z0 −3t
(K2 − K1 z 0 − K1 t)(K2 − K1 z0 − K1 t)2
−
C e−3z0 −3t
(K2 − K1 z 0 − K1 t)2 (K2 − K1 z0 − K1 t)
We therefore need to study the following equation (see (14)):
C e−3z0 −3t
C e−3z0 −3t
(z 0 − z0 )
−
(K2 − K1 z 0 − K1 t)(K2 − K1 z0 − K1 t)2 (K2 − K1 z 0 − K1 t)2 (K2 − K1 z0 − K1 t)
!
′
′
C e−3z0 −3t
C e−3z0 −3t
′
′
.
−
= (z 0 − z0 )
(K2′ − K1′ z ′0 − K1′ t)(K2′ − K1′ z0′ − K1′ t)2 (K2′ − K1′ z ′0 − K1′ t)2 (K2′ − K1′ z0 − K1′ t)
.
(66)
In particular, poles of the left hand side must also be poles of the right hand side. Thus, proceeding in the same
K′
K2
− K2′ , we can deduce the following possibilities
way as earlier and introducing K = K
1
1
(a)
z0′ − z0 = K
z ′0 − z 0 = K
(b)
z ′0 − z0 = K
z0′ − z 0 = K
(c)
z0′ − z0 = K
z0′ − z 0 = K
(d)
z ′0 − z0 = K
z ′0 − z 0 = K
In the case (a), we have z0′ = z0 + K, z ′0 = z 0 + K. From (66) we can then infer the equation
K1′
K1
3
C e−3z0 (K2 − K1 z 0 − K1 t) +
K1′
K1
3
C e−3z0 (K2 − K1 z0 − K1 t)
= e−3K C e−3z0 (K2 − K1 z 0 − K1 t) + e−3K C e−3z0 (K2 − K1 z0 − K1 t).
K′
From this follows e−K = K11 , which is covered by Lemma 9.
Continuing with case (b), we have z ′0 = z0 + K, z0′ = z 0 + K. By the same reasoning as before, we can again
K′
derive e−3K = K11 from it using Equation (66).
Cases (c) and (d) imply z 0 = z0 , so they do not have to be considered.
22
Let us now look for the normal forms of metrics (27). Using the definition of K, we have from (65) that
z−z
C e−3z
C e−3z
2
2
g=
dz .
dz −
K13
(z + ln(KK1 ))2 (z + ln(KK1 ))
(z + ln(KK1 ))(z + ln(KK1 ))2
Performing a transformation z → z − η with η = ln(KK1 ), we obtain (note that KK1 is a positive number)
iϕ −3z
e−iϕ e−3z 2
e e
2
dz
−
g = k (z − z)
,
(67)
dz
z 2z
zz 2
where k = K 3 and C = eiϕ as in Remark 9.
We have already seen that non-isometric metrics require different K. Moreover, the value of ϕ ∈ [0, 2π)
is fixed by C. We have seen in Remark 9 that ϕ mod π specifies the projective class, and in fact a reasoning
similar to Observation 1 holds: By swapping z ↔ z, we transform ϕ → −ϕ, and thus we can w.l.o.g. assume
ϕ ∈ (0, π] and k > 0.
Therefore, metrics (67) are mutually non-isometric for different values of k > 0 and ϕ ∈ R mod π. Summing
up, we have thus proven the following result.
Proposition 10. A metric of the type B(II) in Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric below:
−3z
h e−3z 2
he
2
g = k (z − z)
,
dz
dz
−
z 2z
zz 2
with h ∈ P1 and k > 0.
3.2.3
Normal forms (C.8) of Theorem 1
We now discuss the case of Jordan block metrics, C(II). We can construct any metric g in the projective class,
via Formula (5), from the following two metrics:
g1 = 2(Y + x) dxdy
g2 = −2
Y +x
(Y + x)2 2
dxdy
+
dy
y3
y4
where Y = Y (y) is a function of one variable,
Y =
C2 e
3/2y
p !
|y|
+
y−3
C1 + C2
Z
y
e
3/2ξ
!
p
|ξ|
dξ .
(ξ − 3)2
By grace of a change of coordinates, we may w.l.o.g. assume C1 = 0 and C2 = 1 in the following. Furthermore,
using integration by parts, one can achieve the simplified form
Z y 3/2ξ
e
Y (y) =
dξ .
|ξ|3/2
Remark 10. In Section 3.1.3. of [12], the function Y is obtained as the derivative of a function Y1 , where
Y = Y1′ satisfies the ODE
1
1
(69)
y 2 Y1′′ − (y − 3)Y1′ + Y1 = 0
2
2
This equation is similar to Kummer’s equation. Indeed we can write the solution in terms of the confluent
hypergeometric function 1 F1 ,
√
1 3 3
√ 3/2y
+ 6 ye
, ,
+ C2 (y − 3)
Y1 (y) = C1 2 6y (y − 3) 1 F1
2 2 2y
and w.l.o.g.
1 3 3
, ,
.
Y (y) = y 1 F1
2 2 2y
For y 6= 0, the special function 1 F1 21 , 32 , • can be represented by the (imaginary) error function erf (erfi),
!
√ !
√
√ 3/2y
6
+ 6 ye
+ C2 (y − 3),
for y < 0
6π (y − 3) erfi
Y1 (y) = C1
√
2 y
!
√ !
√
6
√ 3/2y
+ 6 ye
6π (y − 3) erf
+ C2 (y − 3),
for y > 0 .
Y1 (y) = C1
√
2 y
23
Therefore, w.l.o.g. one may assume
Y = erfi(1/√y)
for y < 0
Y = erf(1/√y)
and
for y > 0 .
By formula (5), a metric from the projective class P(g1 , g2 ) can be written as
g=2
Y (y) + x
K2 (Y (y) + x)2 2
dxdy +
dy ,
3
(K1 − K2 y)
(K1 − K2 y)4
(70)
using the same coordinates as in (68). Metrics with K1 = 0 are eigenvectors of Lw and admit a homothetic
vector field. We therefore require K1 6= 0 in the remainder of this section. In view of Lemma 9, we may now
reparametrize the metrics (70), for K1 6= 0, by replacing (K1 , K2 ) → (K, α) according to the definitions
K=
e
K2/K
1
K1
and
α=
K2
.
K1
We thus obtain the metric (70), in the same local coordinates, as
Y (y) + x
α (Y (y) + x)2 2
,
dxdy
+
dy
g = K 3 e−3α 2
(1 − α y)3
(1 − α y)4
(71)
with α ∈ R and K ∈ R \ {0}. In Lemma 9, it is proven that K characterizes orbits of the projective flow. Below
we shall see that the condition is also sufficient, i.e. metrics belonging to different orbits of the projective flow
are non-isometric.
Lemma 10. A metric (71) is locally isometric to a constant multiple of metric g1 of (68), i.e.
Y (y) + x
α (Y (y) + x)2 2
3 −3α
2
K e
dxdy +
dy = 2K 3 (Y (v) + u) dudv .
(1 − α y)3
(1 − α y)4
(72)
in terms of new coordinates u = u(x, y), v = v(y).
Proof. Metrics on the same projective orbit are isometric, with the isometries linking them being given by the
flow of the projective vector field, i.e. solutions to
1
1
(y − 3) x + Y1 (y) ,
2
2
ẏ = y 2 .
ẋ =
The new coordinates are thus given by u = x(t0 ; x, y), v = y(t0 ; y). Now consider the metric on the left hand
side of (72),
Y (y) + x
α (Y (y) + x)2 2
g = K 3 e−3α 2
.
(73)
dxdy
+
dy
(1 − α y)3
(1 − α y)4
Applying a coordinate transformation as just described, the metric g is transformed into another metric of the
same form, as determined by (59). This is because for the metrics g and another metric g̃ isometric to it,
g̃ = φ∗t (g) (for a fixed t where φt is the projective flow), we have with a = ψ −1 (g) and ã = ψ −1 (g̃) that
ã = ψ −1 (g̃) = ψ −1 (φ∗t (g)) = φ∗t (ψ −1 (g)) = φ∗t (a) .
The metric g is determined by two parameters,
α=−
K2
,
K1
K=
e
K2/K
1
K1
,
which are here represented in terms of the initially introduced parameters K1 , K2 . Looking at Figure 2, the
canonical choice for a normal form is the intersection of the (thick=projective orbit) curve with the axis of
abscissa. The value t0 hence is determined by
K2′ = (K1 t0 + K2 ) et0 = 0
⇒
t0 = −
K2
= −α .
K1
The resulting metric, isometric to g, is thus a multiple of g1 , and the multiplicative factor is determined by
K1′ = K1 et0 =
1
eα −α
e =
K
K
and thus, in new coordinates (u, v), the metric (73) assumes the form
g̃ = g[K] = 2K 3 (Y (v) + u) dudv .
24
(74)
K2
K1
Figure 3: The normal forms for metrics of type C(II) of Proposition 3 are chosen to lie on the intersection of
the orbits (lines in the plot) with the axis of abscissa determined by g1 (marked by points). Such metrics are
multiples of the metric g1 .
Figure 3 illustrates geometrically how the normal forms are chosen on the orbits of the projective vector field.
Lemma 11. Two different copies g = g[k] and g̃ = g[k̃] of (74) are non-isometric.
Proof. One only needs to prove that the condition k̃ = k is necessary. Thus assume g and g̃ are isometric.
This means that there exists a diffeomorphism (x, y) 7→ (u, v), u = u(x, y), v = v(x, y) mapping one metric
into the other. There are only two possibilities: u = u(x), v = v(y) or u = u(y), v = v(x). Assume first
u = u(x), v = v(y), so that the following equation holds:
k (Y (y) + x) dxdy = k̃ (Y (v(y)) + u(x)) ux vy dxdy .
The above equation implies
k (Y (y) + x) = k̃ (Y (v(y)) + u(x)) ux vy .
(75)
By differentiating Equation (75) twice w.r.t. x, we obtain
0 = k̃uy (Y (v(y))uxxx + 3ux uxx + uuxxx ) .
(76)
Since v(y), being a coordinate transformation, is invertible, we can infer that Y ′ (v(y))uxxx = 0 (by differentiating
the right hand side of (76)). Thus, uxxx = 0, since Y ′ (v(y)) is not identically zero, so that
u(x) =
e1 x2
+ c1 x + d1 .
2
Substituting this back into (75) yields
e1 x2
+ c1 x + d1 (e1 x + c1 ) vy ,
k (Y (y) + x) = k̃ Y (v(y)) +
2
which is polynomial in x. Analyzing each coefficient w.r.t. x separately, one finds from the cubic term e1 = 0,
and from the linear term that vy is a non-zero constant, i.e. v = c2 y + d2 , c2 ∈ R \ {0}, d2 ∈ R. Taking this into
account, by differentiating Equation (75) w.r.t. y, we obtain
kY ′ = k̃ c22 Yv ux
(77)
implying that ux is a non-zero constant, i.e. u = c1 x + d1 , c1 ∈ R \ {0}, d1 ∈ R, so that equation (76) becomes
k = k̃c21 c2
(78)
whereas Equation (77) assumes the form
3
3
k̃c1 c22 e 2(c2 y+d2 )
k e 2y
=
,
2 |y| 23
2 |c2 y + d2 | 23
(79)
which, on account of (78), reduces to
3
3
c1
e 2y
3
|y| 2
= c2
e 2(c2 y+d2 )
3
|c2 y + d2 | 2
.
(80)
Since Equations (79) and (80) hold for any y, we have that d2 = 0, c2 = 1 and finally c1 = 1, and therefore
k = k̃ in view of (78).
25
Now assume u = u(y) and v = v(x). The analogue of Equation (75) becomes
k(Y (y) + x) = k̃ (Y (v(x)) + u(y)) uy vx .
(81)
Taking two derivatives w.r.t. x, we arrive at
k̃uy Y ′′ (v(x)) vx3 + 2vx vxx Y ′ (v(x)) + Y ′ (v(x))vxx + Y (v(x))vxxx + u(y) vxxx = 0 .
(82)
Dividing by k̃uy (recall that uy is non-zero) and taking a derivative w.r.t. y of this equation, we thus find
2
vxxx = 0, which implies that v(x) is of the form v(x) = e12x + c1 x + d1 , e1 , c1 , d1 ∈ R. Since vxxx = 0, from
(82) we obtain
Y ′′ (v(x))vx3 + (2vx + 1) vxx Y ′ (v(x)) = 0 .
(83)
Differentiating (69) w.r.t. y, we obtain
y 2 Y ′′ (y) +
3
(y + 1) Y ′ (y) = 0 .
2
(84)
Eliminating Y ′′ from (83) by grace of (84), and replacing v(x), a polynomial equation in x is obtained. The
system of equations on e1 , c1 and d1 obtained from its coefficients w.r.t. x can straightforwardly be solved and
yields the solution e1 = c1 = 0, d1 = −1. This, however, is a contradiction since it would imply v(x) = −1, but
v(x), being part of a coordinate transformation, cannot be constant.
Remark 11. Note that in the present case there are no parameters for the projective class. The projective
connection (1) for the projective class defined by a metric g1 = (Y (y) + x) dxdy reads, with y = y(x),
y ′′ =
Y ′ (y) y ′2
y′
−
.
Y (y) + x Y (y) + x
Summing up, we arrive at the following Proposition.
Proposition 11. A metric of the type C(II) in Proposition 3 that admits exactly one essential projective vector
field is isometric to one and only one metric of the form
g = k (Y (y) + x) dxdy ,
where k ∈ R \ {0}.
3.3
Case (III): Lw has 2 complex eigenvalues
Here we are in the case when Lw has the form (III) of (7). So, let (a1 , a2 ) be a basis of A in which Lw assume
such a form. We have that
cos(t) − sin(t)
λt
.
(85)
exp(tLw ) = e
sin(t)
cos(t)
Consider a general metric g = ψ(K1 a1 + K2 a2 ) from the projective class P(g1 , g2 ). By virtue of Equation (85),
we can compute
φ∗t (K1 a1 + K2 a2 ) = eλt (K1 cos(t)a1 − K1 sin(t)a2 + K2 sin(t)a1 + K2 cos(t)a2 ) .
(86)
Thus, the orbits of the projective flow describe logarithmically spiraling curves in A. For what follows, it will
be convenient to reformulate the problem in terms of new parameters (θ, K), i.e. we chose new coordinates in
A,
K1 = Keλθ sin(θ)
and
K2 = Keλθ cos(θ) .
(87)
This reparametrization is invertible,
K1
θ = arctan
K2
and
K=
q
K
K12 + K22 e−λ arctan 1/K2 .
Remark 12. Note that in case λ = 0, the new parametrization is by polar parameters, with K > 0 being the
radius parameter and θ ∈ [0, 2π) being the angle parameter. On the other hand, if λ 6= 0, it is convenient to
take θ ∈ R as parameter along spiraling curves in A, whereas K should be considered as an angle via
K = eλα ,
α ∈ [0, 2π) .
The parametrization (87) is adjusted to the problem in the sense that K is an invariant of the orbits of the
projective flow.
26
K2
K2
K1
K2
K1
K1
Figure 4: The flow of the projective vector field in A, sketched for different values of (K1 , K2 ) and different λ.
The left panel shows orbits for λ < 0, the middle and rightppanel show orbits for λ = 0 and λ > 0, respectively.
Note that for λ = 0, the orbits have constant radius ζ = K12 + K22 , so ζ is invariant under the action of the
projective flow.
Lemma 12. The pairs (K, θ) and (K ′ , θ′ ) are related by the transformation (86) if and only if they satisfy
K ′ = K.
Proof. Consider (86), i.e.
Keλ(t+θ) ((sin(θ) cos(t) + cos(θ) sin(t)) a1 + (− sin(θ) sin(t) + cos(θ) cos(t)) a2 )
′
′
= K ′ eλθ sin(θ′ )a1 + K ′ eλθ cos(θ′ )a2 .
Using standard trigonometrical identities, one finds K ′ = K and θ′ = θ + t (in case λ = 0 we understand the
second equality modulo 2π).
3.3.1
Normal forms (A.3) of Theorem 1
In the Liouville case, A(III), any metric g in the projective class can be constructed, via Formula (5), from the
following two metrics:
−3λx
e
h e−3λy 2
g1 = (tan(x) − tan(y))
,
dx2 +
dy
cos(x)
cos(y)
−3λx
h e−3λy 2
e
.
dx2 +
dy
g2 = (cot(x) − cot(y))
sin(x)
sin(y)
The projective connection is obtained according to (1) as
y ′′ =
1
1
h e3λ (x−y) ′3
e−3λ (y−x)
+ (cot(y − x) − 3λ) y ′ + (cot(y − x) + 3λ) y ′2 +
y ,
2h sin(y − x) 2
2
2 sin(y − x)
so the projective class is determined by the data (h, λ) with h ∈ R \ {0} and λ ∈ R. In the same coordinates
(x, y), the projective vector field, fixed by (13) and case (III) of (7), assumes the form w = ∂x + ∂y . Therefore,
φt (x0 , y0 ) = (x0 + t, y0 + t).
Any metric g ∈ P(g1 , g2 ) is given via (5) by
c1 e−3λx
g = (tan(y) − tan(x))
dx2
cos(x)(K2 tan(y) − K1 )(K2 tan(x) − K1 )2
+
c2 e−3λy
dy 2
cos(y)(K2 tan(y) − K1 )2 (K2 tan(x) − K1 )
Using the parametrization (87) and standard trigonometrical identities, this becomes
h e−3λ y dy 2
sin(y − x)
e−3λ x dx2
+
.
g=
K 3 e3λθ
sin(y − θ) sin2 (x − θ) sin2 (y − θ) sin(x − θ)
(88)
We need to consider Equation (14), which in the current case reads
sin(y0 − x0 )
K 3 e3λθ
h e−3λ (y0 +t)
e−3λ (x0 +t)
+
sin(y0 + t − θ) sin2 (x0 + t − θ) sin2 (y0 + t − θ) sin(x0 + t − θ)
sin(y0′ − x′0 )
=
K ′3 e3λθ′
′
′
e−3λ (x0 +t)
h e−3λ (y0 +t)
+
sin(y0′ + t − θ′ ) sin2 (x′0 + t − θ′ ) sin2 (y0′ + t − θ′ ) sin(x′0 + t − θ′ )
27
!
(89)
As discussed in Step 2 on page 7, and done in the previous sections, poles on the left must correspond to poles
on the right of (89). From this requirement, we infer the following cases:
(a)
x′0 = x0 + H
y0′ = y0 + H
(b)
x′0 = x0 + H
y0′ = x0 + H
(c)
x′0 = y0 + H
y0′ = y0 + H
(d)
x′0 = y0 + H
y0′ = x0 + H
K′
′
1
with H = arctan K1′ − arctan K
K2 = θ − θ.
2
In the case (a), plugging the relations back into (89) we obtain after some computation using standard
trigonometric identities,
3
K
= 1.
(90)
K′
Cases (b) or (c) would imply x′0 − y0′ = 0 which would send the right hand side of (89) to zero for all values of
t (light line). Thus, we can discard these cases. In the case (d), we obtain from (89) that
3
3
1
K
K
=
−1
and
h = −1 .
(91)
K′
h
K′
As a consequence, for case (d) to happen, we must have h = −1 since K ′ , K > 0. In summary we have therefore
verified from (90) and (91) that
K′ = K .
(92)
This corresponds to the expression in Lemma 12 and therefore isometric metrics lie in the orbit of the projective
flow. Equation (92) furthermore implies that the parameter K remains invariant for isometric metrics. With
this in mind, let us rewrite metric (88) as
e−3λ(x+θ)
h e−3λ(y+θ)
2
2
g = K −3 sin(x − y)
(93)
dx
+
dy
sin(y − θ) sin2 (x − θ)
sin2 (y − θ) sin(x − θ)
with h ∈ R \ {0}.
In view of Remark 12, the situations for λ = 0 and λ 6= 0 are manifestly different. In fact, as is also
illustrated in Figure 4, the projective orbits in the case λ = 0 are characterized by a radial parameter K = 0,
while in the case λ 6= 0 the orbits are characterized essentially by an angle α ∈ [0, 2π) as defined in Remark 12.
We shall therefore now discuss the two cases separately.
Assume λ = 0. In this case metric (93) becomes
g=K
−3
sin(x − y)
sin(y − θ) sin(x − θ)
dx2
h dy 2
+
sin(x − θ) sin(y − θ)
,
and by a change of coordinates (x, y) → (x + θ, y + θ) we obtain
sin(x − y)
dx2
h dy 2
g[κ; h] = κ
,
+
sin(y) sin(x) sin(x) sin(y)
(94)
where we introduced κ = K −3 ∈ R \ {0}.
Lemma 13. Two different copies g = g[κ; h], g ′ = g[κ′ ; h] of (94) are non-isometric.
Proof. This follows immediately after realizing that the left and the right side of Equation (89) are, for λ = 0
and θ = 0, the same up to a constant factor (note that H = 0). Therefore these factors must coincide for
isometric metrics.
Assume λ 6= 0. In this case metric (93) can be rewritten as
h e−3λ(y+θ−α)
e−3λ(x+θ−α)
2
2
(95)
dx +
dy
g = ε sin(x − y)
sin(y − θ) sin2 (x − θ)
sin2 (y − θ) sin(x − θ)
where we defined e3λα = K −3 . By a change of coordinates (x, y) → (x − θ + α, y − θ + α) and by renaming
θ → 2θ + α, we thus arrive at the normal form
e−3λ x
h e−3λ y
2
2
g[θ; λ, h] = sin(x − y)
(96)
dx
+
dy
sin(y − θ) sin2 (x − θ)
sin2 (y − θ) sin(x − θ)
with parameters θ ∈ [0, 2π) and h ∈ R \ {0}.
28
Lemma 14. Two different copies g = g[θ; λ, h] and g ′ = g[θ′ ; λ, h] of (96) are non-isometric.
Proof. Consider again the requirement that the length of the projective vector field is preserved by isometries.
h e−3λ (y0 +t)
e−3λ (x0 +t)
+
sin(x0 − y0 )
sin(y0 + t − θ) sin2 (x0 + t − θ) sin2 (y0 + t − θ) sin(x0 + t − θ)
!
(97)
′
′
e−3λ (x0 +t)
h e−3λ (y0 +t)
′
′
= sin(x0 − y0 )
+
sin(y0′ + t − θ′ ) sin2 (x′0 + t − θ′ ) sin2 (y0′ + t − θ′ ) sin(x′0 + t − θ′ )
In order that both sides can be equal, poles of the left hand side must correspond to poles on the right hand
side. Thus, we obtain the following possibilities:
(a)
x′0 = x0 + H
y0′ = y0 + H
(b)
x′0 = x0 + H
y0′ = x0 + H
(c)
x′0 = y0 + H
y0′ = y0 + H
(d)
x′0 = y0 + H
y0′ = x0 + H
K′
′
1
with H = arctan K1′ − arctan K
K2 = θ − θ. Again, we can discard cases (b) and (c).
2
Let us consider case (a): Plugging the relations x′0 = x0 + H and y0′ = y0 + H into (97), we find
′
e−3λ H = e−3λ (θ −θ) = 1 ,
from which we infer θ′ = θ. In case (d), we follow the same procedure and obtain two equations
h e−3λ H = −1
and
h = −e−3λ H .
We find that this case is only relevant if h = 1. Furthermore, we conclude again θ′ = θ, and thus the statement
is proven.
In summary, putting together Lemmas 13 and 14, we conclude:
Proposition 12. A metric of the type A(III) in Proposition 3 that admits exactly one essential projective
vector field is isometric to one and only one metric below:
2
h dy 2
(i) for λ = 0:
g = κ sin(x − y) sin(y)dxsin2 (x) + sin2 (y)
sin(x) ,
(ii) for λ ∈ R \ {0}:
g = sin(x − y)
e−3λ x
sin(y−θ) sin2 (x−θ)
Here, h, κ ∈ R \ {0} and θ ∈ [0, 2π). If λ = 0, h 6= ±1.
3.3.2
dx2 +
h e−3λ y
sin(x−θ)
sin2 (y−θ)
dy 2 .
Normal forms (B.6) of Theorem 1
In case of the complex Liouville metric, B(III), we can construct any metric g in the projective class, via
Formula (5), from the following two metrics, where C ∈ C, |C| = 1:
e−3λz
e−3λz 2
2
dz + C
dz ,
g1 = (tan(z) − tan(z)) C
cos(z)
cos(z)
e−3λz 2
e−3λz 2
g2 = (cot(z) − cot(z)) C
dz + C
dz .
sin(z)
sin(z)
The projective vector field, fixed by condition (13) and the case (III) of (7), assumes the form w = ∂x = ∂z + ∂z ,
and thus its flow reads
φt (z0 , z 0 ) = (z0 + t, z 0 + t).
Let us now analyze the parameters of the projective class. Taking advantage of the fact that a conformal
constant does not affect the Christoffel symbols, we can divide the metric by cos(ϕ) to obtain
sin(z − z)
e−3λ z 2
e−3λ z 2
gnew =
.
(1 + i tan ϕ)
dz + (1 − i tan ϕ)
dz
cos(z) cos(z)
cos(z)
cos(z)
Thus the projective connection should depend on ϕ mod π and λ ∈ R only, and this is indeed confirmed by a
straightforward computation. An arbitrary metric in the projective class P(g1 , g2 ) is
Ce−3λz
g = (tan(z) − tan(z))
dz 2
cos(z)(K2 tan(z) − K1 )(K2 tan(z) − K1 )2
Ce−3λz
2
+
dz
cos(z)(K2 tan(z) − K1 )2 (K2 tan(z) − K1 )
29
We work again with the parametrization (87) and obtain
g=
sin(z − z)
K3
C e−3λ (z+θ) dz 2
C e−3λ (z+θ) dz 2
+
2
sin(z − θ) sin (z − θ) sin2 (z − θ) sin(z − θ)
.
(98)
Now we investigate Equation (14)
sin(z 0 − z0 )
K3
C e−3λ (z0 +θ+t)
C e−3λ (z0 +θ+t)
+
2
2
sin(z 0 + t − θ) sin (z0 + t − θ) sin (z 0 + t − θ) sin(z0 + t − θ)
′
sin(z ′0 − z0′ )
=
K ′3
′
′
′
C e−3λ (z0 +θ +t)
C e−3λ (z0 +θ +t)
+
2
2
′
′
′
sin(z 0 + t − θ′ ) sin (z0 + t − θ′ ) sin (z 0 + t − θ′ ) sin(z0′ + t − θ′ )
!
(99)
As in the case of (real) Liouville metrics, we obtain the following cases from the study of the poles of each side:
(a)
z0′ = z0 + H
z ′0 = z 0 + H
(b)
z0′ = z0 + H
z ′0 = z0 + H
(c)
z0′ = z 0 + H
z ′0 = z 0 + H
(d)
z0′ = z 0 + H
z ′0 = z0 + H
K′
K2
= θ′ − θ. In the cases (a) and (d), plugging the relations back into (99) we
with H = arctan K2′ − arctan K
1
1
obtain
6
K
=1
K′
Cases (b) or (c) would imply z0′ − z ′0 = 0, which would send the right hand side of (99) to infinity for all values
of t. Thus, we can discard these cases. In summary we have therefore verified from (99) that
′
K = 1,
K
which implies K ′ = K, i.e. K is invariant for isometric metrics. With this in mind, let us rewrite the general
metric (98) as
g=K
−3
sin(z − z)
h e−3λ (z+θ)
he−3λ (z+θ)
2
2
dz +
dz .
sin(z − θ) sin2 (z − θ)
sin2 (z − θ) sin(z − θ)
(100)
where h ∈ P1 .
Assume λ = 0. Metric (98) becomes, after the change of coordinates (x, y) → (x + θ, y + θ)
g[κ; h] = κ sin(z − z)
h dz 2
h dz 2
+
sin(z) sin2 (z) sin2 (z) sin(z)
,
(101)
where we introduce κ = K −3 ∈ R \ {0}.
Lemma 15. Two different copies g = g[κ; h] and g ′ = g[κ′ ; h] of (101) are non-isometric.
Proof. This follows from inspection, in the current context, of Equation (14). Both sides of this equation are
the same up to a constant factor.
Assume λ 6= 0. We may rewrite metric (100),
g = sin(z − z)
h e−3λ (z+α)
he−3λ (z+α)
2
dz
+
dz 2
sin(z − θ) sin2 (z − θ)
sin2 (z − θ) sin(z − θ)
,
where h ∈ P1 and e−3λα = K −3 e−λθ . By a change of coordinates z → z + α and redefining θ, we arrive at the
normal form
h e−3λz
h e−3λz
2
2
dz +
dz
(102)
g[θ; h] = sin(z − z)
sin2 (z + θ) sin(z + θ)
sin(z + θ) sin2 (z + θ)
with parameters θ ∈ [0, 2π), h ∈ P1 .
Lemma 16. Two different copies g = g[θ; h] and g ′ = g[θ′ ; h] of (102) are non-isometric.
30
Proof. Consider again the requirement (14),
h e−3λ (z0 +t)
he−3λ (z0 +t)
+
sin(z0 − z 0 )
sin(z 0 + t − θ) sin2 (z0 + t − θ) sin2 (z 0 + t − θ) sin(z0 + t − θ)
′
′
=
sin(z0′
−
z ′0 )
h e−3λ (z0 +t)
he−3λ (z0 +t)
+
sin(z ′0 + t − θ′ ) sin2 (z0′ + t − θ′ ) sin2 (z ′0 + t − θ′ ) sin(z0′ + t − θ′ )
!
(103)
In order that both sides can be equal, poles of the left hand side must correspond to poles on the right hand
side. Thus, we obtain the possibilities:
(a)
z0′ = z0 + H
z ′0 = z 0 + H
(b)
z0′ = z0 + H
z ′0 = z0 + H
(c)
z0′ = z 0 + H
z ′0 = z 0 + H
(d)
z0′ = z 0 + H
z ′0 = z0 + H
K′
′
2
with H = arctan K2′ − arctan K
K1 = θ − θ. Again, in the same way as we did in Section 3.3.1, we can discard
1
cases (b) and (c). Let us consider case (a): Plugging the relations z0′ = z0 + H and z ′0 = z0 + H into (103), we
find
e−3λ H = 1
⇒
θ′ = θ.
In case (d), we follow the same procedure and obtain
h e−3λ H = −h,
and the real part of the requirement amounts to the requirement e−3λ H = −1, which is impossible. Thus, case
(d) does not appear and the statement is proven.
We conclude, putting together Lemma 15 and Lemma 16:
Proposition 13. A metric of the type B(III) in Proposition 3 that admits exactly one essential projective
vector field is isometric to one and only one metric below:
h dz2
h dz 2
(i) for λ = 0:
g = κ sin(z − z) sin2 (z)
+ sin(z)
,
sin(z)
sin2 (z)
(ii) for λ ∈ R \ {0}:
g = sin(z − z)
h e−3λz
sin(z+θ)
sin2 (z+θ)
dz 2 +
h e−3λz
sin(z+θ) sin2 (z+θ)
Here, h ∈ P1 , κ ∈ R \ {0} and θ ∈ [0, 2π). If λ = 0, h 6= ±1.
3.3.3
dz 2 .
Normal forms (C.9) of Theorem 1
This case, C(III), is similar to the case C(II) discussed in Section 3.2.3. Any metric g in the projective class
can be constructed, via Formula (5), from the following two metrics:
g1 = 2(Y + x) dxdy ,
g2 = −2
Y +x
(Y + x)2 2
dxdy +
dy .
3
y
y4
The function Y = Y (y) = Y1′ (y) is given w.l.o.g. by
p
Z
Z y
4
ξ2 + 1
− 23 λ arctan(ξ)
dξ = (y − 3λ)
Y1 (y) = (y − 3λ)
e
(ξ − 3λ)2
The function Y (y) = Y1′ (y) reads, w.l.o.g. [12],
p
Z
4
y2 + 1
− 32 λ arctan(y)
Y (y) = e
+
y − 3λ
y
e
− 23 λ arctan(ξ)
y
3
e− 2 λ arctan(ξ)
dξ .
|ξ 2 + 1|3/4
p
Z
4
ξ2 + 1
dξ
=
(ξ − 3λ)2
y
(105)
3
e− 2 λ arctan(ξ)
dξ .
|ξ 2 + 1|3/4
Remark 13. In [12], Y1 is obtained from the ODE
1
1
(y 2 + 1) Y1′′ − (y − 3λ)Y1′ + Y1 = 0 .
2
2
The solution of this ODE can be obtained and, arguing analogously to Section 3.2.3, we can w.l.o.g. assume it
to be identical to the integral in (105). Another representation, similar to that of Remark 10 in Section 3.2.3,
could be achieved in terms of a confluent hypergeometric function with complex arguments. However, this
representation does not appear to be convenient for our purposes in this paper, so we are not going to discuss it
further.
31
By Formula (5), a metric from the projective class P(g1 , g2 ) can be written
g=2
K2 (Y (y) + x)2 2
Y (y) + x
dxdy +
dy ,
3
(K1 − K2 y)
(K1 − K2 y)4
(106)
using the same coordinates as in (104). Note that we do not have real eigenvalues in the present case, so no
further restrictions on K1 ,K2 apply. Let us again use the parametrization (87). With this parametrization, the
metric (70), in the same local coordinates, reads
(Y (y) + x)2
Y (y) + x
2
−3 −3λθ
,
dxdy + cos(θ)
dy
2
g=K e
(sin θ − y cos θ)3
(sin θ − y cos θ)4
with θ ∈ [0, 2π) and K ∈ R \ {0}. We prove that the orbits of g under isometries and under the action of the
projective flow coincide, i.e. the constant K completely characterizes the orbits under isometries.
Lemma 17. A metric (71) is locally isometric to a constant multiple of the metric g1 of (104), i.e.
(Y (y) + x)2
Y (y) + x
2
= 2 K 3 (Y (v) + u) dudv .
dxdy
+
cos(θ)
dy
K −3 e−3λθ 2
(sin θ − y cos θ)3
(sin θ − y cos θ)4
Proof. The proof is analogous to that of Lemma 10. For the normal forms we make the following choices:
Case λ = 0: We have K = const and thus the projective orbits are concentric circles around the origin in
Figure 4. For the normal forms we can therefore choose freely (by grace of a rotation) the value of θ in (71),
and if we choose θ = π/2, then cos(θ) = 0. The resulting metric is a constant multiple of g1 , with the factor
being a positive real number.
Case λ 6= 0: The orbital invariant is still K, but the orbits are spirals in A, see Figure 4. Since the radial
distance of any of these spirals is strictly growing with increasing θ, each orbit crosses the unit circle exactly
once. The isometry is obtained from the solution of the projective flow, where the parameter value has to be
chosen according to the following computation:
′
ζ ′ = K ′ eλθ = Keλθ eλ t = 1
⇒
t=−
ln K
−θ,
λ
where ζ ′ , ζ denote the radial distance from the origin.
However, it is more convenient to choose the normal forms again on the axis given by multiples of g1 . We use
the parametrization (87), but recall Remark 12, i.e. we let K = eλα where α ∈ [0, 2π). Choosing θ = π/2 + 2πN
(N ∈ Z) for the normal form, we obtain a metric of the form (106) with
K1 = Keλ(
π/2+2πN )
and
K2 = 0 ,
and choosing N appropriately permits us further to have
K1 = eλβ
with
β ∈ [0, 2π) .
Figure 5 illustrates the metrics chosen as representative for each orbit of the projective flow in the (K1 , K2 )space.
Lemma 18. Two metrics g = k (Y (y) + x) dxdy and g̃ = k̃ (Y (v) + u) dudv with k, k̃ ∈ R \ {0} are isomorphic
if and only if k̃ = k.
Proof. The proof is analogous to that of Lemma 11. Instead of (79) we find
3
k
3
2 λ arctan(c2 y+d2 )
e 2 λ arctan(y)
2 e
=
k̃c
c
,
1
2
|y 2 + 1|3/4
|(c2 y + d2 )2 + 1|3/4
and instead of (80) we find
3
3
e 2 λ arctan(c2 y+d2 )
e 2 λ arctan(y)
=
c
.
c1 2
2
|y + 1|3/4
|(c2 y + d2 )2 + 1|3/4
Thus, it follows that d2 = 0, c2 = 1 and finally c1 = 1. Therefore, k̃ = k as claimed.
Putting together Lemma 17 and Lemma 18, we arrive at
32
K2
K2
K1
K2
K1
K1
Figure 5: Points of the thick lines are in 1-1 correspondence with the orbits of the projective flow. The first
picture on the left refers to the case λ = 0, whereas the other two refer to the case λ > 0: in the picture in the
middle we choose the points of thick segment as representatives of the orbits of the projective flow, whereas in
the third picture we choose the points of the thick unit circles.
Proposition 14. A metric of the type C(III) in Proposition 3 that admits exactly one essential projective
vector field is isometric to one and only one metric below:
For λ = 0:
For λ ∈ R \ {0}:
g = k (Y (λ = 0, y) + x) dxdy
g=e
λα
(Y (λ, y) + x) dxdy
with k > 0
with α ∈ [0, 2π) .
The collection of Propositions 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14 form the statement and the proof of
Theorem 1.
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