Spaces of locally homogeneous affine surfaces

Abstract

We examine the topology of various spaces of locally homogeneous affine surfaces which arise from the classification result of Opozda (Differ Geom Appl 21:173–198, 2004) as orbits of the action of $GL(2,{\mathbb {R}})$ (Type ${\mathcal {A}}$) and the $ax+b$ group (Type ${\mathcal {B}}$). We determine the topology of the spaces of Type ${\mathcal {A}}$ models in relation to the rank of the Ricci tensor. We determine the topology of the spaces of Type ${\mathcal {B}}$ models which either are flat or where the Ricci tensor is alternating.

1 Introduction

1.1 Notational conventions

An affine surface is a pair ${\mathcal {M}}=(M,\nabla )$ where M is a smooth surface and where $\nabla$ is a torsion free connection on the tangent bundle of M. Let $x=(x^1,x^2)$ be a system of local coordinates on M. Adopt the Einstein convention and sum over repeated indices to express $\nabla _{\partial _{x^i}}\partial _{x^j}=\Gamma _{ij}{}^k\partial _{x^k}$. The Christoffel symbols$\Gamma =\{\Gamma _{ij}{}^k\}$ determine the connection in the coordinate chart. Let $\rho$ be the associated Ricci tensor. The Ricci tensor carries the geometry in dimension 2; an affine surface is flat if and only if $\rho =0$. Since the Ricci tensor of an affine manifold is not necessarily symmetric, let $\rho _s(X,Y)=\frac{1}{2}\{\rho (X,Y)+\rho (Y,X)\}$ and $\rho _a(X,Y)=\frac{1}{2}\{\rho (X,Y)-\rho (Y,X)\}$ be the symmetric and alternating Ricci tensors.

1.2 Locally homogeneous affine surface geometries

Work of Opozda [11] shows that any locally homogeneous affine surface ${\mathcal {M}}$ is modeled on one of the following geometries.

• ${\mathbf{Type}~{\mathcal {A}}}$.${\mathcal {M}}=({\mathbb {R}}^2,\nabla )$ with constant Christoffel symbols $\Gamma _{ij}{}^k=\Gamma _{ji}{}^k$. This geometry is homogeneous; the Type ${\mathcal {A}}$ connections are the left invariant connections on the Lie group ${\mathbb {R}}^2$.

• ${{\mathbf{Type}~{\mathcal {B}}}}$.${\mathcal {M}}=({\mathbb {R}}^+\times {\mathbb {R}},\nabla )$ with Christoffel symbols $\Gamma _{ij}{}^k=(x^1)^{-1}A_{ij}{}^k$ where $A_{ij}{}^k=A_{ji}{}^k$ is constant. This geometry is homogeneous; the Type ${\mathcal {B}}$ connections are the left invariant connections on the $ax+b$ group.

• ${{\mathbf{Type}~{\mathcal {C}}}}$.${\mathcal {M}}=(M,\nabla )$ where $\nabla$ is the Levi-Civita connection of the round sphere $S^2$.

This result has been applied by many authors. Kowalski and Sekizawa [10] used it to examine Riemannian extensions of affine surfaces, Vanžurová [13] used it to study the metrizability of locally homogeneous affine surfaces, and Dǔsek [5] used it to study homogeneous geodesics. It plays a central role in the study of locally homogeneous connections with torsion of Arias-Marco and Kowalski [1] (see also [2] for a unified treatment independently of the torsion tensor). Although we will work with the local theory, the compact setting has been examined in [8, 12].

The Ricci tensor $\rho$ of an affine surface determines the full curvature tensor. In Sect. 2, we examine the spaces where the Ricci tensor has fixed rank in the Type ${\mathcal {A}}$ setting. In Sect. 3, we consider the spaces where either the Ricci tensor vanishes identically or where the Ricci tensor is alternating and non-trivial in the Type ${\mathcal {B}}$ setting.

1.3 Type ${\mathcal {A}}$ geometries

Let ${\mathcal {M}}(a,b,c,d,e,f):=({\mathbb {R}}^2,\nabla )$ where the Christoffel symbols of $\nabla$ are constant and given by

$$\begin{aligned} \begin{array}{lll} \Gamma _{11}{}^1=a,&{}\Gamma _{11}{}^2=b,&{}\Gamma _{12}{}^1=\Gamma _{21}{}^1=c,\\ \Gamma _{12}{}^2=\Gamma _{21}{}^2=d,&{}\Gamma _{22}{}^1=e,&{}\Gamma _{22}{}^2=f\,. \end{array} \end{aligned}$$

(1.1)

This identifies the set of Type ${\mathcal {A}}$ geometries with ${\mathbb {R}}^6$. The linear transformations $T(x^1,x^2)=(a^1_1 x^1 + a^1_2 x^2, a^2_1 x^1+a^2_2 x^2)$ where $(a_i^j)\in {\text {GL}}(2,{\mathbb {R}})$ act on the set of Type ${\mathcal {A}}$ geometries. We say that two Type ${\mathcal {A}}$ surface models are linearly equivalent if there exists $T\in {\text {GL}}(2,{\mathbb {R}})$ intertwining the two structures. One has that two Type ${\mathcal {A}}$ surfaces with non-degenerate Ricci tensor are affine equivalent if and only if they are linearly equivalent (see [3]). On the contrary, there exist Type ${\mathcal {A}}$ surfaces with degenerate Ricci tensor which are not linearly equivalent but which nevertheless are affine equivalent. We refer to the discussion in [6] for further details.

We consider the induced action of GL(2, R) on ${\mathbb {R}}^6$ and identify the linear orbit of a Type ${\mathcal {A}}$ model ${\mathcal {M}}$ with ${\mathcal {S}}({\mathcal {M}})=GL(2,{\mathbb {R}})/{\mathcal {I}}({\mathcal {M}})$ where ${\mathcal {I}}({\mathcal {M}})$ is the isotropy group ${\mathcal {I}}(M)=\{ T\in GL(2,{\mathbb {R}}); T^*{\mathcal {M}}={\mathcal {M}}\}$.

It was shown in [6] that any flat Type ${\mathcal {A}}$ model is linearly equivalent to one of the following:

$$\begin{aligned} \begin{array}{ll} {\mathcal {M}}_0^0:={\mathcal {M}}(0,0,0,0,0,0),&{} {\mathcal {M}}_1^0:={\mathcal {M}}(1,0,0,1,0,0),\\ {\mathcal {M}}_2^0:={\mathcal {M}}(-1,0,0,0,0,1), &{} {\mathcal {M}}_3^0:={\mathcal {M}}(0,0,0,0,0,1),\\ {\mathcal {M}}_4^0:={\mathcal {M}}(0,0,0,0,1,0), &{} {\mathcal {M}}_5^0:={\mathcal {M}}(1,0,0,1,-1,0). \end{array} \end{aligned}$$

(1.2)

The structure ${\mathcal {M}}_0^0$ is a singular cone point. The next result shows that the remaining orbits ${\mathcal {S}}({\mathcal {M}}_i^0):={\text {GL}}(2,{\mathbb {R}})\cdot {\mathcal {M}}_i^0$ for $1\le i\le 5$ glue together to define a smooth 4-dimensional submanifold of ${\mathbb {R}}^6$. Let $\mathbb {1}$ be the trivial line bundle over the circle $S^1$, let ${\mathbb {L}}$ be the Möbius line bundle over $S^1$, and let ${\mathcal {A}}^0\subset {\mathbb {R}}^6{\setminus }\{0\}$ be the set of all flat Type ${\mathcal {A}}$ geometries other than the cone point ${\mathcal {M}}_0^0$.

Theorem 1.1

${\mathcal {A}}^0$ is a smooth submanifold of ${\mathbb {R}}^6{\setminus }\{0\}$ diffeomorphic to the total space of ${\mathbb {L}}\oplus \mathbb {1}\oplus \mathbb {1}$ minus the zero section.

The Ricci tensor of any Type ${\mathcal {A}}$ model is symmetric. Let ${\mathcal {A}}^{1}_{\pm }\subset {\mathbb {R}}^6$ be the set of all Type ${\mathcal {A}}$ geometries where the Ricci tensor has rank 1 and is positive semi-definite ($+$) or negative semi-definite $(-)$. Any element in ${\mathcal {A}}^{1}_{\pm }$ is linearly equivalent to one of the following, where $c\in {\mathbb {R}}$ and $c_1\in {\mathbb {R}}{\setminus } \{0,-1\}$ (see [3, 6]):

$$\begin{aligned} {\mathcal {M}}_1^1:= & {} {\mathcal {M}}(-1,0,1,0,0,2),\nonumber \\ {\mathcal {M}}_2^1(c_1):= & {} {\mathcal {M}}(-1,0,c_1,0,0,1+2c_1),\nonumber \\ {\mathcal {M}}_3^1(c_1):= & {} {\mathcal {M}}(0,0,c_1,0,0,1+2c_1),\nonumber \\ {\mathcal {M}}_4^1(c):= & {} {\mathcal {M}}(0,0,1,0,c,2),\nonumber \\ {\mathcal {M}}_5^1(c):= & {} {\mathcal {M}}(1,0,0,0,1+c^2,2c). \end{aligned}$$

(1.3)

We will see in Lemma 2.3 that the orbit structure of the action of ${\text {GL}}(2,{\mathbb {R}})$ on ${\mathcal {A}}^1_{\pm }$ is quite complicated. It is therefore, perhaps, a bit surprising that the set of all orbits $\cup _{i,c}{\mathcal {M}}_i^1(c)\cdot {\text {GL}}(2,{\mathbb {R}})$ is smooth as shown in the following result.

Theorem 1.2

${\mathcal {A}}^{1}_{\pm }$ is a smooth submanifold of ${\mathbb {R}}^6$ diffeomorphic to $S^1\times S^1\times {\mathbb {R}}^3$.

The remaining geometries where the Ricci tensor has rank 2 form an open subset ${\mathbb {R}}^6{\setminus }\{\{0\}\cup {\mathcal {A}}^0\cup {\mathcal {A}}^1_+\cup {\mathcal {A}}^{1}_-\}$.

These results should be contrasted with the results in [9] where it is shown that any Type ${\mathcal {A}}$ affine surface is linearly equivalent to a surface determined by at most two non-zero parameters.

1.4 Type ${\mathcal {B}}$ geometries

Let ${\mathcal {N}}(a,b,c,d,e,f):=({\mathbb {R}}^+\times {\mathbb {R}},\nabla )$ where the Christoffel symbols of $\nabla$ are given by

$$\begin{aligned} \begin{array}{lll} \displaystyle \Gamma _{11}{}^1=\frac{a}{x^1},&{}\displaystyle \Gamma _{11}{}^2=\frac{b}{x^1},&{} \displaystyle \Gamma _{12}{}^1=\Gamma _{21}{}^1=\frac{c}{x^1},\\ \Gamma _{12}{}^2=\Gamma _{21}{}^2=\displaystyle \frac{d}{x^1},&{}\Gamma _{22}{}^1=\displaystyle \frac{e}{x^1},&{} \displaystyle \Gamma _{22}{}^2=\frac{f}{x^1}\,. \end{array} \end{aligned}$$

(1.4)

This identifies the space of Type ${\mathcal {B}}$ geometries with ${\mathbb {R}}^6$.

The natural structure group here is not the full general linear group, but rather the $ax+b$ group. We let $T_{a,b}(x^1,x^2):=(x^1,ax^2+bx^1)$ define an action of the $ax+b$ group on ${\mathbb {R}}^+\times {\mathbb {R}}$; this acts on the Type ${\mathcal {B}}$ geometries by reparametrization and defines the natural notion of linear equivalence in this setting. Thus, two Type ${\mathcal {B}}$ models ${\mathcal {N}}_1$ and ${\mathcal {N}}_2$ are said to be linearly equivalent if and only if there exists an affine transformation of the form $\Psi (x^1,x^2)=(x^1, a^2_1 x^1+ a^2_2 x^2)$ for $a^2_2\ne 0$ intertwining the two structures. It follows from the work in [3, 4] that two Type ${\mathcal {B}}$ surfaces which are neither flat nor of Type ${\mathcal {A}}$ are affine isomorphic if and only if they are linearly isomorphic. This is a non-trivial observation as there are non-linear affine transformations from one model to another if the dimension of the space of affine Killing vector fields is 4-dimensional or if the geometry is flat and thus the dimension of the space of affine Killing vector fields is 6-dimensional.

It was shown in [7] that a flat Type ${\mathcal {B}}$ model is linearly equivalent to one of the following models:

$$\begin{aligned} \begin{array}{ll} {\mathcal {N}}^0_0:={\mathcal {N}}(0,0,0,0,0,0), &{}{\mathcal {N}}^0_1(\pm ):={\mathcal {N}}(1,0,0,0,\pm 1,0), \\ {\mathcal {N}}^0_2(c_1):={\mathcal {N}}(c_1-1,0,0,c_1,0,0), \, c_1\ne 0, &{}{\mathcal {N}}^0_3:={\mathcal {N}}(-2,1,0,-1,0,0), \\ {\mathcal {N}}^0_4:={\mathcal {N}}(0,1,0,0,0,0), &{}{\mathcal {N}}^0_5:={\mathcal {N}}(-1,0,0,0,0,0), \\ {\mathcal {N}}^0_6(c_2):={\mathcal {N}}(c_2,0,0,0,0,0),\, c_2\ne 0,-1.&{} \end{array} \end{aligned}$$

Let ${\mathcal {B}}^0\subset {\mathbb {R}}^6$ be the space of flat Type ${\mathcal {B}}$ geometries other than the cone point ${\mathcal {N}}_0^0$ determined by the origin in ${\mathbb {R}}^6$. Unlike the Type ${\mathcal {A}}$ setting described in Theorem 1.1, ${\mathcal {B}}^0$ is not a smooth manifold but consists of the union of 3 smooth submanifolds of ${\mathbb {R}}^6$ which intersect transversally along the union of 3 smooth curves in ${\mathbb {R}}^6$. Define

$$\begin{aligned} \begin{array}{ll} {\mathcal {U}}_1(r,s):={\mathcal {N}}(1+rs^2,-s(1+rs^2),rs,-rs^2,r,-rs),&{}{\mathcal {B}}_1:={\text {Range}}\{{\mathcal {U}}_1\},\\ {\mathcal {U}}_2(u,v):={\mathcal {N}}(u,v,0,0,0,0),&{}{\mathcal {B}}_2:={\text {Range}}\{{\mathcal {U}}_2\},\\ {\mathcal {U}}_3(u,v):={\mathcal {N}}(u,v,0,1+u,0,0),&{}{\mathcal {B}}_3:={\text {Range}}\{{\mathcal {U}}_3\}. \end{array} \end{aligned}$$

(1.5)

Theorem 1.3

${\mathcal {B}}^0 ={\mathcal {B}}_1\cup {\mathcal {B}}_2\cup {\mathcal {B}}_3$. ${\mathcal {B}}_2$ and ${\mathcal {B}}_3$ are closed smooth surfaces in ${\mathbb {R}}^6$ which are diffeomorphic to ${\mathbb {R}}^2$ and which intersect transversally along the curve ${\mathcal {N}}(-1,v,0,0,0,0)$ for $v\in {\mathbb {R}}$. ${\mathcal {B}}_1$ can be completed to a smooth closed surface $\tilde{{\mathcal {B}}}_1$ which intersects ${\mathcal {B}}_2$ transversally along the curve ${\mathcal {N}}(1,v,0,0,0,0)$ and which intersects ${\mathcal {B}}_3$ transversally along the curve ${\mathcal {N}}(0,v,0,1,0,0)$ for $v\in {\mathbb {R}}$.

In the Type ${\mathcal {B}}$ setting, it is possible for the symmetric Ricci tensor $\rho _s$ to vanish without the geometry being flat; this is not possible in the Type ${\mathcal {A}}$ setting. The alternating Ricci tensor, $\rho _a$, carries the geometry in this context.

Let ${\mathcal {B}}_a$ be the set of all Type ${\mathcal {B}}$ structures where $\rho _s=0$ but $\rho _a\ne 0$. Set

$$\begin{aligned} \begin{array}{ll} {\mathcal {V}}_1(r,s,t):={\mathcal {N}}(s, t, r, 0, 0, r),\\ {\mathcal {V}}_2(u,v,w):={\mathcal {N}}( 1 - 2 u w + v w^2, w (1 - u w + v w^2), u - v w, -v w^2, v, u + v w) \end{array}\nonumber \\ \end{aligned}$$

(1.6)

and let ${\mathcal {D}}_1:= {\text {Range}}\{{\mathcal {V}}_1\}$ and ${\mathcal {D}}_2:= {\text {Range}}\{{\mathcal {V}}_2\}$.

Theorem 1.4

${\mathcal {B}}_a={\mathcal {D}}_1\cup {\mathcal {D}}_2$. ${\mathcal {V}}_i$ defines smoothly embedded 3-dimensional submanifolds of ${\mathbb {R}}^6$ for $r\ne 0$ and $u\ne 0$ which intersect transversally along a smooth 2-dimensional submanifold.

2 The space of type ${\mathcal {A}}$ models

Let ${\mathcal {M}}(a,b,c,d,e,f):=({\mathbb {R}}^2,\nabla )$ be given by Eq. (1.1) where the parameters (a, b, c, d, e, f) are real constants. The associated Ricci tensor is symmetric.

2.1 The space of flat type ${\mathcal {A}}$ models

Since the Ricci tensor determines the curvature in dimension two, flat surfaces are determined by a vanishing Ricci tensor. We provide the proof of the first result of the paper as follows.

The proof of Theorem 1.1

Let $\theta \in [0,2\pi ]$ be the usual periodic parameter where we identify 0 with $2\pi$ to define the circle $S^1=(\cos \theta ,\sin \theta )$. Let $(x^1,x^2,x^3)$ be a point of ${\mathbb {R}}^3$. The bundle ${\mathbb {L}}\oplus \mathbb {1}\oplus \mathbb {1}$ is then defined by identifying $(\theta ,x^1,x^2,x^3)$ with $(\theta +\pi ,-x^1,x^2,x^3)$; this puts the necessary half twist in the first x-coordinate. We require that $(x^1,x^2,x^3)$ belongs to ${\mathbb {R}}^3-\{0\}$ to remove the 0-section.

The parametrization of Eq. (1.1) is not a very convenient one for studying the Ricci tensor. We make a linear change of coordinates on ${\mathbb {R}}^6$ and let ${\mathcal {M}}_1(p,q,t,s,v,w)$ be defined by

$$\begin{aligned} \begin{array}{lll} \Gamma _{11}{}^1=2 q,&{}\Gamma _{11}{}^2= p + t,&{}\Gamma _{12}{}^1=\Gamma _{21}{}^1=w,\\ \Gamma _{12}{}^2=\Gamma _{21}{}^2= q + s,&{}\Gamma _{22}{}^1=v,&{}\Gamma _{22}{}^2= p - t\,. \end{array} \end{aligned}$$

We substitute these values in Eq. (2.4) to obtain

$$\begin{aligned} \rho =\left( \begin{array}{cc} p^2 + q^2 - s^2 - t^2 - p w - t w&{}-(p + t) v + (q + s) w\\ -(p + t) v + (q + s) w&{}q v - s v + (p - t - w) w\end{array}\right) \,. \end{aligned}$$

We set $\rho =0$. If $v^2+w^2\ne 0$, we obtain

$$\begin{aligned} \begin{array}{l} p=(v^2+w^2)^{-1}\{2 s v w+t(w^2-v^2)+w^3\},\text { and}\\ q=(v^2+w^2)^{-1}\{s(v^2-w^2)+v w (2 t+w)\}\,. \end{array} \end{aligned}$$

(2.1)

If $v^2+w^2=0$, we obtain a single equation

$$\begin{aligned} p^2+q^2-s^2-t^2=0\,. \end{aligned}$$

(2.2)

We introduce polar coordinates $v=r\cos (\theta )$ and $w=r\sin (\theta )$ to remove the singularity at $(v,w)=(0,0)$ in Eq. (2.1). We may then combine Eqs. (2.1) and Eq. (2.2) into a single expression:

$$\begin{aligned} \begin{array}{l} p=p(\theta ,r,s,t):=r \sin ^3(\theta )+s \sin (2 \theta )-t \cos (2 \theta ),\\ q=q(\theta ,r,s,t):=r \cos (\theta ) \sin ^2 (\theta )+s \cos (2 \theta )+t \sin (2 \theta )\,. \end{array} \end{aligned}$$

(2.3)

We assume $(r,s,t)\ne (0,0,0)$ to avoid the trivial structure ${\mathcal {M}}_0^0$ as the parametrization of Eq. (2.3) is singular there. We have $\theta \in [0,2\pi ]$ and $(r,s,t)\in {\mathbb {R}}^3-\{0\}$; since we are permitting r to be negative in polar coordinates, we must identify $(\theta ,r)$ with $(\theta +\pi ,-r)$ and obtain thereby the bundle ${\mathbb {L}}\oplus \mathbb {1}\oplus \mathbb {1}$ minus the zero section over $[0,\pi ]$. $\square$

Remark 2.1

The isotropy subgroups of the structures ${\mathcal {M}}_i^0$ vary with i and the dimension of the orbit space varies correspondingly. We list below the associated isotropy subgroups.

$$\begin{aligned} {\mathcal {I}}({\mathcal {M}}_0^0)= & {} {\text {GL}}(2,{\mathbb {R}}),\\ {\mathcal {I}}({\mathcal {M}}_1^0)= & {} \left\{ T:T(x^1,x^2)=(x^1,a x^2) \text { for } a\ne 0\right\} ,\\ {\mathcal {I}}({\mathcal {M}}_2^0)= & {} \left\{ id, T\right\} , \text { where } T(x^1,x^2)=(-x^2,-x^1), \\ {\mathcal {I}}({\mathcal {M}}_3^0)= & {} \left\{ T:T(x^1,x^2)=(a x^1,x^2)\text { for } a\ne 0\right\} , \\ {\mathcal {I}}({\mathcal {M}}_4^0)= & {} \left\{ T:T(x^1,x^2)=(a^2 x^1+bx^2,ax^2)\text { for }a\ne 0,\ b\in {\mathbb {R}}\right\} , \\ {\mathcal {I}}({\mathcal {M}}_5^0)= & {} \left\{ T:T(x^1,x^2)=(x^1,\pm x^2)\right\} . \end{aligned}$$

2.2 The space of type ${\mathcal {A}}$ models with rank-one Ricci tensor

If the Ricci tensor has rank 1, we can make a linear change of coordinates to ensure $\rho$ is a multiple of $dx^2\otimes dx^2$. We first establish Theorem 1.2. We then examine the isotropy groups of the models in Eq. (1.3) to determine the orbits of the Type ${\mathcal {A}}$ models which are not Type ${\mathcal {B}}$.

Lemma 2.2

Let ${\mathcal {M}}$ be a Type ${\mathcal {A}}$ model which is not flat. Then $\rho$ is a multiple of $dx^2\otimes dx^2$ if and only if $b=0$ and $d=0$.

Proof

A direct computation shows

$$\begin{aligned} \rho =\left( \begin{array}{cc} (a-d) d+b (f-c) &{} c d-b e \\ c d-b e &{} c(f-c)+(a-d) e \\ \end{array}\right) \,. \end{aligned}$$

(2.4)

Consequently, if $b=0$ and if $d=0$, then $\rho$ is a multiple of $dx^2\otimes dx^2$. Conversely, assume $\rho$ is a multiple of $dx^2\otimes dx^2$ or, equivalently, $-bc+ad-d^2+bf=0$ and $cd-be=0$. We wish to show $b=d=0$.

Case 1. Suppose that $d\ne 0$. The equations are homogeneous so we may assume $d=1$ and hence $c=be$. Substituting these values yields $\rho _{11}=-1+a-b^2e+bf=0$. Thus $a=1+b^2e+bf$. This yields $\rho =0$ so this case is impossible as we assumed ${\mathcal {M}}$ was not flat.

Case 2. Suppose that $b\ne 0$. Again, we may assume $b=1$ so $e=cd$. We compute $\rho _{11}=f-c+ad-d^2$. Setting this to zero again yields $\rho =0$ which is impossible. $\square$

Proof of Theorem 1.2

Let ${\mathcal {A}}^{1}_{\pm ,0}$ be the space of all Type ${\mathcal {A}}$ models where the Ricci tensor is a non-zero multiple of $dx^2\otimes dx^2$ where the ± refers to whether $\rho _{22}$ is positive or negative. By Lemma 2.2, we set $b=d=0$ and obtain $\rho _{22}=-c^2+ae+cf$. We make a change of variables setting

$$\begin{aligned} \begin{array}{llllll} a=q+v,&b=0,&c=u+p,&d=0,&e=q-v,&f=2p. \end{array} \end{aligned}$$

We then have $\rho _{22}=(p^2+q^2-u^2-v^2)dx^2\otimes dx^2$ so we may identify

$$\begin{aligned} {\mathcal {A}}_{+,0}^1= & {} \{\Gamma (p,q,u,v):p^2+q^2>u^2+v^2\},\\ {\mathcal {A}}_{-,0}^1= & {} \{\Gamma (p,q,u,v):p^2+q^2<u^2+v^2\}\,. \end{aligned}$$

We examine ${\mathcal {A}}_{-,0}^1$ as the analysis of ${\mathcal {A}}_{+,0}^1$ is the same after interchanging the roles of (p, q) and (u, v). Let ${\mathcal {D}}^2:=\{(U,V)\in {\mathbb {R}}^2:U^2+V^2<1\}$ be the open disk in ${\mathbb {R}}^2$. Let $-{\mathcal {M}}$ be the Type ${\mathcal {A}}$ model ${\mathcal {M}}(-a,-b,-c,-d,-e,-f)$. We construct a diffeomorphism $\Phi$ from $S^1\times {\mathbb {R}}^+\times {\mathcal {D}}^2$ to ${\mathcal {A}}_{-,0}^1$ by setting $u=r\cos \theta$, $v=r\sin \theta$, $p=rU$, $q=rV$. For $r>0$, $\theta \in S^1$, and $U^2+V^2<1$ we have

$$\begin{aligned} {\mathcal {M}}={\mathcal {M}}(r (\sin (\theta )+V),0,r (\cos (\theta )+U),0,r (V-\sin (\theta )),2 rU)\,. \end{aligned}$$

It is clear that $-{\mathcal {M}}(\theta ,r,U,V)={\mathcal {M}}(\theta +\pi ,r,-U,-V)$.

Let ${{\tilde{{\mathcal {M}}}}}$ be an arbitrary Type ${\mathcal {A}}$ model with ${\text {Rank}}\{\rho _{{\tilde{{\mathcal {M}}}}}\}=1$ and $\rho _{{\tilde{{\mathcal {M}}}}}$ negative semi-definite. We may express

$$\begin{aligned} \rho _{{\tilde{{\mathcal {M}}}}}=\lambda (\cos (\phi )dx^2-\sin (\phi )dx^1)\otimes (\cos (\phi )dx^2-\sin (\phi )dx^1) \end{aligned}$$

for $\lambda <0$. Here $\phi$ is only defined modulo $\pi$ instead of the usual $2\pi$. Let

$$\begin{aligned} T_\phi (x^1,x^2)=(\cos (\phi )x^1+\sin (\phi )x^2,-\sin (\phi )x^1+\cos (\phi )x^2)\,. \end{aligned}$$

be the associated rotation so that $T_\phi ^*(dx^2)=-\sin (\phi )dx^1+\cos (\phi ) dx^2$ and thus $(T_\phi )_*\tilde{{\mathcal {M}}}$ belongs to ${\mathcal {A}}_{-,0}^1$. We then have

$$\begin{aligned} {\mathcal {A}}_-^1=\{{\mathbb {R}}/(2\pi {\mathbb {Z}})\times {\mathcal {A}}_{-,0}^{1}\}/(\phi ,{\mathcal {M}})\sim (\phi +\pi ,-{\mathcal {M}}) \end{aligned}$$

where the gluing reflects the fact that when $\phi =\pi$ we have replaced $(x^1,x^2)$ by $(-x^1,-x^2)$ and thus changed the sign of the Christoffel symbols. Using our previous parametrization of ${\mathcal {A}}_{-,0}^{1}$, this yields

$$\begin{aligned} {\mathcal {A}}_-^1=({\mathbb {R}}^2/(2\pi {\mathbb {Z}})^2)\times {\mathbb {R}}^+\times {\mathcal {D}}^2/ \{(\phi ,\theta ,r,U,V)\sim (\phi +\pi ,\theta +\pi ,r,-U,-V)\}\,. \end{aligned}$$

After setting ${\tilde{\theta }}=\theta +\phi$, we can rewrite this equivalence relation in the form

$$\begin{aligned} (\phi ,{\tilde{\theta }},r,U,V)\sim (\phi +\pi ,{\tilde{\theta }},r,-U,-V)\,. \end{aligned}$$

The variable ${\tilde{\theta }}$ now no longer plays a role in the gluing. After replacing ${\mathbb {R}}^+$ by ${\mathbb {R}}$ and ${\mathcal {D}}^2$ by ${\mathbb {R}}^2$, we see ${\mathcal {A}}_-^1$ is diffeomorphic to $S^1\times S^1\times {\mathbb {R}}^3$ modulo the relation

$$\begin{aligned} (\phi ,{\tilde{\theta }},x_1,x_2,x_3)\sim (\phi +\pi ,{\tilde{\theta }},x_1,-x_2,-x_3)\,. \end{aligned}$$

These gluing relations define the total space of the bundle $\mathbb {1}\oplus {\mathbb {L}}\oplus {\mathbb {L}}$ over $(S^1,\phi )$. Since ${\mathbb {L}}\oplus {\mathbb {L}}$ is diffeomorphic to the trivial 2-plane bundle $\mathbb {1}\oplus \mathbb {1}$, we obtain finally that ${\mathcal {A}}_{-}^1$ is diffeomorphic to $S^1\times S^1\times {\mathbb {R}}^3$. $\square$

We adopt the notation of Eq. (1.3) to describe the orbits of the models ${\mathcal {M}}_i^1(\cdot )$ in the following lemma.

Lemma 2.3

• (1) ${\mathcal {I}}({\mathcal {M}}_1^1)=\{{\text {id}}\}$.

• (2) ${\mathcal {I}}({\mathcal {M}}_2^1(c_1))=\{{\text {id}}\}$ if $c_1\ne -\frac{1}{2}$.

• (3) ${\mathcal {I}}({\mathcal {M}}_2^1(-\frac{1}{2}))=\{{\text {id}},T\}$, where $T(x^1,x^2)=(x^1+x^2,-x^2)$.

• (4) ${\mathcal {I}}({\mathcal {M}}_3^1(c_1))=\{T:T(x^1,x^2)=(v^{-1}x^1,x^2) \text { for } v\in {\mathbb {R}}\backslash \{0\}\}$.

• (5) ${\mathcal {I}}({\mathcal {M}}_4^1(c))=\{T:T(x^1,x^2)=(x^1-wx^2,x^2)\text { for }w\in {\mathbb {R}}\}$, if $c\ne 0$,.

• (6) ${\mathcal {I}}({\mathcal {M}}_4^1(0))=\{T:T(x^1,x^2)=(v^{-1}(x^1-wx^2),x^2)\text { for }w\in {\mathbb {R}}, v\in {\mathbb {R}}\backslash \{0\}\}$.

• (7) ${\mathcal {I}}({\mathcal {M}}_5^1(c))=\{{\text {id}}\}$, if $c\ne 0$.

• (8) ${\mathcal {I}}({\mathcal {M}}_5^1(0))=\{{\text {id}},T\}$ where $T(x^1,x^2)=(x^1,-x^2)$.

Proof

Suppose $T\in {\mathcal {I}}({\mathcal {M}}_i^1(\cdot ))$. The Ricci tensor of ${\mathcal {M}}_i^1(\cdot )$ is a non-zero multiple of $dx^2\otimes dx^2$. Since T must preserve the Ricci tensor, $T(dx^2)=\pm dx^2$. This implies $(y^1,y^2)=T(x^1,x^2)=(v^{-1}(x^1-wx^2),\varepsilon x^2)$ for $\varepsilon =\pm 1$. Then

$$\begin{aligned} dy^1= & {} v^{-1}(dx^1-wdx^2), \quad dy^2=\varepsilon dx^2,\quad \partial _{y^1}=v\partial _{x^1},\quad \partial _{y^2}=\varepsilon (w \partial _{x^1}+\partial _{x^2}), \\ {}^y\Gamma _{11}{}^1:= & {} v ({}^x\Gamma _{11}{}^1 - w\ {}^x\Gamma _{11}{}^2). \\ {}^y\Gamma _{11}{}^2:= & {} v^2\varepsilon {}^x\Gamma _{11}{}^2, \\ {}^y\Gamma _{12}{}^1:= & {} \varepsilon ({}^x\Gamma _{12}{}^1 + w\ ( {}^x\Gamma _{11}{}^1 - {}^x\Gamma _{12}{}^2 - w\ {}^x\Gamma _{11}{}^2)), \\ {}^y\Gamma _{12}{}^2:= & {} v ({}^x\Gamma _{12}{}^2 + w\ {}^x\Gamma _{11}{}^2), \\ {}^y\Gamma _{22}{}^1:= & {} \frac{1}{v}({}^x\Gamma _{22}{}^1 + w (2\ {}^x\Gamma _{12}{}^1 - {}^x\Gamma _{22}{}^2) + w^2 ({}^x\Gamma _{11}{}^1 - 2\ {}^x\Gamma _{12}{}^2) - w^3\ {}^x\Gamma _{11}{}^2), \\ {}^y\Gamma _{22}{}^2:= & {} \varepsilon ({}^x\Gamma _{22}{}^2 + 2 w\ {}^x\Gamma _{12}{}^2 + w^2\ {}^x\Gamma _{11}{}^2). \end{aligned}$$

Case 1.${\mathcal {M}}_1^1={\mathcal {M}}(-1,0,1,0,0,2)$ and $T^*{\mathcal {M}}_1^1={\mathcal {M}}(-v,0,\varepsilon (1-w),0,-\frac{w^2}{v},2\varepsilon )$. Examining $\Gamma _{11}{}^1$ and $\Gamma _{22}{}^2$ yields $\varepsilon =1$ and $v=1$. Examining $\Gamma _{22}{}^1$ yields $w=0$.

Case 2. We have $c\notin \{0,-1\}$, ${\mathcal {M}}_2^1(c)={\mathcal {M}}(-1,0,c,0,0,1+2c)$, and

$$\begin{aligned} \textstyle { T^*{\mathcal {M}}_2^1(c)= {\mathcal {M}}(-v,0,\varepsilon (c-w),0,-\frac{1}{v}(w+w^2),(1+2c)\varepsilon )\,.} \end{aligned}$$

Examining $\Gamma _{11}{}^1$ yields $v=1$. Suppose $c\ne -\frac{1}{2}$. Examining $\Gamma _{22}{}^2$ yields $\varepsilon =1$. Since $\varepsilon =1$, examining $\Gamma _{12}{}^1$ yields $w=0$. Suppose $c=-\frac{1}{2}$. Examining $\Gamma _{12}{}^1$ and $\Gamma _{22}{}^1$ yields $(\varepsilon ,w)=(1,0)$ or $(\varepsilon ,w)=(-1,-1)$.

Case 3. We have $c\notin \{0,-1\}$, ${\mathcal {M}}_3^1(c)={\mathcal {M}}(0,0,c,0,0,1+2c)$, and

$$\begin{aligned} \textstyle { T^*{\mathcal {M}}_3^1(c)={\mathcal {M}}(0,0,c\varepsilon ,0,-\frac{w}{v},(1+2c) \varepsilon )}\,. \end{aligned}$$

Examining $\Gamma _{12}{}^1$ yields $\varepsilon =1$. Examining $\Gamma _{22}{}^1$ yields $w=0$. There is then no condition on v.

Case 4.${\mathcal {M}}_4^1(c)={\mathcal {M}}(0,0,1,0,c,2)$ and $T^*{\mathcal {M}}_4^1(c)={\mathcal {M}}(0,0,\varepsilon ,0,\frac{c}{v},2\varepsilon )$. Examining $\Gamma _{22}{}^2$ yields $\varepsilon =1$. There is no condition on w. If $c\ne 0$, examining $\Gamma _{22}{}^1$ yields $v=1$; if $c=0$, there is no condition on v.

Case 5.${\mathcal {M}}_5^1(c)={\mathcal {M}}(1,0,0,0,1+c^2,2c)$ and

$$\begin{aligned} \textstyle { T^*{\mathcal {M}}_5^1(c)={\mathcal {M}}(v,0,w\varepsilon ,0,\frac{1}{v}(1+(c-w)^2), 2c\varepsilon )\,.} \end{aligned}$$

Examining $\Gamma _{11}{}^1$ shows $v=1$. Examining $\Gamma _{12}{}^1$ shows $w=0$. If $c\ne 0$, examining $\Gamma _{22}{}^2$ shows $\varepsilon =1$. If $c=0$, we obtain $\varepsilon =\pm 1$. $\square$

The general linear group ${\text {GL}}(2,{\mathbb {R}})$ acts on the space ${\mathbb {R}}^6$ of all Type ${\mathcal {A}}$ geometries via change of coordinates. Let ${\text {GL}}_+(2,{\mathbb {R}})$ be the subgroup of matrices with positive determinant. If ${\mathcal {M}}$ is a Type ${\mathcal {A}}$ model with ${\text {Rank}}\{\rho \}({\mathcal {M}})=2$, then the associated space of affine Killing vector fields is 2-dimensional and ${\mathcal {M}}$ does not also admit a Type ${\mathcal {B}}$ structure [3]. But there are Type ${\mathcal {A}}$ models with ${\text {Rank}}\{\rho \}=1$ which also admit Type ${\mathcal {B}}$ structures. Let ${\mathcal {O}}^1_{\pm }\subset {\mathcal {A}}^1_{\pm }$ be the set of Type ${\mathcal {A}}$ models with ${\text {Rank}}\{\rho \}=1$ and which do not admit Type ${\mathcal {B}}$ structures.

Theorem 2.4

• (1) ${\mathcal {O}}^1_{-}$ is empty; every element of ${\mathcal {A}}^1_{-}$ also admits Type ${\mathcal {B}}$ structure.

• (2) ${\text {GL}}_+(2,{\mathbb {R}})$ acts without fixed points on ${\mathcal {O}}^1_{+}$. The action admits a section $s:{\mathbb {R}}\rightarrow {\mathcal {O}}^1_{+}$ so ${\mathcal {O}}^1_{+}={\text {GL}}_+(2,{\mathbb {R}})\times {\mathbb {R}}$ is a principal fiber bundle over ${\mathbb {R}}$.

Proof

Resuts of [3] show that the models ${\mathcal {M}}_i^1(\cdot )$ for $1\le i\le 4$ also admit Type ${\mathcal {B}}$ structures while the models ${\mathcal {M}}_5^1(c)$ do not. The Ricci tensor associated to ${\mathcal {M}}_i^1(\cdot )$ is given by:

$$\begin{aligned} {\begin{array}{lll} \rho ^{{\mathcal {M}}_1^1}= dx^2\otimes dx^2,&{} \,\, \rho ^{{\mathcal {M}}_2^1}=c_1(1+c_1) dx^2\otimes dx^2,\\ \rho ^{{\mathcal {M}}_3^1}=c_1(1+c_1) dx^2\otimes dx^2,&{} \,\, \rho ^{{\mathcal {M}}_4^1}= dx^2\otimes dx^2,\\ \rho ^{{\mathcal {M}}_5^1}=(1+c^2) dx^2\otimes dx^2. \end{array}} \end{aligned}$$

If $\rho \le 0$, then it follows that $i=2$ or $i=3$ and $c\in (-1,0)$. Thus any element of ${\mathcal {A}}^1_{-}$ admits a Type ${\mathcal {B}}$ structure which proves Assertion (1).

Let ${\mathfrak {M}}_5^1=\cup _c{\mathcal {M}}_5^1(c)$; this is a smooth curve in ${\mathbb {R}}^6$. Type ${\mathcal {A}}$ models which are linearly equivalent to ${\mathcal {M}}_1^1$, ${\mathcal {M}}_2^1(c_1)$ for $c_1+c_1^2>0$, ${\mathcal {M}}_3^1(c_1)$ for $c_1+c_1^2>0$, or ${\mathcal {M}}_4^1(c)$ all admit Type ${\mathcal {B}}$ structures and have $\rho \ge 0$. Thus we may identify the structures ${\mathcal {O}}_{1,+}$ which do not admit Type ${\mathcal {B}}$ structures with ${\text {GL}}(2,{\mathbb {R}})\cdot {\mathfrak {M}}_5^1$. Let $T(x^1, x^2):=(x^1, -x^2)$. We have $T{\mathcal {M}}_5^1(c)={\mathcal {M}}_5^1(-c)$. Since $\det (T)=-1$, we conclude therefore that ${\mathcal {O}}^1_{+}={\text {GL}}_+(2,{\mathbb {R}})\cdot {\mathfrak {M}}_5^1$. By Lemma 2.3, the action of ${\text {GL}}_+(2,{\mathbb {R}})$ on ${\mathfrak {M}}_5^1$ is fixed point free. Assertion (2) follows. $\square$

3 The space of type ${\mathcal {B}}$ connections

Let ${\mathcal {N}}(a,b,c,d,e,f):=({\mathbb {R}}^+\times {\mathbb {R}},\nabla )$ where the Christoffel symbols of $\nabla$ are given by (1.4). The Ricci tensor needs not be symmetric in this setting:

$$\begin{aligned} \begin{array}{l} \rho =(x^1)^{-2}\left( \begin{array}{cc} (a-d+1) d+b (f-c) &{} c d-b e+f \\ c (d-1)-b e &{} -c^2+f c+(a-d-1) e \\ \end{array} \right) \end{array} \end{aligned}$$

(3.1)

3.1 The space of flat type ${\mathcal {B}}$ models

The proof of Theorem 1.3

Let ${\mathcal {N}}={\mathcal {N}}(a,b,c,d,e,f)$. We clear denominators in Eq. (3.1) and set ${\tilde{\rho }}_{ij}=(x^1)^2\rho _{ij}$. Adopt the notation of Eq. (1.5). A direct computation shows that the structures ${\mathcal {U}}_i(\cdot )$ are flat. We distinguish cases to establish the converse. We use Eq. (3.1) and set ${\tilde{\rho }}=0$. Since ${\tilde{\rho }}_{12}-{\tilde{\rho }}_{21}=c+f$, $f=-c$.

Case 1. Assume $e\ne 0$. Set $c=rs$, $e=r$, and $f=-rs$ for $r\ne 0$. Then

$$\begin{aligned} {\tilde{\rho }}_{22}=-r (1 - a + d + 2 r s^2 )\text { and } {\tilde{\rho }}_{21}= -r(b+s-ds)\,. \end{aligned}$$

We solve these equations to obtain $a = 1 + d + 2 r s^2$ and $b=(-1 + d) s$. We have ${\tilde{\rho }}_{11}=2(d+ rs^2)$. Thus $d=-rs^2$ which gives the parametrization ${\mathcal {U}}_1$.

Case 2. Suppose $e=0$. Set $a=u$, $b=v$, and $f=-c$ to obtain

$$\begin{aligned} {\tilde{\rho }}=\left( \begin{array}{cc} d(1+u-d)-2 c v &{} c (d-1) \\ c (d-1) &{} -2 c^2 \end{array} \right) \,. \end{aligned}$$

This yields $c=0$ and $d(1+u-d)=0$. If we set $d=0$, we obtain the parametrization ${\mathcal {U}}_2$; if we set $d=1+u$, we obtain the parametrization ${\mathcal {U}}_3$. This establishes the first assertion.

The parametrization ${\mathcal {U}}_2$ and ${\mathcal {U}}_3$ intersect when $u=-1$; the intersection is transversal along the curve ${\mathcal {N}}( -1,v,0,0,0,0)$. We wish to extend the parametrization ${\mathcal {U}}_1$ to study the limiting behavior as $e\rightarrow 0$. We distinguish cases.

Case A. Suppose $\lim _{n\rightarrow \infty }{\mathcal {U}}_1(r_n,s_n)\in {\text {Range}}\{{\mathcal {U}}_2\}$. We have

$$\begin{aligned} \begin{array}{lll} \lim _{n\rightarrow \infty }1+r_ns_n^2=u,&{}\lim _{n\rightarrow \infty }-s_n(1+r_ns_n^2)=v,&{}\lim _{n\rightarrow \infty }-r_ns_n=0,\\ \lim _{n\rightarrow \infty }-r_ns_n^2=0,&{}\lim _{n\rightarrow \infty }r_n=0,&{}\lim _{n\rightarrow \infty }-r_ns_n=0. \end{array} \end{aligned}$$

These equations imply $u=1$, $\lim _{n\rightarrow \infty }r_n=0$, $\lim _{n\rightarrow \infty }s_n=-v$. Thus we may simply set $r=0$ to obtain a transversal intersection along the curve ${\mathcal {N}}(1,v,0,0,0,0)$.

Case B. Suppose $\lim _{n\rightarrow \infty }{\mathcal {U}}_1(r_n,s_n)\in {\text {Range}}\{{\mathcal {U}}_3\}$. We have

$$\begin{aligned} \begin{array}{lll} \lim _{n\rightarrow \infty }1+r_ns_n^2=u,&{}\lim _{n\rightarrow \infty } -s_n(1+r_ns_n^2)=v,&{}\lim _{n\rightarrow \infty }-r_ns_n=0,\\ \lim _{n\rightarrow \infty }-r_ns_n^2=1+u,&{}\lim _{n\rightarrow \infty }r_n=0, &{}\lim _{n\rightarrow \infty }-r_ns_n=0. \end{array} \end{aligned}$$

These equations imply $u=0$, $\lim _{n\rightarrow \infty } r_n=0$, and $\lim _{n\rightarrow \infty }r_ns_n^2=-1$. We change variables setting $r=-t^2$ and $s=\frac{1}{t}+w$ to express

$$\begin{aligned} \begin{array}{llll} {\mathcal {U}}_1 (-t^2,\frac{1}{t}+w)={\mathcal {N}} (&{}- t w(2+t w), &{} w(2 + 3 t w + t^2 w^2), &{}-t(1+ t w), \\ &{}(1 + t w)^2 , &{} -t^2,&{} t(1 + t w))\,. \end{array} \end{aligned}$$

We may now safely set $t=0$ to obtain the intersection with ${\text {Range}}\{{\mathcal {U}}_3\}$ along the curve ${\mathcal {N}}(0,2w,0,1,0,0)$. $\square$

3.2 Type ${\mathcal {B}}$ models with alternating Ricci tensor

It was shown in [3] that any Type ${\mathcal {B}}$ model with alternating Ricci tensor is linearly equivalent to one of the following models:

$$\begin{aligned} \begin{array}{ll} {\mathcal {N}}_1(c):={\mathcal {N}}(0,c,1,0,0,1), &{} \text {for}\,\, c\in {\mathbb {R}}, \\ {\mathcal {N}}_2(c,\pm ):={\mathcal {N}}(1\mp c^2),c,0,\mp c^2,\pm 1,\pm 2c), &{}\text {for}\,\, c>0. \end{array} \end{aligned}$$

The proof of Theorem 1.4

Adopt the notation of Eq. (1.6). It is clear that ${\mathcal {V}}_1$ defines a smooth 3-dimensional submanifold of ${\mathbb {R}}^6$. To see similarly that ${\mathcal {V}}_2$ is smooth, we note that we can recover $u=\frac{1}{2}( c+f)$ and $v=e$. If $v\ne 0$, then $w=\frac{1}{v}(f-u)$ while if $v=0$, $w= \frac{1}{2u}(1-a)$. Thus ${\mathcal {V}}_2$ is 1-1; it is not difficult to verify that the Jacobian determinant is non-zero. This shows that ${\mathcal {V}}_2$ also defines a smooth 3-dimensional submanifold of ${\mathbb {R}}^6$. We set $v=0$ and $u=r$ to see that ${\mathcal {V}}_1$ and ${\mathcal {V}}_2$ intersect along the surface $v=0$, $u=r$, $s=1-2 uw$ and $t=w(1-uw)$. A direct computation shows that the associated Ricci tensors are non-trivial and alternating:

$$\begin{aligned} {\tilde{\rho }}_{{\mathcal {V}}_1}=r\left( \begin{array}{c@{\quad }c}0&{}1\\ -1&{}0\end{array}\right) \text { and } {\tilde{\rho }}_{{\mathcal {V}}_2}=u\left( \begin{array}{c@{\quad }c}0&{}1\\ -1&{}0\end{array}\right) \,. \end{aligned}$$

Let ${\mathcal {N}}$ be a Type ${\mathcal {B}}$ model with $\rho _s=0$ and ${\tilde{\rho }}_{a,12}= \frac{c+f}{2}\ne 0$. We distinguish two cases.

Case 1. Suppose $e=0$. Set $c=2r-f$ for $r\ne 0$. Setting the $\rho _s=0$ yields

$$\begin{aligned} \begin{array}{ll} \rho _{s,11}:\ 0= d(1+a-d) + 2 b (f - r),&{} \rho _{s,12}:\ 0=(1-d ) f + r( 2 d-1),\\ \rho _{s,22}:\ 0=-2 (f^2 - 3 f r + 2 r^2). \end{array} \end{aligned}$$

We solve the equation $-2 (f^2 - 3 f r + 2 r^2)=0$ to obtain $f=r$ or $f=2r$. Setting $f=2r$ yields $\rho _{s12}$: $0=r$ which is false. Thus $f=r$. We obtain $\rho _{s,12}=2dr$ so $d=0$. Set $a=s$ and $b=t$ to obtain the parametrization ${\mathcal {V}}_1$.

Case 2. Set $c=2u-f$ and $e=v$ for $u\ne 0$ and $v\ne 0$. We obtain

$$\begin{aligned}&\rho _{s,11}:\ 0= d(1+a-d) + 2 b( f - u), \\&\rho _{s,12}:\ 0= (1 - d) f - u + 2 d u - b v,\\&\rho _{s,22}:\ 0= -2 f^2 + 6 f u - 4 u^2 - (1 - a + d) v. \end{aligned}$$

Setting $\rho _{s,12}=0$ and $\rho _{s,22}=0$ yields $\textstyle a=\frac{1}{v}(2 f^2 - 6 f u + 4 u^2 + v + d v)$ and $\textstyle b =\frac{1}{v}(f - d f - u + 2 d u)$. We obtain $\rho _{s,11}=\frac{1}{v}(2 (f^2 - 2 f u + u^2 + d v))$. This implies that $d=\textstyle -\frac{(f-u)^2}{v}$. Setting $f=vw+u$ yields the parametrization ${\mathcal {V}}_2$. This parametrization can be extended safely to $v=0$; we require $u\ne 0$ to ensure $\rho _a\ne 0$. $\square$