The Differential Geometry of Curves in $B^1\times \mathbb {R}$

Abstract

The three-dimensional pseudohermitian manifolds with positive, zero, and negative constant pseudohermitian sectional curvature, respectively, with vanishing pseudohermitian torsion are the Heisenberg group $H_1$, the CR sphere $S^3$, and the circle bundle $B^1\times \mathbb {R}$. While the differential geometry of curves in $H_1$ and $S^3$ has been studied before, there has not been any study on the differential geometry of curves in $B^1\times \mathbb {R}$. In this paper, we take the first step in this direction. Among other things, we find the explicit formulas of the p-curvature and T-variation for curves in $B^1\times \mathbb {R}$. We also prove the fundamental theorem of curves in $B^1\times \mathbb {R}$.

1 Introduction

In the three-dimensional Euclidean space $\mathbb {R}^3$, the fundamental theorem of curves asserts that any curve parameterized by arc length is completely determined by its curvature and torsion (c.f. [11]). More precisely, given any two functions $\kappa >0$ and $\tau$, there exists curve in $\mathbb {R}^3$ parameterized by arc length such that its curvature and torsion are, respectively,equal to $\kappa$ and $\tau$. Moreover, such a curve is unique up to a Euclidean rigid motion.

In this paper, we restrict our attention to the three-dimensional smooth manifold M. A contact structure $\xi$ on M is a completely non-integrable plane distribution. M is called a pseudohermitian manifold if there exists a globally defined contact form $\theta$, which is a nondegenerate 1-form, i.e. $\theta \wedge d\theta \ne 0$, and an almost complex structure J defined on $\xi$ such that $J^2=-I$. The contact form $\theta$ induces the unique Reeb vector field T such that $\theta (T)=1$ and $T\lrcorner d\theta =d\theta (T, \cdot )= 0$. It also induces the Levi metric $g_\theta (X,Y)=d\theta (X,JY)$ for any $X,Y\in \xi$ and then is extended to be defined on the whole space TM. Tanaka [20] and Webster [22] independently introduced the unique connection $\nabla$, which is naturally prescribed by the condition $\nabla g_\theta =0$, $\nabla T=0$, and $\nabla J$. Now $\nabla$ is called the Tanaka-Webster connection. Note that the metric $g_\theta$ also induces the Levi-Civita connection on M. We refer the readers to [9, Lemma 1.3] about the relation between $\nabla$ and the Levi-Civita connection.

The Tanaka-Webster connection $\nabla$ has torsion $T_\nabla (X,Y)=\nabla _X Y-\nabla _Y X -[X,Y]$ on M and its partial component $\tau (X)=T_\nabla (T, X)$ for any $X\in TM$ is a TM-valued 1-form, called the pseudohermitian torsion with respect to $\nabla$. A Sasakian manifold, which is an odd-dimensional analogue of Kähler manifolds, is a pseudohermitian manifold with vanishing pseudohermitian torsion. Submanifolds of Sasakian manifolds have been studied intensively not only for the field of geometry, but also for theoretical physics related to AdS/CFT correspondence, which provides a duality between field theory and string theory (see [18] and references therein).

Similar to the Riemann curvature tensor on Riemannian manifolds, the curvature tensor R with respect to the Tanaka-Webster connection is defined by

$$\begin{aligned} R(Z,W,X,Y)=g_\theta (R(X,Y)Z,W), \end{aligned}$$

where $R(X,Y)Z=\nabla _X \nabla _Y Z - \nabla _Y \nabla _X Z-\nabla _{[X,Y]}Z$ for all $X,Y,Z,W\in TM$. However, the symmetric property $R(X,Y,Z,W)=R(Z,W,X,Y)$ does not hold in general, because of the failure of the first Bianchi identity (c.f. [9, Section 1.4]). For any horizontal 2-plane $\sigma =\text{ span}_\mathbb {R}\{X,Y\}\subset \xi$ spanned by the vectors $X,Y\in \xi$, the pseudohermitian sectional curvature of M is defined by

$$\begin{aligned} K_\theta (\sigma )=-\frac{1}{4}\frac{R( X,Y, X,Y)}{g_\theta (X,X)^2}. \end{aligned}$$

(1.1)

Equivalently, if we write $X=Z+\overline{Z}$ for some $Z\in T_{1,0}M$, the eigenvalue space of $\sqrt{-1}$, then (1.1) can be rewritten as

$$\begin{aligned} K_\theta (\sigma )=\frac{1}{4}\frac{R(Z,\overline{Z}, Z,\overline{Z})}{g_\theta (Z, \overline{Z})^2} \end{aligned}$$

(1.2)

(see [9, Section 1.4.3]). Note that $K_\theta (\sigma )$ does not depend on the choice of the basis X, Y.

Of particularly interesting examples among pseudohermitian manifolds are that with constant pseudohermitian sectional curvatures with vanishing pseudohermitian torsion. Those manifolds can also be considered as the space forms of Sasakian manifolds. This includes the CR 3-sphere $S^3$, the Heisenberg group $H_1$, and the circle bundle $B^1\times \mathbb {R}$ defined below, which has,respectively, positive, zero, and negative constant pseudohermitian sectional curvatures (c.f. [21, Theorem 2]). The three spaces are the standard models of the Sasakian space forms (see [1, Section 7.8]). There have been several classical results in Sasakian space forms, for instance, the comparison theorems [12] and Liouville Theorems for holomorphic maps [4].

The second author of this paper together with Chiu and Lai [7] studied the fundamental theorem of curves in $H_1$ (see Theorem 1.1 below) by using the Maurer-Cartan form and the method of moving frame to obtain a complete set of invariants (namely, the set consisting geometric invariants which uniquely determine a given curve). The proofs there rely on the Lie group structure of the Heisenberg group $H_1$. For more about the theory of curves in the Heisenberg group, we refer the readers to [5,6,7,8, 14, 15, 19] and the references therein.

Later, the first author of this paper and Chiu [6] proved the analogous theorems for curves in the CR 3-sphere $S^3$ and gave explicit expression for the geometric invariants, $\kappa$ and $\tau$ (see Theorem 1.3 below). Instead of using the Lie group approach, the authors embedded $S^3$ into $\mathbb {C}^2$ and studied the geometric properties of the curve in $S^3$ by considering it in $\mathbb {C}^2$.

Recall that the Heisenberg group $H_1$ is the set $\mathbb {R}^3$ associated with the group multiplication

$$\begin{aligned} (x_1,y_1,z_1)\circ (x_2,y_2,z_2)=(x_1+x_2,y_1+y_2,z_1+z_2+y_1x_2-x_1y_2), \end{aligned}$$

which is a three-dimensional Lie group. The space of all left invariant vector fields is spanned by the following standard basis:

$$\begin{aligned} \mathring{e}_1=\frac{\partial }{\partial x}+y\frac{\partial }{\partial x},~~ \mathring{e}_2=\frac{\partial }{\partial y}-x\frac{\partial }{\partial z},~~ \text{ and } ~~T=\frac{\partial }{\partial z}. \end{aligned}$$

The standard subbundle $\xi _0$ of the tangent bundle $TH_1$ is spanned by $\mathring{e}_1$ and $\mathring{e}_2$. Equivalently, $\xi _0$ can also be defined as the kernel of the standard contact form

$$\begin{aligned} \begin{aligned} \theta _0=dz+xdy-ydx. \end{aligned} \end{aligned}$$

The standard CR structure on $H_1$ is the endomorphism $J_0: \xi _0\rightarrow \xi _0$ defined by

$$\begin{aligned} J_0(\mathring{e}_1)=\mathring{e}_2~~ \text{ and } ~~ J_0(\mathring{e}_2)=-\mathring{e}_1. \end{aligned}$$

Note that $H_1$ can be viewed as a pseudohermitian manifold with $(J_0,\theta _0)$ as the standard pseudohermitian structure. Let PSH(1) be the group of all pseudohermitian transformations in $H_1$. For more details about the group PSH(1), we refer the readers to [8].

Let $\gamma :(a,b)\subset \mathbb {R}\rightarrow H_1$ be a curve in $H_1$. For any $t\in (a,b)$, the velocity $\gamma '(t)$ has the following decomposition

$$\begin{aligned} \gamma '(t)=\gamma '_{\xi _0}(t)+\gamma '_T(t), \end{aligned}$$

where $\gamma '_{\xi _0}(t)$ and $\gamma '_T(t)$ are, respectively, the orthogonal projection of $\gamma '(t)$ onto $\xi _0$ and the orthogonal projection of $\gamma '(t)$ onto T with respect to the Levi metric $g_\theta (\cdot , \cdot )=d\theta _0(\cdot , J\cdot )$. We say that $\gamma$ is horizontally regular if $\gamma '_{\xi _0}(t)\ne 0$ for all $t\in (a,b)$. Similar to the regular curve in $\mathbb {R}^3$, the horizontally regular curve in $H_1$ can always be reparameterized by a parameter s such that $|\gamma _{\xi _0}'(s)|=1$ for all s (c.f. Proposition 4.1 in [7]). Such a parameter is called the horizontal arc length s.

In the Heisenberg group $H_1$, the p-curvature and T-variation of curves play the same roles as the curvature and torsion of those in $\mathbb {R}^3$. They are defined as follows: For a curve $\gamma (s)$ parameterized by the horizontal arc length s, the p-curvature $\kappa (s)$ and the T-variation $\tau (s)$ are defined by

$$\begin{aligned} \kappa (s)=\left\langle \frac{dX(s)}{ds},Y(s)\right\rangle ~~ \text{ and } ~~\tau =\langle \gamma '(s),T\rangle , \end{aligned}$$

where $X(s)=\gamma '_{\xi _0}(s)$ and $Y(s)=J X(s)$. Both geometric quantities $\kappa$ and $\tau$ are invariant under the pseudohermitian transformations. More precisely, for any element $g\in PSH(1)$, the p-curvature and T-variation of $\gamma (s)$ and $g(\gamma (s))$ are the same. And we have the following fundamental theorem of curves in $H_1$, which was proved in [7]:

Theorem 1.1

(Theorem 1.2 in [7]) Let $\gamma _1$ and $\gamma _2$ be two horizontally regular curves parameterized by horizon arc length. Suppose that they have the same p-curvature $\kappa (s)$ and T-variation $\tau (s)$. Then there exists $g\in PSH(1)$ such that

$$\begin{aligned} \gamma _2(s)=g\circ \gamma _1(s) \text{ for } \text{ all } s. \end{aligned}$$

In addition, given smooth functions $\kappa (s)$ and $\tau (s)$, there exists a horizontally regular curvature $\gamma$ parameterized by the horizontal arc length such that its p-curvature and T-variation equal to $\kappa (s)$ and $\tau (s)$,respectively.

In this paper, we study curves defined in the circle bundle $B^1\times \mathbb {R}$, where $B^1$ is the unit ball in $\mathbb {C}$, i.e. $B^1=\{z\in \mathbb {C}: |z|^2<1\},$ and the contact form defined on $B^1\times \mathbb {R}$ is given by

$$\begin{aligned} \theta =dt+\frac{2\sqrt{-1}}{1-|z|^2}(zd\overline{z}-\overline{z}dz) \end{aligned}$$

(1.3)

for $(z,t)\in B^1\times \mathbb {R}$. The CR structure of $B^1\times \mathbb {R}$ is given by the decomposition of the complexified tangent bundle $\mathbb {C}\otimes T(B^1\times \mathbb {R})=span_{\mathbb {C}}\{ Z_1, Z_{\overline{1}}, T\}$ with $\theta (Z_1)=0=\theta (Z_{\overline{1}})$, $\theta (T)=1$, where

$$\begin{aligned} T=\frac{\partial }{\partial t},~~ Z_1=\frac{1-|z|^2}{2}\frac{\partial }{\partial z}+\sqrt{-1}\overline{z}\frac{\partial }{\partial t},~~Z_{\overline{1}}=\frac{1-|z|^2}{2}\frac{\partial }{\partial \overline{z}}-\sqrt{-1}z\frac{\partial }{\partial t};\end{aligned}$$

(1.4)

the almost complex structure $J: \mathbb {C}\otimes T(B^1\times \mathbb {R})\rightarrow \mathbb {C}\otimes T(B^1\times \mathbb {R})$ is defined such that $J^2=-I$ with $J(Z_1)=\sqrt{-1}Z_1$ and $J(Z_{\overline{1}})=-\sqrt{-1}Z_{\overline{1}}$. (see [9] 0r [12] for more details about the definition of CR structure and the derivation (2.4) in next paragraph).

As mentioned above, $B^1\times \mathbb {R}$ is space form of Sasakian manifold of negative curvature, while the CR 3-sphere $S^3$ and the Heisenberg group $H_1$ are, respectively, space forms of Sasakian manifold of positive and zero curvature. In view of the results about the differential geometry of curves in $H_1$ and $S^3$ mentioned above, we are motivated to study the differential geometry of curves in $B^1\times \mathbb {R}$.

In [7], the authors computed the explicit formulas for the p-curvature and the T-variation of a horizontally regular curve in $H_1$:

Theorem 1.2

(Theorem 1.4 in [7]) Let $\gamma (t)=(x(t),y(t),z(t))$ be a horizontally regular curve in $H_1$, not necessarily parameterized by the horizontal arc length. Then its p-curvature and the T-variation are, respectively, given by

$$\begin{aligned} \kappa (t)=\frac{x'y''-x'' y'}{((x')^2+(y')^2)^{\frac{3}{2}}}~~ \text{ and } ~~ \tau (t)=\frac{xy'-x'y+z'}{((x')^2+(y')^2)^{\frac{1}{2}}}. \end{aligned}$$

Similar to Theorem 1.2, the explicit formulas for the p-curvature and the T-variation of a horizontally regular curve in the CR sphere $S^3$ were obtained in [6]:

Theorem 1.3

(Theorem 1.10 in [6]) Let $\gamma (t)=(z_1(t),z_2(t))$ be a horizontally regular curve in $S^3$, not necessarily parameterized by the horizontal arc length. Then its p-curvature and the T-variation are, respectively, given by

$$\begin{aligned} \begin{aligned} \kappa (t)&=\frac{\sqrt{-1}}{2}\frac{(\overline{z}_2\overline{z}_1''-\overline{z}_1\overline{z}_2'')(z_2z_1'-z_1z_2') -(z_2z_1''-z_1z_2'')(\overline{z}_2\overline{z}_1'-\overline{z}_1\overline{z}_2')}{|z_2z_1'-z_1z_2'|^3}-2\tau (t),\\ \tau (t)&=\frac{-\sqrt{-1}(\overline{z}_1z_1'+\overline{z}_2z_2')}{|z_2z_1'-z_1z_2'|}= \frac{\sqrt{-1}(z_1\overline{z}_1'+z_2\overline{z}_2')}{|z_2z_1'-z_1z_2'|}. \end{aligned} \end{aligned}$$

Inspired by Theorem 1.2 and Theorem 1.3, we compute in Sect. 2 the explicit formulas for the p-curvature and the T-variation of a horizontally regular curve in $B^1\times \mathbb {R}$:

Theorem 1.4

Let $\gamma (\hat{s})=(z(\hat{s}),t(\hat{s}))=(x(\hat{s})+\sqrt{-1}y(\hat{s}),t(\hat{s}))$ be a horizontally regular curve in $B^1\times \mathbb {R}$, not necessarily parameterized by the horizontal arc length. Then its p-curvature and the T-variation are, respectively, given by

$$\begin{aligned} \kappa =\frac{1-x^2-y^2}{2\sqrt{2}\big ((x')^2+(y')^2\big )^{\frac{3}{2}}} \left( x'y''-y'x''-\frac{2(yx'-xy')\big ((x')^2+(y')^2\big )}{1-x^2-y^2}\right) \end{aligned}$$

(1.5)

and

$$\begin{aligned} \tau =\frac{1-x^2-y^2}{2\sqrt{2}\sqrt{(x')^2+(y')^2}}\left( t'+\frac{4(xy'-yx')}{1-x^2-y^2}\right) . \end{aligned}$$

(1.6)

Using the explicit formulas in Theorem 1.4, we are able to write down some examples of curves in $B^1\times \mathbb {R}$ having both constant p-curvature and T-variation. See Sect. 2.1.

As pointed out above in Theorem 1.1, the fundamental theorem of curves in $H_1$ was proved in [7]. Also, the fundamental theorem of curves in higher-dimensional Heisenberg group $H_n$ was proved in [5, Theorem 1.2]. On the other hand, the corresponding fundamental theorem of curves in the CR sphere was proved in [6, Theorem 1.9]. Inspired by these results, we prove the fundamental theorem of curves in $B^1\times \mathbb {R}$ in Sect. 3. In particular, we have the following in Sect. 3:

Theorem 1.5

Given $C^\infty$-functions $\kappa (s)$ and $\tau (s)$ defined in some interval I. Then there exists a horizontally regular curve $\gamma : I\rightarrow \mathbb {B}^1\times \mathbb {R}$ parameterized by horizontal arc such that its p-curvature and T-variation are $\kappa$ and $\tau$,respectively. Moreover, the existence of $\gamma$ is unique up to a pseudohermitian transformation.

In Sect. 3, we also study the symmetry in $B^1\times \mathbb {R}$, which is the set of all pseudohermitian transformations in $B^1\times \mathbb {R}$. We are able to identify all the pseudohermitian transformations in $B^1\times \mathbb {R}$. See Theorem 3.5.

In Sect. 4, we prove some theorems related to the horizontal curves, which are the curves with vanishing T-variation. For a horizontal curve, we are able to related its height to the area of the region bounded by the it projection on $B^1$. See Theorem 4.1 of the precise statement. As an application, we are able to prove that any two distinct points in $B^1\times \mathbb {R}^1$ are joined by a piecewise smooth horizontal curve. See Theorem 4.3.

The following theorem was proved by Bray-Jauregui in [2].

Theorem 1.6

(Bray-Jauregui) Suppose $\gamma$ is a smooth curve in $\mathbb {R}^3$ which is a graph over a simple closed curve $\gamma _0$ in $\mathbb {R}^2$ with positive curvature. If $\gamma$ has nonnegative (or nonpositive) torsion, then $\gamma$ has zero torsion and hence lies in the plane.

The following analogue of Theorem 1.6 in the Heisenberg group $H_1$ was proved in [6]:

Theorem 1.7

(Theorem 1.8 in [6]) Suppose that $\gamma _0$ is a horizontal closed curve in the Heisenberg group $H_1$ and $\gamma$ is a horizontally regular closed curve in the Heisenberg group $H_1$ such that $\pi \circ \gamma =\pi \circ \gamma _0$, where $\pi$ is the projection on the xy-plane along the z-axis. If the T-variation $\tau$ of $\gamma$ is nonnegative (or nonpositive), then $\tau =0$, that is, $\gamma$ is also horizontal.

Unlike Theorem 1.6, the p-curvature of $\gamma _0$ in Theorem 1.7 is not assumed to be positive. In Sect. 5, we prove the following analogue of Theorem 1.7 in $B^1\times \mathbb {R}$.

Theorem 1.8

Suppose that $\gamma _0$ is a horizontal closed curve in $B^1\times \mathbb {R}$ and $\gamma$ is a horizontally regular closed curve in $B^1\times \mathbb {R}$ such that $\pi \circ \gamma =\pi \circ \gamma _0$, where $\pi$ is the projection on $B^1$. If the T-variation $\tau$ of $\gamma$ is nonnegative (or nonpositive), then $\tau =0$, that is, $\gamma$ is also horizontal.

Similar to Theorem  1.7, the p-curvature of $\gamma _0$ in Theorem 1.8 is not assumed to be positive.

Suppose $\gamma (s)$ is a regular curve in $\mathbb {R}^3$ defined on an interval I and parameterized by arc length. Recall that the curve $\overline{\gamma }$ is a Bertrand mate of $\gamma$ if the unit normal vector n of $\gamma$ is on the normal line through the corresponding points of $\gamma (s)$ and $\overline{\gamma }(s)$. One can show that $\gamma$ has a Bertrand mate if and only if there exists a linear relation $A \kappa (s)+B \tau (s)=1$ in I, where A, B are nonzero constants, $\kappa$ and $\tau$ are the curvature and torsion of $\gamma$, respectively. For more details, see [11, 13, 17] for instance.

In Sect. 6, the Bertrand mate is defined for a horizontally regular curve in $B^1\times \mathbb {R}$. See Definition 6.1. Then we prove that any Bertrand mates in $B^1\times \mathbb {R}$ are congruent in the sense that they differ by a pseudohermitian transformation. See Theorem 6.4.

2 Computation of p-Curvature and T-Variation

In this section, we prove Theorem 1.4. If we define

$$\begin{aligned} \theta ^1=\frac{2}{1-|z|^2}dz~~ \text{ and } ~~ \theta ^{\overline{1}}=\frac{2}{1-|z|^2}d\overline{z}, \end{aligned}$$

(2.1)

then it follows from (1.3) and (2.1) that

$$\begin{aligned} d\theta =\frac{4\sqrt{-1}}{(1-|z|^2)^2}dz\wedge d\overline{z} =\sqrt{-1}\theta ^1\wedge \theta ^{\overline{1}}. \end{aligned}$$

(2.2)

Since $d\theta =\sqrt{-1}h_{1\overline{1}}\theta ^1\wedge \theta ^{\overline{1}}$ (c.f. (2.2) in [16]), it follows from (2.2) that

$$\begin{aligned} h_{1\overline{1}}=1. \end{aligned}$$

(2.3)

If we define

$$\begin{aligned} T=\frac{\partial }{\partial t},~~ Z_1=\frac{1-|z|^2}{2}\frac{\partial }{\partial z}+\sqrt{-1}\overline{z}\frac{\partial }{\partial t}, ~~Z_{\overline{1}}= \frac{1-|z|^2}{2}\frac{\partial }{\partial \overline{z}}-\sqrt{-1}z\frac{\partial }{\partial t}, \end{aligned}$$

(2.4)

then the frame $\{T,Z_1,Z_{\overline{1}}\}$ is dual to the coframe $\{\theta ,\theta ^1,\theta ^{\overline{1}}\}$. The Levi metric is defined to be

$$\begin{aligned} g_\theta =\theta \otimes \theta +d\theta (\cdot ,J\cdot ). \end{aligned}$$

Then we have

$$\begin{aligned} g_\theta (Z_1,Z_1)=g_\theta (Z_{\overline{1}},Z_{\overline{1}})=0,~~ g_\theta (Z_1,Z_{\overline{1}})=g_\theta (Z_{\overline{1}},Z_1)=1. \end{aligned}$$

(2.5)

To see this, we compute

$$\begin{aligned} \begin{aligned} g_\theta (Z_1,Z_{\overline{1}})&=d\theta (Z_1,JZ_{\overline{1}}) =d\theta (Z_1,-\sqrt{-1}Z_{\overline{1}}) =-\sqrt{-1}d\theta (Z_1,Z_{\overline{1}})\\&=(\theta ^1\wedge \theta ^{\overline{1}})(Z_1,Z_{\overline{1}})=1, \end{aligned} \end{aligned}$$

where we have used (2.2). The others can be computed similarly.

Note that $Z_1$ and $Z_{\overline{1}}$ can be rewritten as

$$\begin{aligned} \begin{aligned} Z_1&=\frac{1-|z|^2}{2}\cdot \frac{1}{2}\left( \frac{\partial }{\partial x}-\sqrt{-1}\frac{\partial }{\partial y}\right) +\sqrt{-1}(x-\sqrt{-1}y)\frac{\partial }{\partial t}\\&=\left( \frac{1-|z|^2}{4}\frac{\partial }{\partial x}+y\frac{\partial }{\partial t}\right) +\sqrt{-1}\left( -\frac{1-|z|^2}{4}\frac{\partial }{\partial y}+x\frac{\partial }{\partial t}\right) \end{aligned} \end{aligned}$$

(2.6)

and

$$\begin{aligned} \begin{aligned} Z_{\overline{1}}&=\frac{1-|z|^2}{2}\cdot \frac{1}{2}\left( \frac{\partial }{\partial x}+\sqrt{-1}\frac{\partial }{\partial y}\right) -\sqrt{-1}(x+\sqrt{-1}y)\frac{\partial }{\partial t}\\&=\left( \frac{1-|z|^2}{4}\frac{\partial }{\partial x}+y\frac{\partial }{\partial t}\right) +\sqrt{-1}\left( \frac{1-|z|^2}{4}\frac{\partial }{\partial y}-x\frac{\partial }{\partial t}\right) . \end{aligned} \end{aligned}$$

(2.7)

Therefore, if we write

$$\begin{aligned} e_1=\sqrt{2}\left( \frac{1-|z|^2}{4}\frac{\partial }{\partial x}+y\frac{\partial }{\partial t}\right) ~~ \text{ and } ~~e_2=\sqrt{2} \left( \frac{1-|z|^2}{4}\frac{\partial }{\partial y}-x\frac{\partial }{\partial t}\right) \end{aligned}$$

(2.8)

then it follows from (2.6)-(2.8) that

$$\begin{aligned} Z_1=\frac{1}{\sqrt{2}}(e_1-\sqrt{-1} e_2)~~ \text{ and } Z_{\overline{1}}= \frac{1}{\sqrt{2}}(e_1+\sqrt{-1}e_2), \end{aligned}$$

(2.9)

and hence,

$$\begin{aligned} J(e_1)=e_2~~ \text{ and } ~~J(e_2)=-e_1. \end{aligned}$$

(2.10)

It follows from (2.9) that

$$\begin{aligned} e_1=\frac{1}{\sqrt{2}}(Z_1+Z_{\overline{1}})~~ \text{ and } ~~ e_2=\frac{\sqrt{-1}}{\sqrt{2}}(Z_1-Z_{\overline{1}}). \end{aligned}$$

(2.11)

From this, one can easily verify that $\{e_1,e_2,T\}$ is an orthonormal basis with respect to the Levi metric $g_\theta$ by using (2.5). For example, we find

$$\begin{aligned} \begin{aligned} g_\theta (e_2,e_2)&=-\frac{1}{2}g_\theta (Z_1-Z_{\overline{1}},Z_1-Z_{\overline{1}})\\&=-\frac{1}{2}\big (g_\theta (Z_1,Z_1)-g_\theta (Z_{\overline{1}},Z_1)-g_\theta (Z_1,Z_{\overline{1}})+g_\theta (Z_{\overline{1}},Z_{\overline{1}})\big )=1, \end{aligned} \end{aligned}$$

and the others can be computed similarly. On the other hand, the contact plane $\xi$ is spanned by $e_1$ and $e_2$.

Suppose $\gamma :I\rightarrow B^1\times \mathbb {R}$ is a horizontally regular curve in $B^1\times \mathbb {R}$, given by $\gamma (\hat{s})=(z(\hat{s}),t(\hat{s}))$, which may not be parameterized by the horizontal arc length. Here, I is an interval in $\mathbb {R}$. For simplicity, we will hereafter skip writing $\hat{s}$, with the understanding that both t and z depend on $\hat{s}$. Also, the $'$ is the derivative with respect to $\hat{s}$. Then

$$\begin{aligned} \gamma '=z'\frac{\partial }{\partial z}+\overline{z}'\frac{\partial }{\partial \overline{z}} +t'\frac{\partial }{\partial t}. \end{aligned}$$

(2.12)

If we write

$$\begin{aligned} \gamma '=aZ_1+bZ_{\overline{1}}+cT \end{aligned}$$

(2.13)

for some a, b and c, then it follows from (2.12) that

$$\begin{aligned} z'\frac{\partial }{\partial z}+\overline{z}'\frac{\partial }{\partial \overline{z}} +t'\frac{\partial }{\partial t}=aZ_1+bZ_{\overline{1}}+cT. \end{aligned}$$

(2.14)

If the vector field in (2.14) acts on z, we get

$$\begin{aligned} z'=\frac{1-|z|^2}{2}a. \end{aligned}$$

(2.15)

Also, if the vector field in (2.14) acts on $\overline{z}$, we have

$$\begin{aligned} \overline{z}'=\frac{1-|z|^2}{2}b. \end{aligned}$$

(2.16)

On the other hand, if the vector field in (2.14) acts on t, we get

$$\begin{aligned} t'=\sqrt{-1}\overline{z}a-\sqrt{-1}zb+c. \end{aligned}$$

(2.17)

Substituting (2.15)-(2.17) into (2.13) yields

$$\begin{aligned} \gamma '=\frac{2z'}{1-|z|^2}Z_1+\frac{2\overline{z}'}{1-|z|^2}Z_{\overline{1}}+\left( t'-\frac{2\sqrt{-1}\overline{z}z'}{1-|z|^2} +\frac{2\sqrt{-1}z\overline{z}'}{1-|z|^2}\right) T \end{aligned}$$

(2.18)

From (2.18), we obtain

$$\begin{aligned} \gamma _{\xi }'=\frac{2z'}{1-|z|^2}Z_1+\frac{2\overline{z}'}{1-|z|^2}Z_{\overline{1}}~~ \text{ and } ~~\gamma _T'=\left( t'-\frac{2\sqrt{-1}\overline{z}z'}{1-|z|^2} +\frac{2\sqrt{-1}z\overline{z}'}{1-|z|^2}\right) T, \end{aligned}$$

(2.19)

where $\gamma '_\xi$ is the projection of $\gamma '$ on the contact plane $\xi$, and $\gamma '_T$ is the projection of $\gamma '$ on the T direction.

Let s be the horizontal arc length of $\gamma$. By definition, we have

$$\begin{aligned} \frac{ds}{d\hat{s}}=|\gamma _\xi '|, \end{aligned}$$

(2.20)

which is not zero since $\gamma$ is assumed to be horizontally regular. By (2.5), (2.19) and (2.20), we have

$$\begin{aligned} \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}\sqrt{z'\overline{z}'}}{1-|z|^2}. \end{aligned}$$

(2.21)

Let $\tilde{\gamma }(s)$ be the reparametrization of the curve $\gamma$ in terms of the horizontal arc length s. Then we have

$$\begin{aligned} \frac{d}{ds}\tilde{\gamma }(s)=\frac{d\hat{s}}{ds}\gamma '(\hat{s}). \end{aligned}$$

(2.22)

By (2.18) and (2.21), we can compute the T-variation as follows:

$$\begin{aligned} \tau =\left\langle \frac{d}{ds}\tilde{\gamma }(s),T\right\rangle =\frac{d\hat{s}}{ds} \langle \gamma '(\hat{s}),T \rangle =\frac{1-|z|^2}{2\sqrt{2}\sqrt{z'\overline{z}'}}\left( t'-\frac{2\sqrt{-1}\overline{z}z'}{1-|z|^2} +\frac{2\sqrt{-1}z\overline{z}'}{1-|z|^2}\right) \end{aligned}$$

(2.23)

Since $z=x+\sqrt{-1}y$, we obtain (1.6) in Theorem 1.4.

We now proceed to compute the p-curvature $\kappa$. To this end, note that

$$\begin{aligned} X(s):=\frac{d}{ds}\tilde{\gamma }_{\xi }(s) =\frac{d\hat{s}}{ds}\gamma '_{\xi } \end{aligned}$$

(2.24)

It follows from (2.19), (2.21) and (2.24) that

$$\begin{aligned} X(s)=\frac{1}{\sqrt{(x')^2+(y')^2}}(x'e_1+y'e_2). \end{aligned}$$

(2.25)

which together with (2.10) implies that

$$\begin{aligned} Y(s):=JX(s)=\frac{1}{\sqrt{(x')^2+(y')^2}}(x'e_2-y'e_1). \end{aligned}$$

(2.26)

By (2.20) and (2.24), we compute

$$\begin{aligned} \begin{aligned} \frac{d}{ds}X(s)&=\frac{d\hat{s}}{ds}\Bigg [\left( \frac{1}{\sqrt{(x')^2+(y')^2}}\right) '(x'e_1+y'e_2) +\frac{1}{\sqrt{(x')^2+(y')^2}}(x'e_1+y'e_2)'\Bigg ]\\&=\frac{1-x^2-y^2}{2\sqrt{2}\sqrt{(x')^2+(y')^2}}\Bigg [\left( \frac{1}{\sqrt{(x')^2+(y')^2}}\right) '(x'e_1+y'e_2)\\&\quad +\frac{1}{\sqrt{(x')^2+(y')^2}}\big (x''e_1+y''e_2+x'\nabla _{\gamma '}e_1+y'\nabla _{\gamma '}e_2\big )\Bigg ] \end{aligned} \end{aligned}$$

(2.27)

To proceed further, we have to compute $\nabla _{\gamma '}e_1$ and $\nabla _{\gamma '}e_2$. To this end, It follows from (2.1) that

$$\begin{aligned} \begin{aligned} d\theta ^1&=\frac{2z}{(1-|z|^2)^2}d\overline{z}\wedge dz=\theta ^1\wedge \left( -\frac{z}{2}\theta ^{\overline{1}}\right) ,\\ d\theta ^{\overline{1}}&=\frac{2\overline{z}}{(1-|z|^2)^2}dz\wedge d\overline{z}=\theta ^{\overline{1}}\wedge \left( -\frac{\overline{z}}{2}\theta ^1\right) . \end{aligned} \end{aligned}$$

(2.28)

The structure equations imply that (see (4.1) and (4.2) in [16])

$$\begin{aligned} d\theta ^1=\theta ^1\wedge \omega _1^1+\theta \wedge \tau ^1~~ \text{ and } ~~\omega _1^1+\omega _{\overline{1}}^{\overline{1}}=dh_{1\overline{1}} \end{aligned}$$

(2.29)

where $\omega _1^1$ is the connection 1-form and $\tau$ is the torsion form. Comparing this with (2.28) and using (2.3), we obtain

$$\begin{aligned} \omega _1^1=\frac{\overline{z}}{2}\theta ^1-\frac{z}{2}\theta ^{\overline{1}} ~~ \text{ and } ~~\tau \equiv 0. \end{aligned}$$

(2.30)

By (2.30), we find

$$\begin{aligned} \begin{aligned} \nabla _{Z_1}Z_1&=\omega _1^1(Z_1)Z_1=\frac{\overline{z}}{2}Z_1,~~ \nabla _{Z_{\overline{1}}}Z_{\overline{1}}=\overline{(\nabla _{Z_1}Z_1)}=\frac{z}{2}Z_{\overline{1}},\\ \nabla _{Z_{\overline{1}}}Z_1&=\omega _1^1(Z_{\overline{1}})Z_1=-\frac{z}{2}Z_1,~~ \nabla _{Z_1}Z_{\overline{1}}=\overline{(\nabla _{Z_{\overline{1}}}Z_1)}=-\frac{\overline{z}}{2}Z_{\overline{1}},\\ \nabla _{T}Z_1&=\omega _1^1(T)Z_1=0,~~\nabla _{T}Z_{\overline{1}} =\overline{(\nabla _{T}Z_1)}=0. \end{aligned} \end{aligned}$$

(2.31)

An alternative way to compute $\omega _1^1$ is as follows. By (2.4), we find

$$\begin{aligned} \begin{aligned} {[}Z_{\overline{1}},Z_1]&=\left[ \frac{1-|z|^2}{2}\frac{\partial }{\partial \overline{z}}-\sqrt{-1}z\frac{\partial }{\partial t},\frac{1-|z|^2}{2}\frac{\partial }{\partial z}+\sqrt{-1}\overline{z}\frac{\partial }{\partial t}\right] \\&=-\frac{1-|z|^2}{4}\cdot z\frac{\partial }{\partial z} +\frac{1-|z|^2}{4}\cdot \overline{z}\frac{\partial }{\partial \overline{z}} +\sqrt{-1}(1-|z|^2)\frac{\partial }{\partial t}\\&=-\frac{z}{2}Z_1+\frac{\overline{z}}{2}Z_{\overline{1}}+\sqrt{-1}\frac{\partial }{\partial t} \end{aligned} \end{aligned}$$

(2.32)

and

$$\begin{aligned} {[}Z_1,T]=\left[ \frac{1-|z|^2}{2}\frac{\partial }{\partial z}+\sqrt{-1}\overline{z}\frac{\partial }{\partial t},\frac{\partial }{\partial t}\right] =0. \end{aligned}$$

(2.33)

From the last three lines of P.418 in [16], we have

$$\begin{aligned} \begin{aligned} {[}Z_{\overline{1}},Z_1]&=\sqrt{-1}h_{1\overline{1}}T+\omega _1^1(Z_{\overline{1}})Z_1-\omega _{\overline{1}}^{\overline{1}}(Z_1)Z_{\overline{1}},\\ {[}Z_1,T]&={A^{\overline{1}}}_1Z_{\overline{1}}-\omega _1^1(T)Z_1. \end{aligned} \end{aligned}$$

Comparing these with (2.32) and (2.33) and using (2.3), we obtain

$$\begin{aligned} \omega _1^1(Z_{\overline{1}})=-\frac{z}{2},~~\omega _{\overline{1}}^{\overline{1}}(Z_1)=-\frac{\overline{z}}{2},~~ \text{ and } ~~ \omega _1^1(T)=0, \end{aligned}$$

which together with (2.3) and the second equation (2.29) implies that

$$\begin{aligned} \omega _1^1(Z_1)=-\omega _{\overline{1}}^{\overline{1}}(Z_1)=\frac{\overline{z}}{2}. \end{aligned}$$

All these agree with (2.30) and (2.31).

By (2.18) and (2.31), we find

$$\begin{aligned} \begin{aligned} \nabla _{\gamma '}Z_1&= \frac{2z'}{1-|z|^2}\nabla _{Z_1}Z_1+\frac{2\overline{z}'}{1-|z|^2}\nabla _{Z_{\overline{1}}}Z_1+\left( t'-\frac{2\sqrt{-1}\overline{z}z'}{1-|z|^2} +\frac{2\sqrt{-1}z\overline{z}'}{1-|z|^2}\right) \nabla _T Z_1\\&=\frac{\overline{z}z'-z\overline{z}'}{1-|z|^2}Z_1, \end{aligned} \end{aligned}$$

(2.34)

and

$$\begin{aligned} \begin{aligned} \nabla _{\gamma '}Z_{\overline{1}}&= \frac{2z'}{1-|z|^2}\nabla _{Z_1}Z_{\overline{1}}+\frac{2\overline{z}'}{1-|z|^2}\nabla _{Z_{\overline{1}}}Z_{\overline{1}}+\left( t'-\frac{2\sqrt{-1}\overline{z}z'}{1-|z|^2} +\frac{2\sqrt{-1}z\overline{z}'}{1-|z|^2}\right) \nabla _T Z_{\overline{1}}\\&=\frac{z\overline{z}'-\overline{z}z'}{1-|z|^2}Z_{\overline{1}}. \end{aligned} \end{aligned}$$

(2.35)

It follows from (2.11), (2.34) and (2.35) that

$$\begin{aligned} \nabla _{\gamma '} e_1&=\frac{1}{\sqrt{2}}(\nabla _{\gamma '}Z_1+\nabla _{\gamma '}Z_{\overline{1}})=\frac{\overline{z}z'-z\overline{z}'}{1-|z|^2}\cdot \frac{1}{\sqrt{2}}(Z_1-Z_{\overline{1}})\nonumber \\&=-\frac{\sqrt{-1}(\overline{z}z'-z\overline{z}')}{1-|z|^2}e_2=-\frac{2(yx'-xy')}{1-x^2-y^2}e_2,\nonumber \\ \nabla _{\gamma '}e_2&=\frac{\sqrt{-1}}{\sqrt{2}}(\nabla _{\gamma '}Z_1-\nabla _{\gamma '}Z_{\overline{1}})=\frac{\sqrt{-1}(\overline{z}z'-z\overline{z}')}{1-|z|^2}\cdot \frac{1}{\sqrt{2}}(Z_1+Z_{\overline{1}})\nonumber \\&=\frac{\sqrt{-1}(\overline{z}z'-z\overline{z}')}{1-|z|^2}e_1=\frac{2(yx'-xy')}{1-x^2-y^2}e_1. \end{aligned}$$

(2.36)

Substituting (2.36) into (2.27) yields

$$\begin{aligned} \begin{aligned} \frac{d}{ds}X(s)&=\frac{1-x^2-y^2}{2\sqrt{2}\sqrt{(x')^2+(y')^2}}\Bigg [\left( \frac{1}{\sqrt{(x')^2+(y')^2}}\right) '(x'e_1+y'e_2)\\&\quad +\frac{1}{\sqrt{(x')^2+(y')^2}}\left( x''e_1+y''e_2+\frac{2(yx'-xy')}{1-x^2-y^2}(y'e_1-x'e_2)\right) \Bigg ]. \end{aligned} \end{aligned}$$

(2.37)

The p-curvature is defined to be

$$\begin{aligned} \kappa :=\left\langle \frac{d}{ds}X(s),Y(s)\right\rangle . \end{aligned}$$

(2.38)

Substituting (2.26) and (2.37) into the right-hand side of (2.38), and using the fact that $\{e_1,e_2\}$ is orthonormal, we obtain (1.5) in Theorem 1.4. This finishes the proof of Theorem 1.4.

We would like to point out that it follows from the expressions in (1.6) and (1.5) in Theorem 1.4 that the T-variation $\tau$ and the p-curvature $\kappa$ are both real-valued.

2.1 Examples.

In the following, we consider some examples of curves which have constant p-curvature and constant T-variation.

(i) We consider the curve $\gamma :(-1,1)\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=(\hat{s},0,t_0) \end{aligned}$$

for some constant $t_0$. By (2.21), we have $\displaystyle \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}}{1-\hat{s}^2}>0.$ In particular, $\gamma$ is horizontally regular. It follows from (1.6) and (1.5) that both p-curvature and T-variation are zero, i.e. $\kappa =\tau \equiv 0.$ Similarly, the curve $\gamma :(-1,1)\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=(0,\hat{s},t_0), \end{aligned}$$

for some constant $t_0$, is horizontally regular and has vanishing p-curvature and T-variation.

(ii) We consider the curve $\gamma :I\rightarrow B^1\times \mathbb {R}$, where I is an interval in $(-1,1)$, given by

$$\begin{aligned} \gamma (\hat{s})=(a\hat{s},b\hat{s},t_0) \end{aligned}$$

such that $a^2+b^2\ne 0$. It follows from (2.21) that $\displaystyle \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}\sqrt{a^2+b^2}}{1-\hat{s}^2}>0,$ which implies that $\gamma$ is horizontally regular. By (1.6) and (1.5), we see that both p-curvature and T-variation are zero, i.e. $\kappa =\tau \equiv 0$.

(iii) For any constant c, we consider the curve $\gamma :(-1,1)\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=\left( \hat{s},0,\frac{c}{2}\ln \Big (\frac{1+\hat{s}}{1-\hat{s}}\Big )+\tilde{c}\right) \end{aligned}$$

where c and $\tilde{c}$ are constants. It follows from (2.21) that $\displaystyle \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}}{1-\hat{s}^2}>0,$ which shows that $\gamma$ is horizontally regular. Also, it follows from (1.6) and (1.5) that the p-curvature and T-variation are,respectively,given by $\kappa \equiv 0$ and $\tau \displaystyle \equiv \frac{c}{2\sqrt{2}}.$ That is to say, $\gamma$ has constant p-curvature and constant T-variation. Similarly, one can take

$$\begin{aligned} \gamma (\hat{s})=\left( 0,\hat{s},\frac{c}{2}\ln \Big (\frac{1+\hat{s}}{1-\hat{s}}\Big )+\tilde{c}\right) , \end{aligned}$$

which has constant p-curvature and constant T-variation.

(iv) For any fixed $|y_0|<1$, we consider the curve $\gamma :(-\sqrt{1-y_0^2},\sqrt{1-y_0^2})\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=(\hat{s},y_0,t_0) \end{aligned}$$

for some constant $t_0$. By (2.21), we have $\displaystyle \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}}{1-\hat{s}^2-y_0^2}>0,$ which implies that $\gamma$ is horizontally regular. By (1.6) and (1.5), we find that the p-curvature and T-variation are,respectively, given by $\kappa =\displaystyle -\frac{y_0}{\sqrt{2}}$ and $\tau =-\sqrt{2}y_0$. In particular, the p-curvature and T-variation are both constant. Similarly, for any fixed $|x_0|<1$, the curve $\gamma :(-\sqrt{1-x_0^2},\sqrt{1-x_0^2})\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=(x_0,\hat{s},t_0), \end{aligned}$$

for some constant $t_0$, is horizontally regular and has constant p-curvature and T-variation given by $\kappa =\displaystyle \frac{x_0}{\sqrt{2}}$ and $\tau =\sqrt{2}x_0.$

(v) For any fixed $0<r_0<1$, consider the curve $\gamma :\mathbb {R}\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=(r_0\cos (\hat{s}),r_0\sin (\hat{s}),t_1\hat{s}+t_0) \end{aligned}$$

for some constants $t_0$ and $t_1$. By (2.21), we have $\displaystyle \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}r_0}{1-r_0^2}\ne 0,$ which shows that $\gamma$ is horizontally regular. By (1.6) and (1.5), we compute $\kappa =\displaystyle \frac{1+r_0^2}{2\sqrt{2}r_0}$ and $\tau =\displaystyle \frac{1-r_0^2}{2\sqrt{2}r_0}t_1+\sqrt{2}r_0.$ In particular, $\gamma$ has constant p-curvature and T-variation.

(vi) For any fixed $0<r_0<1$, consider the curve $\gamma :\mathbb {R}\rightarrow B^1\times \mathbb {R}$ given by

$$\begin{aligned} \gamma (\hat{s})=\left( r_0\cos (\hat{s}),r_0\sin (\hat{s}),-\frac{4r_0^2}{1-r_0^2}{\hat{s}}+t_0\right) . \end{aligned}$$

It follows from (2.21) that $\displaystyle \frac{ds}{d\hat{s}}=\frac{2\sqrt{2}r_0}{1-r_0^2}\ne 0,$ i.e. $\gamma$ is horizontally regular. By (1.6) and (1.5), we compute $\kappa =\displaystyle \frac{1+r_0^2}{2\sqrt{2}r_0}$ and $\tau =0.$ In particular, $\gamma$ has constant p-curvature and zero T-variation. Therefore, $\gamma$ is an example of horizontal curve, i.e. $\tau \equiv 0$.

(vii) We have the following:

Proposition 2.1

There exist smooth closed horizontal curves in $B^1\times \mathbb {R}$.

Proof

Let a be any nonzero real number and $\epsilon \in (0,1)$. Set $r(\hat{s})=\epsilon \cos (a \hat{s})$, and $\theta (\hat{s})=(1-\epsilon ^2)\frac{\sin (a\hat{s})}{a}+\frac{\epsilon ^2 \sin ^3(a \hat{s})}{3a}$. We consider the curve $\gamma :[0, \frac{2\pi }{a}]\rightarrow B^1\times \mathbb {R}$, where $\gamma (\hat{s})=(x(\hat{s}),y(\hat{s}),t(\hat{s}))$, defined by

$$\begin{aligned} \left\{ \begin{array}{rl} x(\hat{s})&{}= r(\hat{s}) \cos (\theta (\hat{s})),\\ y(\hat{s})&{}= r(\hat{s}) \sin (\theta (\hat{s})),\\ t(\hat{s})&{}=-4\epsilon ^2\Big (\frac{\sin (a \hat{s})}{a}-\frac{\sin ^3(a\hat{s})}{3a}\Big )+c \end{array} \right. \end{aligned}$$

for any constant c. One can easily see that $\gamma$ is smooth. Also, by using (1.6), we can see that $\tau \equiv 0$, i.e. $\gamma$ is horizontal. Moreover, $\gamma (0)=\gamma (\frac{2\pi }{a})=(\epsilon ,0,c)$, which implies that $\gamma$ is closed. Finally, we claim that $\gamma$ is horizontally regular. Indeed, we observe that

$$\begin{aligned} (x')^2+(y')^2&=(r')^2+r^2(\theta ')^2\\&=a^2\epsilon ^2 \sin ^2(a\hat{s})+\epsilon ^2 \cos ^2(a \hat{s})\Big (1-\epsilon ^2 \cos ^2(a \hat{s})\Big )^2 \cos ^2(a \hat{s}), \end{aligned}$$

which is nonzero for any $\hat{s}\in [0, \frac{2\pi }{a}]$ since $\sin ^2(a \hat{s})$ and $\cos ^2(a\hat{s})$ cannot vanish at the same time for any $\hat{s}$. This together with (2.20) and (2.21) implies that $\gamma$ is horizontally regular. This proves Proposition 2.1. $\square$

In Fig. 1, we show an example of the curves appeared in Proposition 2.1 for $a=0.3, \epsilon =0.6, c=0$.

Fig. 1

(1): an example of closed horizontal curves $\gamma$ in $B^1\times \mathbb {R}$ (see Proposition 2.1). The different viewpoints of $\gamma$ based on their projections: (2) on the yz-plane, (3) on the xz-plane, and (4) on the xy-plane

The following lemma gives a characterization of the curve given in Example (ii) above.

Lemma 2.2

Suppose that $\gamma$ is a horizontally regular curve given by

$$\begin{aligned} \gamma (\hat{s})=(x(\hat{s}),y(\hat{s}),t_0) \end{aligned}$$

for some fixed $t_0$ such that it does not touch the x-axis (or y-axis) and its T-variation is zero. Then

$$\begin{aligned} \gamma (\hat{s})=(ax(\hat{s}),bx(\hat{s}),t_0) \end{aligned}$$

(2.39)

for some constants a and b with $a^2+b^2\ne 0$, and its p-curvature is also zero.

Proof

Since $\tau \equiv 0$ and $t\equiv t_0$, it follows from (1.6) that

$$\begin{aligned} xy'-yx'=0. \end{aligned}$$

(2.40)

If $\gamma$ does not touch the x-axis, then $x\ne 0$ and y/x is well defined. In particular, it follows from (2.40) that $(y/x)'=0$. Integrating it gives $y(\hat{s})=c_0x(\hat{s})$ for some constant $c_0$. Similarly, if $\gamma$ does not touch the y-axis, then $y\ne 0$ and x/y is well defined. By (2.40), $(x/y)'=0$, which gives $x(\hat{s})=c_0y(\hat{s})$ for some constant $c_0$. This proves (2.39). Note that $a^2+b^2\ne 0$; otherwise, $\gamma$ would not be horizontally regular. It now follows from (1.5) and (2.39) that p-curvature of $\gamma$ is zero. This proves the assertion. $\square$

We also have the following:

Lemma 2.3

Suppose that $\gamma (s)=(x(s),y(x),t(s))$ is a curve parameterized by the horizontal arc length. If the T-variation of $\gamma$ is equal to $t'$ such that $\gamma$ does not touch the x-axis (or y-axis), then (2.39) holds and the p-curvature is zero.

Proof

Since $\gamma$ is parameterized by the horizontal arc length, we have

$$\begin{aligned} \frac{2\sqrt{2}\sqrt{(x')^2+(y')^2}}{1-x^2-y^2}=1 \end{aligned}$$

(2.41)

by (2.21). Therefore, it follows from (1.6) and (2.41) that

$$\begin{aligned} \tau =t'+\frac{4(xy'-yx')}{1-x^2-y^2}. \end{aligned}$$

Combining this with the assumption that $\tau =t'$, we get (2.40). Now the rest of the proof follows from the proof of Lemma 2.2. $\square$

We have the following:

Lemma 2.4

Suppose that $\gamma$ is a horizontally regular curve given by

$$\begin{aligned} \gamma (\hat{s})=(x(\hat{s}),y_0,t(\hat{s})) \end{aligned}$$

(2.42)

for some constant $y_0$. If the T-variation of $\gamma$ is constant, then

$$\begin{aligned} \gamma (\hat{s})=\left( x(\hat{s}),y_0,c_0\ln \Big (\frac{1+x(\hat{x})}{1-x(\hat{x})}\Big )+c_1\right) \end{aligned}$$

(2.43)

for some constants $c_0$ and $c_1$, and the p-curvature of $\gamma$ is constant.

Proof

By (2.20), (2.21), (2.42) and the assumption that $\gamma$ is horizontally regular, we find $\displaystyle |\gamma '_\xi |=\frac{2\sqrt{2}|x'|}{1-x^2}\ne 0.$ In particular, $|x'|\ne 0$, which gives $x'/|x'|=\epsilon$ where $\epsilon =\pm 1$. Also, by (1.6) and (2.42), we can compute

$$\begin{aligned} \tau =\frac{1-x^2}{2\sqrt{2}|x'|}\left( t'-\frac{4y_0x'}{1-x^2}\right) =\frac{1-x^2}{2\sqrt{2}|x'|}t'-\sqrt{2}\epsilon . \end{aligned}$$

Since $\tau$ is constant by assumption, we therefore have $\displaystyle t'=\frac{cx'}{1-x^2}$ for some constant c. Integrating it with respect to $\hat{s}$ yields (2.43). Finally it follows from (1.5) and (2.42) that the p-curvature is $\kappa =\displaystyle \frac{1-x^2}{2\sqrt{2}|x'|^3}\left( -\frac{2y_0(x')^3}{1-x^2}\right) ,$ which is constant. $\square$

The following lemma can be proved in the same way as in the proof of Lemma 2.4, and its proof is omitted.

Lemma 2.5

Suppose that $\gamma$ is a horizontally regular curve given by

$$\begin{aligned} \gamma (\hat{s})=(x_0,y(\hat{s}),t(\hat{s})) \end{aligned}$$

for some constant $x_0$. If the T-variation of $\gamma$ is constant, then

$$\begin{aligned} \gamma (\hat{s})=\left( x_0,y(\hat{s}),c_0\ln \Big (\frac{1+y(\hat{x})}{1-y(\hat{x})}\Big )+c_1\right) \end{aligned}$$

(2.44)

for some constants $c_0$ and $c_1$, and the p-curvature of $\gamma$ is constant.

Note that the curves given in Example (iv) are Lemma 2.4 and Lemma 2.5 by taking $c_0=0$. Moreover, the first curve in Example (i) (and the second curve in Example (i), respectively) is nothing but Lemma 2.4 with $x_0=0$, $y(\hat{s})=\hat{s}$ and $c_0=0$ (and Lemma 2.5 by taking $y_0=0$, $x(\hat{s})=\hat{s}$ and $c_0=0$, respectively). Furthermore, by taking $x(\hat{s})=\hat{s}$ and $y_0=0$ in Lemma 2.4 (and $y(\hat{s})=\hat{s}$ and $x_0=0$ in Lemma 2.5, respectively), we obtain the first curve (the second curve, respectively) in Example (iii) above. On the other hand, the following lemma gives a characterization of the curve given in Example (v) above.

Lemma 2.6

Suppose that $\gamma$ is a horizontally regular curve given by

$$\begin{aligned} \gamma (\hat{s})=(r_0\cos (\hat{s}),r_0\sin (\hat{s}),t(\hat{s})) \end{aligned}$$

for some fixed $0<r_0<1$. If $\gamma$ has constant T-variation, then

$$\begin{aligned} \gamma (\hat{s})=\left( r_0\cos (\hat{s}),r_0\sin (\hat{s}),\frac{2\sqrt{2}r_0}{1-r_0^2}c_0\hat{s}+c_1\right) \end{aligned}$$

(2.45)

for some constants $c_0$ and $c_1$, and its p-curvature is constant.

Proof

Since $\gamma (\hat{s})=(r_0\cos (\hat{s}),r_0\sin (\hat{s}),t(\hat{s}))$, it follows from (1.6) that the T-variation of $\gamma$ is given by

$$\begin{aligned} \tau =\frac{1-r_0^2}{2\sqrt{2}r_0}t'. \end{aligned}$$

(2.46)

Since $\tau$ is constant by assumption, i.e. $\tau \equiv c_0$ for some constant $c_0$, integrating (2.46) with respect to $\hat{s}$ yields $t=\displaystyle \frac{2\sqrt{2}r_0}{1-r_0^2}c_0\hat{s}+c_1$, which proves (2.45). It follows from (1.5) and (2.45) that the p-curvature of $\gamma$ is constant, which proves the lemma. $\square$

Consider the curve in $B^1\times \mathbb {R}$ when its projection on $B^1$ is a polar curve, i.e.

$$\begin{aligned} \gamma (\theta )=\big (r(\theta )\cos \theta ,r(\theta )\sin \theta , t(\theta )\big ), \end{aligned}$$

(2.47)

i.e. $x(\theta )=r(\theta )\cos \theta$ and $y(\theta )=r(\theta )\sin \theta$. A direct computation gives

$$\begin{aligned} \begin{aligned} x'&=r'\cos \theta -r\sin \theta ~~ \text{ and } ~~y'=r'\sin \theta +r\cos \theta ,\\ x''&=(r''-r)\cos \theta -2r'\sin \theta ~~ \text{ and } ~~y''=(r''-r)\sin \theta +2r'\cos \theta . \end{aligned} \end{aligned}$$

Here $'$ represents the derivative with respect to $\theta$. This together (1.6) and (1.5) gives the following:

Proposition 2.7

Suppose $\gamma$ is a curve in $B^1\times \mathbb {R}$ such that its projection on $B^1$ is a polar curve, i.e. (2.47) holds. Then its T-variation and p-curvature of $\gamma$ are,respectively, given by

$$\begin{aligned} \begin{aligned} \tau&=\frac{1-r^2}{2\sqrt{2}\sqrt{(r')^2+r^2}}\left( t'-\frac{4r^2}{1-r^2}\right) ,\\ \kappa&=\frac{1-r^2}{2\sqrt{2}((r')^2+r^2)^{\frac{3}{2}}} \left( 2(r')^2-rr''+r^2+\frac{2r^2((r')^2+r^2)}{1-r^2}\right) . \end{aligned} \end{aligned}$$

If r is constant and t is a linear function in $\theta$, i.e. $r(\theta )=r_0$ and $t(\theta )=t_1\theta +t_0$ for some constants $r_0, t_0, t_1$, then it follows from Proposition 2.7 that the p-curvature and T-variation of $\gamma$ are constant. This recovers the Examples (v) and (vi) above. On the other hand, from Proposition 2.7, we see that $\gamma$ is horizontal, i.e. $\tau =0$ if and only if $t'=\displaystyle \frac{4r^2}{1-r^2}$. Integrating it yields $t(\theta )=\displaystyle \int ^{\theta }_{\theta _0}\frac{4r^2}{1-r^2}+C$ for some constant C. This gives the following:

Corollary 2.8

Suppose $\gamma$ is a curve in $B^1\times \mathbb {R}$ such that its projection on $B^1$ is a polar curve, i.e. (2.47) holds. Then $\gamma$ is horizontal if and only if $t(\theta )=\displaystyle \int ^{\theta }_{\theta _0}\frac{4r^2}{1-r^2}+C.$

3 Fundamental Theorem of Curves in $B^1\times \mathbb {R}$

In this section, we first study pseudohermitain transformations on $B^1\times \mathbb {R}$. By pseudohermitain transformation, we mean a diffeomorphism $\psi$ on $B^1\times \mathbb {R}$ which satisfies $\phi ^*\theta =\theta$ and $\phi _*J=J\phi _*$, where $\theta$ is defined as in (1.3). The set of all pseudohermitian transformations is denoted by PSH(1) and any element in PSH(1) is called a symmetry.

It follows from Schwarz-Pick Lemma that any isometry of $B^1$ with respect to the Poincarè metric $\displaystyle \frac{|dz|^2}{(1-|z|^2)^2}$ must be Möbius transformation, and any Möbius transformation $\varphi$ of $B^1$ can be written as

$$\begin{aligned} \varphi (z)=e^{\sqrt{-1}\theta }\frac{z-a}{1-\overline{a}z} \end{aligned}$$

for some $\theta \in \mathbb {R}$ and $a\in B^1$. Hence, any Möbius transformation $\varphi$ induces a diffeomorphism $\hat{\varphi }$ on $B^1\times \mathbb {R}$ given by

$$\begin{aligned} \hat{\varphi }(z,t)=(\varphi (z),t)~~ \text{ for } \text{ any } (z,t)\in B^1\times \mathbb {R}. \end{aligned}$$

(3.1)

We wonder when such diffeomorphism is a pseudohermitain transformation. The following lemma, which answers the question above, gives us the first example of symmetries in PSH(1), the rotations in $B^1$ about the origin.

Lemma 3.1

The transformation $\hat{\varphi }$ defined in (3.1) is a pseudohermitian transformation only when $\varphi$ is the rotation at the origin, i.e. $\varphi (z)= e^{\sqrt{-1}\theta }z$.

Proof

If we write $\varphi (z)=e^{\sqrt{-1}\theta }\psi _a(z)$, where $\psi _a(z):=\displaystyle \frac{z-a}{1-\overline{a}z}$, then we have

$$\begin{aligned} \begin{aligned} \hat{\varphi }^*\theta&=dt+\frac{2\sqrt{-1}}{1-|\varphi (z)|^2}\left( \varphi (z)d\overline{\varphi (z)}-\overline{\varphi (z)}d\varphi (z)\right) \\&=dt+\frac{2\sqrt{-1}}{1-|\psi _a(z)|^2}\left( \psi _a(z)d\overline{\psi _a(z)}-\overline{\psi _a(z)}d\psi _a(z)\right) . \end{aligned} \end{aligned}$$

(3.2)

A direct computation shows

$$\begin{aligned} |\psi _a(z)|^2=\psi _a(z)\overline{\psi _a(z)}=\frac{|z|^2+|a|^2-a\overline{z}-\overline{a}z}{1+|a|^2|z|^2-a\overline{z}-\overline{a}z} \end{aligned}$$

which implies that

$$\begin{aligned} 1-|\psi _a(z)|^2=\frac{(1-|a|^2)(1-|z|^2)}{|1-\overline{a}z|^2}. \end{aligned}$$

(3.3)

A direct computation also shows

$$\begin{aligned} d\psi _a(z)=\frac{1-|a|^2}{(1-\overline{a}z)^2}dz, \end{aligned}$$

(3.4)

Substituting (3.3) and (3.4) into (3.2), we obtain

$$\begin{aligned} \hat{\varphi }^*\theta =dt+\frac{2\sqrt{-1}}{1-|z|^2}\left( \frac{z-a}{1-a\overline{z}}d\overline{z}-\frac{\overline{z}-\overline{a}}{1-\overline{a}z}dz\right) . \end{aligned}$$

(3.5)

Therefore, if $\psi ^*\theta =\theta$, it follows from (1.3) and (3.5) that

$$\begin{aligned} \frac{z-a}{1-a\overline{z}}d\overline{z}-\frac{\overline{z}-\overline{a}}{1-\overline{a}z}dz=zd\overline{z}-\overline{z}dz. \end{aligned}$$

This implies that $\displaystyle \frac{z-a}{1-a\overline{z}}=z,$ or equivalently, $a(1-|z|^2)=0$. Since $|z|^2<1$, we must have $a=0$, which proves the assertion. $\square$

Next, the following argument gives us the second example of symmetries in PSH(1) in $B^1\times \mathbb {R}$. Suppose $a=u+iv$, $z=x+iy \in B^1$ and $s,t\in \mathbb {R}$. For any $(a,s)\in B^1\times \mathbb {R}$, define the map $\hat{\varphi }_{(a,s)}:B^1\times \mathbb {R}\rightarrow B^1\times \mathbb {R}$ by

$$\begin{aligned} \hat{\varphi }_{(a,s)}(z, t)=\Big (\frac{z+a}{1+z\overline{a}}, s+t- \frac{4}{1-|z|^2}Im (z\overline{a})\Big ). \end{aligned}$$

(3.6)

Notice that the first component on the right of (3.6) is the Möbius transformation $\varphi _{-a}(z)$ defined above. Also, the inverse and the identity map of ${\hat{\varphi }}_{(a,s)}$ are $\big (\hat{\varphi }_{(a,s)}\big )^{-1}=\hat{\varphi }_{(-a,-s)}$ and ${\hat{\varphi }}_{(0,0)}$,respectively. Therefore, with the compositions of functions as the associative law, the set $\{\hat{\varphi }_{(a,s)}, (a,s)\in B^1\times \mathbb {R}\}$ forms a group.

Recall that given a Lie group $(G, \circ )$ with the group action $\circ$, there exists a natural left translation $L_p:G\rightarrow G$ defined by $L_p(g)=p\circ g$ for any $p\in G$. A left invariant vector field v in G is defined to be $\big ({L_{p^{-1}}}_* \big )v(p)=v(0)$ for any $p\in G$, where $0\in G$ is the identity. However, since the Möbius transformation $\displaystyle z\mapsto \frac{z+a}{1+z\overline{a}}$ is neither commutative nor associative (see [10] for example), the natural map

$$\begin{aligned} \begin{array}{rrrcl} \big (B^1\times \mathbb {R} \big ) \times \big (B^1\times \mathbb {R}\big ) &{} \rightarrow &{} B^1\times \mathbb {R}\\ \big ((a,s) , (z,t)\big )\quad &{} \mapsto &{}{\hat{\varphi }}_{(a,s)}(z,t) \end{array} \end{aligned}$$

(3.7)

dose not form a group action, and so $B^1\times \mathbb {R}$ is not a Lie group under the operation (3.7). Therefore, the concept for left invariant vector fields does not work here. Nevertheless, Proposition 3.2 shows that the vector fields $e_1, e_2$ defined in (2.8) and $T=\frac{\partial }{\partial t}$ all are the invariant in $B^1\times \mathbb {R}$ under the pushforward map $\hat{\varphi }_*$. As a consequence, the map ${\hat{\varphi }}$ defined by (3.6) is the second example of symmetries in PSH(1).

Proposition 3.2

The vector fields $e_1, e_2$ defined in (2.8), and $e_3:=T=\frac{\partial }{\partial t}$ are invariant in the sense that $\big ({\hat{\varphi }}_{-p}\big )_* e_i(p)=e_i(0)$ for $i=1,2,3$.

Proof

Let $p=(z, t)=(x+iy, t)$ be any point in $B^1\times \mathbb {R}$ and set $h(z)=\frac{4}{1-|z|^2}$. Consider the curve

$$\begin{aligned} \gamma (\alpha )=\Big (z+ \frac{\sqrt{2}}{h(z)}\alpha , t+\sqrt{2}y\alpha \Big )\in B^1\times \mathbb {R}, \end{aligned}$$

which realizes $e_1$ at p, namely, $\gamma (0)=p$ and $\gamma '(0)=e_1(p)$. It follows from (3.6) that

$$\begin{aligned} \hat{\varphi }_{-p}\big (\gamma (\alpha )\big )=\left( \frac{ \frac{\sqrt{2}}{h(z)}\alpha }{1+(\frac{\sqrt{2}}{h(z)}\alpha +z)(-\overline{z})}, 0\right) , \end{aligned}$$

so its derivative at $\alpha =0$ gives

$$\begin{aligned} \big ({\hat{\varphi }}_{-p}\big )_* e_1(p)=\frac{d}{d\alpha }\Big |_{\alpha =0}\Big (\hat{\varphi }_{-p}\big (\gamma (\alpha )\big )\Big )=(\frac{\sqrt{2}}{4}, 0)=e_1(0). \end{aligned}$$

This implies that $e_1$ is invariant. One can check that $e_2$ is invariant in a similar way. Similarly, we consider the curve $\gamma (\alpha )=(x+iy, t+\alpha )$ in $B^1\times \mathbb {R}$ realizing the vector field $T=\frac{\partial }{\partial t}$ at p such that $\gamma (0)=p$ and $\gamma '(0)=T$. Note that ${\hat{\varphi }}\big (\gamma (\alpha )\big )=\big (0, \alpha +h(z) Im(z\overline{z})\big )$. Thus,

$$\begin{aligned} \big ({\hat{\varphi }}_{-p}\big )_* T(p)=\frac{d}{d \alpha }\Big |_{\alpha =0} {\hat{\varphi }}\big (\gamma (\alpha )\big )=(0,1)=\frac{\partial }{\partial t}, \end{aligned}$$

and the result follows. $\square$

We see in the following lemma that the p-curvature and T-variation are invariant under the pseudohermitian transformations given in Lemma 3.1.

Lemma 3.3

There holds

$$\begin{aligned} \kappa (\varphi (z),t)=\kappa (z,t)~~ \text{ and } ~~\tau (\varphi (z),t)=\tau (z,t) \end{aligned}$$

for any $(z,t)\in B^1\times \mathbb {R}$ and $\varphi (z)=e^{\sqrt{-1}\theta }z$.

Proof

It follows from (1.6) and (1.5) that

$$\begin{aligned} \tau =\frac{1-|z|^2}{2\sqrt{2}\sqrt{z'\overline{z}'}}\left( t'+\frac{4\text{ Im }(\overline{z}z')}{1-|z|^2}\right) \end{aligned}$$

(3.8)

and

$$\begin{aligned} \kappa =\frac{1-|z|^2}{2\sqrt{2}\big (z'\overline{z}'\big )^{\frac{3}{2}}} \left( \text{ Im }(\overline{z}'z'')+\frac{2z'\overline{z}'\text{ Im }(\overline{z}z')}{1-|z|^2}\right) . \end{aligned}$$

(3.9)

From (3.8) and (3.9), one can easily see that the p-curvature $\kappa$ and the T-variation $\tau$ are invariant under the map $z\mapsto e^{\sqrt{-1}\theta }z$. $\square$

Using Lemma 3.1, we prove the following:

Theorem 3.4

Suppose F is a pseudohermitian transformation in the form $F(z,t)=(f(z),t)$ such that $F(0,0)=(0,0)$. Then f is the rotation at the origin, i.e. $f(z)= e^{\sqrt{-1}\theta }z$.

Proof

Note that $f:B^1\rightarrow B^1$ is a biholomorphism. In particular, we have

$$\begin{aligned} \frac{\partial f}{\partial \overline{z}}=0. \end{aligned}$$

(3.10)

This together with (1.3) implies that

$$\begin{aligned} \begin{aligned} F^*\theta&=dt+\frac{2\sqrt{-1}}{1-|f(z)|^2}\left( f(z)d\overline{f(z)}-\overline{f(z)}df(z)\right) \\&=dt+\frac{2\sqrt{-1}}{1-|f(z)|^2}\left( f(z)\overline{\left( \frac{\partial f}{\partial z}\right) }d\overline{z}-\overline{f(z)}\frac{\partial f}{\partial z}dz\right) . \end{aligned} \end{aligned}$$

(3.11)

Since F is a pseudohermitian transformation, we have $F^*\theta =\theta$. This together with (1.3) and (3.11) implies that

$$\begin{aligned} \frac{2\sqrt{-1}}{1-|f(z)|^2}\left( f(z)\overline{\left( \frac{\partial f}{\partial z}\right) }d\overline{z}-\overline{f(z)}\frac{\partial f}{\partial z}dz\right) =\frac{2\sqrt{-1}}{1-|z|^2}(zd\overline{z}-\overline{z}dz), \end{aligned}$$

which gives $\displaystyle \frac{1}{{1-|f(z)|^2}}\overline{f(z)}\frac{\partial f}{\partial z}=\overline{z}.$ Differentiating it with respect to $\overline{z}$ and using (3.10), we obtain

$$\begin{aligned} \begin{aligned} 1&=\frac{1}{{1-|f(z)|^2}}\overline{\left( \frac{\partial f}{\partial z}\right) }\left( \frac{\partial f}{\partial z}\right) +\frac{1}{(1-|f(z)|^2)^2}\overline{f(z)}\frac{\partial f}{\partial z}\cdot f(z)\overline{\left( \frac{\partial f}{\partial z}\right) }\\&=\frac{1}{(1-|f(z)|^2)^2}\left| \frac{\partial f}{\partial z}\right| ^2. \end{aligned} \end{aligned}$$

This together with the fact that $f:B^1\rightarrow B^1$ is a biholomorphism implies that f isometry of $B^1$ with respect to the Poincarè metric $\displaystyle \frac{|dz|^2}{(1-|z|^2)^2}$. In particular, f must be Möbius transformation. In other words, $F(z,t)=(f(z),t)$ where f is a Möbius transformation. Now the assertion follows immediately from Lemma 3.1. $\square$

Theorem 3.5

The group of pseudohermitian transformation PSH(1) in $B^1\times \mathbb {R}$ consists of the functions of the form ${\hat{\varphi }}_p\circ {\hat{\varphi }}_\theta$, a translation followed by a rotation about the origin, where ${\hat{\varphi }}_\theta ={\hat{\varphi }}_\theta (z,t)=(e^{\sqrt{-1}\theta } z,t)$ for some real number $\theta$ and the translation ${\hat{\varphi }}_p(z,t)$ is defined by (3.6) for some point $p\in B^1\times \mathbb {R}$.

Proof

On the one hand, by Lemma 3.1 and Proposition 3.2, we have known that ${\hat{\varphi }}_\theta$ and ${\hat{\varphi }}_p$ both are in PSH(1). On the other hand, we claim that any symmetry in PSH(1) must be of the form ${\hat{\varphi }}_p\circ {\hat{\varphi }}_\theta$. Indeed, suppose $F\in PSH(1)$ is any symmetry such that $F(0)=p$ for some point $p\in B^1\times \mathbb {R}\setminus \{(0,0)\}$. Then the composition ${\hat{\varphi }}_{-p}\circ F$ is in PSH(1) fixing the origin of $B^1\times \mathbb {R}$. According to Theorem 3.4, ${\hat{\varphi }}_{-p}\circ F$ must be a rotation about the origin, namely, ${\hat{\varphi }}_{-p}\circ F={\hat{\varphi }}_\theta$ for some real $\theta$. Thus, $F=({\hat{\varphi }}_{-p})^{-1}\circ {\hat{\varphi }}_\theta = {\hat{\varphi }}_{p}\circ {\hat{\varphi }}_\theta$ and the result follows. $\square$

Let p be any point in $B^1\times \mathbb {R}$ and $\{X,Y,T\}$ be any orthonormal basis at p such that $Y=JX$ and $T=\frac{\partial }{\partial t}$. Since $\nabla T=0$, the structural equations are of the form

$$\begin{aligned} \begin{array}{rlll} dp&{}=\omega ^1 X &{}+ \omega ^2 Y &{}+\omega ^3 T, \\ dX&{}= &{}+ \omega ^2_1Y&{}+\omega ^3_1 T, \\ dY&{}= \omega ^1_2 X&{} &{} + \omega ^3_2 T,\\ dT&{}=0, \end{array} \end{aligned}$$

(3.12)

for some vector-valued 1-forms $\omega ^i$ and $\omega ^j_k$, $1\le i,j,k\le 3$, with $\omega ^j_k=-\omega ^k_j$. For any horizontally regular curve $\gamma$ parameterized by horizontal arc length s, the tangent vector along $\gamma$ always has the decomposition $\frac{d}{ds}\gamma :=\gamma '_\xi (s) +\gamma '_T(s)$, where $\gamma '_\xi (s)\in \xi$ and $\gamma '_T(s)\in \text{ span}_{\mathbb {R}}\{T\}$. Thus, when restricting the frame $\{p, X(p), Y(p), T(p)\}$ on $\gamma$ with $X=\gamma '_\xi$ and $Y=JX$ and using (2.38), one has

$$\begin{aligned} \omega ^1=ds, \ \ \omega ^2=\kappa ds,\ \ \omega ^3=\tau ds. \end{aligned}$$

(3.13)

Furthermore, we have

$$\begin{aligned} \omega ^2_1=\langle dX, Y\rangle =\langle \omega \otimes Y, Y\rangle =\Big \langle \frac{d}{ds}X, Y\Big \rangle =\kappa ds. \end{aligned}$$

(3.14)

By (2.37), we have

$$\begin{aligned} \begin{aligned} \omega ^3_1&= \langle dX, T\rangle =\Big \langle \frac{d}{ds}X, T\Big \rangle ds=0,\\ \omega ^3_2&=\langle dT, T\rangle = \Big \langle \frac{d}{ds}Y, T\Big \rangle ds =\Big \langle \frac{d}{ds}JX, T\Big \rangle =\Big \langle J\frac{d}{ds}X, T\Big \rangle = 0. \end{aligned} \end{aligned}$$

(3.15)

Combining (3.13)-(3.15) with (3.12), we have reached the structural equations

$$\begin{aligned} \begin{array}{rlll} d\gamma &{}= X ds&{} &{}+\tau T ds, \\ dX&{}= &{} \kappa Y ds, &{} \\ dY&{}= -\kappa X ds,&{} &{}\\ dT&{}=0. \end{array} \end{aligned}$$

(3.16)

Equivalently, we have the following Frenet-type frame formula:

$$\begin{aligned} \begin{array}{rlll} \frac{d\gamma }{ds}&{}= X &{} &{}+\tau T, \\ \frac{dX}{ds}&{}= &{} \kappa Y, &{} \\ \frac{dY}{ds}&{}= -\kappa X ,&{} &{}\\ \frac{dT}{ds}&{}=0. \end{array} \end{aligned}$$

(3.17)

We are now ready to prove Theorem 1.5, the Fundamental Theorem of Curves in $\mathbb {B}^1\times \mathbb {R}$.

Proof of Theorem 1.5

First, we prove the existence. Let $X(s)=(X_1(s), X_2(s), X_3(s))^t$ and $Y(s)=(Y_1(s), Y_2(s), Y_3(s))^t$ be two vectors in $\mathbb {B}^1\times \mathbb {R}$, where $(\cdots )^t$ is the transpose of the vector $(\cdots )$. Given the function $\kappa (s)$, the following first-order ordinary differential equation system locally has the unique solution X(s), Y(s):

$$\begin{aligned} \begin{bmatrix} 0 &{} 0 &{} 0 &{} \kappa &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} \kappa &{} 0 \\ 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \kappa \\ -\kappa &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} \kappa &{} 0 &{} 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} \kappa &{} 0 &{} 0 &{} 0 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \\ Y_1 \\ Y_2 \\ Y_3 \end{bmatrix}= \frac{d}{ds}\begin{bmatrix} X_1 \\ X_2 \\ X_3 \\ Y_1 \\ Y_2 \\ Y_3 \end{bmatrix}; \end{aligned}$$

(3.18)

provided the initial condition X(0) and Y(0) satisfying $\langle X(0), Y(0)\rangle =0$. Note that the system is equivalent to $\displaystyle \frac{dX}{ds}=\kappa Y$ and $\displaystyle \frac{dY}{ds}=-\kappa X$ given in (3.17).

Define the curve $\gamma (s)=\displaystyle \int _{I'}\big ( X(u)+\tau (u)T\big ) du$ on some subinterval $I'\subset I$ containing 0. Clearly, $\displaystyle \frac{d\gamma (s)}{ds}=X(s)+\tau (s) T$. Hence, by definitions of the p-curvature and T-variation in (2.38) and (2.23), we have

$$\begin{aligned} \Big \langle \frac{d\gamma }{ds}, T \Big \rangle =\tau ~~ \text{ and } ~~ \Big \langle \frac{dX}{ds}, Y \Big \rangle =\langle \kappa Y, Y\rangle = \kappa , \end{aligned}$$

as required.

Next, we prove the uniqueness. Suppose $\gamma$ and $\tilde{\gamma }$ both are the curves defined on the interval I mentioned above with the same geometric quantities $\kappa$ and $\tau$. The natural decomposition of the curves gives

$$\begin{aligned} \frac{d\gamma }{ds}= X_\gamma +\tau T ~~ \text{ and } ~~ \frac{d{\tilde{\gamma }}}{ds}=X_{\tilde{\gamma }}+\tau T, \end{aligned}$$

where the vector fields $X_\gamma =X_\gamma (s)$ and $X_{\tilde{\gamma }}=X_{\tilde{\gamma }}(s)$ are contained in the contact plane $\xi _{\gamma (s)}$ and $\xi _{\tilde{\gamma }(s)}$. Denote $Y=JX_\gamma$ and $Y_{\tilde{\gamma }}=JX_{\tilde{\gamma }}$. Without loss of generality, we may assume that $\gamma$ and ${\tilde{\gamma }}$ coincide at $s=0$ and have the same orthonormal frame there, namely, $\gamma (0)={\tilde{\gamma }}(0)$, $X_\gamma (0)=X_{\tilde{\gamma }}(0)$, and $Y_\gamma (0)=Y_{\tilde{\gamma }}(0)$. This can always be done by using the pseudohermitian transformation (namely, the rotations in Lemma 3.1 and translations (3.7)) on both curves. Consider the function

$$\begin{aligned} A(s)=\langle X_\gamma (s), X_{\tilde{\gamma }}(s)\rangle + \langle Y_\gamma (s), Y_{\tilde{\gamma }}(s) \rangle . \end{aligned}$$

Note that $A(0)=2$. Taking the derivative yields

$$\begin{aligned} \frac{d}{ds}A&=\Big \langle \frac{dX_\gamma }{ds}, X_{\tilde{\gamma }}\Big \rangle +\Big \langle X_\gamma , \frac{dX_{\tilde{\gamma }}}{ds}\Big \rangle +\Big \langle \frac{dY_\gamma }{ds}, Y_{\tilde{\gamma }}\Big \rangle +\Big \langle Y_\gamma , \frac{dY_{\tilde{\gamma }}}{ds}\Big \rangle \\&=\langle \kappa Y_\gamma , X_{\tilde{\gamma }}\rangle +\langle X_\gamma , \kappa Y_{\tilde{\gamma }}\rangle +\langle -\kappa X_\gamma , Y_{\tilde{\gamma }}\rangle +\langle Y_\gamma , -\kappa X_{\tilde{\gamma }}\rangle =0 \end{aligned}$$

for any $s \in I$. Therefore, $A(s)=2$ is a constant function. Moreover, since the metric $\langle \cdot , \cdot \rangle$ is Hermitian, we have $\langle Y_\gamma , Y_{\tilde{\gamma }}\rangle =\langle JX_\gamma , JX_{\tilde{\gamma }}\rangle = \langle X_\gamma , X_{\tilde{\gamma }}\rangle$. Thus, $A(s)=2\langle X_\gamma , X_{\tilde{\gamma }}\rangle =2$, which implies that $X_\gamma =X_{\tilde{\gamma }}$ in the interval I, and hence $Y_\gamma =Y_{\tilde{\gamma }}$ in I. By the decomposition of $\gamma$ and ${\tilde{\gamma }}$, we conclude that $\displaystyle \frac{d\gamma }{ds}=\frac{d{\tilde{\gamma }}}{ds}$ in I.

Finally, the uniqueness of the ODE (3.18) implies that $\gamma (s)=\tilde{\gamma }(s)$, and this completes the proof. $\square$

4 Horizontal Curves

Suppose $\gamma :I\rightarrow B^1\times \mathbb {R}$ is a horizontally regular curve, given by $\gamma (\hat{s})=(z(\hat{s}),t(\hat{s}))$, which is not necessarily parameterized by the horizontal arc length. We say that $\gamma$ is horizontal, if the T-variation of $\gamma$ satisfies $\tau \equiv 0$. We have the following:

Theorem 4.1

Suppose that $\gamma :[a,b]\rightarrow B^1\times \mathbb {R}$ is a horizontal curve such that the projection ${\hat{\gamma }}$ of $\gamma$ onto $B^1$ is a simple closed curve in $B^1$. Then the area of the region bounded by ${\hat{\gamma }}$ with respect to the Poincaré metric is equal to $\displaystyle \frac{1}{8}|t(b)-t(a)|$.

Proof

Since $\gamma$ is horizontal, $\tau \equiv 0$. Hence, it follows from (1.6) that

$$\begin{aligned} \frac{t'}{4}=\frac{yx'-xy'}{1-x^2-y^2}~~ \text{ for } \text{ all } \hat{s}\in [a,b]. \end{aligned}$$

Integrating it from $\hat{s}=a$ to $\hat{s}=b$ yields

$$\begin{aligned} \frac{1}{4}\big (t(b)-t(a)\big )=\int _a^b\frac{yx'-xy'}{1-x^2-y^2}d\hat{s}. \end{aligned}$$

(4.1)

Note that the right-hand side of (4.1) is equal to the line integral $\pm \displaystyle \int _{\hat{\gamma }}\omega$, (the sign depends on the orientation of ${\hat{\gamma }}$), where $\omega$ is a 1-form given by $\omega =\displaystyle \frac{ydx-xdy}{1-x^2-y^2}$. By Stokes’ Theorem, we have

$$\begin{aligned} \int _{\hat{\gamma }}\omega =\int _{R}d\omega =-2\int _R\frac{dxdy}{(1-x^2-y^2)^2}, \end{aligned}$$

(4.2)

where R is the region bounded by $\hat{\gamma }$ in $B^1$. Note that the area of R with respect to the Poincaré metric $\displaystyle \frac{|dz|^2}{(1-|z|^2)^2}$ is equal to $\displaystyle \int _R\frac{dxdy}{(1-x^2-y^2)^2}$. Hence, the assertion follows from combining (4.1) and (4.2). $\square$

As an example of Theorem 4.1, Example (vi) in Sect. 2.1 given by $\gamma :[0,2\pi ]\rightarrow B^1\times \mathbb {R}$, where

$$\begin{aligned} \gamma (\hat{s})=\left( r_0\cos (\hat{s}),r_0\sin (\hat{s}),-\frac{4r_0^2}{1-r_0^2}{\hat{s}}\right) , \end{aligned}$$

is a horizontal curve. Note that $t({\hat{s}})=\displaystyle -\frac{4r_0^2}{1-r_0^2}{\hat{s}}$, which implies

$$\begin{aligned} \frac{1}{8}|t(2\pi )-t(0)|=\frac{1}{8}\left| -\frac{8\pi r_0^2}{1-r_0^2}\right| =\frac{2\pi r_0^2}{1-r_0^2}. \end{aligned}$$

(4.3)

On the other hand, the projection ${\hat{\gamma }}:[0,2\pi ]\rightarrow B^1$ of $\gamma$ is given by ${\hat{\gamma }}(\hat{s})=\big (r_0\cos (\hat{s}),r_0\sin (\hat{s})\big )$, which is nothing but a circle centred at the origin with radius $r_0$. Its area in Poincarè metric is $\displaystyle \frac{2\pi r_0^2}{1-r_0^2}$, which is equal to $\displaystyle \frac{1}{8}|t(2\pi )-t(0)|$ by (4.3). This agrees with Theorem 4.1.

Lemma 4.2

Let $\gamma (\hat{s})=(z(\hat{s}), t_0)$ be a curve in $B^1\times \mathbb {R}$, where $t_0$ is a constant. Then $\gamma$ is horizontal if and only if $\gamma ({\hat{s}})=(as, bs, t_0)$ for some $a,b\in \mathbb {R}$.

Proof

Suppose that $\gamma (\hat{s})=(z(\hat{s}), t_0)=(x(\hat{s}),y(\hat{s}),t_0)$ is a horizontal curve. We can write x and y in terms of the polar coordinates, namely, $x({\hat{s}})=\gamma ({\hat{s}})\cos \theta ({\hat{s}})$ and $y({\hat{s}})=r({\hat{s}})\sin \theta ({\hat{s}})$ for some function $r({\hat{s}})>0$. By (1.6), $\gamma$ is horizontal if and only if

$$\begin{aligned} t^\prime =\frac{-4(xy'-yx')}{1-x^2-y^2}=\frac{-4r^2}{1-r^2}\theta ^\prime . \end{aligned}$$

(4.4)

Since t is constant, $\theta$ is constant, which implies that $\gamma$ is a line through the origin contained in the plane $B^1\times \{t_0\}$. Therefore, $\gamma ({\hat{s}})=(a\hat{s},b\hat{s}, t_0)$ as desired. On the other hand, if $\gamma ({\hat{s}})=(as, bs, t_0)$, it follows from (1.6) that $\gamma$ is horizontal. $\square$

Theorem 4.3

Given any two distinct points $p_1, p_2$ in $B^1\times \mathbb {R}^1$, there exists a piecewise smooth horizontal curve joining $p_1$ and $p_2$.

Proof

Let $p_i=(x_i,y_i, t_i)$, where $i=1,2$, be two distinct points in $B^1\times \mathbb {R}$. First, suppose that $p_1, p_2$ both have the same height, namely, $t_1=t_2=t_0$ for some constant $t_0$. Then by Lemma 4.2, the curves

$$\begin{aligned} \begin{array}{rll} c_1({\hat{s}})&{}=\big ( 2{\hat{s}} x_1, 2{\hat{s}} y_1, t_0\big ), &{}{\hat{s}}\in \left[ 0, \frac{1}{2} \right] ,\\ c_2({\hat{s}})&{}=\big ( {\hat{s}} x_2, {\hat{s}}y_2, t_0\big ), &{} {\hat{s}} \in [0, 1], \end{array} \end{aligned}$$

are two horizontal curves from $(0,0,t_0)$ to $p_1$ and from $(0,0,t_0)$ to $p_2$, respectively. Define the curve

$$\begin{aligned} \gamma ({\hat{s}})=\left\{ \begin{array}{rl} c_1\left( \frac{1}{2}-{\hat{s}}\right) ,&{} {\hat{s}}\in \left[ 0, \frac{1}{2}\right] , \\ c_2(2 {\hat{s}}-1), &{} {\hat{s}} \in \left[ \frac{1}{2}, 1\right] . \end{array} \right. \end{aligned}$$

Then $\gamma$ is a piecewise smooth horizontal curve joining $p_1$ and $p_2$.

Now suppose $t_1\ne t_2$. We may assume that $t_1>t_2$ and let $\alpha =t_1-t_2>0$. Denote ${\tilde{p}}_i=(x_i, y_i, 0)$, where $i=1,2$, by the projection of $p_i$ onto $B^1$. We may find a smooth curve ${\tilde{\gamma }}:[0, 1]\rightarrow B^1$ joining ${\tilde{p}}_1, {\tilde{p}}_2$ such that the area bounded by $Op_1, Op_2$, and ${\tilde{\gamma }}$ has signed area $\frac{\alpha }{8}$ (with respect to the Poincaré metric), where $O=(0,0,0)$. The area will be considered positive in the case of a counterclockwise rotation of the curve ${\tilde{\gamma }}$ between $p_1$ and $p_2$. Set ${\tilde{\gamma }}({\hat{s}})=(x({\hat{s}}),y({\hat{s}}))$ and define the function

$$\begin{aligned} t({\hat{s}})=t_1-\int _0^{{\hat{s}}} \frac{4(x'(u)y(u)-y'(u)x(u))}{1-x(u)^2-y(u)^2} du. \end{aligned}$$

(4.5)

We claim that the curve $\gamma :[0,1]\rightarrow B^1\times \mathbb {R}$ defined by $\gamma ({\hat{s}})=\big ({\tilde{\gamma }}({\hat{s}}), t({\hat{s}})\big )$ is a horizontal curve joining $p_1, p_2$. Indeed, it follows from (1.6) and (4.5) that the T-variation of $\gamma$ satisfies $\tau \equiv 0$, i.e. $\gamma$ is horizontal. Moreover, $\gamma (0)=(\tilde{\gamma }(0), t(0))=(\tilde{p}_1, t_1)=p_1.$ According to (4.1) and Theorem 4.1, one has

$$\begin{aligned} \alpha =8(\frac{\alpha }{8})=t_2-t_1=-4\int _0^1\frac{xy'-y'x}{1-x^2-y^2}du. \end{aligned}$$

Putting ${\hat{s}}=1$ in (4.5) yields

$$\begin{aligned} t(1)&=t_1-4\int _0^1\frac{xy'-yx'}{1-x^2-y^2}du.=t_1+\alpha =t_1+(t_2-t_1)=t_2. \end{aligned}$$

In particular, $\gamma (1)=(\tilde{\gamma }(1), t(1))=(\tilde{p}_2, t_2)=p_2$. Therefore, $\gamma$ is a curve joining $p_1$ and $p_2$, which proves the assertion. $\square$

5 Proof of Theorem 1.8

We will prove the following more general theorem.

Theorem 5.1

Suppose $\gamma$ and $\gamma _0$ are horizontally regular closed curve in $B^1\times \mathbb {R}$ such that $\pi \circ \gamma =\pi \circ \gamma _0$, where $\pi$ is the projection on $B^1$. If the T-variation of $\gamma$ and $\gamma _0$ satisfies

$$\begin{aligned} \tau \ge \tau _0~~(\text{ or } \tau \le \tau _0), \end{aligned}$$

(5.1)

then we must have $\tau =\tau _0$.

Note that Theorem 1.8 follows from Theorem 5.1 by taking $\tau _0\equiv 0$.

Proof of Theorem 5.1

Let $\gamma _0:[a,b]\rightarrow B^1\times \mathbb {R}$ be given by $\gamma _0(s)=(x(s),y(s),t(s))$, which is parameterized by the horizontal arc length s. It follows from (2.21) that

$$\begin{aligned} \frac{2\sqrt{2}\sqrt{(x')^2+(y')^2}}{1-x^2-y^2}=1. \end{aligned}$$

(5.2)

Since $\pi \circ \gamma =\pi \circ \gamma _0$ by assumption, $\gamma :[a,b]\rightarrow B^1\times \mathbb {R}$ can be parameterized by $\gamma (s)=(x(s),y(s),z(s))$ for some function z. It follows from (1.6) and (5.2) that the T-variation of $\gamma _0$ and $\gamma$ is, respectively, given by

$$\begin{aligned} \tau _0 =t'+\frac{4(xy'-yx')}{1-x^2-y^2}~~ \text{ and } ~~ \tau =z'+\frac{4(xy'-yx')}{1-x^2-y^2}, \end{aligned}$$

which gives $\tau -\tau _0=z'-t'.$ Integrating this from $s=a$ to $s=b$ yields

$$\begin{aligned} \int _a^b(\tau -\tau _0)ds= \big (z(b)-z(a)\big )-\big (t(b)-t(a)\big ). \end{aligned}$$

(5.3)

Since $\gamma$ and $\gamma _0$ are closed, we have

$$\begin{aligned} z(b)=z(a)~~ \text{ and } ~~t(b)=t(a). \end{aligned}$$

(5.4)

Substituting (5.4) into (5.4) gives

$$\begin{aligned} \int _a^b(\tau -\tau _0)ds=0. \end{aligned}$$

(5.5)

Now the result follows from combining this with the assumption (5.1). $\square$

As one can see from the proof of Theorem 5.1, instead of assuming that $\gamma$ and $\gamma _0$ are closed, we can assume that

$$\begin{aligned} z(b)-z(a)=t(b)-t(a) \end{aligned}$$

to conclude (5.5) from (5.3). Hence, we have the following:

Theorem 5.2

Suppose $\gamma (s)=(x(s),y(s),z(s))$, $\gamma _0(s)=(x(s),y(s),t(s)):[a,b]\rightarrow B^1\times \mathbb {R}$ are two horizontally regular curves parameterized by arc length s, which are not necessarily closed, such that

$$\begin{aligned} z(b)-z(a)=t(b)-t(a). \end{aligned}$$

If the T-variation of $\gamma$ and $\gamma _0$ satisfies

$$\begin{aligned} \tau \ge \tau _0~~(\text{ or } \tau \le \tau _0), \end{aligned}$$

then we must have $\tau =\tau _0$.

By Theorem 5.2, we have the following:

Corollary 5.3

Suppose $\gamma (s)=(x(s),y(s),z(s))$, $\gamma _0(s)=(x(s),y(s),t(s)):[a,b]\rightarrow B^1\times \mathbb {R}$ are two horizontally regular curves parameterized by arc length s, which are not necessarily closed, such that

$$\begin{aligned} z(b)-z(a)=t(b)-t(a). \end{aligned}$$

If $\gamma$ and $\gamma _0$ have the same p-curvature $\kappa =\kappa _0$, and the T-variation of $\gamma$ and $\gamma _0$ satisfy

$$\begin{aligned} \tau \ge \tau _0~~(\text{ or } \tau \le \tau _0), \end{aligned}$$

then there exists a pseudohermitian transformation g of $B^1\times \mathbb {R}$ such that

$$\begin{aligned} \gamma (s)=g\circ \gamma _0(s) \end{aligned}$$

for all s.

Proof

It follows from Theorem 5.2 that $\gamma$ and $\gamma _0$ have the same T-variations. On the other hand, $\gamma$ and $\gamma _0$ have the same p-curvature $\kappa =\kappa _0$ by assumption. Now the assertion follows from Theorem 1.5. $\square$

6 Bertrand Mate

In the following, we will discuss the analogous concept for Bertrand mate in $B^1\times \mathbb {R}$. For any given curve $\gamma (s)$ in $B^1\times \mathbb {R}$, we use the notations $t=X, n=JX=Y, b=T$ for the (horizontal) unit tangent, (horizontal) unit normal, and T of $\gamma$, respectively.

Definition 6.1

Given a horizontally regular curve $\gamma : s\in I \rightarrow B^1\times \mathbb {R}$ parameterized by horizontal arc-length. The distinct curve $\overline{\gamma }$ defined on I is a Bertrand mate of $\gamma (s)$ if it satisfies

• (1) $\overline{\gamma }(s)=\gamma (s)+\lambda (s){n(s)}$ for all $s\in I$ and some scalar function $\lambda (s)$,

• (2) $\overline{n}(s)=\pm n(s)$ for all $s\in I$.

Any curve $\gamma$ is a Bertrand curve if it has a Bertrand mate. Moreover, if any of $\gamma$ and $\overline{\gamma }$ is the Bertrand mate of the other, we call $\gamma$ and $\overline{\gamma }$ are Bertrand mates.

Note that if we take $-\lambda$ instead of $\lambda$, then we may assume that $\overline{n}(s)=n(s)$. We also point out that the second condition (2) above is natural. Indeed, suppose the normal vectors n and $\overline{n}$ of $\gamma$ and $\overline{\gamma }$, respectively, are collinear at the corresponding points, say $\overline{n}(s)=g(s)n(s)$ for some scalar function g(s) to be determined. Since $\overline{n}=gn$ implies that $\overline{t}=-J\overline{n}=-g Jn=g t$. Take the derivative and use the Frenet formula (3.17) to have $-\overline{\kappa }gt=-\overline{\kappa }\overline{t}=\overline{n}'=g' n + g n'=g' n + g (-\kappa t)$. Thus, we have $g'=0$ and $\overline{\kappa }=\kappa$. To simplify our discussion, we may assume $g\equiv \pm 1$ to have the condition (2).

Lemma 6.2

Under the notations in Definition 6.1, if $\gamma$ and $\overline{\gamma }$ are Bertrand mates, then $\lambda (s)=\lambda$ is a nonzero constant.

Proof

Notice that since $\overline{n}=\pm n$ by assumption, we have $\overline{t}=-J\overline{n}=\mp Jn=\pm t$. Taking the derivative on (1) of Definition 6.1 and using the Frenet formula (3.17), on the one hand, one gets

$$\begin{aligned} \overline{\gamma }'=\gamma '+\lambda ' n + \lambda (-\kappa t) =(1-\lambda \kappa )t +\lambda ' n +\tau b; \end{aligned}$$

(6.1)

on the other hand,

$$\begin{aligned} \overline{\gamma }'=\overline{\gamma }'_\xi +\overline{\tau }b=|\overline{\gamma }'_\xi |\overline{t}+\overline{\tau }b=\pm |\overline{\gamma }'_\xi |t+\overline{\tau }b. \end{aligned}$$

(6.2)

Compare (6.1) and (6.2) to have $\lambda '=0$, which implies that $\lambda$ is a nonzero constant since $\gamma$ and $\overline{\gamma }$ are distinct curves. $\square$

Remark 6.3

In (6.2),we notice that the parameter s is the horizontal arc-length parameter for $\gamma$ but is not necessarily horizontally arc length to its Bertrand mate $\overline{\gamma }$. Also, the Bertrand mates $\gamma$ and $\overline{\gamma }$ in $\mathbb {R}^3$ own the same property as Lemma 6.2, namely, both curves keep the constant distance between the corresponding points $\gamma (s)$ and $\overline{\gamma }(s)$. However, the constant $\lambda$ does not play the role of constant distance in $B^1\times \mathbb {R}$ because the projected metric onto $B^1$ is the Poincaré metric.

Theorem 6.4

Any Bertrand mates $\gamma$ and $\overline{\gamma }$ in $B^1\times \mathbb {R}$ are congruent in the sense that they differ from a pseudohermitian transformation defined in Theorem 3.5.

Proof

Suppose $\gamma$ and $\overline{\gamma }$ are Bertrand mates with $\overline{n}=n$ (if $\overline{n}=-n$, we can take $-\lambda$ instead of $\lambda$). Since $\lambda$ is a nonzero constant, by (6.1) and (6.2), we have

$$\begin{aligned} |\overline{\gamma }'_\xi |t+\overline{\tau }b=\overline{\gamma }'=(1-\lambda \kappa )t+\tau b. \end{aligned}$$

So $\overline{\tau }=\tau$. Moreover, since $\overline{n}=n$, we have $\overline{t}=t$. Taking the derivative and using the Frenet formula again, we obtain

$$\begin{aligned} \overline{\kappa } n= \overline{\kappa }\ \overline{n}=\overline{t}'=t'=\kappa n. \end{aligned}$$

Thus, $\overline{\kappa }=\kappa$. As a consequence, since the curves $\gamma$ and $\overline{\gamma }$ both have the same p-curvature $\kappa$ and T-variation $\tau$, by the uniqueness part in the fundamental theorem of curves (see Theorem 1.5), $\gamma$ and $\overline{\gamma }$ are congruent up to a pseudohermitian transformation and the result follows. $\square$

A simple example for Theorem 6.4 is the helix with vanishing T-variation: for any constant $\alpha \in (-1,1){\setminus } \{0\}$, set $A=\frac{1-\alpha ^2}{4\alpha }$. Let

$$\begin{aligned} \gamma (s)=\big (\sqrt{2}\alpha \sin (A s), -\sqrt{2}\alpha \cos (A s), -2 \alpha s\big ) \end{aligned}$$

be a horizontally regular curve in $B^1\times \mathbb {R}$. A direct computation shows that

$$\begin{aligned} \gamma '(s)&=\big (\sqrt{2}A \alpha \cos (As), \sqrt{2}A \alpha \sin (As ), -2 \alpha )= \cos (As) e_1+\sin (As) e_2:=t(s) \end{aligned}$$

with $\tau \equiv 0$, and the unit normal is given by

$$\begin{aligned} n(s)&:=Jt(s)=-\sin (As)e_1+\cos (A s)e_2\\&=(-\sqrt{2}A\alpha \sin (As), \sqrt{2}A\alpha \cos (As), 0). \end{aligned}$$

Then the Bertrand mate is given by

$$\begin{aligned} \overline{\gamma }(s)=\gamma (s)+ 2 n(s)=\big (-\sqrt{2}\alpha \sin (As), \sqrt{2}\alpha \cos (As), -2\alpha s \big ). \end{aligned}$$

Thus, $\overline{\gamma }$ is the curve obtained by rotating $\gamma$ by $\pi$ about the z-axis, namely, $\overline{\gamma }=\hat{\varphi }_\pi (\gamma )$, where $\hat{\varphi }_\theta (z,t)=(e^{\sqrt{-1}\theta }, t)$.