Sub-Riemannian Geometry of Curves and Surfaces in Roto-Translation Group Associated with Canonical Connection

Abstract

The aim of this paper is to obtain the sub-Riemannian properties of the roto-translation group $\mathcal{R}\mathcal{T}$. At the same time, we compute the sub-Riemannian limits of Gaussian curvature associated with two kinds of canonical connections for a $C^2$-smooth surface in the roto-translation group away from characteristic points and signed geodesic curvature associated with two kinds of canonical connections for $C^2$-smooth curves on surfaces. Based on these results, we obtain a Gauss-Bonnet theorem in the $\mathcal{R}\mathcal{T}$.

1. Introduction

The roto-translational group, $\mathcal{R}\mathcal{T}$, is the group comprising rotations and translations of the Euclidean plane. Notice that, usually, in the literature, the roto-translational group is employed to describe rigid body motion in the plane and in space. More precisely, it is a three-dimensional topological manifold diffeomorphic to ${\mathbb{R}}^2\times {\mathbb{S}}^1$, with multiplication given by

$$\begin{array}{c}\hfill \begin{array}{c}\hfill (x,y,\theta )★(\tilde{x},\tilde{y},\tilde{\theta })=(x+\tilde{x}cos\theta -\tilde{y}sin\theta ,y+\tilde{x}sin\theta +\tilde{y}cos\theta ,\theta +\tilde{\theta })\end{array}\end{array}$$

(1)

for all $(x,y,\theta ),(\tilde{x},\tilde{y},\tilde{\theta })\in {\mathbb{R}}^2\times {\mathbb{S}}^1$. Because translation and rotation do not possess nilpotent properties, the group does not come from a nilpotent group, which is in contrast to the situation with the Heisenberg group.

The $\mathcal{R}\mathcal{T}$, as a fundamental concept in mathematics, plays an indispensable role in various fields such as Geometry, Physics, Quantum Mechanics, and Image Processing. For example, in [1], Mueller introduced a special Euclidean group as a Lie group, consisting of a homogeneous transformation matrix and spatial along with spatial rigid-body velocities, to describe the time derivative of the homogeneous matrix and study the motion and posture of robots. In [2], Duits extended the concept of normative frames on images to normative frames on data representations and compared their advantages to the standard left invariant framework on the roto-translation group. Finally, Duits obtained locally adaptive frames in the roto-translation group and their applications in medical imaging. In [3], Robert presented a mathematical model of the perceptual completion and formation of subjective surfaces, and the image was lifted using simple cells to a surface in the roto-translation group. Finally, Robert obtained minimal surfaces in the roto-translation group with applications to a neuro-biological image completion model. In [4], Ryu constructed novel roto-translation equivariant energy-based models to improve the sample efficiency for learning robotic manipulation and experiment with six DoF robotic manipulation tasks to validate the models’ sample efficiency and generalizability.

Among more recent works, orientation scores represented as functions on the roto-translation group have been employed for template matching using cross-correlation. In [5], Bekkers and his colleagues introduced an effective template matching approach based on cross-correlation for identifying patterns that combine orientation and blob features. They also addressed the problem of solving time-integrated hypo-elliptic Brownian motions on the roto-translational group. In [6], Pappas described straight ruled surfaces and proved that a straight ruled surface in G is horizontally minimal. In [7,8,9], the geometric properties on hypersurfaces and Heisenberg groups were given by Barilari, Tan, and Balogh on Riemannian manifolds. In addition, Barilar also provided some examples of induced geometry on Heisenberg groups and hypersurfaces. Based on the above work, in [10,11,12,13,14], the Gauss-Bonnet theorems were proven in BCV spaces, the Lorentzian Heisenberg group, and the Twisted Heisenberg group. In [15,16], the properties of sub-Riemannian geometry and Gauss-Bonnet theorems in the roto-translation group, Lorentzian Sasakian space forms, and the group of rigid motions of the Minkowski plane with the general left-invariant metric were proven. The studies mentioned above indicate that the geometric and topological properties of the roto-translation group play a crucial role in the investigation of sub-Riemannian properties.

Inspired by the above work, in this paper, we focus on solving the sub-Riemannian properties of the roto-translation group with canonical connections and obtain the Gauss-Bonnet theorem on it. We introduce two canonical connections in the roto-translational group. In situations away from characteristic points, we compute the sub-Riemannian limits of Gaussian curvature associated with two kinds of canonical connections for a Euclidean $C^2$-smooth surface in the roto-translation group and signed geodesic curvature for Euclidean $C^2$-smooth curves on surfaces. Furthermore, we prove Gauss-Bonnet theorems associated with two kinds of canonical connections in the roto-translation group.

In Section 2, we provide a short introduction to the structure of the roto-translation group and the notions that we will use throughout the paper, such as the canonical connection and curvature in the Riemannian approximants of the roto-translational group. We provide the definitions of the sub-Riemannian limits of Gaussian curvature associated with the first kind of canonical connection and the signed geodesic curvature of smooth curves on Euclidean $C^2$-smooth surfaces, and then we derive their expressions. We obtain the Gauss-Bonnet theorem associated with the first kind of canonical connections in the roto-translational group. In Section 3, we provide the definitions of the sub-Riemannian limits of Gaussian curvature associated with the second kind of canonical connection and the signed geodesic curvature of smooth curves on Euclidean $C^2$-smooth surfaces, and then we derive their expressions. We obtain the Gauss-Bonnet theorem associated with the second kind of canonical connections in the roto-translational group.

2. The Sub-Riemannian Geometry Associated with the First Kind of Canonical Connections in (${\displaystyle \mathcal{R}\mathcal{T},{\mathit{g}}_{\mathit{L}}}$)

In this section, we introduce some notions on the roto-translation group. The $\mathcal{R}\mathcal{T}$ is given explicitly by the following matrix group:

$$\mathcal{R}\mathcal{T}=\left\{\left(\begin{array}{ccc}cos\theta & -sin\theta & x\\ sin\theta & cos\theta & y\\ 0& 0& 1\end{array}\right)|x,y\in \mathbb{R},\theta \in S^1\right\}.$$

Then, the roto-translation group is isomorphic to ${\mathbb{R}}^2\times S^1$ with multiplication given by

$$(x,y,\theta )★(\tilde{x},\tilde{y},\tilde{\theta })=(x+\tilde{x}cos\theta -\tilde{y}sin\theta ,y+\tilde{x}sin\theta +\tilde{y}cos\theta ,\theta +\tilde{\theta })$$

for all $(x,y,\theta ),(\tilde{x},\tilde{y},\tilde{\theta })\in {\mathbb{R}}^2\times {\mathbb{S}}^1.$ Now, we take positive constants ${\lambda }_1$, ${\lambda }_2$, and ${\lambda }_3$ and a left-invariant frame as follows:

$$F_1=\frac{1}{{\lambda }_1}cos\theta \frac{\partial }{\partial x}+\frac{1}{{\lambda }_1}sin\theta \frac{\partial }{\partial y},F_2={\lambda }_1{\lambda }_2\frac{\partial }{\partial \theta },F_3=-\frac{1}{{\lambda }_2}sin\theta \frac{\partial }{\partial x}+\frac{1}{{\lambda }_2}cos\theta \frac{\partial }{\partial y},$$

(2)

and $span\left\{F_1,F_2,F_3\right\}=T\left(\mathcal{R}\mathcal{T}\right)$. Let $H=span\left\{F_1,F_2\right\}$ be the horizontal distribution on the $\mathcal{R}\mathcal{T}$. To describe the Riemannian approximants to the $\mathcal{R}\mathcal{T},$ let $L>0$ and define a metric $g_L={\omega }_1\otimes {\omega }_1+{\omega }_2\otimes {\omega }_2+L\omega \otimes \omega$, where ${\omega }_1={\lambda }_{1}cos\theta dx+{\lambda }_{1}sin\theta dy,{\omega }_2=\frac{1}{{\lambda }_1{\lambda }_2}d\theta ,$ and $\omega =-{\lambda }_{2}sin\theta dx+{\lambda }_{2}cos\theta dy$. Then, the orthonormal basis on $T\left(\mathcal{R}\mathcal{T}\right)$ with respect to $g_L$ is $F_1,F_2,\widetilde{F_3}:=L^{-\frac{1}{2}}{F}_3$. We have

$$\left[F_1,F_2\right]=-{\lambda }_2^2{F}_3,\left[F_2,F_3\right]=-{\lambda }_1^2{F}_1,\left[F_1,F_3\right]=0.$$

(3)

Then,

$$\frac{\partial }{\partial x}={\lambda }_{1}cos\theta F_1-{\lambda }_{2}sin\theta F_3,\frac{\partial }{\partial y}={\lambda }_{1}sin\theta F_1+{\lambda }_{2}cos\theta F_3,\frac{\partial }{\partial \theta }=\frac{1}{{\lambda }_1{\lambda }_2}{F}_2.$$

(4)

Let ${\nabla }^L$ be the Levi-Civita connection on $\mathcal{R}\mathcal{T}$ with respect to $g_L$, which is determined by Proposition 2.1 in [17]. Let $J{F}_1=F_2,J{F}_2=F_1$ and $J{F}_3=F_3.$ Then, $J^2=id$ and $g_L(JX,JY)=g_L(X,Y)$ for $X,Y\in \Gamma (\mathcal{R}\mathcal{T})$, and J is a product structure. We define the first kind of canonical connections, which is a metric connection in the roto-translation group, as follows: $\mathcal{R}\mathcal{T}$:

$${\nabla }_X^{1}Y=\frac{1}{2}{\nabla }_X^{L}Y+\frac{1}{2}J{\nabla }_X^{L}JY.$$

(5)

Definition 1.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^1$-smooth curve, we say that γ is regular if $\dot{\gamma }\ne 0$ for every $t\in \left[a,b\right]$. Moreover we say that $\gamma (t)$ is a horizontal point of γ if

$$\begin{array}{cc}\hfill \omega \left(\dot{\gamma }\left(t\right)\right)& ={\lambda }_2\left(-sin\theta dx+cos\theta dy\right)\left({\dot{\gamma }}_1\left(t\right)\frac{\partial }{\partial x}+{\dot{\gamma }}_2\left(t\right)\frac{\partial }{\partial y}+{\dot{\theta }}_1\left(t\right)\frac{\partial }{\partial \theta }\right)\hfill \\ \hfill & ={\lambda }_2(-{\dot{\gamma }}_1\left(t\right)sin\theta +{\dot{\gamma }}_2\left(t\right)cos\theta )\hfill \\ \hfill & =0\hfill \end{array}$$

where $\gamma (t)=({\gamma }_1(t),{\gamma }_2(t),{\theta }_1(t))$.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. We can define the curvature associated to ${\nabla }^1$, $k_{\gamma }^{L,{\nabla }^1}$ of $\gamma$ at $\gamma (t)$. We have the following lemma.

Lemma 1.

Let the $\mathcal{R}\mathcal{T}$ be the roto-translation group. Then,

$$\begin{array}{cc}\hfill {\nabla }_{F_1}^1{F}_1& =\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}{F}_3,{\nabla }_{F_1}^1{F}_2=\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}{F}_3,{\nabla }_{F_1}^1{F}_3=\frac{-{\lambda }_1^2+L{\lambda }_2^2}{4L}(F_1+F_2),\hfill \\ \hfill {\nabla }_{F_2}^1{F}_1& =\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}{F}_3,{\nabla }_{F_2}^1{F}_2=\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}{F}_3,{\nabla }_{F_2}^1{F}_3=\frac{-{\lambda }_1^2-L{\lambda }_2^2}{4L}(F_1+F_2),\hfill \\ \hfill {\nabla }_{F_3}^1{F}_1& =0,{\nabla }_{F_3}^1{F}_2=0,{\nabla }_{F_3}^1{F}_3=0.\hfill \\ \hfill \end{array}$$

(6)

Defining the curvature of the connection ${\nabla }^1$ by

$$R^1(X,Y)Z={\nabla }_X^1{\nabla }_Y^{1}Z-{\nabla }_Y^1{\nabla }_X^{1}Z-{\nabla }_{[X,Y]}^{1}Z,$$

we obtain the following proposition.

Proposition 1.

Let $(\mathcal{R}\mathcal{T},g_L)$ be the roto-translation group with the general left-invariant metric. Then,

$$\begin{array}{cc}\hfill & R^1\left(F_2,F_3\right)F_1=\frac{{\lambda }_1^2-L{\lambda }_1^2{\lambda }_2^2}{4L}{F}_3,R^1\left(F_2,F_3\right)F_2=\frac{{\lambda }_1^4-L{\lambda }_1^2{\lambda }_2^2}{4L}{F}_3,\hfill \\ \hfill & R^1\left(F_2,F_3\right)F_3=\frac{-{\lambda }_1^4+L{\lambda }_1^2{\lambda }_2^2}{4}(F_1+F_2),R^1\left(F_i,F_j\right)F_k=0,forotheri,j,k.\hfill \\ \hfill \end{array}$$

(7)

Proof. 

It is a direct computation using

$$R^1(X,Y)Z={\nabla }_X^1{\nabla }_Y^{1}Z-{\nabla }_Y^1{\nabla }_X^{1}Z-{\nabla }_{[X,Y]}^{1}Z.$$

Taking

$$R^1\left(F_1,F_2\right)F_1=\left({\nabla }_{F_1}^1{\nabla }_{F_2}^1-{\nabla }_{F_2}^1{\nabla }_{F_1}^1-{\nabla }_{\left[F_1,F_2\right]}^1\right)F_1,$$

for example, we compute

$${\nabla }_{F_1}^1\left({\nabla }_{F_2}^1{F}_1\right)=\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}{\nabla }_{F_1}^1{F}_3=\frac{(L{\lambda }_2^2+{\lambda }_1^2)(L{\lambda }_2^2-{\lambda }_1^2)}{16L}(F_1+F_2),$$

$${\nabla }_{F_2}^1\left({\nabla }_{F_1}^1{F}_1\right)=\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}{\nabla }_{F_2}^1{F}_3=\frac{(L{\lambda }_2^2+{\lambda }_1^2)(L{\lambda }_2^2-{\lambda }_1^2)}{16L}(F_1+F_2),$$

$${\nabla }_{\left[F_1,F_2\right]}^1{F}_1=0.$$

Hence,

$$R^1\left(F_1,F_2\right)F_1=\frac{(L{\lambda }_2^2+{\lambda }_1^2)(L{\lambda }_2^2-{\lambda }_1^2)}{16L}(F_1+F_2)-\frac{(L{\lambda }_2^2+{\lambda }_1^2)(L{\lambda }_2^2-{\lambda }_1^2)}{16L}(F_1+F_2)=0.$$

□

Let $\gamma (t)=({\gamma }_1(t),{\gamma }_2(t),{\theta }_1(t)).$ Then,

$$\begin{array}{c}\hfill \dot{\gamma }(t)={\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right)F_1+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)F_2+\omega \left(\dot{\gamma }(t)\right)F_3.\end{array}$$

(8)

Let

$$A={\lambda }_1{\dot{\gamma }}_1(t)cos\theta +{\lambda }_1{\dot{\gamma }}_2(t)sin\theta ,$$

$$A^{\prime }={\lambda }_1{\ddot{\gamma }}_{1}cos\theta +{\lambda }_1{\ddot{\gamma }}_{2}sin\theta +{\dot{\theta }}_1\left(t\right)\left(-{\lambda }_1{\dot{\gamma }}_{1}sin\theta +{\lambda }_1{\dot{\gamma }}_{2}cos\theta \right).$$

By (6) and (7), we have

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^L\dot{\gamma }=& \left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]F_1\hfill \\ \hfill +& \left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]F_2\hfill \\ \hfill +& \left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]F_3.\hfill \\ \hfill \end{array}$$

(9)

Definition 2.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. The curvature $k_{\gamma }^{L,{\nabla }^1}$ of γ at $\gamma (t)$ is defined as

$$k_{\gamma }^{L,{\nabla }^1}:=\sqrt{\frac{{∥{\nabla }_{\dot{\gamma }}^1\dot{\gamma }∥}_L^2}{{∥\dot{\gamma }∥}_L^4}-\frac{{〈{\nabla }_{\dot{\gamma }}^1\dot{\gamma },\dot{\gamma }〉}_L^2}{{{∥\dot{\gamma }∥}_L}^6}}.$$

(10)

Lemma 2.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. Then,

$$\begin{array}{cc}\hfill k_{\gamma }^{L,{\nabla }^1}=& \left\{\right\{{\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill +& {\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill +& L{\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]}^2\}\hfill \\ \hfill \times & {\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_1^2}{({\dot{\theta }}_1(t))}^2+L{(\omega (\dot{\gamma }(t)))}^2\right]}^{-2}\hfill \\ \hfill -& \{A{\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill +& {\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill +& L(\omega (\dot{\gamma }(t)))\left[(A+\frac{\dot{{\theta }_1}(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]{\}}^2\hfill \\ \hfill \times & {\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{({\dot{\theta }}_1(t))}^2+L{(\omega (\dot{\gamma }(t)))}^2\right]}^{-3}{\}}^{\frac{1}{2}}.\hfill \end{array}$$

(11)

In particular, if $\gamma (t)$ is a horizontal point of $\gamma$ and $\omega (\dot{\gamma }(t))=0$, we have

$$\begin{array}{cc}\hfill k_{\gamma }^{L,{\nabla }^1}=& \{{(A^{\prime })}^2+\frac{{({\ddot{\theta }}_1(t))}^2}{{\lambda }_1^2{\lambda }_2^2}\hfill \\ \hfill & +L{\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})\times (\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill & \times {(A^2+\frac{{({\dot{\theta }}_1(t))}^2}{{\lambda }_1^2{\lambda }_2^2})}^{-2}-A(A^{\prime }+\frac{{\dot{\theta }}_1(t){\ddot{\theta }}_1(t)}{{\lambda }_1^2{\lambda }_2^2})\times {(A^2+\frac{{({\dot{\theta }}_1(t))}^2}{{\lambda }_1^2{\lambda }_2^2})}^{-3}{\}}^{\frac{1}{2}}.\hfill \end{array}$$

(12)

Proof. 

By (4), we have

$$\begin{array}{cc}\hfill \dot{\gamma }(t)& ={\dot{\gamma }}_1(t)\frac{\partial }{\partial x}+{\dot{\gamma }}_2(t)\frac{\partial }{\partial y}+{\dot{\theta }}_1(t)\frac{\partial }{\partial \theta }\hfill \\ \hfill & ={\dot{\gamma }}_1(t)\left({\lambda }_{1}cos\theta F_1-{\lambda }_{2}sin\theta F_3\right)+{\dot{\gamma }}_2(t)\left({\lambda }_{1}sin\theta F_1+{\lambda }_{2}cos\theta F_3\right)+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}{F}_2\hfill \\ \hfill & ={\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right)F_1+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}{F}_2-\left({\lambda }_2{\dot{\gamma }}_1(t)sin\theta -{\lambda }_2{\dot{\gamma }}_2(t)cos\theta \right)F_3\hfill \\ \hfill & ={\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right)F_1+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)F_2+\omega \left(\dot{\gamma }(t)\right)F_3.\hfill \\ \hfill \end{array}$$

(13)

By (2) and (13), we have

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^1{F}_1& ={\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right){\nabla }_{F_1}^1{F}_1+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t){\nabla }_{F_2}^1{F}_1+\omega \left(\dot{\gamma }(t)\right){\nabla }_{F_3}^1{F}_1\hfill \\ & =\left({\lambda }_1({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta )\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}+\frac{1}{{\lambda }_1{\lambda }_2}\dot{{\theta }_1}(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right)F_3,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^1{F}_2& ={\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right){\nabla }_{F_1}^1{F}_2+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t){\nabla }_{F_2}^1{F}_2+\omega \left(\dot{\gamma }(t)\right){\nabla }_{F_3}^1{F}_2\hfill \\ \hfill & =\left({\lambda }_1({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta )\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}+\frac{1}{{\lambda }_1{\lambda }_2}\dot{{\theta }_1}(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right)F_3,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^1{F}_3=& {\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right){\nabla }_{F_1}^1{F}_3+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t){\nabla }_{F_2}^1{F}_3+\omega \left(\dot{\gamma }(t)\right){\nabla }_{F_3}^1{F}_3\hfill \\ \hfill =& \left({\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right)\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}-\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}\right)(F_1+F_2).\hfill \\ \hfill \end{array}$$

(14)

By (13) and (14), we have

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^1\dot{\gamma }=& {\nabla }_{\dot{\gamma }}^1{\lambda }_1\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right)F_1+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)F_2+\omega \left(\dot{\gamma }(t)\right)F_3\hfill \\ \hfill =& {\lambda }_1\left(\dot{\gamma }\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right)F_1+\left({\dot{\gamma }}_1(t)cos\theta +{\dot{\gamma }}_2(t)sin\theta \right){\nabla }_{\dot{\gamma }}^1{F}_1\right)\hfill \\ \hfill & +\frac{1}{{\lambda }_1{\lambda }_2}\left(\dot{\gamma }({\dot{\theta }}_1(t))F_2+{\dot{\theta }}_1(t){\nabla }_{\dot{\gamma }}^1{F}_2\right)+\dot{\gamma }\left(\omega (\dot{\gamma }(t)\right))F_3+\omega \left(\dot{\gamma }(t)\right){\nabla }_{\dot{\gamma }}^1{F}_3\hfill \\ \hfill =& \left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]F_1\hfill \\ \hfill & +\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]F_2\hfill \\ \hfill & +\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]F_3.\hfill \\ \hfill \end{array}$$

(15)

By (10) and (13), we have

$$\begin{array}{cc}\hfill {∥{\nabla }_{\dot{\gamma }}^1\dot{\gamma }∥}_L^2=& {\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill & +{\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]}^2\hfill \\ \hfill & +L{\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]}^2,\hfill \\ \hfill \end{array}$$

(16)

$$\begin{array}{c}\hfill {∥\dot{\gamma }∥}_L^4={\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{\left({\dot{\theta }}_1\left(t\right)\right)}^2+L{\left(\omega \left(\dot{\gamma }\left(t\right)\right)\right)}^2\right]}^2,\end{array}$$

$$\begin{array}{cc}\hfill {〈{\nabla }_{\dot{\gamma }}^1\dot{\gamma },\dot{\gamma }〉}_L^2=& \{A\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ & +\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ & +L\omega \left(\dot{\gamma }\left(t\right)\right)\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]{\}}^2,\hfill \\ \hfill \end{array}$$

(17)

and

$${∥\dot{\gamma }∥}_L^6={\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{\left({\dot{\theta }}_1\left(t\right)\right)}^2+L{\left(\omega \left(\dot{\gamma }\left(t\right)\right)\right)}^2\right]}^3.$$

By the definition of $k_{\gamma }^{L,{\nabla }^1},$ we obtain Lemma 2. □

Now, we can define the intrinsic curvature associated with the first kind of canonical connection ${\nabla }^1$, $k_{\gamma }^{\infty ,{\nabla }^1}$ of $\gamma$ at $\gamma (t)$.

Definition 3.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. We define the intrinsic curvature ${\kappa }_{\gamma }^{\infty }$ of γ at $\gamma (t)$ to be

$${\kappa }_{\gamma }^{\infty ,{\nabla }^1}:=\underset{L\to \infty }{lim}{\kappa }_{\gamma }^{L,{\nabla }^1}$$

if the limit exists.

We introduce the following notation for continuous functions $f_1,f_2:(0,+\infty )\to \mathbb{R}$:

$$f_1\left(L\right)\sim f_2\left(L\right),asL\to +\infty \iff \underset{L\to \infty }{lim}\frac{f_1\left(L\right)}{f_2\left(L\right)}=1.$$

Lemma 3.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. Then,

$$\begin{array}{c}\hfill k_{\gamma }^{\infty ,{\nabla }^1}=\frac{{\lambda }_2\sqrt{{({\dot{\theta }}_1(t))}^2+A^2{\lambda }_1^2{\lambda }_2^2-2{\lambda }_1{\lambda }_2\dot{{\theta }_1}(t)A}}{2\sqrt{2}{\lambda }_1\left|\omega (\dot{r}(t))\right|},if\omega (\dot{\gamma }(t))\ne 0,\end{array}$$

(18)

$$\begin{array}{c}\hfill \underset{L\to +\infty }{lim}\frac{k_{\gamma }^{L,{\nabla }^1}}{\sqrt{L}}=\frac{\frac{d}{dt}(\omega (\dot{\gamma }(t)))}{\sqrt{A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{{\dot{\theta }}_1(t)}^2}},if\omega (\dot{\gamma }(t))=0and\frac{d}{dt}(\omega (\dot{\gamma }(t)))\ne 0,\end{array}$$

(19)

$$\begin{array}{cc}\hfill k_{\gamma }^{\infty ,{\nabla }^1}=& \{{(A^{\prime })}^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{({\ddot{\theta }}_1(t))}^2\times {[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{({\dot{\theta }}_1(t))}^2]}^{-2}\hfill \\ \hfill -& {\left[A(A^{\prime }+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{\dot{\theta }}_1(t)\ddot{{\theta }_1}(t))\right]}^2\times {[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{\dot{\theta }}_1(t))}^2{]}^{-3}{\}}^{\frac{1}{2}},\hfill \\ \hfill & if\omega (\dot{\gamma }(t))=0and\frac{d}{dt}(\omega (\dot{\gamma }(t)))=0.\hfill \\ \hfill \end{array}$$

(20)

Proof. 

When $\omega \left(\dot{\gamma }\left(t\right)\right)\ne 0$, we have

$$\begin{array}{c}\hfill {∥{\nabla }_{\dot{\gamma }}^1\gamma ∥}_L^2\sim \frac{(w{(\dot{\gamma }(t))}^2({({\dot{\theta }}_1(t))}^2+A^2{\lambda }_1^2{\lambda }_2^2-2{\lambda }_1{\lambda }_2{\dot{\theta }}_1(t)A}{8{\lambda }_1^2}·L^2{\lambda }_1^2,\mathrm{as}L\to +\infty ,\end{array}$$

(21)

$$\begin{array}{c}{∥\dot{\gamma }∥}_L^2\sim L\omega {\left(\dot{\gamma }\left(t\right)\right)}^2,{〈{\nabla }_{\dot{\gamma }}^L\dot{\gamma },\dot{\gamma }〉}_L^2\sim O\left(L^2\right)\mathrm{as}L\to +\infty .\end{array}$$

Therefore,

$$\frac{{∥{\nabla }_{\dot{\gamma }}^1\dot{\gamma }∥}_L^2}{{∥\dot{\gamma }∥}_L^4}\to \frac{{\lambda }_2^2{({\dot{\theta }}_1(t))}^2+A^2{\lambda }_1^2{\lambda }_2^2-2{\lambda }_1{\lambda }_2{\dot{\theta }}_1(t)A}{8{\lambda }_1^2{(w(\dot{\gamma }(t)))}^2}\mathrm{as}L\to +\infty ,$$

$$\frac{{〈{\nabla }_{\dot{\gamma }}^1\dot{\gamma },\dot{\gamma }〉}_L^2}{{∥\dot{\gamma }∥}_L^6}\to \frac{O(L^2)}{L^3\omega {(\dot{\gamma }(t))}^6}=0\mathrm{as}L\to +\infty .$$

If $\omega \left(\dot{\gamma }\left(t\right)\right)\ne 0,$ by (10), we have

$$k_{\gamma }^{\infty ,{\nabla }^1}=\frac{{\lambda }_2\sqrt{{({\dot{\theta }}_1(t))}^2+A^2{\lambda }_1^2{\lambda }_2^2-2{\lambda }_1{\lambda }_2{\dot{\theta }}_1(t)A}}{2\sqrt{2}{\lambda }_1\left|\omega (\dot{r}(t))\right|}.$$

(22)

By (11) and $\frac{d}{dt}\left(\omega \left(\dot{\gamma }\left(t\right)\right)\right)=0,$ we have

$$\begin{array}{cc}\hfill k_{\gamma }^{\infty ,{\nabla }^1}=& \{{(A^{\prime })}^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{({\ddot{\theta }}_{1}t)}^2\times {\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{({\dot{\theta }}_1(t))}^2\right]}^{-2}\hfill \\ \hfill -& {\left[A(A^{\prime }+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{\dot{\theta }}_1(t){\ddot{\theta }}_1(t))\right]}^2\times {\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{\dot{\theta }}_1{(t))}^2\right]}^{-3}{\}}^{\frac{1}{2}}.\hfill \\ \hfill \end{array}$$

(23)

When $\omega \left(\dot{\gamma }\left(t\right)\right)=0$ and $\frac{d}{dt}\left(\omega \left(\dot{\gamma }\left(t\right)\right)\right)\ne 0$, we have

$${∥{\nabla }_{\dot{\gamma }}^1\gamma ∥}_L^2\sim L{\left[\frac{d}{dt}\left(\omega \left(\dot{\gamma }\left(t\right)\right)\right)\right]}^2\mathrm{as}L\to +\infty ,$$

$$\begin{array}{c}{∥\dot{\gamma }∥}_L^2=A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{{\dot{\theta }}_1(t)}^2,{〈{\nabla }_{\dot{\gamma }}^1\dot{\gamma },\dot{\gamma }〉}_L^2\sim O\left(1\right)\mathrm{as}L\to +\infty .\end{array}$$

By (10), we obtain

$$\underset{L\to +\infty }{lim}\frac{k_{\gamma }^{L,{\nabla }^1}}{\sqrt{L}}=\frac{\frac{d}{dt}(\omega (\dot{\gamma }(t)))}{\sqrt{A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{{\dot{\theta }}_1(t)}^2}}.$$

(24)

□

We will say that a surface ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ is regular if ${\Sigma }_{\mathcal{R}\mathcal{T}}$ is a $C^2$-smooth compact and oriented surface. In particular we will assume that there exists a $C^2$-smooth function $u:\mathcal{R}\mathcal{T}\to \mathbb{R}$ such that

$${\Sigma }_{\mathcal{R}\mathcal{T}}=\left\{(x_1,x_2,{\theta }_1)\in \mathcal{R}\mathcal{T}:u(x_1,x_2,{\theta }_1)=0\right\},$$

and $u_{x_1}{\partial }_{x_1}+u_{x_2}{\partial }_{x_2}+u_{{\theta }_1}{\partial }_{{\theta }_1}\ne 0$. Let ${\nabla }_{H}u=F_1(u)F_1+F_2(u)F_2$. A point $x\in {\Sigma }_{\mathcal{R}\mathcal{T}}$ is called characteristic if ${\nabla }_{H}u(x)=0.$ Then, we define the characteristic set

$$C({\Sigma }_{\mathcal{R}\mathcal{T}}):=\{(x_1,x_2,{\theta }_1)\in {\Sigma }_{\mathcal{R}\mathcal{T}}|{\nabla }_{H}u(x_1,x_2,{\theta }_1)=0\}.$$

Our computations will be local and away from the characteristic points of ${\Sigma }_{\mathcal{R}\mathcal{T}}$. Let us define first $p:=F_{1}u,q:=F_{2}u,\mathrm{and}r:=\widetilde{F_3}u.$ We then define

$$\begin{array}{cc}\hfill & \overline{q}:=\frac{q}{l},{\overline{p}}_L:=\frac{p}{l_L},{\overline{q}}_L:=\frac{q}{l_L},{\overline{r}}_L:=\frac{r}{l_L},\hfill \\ \hfill & l:=\sqrt{p^2+q^2},l_L:=\sqrt{p^2+q^2+r^2},\overline{p}:=\frac{p}{l}.\hfill \end{array}$$

(25)

In particular, ${\overline{p}}^2+{\overline{q}}^2=1$. These functions are well defined at every non-characteristic point. Let

$$\begin{array}{cc}\hfill v_L& ={\overline{p}}_L{F}_1+{\overline{q}}_L{F}_2+{\overline{r}}_L\widetilde{F_3},\hfill \\ \hfill e_1& =\overline{q}{F}_1-\overline{p}{F}_2,\hfill \\ \hfill e_2& =\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}\widetilde{F_3},\hfill \end{array}$$

(26)

Then, $v_L$ is the Riemannian unit normal vector to ${\Sigma }_{\mathcal{R}\mathcal{T}}$, and $e_1$ and $e_2$ are the orthonormal bases of ${\Sigma }_{\mathcal{R}\mathcal{T}}$. On $T{\Sigma }_{\mathcal{R}\mathcal{T}}$, we define a linear transformation $J_L:T{\Sigma }_{\mathcal{R}\mathcal{T}}\to T{\Sigma }_{\mathcal{R}\mathcal{T}}$ such that

$$J_L(e_1):=e_2,J_L(e_2):=-e_1.$$

(27)

For every $U,V\in T{\Sigma }_{\mathcal{R}\mathcal{T}}$, we define ${\nabla }_U^{1,{\Sigma }_{\mathcal{R}\mathcal{T}}}V=\pi {\nabla }_U^{1}V$, where $\pi :TG\to T{\Sigma }_{\mathcal{R}\mathcal{T}}$ is the projection. Then, ${\nabla }^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}$ is the Levi-Civita connection on ${\Sigma }_{\mathcal{R}\mathcal{T}}$ with respect to the metric $g_L.$ By (15), (26), and

$${\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }={\langle {\nabla }_{\dot{\gamma }}^1\dot{\gamma },e_1\rangle }_L{e}_1+{\langle {\nabla }_{\dot{\gamma }}^1\dot{\gamma },e_2\rangle }_L{e}_2,$$

(28)

we have

$$\begin{array}{c}\hfill {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }=(\overline{q}{U}_1-\overline{p}{U}_2)e_1+({\overline{r}}_L\overline{p}{U}_1+{\overline{r}}_L\overline{q}{U}_2-\frac{l}{l_L}{L}^{\frac{1}{2}}{U}_3)e_2,\end{array}$$

(29)

where

$$U_1=\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right],$$

$$U_2=\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right],$$

$$U_3=\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right].$$

Moreover if $\omega (\dot{\gamma }(t))=0$, then

$$\begin{array}{c}\hfill {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }=\left[\overline{q}{A}^{\prime }-\overline{p}\frac{1}{{\lambda }_1{\lambda }_2}{\ddot{\theta }}_1(t)\right]e_1+\left[{\overline{r}}_L\overline{p}{A}^{\prime }+{\overline{r}}_L\overline{q}\frac{1}{{\lambda }_1{\lambda }_2}{\ddot{\theta }}_1(t)\right.\left.-\frac{l}{l_L}{L}^{\frac{1}{2}}{U}_3\right]e_2,\end{array}$$

(30)

let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface and $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. We can define the geodesic curvature associated to the ${\nabla }^1$, $k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^1}$ of $\gamma$ at $\gamma (t)$ and the intrinsic geodesic curvature associated to the ${\nabla }^1$, $k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1}$ of $\gamma$ at $\gamma (t)$.

Definition 4.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface and $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a Euclidean $C^2$-smooth regular curve. The geodesic curvature ${\kappa }_{\gamma ,\Sigma }^{L,{\nabla }^1}$ of γ at $\gamma (t)$ is defined as

$${\kappa }_{\gamma ,\Sigma }^{L,{\nabla }^1}:=\sqrt{\frac{\parallel {\nabla }_{\dot{\gamma }}^{\Sigma ,L}\dot{\gamma }{\parallel }_{\Sigma ,L}^2}{\parallel \dot{\gamma }{\parallel }_{\Sigma ,L}^4}-\frac{{\langle {\nabla }_{\dot{\gamma }}^{\Sigma ,L}\dot{\gamma },\dot{\gamma }\rangle }_{\Sigma ,L}^2}{\parallel \dot{\gamma }{\parallel }_{\Sigma ,L}^6}}.$$

Definition 5.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface and $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a Euclidean $C^2$-smooth regular curve. We define the intrinsic geodesic curvature ${\kappa }_{\gamma ,\Sigma }^{\infty ,{\nabla }^1}$ of γ at $\gamma (t)$ to be

$${\kappa }_{\gamma ,\Sigma }^{\infty ,{\nabla }^1}:=\underset{L\to +\infty }{lim}{\kappa }_{\gamma ,\Sigma }^{L,{\nabla }^1}$$

if the limit exists.

Lemma 4.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface and $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. Then,

$$\begin{array}{cc}& k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1}=\frac{{\lambda }_2\sqrt{[{(\overline{q}-\overline{p})}^2+{({\overline{r}}_L\overline{p}+{\overline{r}}_L\overline{q})}^2][{({\dot{\theta }}_1(t))}^2-2{\lambda }_1{\lambda }_{2}A{\dot{\theta }}_1(t)+{\lambda }_1^2{\lambda }_2^2{A}^2]}}{4{\lambda }_1{\left|\omega (\dot{r}(t))\right|}^2},if\omega (\dot{\gamma }(t))\ne 0,\hfill \\ & k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1}=0,if\omega (\dot{\gamma }(t))=0,and\frac{d}{dt}(\omega (\dot{\gamma }(t)))=0,\hfill \\ & \underset{L\to +\infty }{lim}\frac{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^1}}{\sqrt{L}}=\frac{\frac{d}{dt}(\omega (\dot{\gamma }(t)))}{\sqrt{A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{{\dot{\theta }}_1(t)}^2}},if\omega (\dot{\gamma }(t))=0and\frac{d}{dt}(\omega (\dot{\gamma }(t)))\ne 0.\hfill \end{array}$$

(31)

Proof. 

By (13) and $\dot{\gamma }\in T{\Sigma }_{\mathcal{R}\mathcal{T}}$, we have

$$\dot{\gamma }(t)={\lambda }_1(\dot{{\gamma }_1}cos\theta +\dot{{\gamma }_2}sin\theta )F_1+\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)F_2+\omega (\dot{\gamma }(t))F_3.$$

By (30), we have

$$\begin{array}{cc}\hfill \dot{\gamma }(t)& =a{e}_1+b{e}_2\hfill \\ \hfill & =a(\overline{q}{F}_1-\overline{p}{F}_2)+b(\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}\widetilde{F_3})\hfill \\ \hfill & =(a\overline{q}+b\overline{r_L}\overline{p})F_1+(-a\overline{p}+b\overline{r_L}\overline{q})F_2-\frac{bl}{l_L}{L}^{-\frac{1}{2}}{F}_3.\hfill \end{array}$$

Comparing the above equations, we obtain

$$\left\{\begin{array}{c}a\overline{q}+b\overline{r_L}\overline{p}={\lambda }_1\dot{{\gamma }_1}cos\theta +{\lambda }_1\dot{{\gamma }_2}sin\theta ,\hfill \\ -a\overline{p}+b\overline{r_L}\overline{q}=\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t),\hfill \\ -\frac{bl}{l_L}{L}^{-\frac{1}{2}}=\omega (\dot{\gamma }(t)).\hfill \end{array}\right.$$

Solve the following equation:

$$\left\{\begin{array}{c}a={\lambda }_1(\dot{{\gamma }_1}cos\theta +\dot{{\gamma }_2}sin\theta )\overline{q}-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}\dot{{\theta }_1}(t),\hfill \\ b=-L^{\frac{1}{2}}\frac{l_L}{l}\omega (\dot{\gamma }(t)).\hfill \end{array}\right.$$

This proves the following:

$$\dot{\gamma }=\left[-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+{\lambda }_1\overline{q}(\dot{{\gamma }_1}cos\theta +\dot{{\gamma }_2}sin\theta )\right]e_1-\frac{l_L}{l}{L}^{\frac{1}{2}}\omega (\dot{\gamma }(t))e_2.$$

(32)

When $\omega (\dot{\gamma }(t))\ne 0$, we have

$$\begin{array}{cc}\hfill \parallel \dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}& =\sqrt{{[-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+{\lambda }_1\overline{q}(\dot{{\gamma }_1}cos\theta +\dot{{\gamma }_2}sin\theta )]}^2+{(\frac{l_L}{l})}^{2}L{\omega }^2(\dot{\gamma }(t))}\hfill \\ \hfill & \sim L^{\frac{1}{2}}\left|\omega (\dot{\gamma }(t))\right|\mathrm{as}L\to +\infty .\hfill \end{array}$$

(33)

By (14), we have

$$\begin{array}{cc}\hfill \parallel {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}^2=& \{\overline{q}\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ \hfill & -\overline{p}\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]{\}}^2\hfill \\ \hfill & +\{{\overline{r}}_L\overline{p}\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ \hfill & +{\overline{r}}_L\overline{q}\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ \hfill & -\frac{l}{l_L}{L}^{\frac{1}{2}}\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]{\}}^2,\hfill \end{array}$$

(34)

so, we have

$$\begin{array}{cc}\hfill \parallel {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}^2\sim & \{\overline{q}\left(-\frac{\dot{{\theta }_1}(t)\omega (\dot{\gamma }(t))}{4{\lambda }_1{\lambda }_2}·L{\lambda }_2^2+A\frac{\omega (\dot{\gamma }(t))}{4}·L{\lambda }_2^2\right)\hfill \\ \hfill & -\overline{p}\left(-\frac{\dot{{\theta }_1}(t)\omega (\dot{\gamma }(t))}{4{\lambda }_1{\lambda }_2}·L{\lambda }_2^2+A\frac{\omega (\dot{\gamma }(t))}{4}·L{\lambda }_2^2\right){\}}^2\hfill \\ \hfill & +\{{\overline{r}}_L\overline{p}\left(-\frac{\dot{{\theta }_1}(t)\omega (\dot{\gamma }(t))}{4{\lambda }_1{\lambda }_2}·L{\lambda }_2^2+A\frac{\omega (\dot{\gamma }(t))}{4}·L{\lambda }_2^2\right)\hfill \\ \hfill & +{\overline{r}}_L\overline{q}\left(-\frac{\dot{\theta }(t)\omega (\dot{\gamma }(t))}{4{\lambda }_1{\lambda }_2}·L{\lambda }_2^2+A\frac{\omega (\dot{\gamma }(t))}{4}·L{\lambda }_2^2\right){\}}^2\hfill \\ \hfill =& \left[{(\overline{q}-\overline{p})}^2+{({\overline{r}}_L\overline{p}+{\overline{r}}_L\overline{q})}^2\right]\hfill \\ \hfill & \times {(w(\dot{\gamma }(t)))}^2·L^2{\lambda }_2^2·\frac{{(\dot{{\theta }_1}(t))}^2-2\lambda {\lambda }_{2}A\dot{{\theta }_1}(t)+{\lambda }_1^2{\lambda }_2^2{A}^2}{16{\lambda }_1^2}.\hfill \end{array}$$

(35)

By (29) and (32), we have

$$\begin{array}{cc}\hfill {\langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma },\dot{\gamma }\rangle }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}=& \left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+{\lambda }_1\overline{q}(\dot{{\gamma }_1}cos\theta +\dot{{\gamma }_2}sin\theta )\right]\hfill \\ & \times \{\overline{q}\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ & -\overline{p}\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\}\hfill \\ & -\frac{l_L{L}^{\frac{1}{2}}}{l}(\omega (\dot{\gamma }(t)))\hfill \\ & \times \{{\overline{r}}_L\overline{p}\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+\frac{A(L{\lambda }_2^2-{\lambda }_1^2)\omega (\dot{\gamma }(t))}{4}\right]\hfill \\ & +{\overline{r}}_L\overline{q}\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ & -\frac{l}{l_L}{L}^{\frac{1}{2}}\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]\}\hfill \\ \hfill \sim & M_0{L}^{\frac{3}{2}},\hfill \end{array}$$

(36)

where $M_0$ does not depend on $L.$ By Definition 4, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1}=& \underset{L\to +\infty }{lim}\sqrt{\frac{T_1(w{(\dot{\gamma }(t))}^2{L}^2{\lambda }_2^2{T}_2}{16|\omega (\dot{\gamma }(t)){|}^4{\lambda }_1^2{L}^2}}-\underset{L\to +\infty }{lim}\sqrt{\frac{M_0{L}^{\frac{3}{2}}}{L^3{\left|\omega (\dot{\gamma }(t))\right|}^6}}\hfill \\ \hfill =& \frac{{\lambda }_2^2\sqrt{T_1{T}_2}}{4{\lambda }_1{\left|\omega (\dot{\gamma }(t))\right|}^2},\hfill \end{array}\end{array}$$

(37)

where

$$T_1={(\overline{q}-\overline{p})}^2+{({\overline{r}}_L\overline{p}+{\overline{r}}_L\overline{q})}^2,$$

$$T_2={({\dot{\theta }}_1(t))}^2-2{\lambda }_1{\lambda }_{2}A{\dot{\theta }}_1(t)+{\lambda }_1^2{\lambda }_2^2{A}^2$$

if $\omega (\dot{\gamma }(t))\ne 0$.

When $\omega (\dot{\gamma }(t))=0$ and $\frac{d}{dt}(\omega (\dot{\gamma }(t)))=0$, we have

$$\begin{array}{cc}\hfill \parallel {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}^2=& \left\{\overline{q}{A}^{\prime }-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\ddot{\theta }}_1{(t)\}}^2\right\}\times \{\left[\overline{r_L}\overline{p}{A}^{\prime }+\frac{1}{{\lambda }_1{\lambda }_2}\overline{r_L}\overline{q}{\ddot{\theta }}_1(t)\right]\hfill \\ \hfill & -\frac{l}{l_L}{L}^{\frac{1}{2}}\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})\right]\}\hfill \\ \hfill \sim & {\left\{\overline{q}{A}^{\prime }-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\ddot{\theta }}_1(t)\right\}}^2,\hfill \\ \hfill \end{array}$$

(38)

and

$$\begin{array}{c}\hfill \parallel \dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}=|-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A|.\end{array}$$

(39)

Let $P=\left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}A\right]$ and $Q=\left[\overline{q}{A}^{\prime }-\frac{\overline{p}{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}\right]$. Then,

$$\begin{array}{cc}\hfill {\langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma },\dot{\gamma }\rangle }_{\Sigma ,L}& =PQ.\hfill \end{array}$$

(40)

By (37)–(39), we obtain

$${\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty }=\sqrt{\frac{P^2}{Q^4}-\frac{P^2{Q}^2}{Q^6}}=0.$$

When $\omega (\dot{\gamma }(t))=0$ and $\frac{d}{dt}(\omega (\dot{\gamma }(t)))\ne 0$, we have

$$\parallel {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}^2\sim \frac{l^2}{l_L^2}L{[\frac{d}{dt}(\omega (\dot{\gamma }(t)))]}^2\sim L{[\frac{d}{dt}(\omega (\dot{\gamma }(t)))]}^2,$$

$$\parallel \dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}=|-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A|,$$

$${\langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma },\dot{\gamma }\rangle }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}=O(1),$$

$$\begin{array}{cc}\hfill \underset{L\to +\infty }{lim}\frac{{\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^1}}{\sqrt{L}}& =\underset{L\to +\infty }{lim}\frac{1}{\sqrt{L}}\sqrt{\frac{L{\left[\frac{d}{dt}(\omega (\dot{\gamma }(t)))\right]}^2}{|-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A|}}\hfill \\ \hfill & =\frac{|\frac{d}{dt}(\omega (\dot{\gamma }(t)))|}{|-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}{A|}^2},\hfill \end{array}$$

if $\omega (\dot{\gamma }(t))=0$ and $\frac{d}{dt}(\omega (\dot{\gamma }(t)))\ne 0$. So, we obtain (31). □

Definition 6.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface. Let $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. The signed geodesic curvature ${\kappa }_{\gamma ,{\nabla }^1,\Sigma }^{L,s}$ of γ at $\gamma (t)$ is defined as

$${\kappa }_{\gamma ,{\nabla }^1,\Sigma }^{L,s}:=\frac{{\langle {\nabla }_{\dot{\gamma }}^{\Sigma ,1}\dot{\gamma },J_L(\dot{\gamma })\rangle }_{\Sigma ,L}}{\parallel \dot{\gamma }{\parallel }_{\Sigma ,L}^3},$$

where $J_L$ is defined by (27).

Now, we have defined the signed geodesic curvature associated to the ${\nabla }^1$, $k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^1}$ of $\gamma$ at $\gamma (t)$ and the intrinsic geodesic curvature associated to the ${\nabla }^1$, $k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1}$ of $\gamma$ at $\gamma (t)$.

Lemma 5.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface and $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. Then,

$$\begin{array}{cc}\hfill & k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1,s}=\frac{{\lambda }_2^2(-{\dot{\theta }}_1(t)+{\lambda }_1{\lambda }_{2}A)(\overline{q}-\overline{p})}{|\omega (\dot{r}(t))|},if\omega (\dot{\gamma }(t))\ne 0,\hfill \\ \hfill & k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1,s}=0,if\omega (\dot{\gamma }(t))=0,and\frac{d}{dt}(\omega (\dot{\gamma }(t)))=0,\hfill \\ \hfill & \underset{L\to +\infty }{lim}\frac{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^1,s}}{\sqrt{L}}=\frac{-\left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+{\lambda }_1\overline{q}A\right]\frac{d}{dt}\left(\omega (\dot{r}(t))\right)}{{[-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A]}^3},\hfill \\ \hfill & if\omega (\dot{\gamma }(t))=0and\frac{d}{dt}(\omega (\dot{\gamma }(t)))\ne 0.\hfill \\ \hfill \end{array}$$

(41)

Proof. 

By (27) and (32), we obtain

$$\begin{array}{cc}\hfill J_L(\dot{\gamma })& =\left[-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A\right]J_L(e_1)-\frac{l_L}{l}{L}^{\frac{1}{2}}\omega (\dot{\gamma }(t))J_L(e_2)\hfill \\ \hfill & =\frac{l_L}{l}{L}^{\frac{1}{2}}\omega (\dot{\gamma }(t))e_1+\left[-\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A\right]e_2.\hfill \end{array}$$

By (29) and the above equation, we have

$$\begin{array}{cc}\hfill \langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma },J_L(\dot{\gamma })\rangle =& \frac{l_L}{l}{L}^{\frac{1}{2}}\omega (\dot{\gamma }(t))\{\overline{q}\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ \hfill & -\overline{p}\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]{\}}^2\hfill \\ \hfill & +\{{\overline{r}}_L\overline{p}\left[A^{\prime }-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ \hfill & +{\overline{r}}_L\overline{q}\left[\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{{\dot{\theta }}_1(t)\omega (\dot{\gamma }(t))}{{\lambda }_1{\lambda }_2}·\frac{L{\lambda }_2^2+{\lambda }_1^2}{4}+A\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}\omega (\dot{\gamma }(t))\right]\hfill \\ \hfill & -\frac{l}{l_L}{L}^{\frac{1}{2}}\left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d\omega (\dot{\gamma }(t))}{dt}\right]\}\hfill \\ \hfill \sim & \frac{l_L}{l}{L}^{\frac{3}{2}}{\omega }^2(\dot{\gamma }(t)){\lambda }_2^2\frac{(-{\dot{\theta }}_1(t)+{\lambda }_1{\lambda }_{2}A)(\overline{q}-\overline{p})}{4{\lambda }_1{\lambda }_2}4{\lambda }_1{\lambda }_2+\frac{AL{\lambda }_2^2\omega (\dot{\gamma }(t))}{4}]\hfill \\ \hfill \sim & \frac{L^{\frac{3}{2}}{\omega }^2(\dot{\gamma }(t)){\lambda }_2(-\dot{{\theta }_1}(t)+{\lambda }_1{\lambda }_{2}A)(\overline{q}-\overline{p})}{4{\lambda }_1}\mathrm{as}L\to +\infty .\hfill \end{array}$$

So, we obtain

$$\begin{array}{cc}\hfill {\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^1,s}=& \frac{{\langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},L}\dot{\gamma },J_L(\dot{\gamma })\rangle }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}}{\parallel \dot{\gamma }{\parallel }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}^3}\hfill \\ \hfill =& \frac{L^{\frac{3}{2}}{\omega }^2(\dot{\gamma }(t)){\lambda }_2(-\dot{{\theta }_1}(t)+{\lambda }_1{\lambda }_{2}A)(\overline{q}-\overline{p})}{4{\lambda }_1{L}^{\frac{3}{2}}{\left|\omega (\dot{\gamma }(t))\right|}^3}.\hfill \end{array}$$

Furthermore,

$${\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1,s}=\underset{L\to +\infty }{lim}{k}_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,s}=\frac{-{\lambda }_2(\dot{{\theta }_1}(t)+{\lambda }_1{\lambda }_{2}A)(\overline{q}-\overline{p})}{|\omega (\dot{\gamma }(t))|}.$$

When $\omega (\dot{\gamma }(t))=0$ and $\frac{d}{dt}(\omega (\dot{\gamma }(t)))=0$, we obtain

$$\begin{array}{cc}\hfill {\langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma },J_L(\dot{\gamma })\rangle }_{L,\Sigma }=& \left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}A\right]\{\overline{r_L}\overline{p}{A}^{\prime }+\frac{{\overline{r}}_L\overline{q}{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}\hfill \\ \hfill & -\frac{l{L}^{\frac{1}{2}}}{l_L}\left[(A+\frac{\dot{\theta }(t)}{{\lambda }_1{\lambda }_2})(\frac{1}{{\lambda }_1{\lambda }_2}{\dot{\theta }}_1(t)\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})\right]\}\hfill \\ \hfill \sim & M_0{L}^{-\frac{1}{2}}\mathrm{as}L\to +\infty .\hfill \end{array}$$

So, ${\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1,s}=0.$ When $\omega (\dot{\gamma }(t))=0$ and $\frac{d}{dt}(\omega (\dot{\gamma }(t)))\ne 0$, we have

$$\begin{array}{cc}\hfill {\langle {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},1}\dot{\gamma },J_L(\dot{\gamma })\rangle }_{L,{\Sigma }_{\mathcal{R}\mathcal{T}}}=& \left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}A\right]\{\overline{r_L}\overline{p}{A}^{\prime }+\frac{{\overline{r}}_L\overline{q}{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\frac{l}{l_L}{L}^{\frac{1}{2}}\hfill \\ \hfill & \times \left[(A+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2})(\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}\frac{{\lambda }_1^2+L{\lambda }_2^2}{4t}+A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})+\frac{d\omega (\dot{\gamma }(t))}{dt}\right]\}\hfill \\ \hfill \sim & \left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}A\right]\left(-\frac{l}{l_L}{L}^{\frac{1}{2}})\frac{d}{dt}(\omega (\dot{\gamma }(t))\right)\mathrm{as}L\to +\infty \hfill \\ \hfill \sim & -\left[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}A\right]\frac{d}{dt}(\omega (\dot{\gamma }(t)))L^{\frac{1}{2}}\mathrm{as}L\to +\infty .\hfill \end{array}$$

We obtain

$$\begin{array}{cc}\hfill {\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^1,s}=& \underset{L\to +\infty }{lim}\frac{{\kappa }_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^L}{\sqrt{L}}\hfill \\ \hfill =& \underset{L\to +\infty }{lim}\frac{-[-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}A]\frac{d}{dt}(\omega (\dot{\gamma }(t)))L^{\frac{1}{2}}}{|-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}{A|}^3\sqrt{L}}\hfill \\ \hfill =& \frac{[\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}-\overline{q}A]\frac{d}{dt}(\omega (\dot{\gamma }(t)))}{|-\frac{\overline{p}{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+\overline{q}{A|}^3\sqrt{L}}.\hfill \end{array}$$

□

In the following, we will compute the sub-Riemannian limit of the Riemannian Gaussian curvature of surfaces in $(\mathcal{R}\mathcal{T},g_L)$. We define the second fundamental form $I{I}^L$ of the embedding of ${\Sigma }_{\mathcal{R}\mathcal{T}}$ into $(\mathcal{R}\mathcal{T},g_L)$:

$$\begin{array}{c}\hfill I{I}^{L,{\nabla }^1}=\left(\begin{array}{cc}{\langle {\nabla }_{e_1}^L{v}_L,e_1\rangle }_L& {\langle {\nabla }_{e_1}^L{v}_L,e_2\rangle }_L\\ {\langle {\nabla }_{e_2}^L{v}_L,e_1\rangle }_L& {\langle {\nabla }_{e_2}^L{v}_L,e_2\rangle }_L\end{array}\right).\end{array}$$

(42)

We have the following theorem.

Theorem 1.

The second fundamental form $I{I}^{L,{\nabla }^1}$ of the embedding of ${\Sigma }_{\mathcal{R}\mathcal{T}}$ into $(\mathcal{R}\mathcal{T},g_L)$ is given by

$$\begin{array}{c}\hfill I{I}^{L,{\nabla }^1}=\left(\begin{array}{cc}{h}_{11}& h_{12}\\ h_{21}& h_{22}\end{array}\right),\end{array}$$

(43)

where

$$\begin{array}{c}\hfill h_{11}=\frac{l}{l_L}(F_1(\overline{p})+F_2(\overline{q}))-{\overline{r}}_L\frac{{(\overline{p}-\overline{q})}^2{\lambda }_1^2+({\overline{p}}^2-{\overline{q}}^2)L{\lambda }_2^2}{4\sqrt{L}},\end{array}$$

(44)

$$\begin{array}{cc}\hfill h_{12}=h_{21}=& -\frac{l_L}{l}{\langle e_1,{\nabla }_H\overline{r_L}\rangle }_L+(2p_L{\overline{q}}_L+2\overline{p}{r}_L^2\overline{q}+1)\frac{\sqrt{L}}{4}{\lambda }_2^2-({\overline{q}}_L^2-{\overline{p}}_L^2-2{\overline{q}}_L^2{\overline{r}}_L^2)\frac{{\lambda }_1^2}{4\sqrt{L}},\hfill \end{array}$$

(45)

$$\begin{array}{c}\hfill h_{22}=-\frac{l^2}{l_L^2}{\langle e_2,{\nabla }_H^1(\frac{r}{l})\rangle }_L+\widetilde{F_3}(\overline{r_L})-\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L+\frac{{\lambda }_1^2}{2\sqrt{L}}{\overline{r}}_L^2\overline{p}q-\frac{\sqrt{L}}{2}{\lambda }_2^2{\overline{r}}_L({\overline{p}}_L^2+1+{\overline{p}}^2{\overline{r}}_L^2).\end{array}$$

(46)

Proof. 

Since ${\langle e_1,v_L\rangle }_L=0$ and ${\langle e_2,v_L\rangle }_L=0,$ we have

$${\langle {\nabla }_{e_1}^1{v}_L,e_1\rangle }_L=-{\langle {\nabla }_{e_1}^1{e}_1,v_L\rangle }_L,{\langle {\nabla }_{e_2}^1{v}_L,e_2\rangle }_L=-{\langle {\nabla }_{e_2}^1{e}_2,v_L\rangle }_L.$$

Using the definition of the connection, identities in (6), and grouping terms, we have

$$\begin{array}{cc}\hfill {\nabla }_{e_1}^1{e}_1=& {\nabla }_{\overline{q}{F}_1-\overline{p}{F}_2}^1\overline{q}{F}_1-\overline{p}{F}_2\hfill \\ \hfill =& \overline{q}\left(F_1(\overline{q})F_1+\overline{q}{\nabla }_{F_1}^1{F}_1-F_1(\overline{p})F_2-\overline{p}{\nabla }_{F_1}^1{F}_2\right)\hfill \\ \hfill & -\overline{p}\left(F_2(\overline{q})F_1+\overline{q}{\nabla }_{F_2}^1{F}_1-F_2(\overline{p})F_2-\overline{p}{\nabla }_{F_2}^1{F}_2\right)\hfill \\ \hfill =& \overline{q}\left[F_1(\overline{q})F_1-F_1(\overline{p})F_2+(\overline{q}-\overline{p})\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}{F}_3\right]\hfill \\ \hfill & -\overline{p}\left[F_2(\overline{q})F_1-F_2(\overline{p})F_2+(\overline{q}-\overline{p})\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}{F}_3\right]\hfill \\ \hfill =& \left[\overline{q}{F}_1(\overline{q})-\overline{p}{F}_2(\overline{q})\right]F_1-\left[\overline{q}{F}_1(\overline{p})-\overline{p}{F}_2(\overline{p})\right]F_2\hfill \\ \hfill & +\left[{\overline{q}}^2\frac{{\lambda }_1^2-L{\lambda }_1^2}{4L}-\overline{p}\overline{q}(\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}+\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L})+p^2\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right]F_3\hfill \\ \hfill =& \left[\overline{q}{F}_1(\overline{q})-\overline{p}{F}_2(\overline{q})\right]F_1-\left[\overline{q}{F}_1(\overline{p})-\overline{p}{F}_2(\overline{p})\right]F_2\hfill \\ \hfill -& \left[{\overline{q}}^2\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}-p\overline{q}\frac{{\lambda }_1^2}{2L}+{\overline{p}}^2\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right]F_3.\hfill \end{array}$$

Since ${\overline{p}}^2+{\overline{q}}^2=1$, we have $\overline{p}{F}_j\overline{p}+\overline{q}{F}_j\overline{q}=0,j=1,2,3.$ Thus, $-\overline{q}{F}_1\overline{q}=\overline{p}{F}_1\overline{p}$ and

$-\overline{q}{F}_2\overline{q}=\overline{p}{F}_2\overline{p}$, and we have

$$\begin{array}{cc}\hfill {\nabla }_{e_1}^1{e}_1=& -\overline{p}[F_1(\overline{p})+F_2(\overline{q})]F_1-\overline{q}[F_1(\overline{p})+F_2(\overline{q})]F_2\hfill \\ \hfill & -\left[{\overline{q}}^2\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}-p\overline{q}\frac{{\lambda }_1^2}{2L}+{\overline{p}}^2\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right]F_3.\hfill \end{array}$$

Next, we compute the inner product of this with $v_L$, and we obtain

$$\begin{array}{cc}\hfill h_{11}=& -{\langle {\nabla }_{e_1}^1{e}_1,v_L\rangle }_L\hfill \\ \hfill =& \overline{p}{\overline{p}}_L(F_1(\overline{p})+F_2(\overline{q}))+\overline{q}{\overline{q}}_L(F_1(\overline{p})+F_2(\overline{q}))\hfill \\ \hfill & -\overline{r_L}\left[{\overline{q}}^2\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}-p\overline{q}\frac{{\lambda }_1^2}{2L}+{\overline{p}}^2\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right]\sqrt{L}\hfill \\ \hfill =& \frac{p}{l}\frac{p}{l_L}(F_1(\overline{p})+F_2(\overline{q}))+\frac{q}{l}\frac{q}{l_L}(F_1(\overline{p})+F_2(\overline{q}))\hfill \\ \hfill & -\overline{r_L}\sqrt{L}\left[{\overline{q}}^2\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}-p\overline{q}\frac{{\lambda }_1^2}{2L}+{\overline{p}}^2\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right]\hfill \\ \hfill =& \frac{1}{l{l}_L}(p^2+q^2)F_1(\overline{p})+\frac{1}{l{l}_L}(p^2+q^2)F_2(\overline{q})\hfill \\ \hfill & -\overline{r_L}\sqrt{L}\left[{\overline{q}}^2\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}-p\overline{q}\frac{{\lambda }_1^2}{2L}+{\overline{p}}^2\frac{{\lambda }_1^2+L{\lambda }_2^2}{4L}\right]\hfill \\ \hfill =& \frac{l}{l_L}(F_1(\overline{p})+F_2(\overline{q}))-\overline{r_L}\frac{{(\overline{p}-\overline{q})}^2{\lambda }_1^2+({\overline{p}}^2-{\overline{q}}^2)L{\lambda }_2^2}{4L}.\hfill \end{array}$$

To compute $h_{12}$ and $h_{21}$, using the definition of the connection, we obtain

$$\begin{array}{cc}\hfill {\nabla }_{e_1}^1{e}_2=& {\nabla }_{\overline{q}{F}_1-\overline{p}{F}_2}^1\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}{L}^{-\frac{1}{2}}{F}_3\hfill \\ \hfill =& \overline{q}\left({\nabla }_{F_1}^1\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}{L}^{-\frac{1}{2}}{F}_3\right)-\overline{p}\left({\nabla }_{F_2}^1\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}{L}^{-\frac{1}{2}}{F}_3\right)\hfill \\ \hfill =& \overline{q}\left[F_1(\overline{r_L}\overline{p})F_1+\overline{r_L}\overline{p}{\nabla }_{F_1}^1{F}_1+F_1\overline{r_L}\overline{p}{F}_2+\overline{r_L}\overline{p}{\nabla }_{F_1}^1{F}_2\right]\hfill \\ \hfill & -\overline{q}\left[F_1(\frac{l}{l_L})L^{-\frac{1}{2}}{F}_3-\frac{l}{l_L}{L}^{-\frac{1}{2}}{\nabla }_{F_1}^1{F}_3\right]-\overline{p}\left[-F_2(\frac{l}{l_L})L^{-\frac{1}{2}}{F}_3-\frac{l}{l_L}{L}^{-\frac{1}{2}}{\nabla }_{F_2}^1{F}_3\right]\hfill \\ \hfill & -\overline{p}\left[F_2(\overline{r_L}\overline{p})F_1+\overline{r_L}\overline{p}{\nabla }_{F_2}^1{F}_1+F_2\overline{r_L}\overline{p}{F}_2+\overline{r_L}\overline{p}{\nabla }_{F_2}^1{F}_2\right]\hfill \\ \hfill =& \left(\overline{q}{F}_1(\overline{r_L}\overline{p})-\overline{p}{F}_2(\overline{r_L}\overline{p})-\frac{\overline{p}l}{l_L\sqrt{L}}\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}+\frac{\overline{p}l}{l_L\sqrt{L}}\frac{-{\lambda }_1^2-L{\lambda }_2^2}{4}\right)F_1\hfill \\ \hfill & +\left(\overline{q}{F}_1(\overline{r_L}\overline{q})-\overline{p}{F}_2(\overline{r_L}\overline{q})-\frac{\overline{p}l}{l_L\sqrt{L}}\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}+\frac{\overline{p}l}{l_L\sqrt{L}}\frac{-{\lambda }_1^2-L{\lambda }_2^2}{4}\right)F_2\hfill \\ \hfill & +\{\left[(\overline{q}{\overline{r}}_L\overline{p}+\overline{q}{\overline{r}}_L\overline{q})\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}-(\overline{p}{\overline{r}}_L\overline{p}+\overline{p}{\overline{r}}_L\overline{q})\frac{{\lambda }^2+L{\lambda }_2^2}{4L})\right]\hfill \\ \hfill & +\left[-\frac{\overline{q}}{\sqrt{L}}{F}_1(\frac{l}{l_L})+\frac{\overline{p}}{\sqrt{L}}{F}_2(\frac{l}{l_L})\right]\}{F}_3.\hfill \end{array}$$

Next, we compute the inner product of this with $v_L$. Using the product rule and the identity $\overline{q_L}\overline{p}=\overline{p_L}\overline{q}$, we obtain

$$\begin{array}{cc}\hfill {\langle {\nabla }_{e_1}{e}_2,v_L\rangle }_L=& {\overline{q}}_L{F}_1{\overline{r}}_L-\overline{p_L}{F}_2{\overline{r}}_L-(2{\overline{p}}_L{\overline{q}}_L+2\overline{p}{\overline{r}}_L^2\overline{q}+1)\frac{\sqrt{L}}{4}{\lambda }_2^2\hfill \\ \hfill & -({\overline{q}}_L^2-{\overline{p}}_L^2-2{\overline{q}}_L^2{\overline{r}}_L^2)\frac{{\lambda }_1^2}{4\sqrt{L}}-{\overline{r}}_L\overline{q}{F}_1(\frac{1}{l_L})+{\overline{r}}_L\overline{p}{F}_2(\frac{1}{l_L}).\hfill \end{array}$$

To simplify this, we find

$$\overline{p_L}\overline{q}\overline{p}+\overline{q_L}{\overline{q}}^2=(\overline{q_L}{\overline{p}}^2+\overline{q_L}{\overline{q}}^2)=\overline{q_L}({\overline{p}}^2+{\overline{q}}^2)=\overline{q_L},$$

$$\overline{p_L}{\overline{p}}^2+\overline{q_L}\overline{p}\overline{q}=\overline{p_L}{\overline{p}}^2+\overline{p_L}{\overline{q}}^2=\overline{p_L}({\overline{p}}^2+{\overline{q}}^2)=\overline{p_L},$$

$$\overline{p_L}\overline{r_L}\overline{q}{F}_1\overline{p}+\overline{r_L}\overline{q_L}\overline{q}{F}_1\overline{q}=\overline{p_L}\overline{r_L}\overline{q}{F}_1\overline{p}+\overline{r_L}\overline{q_L}(-\overline{p}{F}_1\overline{p})=\overline{r_L}(\overline{p_L}\overline{q}-\overline{q_L}\overline{p})F_1\overline{p}=0,$$

$$\overline{r_L}(\overline{p}\overline{p_L}{F}_2\overline{p}+\overline{p}\overline{q_L}{F}_2\overline{q})=\overline{r_L}(\overline{p_L}\overline{p}{F}_2\overline{p}+\overline{p_L}\overline{q}{F}_2\overline{q})=\overline{r_L}\overline{p_L}(\overline{p}{F}_2\overline{p}+\overline{q}{F}_2\overline{q})=0,$$

and

$$\frac{{\overline{p}}_L\overline{p}l}{l_L{l}_L\sqrt{L}}=\frac{{\overline{p}}_L^2}{\sqrt{L}},\frac{{\overline{q}}_L\overline{q}l}{l_L\sqrt{L}}=\frac{{\overline{q}}_L^2}{\sqrt{L}},\frac{{\overline{p}}_L\overline{q}l}{l_L\sqrt{L}}=\frac{{\overline{q}}_L\overline{p}l}{l_L\sqrt{L}}=\frac{{\overline{p}}_L{\overline{q}}_L}{\sqrt{L}}.$$

Under these simplifications, we obtain

$$\begin{array}{cc}\hfill {\langle {\nabla }_{e_1}{e}_2,v_L\rangle }_L=& {\overline{q}}_L{F}_1{\overline{r}}_L-{\overline{p}}_L{F}_2{\overline{r}}_L-(2{\overline{p}}_L{\overline{q}}_L+2\overline{p}{\overline{r}}_L^2\overline{q}+1)\frac{\sqrt{L}}{4}{\lambda }_2^2\hfill \\ \hfill & -({\overline{q}}_L^2-{\overline{p}}_L^2-2{\overline{q}}_L^2{\overline{r}}_L^2)\frac{{\lambda }_1^2}{4\sqrt{L}}-{\overline{r}}_L\overline{q}{F}_1(\frac{1}{l_L})+{\overline{r}}_L\overline{p}{F}_2(\frac{1}{l_L}).\hfill \\ \hfill =& \frac{l}{l_L}{\langle e_1,{\nabla }_H{\overline{r}}_L\rangle }_L-{\overline{r}}_L{\langle e_1,{\nabla }_H(\frac{l}{l_L})\rangle }_L-(2{\overline{p}}_L{\overline{q}}_L+2\overline{p}{\overline{r}}_L^2\overline{q}+1)\frac{\sqrt{L}}{4}{\lambda }_2^2\hfill \\ \hfill & -({\overline{q}}_L^2-{\overline{p}}_L^2-2{\overline{q}}_L^2{\overline{r}}_L^2)\frac{{\lambda }_1^2}{4\sqrt{L}}.\hfill \end{array}$$

Finally, we use the identity $(\frac{l}{l_L}-\frac{l_L}{l}){\nabla }_H\overline{r_L}=\overline{r_L}{\nabla }_H(\frac{l}{l_L})$ in the above equation:

$$\begin{array}{cc}\hfill {\langle {\nabla }_{e_1}^1{e}_2,v_L\rangle }_L=& \frac{l}{l_L}{\langle e_1,{\nabla }_H{\overline{r}}_L\rangle }_L-(2{\overline{p}}_L{\overline{q}}_L+2\overline{p}{\overline{r}}_L^2\overline{q}+1)\frac{\sqrt{L}}{4}{\lambda }_2^2-({\overline{q}}_L^2-{\overline{p}}_L^2-2{\overline{q}}_L^2{\overline{r}}_L^2)\frac{{\lambda }_1^2}{4\sqrt{L}}.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill h_{12}=h_{21}=& -\frac{l}{l_L}{\langle e_1,{\nabla }_H{\overline{r}}_L\rangle }_L+(2{\overline{p}}_L{\overline{q}}_L+2\overline{p}{\overline{r}}_L^2\overline{q}+1)\frac{\sqrt{L}}{4}{\lambda }_2^2+({\overline{q}}_L^2-{\overline{p}}_L^2-2{\overline{q}}_L^2{\overline{r}}_L^2)\frac{{\lambda }_1^2}{4\sqrt{L}}.\hfill \end{array}$$

Since ${\langle {\nabla }_{e_2}^1{v}_L,e_2\rangle }_L=-{\langle {\nabla }_{e_2}^1{e}_2,v_L\rangle }_L,$ using the definitions of connection, identities in (5), and grouping terms, we have

$$\begin{array}{cc}\hfill {\nabla }_{e_2}^1{e}_2=& {\nabla }_{\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}{L}^{-\frac{1}{2}}{F}_3}^L\left(\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}{L}^{-\frac{1}{2}}{F}_3\right)\hfill \\ \hfill =& \left[{\overline{r}}_L\overline{p}{F}_1({\overline{r}}_L\overline{p})+{\overline{r}}_L\overline{q}{F}_2({\overline{r}}_L\overline{q})-\frac{l}{l_L}{\tilde{F}}_3({\overline{r}}_L\overline{p})+\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{p}}_L+\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{q}}_L\right]F_1\hfill \\ \hfill & +\left[{\overline{r}}_L\overline{p}{F}_1({\overline{r}}_L\overline{q})+{\overline{r}}_L\overline{q}{F}_2({\overline{r}}_L\overline{q})-\frac{l}{l_L}{\tilde{F}}_3({\overline{r}}_L\overline{p})+\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{p}}_L+\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{q}}_L\right]F_2\hfill \\ \hfill & +\{\left[-{\overline{r}}_L\overline{p}{F}_1(\frac{l}{l_L})-{\overline{r}}_L\overline{q}{F}_2(\frac{l}{l_L})+\frac{l}{l_L}{\tilde{F}}_3(\frac{l}{l_L})\right]\hfill \\ \hfill & +\left[\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L^2+\frac{{\lambda }_1^2}{2\sqrt{L}}{\overline{r}}_L^2\overline{p}\overline{q}-\frac{\sqrt{L}{\lambda }_2^2}{2}-\frac{\sqrt{L}}{2}{\lambda }_2^2{\overline{p}}^2{\overline{r}}_L^2\right]\}{\tilde{F}}_3.\hfill \end{array}$$

Taking the inner product with $v_L$ yields

$$\begin{array}{cc}\hfill {\langle {\nabla }_{e_2}^1{e}_2,v_L\rangle }_L=& \overline{p_L}\left(\overline{r_L}{\overline{p}}^2{F}_1\overline{r_L}+{\overline{r_L}}^2\overline{p}{F}_1\overline{p}+\overline{r_L}\overline{q}\overline{p}{F}_2\overline{r_L}+{\overline{r_L}}^2\overline{q}{F}_2\overline{p}\right)\hfill \\ \hfill & +\overline{p_L}\left[-\frac{l}{l_L}{\tilde{F}}_3({\overline{r}}_L)-\frac{l}{l_L}{\overline{r}}_L{\tilde{F}}_3(\overline{p})+\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{p}}_L+\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{q}}_L\right]\hfill \\ \hfill & +\overline{q_L}\left(\overline{r_L}\overline{p}\overline{q}{F}_1\overline{r_L}+{\overline{r_L}}^2\overline{p}{F}_1\overline{q}+{\overline{r_L}}^2\overline{q}{F}_2\overline{r_L}+{\overline{r_L}}^2\overline{q}{F}_2\overline{q}\right)\hfill \\ \hfill & +\overline{q_L}\left[-\frac{l}{l_L}{\tilde{F}}_3({\overline{r}}_L)-\frac{l}{l_L}{\overline{r}}_L{\tilde{F}}_3(\overline{q})+\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{p}}_L+\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L{\overline{q}}_L\right]\hfill \\ \hfill & +{\overline{r}}_L\left[-{\overline{r}}_L\overline{p}{F}_1(\frac{l}{l_L})-{\overline{r}}_L\overline{q}{F}_2(\frac{l}{l_L})+\frac{l}{l_L}{\tilde{F}}_3(\frac{l}{l_L})\right]\hfill \\ \hfill & +{\overline{r}}_L\left[\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L^2+\frac{{\lambda }_1^2}{2\sqrt{L}}{\overline{r}}_L^2\overline{p}\overline{q}-\frac{\sqrt{L}}{2}{\lambda }_2^2(1+{\overline{p}}^2{\overline{r}}_L^2)\right].\hfill \end{array}$$

Under some similar simplifications to those in Theorem 4.3 in [16], we obtain

$$\begin{array}{cc}\hfill h_{22}=& -{\langle {\nabla }_{e_2}^1{e}_2,v_L\rangle }_L\hfill \\ \hfill =& -\frac{l^2}{l_L^2}{\langle e_2,{\nabla }_H^1(\frac{r}{l})\rangle }_L+\widetilde{F_3}(\overline{r_L})-\frac{{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{\overline{r}}_L-\frac{{\lambda }_1^2}{2\sqrt{L}}{\overline{r}}^2\overline{p}\overline{q}+\frac{\sqrt{L}}{2}{\lambda }_2^2{\overline{r}}_L^2(p_L^2+1+{\overline{p}}^2{\overline{r}}_L^2).\hfill \end{array}$$

□

The mean curvature associated with the ${\nabla }^1,{\mathcal{H}}_{{\nabla }^1,L}$ of ${\Sigma }_{\mathcal{R}\mathcal{T}}$ is defined by ${\mathcal{H}}_{{\nabla }^1,L}:=tr(I{I}^{{\nabla }^1,L}).$ Let

$$\begin{array}{c}\hfill {\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},{\nabla }^1}(e_1,e_2)=-{\langle R^{1,{\Sigma }_{\mathcal{R}\mathcal{T}}}(e_1,e_2)e_1,e_2\rangle }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L},{\mathcal{K}}^{{\nabla }^1}(e_1,e_2)=-\langle R^{(}{e}_1,e_2)e_1,e_2{\rangle }_L.\end{array}$$

(47)

By the Gauss equation, we have

$$\begin{array}{c}\hfill {\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},{\nabla }^1}(e_1,e_2)={\mathcal{K}}^{{\nabla }^1}(e_1,e_2)+det(I{I}^{{\nabla }^1,L}).\end{array}$$

(48)

Proposition 2.

Away from characteristic points, the horizontal mean curvature associated with the ${\nabla }^1,{\mathcal{H}}_{{\nabla }^1,\infty }$ of ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ is given by

$$\begin{array}{c}\hfill {\mathcal{H}}_{{\nabla }^1,\infty }=\underset{L\to +\infty }{lim}{\mathcal{H}}_{{\nabla }^1,L}=F_1(\overline{p})+F_2(\overline{q}).\end{array}$$

(49)

By Lemma 1 and (7), we have ${\mathcal{K}}^{{\nabla }^1}(e_1,e_2)=0$. By (27) and Proposition 2, we obtain the following proposition.

Proposition 3.

Away from characteristic points, we have

$$\begin{array}{c}\hfill {\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},{\nabla }^1}(e_1,e_2)\to {\mathcal{K}}^{\Sigma ,{\nabla }^1,\infty }+O(\frac{1}{\sqrt{L}}),\mathrm{as}L\to +\infty ,\end{array}$$

(50)

where

$$\begin{array}{cc}\hfill {\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},{\nabla }^1,\infty }:=& {\lambda }_2^2\widetilde{F_3}u[F_1(\overline{p})+F_2(\overline{q})][-\frac{3}{4}-\frac{1}{2}{\overline{p}}^2(1+{(\widetilde{F_3}u)}^2]\hfill \\ \hfill & -[{(\overline{p}-\overline{q})}^2+1]<e_1,{\nabla }_H\frac{F_{3}u}{|{{\nabla }_H}^u|})>.\hfill \\ \hfill \end{array}$$

(51)

Proof. 

We compute

$$\begin{array}{cc}\hfill R^1\left(e_1,e_2\right)e_1& =R^1\left(\overline{q}{F}_1-\overline{p}{F}_2,{\overline{r}}_L\overline{p}{F}_1+{\overline{r}}_L\overline{q}{F}_2-\frac{l}{l_L\sqrt{L}}{F}_3\right)\left(\overline{q}{F}_1-\overline{p}{F}_2\right)\hfill \\ \hfill & ={\overline{r}}_L\overline{p}{\overline{q}}^2{R}^1\left(F_1,F_1\right)F_1+{\overline{r}}_L{\overline{q}}^3{R}^1\left(F_1,F_2\right)F_1-\frac{l{\overline{q}}^2}{l_L\sqrt{L}}{R}^1\left(F_1,F_3\right)F_1\hfill \\ \hfill & -{\overline{r}}_L{\overline{p}}^2\overline{q}{R}^1\left(F_2,F_1\right)F_1-{\overline{r}}_L\overline{p}{\overline{q}}^2{R}^1(F_2,F_2)F_1+\frac{l\overline{p}\overline{q}}{l_L\sqrt{L}}{R}^1\left(F_2,F_3\right)F_1\hfill \\ \hfill & -{\overline{r}}_L{\overline{p}}^2\overline{q}{R}^1\left(F_1,F_1\right)F_2-{\overline{r}}_L\overline{p}{\overline{q}}^2{R}^1\left(F_1,F_2\right)F_2+\frac{l\overline{p}\overline{q}}{l_L\sqrt{L}}{R}^1\left(F_1,F_3\right)F_2\hfill \\ \hfill & +{\overline{r}}_L{\overline{p}}^3{R}^1\left(F_2,F_1\right)F_2+{\overline{r}}_L{\overline{p}}^2\overline{q}{R}^1\left(F_2,F_2\right)F_2-\frac{l{\overline{p}}^2}{l_L\sqrt{L}}{R}^1\left(F_2,F_3\right)F_2\hfill \\ \hfill & ={\overline{r}}_L\overline{p}{R}^1(F_1,F_2)F_1-\frac{l{\overline{q}}^2}{l_L\sqrt{L}}{R}^1(F_1,F_3)F_1+\frac{l\overline{p}\overline{q}}{l_L\sqrt{L}}{R}^1(F_2,F_3)F_1\hfill \\ \hfill & -{\overline{r}}_L\overline{p}{R}^1(F_1,F_2)F_2-\frac{l\overline{p}\overline{q}}{l_L\sqrt{L}}{R}^1(F_1,F_3)F_2+\frac{l{\overline{p}}^2}{l_L\sqrt{L}}{R}^1(F_2,F_3)F_2\hfill \\ \hfill & =\frac{l\overline{p}\overline{q}}{l_L\sqrt{L}}\frac{{\lambda }_1^4-L{\lambda }_1^2{\lambda }_2^2}{4L}{F}_3-\frac{l{\overline{p}}^2}{l_L\sqrt{L}}\frac{{\lambda }_1^4-L{\lambda }_1^2{\lambda }_2^2}{4L}{F}_3,\hfill \end{array}$$

(52)

and

$$\begin{array}{cc}\hfill {\mathcal{K}}^1\left(e_1,e_2\right)& =-{〈R^1\left(e_1,e_2\right)e_1,e_2〉}_L\hfill \\ \hfill & =\frac{l}{l_L\sqrt{L}}\left(\frac{l\overline{p}\overline{q}}{l_L\sqrt{L}}\frac{{\lambda }_1^4-L{\lambda }_1^2{\lambda }_2^2}{4L}-\frac{l{\overline{p}}^2}{l_L\sqrt{L}}\frac{{\lambda }_1^4-L{\lambda }_1^2{\lambda }_2^2}{4L}\right).\hfill \end{array}$$

(53)

We obtain

$${\mathcal{K}}^1\left(e_1,e_2\right)\sim 0asL\to \infty .$$

(54)

By Theorem 1 and ${\nabla }_H\left({\overline{r}}_L\right)=L^{-\frac{1}{2}}{\nabla }_H\left(\frac{F_{3}u}{|{\nabla }_{H}u|}\right)+O\left(L^{-1}\right)$ as $L\to +\infty$, we obtain

$$\begin{array}{cc}\hfill det\left(I{I}^L\right)=& h_{11}{h}_{22}-h_{12}^2\hfill \\ \hfill =& [F_1(\overline{p})+F_2(\overline{q})]{\lambda }_2^2\widetilde{F_3}u[-\frac{3}{4}-\frac{1}{2}{\overline{p}}^2(1+{(\widetilde{F_3}u)}^2]\hfill \\ \hfill & -<e_1,{\nabla }_H\frac{F_{3}u}{|{{\nabla }_H}^u|})>[{(\overline{p}-\overline{q})}^2+1]\hfill \end{array}$$

(55)

as $L\to +\infty .$ By (48), (53), and (54), we obtain the desired equation. □

Let us first consider the case of a regular curve $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$. We define the Riemannian length measure $\begin{array}{c}\hfill d{s}_L=\left|\right|\dot{\gamma }{\left|\right|}_{L}dt.\end{array}$

Proposition 4.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a Euclidean $C^2$-smooth surface and ${\Sigma }_{\mathcal{R}\mathcal{T}}=\{u=0\}$ and $d{\sigma }_{{\Sigma }_{\mathcal{R}\mathcal{T}},L}$ denote the surface measures on ${\Sigma }_{\mathcal{R}\mathcal{T}}$ with respect to the Riemannian metric $g_L.$ Let

$$d{\sigma }_{\Sigma }:=\left(\overline{p}{\omega }_2-\overline{q}{\omega }_1\right)\wedge \omega ,d{\overline{\sigma }}_{\Sigma }:=\frac{F_{3}u}{l}{\omega }_1\wedge {\omega }_2-\frac{{\left(F_{3}u\right)}^2}{2l^2}\left(\overline{p}{\omega }_2-\overline{q}{\omega }_1\right)\wedge \omega .$$

Then,

$$\frac{1}{\sqrt{L}}d{\sigma }_{\Sigma ,L}=d{\sigma }_{\Sigma }+d{\overline{\sigma }}_{\Sigma }{L}^{-1}+O\left(L^{-2}\right),asL\to +\infty .$$

(56)

Proof. 

It is well known that

$$g_L(F_1,·)={\omega }_1,g_L(F_2,·)={\omega }_2,g_L(F_3,·)=L\omega .$$

We define ${e_1}^{*}:=g_L(e_1,·),{e_2}^{*}:=g_L(e_2,·).$ Then,

$$e_1^{*}=\overline{q}{\omega }_1-\overline{p}{\omega }_2,e_2^{*}={\overline{r}}_L\overline{p}{\omega }_1+{\overline{r}}_L\overline{q}{\omega }_2-\frac{l}{l_L}{L}^{\frac{1}{2}}\omega .$$

Therefore,

$$\frac{1}{\sqrt{L}}d{\sigma }_{\Sigma ,L}=\frac{1}{\sqrt{L}}{e}_1^{*}\wedge e_2^{*}=\frac{l}{l_L}\left(\overline{p}{\omega }_2-\overline{q}{\omega }_1\right)\wedge \omega +\frac{1}{\sqrt{L}}{\overline{r}}_L{\omega }_1\wedge {\omega }_2,$$

recalling

$${\overline{r}}_L=\frac{\left(F_{3}u\right)L^{-\frac{1}{2}}}{\sqrt{p^2+q^2+L^{-1}{\left(F_{3}u\right)}^2}}$$

and the Taylor expansion

$$\frac{1}{l_L}=\frac{1}{l}-\frac{1}{2l^3}{\left(F_{3}u\right)}^2{L}^{-1}+O\left(L^{-2}\right)asL\to +\infty .$$

□

So, we obtain a Gauss-Bonnet theorem in $(\mathcal{R}\mathcal{T},g_L)$ as follows.

Theorem 2.

Let $\Sigma \subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface with finitely many boundary components ${(\partial \Sigma )}_i,i\in \{1,\cdots ,n\}$, given by Euclidean $C^2$-smooth regular and closed curves ${\gamma }_i:[0,2\pi ]\to {(\partial {\Sigma }_{\mathcal{R}\mathcal{T}})}_i.$ Suppose that the characteristic set $C({\Sigma }_{\mathcal{R}\mathcal{T}})$ satisfies ${\mathcal{H}}^1(C({\Sigma }_{\mathcal{R}\mathcal{T}}))=0$, where ${\mathcal{H}}^1(C({\Sigma }_{\mathcal{R}\mathcal{T}}))$ denotes the Euclidean one-dimensional Hausdorff measure of $C({\Sigma }_{\mathcal{R}\mathcal{T}})$ and that $\Vert {\nabla }_H{u\Vert }_H^{-1}$ is locally summable with respect to the Euclidean two-dimensional Hausdorff measure near the characteristic set $C({\Sigma }_{\mathcal{R}\mathcal{T}})$. Then,

$${\int }_{{\Sigma }_{\mathcal{R}\mathcal{T}}}{\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},\infty }d{\sigma }_{{\Sigma }_{\mathcal{R}\mathcal{T}}}+\sum _{i=1}^n{\int }_{{\gamma }_i}{\kappa }_{{\gamma }_i,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,s}ds=0.$$

3. The Sub-Riemannian Geometry Associated with the Second Kind of Canonical Connections in ($\mathcal{R}\mathcal{T},{\mathit{g}}_{\mathit{L}}$)

Let $\widehat{J}{F}_1=\widetilde{F_3},\widehat{J}\widetilde{F_3}=F_1$, and $\widehat{J}{F}_2=F_2$. Then, ${\widehat{J}}^2=id$ and $g_L(\widehat{J}X,\widehat{J}Y)=g_L(X,Y)$ for $X,Y\in \Gamma (\mathcal{R}\mathcal{T},g_L)$, and $\widehat{J}$ is a product structure. We define the second kind of canonical connections, which are metric connections in the roto-translation group $\mathcal{R}\mathcal{T}$:

$${\nabla }_X^{2}Y=\frac{1}{2}{\nabla }_X^{L}Y+\frac{1}{2}\widehat{J}{\nabla }_X^L\widehat{J}Y.$$

(57)

By (57), we have

Lemma 6.

Let $\mathcal{R}\mathcal{T}$ be the roto-translation group. Then,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\nabla }_{F_1}^2{F}_1& =\frac{-{\lambda }_1^2+L{\lambda }_2^2}{4\sqrt{L}}{F}_2,{\nabla }_{F_1}^2{F}_2=\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}(F_1+\widetilde{F_3}),{\nabla }_{F_1}^2{F}_3=\frac{-{\lambda }_1^2+L{\lambda }_2^2}{4}{F}_2,\hfill \\ \hfill {\nabla }_{F_2}^2{F}_1& =0,{\nabla }_{F_2}^2{F}_2=0,{\nabla }_{F_2}^2{F}_3=0,\hfill \\ \hfill {\nabla }_{F_3}^2{F}_1& =\frac{-{\lambda }_1^2+L{\lambda }_2^2}{4}{F}_2,{\nabla }_{F_3}^2{F}_2=\frac{{\lambda }_1^2-L{\lambda }_2^2}{4}(F_1+\widetilde{F_3}),{\nabla }_{F_3}^2{F}_3=\frac{-\sqrt{L}{\lambda }_1^2+L\sqrt{L}{\lambda }_2^2}{4}{F}_2.\hfill \\ \hfill \end{array}\end{array}$$

(58)

By (58) and (8), we have

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^2\dot{\gamma }=& \left[A^{\prime }+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}+\omega (\dot{\gamma }(t))\frac{{\lambda }_1^2-L{\lambda }_2^2}{4}\right]F_1\hfill \\ \hfill +& \left[\frac{1}{{\lambda }_1{\lambda }_2}{\ddot{\theta }}_1(t)+A^2\frac{L{\lambda }_2^2-{\lambda }_1^2}{4\sqrt{L}}+2A\omega (\dot{\gamma }(t))\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}+\omega {(\dot{\gamma }(t))}^2\frac{L^2{\lambda }_2^2-L{\lambda }_1^2}{4\sqrt{L}}\right]F_2\hfill \\ \hfill +& \left[\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}(A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}+\omega (\dot{\gamma }(t))\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}})+\frac{d}{dt}\omega (\dot{\gamma }(t))\right]F_3.\hfill \end{array}$$

(59)

By Lemma 2, we have the following.

Lemma 7.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. Then,

$$\begin{array}{cc}\hfill k_{\gamma }^{L,{\nabla }^2}=& \{{\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_1^2}{({\dot{\theta }}_1(t))}^2+L{(\omega (\dot{\gamma }(t)))}^2\right]}^{-2}\left[{(U_1^{\prime })}^2+{(U_2^{\prime })}^2+L{(U_3^{\prime })}^2\right]\hfill \\ \hfill & -{\left[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{({\dot{\theta }}_1(t))}^2+L{(\omega (\dot{\gamma }(t)))}^2\right]}^{-3}\hfill \\ \hfill & \times \left[A{(U_1^{\prime })}^2+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}{(U_2^{\prime })}^2+L(\omega (\dot{\gamma }(t))){(U_3^{\prime })}^2\right]{\}}^{\frac{1}{2}},\hfill \end{array}$$

(60)

where

$$U_1^{\prime }=A^{\prime }+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}+\omega (\dot{\gamma }(t))\frac{{\lambda }_1^2-L{\lambda }_2^2}{4},$$

$$U_2^{\prime }=\frac{1}{{\lambda }_1{\lambda }_2}{\ddot{\theta }}_1(t)+A^2\frac{L{\lambda }_2^2-{\lambda }_1^2}{4\sqrt{L}}+2A\omega (\dot{\gamma }(t))\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}+\omega {(\dot{\gamma }(t))}^2\frac{L^2{\lambda }_2^2-L{\lambda }_1^2}{4\sqrt{L}},$$

$$U_3^{\prime }=\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}(A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L}+\omega (\dot{\gamma }(t))\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}})+\frac{d}{dt}\omega (\dot{\gamma }(t)).$$

In particular, if $\gamma (t)$ is a horizontal point of γ,

$$\begin{array}{cc}\hfill k_{\gamma }^{L,{\nabla }^2}=& \{[{(A^{\prime }+\frac{\dot{{\theta }_1}(t)}{{\lambda }_1{\lambda }_2}A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}})}^2+{(\frac{{\ddot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}+A^2\frac{L{\lambda }_2^2-{\lambda }_1^2}{4\sqrt{L}})}^2+L(\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}(A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4L})\hfill \\ & +\frac{d}{dt}(\omega (\dot{\gamma }(t))){]}^2\times {[A^2+\frac{1}{{\lambda }_1^2{\lambda }_1^2}{({\dot{\theta }}_1(t))}^2]}^{-2}-{[A^2+\frac{1}{{\lambda }_1^2{\lambda }_1^2}{({\dot{\theta }}_1(t))}^2]}^{-3}\hfill \\ & \times {[A(A^{\prime }+\frac{{\dot{\theta }}_1(t)}{{\lambda }_1{\lambda }_2}A\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}})+\frac{\dot{{\theta }_1}(t)}{{\lambda }_1{\lambda }_2}(\frac{1}{{\lambda }_1{\lambda }_2}{\ddot{\theta }}_1(t)+A^2\frac{L{\lambda }_1^2-{\lambda }_1^2}{4\sqrt{L}})]}^2{\}}^{\frac{1}{2}}.\hfill \\ \hfill \end{array}$$

(61)

We can define the curvature associated to the connection ${\nabla }^2$, $k_{\gamma }^{L,{\nabla }^2}$ of $\gamma$ at $\gamma (t)$. We now have the following lemma.

Lemma 8.

Let $\gamma :\left[a,b\right]\to (\mathcal{R}\mathcal{T},g_L)$ be a $C^2$-smooth regular curve in the Riemannian manifold $(\mathcal{R}\mathcal{T},g_L)$. Then,

$$\begin{array}{c}\hfill \underset{L\to +\infty }{lim}\frac{k_{\gamma }^{L,{\nabla }^2}}{\sqrt{L}}=\frac{{\lambda }_2^2}{4},if\omega (\dot{\gamma }(t))\ne 0,\end{array}$$

$$\begin{array}{c}\hfill \underset{L\to +\infty }{lim}\frac{k_{\gamma }^{L,{\nabla }^2}}{\sqrt{L}}=\sqrt{\frac{A^2{({\dot{\theta }}_1(t))}^2{\lambda }_2^2+A^4{\lambda }_1^2{\lambda }_2^4+16{\lambda }_1^2[\frac{d}{dt}\omega (\dot{\gamma }(t))){]}^2}{16{\lambda }_1^2[A^2+\frac{1}{{\lambda }_1^2{\lambda }_2^2}{(\dot{{\theta }_1}(t))}^2]}},if\omega (\dot{\gamma }(t))=0.\end{array}$$

(62)

For every $U,V\in T{\Sigma }_{\mathcal{R}\mathcal{T}}$, we define ${\nabla }_U^{2,{\Sigma }_{\mathcal{R}\mathcal{T}}}V=\pi {\nabla }_U^{2}V$. By (59) and (26), we have

$$\begin{array}{cc}\hfill {\nabla }_{\dot{\gamma }}^{{\Sigma }_{\mathcal{R}\mathcal{T}},2}\dot{\gamma }=& (\overline{q}{U}_1^{\prime }-\overline{p}{U}_2^{\prime })e_1+({\overline{r}}_L\overline{p}{U}_1^{\prime }+{\overline{r}}_L\overline{q}{U}_2^{\prime }-\frac{l}{l_L}{L}^{\frac{1}{2}}{U}_3^{\prime })e_2.\hfill \end{array}$$

(63)

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface and $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. We can define the geodesic curvature associated to the ${\nabla }^2$, $k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^2}$ of $\gamma$ at $\gamma (t)$.

Lemma 9.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface. Let $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. Then,

$$\begin{array}{c}\hfill \underset{L\to +\infty }{lim}\frac{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^2}}{\sqrt{L}}=\frac{{\lambda }_2^2\sqrt{{\overline{p}}^2+{\overline{r}}_L{\overline{q}}^2}}{4},if\omega (\dot{\gamma }(t))\ne 0,\end{array}$$

$$\begin{array}{c}\hfill \underset{L\to +\infty }{lim}\frac{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^2}}{\sqrt{L}}=\sqrt{-\frac{L{\left[\frac{d}{dx}\omega (\dot{\gamma }(t))\right]}^2}{{\left[-\frac{1}{{\lambda }_1{\lambda }_1}\overline{p}{\dot{\theta }}_1(t)+\overline{q}A\right]}^2}},if\omega (\dot{\gamma }(t))=0.\end{array}$$

(64)

Then, we can define the signed geodesic curvature associated to the ${\nabla }^2$, $k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^2,s}$ of $\gamma$ at $\gamma (t)$, and we obtain

$$\begin{array}{c}\hfill \underset{L\to +\infty }{lim}\frac{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^2,s}}{\sqrt{L}}=\tilde{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^2,s}}.\end{array}$$

(65)

Lemma 10.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset (\mathcal{R}\mathcal{T},g_L)$ be a regular surface. Let $\gamma :[a,b]\to {\Sigma }_{\mathcal{R}\mathcal{T}}$ be a $C^2$-smooth regular curve. Then,

$$\begin{array}{c}\hfill \tilde{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^2,s}}=-\frac{\overline{p}{\lambda }_2^2}{4},if\omega (\dot{\gamma }(t))\ne 0,\end{array}$$

$$\begin{array}{c}\hfill \tilde{k_{\gamma ,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,{\nabla }^2,s}}=\frac{-\frac{d}{dt}\omega (\dot{\gamma }(t))}{|\frac{1}{{\lambda }_1{\lambda }_2}\overline{p}{\dot{\theta }}_1(t)+\overline{q}{A|}^2},if\omega (\dot{\gamma }(t))=0.\end{array}$$

(66)

Example 1.

We assume that there exists a $C^2$-smooth function $u=x_2:\mathcal{R}\mathcal{T}\to \mathbb{R}$ such that

$$\Sigma =\{(x_1,x_2,x_3)\in \mathcal{R}\mathcal{T}:x_2=0\}.$$

Then, ${\nabla }_{\mathcal{R}\mathcal{T}}u=u_{x_1}{\partial }_{x_1}+u_{x_2}{\partial }_{x_2}+u_{x_3}{\partial }_{x_3}={\partial }_{x_2}\ne 0.$ Let

$$F_1=\frac{1}{{\lambda }_1}cos\theta \frac{\partial }{\partial x_1}+\frac{1}{{\lambda }_1}sin\theta \frac{\partial }{\partial x_2},F_2={\lambda }_1{\lambda }_2\frac{\partial }{{\partial }_{x_3}},F_3=-\frac{1}{{\lambda }_2}sin\theta \frac{\partial }{\partial x_1}+\frac{1}{{\lambda }_2}cos\theta \frac{\partial }{\partial x_2},$$

let ${\lambda }_1={\lambda }_2=1$. We have

$$\begin{array}{cc}\hfill p:& =F_{1}u=(\frac{1}{{\lambda }_1}cos\theta \frac{\partial }{\partial x_1}+\frac{1}{{\lambda }_1}sin\theta \frac{\partial }{\partial x_2})(x_2)=sin\theta ,\hfill \\ \hfill q:& =F_{2}u=({\lambda }_1{\lambda }_2\frac{\partial }{{\partial }_{x_3}})(x_2)=0,\hfill \\ \hfill r:& ={\tilde{F}}_{3}u=L^{-\frac{1}{2}}(-\frac{1}{{\lambda }_2}sin\theta \frac{\partial }{\partial x_1}+\frac{1}{{\lambda }_2}cos\theta \frac{\partial }{\partial x_2})(x_2)=0.\hfill \end{array}$$

Therefore, $p^2+q^2={sin}^2\theta >0$, so $\Sigma \subset \mathcal{R}\mathcal{T}$ is a Horizontal spacelike surface. By (25), we have

$$\begin{array}{cc}\hfill & \overline{q}:=\frac{q}{l}=0,{\overline{p}}_L:=\frac{p}{l_L}=1,{\overline{q}}_L:=\frac{q}{l_L}=0,{\overline{r}}_L:=\frac{r}{l_L}=0.\hfill \\ \hfill & l:=\sqrt{p^2+q^2}=sin\theta ,l_L:=\sqrt{p^2+q^2+r^2}=sin\theta ,\overline{p}:=\frac{p}{l}=1.\hfill \end{array}$$

By (26), we have

$$\begin{array}{cc}\hfill & v_L={\overline{p}}_L{F}_1+{\overline{q}}_L{F}_2+{\overline{r}}_L{\tilde{F}}_3=F_1,\hfill \\ \hfill & e_1=\overline{q}{F}_1-\overline{p}{F}_2=-F_2,\hfill \\ \hfill & e_2=\overline{r_L}\overline{p}{F}_1+\overline{r_L}\overline{q}{F}_2-\frac{l}{l_L}{\tilde{F}}_3=-{\tilde{F}}_3.\hfill \end{array}$$

Then, $\{e_1,e_2\}=\{F_2,\widetilde{F_3}\}$ and $T{\Sigma }_{\mathcal{R}\mathcal{T}}=\mathit{span}\{{\partial }_{x_2},{\partial }_{x_3}\}$. Thus, it is concluded that ${\Sigma }_{\mathcal{R}\mathcal{T}}$ is a regular surface in the roto-translation group. Let

$$\gamma :[0,2\pi ]\to {\Sigma }_{\mathcal{R}\mathcal{T}};\theta \to (cos\theta ,0,sin\theta ),$$

be the circle centered at the origin on $x_2=0$. By

$$\omega (\dot{\gamma }(t))={\lambda }_2(-\dot{{\gamma }_1}(t)sin\theta +\dot{{\gamma }_2}(t)cos\theta ),$$

and ${\gamma }_1(\theta )=cos\theta ,{\gamma }_2(\theta )=0,$${\gamma }_3(\theta )=sin\theta ,$ we have

$${\nabla }_{\dot{\gamma }}^{\Sigma ,L}\dot{\gamma }={(\frac{1}{{\lambda }_1{\lambda }_2}sin\theta )}^2.$$

If $cos\theta =0$, then $sin\theta =\pm 1$. In this case, $\omega (\dot{\gamma }(t))=0.$ Then, we have ${\nabla }_{\dot{\gamma }}^{\Sigma ,L}\dot{\gamma }$ is a spacelike vector. By Lemma 8 (62), we have

$$\underset{L\to +\infty }{lim}\frac{k_{\dot{\gamma },{\Sigma }_{\mathcal{R}\mathcal{T}}}^{L,{\nabla }^2}}{\sqrt{L}}=\frac{1}{4}.$$

We know that we can use $k_{\dot{\gamma },{\Sigma }_{\mathcal{RT}}}^{L,{\nabla }^2}$ in the Gauss-Bonnet theorem.

We can define the second fundamental form $I{I}^{{\nabla }^2,L}$ associated with ${\nabla }^2$. We have

Theorem 3.

The second fundamental form $I{I}^{{\nabla }^2,L}$ of the embedding of ${\Sigma }_{\mathcal{R}\mathcal{T}}$ into $(\mathcal{R}\mathcal{T},g_L)$ is given by

$$\begin{array}{c}\hfill I{I}^{{\nabla }^2,L}=\left(\begin{array}{cc}{h}_{11}^2& h_{12}^2\\ h_{21}^2& h_{22}^2\end{array}\right),\end{array}$$

(67)

where

$$\begin{array}{c}\hfill h_{11}^2=\frac{l}{l_L}(F_1(\overline{p})+F_2(\overline{q}))+{\overline{r}}_L\overline{p}\overline{q}\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}},\end{array}$$

$$\begin{array}{cc}\hfill h_{12}=h_{21}=& -\frac{l_L}{l}{\langle e_1,{\nabla }_H\overline{r_L}\rangle }_L-\left[\overline{q}{\overline{r}}_L\overline{p}({\overline{p}}_L-{\overline{q}}_L+{\overline{r}}_L)+{\overline{q}}_L^2\right]\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill h_{22}=-& \frac{l^2}{l_L^2}{\langle e_2,{\nabla }_H^2(\frac{r}{l})\rangle }_L+\widetilde{X_3}({\overline{r}}_L)-\frac{{\lambda }_1^2-L{\lambda }_2^2}{4\sqrt{L}}\left[{\overline{r}}_L^2\overline{p}\overline{q}({\overline{p}}_L-1)-\frac{{\overline{r}}_L{\overline{q}}_L({\overline{p}}_L-{\overline{r}}_L)}{\sqrt{L}}\right]\hfill \\ \hfill -& \frac{L{\lambda }_2^2-{\lambda }_1^2}{4\sqrt{L}}({\overline{r}}_L{\overline{q}}_L{\overline{p}}^2-2{\overline{r}}_L{\overline{p}}_L{\overline{q}}_L-\frac{{\overline{q}}_L}{r^2}).\hfill \end{array}$$

(68)

The mean curvature associated with the ${\nabla }^2,{\mathcal{H}}_{{\nabla }^2,L}$ of ${\Sigma }_{\mathcal{R}\mathcal{T}}$ is defined by ${\mathcal{H}}_{{\nabla }^2,L}:=\mathrm{tr}(I{I}^{{\nabla }^2,L}).$ Similarly, we define the curvature of a connection ${\nabla }^1$ by

$$R^2(X,Y)Z,{\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},{\nabla }^2}(e_1,e_2),{\mathcal{K}}^{{\nabla }^2}(e_1,e_2).$$

We also have the Gauss equation about ${\nabla }^2$. We have the following proposition.

Proposition 5.

Away from characteristic points, the following equality holds

$$\underset{L\to +\infty }{lim}\frac{{\mathcal{H}}_{{\nabla }^2,L}}{\sqrt{L}}=(F_1(\overline{p})+F_2(\overline{q}))+\frac{{\lambda }_2^4}{4}\frac{{(F_{3}u)}^2\overline{p}\overline{q}(\overline{p}-1)}{l^2}.$$

(69)

By Lemma 6, we have the following lemma.

Lemma 11.

Let $\mathcal{R}\mathcal{T}$ be the roto-translation group. Then,

$$\begin{array}{cc}\hfill & R^2(F_1,F_2)F_1={\lambda }_2^2·\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}{F}_2,R^1(F_1,F_2)F_3={\lambda }_2^2·\frac{L\sqrt{L}{\lambda }_2^2-\sqrt{L}{\lambda }_1^2}{4}{F}_2,\hfill \\ \hfill & R^2(F_2,F_3)F_3={\lambda }_1^2·\frac{L{\lambda }_2^2-{\lambda }_1^2}{4}{F}_2,R^2(F_i,F_j)F_k=0,\mathrm{for}\mathrm{other}i,j,k.\hfill \end{array}$$

(70)

So,

$${\mathcal{K}}^{{\nabla }^2}(e_1,e_2)={\overline{r}}_L\overline{p}{\overline{q}}^2(\frac{L}{4}{{\lambda }_2}^4+1).$$

(71)

By Theorem 1, we have

$$det(I{I}^{{\nabla }^2,L})=(F_1(\overline{p})+F_2(\overline{q}))\frac{\sqrt{L}{\lambda }_2^2}{4}-<e_1,{\nabla }_H\frac{F_{3}u}{|{{\nabla }_H}^u|})>,\mathrm{when}L\to \infty .$$

(72)

By (68) and the Gauss equation, we have the following proposition.

Proposition 6.

Away from characteristic points, we have

$$\underset{L\to +\infty }{lim}\frac{{\mathcal{K}}^{{\Sigma }_{\mathcal{R}\mathcal{T}},{\nabla }^1}(e_1,e_2)}{\sqrt{L}}=\widetilde{{\mathcal{K}}^{\infty ,{\nabla }^1}}:=[F_1(\overline{p})+F_2(\overline{q})]\frac{{\lambda }_2^2}{4}+\frac{F_{3}u}{l}\overline{p}{\overline{q}}^2(\frac{{\lambda }_2^2}{4}+1).$$

(73)

So, we obtain a Gauss-Bonnet theorem in $(\mathcal{R}\mathcal{T},g_L)$ as follows.

Theorem 4.

Let ${\Sigma }_{\mathcal{R}\mathcal{T}}\subset ((\mathcal{R}\mathcal{T},g_L({\lambda }_1,{\lambda }_2))$ be a regular surface with finitely many boundary components ${(\partial {\Sigma }_{\mathcal{R}\mathcal{T}})}_i,i\in \{1,\cdots ,n\}$, given by Euclidean $C^2$-smooth regular and closed curves ${\gamma }_i:[0,2\pi ]\to {(\partial {\Sigma }_{\mathcal{R}\mathcal{T}})}_i.$ Suppose that the characteristic set $C({\Sigma }_{\mathcal{R}\mathcal{T}})$ satisfies ${\mathcal{H}}^2(C({\Sigma }_{\mathcal{R}\mathcal{T}}))=0$, where ${\mathcal{H}}^2(C({\Sigma }_{\mathcal{R}\mathcal{T}}))$ denotes the Euclidean one-dimensional Hausdorff measure of $C({\Sigma }_{\mathcal{R}\mathcal{T}})$ and that $\Vert {\nabla }_H{u\Vert }_H^{-1}$ is locally summable with respect to the Euclidean two-dimensional Hausdorff measure near the characteristic set $C({\Sigma }_{\mathcal{R}\mathcal{T}})$. Then,

$${\int }_{{\Sigma }_{\mathcal{R}\mathcal{T}}}{\mathcal{K}}^{\Sigma ,\infty }d{\sigma }_{{\Sigma }_{\mathcal{R}\mathcal{T}}}+\sum _{i=1}^n{\int }_{{\gamma }_i}{\kappa }_{{\gamma }_i,{\Sigma }_{\mathcal{R}\mathcal{T}}}^{\infty ,s}ds=0.$$