Helix surfaces in the special linear group

Abstract

We characterize helix surfaces (constant angle surfaces) in the special linear group $\mathrm {SL}(2,{\mathbb {R}})$. In particular, we give an explicit local description of these surfaces by means of a suitable curve and a 1-parameter family of isometries of $\mathrm {SL}(2,{\mathbb {R}})$.

1 Introduction

In recent years much work has been done to understand the geometry of surfaces whose unit normal vector field forms a constant angle with a fixed field of directions of the ambient space. These surfaces are called helix surfaces or constant angle surfaces and they have been studied in most of the 3-dimensional geometries. In [2] Cermelli and Di Scala analyzed the case of constant angle surfaces in ${\mathbb {R}}^3$ obtaining a remarkable relation with a Hamilton–Jacobi equation and showing their application to equilibrium configurations of liquid crystals. Later, Dillen–Fastenakels–Van der Veken–Vrancken [4], and Dillen–Munteanu [3], classified the surfaces making a constant angle with the ${\mathbb {R}}$-direction in the product spaces ${\mathbb {S}}^2\times {\mathbb {R}}$ and ${\mathbb {H}}^2\times {\mathbb {R}}$, respectively. Moreover, helix submanifolds have been studied in higher dimensional euclidean spaces and product spaces in [5, 6, 10].

The spaces ${\mathbb {S}}^2\times {\mathbb {R}}$ and ${\mathbb {H}}^2\times {\mathbb {R}}$ can be seen as two particular cases of Bianchi–Cartan–Vranceanu spaces (BCV-spaces) which include all 3-dimensional homogeneous metrics whose group of isometries has dimension 4 or 6, except for those of constant negative sectional curvature. A crucial feature of BCV-spaces is that they admit a Riemannian submersion onto a surface of constant Gaussian curvature, called the Hopf fibration, that, in the cases of ${\mathbb {S}}^2\times {\mathbb {R}}$ and ${\mathbb {H}}^2\times {\mathbb {R}}$, it is the natural projection onto the first factor. Consequently, one can consider the angle $\vartheta$ that the unit normal vector field of a surface in a BCV-space forms with the Hopf vector field, which is, by definition, the vector field tangent to the fibers of the Hopf fibration. This angle $\vartheta$ has a crucial role in the study of surfaces in BCV-spaces as shown by Daniel, in [1], where he proved that the equations of Gauss and Codazzi are given in terms of the function $\nu =\cos \vartheta$ and that this angle is one of the fundamental invariants for a surface in BCV-spaces. Consequently, in [7], the authors considered the surfaces in a BCV-space for which the angle $\vartheta$ is constant, giving a complete local classification in the case that the BCV-space is the Heisenberg space ${\mathbb {H}}_3$.

Later, López–Munteanu, in [8], defined and classified two types of constant angle surfaces in the homogeneous 3-manifold $\mathrm {Sol}_3$, whose isometry group has dimension 3. Also, Montaldo–Onnis, in [9], characterized helix surfaces in the 1-parameter family of Berger spheres ${\mathbb {S}}^3_\epsilon$, with $\epsilon >0$, proving that, locally, a helix surface is determined by a suitable 1-parameter family of isometries of the Berger sphere and by a geodesic of a 2-torus in the 3-dimensional sphere.

This paper is a continuation of our work [9] and it is devoted to the study and characterization of helix surfaces in the homogeneous 3-manifold given by the special linear group $\mathrm {SL} (2,{\mathbb {R}})$ endowed with a suitable 1-parameter family $g_{\tau }$ of metrics that we shall describe in sect. 2. Our study of helix surfaces in $(\mathrm {SL}(2,{\mathbb {R}}),g_{\tau })$ will depend on a constant $B:=(\tau ^2+1)\cos ^2\vartheta -1$, where $\vartheta$ is the constant angle between the normal to the surface and the Hopf vector field of $\mathrm {SL} (2,{\mathbb {R}})$. A similar constant appeared also in the study of helix surfaces in the Berger sphere (see [9]) but in that case the constant was always positive. Thus we shall divide our study according to the three possibilities: $B>0,\, B=0$ and $B<0$.

2 Preliminaries

Let ${\mathbb {R}}_2^4$ denote the 4-dimensional pseudo-Euclidean space endowed with the semi-definite inner product of signature (2,2) given by

$$\begin{aligned} \langle v,w\rangle = v_1\,w_1+v_2\,w_2-v_3\,w_3-v_4\,w_4,\quad v,w\in {\mathbb {R}}^4. \end{aligned}$$

We identify the special linear group with

$$\begin{aligned} \mathrm {SL}(2,{\mathbb {R}})=\{(z,w)\in {\mathbb {C}}^2 :|z|^2-|w|^2=1\}=\{v\in {\mathbb {R}}_2^4 :\langle v,v\rangle =1\}\subset {\mathbb {R}}_2^4 \end{aligned}$$

and we shall use the Lorentz model of the hyperbolic plane with constant Gauss curvature $-4$, that is

$$\begin{aligned} {\mathbb {H}}^2(-4)=\{(x,y,z)\in {\mathbb {R}}^3_1:x^2+y^2-z^2=-1/4\}, \end{aligned}$$

where ${\mathbb {R}}^3_1$ is the Minkowski 3-space. Then the Hopf map $\psi :\mathrm {SL}(2,{\mathbb {R}})\rightarrow {\mathbb {H}}^2(-4)$ given by

$$\begin{aligned} \psi (z,w)=\frac{1}{2}\,(2z\bar{w},|z|^2+|w|^2) \end{aligned}$$

is a submersion, with circular fibers, and if we put

$$\begin{aligned} X_1(z,w)=(iz,iw),\quad X_2(z,w)=(i\bar{w},i\bar{z}),\quad X_3(z,w)=(\bar{w},\bar{z}), \end{aligned}$$

we have that $X_1$ is a vertical vector field while $X_2,\, X_3$ are horizontal. The vector field $X_1$ is called the Hopf vector field.

We shall endow $\mathrm {SL}(2,{\mathbb {R}})$ with the 1-parameter family of metrics $g_\tau ,\, \tau >0$, given by

$$\begin{aligned} g_\tau (X_i,X_j)=\delta _{ij},\quad g_\tau (X_1,X_1)=\tau ^2,\quad g_\tau (X_1,X_j)=0,\quad i,j\in \{2,3\}, \end{aligned}$$

which renders the Hopf map $\psi :(\mathrm {SL}(2,{\mathbb {R}}),g_{\tau })\rightarrow {\mathbb {H}}^2(-4)$ a Riemannian submersion.

For those familiar with the notations of Daniel [1], we point out that $(\mathrm {SL}(2,{\mathbb {R}}),g_{\tau })$ corresponds to a model for a homogeneous space $E(k, \tau )$ with curvature of the basis $k=-4$ and bundle curvature $\tau >0$.

With respect to the inner product in ${\mathbb {R}}_2^4$ the metric $g_{\tau }$ is given by

$$\begin{aligned} g_{\tau }(X,Y)=-\langle X,Y\rangle +(1+\tau ^2)\langle X, X_1\rangle \langle Y, X_1\rangle . \end{aligned}$$

(1)

From now on, we denote $(\mathrm {SL}(2,{\mathbb {R}}),g_\tau )$ with ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$. Obviously

$$\begin{aligned} E_1=-\tau ^{-1}\,X_1,\quad E_2=X_2,\quad E_3=X_3, \end{aligned}$$

(2)

is an orthonormal basis on ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ and the Levi-Civita connection ${{\nabla }}^\tau$ of ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ is given by (see, for example, [11]):

$$\begin{aligned} {{\nabla }}^\tau _{E_{1}}E_{1}&= 0,\quad {{\nabla }}^\tau _{E_{2}}E_{2}=0,\quad {{\nabla }}^\tau _{E_{3}}E_{3}=0,\nonumber \\ {{\nabla }}^\tau _{E_{1}}E_{2}&= -\tau ^{-1}(2+\tau ^2)E_{3}, \quad {{\nabla }}^\tau _{E_{1}}E_{3}=\tau ^{-1}(2+\tau ^2)E_{2},\nonumber \\ {{\nabla }}^\tau _{E_{2}}E_{1}&= -\tau E_{3},\quad {{\nabla }}^\tau _{E_{3}}E_{1}=\tau E_{2},\quad {{\nabla }}^\tau _{E_{3}}E_{2}=-\tau E_{1}=-{{\nabla }}^\tau _{E_{2}}E_{3}. \end{aligned}$$

(3)

Finally, we recall that the isometry group of ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ is the 4-dimensional indefinite unitary group $\mathrm {U}_1(2)$ that can be identified with:

$$\begin{aligned} \mathrm {U}_1(2)=\{A\in \mathrm {O}_2(4):AJ_1=\pm J_1A\}, \end{aligned}$$

where $J_1$ is the complex structure of ${\mathbb {R}}^4$ defined by

$$\begin{aligned} J_1 = \left( \begin{matrix}J &{}\quad 0 \\ 0 &{}\quad J\end{matrix}\right) ,\quad J = \left( \begin{matrix}0 &{}\quad -1 \\ 1 &{}\quad 0\end{matrix}\right) , \end{aligned}$$

while

$$\begin{aligned} \mathrm {O}_2(4)=\{A\in \mathrm {GL}(4,{\mathbb {R}}):A^t=\epsilon \,A^{-1}\,\epsilon \},\qquad \epsilon = \begin{pmatrix} I&{}\quad 0\\ 0&{}\quad -I \end{pmatrix},\quad I=\begin{pmatrix} 1&{}\quad 0\\ 0&{}\quad 1 \end{pmatrix} \end{aligned}$$

is the indefinite orthogonal group.

We observe that $\mathrm {O}_2(4)$ is the group of $4\times 4$ real matrices preserving the semi-definite inner product of ${\mathbb {R}}_2^4$.

Suppose now we are given a 1-parameter family $A(v), v\in (a,b)\subset {\mathbb {R}}$, consisting of $4\times 4$ indefinite orthogonal matrices commuting (anticommuting, respectively) with $J_1$. In order to describe explicitly the family $A(v)$, we shall use two product structures of ${\mathbb {R}}^4$, namely

$$\begin{aligned} J_{2} =\begin{pmatrix} 0&{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 0&{}\quad 0 &{}\quad 1 &{}\quad 0 \\ 0&{}\quad 1&{}\quad 0 &{}\quad 0 \\ 1&{}\quad 0 &{}\quad 0 &{}\quad 0 \\ \end{pmatrix},\qquad J_{3} =\begin{pmatrix} 0&{}\quad 0 &{}\quad 1 &{}\quad 0 \\ 0&{}\quad 0 &{}\quad 0 &{}\quad -1 \\ 1&{}\quad 0&{}\quad 0 &{}\quad 0 \\ 0&{}\quad -1 &{}\quad 0 &{}\quad 0 \\ \end{pmatrix}. \end{aligned}$$

Since $A(v)$ is an indefinite orthogonal matrix, the first row must be a unit vector ${\mathbf {r}}_1(v)$ of ${\mathbb {R}}^4_2$ for all $v\in (a,b)$. Thus, without loss of generality, we can take

$$\begin{aligned} {\mathbf {r}}_1(v)&= (\cosh \xi _1(v)\cos \xi _2(v), -\cosh \xi _1(v)\sin \xi _2(v), \sinh \xi _1(v)\cos \xi _3(v),\\&\quad -\sinh \xi _1(v)\sin \xi _3(v)), \end{aligned}$$

for some real functions $\xi _1,\xi _2$ and $\xi _3$ defined in $(a,b)$. Since $A(v)$ commutes (anticommutes, respectively) with $J_1$ the second row of $A(v)$ must be ${\mathbf {r}}_2(v)=\pm J_{1}{\mathbf {r}}_1(v)$. Now, the four vectors $\{{\mathbf {r}}_1, J_1{\mathbf {r}}_1, J_2{\mathbf {r}}_1,J_3{\mathbf {r}}_1\}$ form a pseudo-orthonormal basis of ${\mathbb {R}}^4_2$, thus the third row ${\mathbf {r}}_3(v)$ of $A(v)$ must be a linear combination of them. Since ${\mathbf {r}}_3(v)$ is unit and it is orthogonal to both ${\mathbf {r}}_1(v)$ and $J_1{\mathbf {r}}_1(v)$, there exists a function $\xi (v)$ such that

$$\begin{aligned} {\mathbf {r}}_3(v)=\cos \xi (v) J_2 {\mathbf {r}}_1(v)+\sin \xi (v) J_3 {\mathbf {r}}_1(v). \end{aligned}$$

Finally the fourth row of $A(v)$ is ${\mathbf {r}}_4(v)=\pm J_1{\mathbf {r}}_3(v)=\mp \cos \xi (v) J_3 {\mathbf {r}}_1(v)\pm \sin \xi (v) J_2 {\mathbf {r}}_1(v)$. This means that any 1-parameter family $A(v)$ of $4\times 4$ indefinite orthogonal matrices commuting (anticommuting, respectively) with $J_1$ can be described by four functions $\xi _1,\xi _2,\xi _3$ and $\xi$ as

$$\begin{aligned} A(\xi ,\xi _1,\xi _2,\xi _3)(v)= \left( \begin{array}{ll} &{}{\mathbf {r}}_1(v)\\ &{}\pm J_1{\mathbf {r}}_1(v)\\ &{}\cos \xi (v) J_2 {\mathbf {r}}_1(v)+\sin \xi (v) J_3 {\mathbf {r}}_1(v)\\ &{}\mp \cos \xi (v) J_3 {\mathbf {r}}_1(v)\pm \sin \xi (v) J_2 {\mathbf {r}}_1(v) \end{array} \right) . \end{aligned}$$

(4)

3 Constant angle surfaces

We start this section giving the definition of constant angle surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$.

Definition 3.1

We say that a surface in the special linear group ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ is a helix surface or a constant angle surface if the angle $\vartheta \in [0,\pi )$ between the unit normal vector field and the unit Killing vector field $E_1$ (tangent to the fibers of the Hopf fibration) is constant at every point of the surface.

Let $M^2$ be an oriented helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ and let $N$ be a unit normal vector field. Then, by definition,

$$\begin{aligned} |g_{\tau }(E_1,N)|=\cos \vartheta , \end{aligned}$$

for fixed $\vartheta \in [0,\pi /2]$. Note that $\vartheta \ne 0$. In fact, if it were zero then the vector fields $E_2$ and $E_3$ would be tangent to the surface $M^2$, which is absurd since the horizontal distribution of the Hopf map is not integrable. If $\vartheta =\pi /2$, we have that $E_1$ is always tangent to $M$ and, therefore, $M$ is a Hopf cylinder. Therefore, from now on we assume that the constant angle $\vartheta \ne \pi /2,0$.

The Gauss and Weingarten formulas are

$$\begin{aligned} {{\nabla }}^\tau _X Y&= \nabla _X Y+\alpha (X,Y),\nonumber \\ {{\nabla }}^\tau _X N&= -A(X), \end{aligned}$$

(5)

where with $A$ we have indicated the shape operator of $M$ in $\mathrm {SL}(2,{\mathbb {R}})_\tau$, with $\nabla$ the induced Levi-Civita connection on $M$ and by $\alpha$ the second fundamental form of $M$ in $\mathrm {SL}(2,{\mathbb {R}})_\tau$. Projecting $E_1$ onto the tangent plane to $M$ we have

$$\begin{aligned} E_1=T+\cos \vartheta \, N, \end{aligned}$$

where $T$ is the tangent part which satisfies $g_\tau (T,T)=\sin ^2\vartheta$.

For all $X\in TM$, we have that

$$\begin{aligned} {{\nabla }}^\tau _X E_1&= {{\nabla }}^\tau _X T-\cos \vartheta \,A(X)\nonumber \\&= \nabla _X T+g_\tau (A(X),T)\,N-\cos \vartheta \,A(X). \end{aligned}$$

(6)

On the other hand, if $X=\sum X_i E_i$,

$$\begin{aligned} {{\nabla }}^\tau _X E_1&= \tau \,(X_3 E_2-X_2 E_3)= \tau \, X\wedge E_1\nonumber \\&= \tau \,g_\tau (JX,T)\,N-\tau \,\cos \vartheta JX, \end{aligned}$$

(7)

where $JX=N\wedge X$ denotes the rotation of angle $\pi /2$ on $TM$. Identifying the tangent and normal component of (6) and (7) respectively, we obtain

$$\begin{aligned} \nabla _X T= \cos \vartheta \,(A(X)-\tau \,JX) \end{aligned}$$

(8)

and

$$\begin{aligned} g_\tau (A(X)-\tau \,JX,T)=0. \end{aligned}$$

(9)

Lemma 3.2

Let $M^2$ be an oriented helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ with constant angle $\vartheta$. Then, we have the followings properties.

(i):

With respect to the basis $\{T,JT\}$, the matrix associates to the shape operator $A$ takes the form

$$\begin{aligned} A=\left( \begin{array}{cc} 0 &{}\quad -\tau \\ -\tau &{}\quad \lambda \\ \end{array}\right) , \end{aligned}$$

for some function $\lambda$ on $M$.

(ii):

The Levi-Civita connection $\nabla$ of $M$ is given by

$$\begin{aligned} \nabla _T T=-2\tau \cos \vartheta \, JT,\qquad \nabla _{JT} T=\lambda \cos \vartheta \, JT, \\ \nabla _T JT=2\tau \cos \vartheta \, T,\qquad \nabla _{JT} JT=-\lambda \cos \vartheta \, T. \end{aligned}$$

(iii):

The Gauss curvature of $M$ is constant and satisfies

$$\begin{aligned} K=-4(1+\tau ^2)\,\cos ^2\vartheta . \end{aligned}$$

(iv):

The function $\lambda$ satisfies the equation

$$\begin{aligned} T \lambda +\lambda ^2\,\cos \vartheta +4B\,\cos \vartheta =0, \end{aligned}$$

(10)

where $B:=(\tau ^2+1)\cos ^2\vartheta -1$.

Proof

Point (i) follows directly from (9). From (8) and using

$$\begin{aligned} g_\tau (T,T)=g_\tau (JT,JT)=\sin ^2\vartheta ,\quad g_\tau (T,JT)=0, \end{aligned}$$

we obtain (ii). From the Gauss equation in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ (we refer to the equation in Corollary 3.2 of [1] with $\nu =\cos \theta$ and $k=-4$), and (i), we have that the Gauss curvature of $M$ is given by

$$\begin{aligned} K&= \det A+\tau ^2-4(1+\tau ^2)\,\cos ^2\vartheta \\&= -4(1+\tau ^2)\,\cos ^2\vartheta . \end{aligned}$$

Finally, (10) follows from the Codazzi equation (see [1]):

$$\begin{aligned} \nabla _X A(Y)-\nabla _Y A(X)-A[X,Y]=-4(1+\tau ^2)\,\cos \vartheta \,(g_\tau (Y,T)X-g_\tau (X,T)Y), \end{aligned}$$

putting $X=T,\, Y=JT$ and using (ii). In fact, it is easy to check that

$$\begin{aligned} -4(1+\tau ^2)\,\cos \vartheta \,(g_\tau (JT,T)T-g_\tau (T,T)JT)) =4(1+\tau ^2)\cos \vartheta \,\sin ^2\vartheta \, JT \end{aligned}$$

and

$$\begin{aligned}&\nabla _T A(JT)-\nabla _{JT} A(T)-A[T,JT]\\&=\nabla _T (-\tau \, T+\lambda \, JT)-\nabla _{JT} (-\tau \, JT)-A(2\tau \cos \vartheta \,T-\lambda \cos \vartheta \, JT)\\&=(4\tau ^2\cos \vartheta +T(\lambda )+\lambda ^2\,\cos \vartheta )\,JT. \end{aligned}$$

$\square$

As $g_\tau (E_1,N)=\cos \vartheta$, there exists a smooth function $\varphi$ on $M$ such that

$$\begin{aligned} N=\cos \vartheta E_1+\sin \vartheta \cos \varphi \,E_2+\sin \vartheta \sin \varphi \,E_3. \end{aligned}$$

Therefore

$$\begin{aligned} T=E_1-\cos \vartheta \,N=\sin \vartheta \,[\sin \vartheta \,E_1-\cos \vartheta \cos \varphi \,E_2-\cos \vartheta \sin \varphi \,E_3] \end{aligned}$$

(11)

and

$$\begin{aligned} JT=\sin \vartheta \,(\sin \varphi \,E_2-\cos \varphi \,E_3). \end{aligned}$$

Also

$$\begin{aligned} A(T)&= -{{\nabla }}^\tau _T N=(T \varphi -\tau ^{-1}(2+\tau ^2)\,\sin ^2\vartheta +\tau \cos ^2\vartheta )\,JT,\nonumber \\ A(JT)&= -{{\nabla }}^\tau _{JT} N=(JT\varphi )\,JT-\tau \,T. \end{aligned}$$

(12)

Comparing (12) with (i) of Lemma 3.2, it results that

$$\begin{aligned} \left\{ \begin{array}{l} JT\varphi =\lambda ,\\ T\varphi =-2\tau ^{-1}\,B. \end{array}\right. \end{aligned}$$

(13)

We observe that, as

$$\begin{aligned}{}[T,JT]=\cos \vartheta \,(2\tau \, T-\lambda \,JT), \end{aligned}$$

the compatibility condition of system (13):

$$\begin{aligned} (\nabla _T JT-\nabla _{JT} T)\varphi =[T,JT]\varphi =T (JT\varphi )-JT (T\varphi ) \end{aligned}$$

is equivalent to (10).

We now choose local coordinates $(u,v)$ on $M$ such that

$$\begin{aligned} \partial _u=T. \end{aligned}$$

(14)

Also, as $\partial _v$ is tangent to $M$, it can be written in the form

$$\begin{aligned} \partial _v=a\,T+b\,JT, \end{aligned}$$

(15)

for certain functions $a=a(u,v)$ and $b=b(u,v)$. As

$$\begin{aligned} 0=[\partial _u,\partial _v]=(a_u+2\tau b\cos \vartheta )\,T+(b_u-b\lambda \cos \vartheta )\,JT, \end{aligned}$$

then

$$\begin{aligned} \left\{ \begin{array}{l} a_u=-2\tau b\cos \vartheta ,\\ b_u=b\lambda \cos \vartheta . \end{array}\right. \end{aligned}$$

(16)

Moreover, the Eq. (10) of Lemma 3.2 can be written as

$$\begin{aligned} \lambda _u+\cos \vartheta \,\lambda ^2+4B\,\cos \vartheta =0. \end{aligned}$$

(17)

Depending on the value of $B$, by integration of (17), we have the following three possibilities.

• (i) If $B=0$

  $$\begin{aligned} \lambda (u,v)=\frac{1}{u\,\cos \vartheta +\eta (v)}, \end{aligned}$$

  for some smooth function $\eta$ depending on $v$. Thus the solution of system (16) is given by

  $$\begin{aligned} \left\{ \begin{array}{l} a(u,v)=-\tau \,u\,\cos \vartheta \,(u\,\cos \vartheta +2\,\eta (v)),\\ b(u,v)=u\,\cos \vartheta +\eta (v). \end{array}\right. \end{aligned}$$

• (ii) If $B>0$

  $$\begin{aligned} \lambda (u,v)=2\,\sqrt{B}\tan (\eta (v)-2\cos \vartheta \sqrt{B}\,u), \end{aligned}$$

  for some smooth function $\eta$ depending on $v$ and system (16) has the solution

  $$\begin{aligned} \left\{ \begin{array}{l} a(u,v)=\frac{\tau }{\sqrt{B}}\sin (\eta (v)-2\cos \vartheta \sqrt{B}\,u),\\ b(u,v)=\cos (\eta (v)-2\cos \vartheta \sqrt{B}\,u). \end{array}\right. \end{aligned}$$

• (iii) If $B<0$

  $$\begin{aligned} \lambda (u,v)=2\,\sqrt{-B}\tanh ({\eta }(v)+2\cos \vartheta \sqrt{-B}\,u), \end{aligned}$$

  for some smooth function ${\eta }$ depending on $v$. Solving the system (16), we have

  $$\begin{aligned} \left\{ \begin{array}{l} a(u,v)=-\frac{\tau }{\sqrt{-B}}\sinh ({\eta }(v)+2\cos \vartheta \sqrt{-B}\,u),\\ b(u,v)=\cosh ({\eta }(v)+2\cos \vartheta \sqrt{-B}\,u). \end{array}\right. \end{aligned}$$

Moreover, in the case (i) the system (13) becomes

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _u=0,\\ \varphi _v=1, \end{array}\right. \end{aligned}$$

(18)

and so $\varphi (u,v)=v+c,\, c\in {\mathbb {R}}$. In the cases (ii) and (iii), the system (13) becomes

$$\begin{aligned} \left\{ \begin{array}{l} \varphi _u =-2\tau ^{-1}\,B,\\ \varphi _v =0, \end{array}\right. \end{aligned}$$

of which the general solution is given by

$$\begin{aligned} \varphi (u,v)=-2\tau ^{-1}B\,u+c, \end{aligned}$$

(19)

where $c$ is a real constant.

With respect to the local coordinates $(u,v)$ chosen above, we have the following characterization of the position vector of a helix surface.

Proposition 3.3

Let $M^2$ be a helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }\subset {\mathbb {R}}_2^4$ with constant angle $\vartheta$. Then, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), the position vector $F$ of $M^2$ in ${\mathbb {R}}^4_2$ satisfies the following equation:

(a):

if $B=0$,

$$\begin{aligned} \frac{\partial ^2F}{\partial u^2}=0, \end{aligned}$$

(20)

(b):

if $B\ne 0$,

$$\begin{aligned} \frac{\partial ^4F}{\partial u^4} + (\tilde{b}^2-2\tilde{a})\,\frac{\partial ^2F}{\partial u^2}+\tilde{a}^2\,F=0, \end{aligned}$$

(21)

where

$$\begin{aligned} \tilde{a}=-\tau ^{-2}\sin ^2\vartheta \, B, \qquad \tilde{b}=-2 \tau ^{-1}\, B. \end{aligned}$$

(22)

Proof

Let $M^2$ be a helix surface and let $F$ be the position vector of $M^2$ in ${\mathbb {R}}_2^4$. Then, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), we can write $F(u,v)=(F_1(u,v),\ldots ,F_4(u,v))$. By definition, taking into account (11), we have that

$$\begin{aligned} F_u&= (\partial _uF_1,\partial _uF_2,\partial _uF_3,\partial _uF_4)=T\\&= \sin \vartheta \,[\sin \vartheta \,{E_1}_{|F(u,v)}-\cos \vartheta \cos \varphi \,{E_2}_{|F(u,v)}-\cos \vartheta \sin \varphi \,{E_3}_{|F(u,v)}]. \end{aligned}$$

Using the expression of $E_1,\, E_2$ and $E_3$ with respect to the coordinates vector fields of ${\mathbb {R}}^4_2$, we obtain

$$\begin{aligned} \left\{ \begin{array}{l} \partial _uF_1=\sin \vartheta \,(\tau ^{-1}\sin \vartheta \,F_2-\cos \vartheta \cos \varphi \,F_4-\cos \vartheta \sin \varphi \,F_3),\\ \partial _uF_2=-\sin \vartheta \,(\tau ^{-1}\sin \vartheta \,F_1+\cos \vartheta \cos \varphi \,F_3-\cos \vartheta \sin \varphi \,F_4),\\ \partial _uF_3=\sin \vartheta \,(\tau ^{-1}\sin \vartheta \,F_4-\cos \vartheta \cos \varphi \,F_2-\cos \vartheta \sin \varphi \,F_1),\\ \partial _uF_4=-\sin \vartheta \,(\tau ^{-1}\sin \vartheta \,F_3+\cos \vartheta \cos \varphi \,F_1-\cos \vartheta \sin \varphi \,F_2).\\ \end{array}\right. \end{aligned}$$

(23)

Therefore, if $B=0$, taking the derivative of (23) with respect to $u$ and using (18), we obtain that $F_{uu}=0$.

If $B\ne 0$, taking the derivative of (23) with respect to $u$ and using (19), we find two constants $\tilde{a}$ and $\tilde{b}$ such that

$$\begin{aligned} \left\{ \begin{array}{l} (F_1)_{uu}=\tilde{a}\,F_1+\tilde{b}\,(F_2)_u,\\ (F_2)_{uu}=\tilde{a}\,F_2-\tilde{b}\,(F_1)_u,\\ (F_3)_{uu}=\tilde{a}\,F_3+\tilde{b}\,(F_4)_u,\\ (F_4)_{uu}=\tilde{a}\,F_4-\tilde{b}\,(F_3)_u,\\ \end{array}\right. \end{aligned}$$

(24)

where

$$\begin{aligned} \tilde{a}=\frac{\tau ^{-1}\sin ^2\vartheta }{2}\varphi _u=-\tau ^{-2}\sin ^2\vartheta B, \qquad \tilde{b}=\varphi _u. \end{aligned}$$

Finally, taking twice the derivative of (24) with respect to $u$ and using (23) and (24) in the derivative we obtain the desired Eq. (21). $\square$

Remark 3.4

As $\langle F,F\rangle =1$, using (21), (23) and (24), we find that the position vector $F(u,v)$ and its derivatives must satisfy the relations:

$$\begin{aligned} \begin{array}{lll} \langle F,F\rangle =1,&{}\quad \langle F_u,F_u\rangle =\tilde{a},&{}\quad \langle F,F_u\rangle =0,\\ \langle F_u,F_{uu}\rangle =0,&{}\quad \langle F_{uu},F_{uu}\rangle =D,&{}\quad \langle F,F_{uu}\rangle =-\tilde{a},\\ \langle F_u,F_{uuu}\rangle =-D ,&{}\quad \langle F_{uu},F_{uuu}\rangle =0,&{}\quad \langle F,F_{uuu}\rangle =0,\\ \langle F_{uuu},F_{uuu}\rangle =E, &{}\quad &{}\quad \end{array} \end{aligned}$$

(25)

where

$$\begin{aligned} D=\tilde{a}\,\tilde{b}^2-3\tilde{a}^2,\qquad E=(\tilde{b}^2-2\tilde{a})\,D-\tilde{a}^3. \end{aligned}$$

In addition, as

$$\begin{aligned} J_1F(u,v)={X_1}_{|F(u,v)}=-\tau \,{E_1}_{|F(u,v)}=-\tau \,(F_u+\cos \vartheta \,N), \end{aligned}$$

using (21)–(25), we obtain the following identities

$$\begin{aligned}&\langle J_1F,F_u\rangle =-\tau ^{-1}\sin ^2\vartheta ,\nonumber \\&\langle J_1 F,F_{uu}\rangle =0,\nonumber \\&\langle F_u,J_1F_{uu}\rangle =\tilde{a}\,(\tilde{b}-\tau ^{-1}\,\sin ^2\vartheta ):=I,\nonumber \\&\langle J_1F_u,F_{uuu}\rangle =0,\nonumber \\&\langle J_1F_u,F_{uu}\rangle +\langle J_1F,F_{uuu}\rangle =0,\nonumber \\&\langle J_1F_{uu},F_{uuu}\rangle +\langle J_1F_u,F_{uuuu}\rangle =0. \end{aligned}$$

(26)

Using Remark 3.4 we can prove the following proposition that gives the conditions under which an immersion defines a helix surface.

Proposition 3.5

Let $F:\Omega \rightarrow {{\mathrm {SL}}(2,{\mathbb {R}})_\tau }\subset {\mathbb {R}}_2^4$ be an immersion from an open set $\Omega \subset {\mathbb {R}}^2$, with local coordinates $(u,v)$, such that the projection of $E_1=-\tau ^{-1}J_1F$ to the tangent space of $F(\Omega )\subset {{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ is $F_u$. Then $F(\Omega )\subset {{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ defines a helix surface of constant angle $\vartheta$ if and only if

$$\begin{aligned} g_{\tau }(F_u,F_u)=g_{\tau }(E_1,F_u)=\sin ^2\vartheta , \end{aligned}$$

(27)

and

$$\begin{aligned} g_{\tau }(F_v,E_1)-g_{\tau }(F_u,F_v)=0. \end{aligned}$$

(28)

Proof

Suppose that $F$ is a helix surface of constant angle $\vartheta$. Then

$$\begin{aligned} g_{\tau }(F_u,F_u)&= -\langle F_u,F_u\rangle +(1+\tau ^2) \langle F_u,J_1F\rangle ^2\\&= \tau ^{-2}\sin ^2\vartheta B+(1+\tau ^2)(\tau ^{-2}\sin ^4\vartheta )\\&= \sin ^2\vartheta . \end{aligned}$$

Similarly

$$\begin{aligned} g_{\tau }(E_1,F_u)&= \tau ^{-1}\langle J_1F,F_u\rangle -\tau ^{-1}(1+\tau ^2) \langle J_1F,F_u\rangle \langle J_1F,J_1F\rangle \\&= \tau ^{-1}\langle J_1F,F_u\rangle [1-(1+\tau ^2)]=\sin ^2\vartheta . \end{aligned}$$

Finally, using (15), we have

$$\begin{aligned} g_{\tau }(F_v,E_1)-g_{\tau }(F_u,F_v)&= -\frac{a}{\tau } g_{\tau }(F_u,J_1F)-\frac{b}{\tau } g_{\tau }(J_1F_u,J_1F)\\&\quad -ag_{\tau }(F_u,F_u)-b g_{\tau }(J_1F_u,F_u)\\&= a \sin ^2\vartheta - 0 - a \sin ^2\vartheta -0=0. \end{aligned}$$

For the converse, put

$$\begin{aligned} T_2=F_v-\frac{g_{\tau }(F_v,F_u)F_u}{g_{\tau }(F_u,F_u)}. \end{aligned}$$

Then, if we denote by $N$ the unit normal vector field to the surface $F(\Omega ),\, \{F_u,T_2,N\}$ is an orthogonal bases of the tangent space of ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ along the surface $F(\Omega )$. Now, using (28), we get $g_{\tau }(E_1,T_2)=0$, thus $E_1=a\, F_u+c\, N$. Moreover, using (27) and that $g_{\tau }(E_1,F_u)=a\, g_{\tau }(F_u,F_u)$, we conclude that $a=1$. Finally,

$$\begin{aligned} 1=g_{\tau }(E_1,E_1)=g_{\tau }(F_u+c\, N,F_u+c\, N)=\sin ^2\vartheta +c^2, \end{aligned}$$

which implies that $c^2=\cos ^2\vartheta$. Thus the angle between $E_1$ and $N$ is

$$\begin{aligned} g_{\tau }(E_1,N)=g_{\tau }(F_u+\cos \vartheta N,N)=\cos \vartheta . \end{aligned}$$

$\square$

4 The case $B=0$

Theorem 4.1

Let $M^2$ be a helix surface in the ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }\subset {\mathbb {R}}_2^4$ with constant angle $\vartheta$ such that $B=0$. Then $\cos \vartheta =\frac{1}{\sqrt{1+\tau ^2}}$ and, locally, the position vector of $M^2$ in ${\mathbb {R}}^4_2$, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), is given by

$$\begin{aligned} F(u,v)=A(v)\Big (1,-\frac{\tau \,u}{1+\tau ^2},\frac{\tau \,u}{1+\tau ^2},0\Big ), \end{aligned}$$

(29)

where $A(v)=A(\xi ,\xi _1,\xi _2,\xi _3)(v)$ is a 1-parameter family of $4\times 4$ indefinite orthogonal matrices commuting with $J_1$, as described in (4), with

$$\begin{aligned}&[\xi '(v)+\xi _2'(v)+\xi _3'(v)]\,\sin (\xi _2(v)-\xi _3(v))\,\sinh (2\xi _1(v))\nonumber \\&-\,2(\xi '(v)-\xi _3'(v))\,\sinh ^2\xi _1(v)+2\,[\xi _1'(v)\, \cos (\xi _2(v)-\xi _3(v))+\xi _2'(v)\,\cosh ^2\xi _1(v)]=0.\nonumber \\ \end{aligned}$$

(30)

Conversely, a parametrization

$$\begin{aligned} F(u,v)=A(v)\Big (1,-\frac{\tau \,u}{1+\tau ^2},\frac{\tau \,u}{1+\tau ^2},0\Big ), \end{aligned}$$

with $A(v)$ as above, defines a helix surface in the special linear group with constant angle $\vartheta =\arccos \frac{1}{\sqrt{1+\tau ^2}}$.

Proof

Since $B=0$ we have immediately that $\cos ^2\vartheta =1/({1+\tau ^2})$. Integrating (20), we obtain that

$$\begin{aligned} F(u,v)=h^1(v)+u\,h^2(v), \end{aligned}$$

(31)

where $h^i(v),\, i=1,2$, are vector fields in ${\mathbb {R}}^4_2$, depending only on $v$.

Evaluating in $(0,v)$ the identities:

$$\begin{aligned}&\langle F,F\rangle =1,\quad \langle F_u,F_u\rangle =0,\\&\langle F,F_u\rangle =0,\quad \langle J_1F,F_u\rangle =-\tau ^{-1}\,\sin ^2\vartheta =-\frac{\tau }{1+\tau ^2}, \end{aligned}$$

it results that

$$\begin{aligned}&\langle h^1(v),h^1(v)\rangle =1,\quad \langle h^1(v),h^2(v)\rangle =0,\nonumber \\&\langle h^2(v),h^2(v)\rangle =0,\quad \langle J_1h^1(v),h^2(v)\rangle =-\frac{\tau }{1+\tau ^2}. \end{aligned}$$

(32)

Moreover, using (23) in $(0,v)$, we have that

$$\begin{aligned} h^2(v)=-\frac{\tau }{1+\tau ^2}\,(J_1h^1(v)-h^3(v)), \end{aligned}$$

where $h^3(v)$ is a vector field of ${\mathbb {R}}_2^4$ satisfying

$$\begin{aligned} \langle h^3(v),h^3(v)\rangle =-1,\quad \langle h^1(v),h^3(v)\rangle =0,\quad \langle J_1h^1(v),h^3(v)\rangle =0. \end{aligned}$$

(33)

Consequently, if we fix the orthonormal basis $\{\hat{E}_i\}_{i=1}^4$ of ${\mathbb {R}}^4_2$ given by

$$\begin{aligned} \hat{E}_1=(1,0,0,0),\quad \hat{E}_2=(0,1,0,0),\quad \hat{E}_3=(0,0,1,0),\quad \hat{E}_4=(0,0,0,1), \end{aligned}$$

there must exists a 1-parameter family of matrices $A(v)\in \mathrm {O}_2(4)$, with $J_1\,A(v)=A(v)\,J_1$, such that

$$\begin{aligned} h^1(v)=A(v)\hat{E}_1,\quad J_1h^1(v)=A(v)\hat{E}_2,\quad h^3(v)=A(v)\hat{E}_3,\quad J_1h^3(v)=A(v)\hat{E}_4. \end{aligned}$$

Then (31) becomes

$$\begin{aligned} F(u,v)=h^1(v)-\frac{\tau \,u}{1+\tau ^2}\,(J_1h^1(v)-h^3(v))=A(v) \Big (1,-\frac{\tau \,u}{1+\tau ^2},\frac{\tau \,u}{1+\tau ^2},0\Big ). \end{aligned}$$

Finally, the 1-parameter family $A(v)$ depends, according to (4), on four functions $\xi _1(v),\, \xi _2(v),\, \xi _3(v)$ and $\xi (v)$ and, in this case, condition (28) reduces to $\langle F_u, F_v\rangle =0$ which is equivalent to (30).

For the converse, let

$$\begin{aligned} F(u,v)=A(v)\Big (1,-\frac{\tau \,u}{1+\tau ^2},\frac{\tau \,u}{1+\tau ^2},0\Big ), \end{aligned}$$

be a parametrization where $A(v)=A(\xi (v),\xi _1(v),\xi _2(v),\xi _3(v))$ is a 1-parameter family of indefinite orthogonal matrices with functions $\xi (v),\xi _1(v),\xi _2(v),\xi _3(v)$ satisfying (30). Since $A(v)$ satisfies (30), then $F$ satisfies (28), thus, in virtue of Proposition 3.5, we only have to show that (27) is satisfied for some constant angle $\vartheta$. For this we put

$$\begin{aligned} \gamma (u)=\Big (1,-\frac{\tau \,u}{1+\tau ^2},\frac{\tau \,u}{1+\tau ^2},0\Big ). \end{aligned}$$

Now, using (1) and taking into account that $A(v)$ commutes with $J_1$, we get

$$\begin{aligned} g_{\tau }(F_u,F_u)&= -\langle A(v)\gamma '(u),A(v)\gamma '(u)\rangle +(1+\tau ^2)\langle A(v)\gamma '(u),J_1A(v)\gamma (u)\rangle ^2\\&= (1+\tau ^2)\langle \gamma '(u),J_1\gamma (u)\rangle ^2=\frac{\tau ^2}{1+\tau ^2}, \end{aligned}$$

and we can choose $\vartheta$ such that $\tau ^2/(1+\tau ^2)=\sin ^2\vartheta$. Similarly,

$$\begin{aligned} g_{\tau }(E_1,F_u)&= -\langle E_1,F_u\rangle +(1+\tau ^2)\langle E_1,J_1F\rangle \langle F_u,J_1F\rangle \\&= \frac{\langle F_u,J_1 F\rangle }{\tau }\left[ 1-(1+\tau ^2)\right] \\&= (-\tau )\langle \gamma '(u),J_1\gamma (u)\rangle =\frac{\tau ^2}{1+\tau ^2}. \end{aligned}$$

$\square$

Example 4.2

If we take $\xi _2=\xi _3={{\mathrm{constant}}}$, (30) becomes $\xi '(1-\sinh ^2\xi _1)=0$. Thus, if also $\xi ={{\mathrm{constant}}}$, we find, from (4), a 1-parameter family $A(v)=A(\xi _1(v))$ of indefinite orthogonal matrices such that (29) defines a helix surface for any function $\xi _1$.

5 The case $B>0$

Supposing $B>0$, integrating (21) we have the following

Proposition 5.1

Let $M^2$ be a helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ with constant angle $\vartheta$ so that $B>0$. Then, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), the position vector $F$ of $M^2$ in ${\mathbb {R}}^4_2$ is given by

$$\begin{aligned} F(u,v)=\cos (\alpha _1\, u)\,g^1(v)+\sin (\alpha _1\, u)\,g^2(v)+\cos (\alpha _2\, u)\,g^3(v)+\sin (\alpha _2\, u)\,g^4(v), \end{aligned}$$

where

$$\begin{aligned} \alpha _{1,2}=\frac{1}{\tau }(\tau \sqrt{B} \cos \vartheta \pm B) \end{aligned}$$

are positive real constants, while the $g^i(v),\, i\in \{1,\ldots ,4\}$, are mutually orthogonal vector fields in ${\mathbb {R}}^4_2$, depending only on $v$, such that

$$\begin{aligned} g_{11}&= \langle g^1(v),g^1(v)\rangle =g_{22}=\langle g^2(v),g^2(v)\rangle =-\frac{\tau }{2B}\, \alpha _2,\\ g_{33}&= \langle g^3(v),g^3(v)\rangle =g_{44}=\langle g^4(v),g^4(v)\rangle =\frac{\tau }{2B}\,\alpha _1.\nonumber \end{aligned}$$

(34)

Proof

First, a direct integration of (21), gives the solution

$$\begin{aligned} F(u,v)=\cos (\alpha _1 u)\,g^1(v)+\sin (\alpha _1 u)\,g^2(v)+\cos (\alpha _2 u)\,g^3(v)+\sin (\alpha _2 u)\,g^4(v), \end{aligned}$$

where

$$\begin{aligned} \alpha _{1,2}=\sqrt{\frac{\tilde{b}^2-2\tilde{a}\pm \sqrt{\tilde{b}^4-4\tilde{a}\tilde{b}^2}}{2}} \end{aligned}$$

are two constants, while the $g^i(v),\, i\in \{1,\ldots ,4\}$, are vector fields in ${\mathbb {R}}^4_2$ which depend only on $v$. Now, taking into account the values of $\tilde{a}$ and $\tilde{b}$ given in (22), we get

$$\begin{aligned} \alpha _{1,2}=\frac{1}{\tau }(\tau \sqrt{B} \cos \vartheta \pm B). \end{aligned}$$

(35)

Putting $g_{ij}(v)=\langle g^i(v),g^j(v)\rangle$, and evaluating the relations (25) in $(0,v)$, we obtain:

$$\begin{aligned}&g_{11}+g_{33}+2g_{13}=1, \end{aligned}$$

(36)

$$\begin{aligned}&\alpha _1^2\,g_{22}+\alpha _2^2\,g_{44}+2\alpha _1\alpha _2\,g_{24}=\tilde{a},\end{aligned}$$

(37)

$$\begin{aligned}&\alpha _1\,g_{12}+\alpha _2\,g_{14}+\alpha _1\,g_{23}+\alpha _2\,g_{34}=0, \end{aligned}$$

(38)

$$\begin{aligned}&\alpha _1^3\,g_{12}+\alpha _1\alpha _2^2\,g_{23}+\alpha _1^2\alpha _2\,g_{14} +\alpha _2^3g_{34}=0, \end{aligned}$$

(39)

$$\begin{aligned}&\alpha _1^4\,g_{11}+\alpha _2^4\,g_{33}+2\alpha _1^2\alpha _2^2\,g_{13}=D, \end{aligned}$$

(40)

$$\begin{aligned}&\alpha _1^2\,g_{11}+\alpha _2^2\,g_{33}+(\alpha _1^2+\alpha _2^2)\,g_{13}=\tilde{a},\end{aligned}$$

(41)

$$\begin{aligned}&\alpha _1^4\,g_{22}+\alpha _1^3\alpha _2\,g_{24}+\alpha _1\alpha _2^3\,g_{24}+\alpha _2^4\,g_{44}=D, \end{aligned}$$

(42)

$$\begin{aligned}&\alpha _1^5\,g_{12}+\alpha _1^3\alpha _2^2\,g_{23}+\alpha _1^2\alpha _2^3\,g_{14}+\alpha _2^5\,g_{34}=0, \end{aligned}$$

(43)

$$\begin{aligned}&\alpha _1^3\,g_{12}+\alpha _1^3\,g_{23}+\alpha _2^3\,g_{14}+\alpha _2^3\,g_{34}=0, \end{aligned}$$

(44)

$$\begin{aligned}&\alpha _1^6\,g_{22}+\alpha _2^6\,g_{44}+2\alpha _1^3\alpha _2^3\,g_{24}=E. \end{aligned}$$

(45)

From (38), (39), (43), (44), it follows that

$$\begin{aligned} g_{12}=g_{14}=g_{23}=g_{34}=0. \end{aligned}$$

Also, from (36), (40) and (41), we obtain

$$\begin{aligned} g_{11}=\frac{\tau ^2\,(D+\alpha _2^4)+2B\sin ^2\vartheta \,\alpha _2^2}{\tau ^2 (\alpha _1^2-\alpha _2^2)^2},\qquad g_{13}=0,\qquad g_{33}=\frac{\tau ^2\,(D+\alpha _1^4)+2B\sin ^2\vartheta \,\alpha _1^2}{\tau ^2 (\alpha _1^2-\alpha _2^2)^2}. \end{aligned}$$

Finally, using (37), (42) and (45), we obtain

$$\begin{aligned} g_{22}\,{=}\,\frac{\tau ^2\,(E-2D\,\alpha _2^2)-B\sin ^2\vartheta \,\alpha _2^4}{\tau ^2 \alpha _1^2\,(\alpha _1^2-\alpha _2^2)^2},\quad g_{24}\,{=}\,0,\qquad g_{44}\,{=}\,\frac{\tau ^2\,(E-2D\,\alpha _1^2)-B\sin ^2\vartheta \, \alpha _1^4}{\tau ^2\alpha _2^2\,(\alpha _1^2-\alpha _2^2)^2}. \end{aligned}$$

We observe that

$$\begin{aligned} g_{11}=g_{22}=\frac{\sqrt{B}-\tau \,\cos \vartheta }{2\sqrt{B}}<0,\qquad g_{33}=g_{44}=\frac{\sqrt{B}+\tau \,\cos \vartheta }{2\sqrt{B}}>0. \end{aligned}$$

Therefore, taking into account (35), we obtain the expressions (34). $\square$

We are now in the right position to state the main result of this section.

Theorem 5.2

Let $M^2$ be a helix surface in the ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }\subset {\mathbb {R}}_2^4$ with constant angle $\vartheta \ne \pi /2$ so that $B>0$. Then, locally, the position vector of $M^2$ in ${\mathbb {R}}^4_2$, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), is

$$\begin{aligned} F(u,v)=A(v)\,\gamma (u), \end{aligned}$$

(46)

where

$$\begin{aligned} \gamma (u)=(\sqrt{g_{33}}\,\cos (\alpha _2\, u),-\sqrt{g_{33}}\,\sin (\alpha _2\, u),\sqrt{-g_{11}}\,\cos (\alpha _1\, u),\sqrt{-g_{11}}\,\sin (\alpha _1\, u)) \end{aligned}$$

(47)

is a curve in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau },\, g_{11},\, g_{33},\, \alpha _1,\, \alpha _2$ are the four constants given in Proposition 5.1, and $A(v)=A(\xi ,\xi _1,\xi _2,\xi _3)(v)$ is a 1-parameter family of $4\times 4$ indefinite orthogonal matrices commuting with $J_1$, as described in (4), with $\xi ={{\mathrm{constant}}}$ and

$$\begin{aligned} \cosh ^2(\xi _1(v)) \,\xi _{2}'(v)+\sinh ^2(\xi _1(v))\, \xi _{3}'(v)=0. \end{aligned}$$

(48)

Conversely, a parametrization $F(u,v)=A(v)\,\gamma (u)$, with $\gamma (u)$ and $A(v)$ as above, defines a constant angle surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ with constant angle $\vartheta \ne \pi /2$.

Proof

With respect to the local coordinates $(u,v)$ on $M$ defined in (14), Proposition 5.1 implies that the position vector of the helix surface in ${\mathbb {R}}^4_2$ is given by

$$\begin{aligned} F(u,v)=\cos (\alpha _1 u)\,g^1(v)+\sin (\alpha _1 u)\,g^2(v)+\cos (\alpha _2 u)\,g^3(v)+\sin (\alpha _2 u)\,g^4(v), \end{aligned}$$

where the vector fields $\{g^i(v)\}_{i=1}^4$ are mutually orthogonal and

$$\begin{aligned} ||g^1(v)||&= ||g^2(v)||=\sqrt{-g_{11}}=\text {constant}, \\ ||g^3(v)||&= ||g^4(v)||=\sqrt{g_{33}}=\text {constant}. \end{aligned}$$

Thus, if we put $e_i(v)=g^i(v)/||g^i(v)||,\, i\in \{1,\ldots ,4\}$, we can write:

$$\begin{aligned} F(u,v)&= \sqrt{-g_{11}}\,(\cos (\alpha _1\,u)\,e_1(v)+\sin (\alpha _1\,u)\,e_2(v))\nonumber \\&+\sqrt{g_{33}}\,(\cos (\alpha _2\,u)\,e_3(v)+\sin (\alpha _2\,u)\,e_4(v)). \end{aligned}$$

(49)

Now, the identities (26), evaluated in $(0,v)$, become respectively:

$$\begin{aligned}&\alpha _2\,g_{33}\langle J_1e_3,e_4\rangle -\alpha _1 g_{11}\langle J_1e_1,e_2\rangle \nonumber \\&\quad +\sqrt{-g_{11}g_{33}}\,(\alpha _1\langle J_1e_3,e_2\rangle +\alpha _2\langle J_1e_1,e_4\rangle ) =-\tau ^{-1}\sin ^2\vartheta , \end{aligned}$$

(50)

$$\begin{aligned}&\langle J_1e_1,e_3\rangle =0, \end{aligned}$$

(51)

$$\begin{aligned}&\alpha _2^3\,g_{33}\langle J_1e_3,e_4\rangle -\alpha _1^3\,g_{11}\langle J_1e_1,e_2\rangle \nonumber \\&\quad +\sqrt{-g_{11}g_{33}}\,(\alpha _1\alpha _2^2\langle J_1e_3,e_2\rangle +\alpha _1^2\alpha _2\langle J_1e_1,e_4\rangle )=-I, \end{aligned}$$

(52)

$$\begin{aligned}&\langle J_1e_2,e_4\rangle =0, \end{aligned}$$

(53)

$$\begin{aligned}&\alpha _1\langle J_1e_2,e_3\rangle +\alpha _2\langle J_1e_1,e_4\rangle =0, \end{aligned}$$

(54)

$$\begin{aligned}&\alpha _2\langle J_1 e_2,e_3\rangle +\alpha _1\langle J_1e_1,e_4\rangle =0. \end{aligned}$$

(55)

We point out that to obtain the previous identities we have divided by $\alpha _1^2-\alpha _2^2=4 \tau ^{-1} \sqrt{B^3} \cos \vartheta$ which is, by the assumption on $\vartheta$, always different from zero. From (54) and (55), taking into account the $\alpha _1^2-\alpha _2^2\ne 0$, it results that

$$\begin{aligned} \langle J_1e_3,e_2\rangle =0,\qquad \langle J_1e_1,e_4\rangle =0. \end{aligned}$$

(56)

Therefore

$$\begin{aligned} |\langle J_1e_1,e_2\rangle |=1=|\langle J_1e_3,e_4\rangle |. \end{aligned}$$

Substituting (56) in (50) and (52), we obtain the system

$$\begin{aligned} \left\{ \begin{array}{l} \alpha _1\, g_{11}\langle J_1e_1,e_2\rangle -\alpha _2\,g_{33}\langle J_1e_3,e_4\rangle =\tau ^{-1}\sin ^2\vartheta \\ \alpha _1^3\, g_{11}\langle J_1e_1,e_2\rangle -\alpha _2^3\,g_{33}\langle J_1e_3,e_4\rangle =I, \end{array}\right. \end{aligned}$$

a solution of which is

$$\begin{aligned} \langle J_1e_1,e_2\rangle =\frac{\tau I-\alpha _2^2\sin ^2\vartheta }{\tau g_{11}\,\alpha _1 (\alpha _1^2-\alpha _2^2)},\qquad \langle J_1e_3,e_4\rangle =\frac{\tau I-\alpha _1^2\sin ^2\vartheta }{\tau g_{33}\,\alpha _2(\alpha _1^2-\alpha _2^2)}. \end{aligned}$$

Now, as

$$\begin{aligned} g_{11}\,g_{33}=-\frac{\sin ^2\vartheta }{4B},\qquad \alpha _1\,\alpha _2=\frac{B}{\tau ^2}\sin ^2\vartheta ,\qquad \alpha _1^2-\alpha _2^2=\frac{4\sqrt{B^3}}{\tau }\cos \vartheta , \end{aligned}$$

it results that

$$\begin{aligned} \langle J_1e_1,e_2\rangle \langle J_1e_3,e_4\rangle =1. \end{aligned}$$

Moreover, as

$$\begin{aligned} \tau I-\alpha _2^2\sin ^2\vartheta =2\tau ^{-1}\,\sqrt{B^3}\cos \vartheta \sin ^2\vartheta , \end{aligned}$$

it results that $\langle J_1e_1,e_2\rangle <0$. Consequently, $\langle J_1e_1,e_2\rangle =\langle J_1e_3,e_4\rangle =-1$ and $J_1e_1=e_2,\, J_1e_3=-e_4$.

Then, if we fix the orthonormal basis of ${\mathbb {R}}^4_2$ given by

$$\begin{aligned} \tilde{E}_1=(0,0,1,0),\quad \tilde{E}_2=(0,0,0,1),\quad \tilde{E}_3=(1,0,0,0),\quad \tilde{E}_4=(0,-1,0,0), \end{aligned}$$

there must exists a 1-parameter family of $4\times 4$ indefinite orthogonal matrices $A(v)\in \mathrm {O}_2(4)$, with $J_1A(v)=A(v)J_1$, such that $e_i(v)=A(v)\tilde{E}_i$. Replacing $e_i(v)=A(v)\tilde{E}_i$ in (49) we obtain

$$\begin{aligned} F(u,v)=A(v)\gamma (u), \end{aligned}$$

where

$$\begin{aligned} \gamma (u)=(\sqrt{g_{33}}\,\cos (\alpha _2\, u),-\sqrt{g_{33}}\,\sin (\alpha _2\, u),\sqrt{-g_{11}}\,\cos (\alpha _1\, u),\sqrt{-g_{11}}\,\sin (\alpha _1\, u)) \end{aligned}$$

is a curve in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$.

Let now examine the 1-parameter family $A(v)$ that, according to (4), depends on four functions $\xi _1(v),\xi _2(v),\xi _3(v)$ and $\xi (v)$. From (15), it results that $\langle F_v, F_v\rangle =-\sin ^2 \vartheta ={{\mathrm{constant}}}$. The latter implies that

$$\begin{aligned} \frac{\partial }{\partial u}\langle F_v, F_v\rangle _{| u=0}=0. \end{aligned}$$

(57)

Now, if we denote by ${\mathbf {c}_1},{\mathbf {c}_2},{\mathbf {c}_3},{\mathbf {c}_4}$ the four colons of $A(v)$, (57) implies that

$$\begin{aligned} {\left\{ \begin{array}{ll} \langle {\mathbf {c}_2}',{\mathbf {c}_3}'\rangle =0\\ \langle {\mathbf {c}_2}',{\mathbf {c}_4}'\rangle =0, \end{array}\right. } \end{aligned}$$

(58)

where with $^\prime$ we means the derivative with respect to $v$. Replacing in (58) the expressions of the ${\mathbf {c}_i}$’s as functions of $\xi _1(v),\xi _2(v),\xi _3(v)$ and $\xi (v)$, we obtain

$$\begin{aligned} {\left\{ \begin{array}{ll} \xi ^{\prime }\, h(v)=0\\ \xi ^{\prime }\, k(v)=0, \end{array}\right. } \end{aligned}$$

(59)

where $h(v)$ and $k(v)$ are two functions such that

$$\begin{aligned} h^2+k^2=4 (\xi _1^{\prime })^2+\sinh ^2(2\xi _1)\, (-\xi ^{\prime }+\xi _2^{\prime }+\xi _3^{\prime })^2. \end{aligned}$$

From (59) we have two possibilities:

• (i) $\xi ={{\mathrm{constant}}}$;

  or

• (ii) $4 (\xi _1^{\prime })^2+\sinh ^2(2\xi _1)\, (-\xi ^{\prime }+\xi _2^{\prime }+\xi _3^{\prime })^2=0$.

We will show that case (ii) cannot occur, more precisely we will show that if (ii) happens then the parametrization $F(u,v)=A(v)\gamma (u)$ defines a Hopf tube, that is the Hopf vector field $E_1$ is tangent to the surface. To this end, we write the unit normal vector field $N$ as

$$\begin{aligned} N=\frac{N_1 E_1+N_2E_2+N_3 E_3}{\sqrt{N_1^2+N_2^2+N_3^2}}. \end{aligned}$$

A long but straightforward computation (that can be also made using a software of symbolic computations) gives

$$\begin{aligned} N_1&= 1/2 (\alpha _1 + \alpha _2) \sqrt{-g_{11}} \sqrt{g_{33}}\,[2\xi _1^{\prime }\,\cos (\alpha _1 u+\alpha _2 u-\xi _2+\xi _3)\\&+\sinh (2\xi _1)\sin (\alpha _1 u+\alpha _2 u-\xi _2+\xi _3)( -\xi ^{\prime }+\xi _2^{\prime }+\xi _3^{\prime })]. \end{aligned}$$

Now case (ii) occurs if and only if $\xi _1={{\mathrm{constant}}}=0$, or if $\xi _1={{\mathrm{constant}}}\ne 0$ and $-\xi ^{\prime }+\xi _2^{\prime }+\xi _3^{\prime }=0$. In both cases $N_1=0$ and this implies that $g_{\tau }(N,J_1F)=-\tau g_{\tau }(N,E_1)=0$, i.e. the Hopf vector field is tangent to the surface. Thus we have proved that $\xi ={{\mathrm{constant}}}$.

Finally, in this case, (28) is equivalent to

$$\begin{aligned} \tau \cos \vartheta \sqrt{B}[\cosh ^2(\xi _1(v)) \,\xi _{2}^{\prime }(v)+\sinh ^2(\xi _1(v))\, \xi _{3}^{\prime }(v)]=0. \end{aligned}$$

Since $\vartheta \ne \pi /2$ we conclude that condition (48) is satisfied.

The converse follows immediately from Proposition 3.5 since a direct calculation shows that $g_{\tau }(F_u,F_u)=g_{\tau }(E_1,F_u)=\sin ^2\vartheta$ which is (27), while (48) is equivalent to (28). $\square$

6 The case $B<0$

In this section we study the case $B<0$. Integrating (21) we have the following result:

Proposition 6.1

Let $M^2$ be a helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ with constant angle $\vartheta$ and $B<0$. Then, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), the position vector $F$ of $M^2$ in ${\mathbb {R}}^4_2$ is given by

$$\begin{aligned} F(u,v)&= \cos \big (\frac{\tilde{b}}{2}\, u\big )\,[\cosh (\beta \, u)\,w^1(v)+\sinh (\beta \, u)\,w^3(v)]\nonumber \\&+\sin \big (\frac{\tilde{b}}{2}\, u\big )\,[\cosh (\beta \, u)\,w^2(v)+\sinh (\beta \, u)\,w^4(v)], \end{aligned}$$

(60)

where

$$\begin{aligned} \beta =\sqrt{-B}\,\cos \vartheta \end{aligned}$$

is a real constant, $\tilde{b}=-2 \tau ^{-1}\, B$, while the $w^i(v),\, i\in \{1,\ldots ,4\}$, are vector fields in ${\mathbb {R}}^4_2$, depending only on $v$, such that

$$\begin{aligned} \langle w^1(v),w^1(v)\rangle&= \langle w^2(v),w^2(v)\rangle =-\langle w^3(v),w^3(v)\rangle =-\langle w^4(v),w^4(v)\rangle =1,\nonumber \\ \langle w^1(v),w^2(v)\rangle&= \langle w^1(v),w^3(v)\rangle =\langle w^2(v),w^4(v)\rangle =\langle w^3(v),w^4(v)\rangle =0,\\ \langle w^1(v),w^4(v)\rangle&= -\langle w^2(v),w^3(v)\rangle =-\frac{2\beta }{\tilde{b}}.\nonumber \end{aligned}$$

(61)

Proof

A direct integration of (21), gives the solution

$$\begin{aligned} F(u,v)&= \cos \big (\frac{\tilde{b}}{2}\, u\big )\,[\cosh (\beta \, u)\,w^1(v)+\sinh (\beta \, u)\,w^3(v)]\\&+\sin \big (\frac{\tilde{b}}{2}\, u\big )\,[\cosh (\beta \, u)\,w^2(v)+\sinh (\beta \, u)\,w^4(v)], \end{aligned}$$

where

$$\begin{aligned} \beta =\frac{\sqrt{4\tilde{a}-\tilde{b}^2}}{2}=\sqrt{-B}\,\cos \vartheta \end{aligned}$$

is a constant, while the $w^i(v),\, i\in \{1,\ldots ,4\}$, are vector fields in ${\mathbb {R}}^4$ which depend only on $v$. If $w_{ij}(v):=\langle w^i(v),w^j(v)\rangle$, evaluating the relations (25) in $(0,v)$, we obtain

$$\begin{aligned}&w_{11}=1, \end{aligned}$$

(62)

$$\begin{aligned}&\frac{\tilde{b}^2}{4}\,w_{22}+\beta ^2\,w_{33}+\beta \,\tilde{b}\,w_{23}=\tilde{a}, \end{aligned}$$

(63)

$$\begin{aligned}&\frac{\tilde{b}}{2}\,w_{12}+\beta \,w_{13}=0, \end{aligned}$$

(64)

$$\begin{aligned}&\frac{\tilde{b}}{2}\,\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{12}+\beta ^2 \,\tilde{b}\,w_{34}+\beta \,\frac{\tilde{b}^2}{2}\,w_{24} +\beta \,\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{13}=0, \end{aligned}$$

(65)

$$\begin{aligned}&\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )^2\,w_{11}+\beta ^2 \,\tilde{b}^2\,w_{44} +2\beta \,\tilde{b}\,\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{14}=D, \end{aligned}$$

(66)

$$\begin{aligned}&\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{11}+\beta \,\tilde{b}\,w_{14}=-\tilde{a}, \end{aligned}$$

(67)

$$\begin{aligned}&\frac{\tilde{b}^2}{4}\, \Big (3\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{22}+\beta ^2 \,\Big (\beta ^2-3\frac{\tilde{b}^2}{4}\Big )\,w_{33} +\beta \,\frac{\tilde{b}}{2}\,(4\beta ^2-\tilde{b}^2)\,w_{23}=-D, \end{aligned}$$

(68)

$$\begin{aligned}&\frac{\tilde{b}}{2}\, \Big (3\beta ^2-\frac{\tilde{b}^2}{4}\Big )\, \Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{12}+\tilde{b}\,\beta ^2 \,\Big (\beta ^2-3\frac{\tilde{b}^2}{4}\Big )\,w_{34}\nonumber \\&+\beta \,\Big (\beta ^2-3\frac{\tilde{b}^2}{4}\Big )\,\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )w_{13}+\beta \, \frac{\tilde{b}^2}{2}\,\Big (3\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{24}=0,\end{aligned}$$

(69)

$$\begin{aligned}&\frac{\tilde{b}}{2}\,\Big (3\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w_{12} +\beta \,\Big (\beta ^2-3\frac{\tilde{b}^2}{4}\Big )\,w_{13}=0,\end{aligned}$$

(70)

$$\begin{aligned}&\frac{\tilde{b}^2}{4}\, \Big (3\beta ^2-\frac{\tilde{b}^2}{4}\Big )^2\,w_{22}+\beta ^2 \,\Big (\beta ^2-3\frac{\tilde{b}^2}{4}\Big )^2\,w_{33}\nonumber \\&+\beta \,\tilde{b}\,\Big (3\beta ^2-\frac{\tilde{b}^2}{4}\Big )\, \Big (\beta ^2-3\frac{\tilde{b}^2}{4}\Big )\,w_{23}=E. \end{aligned}$$

(71)

From (62), (66) and (67), it follows that

$$\begin{aligned} w_{11}=-w_{44}=1, \qquad w_{14}=-\frac{2\,\beta }{\tilde{b}}. \end{aligned}$$

Also, from (64) and (70), we obtain

$$\begin{aligned} w_{12}=w_{13}=0 \end{aligned}$$

and, therefore, from (65) and (69),

$$\begin{aligned} w_{24}=w_{34}=0. \end{aligned}$$

Moreover, using (63), (68) and (71), we get

$$\begin{aligned} w_{22}=-w_{33}=1, \qquad w_{23}=\frac{2\,\beta }{\tilde{b}}. \end{aligned}$$

$\square$

Theorem 6.2

Let $M^2$ be a helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ with constant angle $\vartheta \ne \pi /2$ so that $B<0$. Then, locally, the position vector of $M^2$ in ${\mathbb {R}}^4_2$, with respect to the local coordinates $(u,v)$ on $M$ defined in (14), is given by

$$\begin{aligned} F(u,v)=A(v)\,\gamma (u), \end{aligned}$$

where the curve $\gamma (u)=(\gamma _1(u),\gamma _2(u),\gamma _3(u),\gamma _4(u))$ is given by

$$\begin{aligned} \left\{ \begin{array}{l} \gamma _1(u)=\cos \Big (\frac{\tilde{b}}{2}u\Big )\,\cosh (\beta \,u)-\frac{2\beta }{\tilde{b}}\,\sin \Big (\frac{\tilde{b}}{2}u\Big )\,\sinh (\beta \,u),\\ \gamma _2(u)=\sin \Big (\frac{\tilde{b}}{2}u\Big )\,\cosh (\beta \,u)+\frac{2\beta }{\tilde{b}}\,\cos \Big (\frac{\tilde{b}}{2}u\Big )\,\sinh (\beta \,u),\\ \gamma _3(u)=\frac{\sin \vartheta }{\sqrt{-B}}\,\cos \Big (\frac{\tilde{b}}{2}u\Big )\,\sinh (\beta \,u),\\ \gamma _4(u)=\frac{\sin \vartheta }{\sqrt{-B}}\,\sin \Big (\frac{\tilde{b}}{2}u\Big )\,\sinh (\beta \,u),\\ \end{array}\right. \end{aligned}$$

(72)

$\beta =\sqrt{-B}\,\cos \vartheta ,\, \tilde{b}=-2 \tau ^{-1}\, B$ and $A(v)=A(\xi ,\xi _1,\xi _2,\xi _3)(v)$ is a 1- parameter family of $4\times 4$ indefinite orthogonal matrices anticommuting with $J_1$, as described in (4), with $\xi ={{\mathrm{constant}}}$ and

$$\begin{aligned}&\sin \vartheta \,[2\cos (\xi _2(v)-\xi _3(v))\,\xi _1^{\prime }(v)+(\xi _2^{\prime }(v)+\xi _3^{\prime }(v)) \sin (\xi _2(v)-\xi _3(v))\,\sinh (2\xi _1(v))]\nonumber \\&\quad -2\tau \cos \vartheta \,[\cosh ^2(\xi _1(v)) \,\xi _{2}^{\prime }(v)+\sinh ^2(\xi _1(v))\, \xi _{3}^{\prime }(v)] =0. \end{aligned}$$

(73)

Conversely, a parametrization $F(u,v)=A(v)\,\gamma (u)$, with $\gamma (u)$ and $A(v)$ as above, defines a helix surface in ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ with constant angle $\vartheta \ne \pi /2$.

Proof

From (61), we can define the following orthonormal basis in ${\mathbb {R}}^4_2$:

$$\begin{aligned} \left\{ \begin{array}{l} e_1(v)=w^1(v),\\ e_2(v)=w^2(v),\\ e_3(v)=\frac{1}{\sin \vartheta }[\sqrt{-B}\,w^3(v)-\tau \cos \vartheta \,w^2(v)],\\ e_4(v)=\frac{1}{\sin \vartheta }[\sqrt{-B}\,w^4(v)+\tau \cos \vartheta \,w^1(v)], \end{array}\right. \end{aligned}$$

(74)

with $\langle e_1,e_1\rangle =1=\langle e_2,e_2\rangle$ and $\langle e_3,e_3\rangle =-1=\langle e_4,e_4\rangle$.

Evaluating the identities (26) in $(0,v)$, and taking into account that:

$$\begin{aligned}&F(0,v)=w^1(v),\\&F_u(0,v)=\frac{\tilde{b}}{2}\,w^2(v)+\beta \,w^3(v),\\&F_{uu}(0,v)=\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w^1(v)+\beta \, \tilde{b}\,w^4(v),\\&F_{uuu}(0,v)=\frac{\tilde{b}}{2}\,\Big (3\beta ^2 -\frac{\tilde{b}^2}{4})\,w^2(v)+\beta \,\Big (\beta ^2 -\frac{3}{4}\tilde{b}^2\Big )\,w^3(v),\\&F_{uuuu}(0,v)=\Big (\beta ^4-\frac{3}{2}\beta ^2\,\tilde{b}^2 +\frac{\tilde{b}^4}{16}\Big )\,w^1(v)+ 2\beta \,\tilde{b}\,\Big (\beta ^2-\frac{\tilde{b}^2}{4}\Big )\,w^4(v), \end{aligned}$$

we conclude that

$$\begin{aligned} \langle J_1w^3,w^4\rangle&= -\langle J_1w^1,w^2\rangle =1,\\ \langle J_1w^3,w^2\rangle&= \langle J_1w^1,w^4\rangle =0,\\ \langle J_1w^2,w^4\rangle&= \langle J_1w^1,w^3\rangle =-\frac{\tau \,\cos \vartheta }{\sqrt{-B}}. \end{aligned}$$

Then,

$$\begin{aligned} -\langle J_1e_1,e_2\rangle&= \langle J_1e_3,e_4\rangle =1,\\ \langle J_1e_1,e_4\rangle =\langle J_1e_1,e_3\rangle&= \langle J_1e_2,e_3\rangle =\langle J_1e_2,e_4\rangle =0. \end{aligned}$$

Therefore, we obtain that

$$\begin{aligned} J_1e_1=-e_2,\qquad J_1e_3=-e_4. \end{aligned}$$

Consequently, if we consider the orthonormal basis $\{\hat{E}_i\}_{i=1}^4$ of ${\mathbb {R}}^4_2$ given by

$$\begin{aligned} \hat{E}_1=(1,0,0,0),\quad \hat{E}_2=(0,1,0,0),\quad \hat{E}_3=(0,0,1,0),\quad \hat{E}_4=(0,0,0,1), \end{aligned}$$

there must exists a 1-parameter family of matrices $A(v)\in \mathrm {O}_2(4)$, with $J_1A(v)=-A(v)J_1$, such that $e_i(v)=A(v)\hat{E}_i,\, i\in \{1,\ldots ,4\}$. As

$$\begin{aligned} F=\langle F,e_1\rangle \,e_1+\langle F,e_2\rangle \,e_2-\langle F,e_3\rangle \,e_3-\langle F,e_4\rangle \,e_4, \end{aligned}$$

computing $\langle F,e_i\rangle$ and substituting $e_i(v)=A(v)\hat{E}_i$, we obtain that $F(u,v)=A(v)\,\gamma (u)$, where the curve $\gamma (u)$ of ${{\mathrm {SL}}(2,{\mathbb {R}})_\tau }$ is given in (72).

Let now examine the 1-parameter family $A(v)$ that, according to (4), depends on four functions $\xi _1(v),\xi _2(v),\xi _3(v)$ and $\xi (v)$. Similarly to what we have done in the proof of Theorem 5.2 we have that the condition

$$\begin{aligned} \frac{\partial }{\partial u}\langle F_v, F_v\rangle _{| u=0}=0 \end{aligned}$$

implies that the functions $\xi _1(v),\xi _2(v),\xi _3(v)$ and $\xi (v)$ satisfy the equation

$$\begin{aligned} \xi ^{\prime }\,[2\sin (\xi _2-\xi _3)\,\xi _1^{\prime }-(\xi _2^{\prime }+\xi _3^{\prime }-\xi ^{\prime })\cos (\xi _2-\xi _3)\, \sinh (2\,\xi _1)]=0. \end{aligned}$$

Then we have two possibilities:

• (i) $\xi ={{\mathrm{constant}}}$;

  or

• (ii) $2\sin (\xi _2-\xi _3)\,\xi _1^{\prime }-(\xi _2^{\prime }+\xi _3^{\prime }-\xi ^{\prime })\cos (\xi _2-\xi _3)\,\sinh (2\,\xi _1)=0$.

Also in this case, using the same argument as in Theorem 5.2, condition (ii) would implies that the surface is a Hopf tube, thus we can assume that $\xi ={{\mathrm{constant}}}$.

Finally, a long but straightforward computation shows that, in the case $\xi ={{\mathrm{constant}}}$, (28) is equivalent to (73).

The converse of the theorem follows immediately from Proposition 3.5 since a direct calculation shows that $g_{\tau }(F_u,F_u)=g_{\tau }(E_1,F_u)=\sin ^2\vartheta$ which is (27) while (73) is equivalent to (28). $\square$