J-Trajectories in 4-Dimensional Solvable Lie Group $\textrm{Sol}_1^4$

Abstract

J-trajectories are arc-length-parameterized curves in almost Hermitian manifolds, which satisfy the equation $\nabla _{{\dot{\gamma }}}{\dot{\gamma }}=q J {\dot{\gamma }}$. In this paper, J-trajectories in the solvable Lie group $\textrm{Sol}_1^4$ are investigated. J-trajectories of osculating order 2 and 3, homogeneous J-trajectories and J-trajectories in subspace$\textrm{Nil}_3$ are examined.

Appendices

Appendix A: Curves in LCK Manifolds

Definition 3

If $\gamma$ is a curve in a Riemannian manifold M, parametrized by arc length s, we say that $\gamma$ is a Frenet curve of osculating order r if there exist orthonormal vector fields $E_1$, $E_2, \ldots ,E_r$ along $\gamma$ such that

$$\begin{aligned}&{\dot{\gamma }}=E_1 ,\,\, \nabla _{{\dot{\gamma }}} E_1 =\kappa _1 E_2 ,\,\, \nabla _{{\dot{\gamma }}} E_2 =-\kappa _1 E_1 +\kappa _2 E_3, \ldots ,\nonumber \\&\nabla _{{\dot{\gamma }}} E_{r-1} =-\kappa _{r-2} E_{r-2} +\kappa _{r-1} E_r ,\,\, \nabla _{{\dot{\gamma }}} E_r =-\kappa _{r-1} E_{r-1}, \end{aligned}$$

(17)

where $\kappa _1, \kappa _2, \ldots , \kappa _{r-1}$ are positive $C^\infty$ functions of s. The function $\kappa _j$ is called the j-th curvature of $\gamma$.

A geodesic is regarded as a Frenet curve of osculating order 1. A circle is defined as a Frenet curve of osculating order 2 with constant $\kappa _1$. A helix of order r is a Frenet curve of osculating order r, such that all the curvatures $\kappa _1,\kappa _2, \ldots , \kappa _{r-1}$ are constant.

For Frenet curves in almost Hermitian manifolds, we recall the following notion:

Definition 4

Let $\gamma (s)$ be a Frenet curve of osculating order $r>0$ in an almost Hermitian manifold (M, J, g). The complex torsions $\tau _{ij}$ ($1\le i<j\le r$) are smooth functions along $\gamma$ defined by $\tau _{ij}=g(E_{i},JE_{j})$. A helix of order r in (M, J, g) is said to be a holomorphic helix of order r if all complex torsions are constant. In particular holomorphic helices of order 2 are called holomorphic circles.

Now let $\gamma (s)$ be a normal J-trajectory of charge q in an LCK surface $M=(M,J,g)$. First we observe that the first curvature $\kappa _1$ is constant $\vert q \vert$ by comparing the J-trajectory equation and the Frenet formulas (17). The Frenet formulas imply that the first normal vector field $E_2$ is given by $E_2=\pm J{\dot{\gamma }}$. Let $\varepsilon =q/ \vert q \vert$, then we have $E_2=\varepsilon \,J{\dot{\gamma }}$ and $\kappa _1=\varepsilon q>0$.

Remark A.1

If a Frenet curve $\gamma$ in an almost Hermitian manifold (M, J, g) is a J-trajectory, then

$$\begin{aligned} \tau _{12}=g(E_1,JE_2) =-\varepsilon . \end{aligned}$$

If M is a Kähler manifold, then every J-trajectory is a holomorphic circle. This fact is closely related to the so-called Yano-Obata conjecture. In the complex projective space $\mathbb {C}P^n(c)$ of constant holomorphic sectional curvature $c>0$, every h-planar curve lies in a complex projective line. Moreover, the Lie group of h-projective transformations is strictly larger than holomorphic isometry group. In Matveev and Rosemann (2012) and Bolsinov et al. (2021) is proved that the identity components of h-projective transformations on a compact Kähler manifold (M, J, g) of complex dimension $>1$ coincides with that of holomorphic isometry group unless M is isomorphic to $\mathbb {C}P^n(c)$. On the other hand, in Kiyohara and Topalov (2010) is showed that under suitable non-degeneracy condition, compact Kähler manifolds admitting h-projectively equivalent metrics are biholomorphically equivalent to complex projective space.

by using the formula (2) and Frenet equations, we have

$$\begin{aligned} 2\kappa _{2}E_{3}= 2\varepsilon (\nabla _{{\dot{\gamma }}}J){\dot{\gamma }}= \varepsilon (\omega (E_2)E_1-\omega (E_1)E_2+A). \end{aligned}$$

(18)

Equation (18) implies that if M is Kähler, then $\kappa _2=0$. This conclusion is consistent with the fact “every J-trajectory of a Kähler manifold is a holomorphic circle" mentioned in Remark A.1. In addition, we notice that $\kappa _2=0$ along $\gamma$ if and only if $(\nabla _{{\dot{\gamma }}}J){\dot{\gamma }}=0$ holds. Note that (18) is rewritten as

$$\begin{aligned} 2\kappa _{2}E_{3}=\varepsilon (A-g(A,E_1)E_1-g(A,E_2)E_2). \end{aligned}$$

(19)

Equation (19) implies that every J-trajectory satisfies $\kappa _2\tau _{13}=\kappa _2\tau _{23}=0$. Moreover from (19) we notice that $\gamma$ is of order 2 if and only if

$$\begin{aligned} A=g(A,E_1)E_1+g(A,E_2)E_2 \end{aligned}$$

along $\gamma$. In other words, $\gamma$ is of osculating order 2 if and only of $A_\gamma$ lies in the osculating plane of $\gamma$.

Proposition A.1

(Erjavec and Inoguchi 2022) Let $\gamma$ be a non-geodesic J-trajectory with strength $q\not =0$ parametrized by arc length in an LCK manifold M whose anti-Lee field has constant length. Then, the first and the second curvatures of $\gamma$ are given by:

$$\begin{aligned} \kappa _1=\vert q \vert ,\quad \kappa _2=\frac{1}{2} \sqrt{ \vert A \vert ^2-\omega ({\dot{\gamma }})^2- \omega (J{\dot{\gamma }})^2 }. \end{aligned}$$

(20)

In particular, $\kappa _1$ is constant.

Proposition A.2

(Erjavec and Inoguchi 2022) Let $\gamma$ be a non-geodesic J-trajectory with strength $q\not =0$ parametrized by arc length in an LCK manifold M whose anti-Lee field has constant length. Assume that $\gamma$ is a curve of osculating order $\ge 3$. Then, we have

$$\begin{aligned} 4\kappa _{2}{\dot{\kappa }}_{2} = \omega (J{\dot{\gamma }}) g(\nabla _{{\dot{\gamma }}}A,{\dot{\gamma }}) -\omega ({\dot{\gamma }}) g(\nabla _{{\dot{\gamma }}}A,J{\dot{\gamma }}) -2\omega ({\dot{\gamma }})\kappa _2^2. \end{aligned}$$

(21)

Assuming that the order of $\gamma$ is $r\ge 3$ and $\kappa _2\not =0$, we obtain (see Erjavec and Inoguchi 2022):

$$\begin{aligned} \nabla _{{\dot{\gamma }}}A =\varepsilon (2{\dot{\kappa }}_{2}+\kappa _{2}\omega ({\dot{\gamma }}))E_3 +2\varepsilon \kappa _{2}\kappa _{3}E_4. \end{aligned}$$

Appendix B: Circles in the Poincaré Upper Half Plane

1.1 B.1 Signed Curvature

Let us consider arc-length-parametrized curve ${\underline{\gamma }}(s)=(x(s),y(s))$ in the upper half plane $\mathbb {H}$ equipped with Poincaré metric ${\bar{g}}=(dx^2+dy^2)/y^2$ and complex structure

$$\begin{aligned} J\frac{\partial }{\partial x}=-\frac{\partial }{\partial y}, \quad J\frac{\partial }{\partial y}=\frac{\partial }{\partial x}. \end{aligned}$$

Take a global orthonormal frame field

$$\begin{aligned} {\bar{e}}_1=y\frac{\partial }{\partial x}, \quad {\bar{e}}_2=y\frac{\partial }{\partial y}. \end{aligned}$$

Then, its dual coframe field is given by $\{{\bar{\vartheta }}^1=dx/y,{\bar{\vartheta }}^2=dy/y\}$. The connection form is computed as

$$\begin{aligned} {\bar{\omega }}_{2}^{\>\>1}=-\frac{1}{y}dx, \end{aligned}$$

and the Levi-Civita connection ${\overline{\nabla }}$ of $\mathbb {H}$ is given by

$$\begin{aligned} {\overline{\nabla }}_{{\bar{e}}_1}{\bar{e}}_1={\bar{e}}_2, \qquad {\overline{\nabla }}_{{\bar{e}}_1}{\bar{e}}_2=-{\bar{e}}_1. \end{aligned}$$

Since $\gamma (s)$ is arc-length-parameterized, we have ${\dot{x}}(s)^2+{\dot{y}}(s)^2=y(s)^2$. The unit tangent vector field ${\overline{T}}(s)$ and unit normal vector field ${\bar{N}}(s)=J{\overline{T}}(s)$ are given by

$$\begin{aligned}{} & {} {\overline{T}}(s)={\dot{x}}(s)\frac{\partial }{\partial x}+ {\dot{y}}(s)\frac{\partial }{\partial y}=X(s){\bar{e}}_1+Y(s){\bar{e}}_2, \\{} & {} {\overline{N}}(s)={\dot{y}}(s)\frac{\partial }{\partial x}- {\dot{x}}(s)\frac{\partial }{\partial y}=Y(s){\bar{e}}_1-X(s){\bar{e}}_2, \end{aligned}$$

where

$$\begin{aligned} X(s)=\frac{{\dot{x}}(s)}{y(s)}, \quad Y(s)=\frac{{\dot{y}}(s)}{y(s)}. \end{aligned}$$

Denote by ${\overline{\nabla }}$ the Levi-Civita connection of $\mathbb {H}$, we have

$$\begin{aligned} {\overline{\nabla }}_{ \dot{ {\underline{\gamma }}}}{\overline{T}}= \left( {\dot{X}}(s)-X(s)Y(s)\right) \,{\bar{e}}_1 + \left( {\dot{Y}}(s)+X(s)^2\right) \,{\bar{e}}_2 \end{aligned}$$

Thus, the signed geodesic curvature ${\bar{\kappa }}(s)$ is computed as

$$\begin{aligned} {\bar{\kappa }}(s)=-X(s){\dot{Y}}(s)+{\dot{X}}(s)Y(s)-X(s). \end{aligned}$$

(22)

Hence,

$$\begin{aligned} {\bar{\kappa }}(s)=-\frac{\>\>x^{\prime }(s)y^{\prime \prime }(s) -x^{\prime \prime }(s)y^{\prime }(x)\>\>}{y(s)^2} -\frac{\>x^{\prime }(s)\>}{y(s)}. \end{aligned}$$

Since $X(s)^2+Y(s)^2=1$, we may put

$$\begin{aligned} X(s)=\cos \phi (s), \quad Y(s)=\sin \phi (s) \end{aligned}$$

for some function $\phi (s)$. From (22), the angle function $\phi (s)$ satisfies

$$\begin{aligned} \frac{d\phi }{ds}(s)+\cos \phi (s)=-{\bar{\kappa }}(s). \end{aligned}$$

(23)

1.2 B.2 Magnetic Curves

The Kähler form ${\bar{\Omega }}={\bar{g}}(\cdot ,J)$ of $\mathbb {H}$ is a magnetic field on $\mathbb {H}$. The magnetic curve equation with respect to the magnetic field (called the Kähler magnetic field) $-{\bar{\Omega }}$ is:

$$\begin{aligned} {\overline{\nabla }}_{ \dot{ {\underline{\gamma }}}} \dot{{\underline{\gamma }}}=qJ\dot{{\underline{\gamma }}}. \end{aligned}$$

Since ${\overline{N}}(s))=J\dot{{\underline{\gamma }}}$, the magnetic curve equation is rewritten as:

$$\begin{aligned} {\overline{\nabla }}_{ \dot{ {\underline{\gamma }}}} {\overline{T}}=q{\overline{N}}. \end{aligned}$$

This shows that Kähler magnetic curves are nothing but Riemannian circles of signed curvature q.

Note that the system of Kähler magnetic curve equations is rewritten as:

$$\begin{aligned} {\dot{X}}(s)-X(s)Y(s)=qY(s), \quad {\dot{Y}}(s)+X(s)^2=-qX(s) \end{aligned}$$

(24)

Now let us determine Riemannian circles.

1.2.1 B.2.1 The Case $\phi$ is Constant

Assuming ${\bar{\kappa }}=q$ is a nonzero constant, from (23) we have $q=-\cos \phi$. Hence, we get $0<\vert q \vert <1$ and $\sin \phi =\pm \sqrt{1-q^2}$. Thus,

$$\begin{aligned} X=-q, \quad Y=\pm \sqrt{1-q^2}. \end{aligned}$$

In case $\vert q \vert =1$, we have $\sin \phi =0$. This implies that $y(s)>0$ is a constant. Namely ${\bar{\gamma }}$ is a horizontal line. The arc length parametrization is given by

$$\begin{aligned} x(s)=\pm y_{0}s+x_{0}, \quad y(s)=y_0>0. \end{aligned}$$

Thus, ${\underline{\gamma }}$ is a horizontal line in the upper half plane.

Next, if $0<\vert q \vert <1$, then we have $\sin \phi \not =0$. Moreover since

$$\begin{aligned} \frac{dx}{ds}=y\cos \phi , \quad \frac{dy}{ds}=y\sin \phi , \end{aligned}$$

we get

$$\begin{aligned} \frac{dy}{ds}=\pm y \sqrt{1-q^2},\quad \frac{dx}{dy}=\cot \phi =\frac{\mp q}{\sqrt{1-q^2}}. \end{aligned}$$

Thus, we have the arc length parametrization

$$\begin{aligned} x=\frac{\mp q y_0}{\sqrt{1-q^2}}\,\left( \exp (\pm \sqrt{1-q^2}\,s)-1 \right) +x_0, \quad y=y_0 \exp (\pm \sqrt{1-q^2}\,s),\quad y_0>0. \end{aligned}$$

where $x(0)=x_0$ and $y(0)=y_0$. The Kähler magnetic trajectory is an oblique half line $\sqrt{1-q^2}x=\pm q(y-y_0)$.

1.2.2 B.2.2 The Case $\phi$ is Non-constant

Assuming again ${\bar{\kappa }}=q=const\ne 0$, from (23) we have

$$\begin{aligned} \frac{d\phi }{ds}=-q-\cos \phi \end{aligned}$$

Combining this with

$$\begin{aligned} \frac{dx}{ds}=y\cos \phi , \quad \frac{dy}{ds}=y\sin \phi , \end{aligned}$$

we get

$$\begin{aligned} \frac{dy}{d\phi } =\frac{dy}{ds}\frac{ds}{d\phi } =-\frac{y\sin \phi }{q+\cos \phi }. \end{aligned}$$

This is rewritten as

$$\begin{aligned} \frac{dy}{y} =-\frac{\sin \phi }{q+\cos \phi }\,d\phi . \end{aligned}$$

Integrating this equation, we get

$$\begin{aligned} \ln y=\int \frac{-\sin \phi }{q+\cos \phi }\,d\phi =\ln \vert q+\cos \phi \vert +C. \end{aligned}$$

Thus, we have

$$\begin{aligned} y=\pm r(q+\cos \phi ),\quad r>0. \end{aligned}$$

Next, the x-coordinate is computed as

$$\begin{aligned} x= & {} \int \frac{dx}{d\phi }\,d\phi = \int \frac{dx}{ds}\frac{ds}{d\phi }\,d\phi =\int \frac{y\cos \phi }{-q-\cos \phi }d\phi = \\= & {} \int \frac{r(-q-\cos \phi )\cos \phi }{-q-\cos \phi }d\phi =r\sin \phi +a. \end{aligned}$$

Henceforth, the Riemannian circle ${\bar{\gamma }}(s)=(x(s),y(s))$ is parameterized as:

$$\begin{aligned} x=a+r\sin \phi ,\quad y=\pm r(q+\cos \phi ). \end{aligned}$$

The image of this curve is a part of Euclidean circle

$$\begin{aligned} (x-a)^2+(y\pm rq)^{2}=r^2. \end{aligned}$$

Since $y>0$, the whole image of this curve is contained in $\mathbb {H}$ when and only when $\vert q \vert >1$. In case $\vert q \vert =1$, the curve is a horocycle, that is, a circle tangent to the ideal boundary without the tangent point.