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J-Trajectories in 4-Dimensional Solvable Lie Group \(\mathrm {Sol}_0^4\)

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Abstract

J-trajectories are arc length parameterized curves in almost Hermitian manifold which satisfy the equation \(\nabla _{{\dot{\gamma }}}{\dot{\gamma }}=q J {\dot{\gamma }}\). In this paper J-trajectories in the solvable Lie group \(\mathrm {Sol}_0^4\) are investigated. The first and the second curvature of a non-geodesic J-trajectory in an arbitrary LCK manifold whose anti Lee field has constant length are examined. In particular, the curvatures of non-geodesic J-trajectories in \(\mathrm {Sol}_0^4\) are characterized.

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Acknowledgements

The second named author is partially supported by JSPS KAKENHI Grant Number 19K03461. The authors would like to thank the referee for her/his careful reading of the manuscript and valuable suggestions for improving this article.

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Correspondence to Zlatko Erjavec.

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Communicated by F W Nijhoff.

Appendix A Almost Hamiltonian Systems

Appendix A Almost Hamiltonian Systems

1.1 Almost Symplectic Manifolds

Let N be a manifold. A two-form \(\varPsi \) is said to be an almost symplectic form if it is non-degenerate. A manifold N equipped with an almost symplectic form is called an almost symplectic manifold. It should be remarked that almost symplectic manifolds are even-dimensional. In particular, almost symplectic form is called a symplectic form if it is closed. A manifold N together with a symplectic form \(\varPsi \) is called a symplectic manifold.

On an almost symplectic manifold \((N,\varPsi )\), one can define a vector field \(\xi _H\) uniquely for any prescribed smooth function H on N by

$$\begin{aligned} dH(Y)=\varPsi (\xi _{H},Y) \end{aligned}$$

for all vector fields Y on N. The vector field \(\xi _H\) is called the Hamiltonian vector field with Hamiltonian H.

The dynamical system \((N,\xi _H)\) determined by the Hamiltonian vector field \(\xi _H\) will be called an almost Hamiltonian system. Let us consider a local flow \(\{\psi _t\}\) of the Hamiltonian vector field (called a local Hamiltonian flow). Then one can see that the Hamiltonian H is preserved under local Hamiltonian flows, since \(dH(\xi _H)=(\iota _{\xi _H}\varPsi )(\xi _H)=\varPsi (\xi _H,\xi _H)=0\).

An almost Hamiltonian system is called a Hamiltonian system if its almost symplectic form is closed. On a Hamiltonian system \((N,\varPsi ,H)\), Hamiltonian vector field \(\xi _H\) satisfies

$$\begin{aligned} \pounds _{\xi _H}\varPsi =0. \end{aligned}$$

Here \(\pounds _{\xi _H}\) denotes the Lie differentiation by \(\xi _H\). Namely \(\xi _H\) is an infinitesimal symmetry of \((N,\varPsi ,H)\). In case \(d\varPsi \not =0\), this property does not hold, in general. By the Cartan’s formula

$$\begin{aligned} \pounds _{\xi _H}=\iota _{\xi _H}\circ d+d\circ \iota _{\xi _H}, \end{aligned}$$

where \(\iota _{\xi _H}\) denotes the interior product by \(\xi _H\), we deduce that \(\pounds _{\xi _H}\varPsi =0\) holds if and only if

$$\begin{aligned} \iota _{\xi _H}(d\varPsi )=0. \end{aligned}$$

If \(\xi _H\) satisfies this condition, then \(\xi _H\) is said to be a strongly Hamiltonian. Note that Vaisman [41] requires “strongly Hamiltonian” for \(\xi _H\) of almost Hamiltonian systems.

Remark 15

On an almost symplectic manifold \((N,\varPsi )\), a vector field X is said to be locally Hamiltonian if \(d(\iota _{X}\varPsi )=0\). A locally Hamiltonian vector field X admits a local potential H, i.e., locally defined smooth function H so that \(dH=\iota _{X}\varPsi \). Thus X is locally expressed as \(X=\xi _H\).

Lemma 1

([17]) Let Y be a locally Hamiltonian vector field and Z is an infinitesimal symmetry of \(\varPsi \) on an almost symplectic manifold \((N,\varPsi )\) then \([Y,Z]=-\xi _{\varPsi (Y,Z)}\).

The (almost) Poisson bracket \(\{f,h\}\) of functions f and h on N is defined by

$$\begin{aligned} \{f,h\}=\varPsi (\xi _f,\xi _h). \end{aligned}$$

The bracket operation \(\{\cdot ,\cdot \}\) satisfies the Jacobi identity if and only if \(d\varPsi =0\).

For any vector fields X and Y on an almost symplectic manifold \((N,\varPsi )\), by Cartan’s formula, we have

$$\begin{aligned} \iota _{[X,Y]}\varPsi&= \pounds _{X}(\iota _{Y}\varPsi ) -\iota _{Y}(\pounds _{X}\varPsi ) \\&= d(\iota _{X}\iota _{Y}\varPsi ) +\iota _{X}d(\iota _{Y}\varPsi ) -\iota _{Y}(d(\iota _{X}\varPsi )) -\iota _{Y}(\iota _{X}(d\varPsi )). \end{aligned}$$

In particular, if X and Y are locally Hamiltonian, we have

$$\begin{aligned} \iota _{[X,Y]}\varPsi =-d(\varPsi (X,Y)) -\iota _{Y}(\iota _{X}(d\varPsi )). \end{aligned}$$

As a result, for strongly Hamiltonian vector fields \(\xi _f\) and \(\xi _h\),

$$\begin{aligned} \iota _{[\xi _f,\xi _h]}\varPsi =-d(\varPsi (\xi _f,\xi _h)) =-d(\{f,h\}). \end{aligned}$$

Namely we have

$$\begin{aligned} \xi _{\{f,h\}}=-[\xi _f,\xi _h]. \end{aligned}$$

1.2 The Canonical Two-Form

Let M be an m-manifold. Denote by \(T^{*}M\) its cotangent bundle. Take a local coordinate system \((x^1,x^2,\ldots ,x^m)\) of M, then it induces a fiber coordinates \((p_1,p_2,\ldots ,p_m)\). Then \(\vartheta =p_{j}\,dx^j\) is globally defined one-form on \(T^{*}M\) and called the canonical one-form (also called the Liouville form). The two-form \(\varPsi =-d\vartheta =dx^{j}\wedge dp_{j}\) is a symplectic form on \(T^{*}M\). The symplectic form \(\varPsi \) is called the canonical two-form on \(T^{*}M\). For a prescribed Hamiltonian H, the Hamiltonian vector field \(\xi _H\) is locally expressed as

$$\begin{aligned} \xi _{H}=\sum _{i=1}^{m}\frac{\partial H}{\partial p_i}\frac{\partial }{\partial x^i} -\sum _{i=1}^{m}\frac{\partial H}{\partial x^i}\frac{\partial }{\partial p_i}. \end{aligned}$$

Take a curve \({\bar{\gamma }}(t)\) in \(T^{*}M\). Then \({\bar{\gamma }}(t)=(x^i(t);p_i(t))\) is an integral curve of \(\xi _H\) if and only if it satisfies the Hamilton equation:

$$\begin{aligned} \frac{dx^i}{dt}=\frac{\partial H}{\partial p_i}, \quad \frac{dp_i}{dt}=-\frac{\partial H}{\partial x^i}, \quad 1\le i\le m. \end{aligned}$$

1.3 Geodesic Flows

Let us consider a Riemannian m-manifold (Mg), then its cotangent bundle \(T^{*}M\) is identified with the tangent bundle TM via the metric g. The so-called musical isomorphism

$$\begin{aligned} \flat :TM\rightarrow T^{*}M;\quad \flat v=g_{p}(v,\cdot ),\quad v\in T_{p}M \end{aligned}$$

is a vector bundle isomorphism. We denote by \(\pi \) the natural projection of TM onto M. Take a local coordinate system \((x^1,x^2,\dots ,x^m)\) with fiber coordinates \((u^1,u^2,\dots ,u^m)\). Then the pull-backed one-form \(\flat ^{*}\vartheta \) is expressed as \(\flat ^{*}\vartheta =g_{ij}u^{j}dx^{i}\). Then the pull-backed two-form \(\varPhi :=\flat ^{*}\varPsi \) is computed as

$$\begin{aligned} \flat ^{*}\varPsi = g_{ij}dx^{i}\wedge du^{j} +\frac{\partial g_{ji}}{\partial x^k}\,u^{j}\, dx^{i}\wedge dx^{k}. \end{aligned}$$

The pull-backed two-form \(\varPhi \) is a symplectic form on TM. Let us consider the kinetic energy on TM:

$$\begin{aligned} E(p;v)=\frac{1}{2}g_{p}(v,v),\quad v\in T_{p}M. \end{aligned}$$

Then we get a Hamiltonian system \((TM,\varPhi , E)\). The Hamiltonian vector field on TM with Hamiltonian E is called the geodesic spray and has local expression:

$$\begin{aligned} \xi _{E}=u^{i}\frac{\partial }{\partial x^i} -\varGamma _{ij}^{\>k}u^iu^j\frac{\partial }{\partial u^k}. \end{aligned}$$

The integral curves of \(\xi _E\) are solutions to the system

$$\begin{aligned} \frac{dx^k}{dt}=u^{k},\quad \frac{du^k}{dt}=-\varGamma _{ij}^{\>k}u^iu^j. \end{aligned}$$

One can see that for every trajectory \({\bar{\gamma }}(t)\) of the Hamiltonian system \((TM,\varPhi ,E)\), the projected curve \(\gamma (t)=\pi (\gamma (t))\) in M is a geodesic.

1.4 Magnetic Trajectories

Let us consider a two-form F on a Riemannian manifold (Mg). Express F as

$$\begin{aligned} F=\sum _{i<j}F_{ij}dx^{i}\wedge dx^{j}= \frac{1}{2}\sum _{i,j=1}^{m}F_{ij}dx^{i}\wedge dx^{j}. \end{aligned}$$

Then we introduce an endomorphism field \(\phi \) by

$$\begin{aligned} g(\phi X,Y)=F(X,Y) \end{aligned}$$

and represent it locally as

$$\begin{aligned} \phi \frac{\partial }{\partial x^i}=\phi _{i}^{\>k}\frac{\partial }{\partial x^k}, \end{aligned}$$

then

$$\begin{aligned} F_{ij}=F\left( \frac{\partial }{\partial x^i}, \frac{\partial }{\partial x^j} \right) =g\left( \phi \frac{\partial }{\partial x^i}, \frac{\partial }{\partial x^j} \right) =\phi _{i}^{\>k}g_{kj}. \end{aligned}$$

Equivalently we have

$$\begin{aligned} \phi _{i}^{\>l}=-g^{lj}F_{ji}. \end{aligned}$$

We deform the symplectic form \(\varPhi \) of TM as

$$\begin{aligned} \varPhi _F=\varPhi +q\pi ^{*}F \end{aligned}$$

for some constant q. Then \(\varPhi _F\) has local expression

$$\begin{aligned} \varPhi _{F}=g_{ij}dx^{i}\wedge du^{j} +\frac{\partial g_{ji}}{\partial x^k}u^{j}\, dx^{i}\wedge dx^{k} +q\,\sum _{i<j}F_{ij}dx^{i}\wedge dx^{j}. \end{aligned}$$

The deformed two-form \(\varPhi _F\) is still non-degenerate. When F is a magnetic field, then \(\varPhi _F\) is a symplectic form on TM (and hence \(\varPhi \) itself is a magnetic field).

Let us consider an almost Hamiltonian system \((TM,\varPhi _F,E)\). The Hamiltonian vector field is given by

$$\begin{aligned} \xi _E^F=\xi _{E}+q\mathrm {v}\{\phi (u)\}, \end{aligned}$$

where \(\mathrm {v}\{\phi (u)\}\) is a vertical vector field on TM globally defined by (cf. [35])

$$\begin{aligned} \mathrm {v}\{\phi (u)\}=\phi _{i}^{\>k}u^i\frac{\partial }{\partial u^k}. \end{aligned}$$

The Hamilton equation is

$$\begin{aligned} \frac{dx^k}{dt}=u^{k},\quad \frac{du^k}{dt}=-\varGamma _{ij}^{\>k}u^iu^j+q\phi _{i}^{\>k}u^i. \end{aligned}$$

This is a second order ODE

$$\begin{aligned} \frac{d^2x^k}{dt^2} +\varGamma ^{\>k}_{ij} \frac{dx^i}{dt} \frac{dx^j}{dt} =q\,\phi _{i}^{k}\frac{dx^i}{dt}. \end{aligned}$$

This system has coordinate-free expression

$$\begin{aligned} \nabla _{{\dot{\gamma }}}{\dot{\gamma }}=q\phi {\dot{\gamma }}. \end{aligned}$$

Proposition 8

Let (Mg) be a Riemannian manifold with a magnetic field F. Then the magnetic trajectory equation is the Hamilton equation of the Hamiltonian system \((TM,\varPhi _F,E)\).

Example 5

Let (MJg) be an almost Hermitian manifold with fundamental two-form \(\Omega \). Then the J-trajectory equation

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

is the Hamilton equation of the almost Hamiltonian system \((TM,\varPhi _{-\Omega },E)\).

Now let (MJg) be a LCK manifold. We choose \(F=-\varOmega \). Then the perturbed two-form \(\varPhi ^J:=\varPhi _{-\varOmega }\) is given by

$$\begin{aligned} \varPhi ^{J}=\varPhi -q\pi ^{*}\varOmega \end{aligned}$$

and hence

$$\begin{aligned} d\varPhi ^{J}=-q\,d(\pi ^{*}\varOmega ) =-q\pi ^{*}(d\varOmega )=-q\pi ^{*}(\omega \wedge \varOmega ). \end{aligned}$$

Thus the Hamiltonian vector field

$$\begin{aligned} \xi ^J_E:=\xi _E^{-\Omega }=\xi _{E}+qJ_{i}^{\>k}u^{i} \frac{\partial }{\partial u^k} \end{aligned}$$

is strongly Hamiltonian if and only if

$$\begin{aligned} (\omega \wedge \Omega )(\pi _{*}\xi ^{J},\pi _{*}{\bar{X}},\pi _{*}{\bar{Y}})=0 \end{aligned}$$

for all vector fields \({\bar{X}}\) and \({\bar{Y}}\) on TM. Here we note that any point \((p;v)\in TM\), we have

$$\begin{aligned} \pi _{*v}(\xi ^J_E)_{v}=v+qJ_{p}v. \end{aligned}$$

1.5 The Almost Hamiltonian Systems Derived from J-Trajectories on \(\mathrm {Sol}^4_0\)

Set

$$\begin{aligned} x^1=x,\quad x^2=y,\quad x^3=z,\quad x^4=t \end{aligned}$$

on \(\mathrm {Sol}_0^4\), then the fiber coordinates are given by

$$\begin{aligned} u^1=dx,\quad u^2=dy,\quad u^3=dz,\quad u^4=dt. \end{aligned}$$

The Christoffel symbols of \(\mathrm {Sol}_0^4\) are

$$\begin{aligned} \varGamma ^{\>4}_{11}=e^{-2t},\quad \varGamma ^{\>1}_{14}=-1,\quad \varGamma ^{\>2}_{24}=-1,\quad \varGamma ^{\>4}_{22}=e^{-2t},\quad \varGamma ^{\>3}_{34}=2,\quad \varGamma ^{\>4}_{33}=-2e^{4t}. \end{aligned}$$

Thus the geodesic spray is given by

$$\begin{aligned} \xi _{E}&= {\bar{x}}\frac{\partial }{\partial x} +{\bar{y}}\frac{\partial }{\partial y} +{\bar{z}}\frac{\partial }{\partial z} + {\bar{t}}\frac{\partial }{\partial t} \\&\quad +2{\bar{x}}{\bar{t}}\frac{\partial }{\partial {\bar{x}}} +2{\bar{y}}{\bar{t}}\frac{\partial }{\partial {\bar{y}}} -4{\bar{z}}{\bar{t}}\frac{\partial }{\partial {\bar{z}}} -\left( e^{-2t}({\bar{x}}^2+{\bar{y}}^2)-2e^{4t}{\bar{z}}^2 \right) \frac{\partial }{\partial {\bar{t}}}, \end{aligned}$$

where

$$\begin{aligned} {\bar{x}}=dx,\quad {\bar{y}}=dy,\quad {\bar{z}}=dz,\quad {\bar{t}}=dt. \end{aligned}$$

The vertical vector field \(\mathrm {v}\{J(u)\}\) is given by

$$\begin{aligned} \mathrm {v}\{J(u)\}=J_{i}^{\>k}u^{i}\frac{\partial }{\partial u^k} ={\bar{y}}\frac{\partial }{\partial {\bar{x}}} -{\bar{x}}\frac{\partial }{\partial {\bar{y}}} -e^{-2t}{\bar{t}}\frac{\partial }{\partial {\bar{z}}} +e^{2t}{\bar{z}}\frac{\partial }{\partial {\bar{t}}}. \end{aligned}$$

Thus we obtain

$$\begin{aligned} \xi _{E}^{J}&= {\bar{x}}\frac{\partial }{\partial x} +{\bar{y}}\frac{\partial }{\partial y} +{\bar{z}}\frac{\partial }{\partial z} + {\bar{t}}\frac{\partial }{\partial t} \\&\quad +(2{\bar{x}}{\bar{t}}+q{\bar{y}})\frac{\partial }{\partial {\bar{x}}} +(2{\bar{y}}{\bar{t}}-q{\bar{x}})\frac{\partial }{\partial {\bar{y}}} -(4{\bar{z}}+qe^{-2t}){\bar{t}}\frac{\partial }{\partial {\bar{z}}} \\&\quad -\left( e^{-2t}({\bar{x}}^2+{\bar{y}})^2-2e^{4t}{\bar{z}}^2 -qe^{2t}{\bar{z}}\right) \frac{\partial }{\partial {\bar{t}}}. \end{aligned}$$

The symplectic form \(\varPhi \) of \(T\mathrm {Sol}_0^4\) is computed as

$$\begin{aligned} \varPhi&=g_{ij}dx^{i}\wedge du^{j} +\frac{\partial g_{ji}}{\partial x^k}\,u^{j}\, dx^{i}\wedge dx^{k}\\&=e^{-2x^4}(dx^1\wedge du^1+dx^2\wedge du^2) +e^{4x^4}dx^3\wedge du^3+dx^4\wedge du^4 \\&\quad -2u^1e^{-2x^4}dx^1\wedge dx^4 -2u^2e^{-2x^4}dx^2\wedge dx^4 +4u^3e^{4x^4}dx^3\wedge dx^4. \end{aligned}$$

Next, we have

$$\begin{aligned} \pi ^{*}\varOmega =e^{-2x^4}dx^1\wedge dx^2 -e^{2x^4}dx^3\wedge dx^4, \quad \pi ^{*}\omega =-2dx^4. \end{aligned}$$

Hence we have

$$\begin{aligned} \varPhi ^J&=e^{-2x^4}(dx^1\wedge du^1+dx^2\wedge du^2) +e^{4x^4}dx^3\wedge du^3+dx^4\wedge du^4 \\&\quad -2u^1e^{-2x^4}dx^1\wedge dx^4 -2u^2e^{-2x^4}dx^2\wedge dx^4 +4u^3e^{4x^4}dx^3\wedge dx^4 \\&\quad -q(e^{-2x^4}dx^1\wedge dx^2 -e^{2x^4}dx^3\wedge dx^4). \end{aligned}$$

In the original notation we have

$$\begin{aligned} \varPhi ^J&= e^{-2t}(dx\wedge d{\bar{x}}+dy\wedge d{\bar{y}}) +e^{4t}dz\wedge d{\bar{z}}+dt\wedge d{\bar{t}} \\&\quad -2e^{-2t}({\bar{x}}dx\wedge dt+{\bar{y}}dy\wedge dt) +4{\bar{z}}e^{4t}dz\wedge dt \\&\quad -q(e^{-2t}dx\wedge dy -e^{2t}dz\wedge dt). \end{aligned}$$

The Hamiltonian vector field \(\xi _f^J\) is given by the following Proposition.

Proposition 9

The Hamiltonian vector field \(\xi ^J_f\) corresponding to a function f is given by

$$\begin{aligned} \xi ^J_f&=e^{2t}f_{{{\bar{x}}}}\frac{\partial }{\partial x} +e^{2t}f_{{{\bar{y}}}}\frac{\partial }{\partial y} +e^{-4t}f_{{{\bar{z}}}}\frac{\partial }{\partial z} +f_{{{\bar{t}}}}\frac{\partial }{\partial t} \\&\quad -e^{2t} (f_{x}-2{\bar{x}}e^{-2t}f_{{{\bar{t}}}}-qf_{{{\bar{y}}}}) \frac{\partial }{\partial {\bar{x}}} -e^{2t} (f_{y}-2{\bar{y}}e^{-2t}f_{{{\bar{t}}}}+qf_{{{\bar{x}}}}) \frac{\partial }{\partial {\bar{y}}} \\&\quad -e^{-4t} (f_{z}+(4{\bar{z}}e^{4t}+qe^{2t})f_{{{\bar{t}}}}) \frac{\partial }{\partial {\bar{z}}} \\&\quad -(f_{t} +2({\bar{x}}f_{{{\bar{x}}}}+{\bar{y}}f_{{{\bar{y}}}}) -(4z+qe^{-2t}) f_{{{\bar{z}}}}) \frac{\partial }{\partial {\bar{t}}}. \end{aligned}$$

Example 6

The dual one-forms of the right invariant vector fields \(e_1^R\), \(e_2^R\), \(e_3^R\), \(e_4^R\) are given by:

$$\begin{aligned} \vartheta ^1_R=dx-xdt, \quad \vartheta ^2_R=dy-ydt, \quad \vartheta ^3_R=dz+2zdt, \quad \vartheta ^4_R=dt. \end{aligned}$$

These one-forms are regarded as right invariant functions

$$\begin{aligned} f_1={\bar{x}}-x{\bar{t}}, \quad f_2={\bar{y}}-y{\bar{t}}, \quad f_3={\bar{z}}+2z{\bar{t}}, \quad f_4={\bar{t}} \end{aligned}$$

on \(T\mathrm {Sol}_0^4\). The Hamiltonian vector fields \(\xi _1:=\xi ^J_{f_1}\), \(\xi _2:=\xi ^J_{f_2}\), \(\xi _3:=\xi ^J_{f_3}\), \(\xi _4:=\xi ^J_{f_4}\) are given by

$$\begin{aligned} \xi _1&=e^{2t}\frac{\partial }{\partial x} -x\frac{\partial }{\partial t} -(2x{\bar{x}}-{\bar{t}}e^{2t})\frac{\partial }{\partial {\bar{x}}}\\&\quad -(2x{\bar{y}}+qe^{2t})\frac{\partial }{\partial {\bar{y}}} +x(4{\bar{z}}+qe^{-2t})\frac{\partial }{\partial {\bar{z}}} -2{\bar{x}}\frac{\partial }{\partial {\bar{t}}}, \\ \xi _2&=e^{2t}\frac{\partial }{\partial y} -y\frac{\partial }{\partial t} -(2y{\bar{x}}-qe^{2t})\frac{\partial }{\partial {\bar{x}}}\\&\quad -(2y{\bar{y}}-{\bar{t}}e^{2t})\frac{\partial }{\partial {\bar{y}}} +y(4{\bar{z}}+qe^{-2t})\frac{\partial }{\partial {\bar{z}}} -2{\bar{y}}\frac{\partial }{\partial {\bar{t}}}, \\ \xi _3&= e^{-4t}\frac{\partial }{\partial z} +2z\frac{\partial }{\partial t} +4z{\bar{x}}\frac{\partial }{\partial {\bar{x}}} +4z{\bar{y}}\frac{\partial }{\partial {\bar{y}}}\\&\quad -(8z{\bar{z}}+2{\bar{t}}e^{-4t}+2qze^{-2t})\frac{\partial }{\partial {\bar{z}}} -(4z+qe^{-2t})\frac{\partial }{\partial {\bar{t}}}, \\ \xi _4&=\frac{\partial }{\partial t} +2{\bar{x}}\frac{\partial }{\partial {\bar{x}}} +2{\bar{y}}\frac{\partial }{\partial {\bar{y}}} -(4{\bar{z}}+qe^{-2t})\frac{\partial }{\partial {\bar{z}}}. \end{aligned}$$

Now let us investigate the strong Hamiltonian property of \((T\mathrm {Sol}_0^4,\varPhi ^J,E)\). Since \(\varPhi \) is closed, we have

$$\begin{aligned} d\varPhi ^{J}= 2qe^{-2t}dt \wedge dx\wedge dy. \end{aligned}$$

One can see that

$$\begin{aligned} \iota _{\xi ^J_E}(d\varPhi ^j)=2qe^{-2t}({\bar{t}}dx\wedge dy-{\bar{y}}dx\wedge dt -{\bar{x}}dy\wedge dt)\not =0. \end{aligned}$$

This shows that \(\xi ^J_E\) is not strongly Hamiltonian.

For any smooth functions f and h on \(T\mathrm {Sol}_0^4\), we introduce the (almost) Poisson bracket by

$$\begin{aligned} \{f,h\}^{J}:=\varPhi ^{J}(\xi _f,\xi _h) . \end{aligned}$$

Since \(d\varPhi ^J\not =0\), this bracket does not satisfy Jacobi identity.

1.6 Tangent Group

Let G be a Lie group, then its tangent bundle is a Lie group with multiplication:

$$\begin{aligned} (a;A_a)(b;B_b)=(ab;L_{a*b}B_{b}+R_{b*a}A_a). \end{aligned}$$

The resulting Lie group TG is called the tangent group of G. The tangent group is identified with the semi-direct product \(G\ltimes {\mathfrak {g}}\) via the left Maurer–Cartan form as:

$$\begin{aligned} (a,A_a)\longmapsto (a,L_{a*}^{-1}A_a)\in G\times {\mathfrak {g}}. \end{aligned}$$

Note that under this identification the Lie algebra \({\mathfrak {g}}=T_{e}G\) is regarded as the Lie algebra of all left invariant smooth vector fields of G. The multiplication law is transformed as

$$\begin{aligned} (a,A)(b,B)=(ab,\mathrm {Ad}(b^{-1})A+B). \end{aligned}$$

Remark 16

If we identify TG with \(G\times {\mathfrak {g}}\) via the right Maurer–Cartan form, the multiplication law becomes

$$\begin{aligned} (a,A)(b,B)=(ab,A+\mathrm {Ad}(a)B). \end{aligned}$$

The Lie algebra \({\mathfrak {g}}\) is regarded as the Lie algebra of right invariant smooth vector fields of G.

In the case of \(\mathrm {Sol}_0^4\), the tangent group \(T\mathrm {Sol}_0^4=\mathrm {Sol}_0^4\ltimes \mathfrak {sol}_0^4\) is \({\mathbb {R}}^8\) with multiplication:

$$\begin{aligned} \left( \begin{array}{c} x_1\\ y_1\\ z_1\\ t_1\\ X_1\\ Y_1\\ Z_1\\ T_1 \end{array} \right) \cdot \left( \begin{array}{c} x_2\\ y_2\\ z_2\\ t_2\\ X_2\\ Y_2\\ Z_2\\ T_2 \end{array} \right) =\left( \begin{array}{c} x_1+e^{t_1}x_2 \\ y_1+e^{t_1}y_2 \\ z_1+e^{-2t_1}z_2 \\ t_1+t_2 \\ e^{-t_2}(X_1+x_2T_1)+X_2 \\ e^{-t_2}(Y_1+y_2T_1)+Y_2 \\ e^{2t_2}(Z_1-2z_2T_1)+Z_2 \\ T_1+T_2 \end{array} \right) . \end{aligned}$$

The abelian subgroup \(G(1,2,3)\subset T\mathrm {Sol}_0^4\) given in Example 3 acts on \(T\mathrm {Sol}_0^4\) by right as

$$\begin{aligned} \left( \begin{array}{c} x\\ y\\ z\\ t\\ X\\ Y\\ Z\\ T \end{array} \right) \cdot \left( \begin{array}{c} a\\ b\\ c \end{array} \right) = \left( \begin{array}{c} x+e^{t}a \\ y+e^{t}b \\ z+e^{-2t}c \\ t \\ X+aT \\ Y+bT \\ Z-2cT\\ T \end{array} \right) . \end{aligned}$$

Note that the left action is just a translation

$$\begin{aligned} \left( \begin{array}{c} a\\ b\\ c \end{array} \right) \cdot \left( \begin{array}{c} x\\ y\\ z\\ t\\ X\\ Y\\ Z\\ T \end{array} \right) = \left( \begin{array}{c} x+a \\ y+b \\ z+c \\ t \\ X \\ Y \\ Z\\ T \end{array} \right) . \end{aligned}$$

Let us consider an arc length parametrized J-trajectory \(\gamma (s)=(x(s),y(s),z(s),t(s))\) in \(\mathrm {Sol}_0^4\). Then its velocity vector field \({\dot{\gamma }}(s)\) is a smooth curve in \(T\mathrm {Sol}_0^4\) with parametrization

$$\begin{aligned} (x(s),y(s),z(s),t(s),e^{-t(s)}{\dot{x}}(s), e^{-t(s)}{\dot{y}}(s),e^{2t(s)}{\dot{z}}(s), {\dot{t}}(s) ). \end{aligned}$$

Let us represent \(\gamma (s)^{-1}\) as \(\gamma (s)^{-1}=({\underline{x}}(s),{\underline{y}}(s), {\underline{z}}(s),{\underline{t}}(s))\). We define a curve \({\underline{\gamma }}(s)\) of G(1, 2, 3) by

$$\begin{aligned} {\underline{\gamma }}(s)=({\underline{x}}(s),{\underline{y}}(s), {\underline{z}}(s) ). \end{aligned}$$

Then we get

$$\begin{aligned} {\dot{\gamma }}(s)\cdot {\underline{\gamma }}(s)= \left( \begin{array}{c} 0 \\ 0 \\ 0 \\ t \\ e^{-t}({\dot{x}}-x{\dot{t}}) \\ e^{-t}({\dot{y}}-y{\dot{t}}) \\ e^{2t}({\dot{z}}+2z{\dot{t}}) \\ {\dot{t}} \end{array} \right) . \end{aligned}$$

We know that

$$\begin{aligned} {\dot{x}}-x{\dot{t}}, \quad {\dot{y}}-y{\dot{t}}, \quad {\dot{z}}+2z{\dot{t}} \end{aligned}$$

are right invariant functions on \(\mathrm {Sol}_0^4\). This result implies that to solve the ODE system of J-trajectories, one needs to obtain t-coordinate t(s) first.

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Erjavec, Z., Inoguchi, Ji. J-Trajectories in 4-Dimensional Solvable Lie Group \(\mathrm {Sol}_0^4\). Math Phys Anal Geom 25, 8 (2022). https://doi.org/10.1007/s11040-022-09418-5

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