Abnormal Extremals of Left-Invariant Sub-Finsler Quasimetrics on Four-Dimensional Lie Groups

Abstract

We find the abnormal extremals on four-dimensional connected Lie groups with left-invariant sub-Finsler quasimetric defined by a seminorm on a two-dimensional subspace of the Lie algebra generating the algebra. In terms of the structure constants of a Lie algebra and the Minkowski support function of the unit ball of the seminorm on the two-dimensional subspace of a Lie algebra which defines a quasimetric, we establish a criterion for the strict abnormality of these extremals.

1. Introduction

The first author indicated in [1] that the shortest arcs of a left-invariant (sub-)Finsler metric $d$ on each connected Lie group $G$ defined by a left-invariant bracket generating distribution $D$ and a norm $F$ on $D(e)={\mathfrak{p}}\subset TG_{e}={\mathfrak{g}}$ are solutions of a left-invariant time-optimal problem with the closed unit ball $U$ with center zero of the normed vector space $(D(e),F)$ as the control domain. The distribution $D$ is bracket generating if and only if the subspace ${\mathfrak{p}}$ generates the Lie algebra $({\mathfrak{g}},[\cdot,\cdot])$ by the Lie bracket $[\cdot,\cdot]$. Moreover, the statements about shortest arcs are also true for the pair $(D(e),F)$ with a seminorm $F$ satisfying $F(u)>0$ for $0\neq u\in D(e)$ and defining a left-invariant sub-Finsler quasimetric $d$ on $G$.

The Pontryagin Maximum Principle (PMP) [2] gives some necessary conditions for solutions to the time-optimal control problem. An extremal is a curve in $G$ which is parametrized by the arclength and satisfies the PMP.

An extremal can be normal or abnormal. Some extremals can be both normal and abnormal with respect to different controls (control functions); these extremals are called nonstrictly abnormal. An abnormal but not nonstrictly abnormal extremal is called strictly abnormal.

In this paper, we solve the search problem for abnormal extremals on four-dimensional connected Lie groups with a left-invariant sub-Finsler quasimetric, establish a criterion for the nonstrict abnormality of these extremals which allows us to formulate the criterion of their strict abnormality. Since the quasimetric is left-invariant, we can assume that extremals issue from the unity of the group. Each of these abnormal extremals is some one-parametric subgroup of the Lie group.

A four-dimensional Lie group $G$ has abnormal extremals for all seminorms of the above-indicated form $F$ on ${\mathfrak{p}}\subset{\mathfrak{g}}$ if $\dim({\mathfrak{p}})=2$ and ${\mathfrak{p}}$ generates ${\mathfrak{g}}$. In this paper we mainly consider these Lie groups $G$ and ${\mathfrak{p}}\subset{\mathfrak{g}}$.

We were inspired to study these problems by the memoir of Liu and Sussmann [3] on strictly abnormal extremals of the sub-Riemannian metrics defined by the bracket generating distributions of rank 2. In particular, they proved in [3] that every abnormal sub-Riemannian extremal regular in their sense is locally optimal; i.e. it is locally a shortest arc. Also, they presented an example of left-invariant metric on the Lie group $SO(3,{𝕉})\times{𝕉}$ with a strictly abnormal extremal. Notice that all abnormal extremals of left-invariant sub-Finsler quasimetrics on four-dimensional connected Lie groups are regular in this sense.

Our solution of the problems under consideration splits naturally into the three steps: At first, we need to select all four-dimensional real Lie algebras with two-dimensional subspaces generating them by Lie brackets. Then, given such a Lie algebra, we have to find all pairwise nonequivalent classes of the two-dimensional generating subspaces of this Lie algebra which are equivalent, i.e. transformable into each other by some automorphism of the Lie algebra. Finally, given a seminorm on an arbitrary representative of such an equivalence class, we must find the abnormal extremals of the corresponding left-invariant sub-Finsler quasimetric on the Lie group with a given Lie algebra and detect strictly abnormal extremals among them, if the last exists.

To implement these three steps we need some classification (up to isomorphism) of all four-dimensional real Lie algebras. This classification was obtained by Mubarakzyanov in [4]. We will use in this paper the modified version of Mubarakzyanov’s classification from the paper [5] by Biggs and Remsing. In the paper, they indicate all automorphisms of Lie algebras under consideration and classify all connected Lie groups with these Lie algebras.

Agrachev and Barilari obtained in [6] the isometric classification of left-invariant sub-Riemannian metrics on three-dimensional Lie groups. They proved simultaneously that, up to equivalence, all three-dimensional real Lie algebras but ${\mathfrak{sl}}(2,{𝕉})$ include at most one two-dimensional generating subspace, and ${\mathfrak{sl}}(2,{𝕉})$ has two such subspaces. We prove that the same result is true for the one-dimensional extensions ${\mathfrak{g}}_{3}\oplus{𝕉}$ of these algebras for ${\mathfrak{g}}_{3}\neq{\mathfrak{sl}}(2,{𝕉})$. Moreover, we prove that, up to equivalence, ${\mathfrak{sl}}(2,{𝕉})\oplus{𝕉}$ possesses four two-dimensional generating subspaces, and each indecomposable Lie algebra ${\mathfrak{g}}_{4}$ includes at most one two-dimensional generating subspace. All other (main) results of the paper are obtained by using the results of [1, 2, 5, 7, 8].

2. Preliminaries

Recall the Engel Theorem and the Cartan Theorem which characterize nilpotent and solvable Lie algebras; see [9, 10].

A Lie algebra $({\mathfrak{g}},[\cdot,\cdot])$ is nilpotent if and only if every linear operator $\operatorname{ad}x:=[x,\cdot]:{\mathfrak{g}}\rightarrow{\mathfrak{g}}$, $x\in{\mathfrak{g}}$, is nilpotent; i.e., $(\operatorname{ad}x)^{k}=0$ for some natural $k$. A Lie algebra $({\mathfrak{g}},[\cdot,\cdot])$ is solvable if and only if its derived ideal ${\mathfrak{g}}^{\prime}=[{\mathfrak{g}},{\mathfrak{g}}]$ is nilpotent.

A Lie algebra ${\mathfrak{g}}$ is semisimple (respectively, simple) if ${\mathfrak{g}}$ has no nonzero solvable (respectively, improper) ideals (and ${\mathfrak{g}}^{\prime}\neq 0$). Every simple Lie algebra is semisimple.

The Lie algebras of interest to us are of the two types: ${\mathfrak{g}}_{3}\oplus{𝕉}$ and ${\mathfrak{g}}_{4}$, where ${\mathfrak{g}}_{3}$ and ${\mathfrak{g}}_{4}$ are indecomposable three-dimensional and four-dimensional Lie algebras. Moreover, all four-dimensional Lie algebras are solvable, with the exception of the two algebras: ${\mathfrak{g}}_{3,7}\oplus{𝕉}={\mathfrak{so}}(3,{𝕉})\oplus{𝕉}={\mathfrak{s}u}(2)\oplus{𝕉}$ and ${\mathfrak{g}}_{3,6}\oplus{𝕉}={\mathfrak{sl}}(2,{𝕉})\oplus{𝕉}$. The simply connected Lie groups with the latter Lie algebras are the multiplicative group of nonzero quaternions and the universal covering of the group $\operatorname{GL}_{e}(2,{𝕉})$ respectively.

Table 1 classifies the real Lie algebras of type ${\mathfrak{g}}_{3}\oplus{\mathfrak{g}}_{1}$, where ${\mathfrak{g}}_{1}={𝕉}$, and ${\mathfrak{g}}_{4}$ from [5]. In this table the authors of [5] chose some basis $E_{1}$, $E_{2}$, $E_{3}$, $E_{4}$ for these algebras and indicate only the nonzero brackets of the basis vectors. Let us give some comments about the Lie algebras from this table.

The nilpotent algebras are only ${\mathfrak{g}}_{3,1}\oplus{\mathfrak{g}}_{1}={\mathfrak{h}}\oplus{\mathfrak{g}}_{1}$ and ${\mathfrak{g}}_{4,1}={\mathfrak{eng}}$, where ${\mathfrak{h}}={\mathfrak{nil}}$ and ${\mathfrak{eng}}$ are the Lie algebras of the Heisenberg group and the Engel group. Moreover, the Heisenberg and Engel groups are the so-called Carnot groups. Maximal nilpotent ideals has dimension 3 for the Lie algebra ${\mathfrak{g}}_{3,1}$ and dimension 2 for all other solvable algebras of type ${\mathfrak{g}}_{3}$. In the latter case ${\mathfrak{g}}_{3,1}$ is commutative and equal to ${\mathfrak{g}}_{3}^{\prime}=[{\mathfrak{g}}_{3},{\mathfrak{g}}_{3}]$. Although the Heisenberg algebra has a two-dimensional generating subspace, this is wrong for ${\mathfrak{g}}_{3,1}\oplus{\mathfrak{g}}_{1}$. The Lie algebra ${\mathfrak{g}}_{3,3}$ and, consequently, ${\mathfrak{g}}_{3,3}\oplus{\mathfrak{g}}_{1}$ have no two-dimensional generating subspaces, in contrast to all other Lie algebras of type ${\mathfrak{g}}_{3}\oplus{\mathfrak{g}}_{1}$.

The maximal nilpotent ideals of Lie algebras of type ${\mathfrak{g}}_{4}\neq{\mathfrak{g}}_{4,1}$ have dimension 3 for the Lie algebras ${\mathfrak{g}}_{4,2}$–${\mathfrak{g}}_{4,9}$, and dimension 2 for the Lie algebra ${\mathfrak{g}}_{4,10}$; moreover, the maximal nilpotent ideals of Lie algebras ${\mathfrak{g}}_{4,2}$–${\mathfrak{g}}_{4,6}$ are commutative, while the maximal nilpotent ideals of the Lie algebras ${\mathfrak{g}}_{4,7}$–${\mathfrak{g}}_{4,9}$ are isomorphic to the Heisenberg algebra [4]. All Lie algebras ${\mathfrak{g}}_{4}$, with possible exception of some parameters in case of their parametric families, have two-dimensional generating subspaces.

The descriptions of the Lie algebras ${\mathfrak{g}}_{4}$, the choice of their bases, and the parameters in [4, 5] coincide up to notation, except that for all families with parameters in [5] the cases of extreme parameter values are highlighted, while for the family ${\mathfrak{g}}_{4,5}$ other parameter values are indicated. For the Lie algebras ${\mathfrak{g}}_{3}$ it is true only for ${\mathfrak{g}}_{3,1}$ and ${\mathfrak{g}}_{3,7}$.

Table 1 Four-dimensional Lie algebras ${\mathfrak{g}}_{3}\oplus{\mathfrak{g}}_{1}$, where ${\mathfrak{g}}_{1}={𝕉}$, and ${\mathfrak{g}}_{4}$.

Every connected Lie group $G_{3,3}$ with Lie algebra ${\mathfrak{g}}_{3,3}$ as well as every connected Lie group $G$ with Lie algebra ${\mathfrak{g}}^{1,1}_{4,5}$ is simply connected and characterized by the property that its every left-invariant Riemannian metric has constant negative sectional curvature; i.e., the metric gives three-dimensional and four-dimensional Lobachevsky spaces [11]. The Lie groups $G_{3,3}$ and $G$ are isomorphic to the groups generated by homotheties (without rotations) and parallel translations of two-dimensional and three-dimensional Euclidean spaces.

Table 2 Automorphisms of the needed indecomposable four-dimensional Lie algebras.

3. Algebraic Results

Lemma 1

Let $({\mathfrak{g}},[\cdot,\cdot])$ be a four-dimensional real Lie algebra, and let ${\mathfrak{p}}\subset{\mathfrak{g}}$ be a two-dimensional subspace generating ${\mathfrak{g}}$ by the Lie bracket $[\cdot,\cdot]$. Then there exists a basis $(e_{1},e_{2},e_{3},e_{4})$ for ${\mathfrak{g}}$ such that $e_{1},e_{2}\in{\mathfrak{p}}$ and in this basis

$$ [e_{1},e_{2}]=e_{3},\quad[e_{1},e_{3}]=e_{4},\quad C_{23}^{4}=0, $$

(1)

where $C_{ij}^{k}$, $i,j,k=1,\dots,4$, are the structure constants of the Lie algebra ${\mathfrak{g}}$. For given vector $e_{1}\in{\mathfrak{p}}$ with this property, there exist linearly independent vectors $e_{1}$, $e_{2}$, $e_{3}$, and $e_{4}$ in ${\mathfrak{g}}$ satisfying (1) which are unique up to multiplication by nonzero reals. Furthermore, up to multiplication by nonzero reals, $e_{2}$ and $e_{3}$ do not depend on the choice of $e_{1}$. Moreover, if $[e_{2},e_{3}]$ is not a linear combination of $e_{2}$ and $e_{3}$, then $e_{1}$ can be chosen so that $C^{2}_{23}=0$.

Proof

At first, let us choose a basis $(e_{1},e_{2})$ for ${\mathfrak{p}}$. Then, due to the condition on ${\mathfrak{p}}$, the vectors $e_{1},e_{2},e_{3}:=[e_{1},e_{2}]$ are linearly independent; moreover, $e_{1}$, $e_{2}$, $e_{3}$ and $[e_{1},e_{3}]$ or $[e_{2},e_{3}]$ are linearly independent. We can assume that $(e_{1},e_{2},e_{3},e_{4}:=[e_{1},e_{3}])$ is a basis for the Lie algebra ${\mathfrak{g}}$. If $C_{23}^{4}=0$ then the equalities in (1) are satisfied. Otherwise, replacing $e_{2}$ with $e_{2}^{\prime}:=e_{2}-C_{23}^{4}e_{1}$ and denoting $e_{2}^{\prime}$ by $e_{2}$, we get the required basis for ${\mathfrak{g}}$.

The second statement follows from the construction of the basis $(e_{1},e_{2},e_{3},e_{4})$.

It is clear that the third statement is true for $e_{3}$.

Given such a basis, let us take a vector of the form $e^{\prime}_{1}:=e_{1}+xe_{2}$, $x\in{𝕉}$. Then

$$ [e_{2},e_{3}]=\sum\limits_{k=1}^{3}a_{k}e_{k} $$

for some reals $a_{k}\in{𝕉}$, $k=1,2,3$, and

$$ [e^{\prime}_{1},e_{2}]=e_{3},\ e^{\prime}_{4}=[e^{\prime}_{1},e_{3}]=e_{4}+x[e_{2},e_{3}], $$

$$ [e_{2},e_{3}]=\sum\limits_{k=1}^{3}a_{k}e_{k}=a_{1}e^{\prime}_{1}+(a_{2}-xa_{1})e_{2}+a_{3}e_{3}; $$

i.e. the equalities in (1) hold after replacing $e_{1}$ with $e^{\prime}_{1}$. Moreover, by (1), no vector collinear to $e_{2}$ have the above properties of $e_{1}$. This completes the proof of the third claim.

By definition, $[e_{2},e_{3}]=C^{1}_{23}e_{1}+C^{2}_{23}e_{2}+C^{3}_{23}e_{3}$. Therefore, if $[e_{2},e_{3}]$ is not a linear combination of $e_{2}$ and $e_{3}$, then $C^{1}_{23}\neq 0$. If $C^{2}_{23}\neq 0$, then we replace $e_{1}$ with $e^{\prime}_{1}=e_{1}+\bigl{(}C^{2}_{23}/C^{1}_{23}\bigr{)}e_{2}$ and get

$$ C^{1}_{23}e_{1}+C^{2}_{23}e_{2}=C^{1}_{23}\bigl{(}e^{\prime}_{1}-\bigl{(}C^{2}_{23}/C^{1}_{23}\bigr{)}e_{2}\bigr{)}+C^{2}_{23}e_{2}=C^{1}_{23}e^{\prime}_{1}.\quad\text{☐} $$

Proposition 1

For the basis $(e_{1},e_{2},e_{3},e_{4})$ for ${\mathfrak{g}}$ from Lemma 1, we have

$$ C_{24}^{1}=C_{24}^{2}=0,\quad C_{24}^{3}=C_{23}^{2},\quad C_{24}^{4}=C_{23}^{3}. $$

Proof

Owing to (1), the Jacobi identity

$$ [e_{1},[e_{2},e_{3}]]+[e_{2},[e_{3},e_{1}]]+[e_{3},[e_{1},e_{2}]]=0 $$

is equivalent to the equality $[e_{2},e_{4}]=[e_{1},[e_{2},e_{3}]]$; i.e.,

$$ \sum\limits_{k=1}^{4}C_{24}^{k}e_{k}=\bigg{[}e_{1},\sum\limits_{m=1}^{4}C_{23}^{m}e_{m}\bigg{]}=C_{23}^{2}e_{3}+C_{23}^{3}e_{4}+C_{23}^{4}\sum\limits_{k=1}^{4}C_{14}^{k}e_{k}=C_{23}^{2}e_{3}+C_{23}^{3}e_{4}. $$

This yields Proposition 1.  ☐

Proposition 2

A real Lie algebra ${\mathfrak{g}}={\mathfrak{g}}_{3}\oplus{\mathfrak{g}}_{1}$ with solvable indecomposable subalgebra ${\mathfrak{g}}_{3}$ has a two-dimensional generating subspace if and only if the operator $\operatorname{ad}E_{3}:{\mathfrak{g}}^{\prime}\rightarrow{\mathfrak{g}}^{\prime}$ is nondegenerate and has noneigenvectors; in other words, ${\mathfrak{g}}_{3}\neq{\mathfrak{g}}_{3,1}$ and ${\mathfrak{g}}_{3}\neq{\mathfrak{g}}_{3,3}$. Moreover, a two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ if and only if ${\mathfrak{p}}$ has a basis of the form $(e_{1},e_{2}=a+b)$, where $e_{1}\in{\mathfrak{g}}_{3}$, $e_{1}\notin{\mathfrak{g}}^{\prime}$, $a=\beta E_{4}$, $\beta\neq 0$, and $b\in{\mathfrak{g}}^{\prime}$ is a noneigenvector of $\operatorname{ad}E_{3}$. The claims of Lemma 1 are true for $e_{1}$, $e_{2}$, $[e_{2},e_{3}]=0$, and the vectors $e_{1}$ and $e_{2}^{\prime}=b$ generate the Lie algebra ${\mathfrak{g}}_{3}$.

Proof

It follows from Table 1 that all these Lie algebras ${\mathfrak{g}}$ have a central subalgebra $\mathfrak{g}_{1}$, the two-dimensional commutative ideal ${\mathfrak{I}}$ spanned by the vectors $E_{1}$ and $E_{2}$, and ${\mathfrak{g}}^{\prime}=[E_{3},{\mathfrak{I}}]\subset{\mathfrak{I}}$. Moreover, ${\mathfrak{g}}^{\prime}$ is one-dimensional (and the center of ${\mathfrak{g}}$) only in case ${\mathfrak{g}}_{3}={\mathfrak{g}}_{3,1}$.

Assume that a subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ and $(e,f)$ is some basis for ${\mathfrak{p}}$. Then by the above none of the vectors $e$ and $f$ can lie in ${\mathfrak{I}}$ or be collinear to $E_{4}$. Moreover, since each vector collinear to a basis vector can be added to another basis vector, we can assume that $e\in{\mathfrak{g}}_{3}$, $e\notin{\mathfrak{J}}$, and $f=a+b$, where $a=\beta E_{4}$, $\beta\neq 0$, and $b\in{\mathfrak{I}}$; moreover $b\neq 0$. Furthermore, $b$ cannot be an eigenvector of $\operatorname{ad}E_{3}:{\mathfrak{J}}\rightarrow{\mathfrak{J}}$ and this operator is nondegenerate; otherwise ${\mathfrak{p}}$ is a subalgebra or lies in some three-dimensional subalgebra ${\mathfrak{f}}\subset{\mathfrak{g}}$ and does not generate ${\mathfrak{g}}$.

As a consequence of the last statement, no two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ if ${\mathfrak{g}}_{3}={\mathfrak{g}}_{3,3}$ or ${\mathfrak{g}}_{3}={\mathfrak{g}}_{3,1}$.

In all other cases under consideration, ${\mathfrak{g}}^{\prime}={\mathfrak{I}}$, the operator $\operatorname{ad}E_{3}:{\mathfrak{g}}^{\prime}\rightarrow{\mathfrak{g}}^{\prime}$ is nondegenerate and has noneigenvectors. Let a subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ have a basis $(e_{1}=e,e_{2}=f)$ as in the claim of Proposition 2. Then $[e_{1},e_{2}]=a_{3}[E_{3},b]$ for some $a_{3}\neq 0$ and $[E_{3},b]\in{\mathfrak{g}}^{\prime}$ is also a noneigenvector of $\operatorname{ad}E_{3}$; otherwise $b=(\operatorname{ad}E_{3})^{-1}([E_{3},b])$ would be an eigenvector. Thus, $e_{3}:=[e_{1},e_{2}]\in{\mathfrak{g}}^{\prime}$ and $e_{4}:=[e_{1},e_{3}]\in{\mathfrak{g}}^{\prime}$, while $e_{3}$ and $e_{4}$ are not collinear. Then $e_{1},\dots,e_{4}$ are linearly independent; i.e., $e_{1}$ and $e_{2}$ generate ${\mathfrak{g}}$. Moreover, $[e_{2},e_{3}]=[b,e_{3}]=0$, because $b,e_{3}\in{\mathfrak{g}}^{\prime}$. So all claims of Lemma 1 are true for $e_{1}$ and $e_{2}$.

The last claim of Proposition 2 is obvious.  ☐

Proposition 2 and the paper [6] imply the following

Corollary 1

Every two of the two-dimensional generating subspaces of the Lie algebra ${\mathfrak{g}}$ from Proposition $2$ are transformed into each other by some automorphism of ${\mathfrak{g}}$.

Theorem 1

A real Lie algebra ${\mathfrak{g}}={\mathfrak{g}}_{3}\oplus{\mathfrak{g}}_{1}$ with a simple subalgebra ${\mathfrak{g}}_{3}$ has a two-dimensional generating subspace. Moreover, a two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates the Lie algebra ${\mathfrak{g}}$ if and only if

(1) ${\mathfrak{p}}_{1}=p({\mathfrak{p}})\neq{\mathfrak{p}}$ for the projection $p:{\mathfrak{g}}\rightarrow{\mathfrak{g}}_{3}$;

(2) ${\mathfrak{p}}_{1}$ is two-dimensional;

(3) the restriction of the Killing form $k$ of the Lie algebra ${\mathfrak{g}}_{3}$ to ${\mathfrak{p}}_{1}$ is nondegenerate.

For ${\mathfrak{g}}_{3,7}\oplus{\mathfrak{g}}_{1}$ (respectively, ${\mathfrak{g}}_{3,6}\oplus{\mathfrak{g}}_{1}$) there exists exactly one (respectively, exist four) equivalence class(es) of such subspaces ${\mathfrak{p}}\subset{\mathfrak{g}}$. Moreover, there exist basis vectors in ${\mathfrak{p}}$ unique up to multiplication by nonzero reals such that either $e_{1}\in{\mathfrak{g}}_{3}$, $e_{2}\notin{\mathfrak{g}}_{3}$, and $k(e_{1},p(e_{2}))=0$, or $e_{1}\notin{\mathfrak{g}}_{3}$, $e_{2}\in{\mathfrak{g}}_{3}$ and both vectors $p(e_{1})$ and $e_{2}$ are isotropic. Moreover, $e_{1}$ and $e_{2}$ satisfy Lemma $1$ with conditions $0\neq[e_{2},e_{3}]||e_{1}$, i.e. $C^{1}_{23}\neq 0$ and $C^{2}_{23}=C^{3}_{23}=0$ in the first case and $0\neq[e_{2},e_{3}]||e_{2}$; i.e., $C^{2}_{23}\neq 0$ and $C^{1}_{23}=C^{3}_{23}=0$ in the second case.

Proof

According to Table 1

$$ {\mathfrak{g}}_{3}={\mathfrak{g}}_{3,7}={\mathfrak{s}u}(2)={\mathfrak{so}}(3)\quad\text{or}\quad{\mathfrak{g}}_{3}={\mathfrak{g}}_{3,6}={\mathfrak{sl}}(2,{𝕉}). $$

By the Cartan criterion, a Lie algebra ${\mathfrak{g}}$ is semisimple if and only if its Killing form

$$ k_{{\mathfrak{g}}}(X,Y)=k(X,Y):=\operatorname{tr}(\operatorname{ad}X\operatorname{ad}Y),\quad X,Y\in{\mathfrak{g}}, $$

is nondegenerate on ${\mathfrak{g}}$. Moreover, $k$ is invariant under the automorphism group $\operatorname{Aut}({\mathfrak{g}})$ of ${\mathfrak{g}}$, the connected component of the identity $\operatorname{Aut}_{e}({\mathfrak{g}})$ of the group $\operatorname{Aut}({\mathfrak{g}})$ coincides with $\operatorname{Ad}G$ in case of a connected Lie group $G$ with the Lie algebra ${\mathfrak{g}}$, and so the Lie algebra $L(\operatorname{Aut}({\mathfrak{g}}))$ is equal to $\operatorname{ad}{\mathfrak{g}}$ [9, 10].

In our case, the symmetric form $b:=-\frac{1}{2}k$ is more usual. Then in consequence of Table 1, the vectors $E_{1}$, $E_{2}$, and $E_{3}$ are mutually orthogonal with respect to $b$. Also, $b(E_{i},E_{i})=1$, $i=1,2,3$, for ${\mathfrak{g}}_{3,7}$ and $b(E_{1},E_{1})=b(E_{2},E_{2})=-b(E_{3},E_{3})=1$ for ${\mathfrak{g}}_{3,6}$. By the above, $\operatorname{Aut}_{e}({\mathfrak{so}}(3))=\operatorname{SO(3)}$ and $\operatorname{Aut}_{e}({\mathfrak{sl}}(2,{𝕉}))=\operatorname{SO_{0}(2,1)}$. The last Lie group is the so-called restricted Lorentz group, its two-sheeted covering is isomorphic to the Lie group $\operatorname{SL}(2,{𝕉})$ [12]. Moreover, $\operatorname{Aut}({\mathfrak{so}}(3))=\operatorname{SO}(3)$ and $\operatorname{Aut}({\mathfrak{sl}}(2,{𝕉}))=\operatorname{SO(2,1)}$; the last group has two connected components, one of which is $\operatorname{SO_{0}(2,1)}$ (see Table 6 in [5]).

The spheres $S^{2}(r)=\{X\in{\mathfrak{so}}(3):b(X,X)=r^{2}\}$, $r\geq 0$, are action orbits of $\operatorname{SO(3)}$ on ${\mathfrak{so}}(3)$. Then $\operatorname{SO(3)}$ acts transitively on a two-dimensional manifold of two-dimensional subspaces in ${\mathfrak{so}}(3)$ with corresponding orientation, $b$-orthogonal to vectors $X\in S^{2}(1)$, the two-sheeted covering of the Grassmann manifold $\operatorname{Gr}(2,3)$ of two-dimensional subspaces. Furthermore, all such subspaces generate ${\mathfrak{so}}(3)$. In fact, by Table 1, the bracket $[X,Y]$ is the vector product for the Euclidean space $({\mathfrak{so}}(3),b)$; if $(X,Y)$ is an orthonormal basis for any two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{so}}(3)$, then $(X,Y,Z=[X,Y])$ is an orthonormal basis for $({\mathfrak{so}}(3),b)$. Furthermore,

$$ [X,Y]=Z,\quad[Z,X]=Y,\quad[Y,Z]=X. $$

(2)

Moreover, $\operatorname{SO(3)}$ acts simply transitively on the manifold of ordered orthonormal pairs of vectors $(X,Y)$ in $({\mathfrak{so}}(3),b)$.

The action orbits of the group $\operatorname{SO_{0}(2,1)}$ on ${\mathfrak{sl}}(2,{𝕉})$ are hyperboloids of one sheet $S^{2}(r)=\{X:b(X,X)=r^{2}\},r>0$, consisting of spacelike vectors, the sheets in hyperboloids of two sheets $S^{2}(-r)=\{X:b(X,X)=-r^{2}\}$, $r>0$, consisting of timelike vectors, zero $\{0\}$, two connected components in $S^{2}(0)-\{0\}$ in the cone $S^{2}(0)=\{X:b(X,X)=0\}$ of isotropic vectors.

In consequence, the two-dimensional Grassmann manifold $\operatorname{Gr}(2,3)$ of all two-dimensional subspaces ${\mathfrak{p}}\subset{\mathfrak{sl}}(2,{𝕉})$ splits into three $\operatorname{SO}_{0}(2,1)$-orbits:

(I) $\operatorname{Gr}^{+}(2,3)$, where the form $k$ is nondegenerate and sign-definite ($b$ is positive definite), its every element is $b$-orthogonal to exactly one vector from the connected component of $S^{2}(-1)$;

(II) $\operatorname{Gr}^{-}(2,3)$, where forms $k$ and $b$ are nondegenerate and non-sign-definite, its every element is $b$-orthogonal to exactly one vector from $S^{2}(1)$;

(III) $\operatorname{Gr}^{0}(2,3)$, where forms $k$ and $b$ are degenerate, its elements touch the cone $S^{2}(0)$. Moreover, $\operatorname{Gr}^{0}(2,3)$ is a one-dimensional closed submanifold in $\operatorname{Gr}(2,3)$ dividing $\operatorname{Gr}(2,3)$ into the open submanifolds $\operatorname{Gr}^{+}(2,3)$ and $\operatorname{Gr}^{-}(2,3)$.

For each element in an orbit of the form (I)–(III), we can choose a $b$-orthogonal basis of one of the types (I) $(X,Y)$, (II) $(X,Z)$ or $(Z,X)$, and (III) $(X,W)$, where

$$ b(X,X)=b(Y,Y)=-b(Z,Z)=1,\quad b(X,Y)=b(X,Z)=0; $$

(3)

$$ b(X,W)=b(W,W)=0,\quad W\neq 0. $$

(4)

The group $\operatorname{SO}(2,1)$ acts simply transitively on the manifolds of ordered vector pairs of each of the four mentioned types $(X,Y)$, $(X,Z)$, $(Z,X)$, or $(X,W)$ from (3) and (4).

We state that each element of the first two orbits generates ${\mathfrak{sl}}(2,{𝕉})$, while each element from $\operatorname{Gr}^{0}(2,3)$ is a subalgebra in ${\mathfrak{sl}}(2,{𝕉})$. In consequence of the above, to prove the last statement it is enough to consider two-dimensional subspaces with bases of a special form.

(I) Let $(X=E_{1},Y=E_{2})$ be a $b$-orthogonal basis of a two-dimensional subspace ${\mathfrak{p}}$. Then according to Table 1

$$ Z:=[X,Y],\quad[Z,X]=-Y,\quad[Y,Z]=-X,\quad b(Y,Z)=0, $$

(5)

with fulfillment of all equalities from (3).

(II) Defining $Z$ by the first formula from (5) for the vectors $X$ and $Y$ in (I), we get $Z=-E_{3}$ and the bases of the form $(X,Z)$ or $(Z,X)$ for a two-dimensional subspace ${\mathfrak{p}}$ with conditions from (3) and (5).

(III) Let $(X=E_{1},W=E_{2}+E_{3})$ be a $b$-orthogonal basis of a two-dimensional subspace $\mathfrak{p}$. Then according to Table 1

$$ Z:=[X,W]=-W,\quad b(X,X)=1,\quad b(W,W)=0. $$

(6)

Suppose that ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ and prove that conditions (1)–(3) are satisfied.

If (1) is not satisfied, then the subalgebra generated by ${\mathfrak{p}}$ lies in ${\mathfrak{g}}_{3}$. If (2) is not satisfied, then ${\mathfrak{p}}$ is a two-dimensional commutative subalgebra in ${\mathfrak{g}}$. In the case that ${\mathfrak{g}}={\mathfrak{g}}_{3,7}\oplus{\mathfrak{g}}_{1}$, (3) is a consequence of (2). If ${\mathfrak{g}}={\mathfrak{g}}_{3,6}\oplus{\mathfrak{g}}_{1}$ and (3) is not satisfied, then, due to (1) and (2), as was shown, ${\mathfrak{p}}_{1}$ is a two-dimensional subalgebra in ${\mathfrak{g}}_{3}$ and so ${\mathfrak{p}}$ generates a three-dimensional subalgebra in ${\mathfrak{g}}$.

Let all conditions (1)–(3) be satisfied.

Then (I) ${\mathfrak{p}}_{1}\in\operatorname{Gr}^{+}(2,3)$ or (II) ${\mathfrak{p}}_{1}\in\operatorname{Gr}^{-}(2,3)$ for ${\mathfrak{g}}={\mathfrak{g}}_{3,6}\oplus{\mathfrak{g}}_{1}$.

(I) The space ${\mathfrak{s}}={\mathfrak{p}}\cap{\mathfrak{g}}_{3}={\mathfrak{p}}\cap{\mathfrak{p}}_{1}$ is one-dimensional and its nonzero vectors are $b$-spacelike. Let us choose some $b$-orthonormal vector $e_{1}\in{\mathfrak{s}}$. It is easy to see that there exists a vector $e_{2}\in{\mathfrak{p}}$ unique up to multiplication by $\pm 1$ such that $(e_{1},p(e_{2})=Y)$ is a $b$-orthonormal basis for ${\mathfrak{p}}_{1}$. Clearly, $e_{2}\notin{\mathfrak{g}}_{3}$.

(II) The three cases are possible: (a) nonzero vectors from ${\mathfrak{s}}$ are $b$-spacelike; (b) nonzero vectors from ${\mathfrak{s}}$ are $b$-timelike; and (c) the space ${\mathfrak{s}}$ consists of $b$-isotropic vectors.

Arguing by analogy to case (I), we infer that

(a) there exists a basis $(e_{1},e_{2})$ for ${\mathfrak{p}}$ such that $e_{1}=X\in{\mathfrak{s}}$ and $e_{2}\notin{\mathfrak{g}}_{3}$, while $(X,Z=p(e_{2}))$ is a basis for ${\mathfrak{p}}_{1}$ with conditions (3);

(b) there exists a basis $(e_{1},e_{2})$ for $\mathfrak{p}$ such that $e_{1}=Z\in{\mathfrak{s}}$ and $e_{2}\notin{\mathfrak{g}}_{3}$, while $(Z,X=p(e_{2}))$ is a basis for ${\mathfrak{p}}_{1}$ with conditions (3);

(c) there exists a basis $(e_{1},e_{2})$ for ${\mathfrak{p}}$ such that $e_{1}\notin{\mathfrak{g}}_{3}$ and $e_{2}=W_{2}\in{\mathfrak{s}}$. Also, $(W_{1}=p(e_{1}),W_{2}=e_{2})$ is a basis of isotropic vectors in ${\mathfrak{p}}_{1}$, transformed by some element $\xi^{-1}\in SO(2,1)$ to the pair $(E_{3}+E_{1}$, $E_{3}-E_{1})$.

Every element $\xi$ of the automorphism group of the Lie algebra ${\mathfrak{g}}$ transforms the subalgebra ${\mathfrak{g}}_{3}$ into itself isomorphically. Consequently, for ${\mathfrak{g}}={\mathfrak{g}}_{3,6}\oplus{\mathfrak{g}}_{1}$ it saves two types (I) and (II) of subspaces ${\mathfrak{p}}_{1}$ and three types (a), (b), and (c) of subspaces $\mathfrak{s}\subset\mathfrak{p}_{1}$ for type (II) and there exist at least four nonequivalent types of subspaces ${\mathfrak{p}}\subset{\mathfrak{g}}$.

Now, to complete the proof of Theorem 1, it remains to show that all constructed bases $(e_{1},e_{2})$ have the properties indicated in Theorem 1, and these bases for different subspaces ${\mathfrak{p}}$ of one and the same among four indicated types are transformed into each other by some automorphism of the Lie algebra ${\mathfrak{g}}$.

(I) Let ${\mathfrak{g}}_{3}={\mathfrak{g}}_{3,7}$. Then, in consequence of (2), we have

$$ e_{3}=[e_{1},e_{2}]=[X,Y]=Z,\quad e_{4}=[e_{1},e_{3}]=[X,Z]=-Y,\quad[e_{2},e_{3}]=[Y,Z]=X=e_{1} $$

subject to equalities (3) with replacement $-b(Z,Z)$ by $b(Z,Z)$.

Let ${\mathfrak{g}}_{3}={\mathfrak{g}}_{3,6}$. Then, in consequence of (5), we have

$$ e_{3}=[e_{1},e_{2}]=[X,Y]=Z,\quad e_{4}=[e_{1},e_{3}]=[X,Z]=-Y,\quad[e_{2},e_{3}]=[Y,Z]=-X=-e_{1} $$

with fulfillment of (3).

(II) (a) In consequence of (5), we have

$$ \begin{gathered}\displaystyle e_{3}=[e_{1},e_{2}]=[X,Z]=-Y,\quad e_{4}=[e_{1},e_{3}]=[X,-Y]=-Z,\\ \displaystyle[e_{2},e_{3}]=[Z,-Y]=-X=-e_{1}\end{gathered} $$

with fulfillment of (3).

(b) In consequence of (5), we have

$$ e_{3}=[e_{1},e_{2}]=[Z,X]=Y,\quad e_{4}=[e_{1},e_{3}]=[Z,Y]=Y,\quad[e_{2},e_{3}]=[X,Y]=Z=e_{1} $$

with fulfillment of (3).

(c) We have

$$ \begin{gathered}\displaystyle e_{3}=[e_{1},e_{2}]=[F(E_{3}+E_{1}),F(E_{3}-E_{1})]=F(-2E_{2}),\\ \displaystyle e_{4}=F[E_{3}+E_{1},-2E_{2}]=2F(E_{3}+E_{1}),\\ \displaystyle[e_{2},e_{3}]=[F(E_{3}-E_{1}),-2F(E_{2})]=-2F(E_{3}-E_{1})=-2e_{2}.\end{gathered} $$

It is clear from the above that for indicated bases $(e_{1},e_{2})$ of various subspaces ${\mathfrak{p}}$ of the same type among four mentioned, the bases $(p(e_{1}),p(e_{2}))$ are translated into each other by some automorphism $\xi$ of ${\mathfrak{g}}_{3}$. If at the same time $e_{i}\neq p(e_{i})$ for $i=2$ or $i=1$, then $e_{i}-p(e_{i})\in{\mathfrak{g}}_{1}$ and $\xi(e_{i})$ is defined by the formula $\xi(e_{i})=\xi(p(e_{i}))+\xi(e_{i}-p(e_{i}))$, where $\xi(e_{i}-p(e_{i}))$ is a corresponding vector from ${\mathfrak{g}}_{1}$.  ☐

Proposition 3

All real Lie algebras ${\mathfrak{g}}={\mathfrak{g}}_{4}$ with three-dimensional commutative ideal ${\mathfrak{I}}$ but ${\mathfrak{g}}^{1}_{4,2}$, ${\mathfrak{g}}^{\alpha,1}_{4,5}$, $-1\leq\alpha\leq 1$, $\alpha\neq 0$, and ${\mathfrak{g}}^{\alpha,\alpha}_{4,5}$, $-1<\alpha<1$, $\alpha\neq 0$, have two-dimensional generating subspaces. A two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ if and only if ${\mathfrak{p}}$ does not belong to ${\mathfrak{I}}$ and intersects by zero with every ideal ${\mathfrak{J}}\subset{\mathfrak{I}}$ of ${\mathfrak{g}}$, where ${\mathfrak{J}}\neq{\mathfrak{I}}$. Moreover, for every basis $(e_{1},e_{2})$ for ${\mathfrak{p}}$ with $e_{2}\in{\mathfrak{p}}\cap{\mathfrak{I}}$, the claims of Lemma $1$ hold and $[e_{2},e_{3}]=0$. Every two of the two-dimensional generating subspaces of the Lie algebra ${\mathfrak{g}}$ are translated into each other by an automorphism of ${\mathfrak{g}}$.

Proof

It follows from Table 1 that the three-dimensional commutative ideal ${\mathfrak{I}}$ of ${\mathfrak{g}}$ is spanned by the vectors $E_{1}$, $E_{2}$, and $E_{3}$. Note that in case ${\mathfrak{g}}={\mathfrak{g}}^{1}_{4,2}$ each one-dimensional subspace with a basis vector of the form $xE_{1}+yE_{2}$, $x,y\in{𝕉}$, is a one-dimensional ideal of ${\mathfrak{g}}$. In case ${\mathfrak{g}}={\mathfrak{g}}^{\alpha,1}_{4,5}$, $-1\leq\alpha<1$, $\alpha\neq 0$ (respectively, ${\mathfrak{g}}={\mathfrak{g}}^{\alpha,\alpha}_{4,5}$, $-1<\alpha<1$, $\alpha\neq 0$) each two-dimensional subspace with basis vectors $E_{3}$, $xE_{1}+yE_{2}$ (respectively, $E_{1}$, $xE_{2}+yE_{3}$), where $x,y\in{𝕉}$, is a two-dimensional ideal of ${\mathfrak{g}}$ which lies in ${\mathfrak{I}}$, and each of these basis vectors generates a one-dimensional ideal of ${\mathfrak{g}}$. Each one-dimensional (two-dimensional) subspace in ${\mathfrak{I}}\subset{\mathfrak{g}}^{1,1}_{4,5}$ is a one-dimensional (two-dimensional) ideal of ${\mathfrak{g}}^{1,1}_{4,5}$.

Let us suppose that the subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ and $(e,f)$ is some basis for ${\mathfrak{p}}$. It is clear that at least one of the vectors $e$ and $f$ does not belong to ${\mathfrak{I}}$. Since each vector collinear to one basis vector can be added to another vector of the basis, we can assume that $e\notin{\mathfrak{I}}$ and $f\in{\mathfrak{I}}$. Moreover, $f$ belongs to no ideal ${\mathfrak{J}}\subset{\mathfrak{I}}$ of ${\mathfrak{g}}$ different from ${\mathfrak{I}}$; otherwise, ${\mathfrak{p}}$ is a subalgebra or ${\mathfrak{p}}$ lies in some three-dimensional subalgebra and so ${\mathfrak{p}}$ does not generate ${\mathfrak{g}}$. Therefore, ${\mathfrak{p}}\cap{\mathfrak{J}}=\{0\}$. From here and the previous paragraph it follows that the algebras ${\mathfrak{g}}_{4,2}^{1}$, ${\mathfrak{g}}^{\alpha,1}_{4,5}$, $-1\leq\alpha\leq 1$, $\alpha\neq 0$, and ${\mathfrak{g}}^{\alpha,\alpha}_{4,5}$, $-1<\alpha<1$, $\alpha\neq 0$, have no two-dimensional generating subspace.

1. Let ${\mathfrak{g}}={\mathfrak{g}}_{4,k}$, $k=1,4$. At first, consider the case when the basis from Proposition 3 has the special form $(e_{1},e_{2})=(E_{4},E_{3})$. Then Table 1 implies that

$$ {\mathfrak{g}}_{4,1}:\quad e_{3}=[e_{1},e_{2}]=-E_{2},\quad e_{4}=[e_{1},e_{3}]=E_{1}; $$

$$ {\mathfrak{g}}_{4,4}:\quad e_{3}=[e_{1},e_{2}]=-E_{2}-E_{3},\quad e_{4}=[e_{1},e_{3}]=E_{1}+2E_{2}+E_{3}. $$

It is easy to see that in all cases $(e_{1},e_{2},e_{3},e_{4})$ is a basis for the Lie algebra ${\mathfrak{g}}$ and $[e_{2},e_{3}]=0$, and so the vectors $e_{1}$ and $e_{2}$ satisfy Lemma 1.

Now it follows from Table 2 that for the basis $(e_{1},e_{2})$ specified in Proposition 3 there exists an automorphism $\xi$ of the Lie algebra ${\mathfrak{g}}$ such that $e_{1}=\xi(\beta E_{4})$ and $e_{2}=\xi(E_{3})$, where $\beta\neq 0$. Consequently, the two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ with basis $(e_{1},e_{2})$ generates the Lie algebra ${\mathfrak{g}}$, and for the vectors $e_{1}$ and $e_{2}$ all claims of Lemma 1 are true. Hence, all claims of Proposition 3 are true.

2. Let ${\mathfrak{g}}={\mathfrak{g}}_{4,3}$, ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,2}$, $\alpha\notin\{0,1\}$, or ${\mathfrak{g}}={\mathfrak{g}}^{\alpha,\beta}_{4,6}$, $\alpha>0$, $\beta\in{𝕉}$. At first, consider the case when the basis from Proposition 3 has the special form $(e_{1},e_{2})=(E_{4},E_{1}+E_{3})$. Then it follows from Table 1 that

$$ {\mathfrak{g}}^{\alpha}_{4,2}:\quad e_{3}=[e_{1},e_{2}]=-\alpha E_{1}-E_{2}-E_{3},\quad e_{4}=[e_{1},e_{3}]=\alpha^{2}E_{1}+2E_{2}+E_{3}; $$

$$ {\mathfrak{g}}_{4,3}:\quad e_{3}=[e_{1},e_{2}]=-E_{1}-E_{2},\quad e_{4}=[e_{1},e_{3}]=E_{1}; $$

$$ {\mathfrak{g}}^{\alpha,\beta}_{4,6}:\quad e_{3}=[e_{1},e_{2}]=-\alpha E_{1}-E_{2}-\beta E_{3},\quad e_{4}=[e_{1},e_{3}]=\alpha^{2}E_{1}+2\beta E_{2}+(\beta^{2}-1)E_{3}. $$

It is easy to see that in all cases $(e_{1},e_{2},e_{3},e_{4})$ is a basis for the Lie algebra ${\mathfrak{g}}$ and $[e_{2},e_{3}]=0$, so that the vectors $e_{1}$ and $e_{2}$ satisfy Lemma 1.

Now it follows from Table 2 that for the basis $(e_{1},e_{2})$, specified in Proposition 3, there exists an automorphism $\xi$ of the Lie algebra ${\mathfrak{g}}$ such that $e_{1}=\xi(\beta E_{4})$ and $e_{2}=\xi(E_{1}+E_{3})$, where $\beta\neq 0$. Consequently, the two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ with basis $(e_{1},e_{2})$ generates the Lie algebra ${\mathfrak{g}}$, and for vectors $e_{1}$ and $e_{2}$ all claims of Lemma 1 are true. Hence, all claims of Proposition 3 are true.

3. Let ${\mathfrak{g}}={\mathfrak{g}}^{\alpha,\beta}_{4,5}$, $-1<\alpha<\beta<1$, $\alpha\beta\neq 0$ or $\alpha=-1$, $0<\beta\leq 1$. At first, consider the case when the basis from Proposition 3 has the special form $(e_{1},e_{2})=(E_{4},E_{1}+E_{2}+E_{3})$. Then it follows from Table 1 that

$$ e_{3}=[e_{1},e_{2}]=-E_{1}-\beta E_{2}-\alpha E_{3},\quad e_{4}=[e_{1},e_{3}]=E_{1}+\beta^{2}E_{2}+\alpha^{2}E_{3}. $$

It is easy to see that $(e_{1},e_{2},e_{3},e_{4})$ is a basis for the Lie algebra ${\mathfrak{g}}$ and $[e_{2},e_{3}]=0$, so that the vectors $e_{1}$ and $e_{2}$ satisfy Lemma 1.

Now it follows from Table 2 that for the basis $(e_{1},e_{2})$, specified in Proposition 3, there is an automorphism $\xi$ of the Lie algebra ${\mathfrak{g}}$ such that $e_{1}=\xi(\beta E_{4})$ and $e_{2}=\xi(E_{1}+E_{2}+E_{3})$, where $\beta\neq 0$. Consequently, the two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ with the basis $(e_{1},e_{2})$ generates ${\mathfrak{g}}$, and for the vectors $e_{1}$ and $e_{2}$ all claims of Lemma 1 are true. Hence, all claims of Proposition 3 are true.  ☐

Theorem 2

The real Lie algebras ${\mathfrak{g}}$, where ${\mathfrak{g}}={\mathfrak{g}}_{4,7}$, ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,9}$, $\alpha\geq 0$, ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,8}$, $-1\leq\alpha\leq 1$, but ${\mathfrak{g}}^{1}_{4,8}$ have two-dimensional generating subspaces. A two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ if and only if ${\mathfrak{p}}$ does not lie in the three-dimensional ideal ${\mathfrak{I}}$ and intersects by zero every ideal ${\mathfrak{J}}\subset{\mathfrak{I}}$, where ${\mathfrak{J}}\neq{\mathfrak{I}}$, of ${\mathfrak{g}}$. Every two of the two-dimensional generating subspaces of the Lie algebra ${\mathfrak{g}}$ are translated to each other by an automorphism of ${\mathfrak{g}}$. Moreover, Lemma $1$ is valid for the basis $(e_{1},e_{2})\in{\mathfrak{p}}$ with $e_{1}\in{\mathfrak{p}}\cap{\mathfrak{I}}$. Furthermore, $C^{1}_{23}\neq 0$ and $C^{2}_{23}=0$ for ${\mathfrak{g}}_{4,7}$, ${\mathfrak{g}}^{\alpha}_{4,9}$, ${\mathfrak{g}}^{\alpha}_{4,8}$, $\alpha\neq 0$, and $C^{1}_{23}=C^{2}_{23}=0$ for ${\mathfrak{g}}^{0}_{4,8}$.

Proof

It follows from Table 1 that the three-dimensional ideal ${\mathfrak{I}}$ of the Lie algebra ${\mathfrak{g}}$ is spanned by the vectors $E_{1}$, $E_{2}$, and $E_{3}$. Moreover, the algebra ${\mathfrak{g}}={\mathfrak{g}}_{4,7}$ has unique (two-dimensional) ideal ${\mathfrak{J}}_{1}=\operatorname{span}(E_{1},E_{2})\subset{\mathfrak{I}}$, the algebra ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,9}$ has unique (one-dimensional) ideal ${\mathfrak{J}}_{2}\subset{\mathfrak{I}}$ spanned on $E_{1}$, the algebra ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,8}$, $\alpha\neq 1$, has two two-dimensional commutative ideals ${\mathfrak{J}}_{3}=\operatorname{span}(E_{1},E_{3})\subset{\mathfrak{I}}$, ${\mathfrak{J}}_{4}=\operatorname{span}(E_{2},E_{3})\subset{\mathfrak{I}}$, and one one-dimensional ideal ${\mathfrak{J}}_{2}\subset{\mathfrak{I}}$ spanned on $E_{1}$. Note also that every two-dimensional subspace with basis vectors $E_{1}$, $xE_{2}+yE_{3}$, where $x,y\in{𝕉}$, is a two-dimensional ideal of ${\mathfrak{g}}^{1}_{4,8}$.

Assume that a subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates the algebra ${\mathfrak{g}}$ and $(e,f)$ is a basis for ${\mathfrak{p}}$. It is clear that at least one of the vectors $e$ and $f$ does not belong to ${\mathfrak{I}}$; otherwise the subspace ${\mathfrak{p}}\subset{\mathfrak{I}}$ does not generate ${\mathfrak{g}}$. Since to a vector of the basis we can add an arbitrary vector collinear to another, we can suppose that $e\in{\mathfrak{I}}$ and $f\notin{\mathfrak{I}}$, and also the component of $f$ for $E_{4}$ in the basis $(E_{1},E_{2},E_{3},E_{4})$ is equal to 1. If $e$ belongs to some two-dimensional or one-dimensional subideal of ${\mathfrak{I}}$, then ${\mathfrak{p}}$ is a subalgebra or ${\mathfrak{p}}$ lies in a three-dimensional subalgebra and hence ${\mathfrak{p}}$ does not generate ${\mathfrak{g}}$. Then ${\mathfrak{p}}\cap{\mathfrak{J}}=\{0\}$. This and the previous paragraph imply that ${\mathfrak{g}}^{1}_{4,8}$ has no two-dimensional generating subspace.

1. Assume that ${\mathfrak{g}}={\mathfrak{g}}_{4,7}$. At first consider the special basis $(e_{1},e_{2})=(E_{3},E_{4})$. Then Table 1 implies that

$$ e_{3}=[e_{1},e_{2}]=[E_{3},E_{4}]=E_{2}+E_{3},\quad e_{4}=[e_{1},e_{3}]=-E_{1},\quad[e_{2},e_{3}]=e_{1}-2e_{3}. $$

It is easy to see that $(e_{1},e_{2},e_{3},e_{4})$ is a basis for ${\mathfrak{g}}$, while $C_{23}^{1}\neq 0$, $C_{23}^{2}=C^{4}_{23}=0$; and so the last claim of Theorem 2 is true for the basis $(e_{1},e_{2})$.

Since $e\notin{\mathfrak{J}}_{2}$, the component $z$ of $e$ for $E_{3}$ is not zero. Put $e_{1}=e/z$ and $e_{2}=f$. The conditions for $e_{1}$ and $e_{2}$ imply that

$$ e_{1}=xE_{1}+a_{2}E_{2}+E_{3},\quad e_{2}=a_{5}E_{1}+a_{4}E_{2}+a_{3}E_{3}+E_{4}. $$

Let us show that for some change of the vector $e_{2}$ by $e_{2}^{\prime}=e_{2}+ye_{1}$, $y\in{𝕉}$, there exists an automorphism $\xi$ of ${\mathfrak{g}}$ such that $\xi(E_{3})=e_{1}$ and $\xi(E_{4})=e^{\prime}_{2}$. According to Table 2, the corresponding components for $e_{2}^{\prime}$ are equal to $a_{4}^{\prime}=a_{4}+ya_{2}$, $a_{3}^{\prime}=a_{3}+y$, and so we must have

$$ x=a^{\prime}_{4}-(1+a_{2})a^{\prime}_{3}=a_{4}-(1+a_{2})a_{3}-y. $$

It is clear that there exists exactly one solution $y$ of this equation.

This, together with $[e_{1},e_{2}^{\prime}]=[e_{1},e_{2}]$, implies that the last claim of Theorem 2 is valid for an appropriate basis $(e_{1},e^{\prime}_{2})$ for ${\mathfrak{p}}$ of the above-indicated form.

2. Assume that ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,9}$, $\alpha\geq 0$.

Let us consider at first the case of the special basis $(e_{1},e_{2})=(E_{3},E_{4})$ with the properties from Theorem 2. Then it follows from Table 1 that

$$ e_{3}=[e_{1},e_{2}]=E_{2}+\alpha E_{3},\quad e_{4}=[e_{1},e_{3}]=-E_{1},\quad[e_{2},e_{3}]=(1+\alpha^{2})e_{1}-2\alpha e_{3}. $$

It is easy to see that $(e_{1},e_{2},e_{3},e_{4})$ is a basis for the Lie algebra ${\mathfrak{g}}$ and $C_{23}^{1}\neq 0$, $C_{23}^{2}=C^{4}_{23}=0$. So the last claim of Theorem 2 is true for $e_{1}$ and $e_{2}$.

Put $e_{1}=e$ and $e_{2}=f$. By conditions on $e_{1}$ and $e_{2}$, we have

$$ e_{1}=xE_{1}+a_{2}E_{2}+a_{1}E_{3},\quad e_{2}=a_{3}E_{1}+a_{4}E_{2}+a_{5}E_{3}+E_{4},\quad a_{1}^{2}+a_{2}^{2}\neq 0. $$

Let us show that under some change of $e_{2}$ by $e_{2}^{\prime}=e_{2}+ye_{1}$, $y\in{𝕉}$, there exists an automorphism $\xi$ of ${\mathfrak{g}}$ such that $\xi(E_{3})=e_{1}$ and $\xi(E_{4})=e^{\prime}_{2}$. In view of Table 3, the corresponding components of $e_{2}^{\prime}$ are equal to $a^{\prime}_{5}=a_{5}+ya_{1}$ and $a^{\prime}_{4}=a_{4}+ya_{2}$.

The following must hold:

$$ \begin{gathered}\displaystyle x=\frac{1}{1+\alpha^{2}}[a_{1}(\alpha a^{\prime}_{4}-a^{\prime}_{5})-a_{2}(a^{\prime}_{4}+\alpha a^{\prime}_{5})]\\ \displaystyle=\frac{1}{1+\alpha^{2}}\bigl{[}a_{1}(\alpha a_{4}-a_{5})-a_{2}(a_{4}+\alpha a_{5})-y\bigl{(}a_{1}^{2}+a_{2}^{2}\bigr{)}\bigr{]}.\end{gathered} $$

Because $a_{1}^{2}+a_{2}^{2}>0$, there exists exactly one solution $y$ of this equation.

It follows from here and the equality $[e_{1},e_{2}^{\prime}]=[e_{1},e_{2}]$ that the last claim of Theorem 2 is true for an appropriate basis $(e_{1},e^{\prime}_{2})$ for ${\mathfrak{p}}$ of the above form.

3. Assume that ${\mathfrak{g}}={\mathfrak{g}}^{\alpha}_{4,8}$, $-1\leq\alpha<1$.

Let us consider at first the case when the basis from Theorem 2 is of the special form: $(e_{1},e_{2})=(E_{2}+E_{3},E_{4})$. Then, in consequence of Table 1, we have

$$ e_{3}=[e_{1},e_{2}]=E_{2}+\alpha E_{3},\quad e_{4}=[e_{1},e_{3}]=(\alpha-1)E_{1},\quad[e_{2},e_{3}]=-\alpha e_{1}-(1+\alpha)e_{3}. $$

Since $\alpha\neq 1$, it is easy to see that $(e_{1},e_{2},e_{3},e_{4})$ is a basis for the Lie algebra ${\mathfrak{g}}$, $C_{23}^{2}=C_{23}^{4}=0$; and, moreover, $C_{23}^{1}=0$ only for $\alpha=0$. Thus, the last claim of Theorem 2 is valid for $e_{1}$ and $e_{2}$.

Put $e_{1}=e$ and $e_{2}=f$. By the conditions on $e_{1}$, the latter has a form $e_{1}=xE_{1}+a_{1}E_{2}+a_{2}E_{3}$, where $a_{1}a_{2}\neq 0$. We will show that after some change of $e_{2}$ by $e_{2}^{\prime}=e_{2}+ye_{1}$, $y\in{𝕉}$, there exists an automorphism $\xi$ of ${\mathfrak{g}}$ such that $\xi(E_{2}+E_{3})=e_{1}$ and $\xi(E_{4})=e^{\prime}_{2}$.

Let $-1\leq\alpha<1$, $\alpha\neq 0$, and $e_{2}=a_{5}E_{1}+a_{4}E_{2}+\alpha a_{3}E_{3}+E_{4}$. By Table 2, the corresponding components of $e_{2}^{\prime}$ are equal to $a_{4}^{\prime}=a_{4}+ya_{1}$ and $a_{3}^{\prime}=a_{3}+ya_{2}/\alpha$, while

$$ x=a_{2}a_{4}^{\prime}-a_{1}a^{\prime}_{3}=a_{2}a_{4}-a_{1}a_{3}+a_{1}a_{2}(1-1/\alpha)y. $$

Since $a_{1}a_{2}\neq 0$, there exists exactly one solution $y$ of this equation.

Assume that $\alpha=0$ and $b_{3}$ is the appropriate component of $e_{2}$ for $E_{3}$. Since $a_{2}\neq 0$, there exists a unique $y\in{𝕉}$ such that $ya_{2}+b_{3}=0$. Then we can suppose that $e^{\prime}_{2}=e_{2}+ye_{1}$ has components as in the last column of the automorphism matrix in Table 2. There exists a unique $a_{3}$ such that $x=a_{3}+a_{2}a_{4}$. Then $\xi(E_{2}+E_{3})=e_{1}$ and $\xi(E_{4})=e^{\prime}_{2}$ for an automorphism $\xi$ of ${\mathfrak{g}}$ with a matrix from Table 2. Hence, since $[e_{1},e_{2}^{\prime}]=[e_{1},e_{2}]$, the last claim of Theorem 2 is valid for an appropriate basis $(e_{1},e^{\prime}_{2})$ for ${\mathfrak{p}}$ of the indicated form.  ☐

Proposition 4

The real Lie algebra ${\mathfrak{g}}={\mathfrak{g}}_{4,10}$ has two-dimensional generating subspaces. A two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ if and only if ${\mathfrak{p}}$ lies in no three-dimensional ideal of ${\mathfrak{g}}$ and intersects the (unique) two-dimensional ideal of ${\mathfrak{g}}$ by zero. Every two of the two-dimensional generating subspaces of the Lie algebra ${\mathfrak{g}}$ can be translated to each other by an automorphism of ${\mathfrak{g}}$. Moreover, $C^{1}_{23}=C^{2}_{23}=0$ for the basis $(e_{1},e_{2},e_{3},e_{4})$ from Lemma $1$.

Proof

It follows from Table 1 that the Lie algebra ${\mathfrak{g}}={\mathfrak{g}}_{4,10}$ has two three-dimensional ideals ${\mathfrak{I}}_{1}=\operatorname{span}(E_{1},E_{2},E_{4})$ and ${\mathfrak{I}}_{2}=\operatorname{span}(E_{1},E_{2},E_{3})$. Moreover, ${\mathfrak{J}}={\mathfrak{I}}_{1}\cap{\mathfrak{I}}_{2}=\operatorname{span}(E_{1},E_{2})$ is the two-dimensional commutative ideal of ${\mathfrak{g}}$.

Assume that ${\mathfrak{p}}\subset{\mathfrak{g}}$ generates ${\mathfrak{g}}$ and $(e,f)$ is a basis for ${\mathfrak{p}}$. It is obvious that at least one of the vectors $e$ and $f$ does not belong to ${\mathfrak{I}}_{1}$; otherwise the subspace ${\mathfrak{p}}\subset{\mathfrak{I}}_{1}$ does not generate ${\mathfrak{g}}$. Since one vector of this basis could be summed with an arbitrary vector collinear with another vector of the basis, we can suppose that $f\notin{\mathfrak{I}}_{1}$ and $e\in{\mathfrak{I}}_{1}$. Moreover, $e\notin{\mathfrak{J}}$; in the opposite case ${\mathfrak{p}}$ is a part of some three-dimensional subalgebra and so ${\mathfrak{p}}$ does not generate ${\mathfrak{g}}$. Without loss of generality we can suppose that the component of $e$ (respectively, $f$) for $E_{4}$ (respectively, $E_{3}$) in the basis $(E_{1},E_{2},E_{3},E_{4})$ is equal to 1. If $x$ is the component of $f$ for $E_{4}$ then $f^{\prime}=f-xe\in{\mathfrak{I}}_{2}$.

At first consider the special case $(e_{1},e_{2})=(E_{4},E_{1}+E_{3})$ of bases from Proposition 4. It follows from Table 1 that

$$ e_{3}=[e_{1},e_{2}]=E_{2},\quad e_{4}=[e_{1},e_{3}]=-E_{1},\quad[e_{2},e_{3}]=-e_{3}. $$

Since $\alpha\neq 1$, it is easy to see that $(e_{1},e_{2},e_{3},e_{4})$ is a basis for the Lie algebra ${\mathfrak{g}}$ and $C_{23}^{4}=0$. Thus, Lemma 1 is true for $e_{1}$ and $e_{2}$ and, moreover $C_{23}^{1}=C_{23}^{2}=0$.

Put $e_{1}=e$ and $e_{2}=f^{\prime}$. We can consider that the vector $e_{1}$ has components as in the last column of an automorphism matrix in Table 3 and $\sigma=1$. There exists a unique pair of reals $(a_{1},a_{2})\in{𝕉}^{2}$ such that the components of $e_{2}$ are equal to the sum of the corresponding components of the first and the third columns of an automorphism matrix in Table 3. Moreover, $a_{1}^{2}+a_{2}^{2}\neq 0$; in the opposite case $[e_{1},e_{2}]=0$. Therefore there exists an automorphism $\xi$ of ${\mathfrak{g}}$ such that $\xi(E_{4})=e_{1}$ and $\xi(E_{1}+E_{3})=e_{2}$.

Thus, since $[e_{1},e_{2}^{\prime}]=[e_{1},e_{2}]$, for an appropriate basis $(e_{1},e_{2})$ for ${\mathfrak{p}}$, Lemma 1 is true; and, moreover, $C^{1}_{23}=C^{2}_{23}=0$.  ☐

4. The Main Results

It is indicated in [1] that the shortest curves, parametrized by arclength, of a left-invariant sub-Finsler metric $d$ on each connected Lie group $G$ defined by a left-invariant bracket generating distribution $D$ and a norm $F$ on $D(e)$ coincide with solutions to the time-optimal control problem for the system

$$ \dot{g}(t)=dl_{g(t)}(u(t)),\quad u(t)\in U, $$

(7)

with a measurable control $u=u(t)$. Here $l_{g}(h)=gh$, the control domain is the unit ball $U=\{u\in D(e)\mid F(u)\leq 1\}$, while $D$ is bracket generating if and only if the corresponding subspace ${\mathfrak{p}}:=D(e)\subset{\mathfrak{g}}$ satisfies the hypotheses of Lemma 1. It is clear that every shortest curve $g(t)$, $0\leq t\leq a$, parametrized by arclength, satisfies (7) and $F(u(t))=1$ for almost all $t\in[0,a]$.

Moreover, the statements about shortest curves are true also for every pair $(D(e),F)$ with a seminorm $F$ such that $F(u)>0$ for $u\neq 0$ defining a left-invariant sub-Finsler quasimetric $d$ on $G$.

By the PMP [2], for the time-optimality of a control $u(t)$ and the corresponding trajectory $g(t)$, $t\in[0,a]$, it is necessary the existence of a nowhere zero absolutely continuous covector-function $\psi(t)\in T^{\ast}_{g(t)}G$ such that for almost all $t\in[0,a]$ the function $\mathcal{H}(g(t);\psi(t);u)=\psi(t)(dl_{g(t)}(u))$ of $u\in U$ attains the maximum at the point $u=u(t)$; i.e.,

$$ M(t)=\psi(t)(dl_{g(t)}(u(t)))=\max\limits_{u\in U}\psi(t)(dl_{g(t)}(u)); $$

(8)

and there is fulfilled an analog of the Hamilton–Jacobi system of ODEs. Moreover, the function $M(t)$, $t\in[0,a]$, is constant and nonnegative; i.e., $M(t)\equiv M\geq 0$.

By an extremal we will mean a parametrized curve $g(t)$ in $G$ with a maximally admissible connected domain $\Omega\subset{𝕉}$ which satisfies almost everywhere on the maximal subset in $\Omega$ the PMP and conditions (7), $F(u(t))=1$ with a measurable function $u(t)$. In the case $M=0$ (respectively, $M>0$) an extremal is called abnormal (respectively, normal). In the normal case, proportionally changing $\psi=\psi(t)$, $t\in 𝕉$, if need be, we can assume that $M=1$.

Proposition 5

Each abnormal extremal in $(G,d)$ is one of the two one-parameter subgroups

$$ g(t)=\exp\left(\frac{ste_{2}}{F(se_{2})}\right),\quad t\in{𝕉},\ s=\pm 1, $$

(9)

or its left shift on $(G,d)$.

Proof

We can consider (each value of) the covector function $\psi(t)\in T^{\ast}_{g(t)}G$ from the PMP as a left-invariant $1$-form on $(G,\cdot)$ and so naturally identify the latter with a covector function $\psi(t)\in{\mathfrak{g}}^{\ast}=T^{\ast}_{e}G$.

In [7] for an extremal $g(t)\in G$, the following relations are proved which are fulfilled for almost all $t$ in the domain:

$$ \dot{g}(t)=dl_{g(t)}(u(t)),\quad(\psi(t)(v))^{\prime}=\psi(t)([u(t),v]),\ u(t),v\in{\mathfrak{g}},\quad F(u(t))=1. $$

(10)

Omitting for brevity the parameter $t$, we can write the last equation in (10) as $\psi^{\prime}(v)=\psi([u,v])$. In particular, for $\psi_{i}:=\psi(e_{i})$, $i=1,2,3,4$, we have

$$ \psi_{i}^{\prime}=\psi([u,e_{i}]). $$

(11)

Let $u=u_{1}e_{1}+u_{2}e_{2}\in U$. We get from (11), (1), and Proposition 1:

$$ \psi_{1}^{\prime}=\psi(-u_{2}e_{3})=-u_{2}\psi_{3}, $$

(12)

$$ \psi_{2}^{\prime}=\psi(u_{1}e_{3})=u_{1}\psi_{3}, $$

(13)

$$ \psi_{3}^{\prime}=\psi(u_{1}e_{4}+u_{2}[e_{2},e_{3}])=u_{1}\psi_{4}+u_{2}\sum\limits_{k=1}^{3}C_{23}^{k}\psi_{k}, $$

(14)

$$ \psi_{4}^{\prime}=\psi(u_{1}[e_{1},e_{4}]+u_{2}[e_{2},e_{4}])=u_{1}\sum\limits_{k=1}^{4}C_{14}^{k}\psi_{k}+u_{2}\sum\limits_{k=3}^{4}C_{24}^{k}\psi_{k}. $$

(15)

In the abnormal case it must clearly be $\psi_{1}=\psi_{2}\equiv 0$. Then (12) and (13) imply that $\psi_{3}\equiv 0$. It follows from $\psi_{4}\neq 0$, (14), (1), and the equality $F(u)=1$ that

$$ u_{1}\equiv 0,\quad u_{2}\equiv\frac{s}{F(se_{2})},\quad s=\pm 1. $$

(16)

It follows from (16), Proposition 1, and (15) that

$$ \psi_{4}(t)=\varphi_{4}\exp\left(\frac{C_{24}^{4}st}{F(se_{2})}\right)=\varphi_{4}\exp\left(\frac{C_{23}^{3}st}{F(se_{2})}\right),\quad s=\pm 1, $$

(17)

is a solution to equation (15) with the initial condition $\psi_{4}(0)=\varphi_{4}\neq 0$. Obviously, it is possible to find $u(t)$ and $\psi(t)$ by the above formulas for all $t\in{𝕉}$.

Now Proposition 5 is a direct corollary of (16) and the first equation in (10).  ☐

In view of Proposition 1 and (16), every system (12)–(15) fulfilled for almost all $t\in{𝕉}$ has the form

$$ \psi_{1}^{\prime}=-u_{2}\psi_{3},\ \psi_{2}^{\prime}=0,\ \psi_{3}^{\prime}=u_{2}\sum\limits_{k=1}^{3}C_{23}^{k}\psi_{k},\ \psi_{4}^{\prime}=u_{2}\bigl{(}C_{23}^{2}\psi_{3}+C_{23}^{3}\psi_{4}\bigr{)},\ u_{2}\equiv{s\over F(se_{2})}. $$

(18)

Then (18) holds for all $t\in 𝕉$ and (18) is a system of linear ordinary differential equations with constant coefficients, and so all solutions of (18) are real-analytic.

Below $F(u_{1},u_{2}):=F(u)$, while $F_{U}$ is the Minkowski support function of $U$:

$$ F_{U}(x,y)=\max\limits_{(u_{1},u_{2})\in U}(xu_{1}+yu_{2}). $$

(19)

The following theorem is valid:

Theorem 3

Abnormal extremal (9) is nonstrictly abnormal if and only if for the basis $(e_{1},e_{2},e_{3},e_{4})$ from Lemma $1$, either $C_{23}^{1}=C_{23}^{2}=0$, or $C_{23}^{1}\neq 0$ and $F_{U}\left(0,s\right)=1/F(0,s)$, $s=\pm 1$.

Proof. Necessity

Assume that (9) is nonstrictly abnormal. Then by the above there exists a nowhere vanishing real-analytic covector-function $\psi(t)$ which is a solution of system (18), and $\psi(t)(u(t))=F_{U}(\psi_{1}(t),\psi_{2}(t))\equiv 1$ for almost all $t\in{𝕉}$. This and (16) imply that

$$ \psi_{2}(t)\equiv 1/u_{2},\quad F_{U}\left(\psi_{1}(t),1/u_{2}\right)\equiv 1, $$

(20)

and the points $(\psi_{1}(t),1/u_{2})$, $t\in{𝕉}$, are dual for the point $(0,u_{2})$. Therefore, the range of $\psi_{1}(t)$, $t\in{𝕉}$, is a segment (degenerating to a point if $F$ is differentiable at $(0,u_{2})$) because the body $U^{\ast}$, dual to $U$, is convex and bounded.

System (18) implies the ordinary differential equation

$$ \psi_{1}^{\prime\prime}-u_{2}C_{23}^{3}\psi_{1}^{\prime}+u_{2}^{2}C_{23}^{1}\psi_{1}+u_{2}C_{23}^{2}=0. $$

(21)

Assume that $C_{23}^{1}\neq 0$. By Lemma 1, we can suppose that $C_{23}^{2}=0$. Let us set $B:=\left(C_{23}^{3}\right)^{2}-4C_{23}^{1}$. Taking into account the theory in [8], it is easy to see that general solution of (21) has the form

$$ \psi_{1}(t)=\begin{cases}A_{1}e^{\lambda_{1}t}+A_{2}e^{\lambda_{2}t},\ \lambda_{1,2}=u_{2}\bigl{(}C_{23}^{3}\pm\sqrt{B}\bigr{)}/2,&\text{if }B>0,\\ (A_{1}t+A_{2})e^{\frac{1}{2}C_{23}^{3}u_{2}t},&\text{if }B=0,\\ e^{\frac{1}{2}C_{23}^{3}u_{2}t}\bigl{(}A_{1}\cos\frac{u_{2}\sqrt{-B}t}{2}+A_{2}\sin\frac{u_{2}\sqrt{-B}t}{2}\bigr{)},&\text{if }B<0,\end{cases} $$

where $A_{1}$ and $A_{2}$ are arbitrary reals. The function $\psi_{1}(t)$, $t\in{𝕉}$, is bounded; therefore, either $\psi_{1}(t)\equiv 0$ or

$$ \psi_{1}(t)=A_{1}\cos\frac{u_{2}\sqrt{-B}t}{2}+A_{2}\sin\frac{u_{2}\sqrt{-B}t}{2},\quad C_{23}^{1}<0,\ C_{23}^{3}=0. $$

(22)

Notice that for all $A_{1},A_{2}\in{𝕉}$ the range of $\psi_{1}(t)$, defined by (22), includes $0$. Thus, in consequence of (20), $F_{U}\left(0,1/u_{2}\right)=1$ which is equivalent to the corresponding equality in Theorem 3.

Assume that $C_{23}^{1}=0$. By the theory in [8], the general solution of (21) has the form

$$ \psi_{1}(t)=\begin{cases}A_{1}e^{C_{23}^{3}u_{2}t}+C_{23}^{2}t/C_{23}^{3}+A_{2},&\text{if }C_{23}^{3}\neq 0,\\ -\frac{1}{2}C_{23}^{2}u_{2}t^{2}+A_{1}t+A_{2},&\text{if }C_{23}^{3}=0,\end{cases} $$

where $A_{1}$ and $A_{2}$ are arbitrary reals. Since $\psi_{1}(t)$, $t\in{𝕉}$, is bounded, this implies that $C_{23}^{2}=0$ and $A_{1}=0$.

Sufficiency

Since $U$ is a bounded convex set in ${𝕉}^{2}$, $0\in\operatorname{Int}(U)$, and $F(0,u_{2})=1$, where $u_{2}$ is defined in (16) for $s=1$ or $s=-1$, there exists at least one $k\in{𝕉}$ such that $F_{U}(k,1/u_{2})=1$.

If $C_{23}^{1}=C_{23}^{2}=0$, then we put

$$ \psi_{1}(t)\equiv k,\quad\psi_{2}(t)\equiv 1/u_{2},\quad\psi_{3}(t)\equiv 0,\quad\psi_{4}(t)=e^{C_{23}^{3}u_{2}t}. $$

(23)

It is easy to see that $\psi_{i}(t)$, $i=1,\dots,4$, satisfy system (18) and, moreover,

$$ F_{U}(\psi_{1}(t),\psi_{2}(t))\equiv 1. $$

(24)

Hence, abnormal extremal (9) satisfies the PMP with $M(t)\equiv 1$ (see (8)), and so (9) is nonstrictly abnormal.

If $C_{23}^{1}\neq 0$ then according to Lemma 1 we can suppose that $C_{23}^{2}=0$. Let us define the functions $\psi_{i}(t)$, $i=1,\dots,4$, by formulas (23), setting $k=0$. It is easy to see that $\psi_{i}(t)$, $i=1,\dots,4$, satisfy system (18). Furthermore, the conditions $F_{U}(0,s)=1/F(0,s)$, $s=\pm 1$, and (16) imply (23). Thus, (9) satisfies the PMP with $M(t)\equiv 1$ (see (8)) and hence (9) is nonstrictly abnormal.

Theorem 3 is proved.  ☐

Remark 1

If $d$ is a left-invariant sub-Riemannian metric on a four-dimensional connected Lie group $G$ with the Lie algebra ${\mathfrak{g}}$ defined by an inner product $(\cdot,\cdot)$ on a two-dimensional subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$, $C^{1}_{23}\neq 0$, and $(e_{1},e_{2},e_{3},e_{4})$ is a basis for ${\mathfrak{p}}$ from Lemma 1, then the conditions $F_{U}(0,s)=1/F(0,s)$ are equivalent to the equality $(e_{1},e_{2})=0$.

5. Abnormal Extremals in the Case $\dim({\mathfrak{p}})=3$

Proposition 6

A four-dimensional connected Lie group $G$ with Lie algebra ${\mathfrak{g}}$ and a three-dimensional generating subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ has abnormal extremals (for an arbitrary left-invariant quasimetric $d$ on $G$ defined by a seminorm $F$ on ${\mathfrak{p}}$) if and only if ${\mathfrak{p}}_{1}={\mathfrak{p}}\cap{\mathfrak{N}}({\mathfrak{p}})\neq\{0\}$, where ${\mathfrak{N}}({\mathfrak{p}})$ is the normalizer of ${\mathfrak{p}}$ in ${\mathfrak{g}}$. Furthermore, $\dim({\mathfrak{p}}_{1})=1$ and each one-parameter subgroup $g=g(t)=\exp(tX)$, where $X\in{\mathfrak{p}}_{1}$, $F(X)=1$, is an abnormal extremal for $(G,d)$; there is no other abnormal extremal with origin $e\in G$. Moreover, the extremal $g$ is strictly abnormal (nonstrictly abnormal) for every quasimetric $d$ if and only if ${\mathfrak{p}}_{1}\subset[{\mathfrak{p}}_{1},{\mathfrak{p}}]$ (respectively, ${\mathfrak{p}}_{1}={\mathfrak{p}}\cap{\mathfrak{C}}({\mathfrak{p}})$, where ${\mathfrak{C}}({\mathfrak{p}})$ is the centralizer of ${\mathfrak{p}}$ in ${\mathfrak{g}})$.

Proof

Let we have an abnormal extremal $g(t)$, $t\in I$, where $I\subset{𝕉}$ is some nonempty connected open subset. Note that $g(t)$ is defined by conditions (10) for $u(t)\in U$, $0\neq\psi(t)\in{\mathfrak{g}}^{\ast}$, $t\in I$, where $u(t)$ and $\psi(t)$ are respectively measurable and absolutely continuous functions such that $\psi(t)|{\mathfrak{p}}\equiv 0$, $t\in I$. Consequently, for almost all $t\in I$, $\operatorname{ad}u(t)({\mathfrak{p}})\subset{\mathfrak{p}}$, i.e. $u(t)\in{\mathfrak{p}}_{1}$ taken from the claim of Proposition 6. Since $\dim({\mathfrak{p}})=3$, ${\mathfrak{p}}$ generates ${\mathfrak{g}}$, and $F(u(t))=1$, it is clear that $\dim({\mathfrak{p}}_{1})=1$, $u(t)$ is defined uniquely for these $t$, and we may suppose that $u(t)\equiv X$ for some $0\neq X\in{\mathfrak{p}}_{1}$.

Assume now that ${\mathfrak{p}}_{1}\neq\{0\}$, and so $\dim({\mathfrak{p}}_{1})=1$. Let $X\in{\mathfrak{p}}_{1}$, $F(X)=1$ and $\psi_{0}|{\mathfrak{p}}\equiv 0$ for some nonzero covector $\psi_{0}\in{\mathfrak{g}}^{\ast}$. It is evident that if $u(t)\equiv X$ then there exist unique solutions $g(t)=\exp(tX)$, $\psi(t)$, $t\in{𝕉}$, of system (10) with the initial conditions $g(0)=e$ and $\psi(0)=\psi_{0}$. Moreover, $\psi(t)|{\mathfrak{p}}\equiv 0$, $t\in{𝕉}$, because $X\in{\mathfrak{p}}_{1}$. Furthermore, $\psi(t)\equiv\psi_{0}$ if $\operatorname{ad}X({\mathfrak{g}})\subset{\mathfrak{p}}$. Consequently, $g(t)$, $t\in{𝕉}$, is an abnormal extremal with respect to the covector function $\psi(t)$, $t\in{𝕉}$.

Let us suppose that the above extremal $g(t)$ is nonstrictly abnormal for $(G,d)$; i.e., for some function $\psi(t)\in{\mathfrak{g}}^{\ast}$ we have $1\equiv\psi(t)(X)=\max\nolimits_{u\in U}\psi(t)(u)$ and (10) is fulfilled. Then $\psi(t)(v)=0$ for all $t\in{𝕉}$ and all $v\in[{\mathfrak{p}}_{1},{\mathfrak{p}}]$. This may happen only in the two cases: (1) $[{\mathfrak{p}}_{1},{\mathfrak{p}}]=\{0\}$, i.e. ${\mathfrak{p}}_{1}={\mathfrak{p}}\cap{\mathfrak{C}}({\mathfrak{p}})$; (2) $[{\mathfrak{p}}_{1},{\mathfrak{p}}]\neq\{0\}$, but this set is parallel in ${\mathfrak{p}}$ to the set $W=\{w\in{\mathfrak{p}}:\psi(0)(w)=1\}$, which is one of the supporting planes to $U$ at a boundary point $X$ of $U$, and so $X\notin\operatorname{ad}X({\mathfrak{p}})$. It is clear from here that $g(t)$, $t\in{𝕉}$, is a nonstrictly abnormal extremal for all $d$ in case 1), and only for some $d$ if $[{\mathfrak{p}}_{1},{\mathfrak{p}}]\neq\{0\}$; and $g$ is strictly abnormal extremal for all $d$ if and only if ${\mathfrak{p}}_{1}\subset[{\mathfrak{p}}_{1},{\mathfrak{p}}]$.  ☐

Remark 2

Proposition 6 is valid for a connected Lie group $G$ of dimension $n\geq 5$ and a subspace ${\mathfrak{p}}\subset{\mathfrak{g}}$ of codimension 1 with possible violation of the equality $d=\dim({\mathfrak{p}}_{1})=1$, if ${\mathfrak{p}}_{1}\neq\{0\}$ (we can guarantee only the inequalities $1\leq d\leq n-3$).

6. Examples

The next theorem is a compendium of Propositions 2–5 and Theorems 1–3.

Theorem 4

Let $G$ be a four-dimensional connected Lie group with Lie algebra $({\mathfrak{g}},[\cdot,\cdot])$, let ${\mathfrak{p}}\subset{\mathfrak{g}}$ be a two-dimensional subspace generating ${\mathfrak{g}}$ by the operation $[\cdot,\cdot]$, and let $d$ be an arbitrary left-invariant quasimetric on $G$ defined by some seminorm $F$ on ${\mathfrak{p}}$. Then

1. Every abnormal extremal in $(G,d)$ is nonstrictly abnormal in each of the following cases: (1) ${\mathfrak{g}}={\mathfrak{g}}^{0}_{4,8}$; (2) ${\mathfrak{g}}={\mathfrak{g}}_{4,10}$; (3) ${\mathfrak{g}}$ is decomposable and has a solvable indecomposable three-dimensional subalgebra; (4) ${\mathfrak{g}}$ is indecomposable and has a three-dimensional commutative ideal.

2. Let ${\mathfrak{g}}$ be one of the Lie algebras ${\mathfrak{g}}_{3,7}\oplus{\mathfrak{g}}_{1}$, ${\mathfrak{g}}_{4,7}$, ${\mathfrak{g}}^{\alpha}_{4,8}$, $-1\leq\alpha<1$, $\alpha\neq 0$, ${\mathfrak{g}}^{\alpha}_{4,9}$, $\alpha\geq 0$, ${\mathfrak{g}}={\mathfrak{g}}_{3,6}\oplus{\mathfrak{g}}_{1}$ and the restriction of the Killing form $k_{{\mathfrak{g}_{3,6}}}$ to the projection of the subspace ${\mathfrak{p}}$ onto ${\mathfrak{g}}_{3,6}$ is negative definite. Then abnormal extremal (9) (and its every left shift) in $(G,d)$ is nonstrictly abnormal if and only if $F_{U}(0,s)=1/F(0,s)$, $s=\pm 1$.

3. If ${\mathfrak{g}}={\mathfrak{g}}_{3,6}\oplus{\mathfrak{g}}_{1}$ and the restriction of the Killing form $k_{{\mathfrak{g}_{3,6}}}$ to the projection of ${\mathfrak{p}}$ onto ${\mathfrak{g}}_{3,6}$ is nondegenerate and with alternating signs, then every abnormal extremal in $(G,d)$ is strictly abnormal.

Example 1

Each abnormal extremal on the Engel group with nilpotent Lie algebra ${\mathfrak{g}}_{4,1}={\mathfrak{eng}}$ as well as each left-invariant quasimetric $d$ defined by some seminorm $F$ on ${\mathfrak{p}}=\operatorname{span}(E_{1},E_{3},E_{4})\subset{\mathfrak{eng}}$, is one of the two 1-parameter subgroups $g(t)=\exp(tsE_{1}/F(sE_{1}))$, $s=\pm 1$, or its left shift. In consequence of Proposition 6 and the centrality of $E_{1}$ these extremals are nonstrictly abnormal.

Example 2

Each abnormal extremal on the (solvable) connected Lie group $G$ with Lie algebra ${\mathfrak{g}}={\mathfrak{g}}_{4,3}$ as well as each left-invariant quasimetric $d$ defined by some seminorm $F$ on ${\mathfrak{p}}=\operatorname{span}(E_{1},E_{3},E_{4})\subset{\mathfrak{g}}$ is one of two 1-parameter subgroups $g(t)=\exp(tsE_{1}/F(sE_{1}))$, $s=\pm 1$, or its left shift. In consequence of Proposition 6 and the equality $[E_{1},E_{4}]=E_{1}$ these extremals are strictly abnormal.

The interesting partial case of the last example is $(G,d_{1})$ with left-invariant sub-Riemannian metric $d_{1}$ defined by the inner product $(\cdot,\cdot)$ on ${\mathfrak{p}}$ with the orthonormal basis $(E_{1},E_{3},E_{4})$.

Proposition 7

Strictly abnormal extremals $g(t)=\exp(stE_{1})$, $s=\pm 1$, in $(G,d_{1})$ are not geodesics; i.e., none of their segments $g(t)$, $t\in[t_{0},t_{1}]$, where $t_{0}<t_{1}$, is shortest.

Proof

The proposition is true because of the equality ${\mathfrak{p}}+[{\mathfrak{p}},{\mathfrak{p}}]={\mathfrak{g}}$ and the so-called Goh optimality condition for abnormal geodesics (see [13]) which implies that every geodesic in $(G,d_{1})$ parametrized by arclength is normal. We will give an independent geometric proof in this case, reducing the matter to the question about the geodesics of a left-invariant Riemannian metric on a simply connected noncompact noncommutative two-dimensional Lie group.

The subgroup $N=\{\exp(tE_{2}),t\in{𝕉}\}\subset G$ is a central subgroup of the isometry group of $(G,d_{1})$, consisting of the left shifts $l_{g}$, $g\in G$. Therefore, the canonical projection $p_{3}:(G,d_{1})\rightarrow(G_{3}=G/N,d_{3})$ onto $G_{3}$ with the left-invariant Riemannian metric $d_{3}$ defined by the inner product $(\cdot,\cdot)$ with the orthonormal basis $(E_{1},E_{3},E_{4})$ for the Lie algebra ${\mathfrak{g}}_{3}$ of this group with the relation $[E_{1},E_{4}]=E_{1}$ and the central element $E_{3}$, is a submetry [14]. Notice that $G_{3}\cong G_{2}\times H$, where $G_{2}$ is a two-dimensional noncommutative Lie group, while $H=\{\exp(tE_{3}),t\in{𝕉}\}$ is a central subgroup of $G_{3}$. Thus, the canonical projection $p_{2}:(G_{3},d_{3})\rightarrow(G_{2}=G_{3}/H,d_{2})$ onto $G_{2}$ with the left-invariant Riemannian metric $d_{2}$ defined by the inner product $(\cdot,\cdot)$ with the orthonormal basis $(E_{1},E_{4})$ for the Lie algebra ${\mathfrak{g}}_{2}$ of this Lie group with the relation $[E_{1},E_{4}]=E_{1}$, is a submetry, moreover, a Riemannian submersion [14]. Consequently, the composition $p=(p_{2}\circ p_{3}):(G,d_{1})\rightarrow(G_{2},d_{2})$ is a submetry and so $g_{2}(t)=p(g(t))=\exp(tE_{1})\in G_{2}$, $t\in{𝕉}$, is geodesic if $g(t)$, $t\in{𝕉}$, is geodesic. But this is impossible because $(G_{2},d_{2})$ is isometric to the Lobachevsky plane with constant sectional curvature $-1$, and in the well-known conformal model of the plane in the upper semiplane, $g_{2}(t)$, $t\in{𝕉}$, is a horizontal straight line which is not geodesic [11, 7]. The so-obtained contradiction proves Proposition 7.  ☐

Example 3 [3, Section 9.5]

Let us consider the Lie group $G=\operatorname{SO(3)}\times{𝕉}$ with the Lie algebra ${\mathfrak{g}}_{3,7}\oplus{\mathfrak{g}}_{1}$ and the left-invariant sub-Riemannian metric $d$ defined by the inner product $(\cdot,\cdot)$ on the subspace ${\mathfrak{p}}=\operatorname{span}(f,g)$ with the orthonormal basis $(f,g)$, where $f=E_{1}+E_{4}$ and $g=E_{1}+E_{2}+2E_{4}$. It is stated that the one-parameter subgroups $g_{s}(t)=\exp(stg)$, $t\in{𝕉}$, $s=\pm 1$, are unique (strictly) abnormal geodesics parametrized by arclength with origin $e$.

Let us compare this with our results. It is clear that $e_{1}=f$, $e_{2}=g$,

$$ e_{3}=[e_{1},e_{2}]=E_{3},\quad e_{4}=[e_{1},e_{3}]=-E_{2},\quad[e_{2},e_{3}]=E_{1}-E_{2}=2e_{1}-e_{2}. $$

Thus, $C_{23}^{4}=0$ and $C_{23}^{1}=2\neq 0$, but $C_{23}^{2}=-1\neq 0$. In consequence of Proposition 5, $g_{s}$, $s=\pm 1$, are unique abnormal extremals with origin $e$. Following Lemma 1, in order that $C^{2}_{23}$ become zero, we change $e_{1}$ by

$$ e^{\prime}_{1}=e_{1}+\bigl{(}C^{2}_{23}/C^{1}_{23}\bigr{)}e_{2}=e_{1}-e_{2}/2=E_{1}+E_{4}-(E_{1}+E_{2}+2E_{4})/2=(E_{1}-E_{2})/2. $$

Since $(e^{\prime}_{1},e_{2})\neq 0$, it follows from claim 2 of Theorem 4 and Remark 1 that abnormal extremals $g_{s}$, $s=\pm 1$, are strictly abnormal.