Geodesic vectors and flat totally geodesic subalgebras of six-dimensional filiform metric Lie algebras

Abstract

A nilpotent Lie algebra $\mathfrak {n}$ equipped with an Euclidean inner product is called nilpotent metric Lie algebra. In this paper we describe the sets of the geodesic vectors and the flat totally geodesic subalgebras of the six-dimensional filiform metric Lie algebras. In this class with the exception of the metric Lie algebras corresponding to the standard filiform Lie algebra the flat totally geodesic subalgebras of every metric Lie algebra have dimension at most two.

1 Introduction

Let $\mathfrak {n}$ be a real nilpotent Lie algebra and N be the connected and simply connected Lie group having Lie algebra $\mathfrak {n}$. We call $(\mathfrak {n}, \langle .,.\rangle )$a metric nilpotent Lie algebra if it is given an Euclidean inner product $\langle .,.\rangle$on $\mathfrak {n}$. An inner product $\langle .,.\rangle$ on $\mathfrak {n}$ determines a left-invariant metric $\langle .,.\rangle _N$ on N and conversely. The corresponding nilpotent Lie group N endowed with the left-invariant metric $\langle .,.\rangle _N$ arising from $\langle .,.\rangle$ is a Riemannian nilmanifold.

Riemannian nilmanifolds have been studied extensively in the last decades. The first general studies for 2-step nilmanifolds were done by P. Eberlein (see [3, 4]).

We call a subalgebra $\mathfrak { h}$ of a metric Lie algebra $(\mathfrak {n}, \langle .,.\rangle )$ flat, respectively totally geodesic if its exponential image H in the Lie group N with the left invariant Riemann metric $\langle .,.\rangle _N$ is flat, respectively totally geodesic submanifold. A subalgebra $\mathfrak { h}$ of a metric Lie algebra $(\mathfrak {n}, \langle .,. \rangle )$ is totally geodesic if and only if it satisfies

$$\begin{aligned} \left\langle [X,Y],Z \right\rangle + \left\langle [X,Z],Y \right\rangle =0, \end{aligned}$$

(1.1)

for all $Y,Z \in {\mathfrak { h}}$ and X in the orthogonal complement $\mathfrak { h}^\perp$ of ${\mathfrak { h}}$ (cf. Lemma 1.2 in [1]). Moreover, a non-zero vector $Y \in (\mathfrak {n}, \langle .,. \rangle )$ is geodesic precisely if for all $X \in (\mathfrak {n}, \langle .,. \rangle )$ one has

$$\begin{aligned} \left\langle [X,Y],Y \right\rangle =0. \end{aligned}$$

(1.2)

Cairns, Hinić Galić and Nikolayevsky presented a comprehensive study of totally geodesic subgroups of Riemannian nilmanifolds and the corresponding subalgebras of nilpotent metric Lie algebras. In particular the authors gave several results on the possible dimensions of totally geodesic subalgebras. They also found examples, where the obtained bounds on the dimensions of totally geodesic subalgebras are attained. Furthermore they gave an example of a 6-dimensional filiform nilpotent Lie algebra that has no totally geodesic subalgebra of dimension $> 2$, for any choice of inner product (see [1, 2]).

Nagy and Homolya in [8] studied geodesic vectors and flat totally geodesic subalgebras in 2-step nilpotent metric Lie algebras and proved that for isomorphic Lie algebras $\mathfrak {n}$ and $\mathfrak {n}^{*}$ there exists a bijective linear map $\mathfrak {n} \rightarrow \mathfrak {n}^{*}$ preserving the geodesic, respectively flat totally geodesic property of vectors, respectively subalgebras. Moreover they determined the sets of the geodesic vectors and the flat totally geodesic subalgebras in the 2-step nilpotent metric Lie algebras of dimension $\le 6$.

In [5] Figula and Nagy classified the isometry equivalence classes and determined the isometry groups of connected and simply connected Riemannian nilmanifolds on filiform Lie groups of arbitrary dimension and on five dimensional nilpotent Lie groups of nilpotency class $> 2$. Applying their approach Abbas and Figula proved that up to isometric isomorphism any six-dimensional filiform metric Lie algebra is of the form $\mathfrak {n}_{6,14}(\alpha _i, \beta _j)$, $\mathfrak {n}_{6,15}(\alpha _i, \beta _j)$, $\mathfrak {n}_{6,16}(\alpha _i, \beta _j)$, $\mathfrak {n}_{6,17}(\alpha _i, \beta _j)$ and $\mathfrak {n}_{6,18}(\alpha _i, \beta _j)$. These metric Lie algebras are defined by the following non-vanishing commutators (see [6]):

• (1) $\mathfrak {n}_{6,14}(\alpha _i,\beta _j )$

  $$\begin{aligned} \begin{aligned}&[E_1,E_2]=\alpha _1E_3+\beta _1 E_4+ \beta _2 E_5 +\beta _3 E_6,\\ {}&[E_1,E_3]=\alpha _2E_4- \frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4} E_5+\beta _4E_6, \\ {}&[E_1,E_4]=\frac{\alpha _1\alpha _5}{\alpha _4}E_5+\beta _5 E_6,&[E_2,E_3]&=\alpha _3 E_5+\beta _6 E_6, \\ {}&[E_2,E_4]=\beta _7 E_6,&[E_2,E_5]&=\alpha _4E_6,\\ {}&[E_4,E_3]=\alpha _5E_6, \end{aligned} \end{aligned}$$

  (1.3)

  such that $\alpha _i >0,i=1,\dots ,5$, $\beta _j \in \mathbb {R}, j=1,\dots ,7$, and if the set $J=\{ j \in \{1,4,7\}:\beta _j \ne 0 \}\ne \emptyset$, then $\beta _{j_\circ } >0$ for the minimal element $j_\circ \in J$,

• (2) $\mathfrak {n}_{6,15}(\alpha _i,\beta _j )$

  $$\begin{aligned} \begin{aligned}&[E_1,E_2]=\alpha _1E_3+\beta _1 E_4+ \beta _2 E_5 +\beta _3 E_6,&[E_1,E_3]&=\alpha _2E_4+\beta _4 E_5+\beta _5E_6,\\&[E_1,E_4]=\alpha _3E_5+\beta _6 E_6,&[E_2,E_3]&=\frac{\alpha _2\alpha _5}{\alpha _4}E_5+ \beta _7 E_6, \\&[E_1,E_5]=\alpha _4E_6,&[E_2,E_4]&=\alpha _5E_6, \end{aligned} \end{aligned}$$

  (1.4)

  such that $\alpha _i> 0,i=1,\dots ,5$, $\beta _j \in \mathbb {R}, j=1,\dots ,7$, and if the set $J=\{ j \in \{1,3,4,6,7\}:\beta _j \ne 0 \}\ne \emptyset$, then $\beta _{j_\circ } >0$ for the minimal element $j_\circ \in J$,

• (3) $\mathfrak {n}_{6,16}(\alpha _i,\beta _j )$

  $$\begin{aligned} \begin{aligned}&[E_1,E_2]=\alpha _1E_3+\beta _1 E_4+ \beta _2 E_5 +\beta _3 E_6,\\ {}&[E_1,E_3]=\alpha _2E_4-\left( \frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4} \right) E_5+\beta _4E_6,\\ {}&[E_1,E_4]=\alpha _3E_5+\beta _5 E_6,&[E_1,E_5]&=\beta _6 E_6, \\ {}&[E_2,E_3]=\beta _7E_6,&[E_2,E_4]&=\beta _8E_6, \\ {}&[E_2,E_5]=\alpha _4E_6,&[E_4,E_3]&=\frac{\alpha _3\alpha _4}{\alpha _1}E_6, \end{aligned} \end{aligned}$$

  (1.5)

  such that $\alpha _i > 0, i=1,\dots ,4$, $\beta _j \in \mathbb {R}, j=1,\dots ,8$, and one of the following cases is satisfied:

  • (a) $\beta _1=\beta _3=\beta _4=\beta _5=\beta _6=\beta _8=0$,

  • (b) $\beta _3>0$ or $\beta _5>0,$ $\beta _1=\beta _4=\beta _6=\beta _8=0$,

  • (c) $\beta _6>0$ or $\beta _4>0,$ $\beta _1=\beta _3=\beta _5=\beta _8=0$,

  • (d) $\beta _1>0$ or $\beta _8>0,$ $\beta _3=\beta _4=\beta _5=\beta _6=0$,

  • (e) at least two elements of the set $\{\beta _1, \beta _3, \beta _4, \beta _5, \beta _6, \beta _8 \}$ are positive with the exceptions $(\beta _1>0, \beta _8>0)$, $(\beta _3>0, \beta _5>0)$, $(\beta _4>0, \beta _6>0)$,

• (4) $\mathfrak {n}_{6,17} (\alpha _i,\beta _j )$

  $$\begin{aligned} \begin{aligned}&[E_1,E_2]=\alpha _1E_3+\beta _1 E_4+ \beta _2 E_5 +\beta _3 E_6,&[E_1,E_3]&=\alpha _2E_4+\beta _4 E_5+\beta _5E_6, \\&[E_1,E_4]=\alpha _3E_5+\beta _6 E_6,&[E_1,E_5]&=\alpha _4E_6,\\&[E_2,E_3]=\alpha _5E_6, \end{aligned} \end{aligned}$$

  (1.6)

  such that $\alpha _i > 0, i=1,\dots ,5$, $\beta _j \in \mathbb {R}, j=1,\dots ,6$ and if the set $J=\{ j \in \{1,3,4,6\}:\beta _j \ne 0 \}\ne \emptyset$, then $\beta _{j_\circ } >0$ for the minimal element $j_\circ \in J$. Moreover, the Lie algebra $\mathfrak {n}_{6,18} (\alpha _i,\beta _j )$ is defined by the same commutators with $\alpha _5=0$.

Our aim in this work is to determine the sets of the geodesic vectors and the flat totally geodesic subalgebras in the class $\mathcal {C}$ of the six-dimensional filiform metric Lie algebras. Our investigation shows that in $\mathcal {C}$ only metric Lie algebras corresponding to the standard filiform Lie algebra allow a flat totally geodesic subalgebra of dimension $> 2$ (cf. Theorems 3.2, 3.4, 3.8, 4.2, 4.3).

2 Preliminaries

For any Lie algebra $\mathfrak {n}$, the lower central series of $\mathfrak {n}$ is defined by $\mathfrak {n}^{(0)}:=\mathfrak {n}, \mathfrak {n}^{(1)}=[\mathfrak {n}, \mathfrak {n}]$ and $\mathfrak {n}^{(i+1)}=[\mathfrak {n}, \mathfrak {n}^{(i)}], \, i\ge 1$. If $\mathfrak {n}^{(k)}$ is trivial for some integer k, then the Lie algebra $\mathfrak {n}$ is called nilpotent. Let k be the smallest integer so that $\mathfrak {n}^{(k)}$ is trivial, then $\mathfrak {n}$ is said to be k-step nilpotent. An n-dimensional nilpotent Lie algebra is called filiform if it is $n-1$-step nilpotent. We say an n-dimensional nilpotent Lie algebra $\mathfrak {n}$ is $\mathbb N$-graded filiform, if it can be decomposed in a direct sum of one-dimensional subspaces $\mathfrak {n}=\oplus _{i=1}^n V_i$ with $[V_1, V_i] = V_{i+1}$ for all $i > 1$ and $[V_i, V_j ] \subset V_{i+j}$ for all $i, j \in \mathbb N$, where for convenience we set $V_i = 0$ for $i > n$.

Let $\mathbb E^6$ be a 6-dimensional Euclidean vector space with a distinguished orthonormal basis $\{E_1, E_2, E_3, E_4, E_5, E_6\}$. Let $(\mathfrak {n}, \langle .,.\rangle )$ be a filiform metric Lie algebra of dimension 6 defined on $\mathbb E^6$ by the commutators and relations described in cases (1.3), (1.4), (1.5), (1.6). The lower central series of the 6-dimensional filiform Lie algebra $\mathfrak {n}$ is given by $\mathfrak {n}^{(1)}=\text {span}(E_3, E_4, E_5, E_6)$, $\mathfrak {n}^{(2)}=\text {span}(E_4, E_5, E_6)$, $\mathfrak {n}^{(3)}=\text {span}(E_5, E_6)$, $\mathfrak {n}^{(4)}=\text {span}(E_6)$ which is the center $\zeta$ of $\mathfrak {n}$, and $\mathfrak {n}^{(5)}=\{0\}$. Denote by the orthogonal complement of $\mathfrak {n}^{(i)}$ in $\mathfrak {n}^{(i-1)}, \, i=1, 2, 3, 4$. Hence we have , , , , and .

According to [8, Lemmas 1, 2], we have the following:

Lemma 2.1

A subalgebra $\mathfrak { h}$ in a nilpotent metric Lie algebra $(\mathfrak {n}, \langle .,. \rangle )$ is flat if and only if it is abelian. A subalgebra ${\mathfrak { h}}$ in $(\mathfrak {n}, \langle .,. \rangle )$ is flat totally geodesic precisely if each non-zero element of ${\mathfrak { h}}$ is geodesic.

Lemma 2.2

Let $(\mathfrak {n}, \langle .,. \rangle )$ be a 6-dimensional filiform metric Lie algebra. Each non-zero vector in is geodesic. Any subalgebra contained in is flat totally geodesic.

Proof

If $Y \in \zeta$, then for every $X \in \mathfrak {n}$ we have $[X,Y]=0$ and hence $\left\langle [X, Y], Y \right\rangle =0$. Let , $i=1,2,3,4$. Since [X, Y] lies in $\mathfrak {n}^{(i)}$ for any $X \in \mathfrak {n}$ we obtain $\left\langle [X, Y], Y \right\rangle =0$ which proves the first assertion. According to Lemma 2.1 the second claim follows from the first assertion. $\square$

We often use the following Proposition (see [1, Proposition 1.13, Theorems 1.17, 1.18 ]):

Proposition 2.3

(a) Filiform nilpotent metric Lie algebras do not possess any totally geodesic subalgebra of codimension one.

(b) A filiform nilpotent metric Lie algebra $(\mathfrak {n}, \langle .,. \rangle )$ possesses a totally geodesic subalgebra of codimension two if and only if $\mathfrak {n}$ is isomorphic to the standard filiform Lie algebra of dimension $n \ge 3$.

3 Geodesic vectors and flat totally geodesic subalgebras of metric Lie algebras $\mathfrak {n}_{6,14}(\alpha _i,\beta _j )$, $\mathfrak {n}_{6,15}(\alpha _i,\beta _j )$ and $\mathfrak {n}_{6,16}(\alpha _i,\beta _j )$

In this section, we consider the metric Lie algebras $\left( \mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$, $\left( \mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$, and $\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$. Our goal is to describe the sets of geodesic vectors and to compute the flat totally geodesic subalgebras.

Let us start with the metric Lie algebra $\left( \mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$.

Theorem 3.1

Let $(\mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.3). The geodesic vectors of $(\mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle )$ not belonging to are the non-zero elements of the set $C_1 \cup C_2 \cup C_4$ in the case $\alpha _5 \beta _1+\alpha _2 \beta _7=0=\beta _4$, for $\alpha _5 \beta _1+\alpha _2 \beta _7 \ne 0$ these are the non-zero elements of the set $C_1 \cup C_3 \cup C_4$, where

$$\begin{aligned} \begin{aligned} C_1:=&\left\{ bE_2+cE_3+dE_4: b \left( \alpha _1 c+ \beta _1d\right) +c\alpha _2d=0 \right\} ,\\ {}&\text{ at } \text{ least } \text{ two } \text{ of } \text{ the } \text{ numbers } \ b, c, d \ \text{ are } \text{ non-zero } \text{ with } \text{ exception } \\ {}&\text{ of } \text{ the } \text{ cases: } \\ {}&1. \ b=0, \\ {}&2. \ d=0, \\ {}&3.\ c=0 \text{ with } \beta _1 \ne 0, \, \\ C_2:=&\big \{ a(E_1-\frac{\alpha _2}{\alpha _5} E_6)+cE_3+dE_4+eE_5: a\ne 0, \, \alpha _5 \beta _1+\alpha _2 \beta _7=0=\beta _4, \\ {}&e=(\beta _5a -\alpha _5c ) \frac{\alpha _2 \alpha _4}{\alpha _1 \alpha _5^2}, \, a\left( \alpha _1c+\beta _1d+\beta _2e+\beta _3f\right) -c \left( \alpha _3e+\beta _6 f\right) \\ {}&\quad -f (\beta _7d +\alpha _4e )=0\big \},\\ C_3:=&\big \{ a\left( E_1-\frac{\alpha _2}{\alpha _5} E_6 + (\frac{\beta _5}{\alpha _5}+\frac{\alpha _1\beta _4 }{\alpha _5 \beta _1+\alpha _2 \beta _7}) E_3- \frac{\alpha _2 \alpha _4 \beta _4 }{\alpha _5(\alpha _5 \beta _1+\alpha _2 \beta _7)} E_5 \right) \\ {}&\quad + dE_4: a\ne 0,\, \\ {}&a\left( \alpha _1c+\beta _1d+\beta _2e+\beta _3f\right) -c \left( \alpha _3e+\beta _6 f\right) -f (\beta _7d +\alpha _4e )=0\big \}, \\ C_4:=&\big \{ aE_1+cE_3+dE_4+eE_5+fE_6: a\ne 0,\, f \ne -a \frac{\alpha _2}{\alpha _5},\, f \ne 0, \\ {}&\quad e=(c \alpha _5 - a \beta _5) \frac{f \alpha _4}{a \alpha _1 \alpha _5}, \\ {}&d=\frac{a}{\alpha _5 f+ \alpha _2 a}(\frac{ \alpha _5 \beta _1+\alpha _2 \beta _7}{\alpha _4}e-\beta _4 f),\quad a\left( \alpha _1c+\beta _1d+\beta _2e+\beta _3f\right) \\ {}&-c \left( \alpha _3e+\beta _6 f\right) - f (\beta _7d +\alpha _4e )=0 \big \}. \end{aligned} \end{aligned}$$

Proof

According to (1.2) a non-zero vector $Y=a E_1+b E_2+cE_3+d E_4+e E_5+fE_6 \in \mathfrak {n}_{6,14}(\alpha _i,\beta _j )$ is geodesic if and only if one has $\langle [X,Y],Y \rangle =0$ for all $X=\sum _{i=1}^{5} x_iE_i \in \mathfrak {n}_{6,14}(\alpha _i,\beta _j )$ or equivalently if and only if the system of equations

$$\begin{aligned} \left\{ \begin{array}{ll} &{}b \alpha _4 f =0,\\ &{}a(\frac{\alpha _1\alpha _5}{\alpha _4} e+\beta _5 f)-c \alpha _5 f=0,\\ &{}a(\alpha _2d-\frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4} e+\beta _4 f)+ b\alpha _3e+d\alpha _5f=0,\\ &{}a\left( \alpha _1c+\beta _1d+\beta _2e+\beta _3f\right) -c \left( \alpha _3e+\beta _6 f\right) -f (\beta _7d +\alpha _4e )=0,\\ &{}b \left( \alpha _1 c+ \beta _1d+\beta _2e\right) +c (\alpha _2d-\frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4}e+\beta _4f)+d(\frac{\alpha _1\alpha _5}{\alpha _4}e+\beta _5f)=0 \end{array} \right. \end{aligned}$$

(3.1)

is satisfied. Taking into account that $\alpha _i \ne 0, \, i=\{1, \dots ,5\}$ and assume that $f=0$ the system of equations (3.1) reduces to

$$\begin{aligned} \left\{ \begin{array}{ll} &{}ae=0,\\ &{}a\alpha _2d+b\alpha _3e=0,\\ &{}a (\alpha _1c+\beta _1d)-c\alpha _3e=0,\\ &{}b\left( \alpha _1 c+ \beta _1d+\beta _2e\right) +c (\alpha _2d-\frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4}e) + d \frac{\alpha _1\alpha _5}{\alpha _4}e =0. \end{array} \right. \end{aligned}$$

(3.2)

If $a=0$, then from (3.2) we obtain

$$\begin{aligned} \left\{ \begin{array}{ll} &{}be=0=ce,\\ &{}b\left( \alpha _1 c+ \beta _1d\right) +c \alpha _2d +d \frac{\alpha _1\alpha _5}{\alpha _4}e =0. \end{array} \right. \end{aligned}$$

In the case $e=0=a=f$ the geodesic vectors of $\left( \mathfrak {n}_{6,14}(\alpha _{i},\beta _{j} ), \langle .,. \rangle \right)$ are the non-zero elements of the set $\overline{C_1}:=\left\{ bE_2+cE_3+dE_4: b \left( \alpha _1 c+ \beta _1d\right) +c\alpha _2d=0 \right\}$. For $b=0$ the element $Y=c E_3+ d E_4 \in \overline{C_1}$, $c, d \in \mathbb R$, satisfies the condition $\alpha _2 c d=0$. Since $\alpha _2 \ne 0$ we obtain that either $c=0$ and the element $Y=d E_4$ is in , or $d=0$ and the element $Y=c E_3$ is in . If $d=0$, then for the element $Y=b E_2+ c E_3 \in \overline{C_1}$, $b, c \in \mathbb R$, the condition $\alpha _1 b c=0$ holds. Since $\alpha _1 \ne 0$ we receive that either $c=0$ and the element $Y=b E_2$ is in , or $b=0$ and the element $Y=c E_3$ is in . If $c=0$ and $\beta _1 \ne 0$, then for the element $Y=b E_2+ d E_4 \in \overline{C_1}$, $b, d \in \mathbb R$ the condition $b d=0$ is satisfied. Hence we get either $b=0$ and the element $Y=d E_4$ is in , or $d=0$ and the element $Y=b E_2$ is in . Therefore the conditions for the set $C_1$ are proved.

In the case $b=c=0=a=f$ the vector $Y=d E_4+e E_5$ is geodesic if and only if either $d=0$ or $e=0$, and hence it lies either in or in .

If $e=0$, then from (3.2) we get

$$\begin{aligned} \left\{ \begin{array}{ll} &{}ad=0=ac,\\ &{}b\left( \alpha _1 c+ \beta _1d\right) +c\alpha _2d=0. \end{array} \right. \end{aligned}$$

The case $a=0=e=f$ is discussed above. In the case $d=c=0=e=f$ we receive that any non-zero vector $Y=a E_1+ b E_2$ is geodesic since it lies in the set .

Now, we suppose that $b=0$. In this case the system of equations (3.1) reduces to the following

$$\begin{aligned} \left\{ \begin{array}{ll} &{}a\frac{\alpha _1\alpha _5}{\alpha _4} e=f(c\alpha _5-a\beta _5),\\ &{}d\alpha _5f=- a(\alpha _2d-\frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4} e+\beta _4f),\\ &{}a\left( \alpha _1c+\beta _1d+\beta _2e+\beta _3f\right) -c \left( \alpha _3e+\beta _6 f\right) -f (\beta _7d +e\alpha _4 )=0,\\ &{}c (\alpha _2d-\frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4}e+\beta _4f)+d (\frac{\alpha _1\alpha _5}{\alpha _4}e+\beta _5f)=0. \end{array} \right. \end{aligned}$$

(3.3)

Assume that $a=0$. The first and second equations of (3.3) give $cf=0=df$. In the case $f=0= a$ we obtain from the third equation of (3.3) that $ce=0$. For $c=0=f=a=b$ the vector $Y=d E_4 +e E_5$ is geodesic if and only if it lies in or in . For $e=0=f=a$ the vector $Y=c E_3+d E_4$ is geodesic if and only if it is either in or in .

In the case $c=d=0=a$ the third equation of (3.3) gives $fe=0$. In this case the geodesic vector Y has either the form or the form $f E_6 \in \zeta$.

If $a \ne 0$, then the system of equations (3.3) is equivalent to the following system:

$$\begin{aligned} \left\{ \begin{array}{ll} &{}\frac{\alpha _1\alpha _5}{\alpha _4}e=\frac{f}{a}(c\alpha _5-a\beta _5),\\ &{}d (\alpha _5f+\alpha _2a) =a(\frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4} e-\beta _4f),\\ &{}a(\alpha _1c+\beta _1d+\beta _2e+\beta _3f)-c \left( \alpha _3e+\beta _6 f\right) -f (\beta _7d +e\alpha _4 )=0. \end{array} \right. \end{aligned}$$

(3.4)

Let $f=-a \frac{\alpha _2}{\alpha _5}$. Putting this expression into the first and second equations of (3.4) we receive

$$\begin{aligned} e=(a \beta _5-c\alpha _5) \frac{\alpha _2\alpha _4}{\alpha _1\alpha _5^2}, \end{aligned}$$

(3.5)

$$\begin{aligned} \frac{\alpha _5\beta _1+\alpha _2\beta _7}{\alpha _4}e+ \frac{\alpha _2}{\alpha _5} a \beta _4 =0. \end{aligned}$$

(3.6)

The substitution of (3.5) into (3.6) yields that

$$\begin{aligned} c (\alpha _5\beta _1+\alpha _2\beta _7)= \frac{a}{ \alpha _5}\Big ({(\alpha _5\beta _1+\alpha _2\beta _7)}\beta _5+ \alpha _1 \alpha _5 \beta _4 \Big ). \end{aligned}$$

(3.7)

If $\alpha _5\beta _1+\alpha _2\beta _7=0$, then the equation (3.7) gives $\beta _4=0$. In this case the geodesic vectors are the non-zero elements of the set $C_2$.

If $\alpha _5\beta _1+\alpha _2\beta _7\ne 0$, then from equation (4.23) we obtain:

$$\begin{aligned} c=a\Big (\frac{\beta _5}{\alpha _5}+\frac{ \alpha _1\beta _4}{\alpha _5\beta _1+\alpha _2\beta _7}\Big ). \end{aligned}$$

(3.8)

Putting (3.8) into (3.5) we receive

$$\begin{aligned} e=- \frac{\alpha _2 \alpha _4\beta _4 }{\alpha _5(\alpha _5 \beta _1+\alpha _2 \beta _7)}a. \end{aligned}$$

Hence the vector $Y=aE_1+cE_3+dE_4+eE_5+fE_6$ is a geodesic vector if and only if it lies in the set $C_3$.

If $f\ne -a \frac{\alpha _2}{\alpha _5}$, then from the first and second equations of (3.4) we have

$$\begin{aligned} e=\Big (c \alpha _5 - a \beta _5\Big ) \frac{f \alpha _4}{a \alpha _1 \alpha _5}, \quad d=\frac{a}{\alpha _5 f+ \alpha _2 a }\Big (\frac{ \alpha _5 \beta _1+\alpha _2 \beta _7}{\alpha _4}e-\beta _4 f\Big ). \end{aligned}$$

In this case the geodesic vectors of $\left( \mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ are the non-zero vectors of the set $C_4$. For $f=0$ the elements in $C_4$ have the shape $Y=a E_1+c E_3$, $a \ne 0$, and they satisfy the condition $\alpha _1 a c =0$. Therefore we have $c=0$ and hence the element $Y=a E_1$ lies in . The intersection $C_1 \cap C_2 \cap C_4$ is empty because for the elements of $C_1$ one has $a =0$ but for the elements in $C_2 \cup C_4$ we have $a \ne 0$. Moreover, for the elements in $C_2$ one gets $f=-a \frac{\alpha _2}{\alpha _5}$, in contrast to this for the elements in $C_4$ we have $f \ne -a \frac{\alpha _2}{\alpha _5}$. Hence the sets $C_2$ and $C_4$ are disjoint. Similarly we can see that the sets $C_1$, $C_3$ and $C_4$ are disjoint. This completes the proof. $\square$

Now we determine the flat totally geodesic subalgebras of dimension $>1$ in the metric Lie algebra $(\mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle )$.

Theorem 3.2

Let $(\mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.3). The flat totally geodesic subalgebras of dimension $>1$ in the metric Lie algebra $(\mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle )$ are the 2-dimensional subalgebras:

• (1) $\mathfrak {h}_{2}= \text {span} (E_1, E_6)$ in the case $\beta _3=\beta _4=\beta _5=0$,

• (2) $\mathfrak {h}_{3}=\text {span}(E_2, E_4)$ in the case $\beta _1=\beta _7=0$.

Proof

In view of Proposition 2.3 any flat totally geodesic subalgebra of $(\mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle )$ has dimension at most 3. Hence firstly we compute the 2- and 3-dimensional abelian subalgebras in the Lie algebra $\mathfrak {n}_{6,14}(\alpha _i,\beta _j )$.

The 2- and 3- dimensional abelian subalgebras of the Lie algebra $\mathfrak {n}_{6,14}(\alpha _i, \beta _j )$ are:

$$\begin{aligned}&\mathfrak {h}_{1}= \text{ span } \left( E_1+k_2E_3+k_3E_4+k_4E_6, \quad E_5+l_1E_6 \right) ,\\ {}&\mathfrak {h}_{2}= \text{ span } \left( E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5, \quad E_6\right) ,\\ {}&\mathfrak {h}_{3}= \text{ span } \left( E_2+k_1E_3+k_2E_5+k_3E_6, \quad E_4+l_1E_5+l_2E_6 \right) \\ {}&\hspace{0.80cm} \text{ with } \beta _7+l_1\alpha _4 -k_1\alpha _5=0,\\ {}&\mathfrak {h}_{4}= \text{ span } \left( E_2+k_1E_3+k_2E_4+k_3E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{5}= \text{ span } \left( E_3+k_1E_4+k_2E_6, \quad E_5+l_1E_6 \right) ,\\ {}&\mathfrak {h}_{6}= \text{ span } \left( E_3+k_1E_4+k_2E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{7}= \text{ span } \left( E_4+k_1E_6, \quad E_5+l_1E_6 \right) ,\\ {}&\mathfrak {h}_{8}= \text{ span } \left( E_4+l_1E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{9}= \text{ span } \left( E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{10}= \text{ span } \left( E_1+k_2E_3+k_3E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{11}= \text{ span } \left( E_3+k_1E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{12}= \text{ span } \left( E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{13}= \text{ span } \left( E_2+k_1E_3+k_2E_5, \quad E_4+l_1E_5, \quad E_6 \right) \\ {}&\hspace{0.80cm} \text{ with } \beta _7+ l_1 \alpha _4 -k_1\alpha _5=0, \end{aligned}$$

where $k_1, k_2, k_3, k_4, l_1, l_2 \in \mathbb {R}$.

The subalgebras $\mathfrak {h}_{9}$, $\mathfrak {h}_{10}$, $\mathfrak {h}_{11}$, and $\mathfrak {h}_{12}$ are not flat totally geodesic because for the vector $E_5+E_6$ the fourth equation of (3.1) gives the contradiction $\alpha _4=0$. Hence the vector $E_5+E_6 \in \mathfrak {h}_{9} \cap \mathfrak {h}_{10} \cap \mathfrak {h}_{11} \cap \mathfrak {h}_{12}$ is not geodesic. Therefore the subalgebras $\mathfrak {h}_{9}$, $\mathfrak {h}_{10}$, $\mathfrak {h}_{11}$, $\mathfrak {h}_{12}$ are excluded. The subalgebras $\mathfrak {h}_{8}$ and $\mathfrak {h}_{13}$ are not flat totally geodesic since for the vector $E_4+l_1E_5+E_6 \in \mathfrak {h}_{8} \cap \mathfrak {h}_{13}$ the third equation of the system (3.1) yields $\alpha _5=0$ which is a contradiction. Therefore the vector $E_4+l_1E_5+E_6$ is not geodesic and hence by Lemma 2.1 the subalgebras $\mathfrak {h}_{8}$, $\mathfrak {h}_{13}$ are excluded, too. The vector $E_2+k_1E_3+k_2E_4+k_3E_5+E_6 \in \mathfrak {h}_{4}$ is geodesic if and only if for $a=0, b=f=1$, $c=k_1$, $d=k_2$, $e=k_3$ the system (3.1) of equations holds. From the first equation of (3.1) we get the contradiction $\alpha _4=0$. Hence the vector $E_2+k_1E_3+k_2E_4+k_3E_5+E_6 \in \mathfrak {h}_{4}$ is not geodesic. Thus the subalgebra $\mathfrak {h}_{4}$ is not flat totally geodesic. Therefore the case $\mathfrak {h}_{4}$ is excluded. The element $E_3+k_1 E_4 + k_2 E_5+E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $a=b=0, c=f=1$, $d=k_1$, $e=k_2$, the system (3.1) of equations is satisfied. The second equation of (3.1) yields the contradiction $\alpha _5=0$. Therefore the subalgebra $\mathfrak {h}_{6}$ is not flat totally geodesic and hence it is excluded.

The non-zer vector $E_5+l_1E_6 \in \mathfrak {h}_{1} \cap \mathfrak {h}_{5} \cap \mathfrak {h}_{7}$ is geodesic if and only if for $a=b=c=d=0, e=1, f=l_1$, the system (3.1) of equations holds. The fourth equation of (3.1) gives $\alpha _4 l_1=0$. As $\alpha _4 \ne 0$ we receive that $l_1=0$.

We treat the subalgebra $\mathfrak {h}_{1}$. The vector $E_1+k_2E_3+k_3E_4+k_4E_6 \in \mathfrak {h}_{1}$ is geodesic if and only if for $b=e=0, a=1, c=k_2, d=k_3, f=k_4$ the system (3.1) of equations holds. From the second equation of (3.1) we receive

$$\begin{aligned} k_4(\beta _5-\alpha _5k_2)=0. \end{aligned}$$

(3.9)

Furthermore, the element $E_1+k_2E_3+k_3E_4+k_4E_6+E_5 \in \mathfrak {h}_{1}$ is geodesic if and only if for $b=0, a=e=1, c=k_2, d=k_3, f=k_4$ the system (3.1) of equations is satisfied. The second equation of (3.1) gives

$$\begin{aligned} \frac{\alpha _1 \alpha _5}{\alpha _4}+ k_4(\beta _5-\alpha _5k_2)=0. \end{aligned}$$

(3.10)

Taking into account (3.9), equation (3.10) yields the contradiction $\frac{\alpha _1 \alpha _5}{\alpha _4}=0$. Hence the subalgebra $\mathfrak {h}_{1}$ is not flat totally geodesic and therefore it is excluded.

Next we consider the subalgebra $\mathfrak {h}_{5}$. The vector $E_3+ k_1 E_4+ k_2 E_6 \in \mathfrak {h}_{5}$ is geodesic if and only if for $a=b=e=0, c=1, d=k_1, f=k_2$, the system (3.1) of equations holds. From the second equation of (3.1) one obtains $\alpha _5 k_2=0$. As $\alpha _5\ne 0$ we have $k_2=f=0$. Using this the fifth equation of gives $\alpha _2 k_1=0$. Since $\alpha _2 \ne 0$ we receive $k_1=0$. But the vector $E_3+E_5 \in \mathfrak {h}_{5}$ is not geodesic because the fourth equation of the system (3.1) gives the contradiction $\alpha _3=0$. Therefore the subalgebra $\mathfrak {h}_{5}$ is excluded, too.

Here we deal with subalgebra $\mathfrak {h}_{7}$. The element $E_4+ k_1 E_6 \in \mathfrak {h}_{7}$ is geodesic if and only if for $a=b=c=e=0, d=1, f=k_1$, the system (3.1) of equations is satisfied. From the third equation of (3.1) one obtains $\alpha _5 k_1=0$. Since $\alpha _5 \ne 0$ we get $k_1=0$. But the vector $E_4+E_5 \in \mathfrak {h}_{7}$ is not geodesic because from the fifth equation of the system (3.1) we receive the contradiction $\frac{\alpha _1 \alpha _5}{\alpha _4}=0$. Hence the subalgebra $\mathfrak {h}_{7}$ is excluded.

Now we treat the subalgebra $\mathfrak {h}_{2}$. The vector $E_1+k_1 E_2+k_2 E_3+k_3 E_4+k_4 E_5 \in \mathfrak {h}_{2}$ is geodesic if and only if for $f=0, a=1, b=k_1, c=k_2, d=k_3, e=k_4$ the system (3.1) of equations is satisfied. The second equation of (3.1) gives $\alpha _1 \alpha _5 k_4=0$. As $\alpha _1 \alpha _5 \ne 0$ we get $k_4=e=0$. Using this from the third equation of (3.1) we receive $\alpha _2 k_3=0$. Since $\alpha _2 \ne 0$ we have $k_3=d=0$. Applying this the fourth equation of (3.1) yields $\alpha _1 k_2=0$. As $\alpha _1 \ne 0$ we receive $k_2=c=0$. The element $E_1+k_1E_2+E_6 \in \mathfrak {h}_{2}$ is geodesic if and only if for $e=c=d=0, a=f=1, b=k_1$ the system (3.1) of equations holds. From the first equation of (3.1) one has $\alpha _4 k_1=0$. The condition $\alpha _4 \ne 0$ implies that $k_1=0$. Additionally, the vector $E_1+E_6 \in \mathfrak {h}_{2}$ is geodesic if and only if for $b=c=d=e=0, a=f=1$ the system (3.1) of equations is satisfied. From the second, third, and fourth equations we get $\beta _5=0$, $\beta _4=0$, and $\beta _3=0$. Hence the assertion (1) follows.

Finally we consider the subalgebra $\mathfrak {h}_{3}$. The vector $E_2+k_1 E_3+k_2E_5+k_3E_6 \in \mathfrak {h}_{3}$ is geodesic if and only if for $a=d=0, b=1, c=k_1, e=k_2, f=k_3$ the system (3.1) of equations holds. The first equation of (3.1) gives $\alpha _4 k_3=0$. As $\alpha _4 \ne 0$ one has $k_3=f=0$. Using this from the third equation of (3.1) we obtain $\alpha _3 k_2=0$. Since $\alpha _3 \ne 0$ we receive $k_2=e=0$. Applying this from the fifth equation of (3.1) one gets $\alpha _1 k_1=0$. Since $\alpha _1 \ne 0$ yields that $k_1=c=0$. The element $E_4+l_1 E_5+l_2 E_6 \in \mathfrak {h}_{3}$ is geodesic if and only if for $a=b=c=0, d=1, e=l_1, f=l_2$ the system (3.1) of equations is satisfied. From the third equation of (3.1) we obtain $\alpha _5 l_2=0$ which implies that $l_2=f=0$. Using this the fifth equation of (3.1) gives $\alpha _1 \alpha _5 l_1=0$. This yields $l_1=0$. In addition, the vector $E_2+E_4 \in \mathfrak {h}_{3}$ is geodesic if and only if for $a=e=c=f=0, b=d=1$ the system (3.1) of equations holds. The fifth equation of (3.1) gives $\beta _1=0$. Taking into account the abelian condition when $l_1=k_1=0$ we receive that $\beta _7=0$. Therefore the case (2) is proved. This proves Theorem 3.2. $\square$

Next we deal with the $\mathbb N$-graded filiform metric Lie algebra $\left( \mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ defined by non-vanishing commutators given by (1.4). Firstly we give the set of its geodesic vectors.

Theorem 3.3

Let $(\mathfrak {n}_{6,15}(\alpha _i,\beta _j ),\langle .,.\rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.4). The geodesic vectors of $(\mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle )$ not belonging to are the non-zero elements of the set $C_1 \cup C_2$, where

$$\begin{aligned} \begin{aligned} C_1:=&\big \{ cE_3+dE_4+eE_5+f E_6: f \ne 0, \, c \ne 0, \, d=\frac{-c}{\alpha _5 f}\Big ( \frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7 f \Big ), \\ {}&\frac{-c}{\alpha _5 f}\Big ( \frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7 f \Big ) (\alpha _3e+\beta _6 f+c \alpha _2) +c (\beta _4e+\beta _5 f)+e\alpha _4 f=0 \big \},\\ C_2:=&\left\{ bE_2+cE_3+dE_4: b(\alpha _1c+ \beta _1d)+c \alpha _2d=0 \right\} \\ {}&\text{ at } \text{ least } \text{ two } \text{ of } \text{ the } \text{ numbers } \ b, c, d \text { are non-zero with exception} \\ {}&\text{ of } \text{ the } \text{ cases: } \\ {}&1. \ b=0, \\ {}&2. \ d=0, \\ {}&3. \ \text{ for } \, \beta _1 \ne 0, \, c=0. \end{aligned} \end{aligned}$$

Proof

Applying the commutators (1.4) and (1.2) we obtain that the non-zero element $Y=aE_1+bE_2+cE_3+dE_4+eE_5+fE_6 \in \mathfrak {n}_{6,15}(\alpha _i,\beta _j )$ is geodesic if and only if for the real numbers a, b, c, d, e, f with respect to a basis $\{E_1, E_2, E_3, E_4, E_5, E_6\}$ the system of equations

$$\begin{aligned} \left\{ \begin{array}{ll} &{}a f=0,\\ &{}a \alpha _3 e+b \alpha _5 f=0,\\ &{}a(\alpha _2d+\beta _4e)+b(\frac{\alpha _2 \alpha _5}{\alpha _4}e+\beta _7f)=0,\\ &{}a(\alpha _1c+\beta _1d+\beta _2e)-c(\frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7f)-d\alpha _5f=0,\\ &{}b(\alpha _1c+\beta _1d+\beta _2e+\beta _3f)+c (\alpha _2 d+\beta _4e+\beta _5f)+d(\alpha _3e+\beta _6f)+e\alpha _4f =0 \end{array} \right. \end{aligned}$$

(3.11)

is satisfied. If $a=0$, then the second and the third equations of (3.11) gives $bf=0 =be$. In the case $b=0=a$ the system (3.11) reduces to

$$\begin{aligned} \left\{ \begin{array}{ll} &{}c(\frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7f)+d\alpha _5f=0,\\ &{}c (\alpha _2 d+\beta _4e+\beta _5f)+d(\alpha _3e+\beta _6f)+e\alpha _4f =0. \end{array} \right. \end{aligned}$$

(3.12)

Suppose that $f=0$. From the first equation of (3.12) we receive either $c=0$ or $e=0$. In the case $c=0$ the vector $Y=d E_4+eE_5$ is geodesic of $\left( \mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ if and only if it lies either in or in , whereas if $e=0$ the vector $Y=c E_3+dE_4$ is geodesic of $\left( \mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ precisely if it lies either in or in .

Assume that $f\ne 0$. From the first equation of (3.12) we obtain

$$\begin{aligned} d=\frac{-c}{\alpha _5 f}\Big ( \frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7 f \Big ). \end{aligned}$$

(3.13)

Substituting (3.13) into the second equation of (3.12) we recieve that

$$\begin{aligned} \frac{-c}{\alpha _5 f}\Big ( \frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7 f \Big ) (c\alpha _2+\alpha _3e+\beta _6 f) +c(\beta _4e+\beta _5 f)+e\alpha _4 f=0. \end{aligned}$$

In this case the vector $Y=cE_3+dE_4+eE_5+fE_6$, $f \ne 0$ is geodesic of $\left( \mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ if and only if it lies in the set

$$\begin{aligned} \begin{aligned} \overline{C_1}:=&\big \{ cE_3+dE_4+eE_5+f E_6: f \ne 0, \, d=\frac{-c}{\alpha _5 f}\Big ( \frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7 f \Big ), \\ {}&\frac{-c}{\alpha _5 f}\Big ( \frac{\alpha _2\alpha _5}{\alpha _4}e+\beta _7 f \Big ) (\alpha _3e+\beta _6 f+c \alpha _2) +c (\beta _4e+\beta _5 f)+e\alpha _4 f=0 \big \}. \end{aligned} \end{aligned}$$

If $c=0$, then any vector $Y=e E_5 + f E_6 \in \overline{C_1}$ lies either in or in $\zeta$. Hence we get the set $C_1$ in the assertion.

In the case $f=e=0=a$, the vector $Y=bE_2+cE_3+dE_4$ is geodesic of $\left( \mathfrak {n}_{6,15}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ if and only if it lies in the set $C_2$. The set $C_2$ coincides with the set $C_1$ in the filiform metric Lie algebra $\left( \mathfrak {n}_{6,14}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$. Therefore the validity of the condition for the numbers b, c, d in the case of the set $C_2$ in the assertion can be proved in the same manner as in the proof of Theorem 3.1 in the case $C_1$.

Now, if $f=0$, the system (3.11) of equations reduces to

$$\begin{aligned} \left\{ \begin{array}{ll} &{}a e=0,\\ &{}ad+\frac{\alpha _5}{\alpha _4}be=0,\\ &{}a(\alpha _1c+\beta _1d)-\frac{\alpha _2\alpha _5}{\alpha _4}ce=0,\\ &{}b(\alpha _1c+\beta _1d+\beta _2e)+c (\alpha _2 d+\beta _4e)+d\alpha _3e=0. \end{array} \right. \end{aligned}$$

(3.14)

We discussed the case $a=0$ above. In the case $e=0$, from the second and the third equations of (3.14) one has $ad=0=ac$. The case $f=e = a=0$ is discussed above. In the case $d=c=0=e$ any vector $Y=aE_1+bE_2$ is geodesic because it lies in . The intersection $C_1 \cap C_2$ is empty since for the elements of $C_2$ one has $f =0$ but for the elements in $C_1$ we have $f \ne 0$. This proves Theorem 3.3. $\square$

According to [2, Theorem 5.17], the metric Lie algebra $(\mathfrak {n}_{6,15}(\alpha _i,\beta _j ),\langle .,.\rangle )$ is isomorphic to the $\mathbb N$-graded filiform Lie algebra .

Theorem 3.4

Let $(\mathfrak {n}_{6,15}(\alpha _i,\beta _j ),\langle .,.\rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.4). Then the unique flat totally geodesic subalgebra of dimension $>1$ in $(\mathfrak {n}_{6,15}(\alpha _i,\beta _j ),\langle .,.\rangle )$ is $\mathfrak {h}_{6}=\text {span}(E_3, E_6)$ in the case $\beta _5=\beta _7=0$.

Proof

In view of Proposition 2.3 and [2, Theorem 4.2], the dimension of flat totally geodesic subalgebras of $(\mathfrak {n}_{6,15}(\alpha _i,\beta _j ),\langle .,.\rangle )$ is at most 3. Hence firstly we determine the 2- and 3-dimensional abelian subalgebras in the Lie algebra $\mathfrak {n}_{6,15}(\alpha _i,\beta _j )$.

The 3- and 2-dimensional abelian subalgebras of $\mathfrak {n}_{6,15}(\alpha _i,\beta _j )$ are:

$$\begin{aligned}&\mathfrak {h}_{1}= \text {span} \left( E_2+k_1E_3+k_2E_4, \quad E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{2}= \text {span} \left( E_3+k_1E_4, \quad E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{3}= \text {span} \left( E_4, \quad E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{4}= \text {span} \left( E_3+k_1E_6, \quad E_4+l_2 E_6, \quad E_5+s_1 E_6 \right) ,\\&\mathfrak {h}_{5}= \text {span} \left( E_3+k_1E_5, \quad E_4+l_1E_5, \quad E_6 \right) , \\&\mathfrak {h}_{6}= \text {span} \left( E_3+k_1E_4+k_2E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{7}= \text {span} \left( E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5, \quad E_6\right) ,\\&\mathfrak {h}_{8}= \text {span} \left( E_2+k_1E_3+k_2E_4+k_3E_6, \quad E_5+ s_1E_6 \right) ,\\&\mathfrak {h}_{9}= \text {span} \left( E_2+k_1E_3+k_2E_4+k_3E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{10}= \text {span} \left( E_3+k_1E_5+k_2E_6, \quad E_4+l_1E_5+l_2E_6 \right) ,\\&\mathfrak {h}_{11}= \text {span} \left( E_3+k_1E_4+k_2E_6, \quad E_5+s_1E_6 \right) ,\\&\mathfrak {h}_{12}= \text {span} \left( E_4+ l_2E_6, \quad E_5+s_1E_6 \right) ,\\&\mathfrak {h}_{13}= \text {span} \left( E_4+ l_1E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{14}= \text {span} \left( E_5, E_6 \right) , \end{aligned}$$

where $k_1, k_2, k_3, k_4, l_1, l_2, s_1 \in \mathbb {R}$.

The subalgebras $\mathfrak {h}_{1}$, $\mathfrak {h}_{2}$, $\mathfrak {h}_{3}$, and $\mathfrak {h}_{14}$ are not flat totally geodesic because for the vector $E_5+E_6$ the fifth equation of (3.11) gives the contradiction $\alpha _4=0$. Hence the vector $E_5+E_6 \in \mathfrak {h}_{1} \cap \mathfrak {h}_{2} \cap \mathfrak {h}_{3} \cap \mathfrak {h}_{14}$ is not geodesic. Therefore the subalgebras $\mathfrak {h}_{1}$, $\mathfrak {h}_{2}$, $\mathfrak {h}_{3}$, $\mathfrak {h}_{14}$ are excluded. By the system (3.11) of equations the vector $E_4+ l_2 E_6 \in \mathfrak {h}_{4} \cap \mathfrak {h}_{12}$, respectively $E_5+ s_1 E_6 \in \mathfrak {h}_{4} \cap \mathfrak {h}_{8} \cap \mathfrak {h}_{11} \cap \mathfrak {h}_{12}$, respectively $E_4+ l_1 E_5 \in \mathfrak {h}_{5} \cap \mathfrak {h}_{13}$ is totally geodesic precisely if one has $\alpha _5 l_2=0=\beta _6 l_2$, respectively $\alpha _4 s_1=0$, respectively $\alpha _3 l_1=0$. Since $\alpha _5 \alpha _4 \alpha _3 \ne 0$ we receive that $l_2=l_1=s_1=0$. Using this for the vector $E_4+E_5 \in \mathfrak {h}_{4} \cap \mathfrak {h}_{12}$ the fifth equation of (3.11) leads to the contradiction $\alpha _3=0$, whereas for the vector $E_4+E_6 \in \mathfrak {h}_{5} \cap \mathfrak {h}_{13}$ the fourth equation (3.11) gives the contradiction $\alpha _5=0$. Therefore the vectors $E_4+E_5 \in \mathfrak {h}_{4} \cap \mathfrak {h}_{12}$, $E_4+E_6 \in \mathfrak {h}_{5} \cap \mathfrak {h}_{13}$ are not geodesic. Hence the subalgebras $\mathfrak {h}_{4}$, $\mathfrak {h}_{5}$, $\mathfrak {h}_{12}$, $\mathfrak {h}_{13}$ are excluded, too.

The vector $E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5+ E_6 \in \mathfrak {h}_{7}$ is not geodesic because the first equation of (3.11) leads to the contradiction $1=0$. Therefore the subalgebra $\mathfrak {h}_{7}$ is not flat totally geodesic and hence it is excluded.

The element $E_2+k_1E_3+k_2E_4+k_3E_6+E_5 \in \mathfrak {h}_{8}$ is geodesic if and only if for $a=0, b=e=1$, $c=k_1$, $d=k_2$, $f=k_3$ the system (3.11) of equations is satisfied. From the second equation of (3.11) we get $\alpha _5 k_3=0$. As $\alpha _5 \ne 0$ we have $k_3=f=0$. Using this the third equation of (3.11) gives the contradiction $\frac{\alpha _2 \alpha _5}{\alpha _4}=0$. Hence the vector $E_2+k_1E_3+k_2E_4+k_3E_6+E_5 \in \mathfrak {h}_{8}$ is not geodesic. Therefore the subalgebra $\mathfrak {h}_{8}$ is excluded.

The vector $E_2+k_1E_3+k_2E_4+k_3E_5+E_6 \in \mathfrak {h}_{9}$ is geodesic if and only if for $a=0, b=f=1$, $c=k_1$, $d=k_2$, $e=k_3$ the system (3.11) of equations holds. From the second equation of (3.11) we get the contradiction $\alpha _5=0$. Hence the vector $E_2+k_1E_3+k_2E_4+k_3E_5+E_6 \in \mathfrak {h}_{9}$ is not geodesic. So the subalgebra $\mathfrak {h}_{9}$ is excluded.

The element $E_3+k_1E_4+k_2E_6 \in \mathfrak {h}_{11}$ is geodesic if and only if for $a=b=e=0$, $c=1$, $d= k_1$, $f=k_2$ the system (3.11) of equations is satisfied. The fourth equation gives

$$\begin{aligned} \beta _7 k_2+ \alpha _5 k_1 k_2=0. \end{aligned}$$

(3.15)

In addition the vector $E_3+k_1E_4+k_2E_6+E_5 \in \mathfrak {h}_{11}$ is geodesic if and only if for $a=b=0, c=e=1$, $d= k_1$, $f=k_2$ the system (3.11) of equations holds. The fourth equation of (3.11) yields

$$\begin{aligned} \frac{\alpha _2 \alpha _5}{\alpha _4} +\beta _7 k_2+ \alpha _5 k_1 k_2=0. \end{aligned}$$

(3.16)

Comparing (3.15) with (3.16) we obtain the contradiction $\frac{\alpha _2 \alpha _5}{\alpha _4}=0$. Hence the vector $E_3+k_1E_4+k_2E_6+E_5 \in \mathfrak {h}_{11}$ is not geodesic and the subalgebra $\mathfrak {h}_{11}$ is excluded.

The vector $E_4+ l_1 E_5 + l_2 E_6 \in \mathfrak {h}_{10}$ is geodesic if for $a=b=c=0$, $d=1$, $e=l_1$, $f=l_2$ the system (3.11) of equations holds. The fourth and the fifth equations of (3.11) yield $\alpha _5 l_2=0$, $\alpha _3 l_1+ \beta _6 l_2+ \alpha _4 l_1 l_2=0$. As $\alpha _5 \alpha _3 \ne 0$ we obtain that $l_1=l_2=0$. The element $E_3+k_1E_5+k_2E_6 \in \mathfrak {h}_{10}$ is geodesic if for $a=b=d=0$, $c=1$, $e=k_1$, $f=k_2$ the system (3.11) of equations is satisfied. From the fourth and fifth equations of (3.11) we get

$$\begin{aligned} \frac{\alpha _2 \alpha _5}{\alpha _4} k_1+ \beta _7 k_2=0, \quad \beta _4 k_1+ \beta _5 k_2+ \alpha _4 k_1 k_2=0. \end{aligned}$$

(3.17)

The element $E_3+E_4+k_1E_5+k_2E_6 \in \mathfrak {h}_{10}$ is geodesic if for $a=b=0$, $c=d=1$, $e=k_1, f=k_2$ the system (3.11) of equations is valid. From the fourth equation of (3.11) one has

$$\begin{aligned} \frac{\alpha _2 \alpha _5}{\alpha _4} k_1+ \beta _7 k_2+ \alpha _5 k_2=0. \end{aligned}$$

(3.18)

Comparing it with the first equation of (3.17) we obtain $\alpha _5 k_2=0$. Since $\alpha _5 \ne 0$ one gets $k_2=0$. Using this from (3.18) we receive that also $k_1=0$. But the vector $E_3+ E_4 \in \mathfrak {h}_{10}$ is not geodesic since for $a=b=e=f=0$, $c=d=1$ the fifth equation of (3.11) gives the contradiction $\alpha _2=0$. Therefore the subalgebra $\mathfrak {h}_{10}$ is excluded.

Finally we treat the subalgebra $\mathfrak {h}_{6}$. The element $E_3+k_1E_4+k_2E_5 \in \mathfrak {h}_{6}$ is geodesic if and only if the system (3.11) of equations are satisfied for $a=b=f=0$, $c=1$, $d=k_1$, $e=k_2$. From the fourth and fifth equations of (3.11) we obtain $\frac{\alpha _2 \alpha _5}{\alpha _4} k_2=0, \, \alpha _2 k_1+ \beta _4 k_2+ k_1 \alpha _3 k_2=0$. As $\alpha _2 \alpha _5 \ne 0$ we get $k_2=k_1=0$. Using this the vector $E_3+E_6$ lies in $\mathfrak {h}_{6}$, which is geodesic if for $a=b=d=e=0$, $c=f=1$ the system (3.11) is valid. From the fourth and fifth equations of (3.11) we receive $\beta _7=\beta _5=0$. Therefore the subalgebra $\mathfrak {h}_{6}=\text {span}(E_3, E_6)$ is flat totally geodesic in the case $\beta _5=\beta _7=0$. Hence Theorem 3.4 is proved. $\square$

Now we treat the filiform Lie algebra $\ell _{6,16}$ defined by the non-vanishing Lie brackets

$$\begin{aligned}{}[ G_1,G_2] =G_3, \, [ G_1,G_3] =G_4,\, [ G_1,G_4] =G_5,\, [ G_2,G_5] =G_6, \, [ G_4,G_3] =G_6. \end{aligned}$$

The following theorem describes the isometric isomorphism classes of the metric Lie algebras $(\ell _{6,16}, \langle .,.\rangle )$ and the group of orthogonal automorphisms of $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$.

Theorem 3.5

Let $\langle .,. \rangle$ be an inner product on the 6-dimensional filiform Lie algebra $\ell _{6,16}$.

• (1) There is a unique metric Lie algebra $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ which is isometrically isomorphic to the metric Lie algebra $(\ell _{6,16}, \langle .,. \rangle )$ with $\alpha _i>0, i=1,\dots ,4$ and one of the following cases is satisfied:

  • (i) $\beta _1=\beta _3=\beta _4=\beta _5=\beta _6=\beta _8$,

  • (ii) $\beta _3>0$ or $\beta _5>0,$ $\beta _1=\beta _4=\beta _6=\beta _8=0,$

  • (iii) $\beta _6>0$ or $\beta _4>0,$ $\beta _1=\beta _3=\beta _5=\beta _8=0,$

  • (iv) $\beta _1>0$ or $\beta _8>0,$ $\beta _3=\beta _4=\beta _5=\beta _6=0,$

  • (v) at least two of the elements of the set $\{\beta _1, \beta _3, \beta _4, \beta _5, \beta _6, \beta _8 \}$ are positive with the exception of the cases $(\beta _1>0, \beta _8>0),\, (\beta _3>0, \beta _5>0), \, (\beta _4>0, \beta _6>0)$.

• (2) The group $\mathcal{O}\mathcal{A}\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j )\right)$ of orthogonal automorphisms of the metric Lie algebra $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ is the following group:

  • (a) in case (i) one has $\mathcal{O}\mathcal{A}\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j )\right) =\{TE_i=\varepsilon _1 E_i,i=1,6, TE_i=\varepsilon _2 E_i, i=2, 4, TE_i=\varepsilon _1\varepsilon _2 E_i,i=3,5, \, \varepsilon _1,\varepsilon _2=\pm 1\} \simeq \mathbb {Z}_2 \times \mathbb {Z}_2$,

  • (b) in case (ii) one has $\mathcal{O}\mathcal{A}\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j )\right) =\{TE_2= E_2,TE_4= E_4, TE_i=\varepsilon E_i,i=1,3,5,6, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$,

  • (c) in case (iii) one has $\mathcal{O}\mathcal{A}\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j )\right) =\{TE_3= E_3,TE_5= E_5, TE_i=\varepsilon E_i,i=1,2,4,6, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$,

  • (d) in case (iv) one has $\mathcal{O}\mathcal{A}\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j )\right) =\{TE_1= E_1,TE_6= E_6, TE_i=\varepsilon E_i,i=2,3,4,5, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$,

  • (e) in case (v) the group $\mathcal{O}\mathcal{A}\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j )\right)$ is trivial.

Remark 3.6

The proof of Theorem 3.5 can be found in [6, Theorem 3.8, p. 8.] with the exception of the case (ii) and its group of orthogonal automorphism given by (b) which are missing from there.

Firstly we determine the geodesic vectors of the filiform metric Lie algebra $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$.

Theorem 3.7

Let $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.5). The geodesic vectors of $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ not belonging to are the non-zero elements of the set $C_1 \cup C_2$, where

$$\begin{aligned} \begin{aligned} C_1:=&\, \Big \{ b E_2+c E_3+d E_4 +e E_5: b (\alpha _1c+\beta _1 d+\beta _2 e)\\ {}&+\, c \Big (\alpha _2d- (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})e\Big )+ d\alpha _3e=0 \Big \},\\ {}&\text{ at } \text{ least } \text{ two } \text{ of } \text{ the } \text{ numbers } \ b, c, d, e \text { are non-zero with exception} \\ {}&\text{ of } \text{ the } \text{ cases: }\\ {}&1. \ b=c=0, \\ {}&2. \, b=e=0, \\ {}&3.\ d=e=0, \\ {}&4.\ b=d=0 \text{ with } \alpha _3 \alpha _4 \beta _1+ \alpha _1 \alpha _2 \beta _8 \ne 0, \\ {}&5. \, c=d=0 \, \text{ with } \, \beta _2 \ne 0,\\ {}&6.\ c=e=0 \text{ with } \beta _1 \ne 0, \\ C_2:=&\, \Big \{ a \Big (E_1-\frac{\beta _6}{\alpha _4} E_2\Big )+c E_3+d E_4 +e E_5+f E_6: a f \ne 0,\\ {}&a e= \frac{f}{\alpha _3} \Big (\frac{a \beta _6 \beta _8}{\alpha _4}+c\frac{\alpha _3 \alpha _4}{\alpha _1} -a\beta _5\Big ),\\ {}&a c=\frac{f}{\alpha _1} (c\beta _7+d\beta _8+e\alpha _4)-\frac{a}{\alpha _1}\left( \beta _1 d +\beta _2 e+\beta _3 f\right) ,\\ {}&ad = \frac{a}{\alpha _2}\Big ((\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2\beta _8}{\alpha _4})e- \beta _4f\Big )+\frac{f}{\alpha _2}\Big (\frac{a \beta _6 \beta _7}{\alpha _4}-d\frac{\alpha _3 \alpha _4}{\alpha _1}\Big ),\\ {}&d(\alpha _2c+ \alpha _3e)= a \frac{\beta _6}{\alpha _4}\left( \alpha _1 c+\beta _1 d+\beta _2 e+\beta _3 f\right) + c e \Big (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2\beta _8}{\alpha _4}\Big )\\ {}&-f \left( c \beta _4+d\beta _5+e\beta _6 \right) \Big \}. \end{aligned} \end{aligned}$$

Proof

According to (1.2) the vector $Y=a E_1+b E_2+c E_3+d E_4+e E_5+f E_6$ is geodesic of $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ if and only if the following system of equations

$$\begin{aligned} \left\{ \begin{array}{ll} &{}f\left( a \beta _6+ b \alpha _4 \right) =0, \\ &{}a \left( \alpha _3 e+\beta _5f\right) +f(b\beta _8-c\frac{\alpha _3\alpha _4}{\alpha _1})=0,\\ &{}a\Big (\alpha _2d-(\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2\beta _8}{\alpha _4})e+\beta _4 f\Big )+f (b\beta _7+d\frac{\alpha _3\alpha _4}{\alpha _1})=0,\\ &{}a\left( \alpha _1 c+\beta _1 d+\beta _2e+\beta _3 f\right) -f\left( c\beta _7+d\beta _8+e\alpha _4\right) =0,\\ &{}b (\alpha _1c+\beta _1 d+\beta _2 e+\beta _3 f)+c \Big (\alpha _2d- (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})e+\beta _4f\Big )\\ {} &{}+ d(\alpha _3e+\beta _5f)+e \beta _6f =0 \end{array} \right. \end{aligned}$$

(3.19)

is satisfied for $a,b,c,d,e,f \in \mathbb R$.

Firstly, we suppose that $f=0$. Take into account that $\alpha _{i} >0, \, i=1, \dots , 4$, the system (3.19) of equations gives $ae=0=ad=ac$.

For $a=0=f$ the vector $Y=bE_2+cE_3+dE_4+eE_5$ is geodesic of $\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j ),.\langle .,.\rangle \right)$ if and only if it lies in set $\overline{C_1}= \{ b E_2+c E_3+d E_4 +e E_5: b (\alpha _1c+\beta _1 d+\beta _2 e)+c (\alpha _2d- (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})e)+ d\alpha _3e=0\}$. If $b=c=0$, then for the element $Y=d E_4+ e E_5 \in \overline{C_1}$, $d, e \in \mathbb R$ one has $d \alpha _3 e=0$. As $\alpha _3 \ne 0$ we obtain that Y lies either in or in . If $b=e=0$, then for the element $Y=c E_3+ d E_4 \in \overline{C_1}$, $c, d \in \mathbb R$ we have $c \alpha _2 d=0$. As $\alpha _2 \ne 0$ we obtain that Y lies either in or in . In the case $d=e=0$ for the element $Y=b E_2+ c E_3 \in \overline{C_1}$, $b, c \in \mathbb R$ one gets $b \alpha _1 c=0$. Since $\alpha _1 \ne 0$ we receive that Y lies either in or in . In the case $b=d=0$ for the element $Y=c E_3+ e E_5 \in \overline{C_1}$, $c, e \in \mathbb R$ one obtains $c e (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})=0$. If $\alpha _3 \alpha _4 \beta _1+ \alpha _1 \alpha _2 \beta _8 \ne 0$, then we have $ce=0$, or equivalently the element Y is either in or in . In the case $c=d=0$ for the element $Y=b E_2+ e E_5 \in \overline{C_1}$, $b, e \in \mathbb R$ we receive $b \beta _2 e=0$. If $\beta _2 \ne 0$, then the element Y lies either in or in . In the case $c=e=0$ for the element $Y=b E_2+ d E_4 \in \overline{C_1}$, $b, d \in \mathbb R$ we receive $b \beta _1 d=0$. If $\beta _1 \ne 0$, then the element Y lies either in or in . This proves the conditions for the set $C_1$.

For $e=d=c=0=f$ any vector $Y=a E_1+ b E_2$ is geodesic of $\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j ),.\langle .,.\rangle \right)$ because it lies in .

Secondly, assume that $a \beta _6+b\alpha _4=0$. Hence we receive $b=-\frac{a\beta _6}{\alpha _4}$. Putting this expression into (3.19) one obtains that

$$\begin{aligned} \left\{ \begin{array}{ll} &{}a e= \frac{f}{\alpha _3} (\frac{a \beta _6 \beta _8}{\alpha _4}+c\frac{\alpha _3 \alpha _4}{\alpha _1} -a\beta _5),\\ &{}ad = \frac{a}{\alpha _2}\Big ( (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2\beta _8}{\alpha _4})e- \beta _4f\Big )+\frac{f}{\alpha _2}(\frac{a \beta _6 \beta _7}{\alpha _4}-d\frac{\alpha _3 \alpha _4}{\alpha _1}),\\ &{}a c=\frac{f}{\alpha _1} \left( c\beta _7+d\beta _8+e\alpha _4\right) -\frac{a}{\alpha _1}\left( \beta _1 d +\beta _2 e+\beta _3 f\right) ,\\ &{}a \frac{ \beta _6}{\alpha _4} \left( \alpha _1 c+\beta _1 d+\beta _2 e+\beta _3 f\right) + ce(\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})\\ {} &{}\quad -f(c \beta _4+d \beta _5+e \beta _6)=d(\alpha _2c+\alpha _3e). \end{array} \right. \end{aligned}$$

(3.20)

Suppose that $a=0$. Hence one has $b=0$. From system (3.20) of equations we receive $fc=0=fd=fe$. The case $f=0=a=b$ is discussed above. If $c=d=e=0=a=b$, then any vector $Y= f E_6$, $f \in \mathbb R$ is geodesic since it lies in $\zeta$.

In the case $a \ne 0$ the geodesic vectors of $\left( \mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle \right)$ are in the set

$$\begin{aligned} \begin{aligned} \overline{C_2}=&\Big \{ a (E_1-\frac{\beta _6}{\alpha _4} E_2)+c E_3+d E_4 +e E_5+f E_6: a \ne 0,\\&a e= \frac{f}{\alpha _3} \Big (\frac{a \beta _6 \beta _8}{\alpha _4}+c\frac{\alpha _3 \alpha _4}{\alpha _1} -a\beta _5\Big ),\\&a c=\frac{f}{\alpha _1} \left( c\beta _7+d\beta _8+e\alpha _4\right) -\frac{a}{\alpha _1}\left( \beta _1 d +\beta _2 e+\beta _3 f\right) ,\\&ad = \frac{a}{\alpha _2}\Big ((\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2\beta _8}{\alpha _4})e- \beta _4f\Big )+\frac{f}{\alpha _2}\Big (\frac{a \beta _6 \beta _7}{\alpha _4}-d\frac{\alpha _3 \alpha _4}{\alpha _1}\Big ),\\&d( \alpha _2c+ \alpha _3e)= a \frac{\beta _6}{\alpha _4}\left( \alpha _1 c+\beta _1 d+\beta _2 e+\beta _3 f\right) + c e \Big (\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2\beta _8}{\alpha _4}\Big )\\ {}&-f \left( c \beta _4+d\beta _5+e\beta _6 \right) \Big \}. \end{aligned} \end{aligned}$$

(3.21)

If $f=0$, then from (3.21) it follows that $e=d=c=0$. Hence any element $Y=a(E_1-\frac{\beta _6}{\alpha _4} E_2)$ lies in . Therefore we may assume that $f \ne 0$. The intersection $C_1 \cap C_2$ is empty since for the elements of $C_1$ one has $a=0$ and $f =0$ but for the elements in $C_2$ we have $a f \ne 0$. This completes the proof. $\square$

Theorem 3.8

Let $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ be a metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.5). The flat totally geodesic subalgebras of dimension $>1$ in $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ are the 2-dimensional subalgebras:

• (1) $\mathfrak {h}_{10}= \text {span}(E_3, E_5)$ in the case $\alpha _3 \alpha _4 \beta _1+ \alpha _2 \alpha _1 \beta _8=0$,

• (2) $\mathfrak {h}_{7}= \text {span} (E_1-\frac{\beta _6}{\alpha _4}E_2, \quad E_6)$ in the case $\beta _3=0$,   $\beta _4=\frac{\beta _6 \beta _7}{\alpha _4}$,  $\beta _5=\frac{\beta _6 \beta _8}{\alpha _4}$,

• (3) $\mathfrak {h}_{8}= \text {span}(E_2+k_1E_4+k_2E_5, \quad E_3+l_1E_4+l_2E_5)$ if and only if the equations

  $$\begin{aligned} \begin{aligned} \beta _7+l_1\beta _8+l_2\alpha _4+k_1\frac{\alpha _3 \alpha _4}{\alpha _1}=0, \\ \beta _1 k_1+\beta _2 k_2+k_1 \alpha _3 k_2=0,\\ \alpha _2 l_1 -(\frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})l_2+l_1 \alpha _3 l_2=0,\\ \alpha _1+\beta _1 l_1+\beta _2 l_2+\alpha _2 k_1-(\frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})k_2+\alpha _3(k_1l_2+l_1 k_2)=0 \end{aligned} \end{aligned}$$

  (3.22)

  are satisfied,

• (4) $\mathfrak {h}_{9}= \text{ span }(E_2+\frac{\beta _8 \alpha _1}{\alpha _3 \alpha _4} E_3- \frac{1}{\alpha _3}(\beta _1+ \frac{\beta _8 \alpha _1 \alpha _2}{\alpha _3 \alpha _4} ) E_5, \quad E_4)$ if and only if for $\alpha _i, \beta _j$, $i=1,2,3,4$, $j=1,2,8$ the equation

  $$\begin{aligned} \frac{(\alpha _1)^2 \beta _8}{\alpha _3 \alpha _4} + \Big (\beta _1 + \frac{\beta _8 \alpha _1 \alpha _2}{\alpha _3 \alpha _4}\Big )\Big (- \frac{\beta _2}{\alpha _3}+\frac{\beta _8 \alpha _1}{(\alpha _3)^2 \alpha _4}( \frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})\Big )=0 \end{aligned}$$

  holds.

Proof

According to Proposition 2.3 any totally geodesic subalgebra $\mathfrak {h}$ of $(\mathfrak {n}_{6,16}(\alpha _i, \beta _j ),\langle .,.\rangle )$ has dimension less or equal to 3. Firstly we determine the 2- and the 3-dimensional abelian subalgebras of the Lie algebra $\mathfrak {n}_{6,16}(\alpha _i, \beta _j )$.

The 3- and the 2-dimensional abelian subalgebras of $\left( \mathfrak {n}_{6,16}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ are:

$$\begin{aligned}&\mathfrak {h}_{1}= \text{ span } \left( E_1+k_1E_2+k_2E_3+k_3E_4, \quad E_5, \quad E_6 \right) \text{ with } \beta _6+k_1 \alpha _4=0, \\ {}&\mathfrak {h}_{2}= \text{ span } \left( E_3+k_1E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{3}= \text{ span } \left( E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{4}= \text{ span } \left( E_2+k_1E_4+k_2E_5, \quad E_3+l_1E_4+l_2E_5, \quad E_6 \right) \\ {}&\qquad \text{ with } \beta _7+l_1\beta _8+ l_2\alpha _4+k_1 \frac{\alpha _3 \alpha _4}{\alpha _1}=0,\\ {}&\mathfrak {h}_{5}= \text{ span } \left( E_2+k_1E_3+k_2E_5, \quad E_4+l_1E_5, \quad E_6 \right) \\ {}&\qquad \text{ with } \beta _8+l_1 \alpha _4-k_1\frac{\alpha _3 \alpha _4}{\alpha _1}=0,\\ {}&\mathfrak {h}_{6}= \text{ span } \left( E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_6, \quad E_5+ s_1E_6 \right) \text{ with } k_1=-\frac{\beta _6}{\alpha _4},\\ {}&\mathfrak {h}_{7}= \text{ span } \left( E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5, \quad E_6\right) ,\\ {}&\mathfrak {h}_{8}= \text{ span } \left( E_2+k_1E_4+k_2E_5+k_3E_6, \quad E_3+l_1E_4+l_2E_5+l_3E_6 \right) \\ {}&\qquad \text{ with } \beta _7+l_1\beta _8+l_2\alpha _4+k_1\frac{\alpha _3 \alpha _4}{\alpha _1}=0,\\ {}&\mathfrak {h}_{9}= \text{ span } \left( E_2+k_1E_3+k_2E_5+k_3E_6, \quad E_4+l_1E_5+l_2E_6 \right) \\ {}&\qquad \text{ with } \beta _8+l_1\alpha _4-k_1\frac{\alpha _3 \alpha _4}{\alpha _1}=0,\\ {}&\mathfrak {h}_{10}= \text{ span } \left( E_3+k_1E_4+k_2E_6, \quad E_5+ s_1E_6 \right) ,\\ {}&\mathfrak {h}_{11}= \text{ span } \left( E_2+k_1E_3+k_2E_4+k_3E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{12}= \text{ span } \left( E_3+l_1E_4+l_2E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{13}= \text{ span } \left( E_4+l_2E_6, \quad E_5+ s_1E_6 \right) ,\\ {}&\mathfrak {h}_{14}= \text{ span } \left( E_4+l_1E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{15}= \text{ span } \left( E_5, \quad E_6 \right) , \end{aligned}$$

where $k_1, k_2, k_3, k_4, l_1, l_2, l_3, s_1 \in \mathbb {R}$.

The vector $E_5+E_6 \in \mathfrak {h}_{1} \cap \mathfrak {h}_{2} \cap \mathfrak {h}_{3} \cap \mathfrak {h}_{15}$ is not geodesic, because putting $a=b=c=d=0$, $e=f=1$ into the fourth equation of the system (3.19) we obtain the contradiction $\alpha _4 =0$. Therefore the subalgebras $\mathfrak {h}_{1}$, $\mathfrak {h}_{2}$, $\mathfrak {h}_{3}$ and $\mathfrak {h}_{15}$ are not flat totally geodesic. Hence they are excluded. The vector $E_3+l_1E_4+l_2E_5+E_6 \in \mathfrak {h}_{4} \cap \mathfrak {h}_{12}$ is not geodesic, since putting $a=b=0$, $c=f=1$, $d=l_1$, $e=l_2$ into the second equation of the system (3.19) we get the contradiction $\frac{\alpha _3 \alpha _4}{\alpha _1} =0$. Hence the subalgebras $\mathfrak {h}_{4}$ and $\mathfrak {h}_{12}$ are not flat totally geodesic. The vector $E_4+ l_1 E_5+E_6 \in \mathfrak {h}_{5} \cap \mathfrak {h}_{14}$ is not geodesic, because substituting $a=b=c=0$, $d=1=f$, $e=l_1$ into the third equation of the system (3.19) we receive the contradiction $\frac{\alpha _3 \alpha _4}{\alpha _1} =0$. Hence the subalgebras $\mathfrak {h}_{5}$ and $\mathfrak {h}_{14}$ are not flat totally geodesic. The vector $E_5+ s_1 E_6 \in \mathfrak {h}_{6} \cap \mathfrak {h}_{10} \cap \mathfrak {h}_{13}$ is geodesic if and only if for $a=b=c=d=0$, $e=1$, $f=s_1$ the system (3.19) of equations holds. From the fourth and fifth equations of (3.19) we have $\alpha _4 s_1=0$, $\beta _6 s_1=0$. Since $\alpha _4 \ne 0$ we get $s_1=0$. The vector $E_4+ l_2 E_6 \in \mathfrak {h}_{13}$ is geodesic precisely if for $a=b=c=e=0$, $d=1$, $f=l_2$ the system (3.19) of equations is satisfied. From the third, fourth and fifth equations of (3.19) we have $\frac{\alpha _3 \alpha _4}{\alpha _1} l_2=0$, $\beta _8 l_2=0$, $\beta _5 l_2=0$. Since $\frac{\alpha _3 \alpha _4}{\alpha _1} \ne 0$ one has $l_2=0$. But the vector $E_4+E_5 \in \mathfrak {h}_{13}$ is not geodesic because the substitution $a=b=c=f=0$, $d=e=1$ into the fifth equation of the system (3.19) leads to the contradiction $\alpha _3=0$. Therefore the subalgebra $\mathfrak {h}_{13}$ is not flat totally geodesic. The vector $E_2+ k_1 E_3+ k_2 E_4+ k_3 E_5+E_6 \in \mathfrak {h}_{11}$ is not geodesic, since putting $a=0$, $b=f=1$, $c=k_1$, $d=k_2$, $e=k_3$ into the first equation of the system (3.19) we obtain the contradiction $\alpha _4=0$. Hence the subalgebra $\mathfrak {h}_{11}$ is not flat totally geodesic. The vector $E_3+ k_1 E_4+ k_2 E_6 \in \mathfrak {h}_{10}$ is geodesic if and only if for $a=b=e=0$, $c=1$, $d=k_1$, $f=k_2$ the system (3.19) of equations holds. The second, fourth and fifth equations yield

$$\begin{aligned} k_2 \frac{\alpha _3 \alpha _4}{\alpha _1}=0, \quad k_2(\beta _7 + k_1 \beta _8)=0, \quad \alpha _2 k_1+ \beta _4 k_2 +k_1 \beta _5 k_2=0. \end{aligned}$$

(3.23)

As $\frac{\alpha _3 \alpha _4}{\alpha _1} \ne 0, \alpha _2 \ne 0$ we receive that $k_2=k_1=0$. The vector $E_3+ E_5 \in \mathfrak {h}_{10}$ is geodesic precisely if for $a=b=d=f=0$, $c=e=1$, the system (3.19) of equations is satisfied. The fifth equation gives $\frac{\alpha _3\beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4}=0$. Hence the subalgebra $\mathfrak {h}_{10}=\text {span}(E_3, E_5)$ is flat totally geodesic if and only if $\alpha _3 \alpha _4 \beta _1+ \alpha _2 \alpha _1 \beta _8=0$. This proves the case (1).

The vector $E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $a=1, e=0$, $b=k_1=-\frac{\beta _6}{\alpha _4}, c=k_2, d=k_3, f=k_4$ the system (3.19) of equations holds. The second equation of (3.19) gives

$$\begin{aligned} k_4(\beta _5+\beta _8 k_1-\frac{\alpha _3 \alpha _4}{\alpha _1}k_2)=0 \end{aligned}$$

(3.24)

Furthermore, the element $E_1+E_5+ k_1E_2+k_2E_3+k_3E_4+k_4E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $a=1, e=1$, $b=k_1=-\frac{\beta _6}{\alpha _4}, c=k_2, d=k_3, f=k_4$ the system (3.19) of equations is valid. From the second equation of (3.19) one has

$$\begin{aligned} \alpha _3+ k_4(\beta _5+\beta _8 k_1-\frac{\alpha _3 \alpha _4}{\alpha _1}k_2)=0. \end{aligned}$$

(3.25)

Comparing (3.25) with (3.24) we obtain the contradiction $\alpha _3=0$, which excludes the subalgebra $\mathfrak {h}_{6}$.

The vector $E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5 \in \mathfrak {h}_{7}$ is geodesic precisely if for $a=1, f=0$, $b=k_1, c=k_2, d=k_3, e=k_4$ the system (3.19) of equations holds. From the second equation of (3.19) we get $\alpha _3 k_4=0$. As $\alpha _3 \ne 0$ we have $e=k_4=0$. Using this from the third equation of (3.19) we obtain $\alpha _2 k_3=0$. Since $\alpha _2 \ne 0$ we receive that $d=k_3=0$. Taking into account that $e=d=0$ from the fourth equations of (3.19) we have $\alpha _1 k_2=0$. As $\alpha _1 \ne 0$ we obtain $k_2=0$. Using this the vector $E_1+k_1 E_2+E_6 \in \mathfrak {h}_{7}$ is geodesic if for $a=1, b=k_1, c=d=e=0, f=1$ the system (3.19) is valid. From the first equation of (3.19) we receive that $k_1=-\frac{\beta _6}{\alpha _4}$. Using this it follows from the second equation of (3.19) that $\beta _5=\frac{\beta _6 \beta _8 }{\alpha _4}$, whereas from the third equation of (3.19) that $\beta _4=\frac{\beta _6\beta _7}{\alpha _4}$. From the fourth equation of (3.19) we obtain that $\beta _3=0$. This proves case (2) of the Theorem.

Now, we consider the subalgebra $\mathfrak {h}_{8}$. The element $E_2+k_1E_4+k_2E_5+k_3E_6 \in \mathfrak {h}_{8}$ is geodesic if and only if for $a=c=0$, $b=1, d=k_1, e=k_2, f=k_3$ the system (3.19) of equations holds. The first equation of (3.19) yields $\alpha _4 k_3=0$. As $\alpha _4 \ne 0$ one gets $f=k_3=0$. Applying this the fifth equation of (3.19) gives

$$\begin{aligned} \beta _1 k_1+\beta _2 k_2+k_1 \alpha _3 k_2=0. \end{aligned}$$

(3.26)

The vector $E_3+l_1E_4+l_2E_5+l_3E_6 \in \mathfrak {h}_{8}$ is geodesic if and only if for $a=b=0$, $c=1, d=l_1, e=l_2, f=l_3$ the system (3.19) of equations is satisfied. It follows from the second equation of (3.19) that $\frac{\alpha _3 \alpha _4}{\alpha _1} l_3=0$. Since $\frac{\alpha _3 \alpha _4}{\alpha _1} \ne 0$ we receive $f=l_3=0$. Using this the fifth equation of (3.19) gives

$$\begin{aligned} \alpha _2 l_1 -\Big (\frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4}\Big )l_2+l_1 \alpha _3 l_2=0. \end{aligned}$$

(3.27)

Let us consider the vector $E_2+E_3+(k_1+l_1) E_4+(k_2+l_2) E_5 \in \mathfrak {h}_{8}$. It is geodesic if and only if and only if for $a=f=0, b=c=1, d=k_1+l_1, e=k_2+l_2$ the system (3.19) of equations is valid. From the fifth equation of (3.19) one has

$$\begin{aligned}{} & {} \alpha _1+\beta _1 (k_1+ l_1) +\beta _2 (k_2+ l_2) +\alpha _2( k_1+ l_1)\nonumber \\{} & {} \qquad -\Big (\frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4}\Big )(k_2+l_2)+(k_1+l_1) \alpha _3 (k_2+l_2)=0. \end{aligned}$$

(3.28)

Taking into account (3.26) and (3.27) equation (3.28) reduces to

$$\begin{aligned} \alpha _1+\beta _1 l_1 +\beta _2 l_2 +\alpha _2 k_1 -\Big (\frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4}\Big ) k_2+ \alpha _3 (k_1 l_2+ l_1 k_2)=0. \nonumber \\ \end{aligned}$$

(3.29)

Therefore the subalgebra $\mathfrak {h}_{8}$ is flat totally geodesic if and only if it satisfies the conditions of the case (3.22) in the Theorem.

Finally we treat the subalgebra $\mathfrak {h}_{9}$. The vector $E_4+ l_1 E_5+ l_2 E_6 \in \mathfrak {h}_{9}$ is geodesic precisely if for $a=b=c=0$, $d=1$, $e=l_1$, $f=l_2$ the system (3.19) of equations is satisfied. From the third equation we obtain $\frac{\alpha _3 \alpha _4}{\alpha _1} l_2=0$. As $\frac{\alpha _3 \alpha _4}{\alpha _1} \ne 0$ we have $f=l_2=0$. Using this the fifth equation of (3.19) gives $\alpha _3 l_1=0$. Since $\alpha _3 \ne 0$ one gets $l_1=0$. Using this from the condition $\beta _8+l_1\alpha _4-k_1\frac{\alpha _3 \alpha _4}{\alpha _1}=0$ to be abelian the subalgebra $\mathfrak {h}_{9}$ we receive that $k_1=\frac{\beta _8 \alpha _1}{\alpha _3 \alpha _4}$. The element $E_2+ k_1 E_3+ k_2 E_5 + k_3 E_6 \in \mathfrak {h}_{9}$ is geodesic if and only if for $a=d=0$, $b=1$, $c=k_1$, $e=k_2$, $f=k_3$ the system (3.19) of equations holds. From the first equation of (3.19) we have $k_3 \alpha _4=0$. As $\alpha _4 \ne 0$ it follows that $f=k_3=0$. Using this the fifth equation of (3.19) yields

$$\begin{aligned} \alpha _1 k_1+\beta _2 k_2-k_1 k_2(\frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})=0. \end{aligned}$$

(3.30)

The element $E_2+ k_1 E_3+ k_2 E_5+ E_4 \in \mathfrak {h}_{9}$ is geodesic precisely if for $a=f=0$, $b=d=1$, $c=k_1$, $e=k_2$ the system (3.19) of equations is valid. From the fifth equation of (3.19) we obtain

$$\begin{aligned} \alpha _1 k_1+ \beta _1 +\beta _2 k_2+ k_1 \Big (\alpha _2 -( \frac{\alpha _3 \beta _1}{\alpha _1}+\frac{\alpha _2 \beta _8}{\alpha _4})k_2 \Big )+ \alpha _3 k_2=0. \end{aligned}$$

(3.31)

Taking into account equation (3.29) equation (3.31) reduces to

$$\begin{aligned} \beta _1 + k_1 \alpha _2 + \alpha _3 k_2=0. \end{aligned}$$

(3.32)

Putting the expression for $k_1$ into (3.32) we obtain that $k_2=-\frac{1}{\alpha _3}\Big (\beta _1 + \frac{\beta _8 \alpha _1 \alpha _2}{\alpha _3 \alpha _4}\Big )$. Substituting the expressions of $k_1$ and $k_2$ into (3.30) for the parameters $\alpha _i$, $\beta _j$, $i=1,2,3,4$, $j=1,2,8$ of the filiform metric Lie algebra $(\mathfrak {n}_{6,16}(\alpha _i,\beta _j ), \langle .,.\rangle )$ we obtain equation (3.31). This proves case (4) of the Theorem. Thus Theorem 3.8 is shown. $\square$

4 Geodesic vectors and flat totally geodesic subalgebras of Lie algebras $\mathfrak {n}_{6,17}(\alpha _{i},\beta _j )$ and $\mathfrak {n}_{6,18}(\alpha _{i},\beta _j )$

In this section we deal with the metric Lie algebras $\left( \mathfrak {n}_{6,17} (\alpha _{i},\beta _j ), \langle .,.\rangle \right)$ and $\left( \mathfrak {n}_{6,18} (\alpha _{i},\beta _j ), \langle .,.\rangle \right)$. The next result describes the sets of geodesic vectors of the metric Lie algebras $(\mathfrak {n}_{6,j}(\alpha _i,\beta _j ),\langle .,.\rangle ), j \in \{17,18\}$.

Theorem 4.1

Let $(\mathfrak {n}_{6,17} (\alpha _{i},\beta _j ),\langle .,. \rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.6). The geodesic vectors of $(\mathfrak {n}_{6,17} (\alpha _{i},\beta _j ),\langle .,. \rangle )$ not belonging to are the non-zero elements of the set $C_1 \cup C_2$, where

$$\begin{aligned} \begin{aligned} C_1:=&\{dE_4+eE_5+fE_6: d(\alpha _3e+\beta _6f)+e\alpha _4f=0 \} \\ {}&\text{ such } \text{ that } \text{ at } \text{ least } \text{ two } \text{ of } \text{ the } \text{ numbers }\ d, e, f \text {are non-zero with exception} \\ {}&\text{ of } \text{ the } \text{ cases: }\\ {}&1. \ f=0, \\ {}&2. \ d=0, \\ {}&3.\ e=0 \text{ with } \beta _6 \ne 0, \\C_2:=&\{bE_2+cE_3+dE_4+eE_5: b(\alpha _1 c+ \beta _1 d+\beta _2e)+c(\alpha _2 d+\beta _4 e)+d\alpha _3 e =0 \}\\ {}&\text{ such } \text{ that } \text{ at } \text{ least } \text{ two } \text{ of } \text{ the } \text{ numbers } b, c, d, e \text { are non-zero with exception}\\ {}&\text{ of } \text{ the } \text{ cases: }\\ {}&1. \ b=c=0,\qquad \\ {}&2. \, b=e=0, \\ {}&3.\ d=e=0, \\ {}&4. \, b=d=0 \, \text{ with } \beta _4 \ne 0,\\ {}&5. \ c=d=0 \text{ with } \beta _2 \ne 0, \\ {}&6.\ c=e=0 \text{ with } \beta _1 \ne 0. \end{aligned} \end{aligned}$$

Let $(\mathfrak {n}_{6,18} (\alpha _{i},\beta _j ),\langle .,. \rangle )$ be the metric Lie algebra defined on $\mathbb {E}^6$ by the non-vanishing commutators given by (1.6) with $\alpha _5=0$. The geodesic vectors of $(\mathfrak {n}_{6,18} (\alpha _{i},\beta _j ),\langle .,. \rangle )$ not belonging to are the non-zero elements of the set

$$\begin{aligned} C_3&:=\{bE_2+cE_3+dE_4+eE_5+fE_6: b( \alpha _1 c+ \beta _1 d+ \beta _2e+ \beta _3f)\\ {}&\quad +c(\alpha _2 d+\beta _4 e+\beta _5 f)+d(\alpha _3e+\beta _6f)+e\alpha _4 f =0, b, c, d, e, f \in \mathbb R \},\\ {}&\quad \text{ such } \text{ that } \text{ at } \text{ least } \text{ two } \text{ of } \text{ the } \text{ numbers } \ b, c, d, e, f \\ {}&\quad \text{ are } \text{ non-zero } \text{ with } \text{ exception } \text{ of } \text{ the } \text{ cases: } \\ {}&1.\ b=c=d=0,\qquad \\ {}&2.\ b=c=f=0,\qquad \\ {}&3.\ d=e=f=0, \\ {}&4.\ b=c=e=0 \text{ with } \beta _6 \ne 0,\qquad \\ {}&5. \, c=d=f=0 \text{ with } \beta _2 \ne 0, \\ {}&6.\ c=d=e=0 \text{ with } \beta _3 \ne 0. \end{aligned}$$

Proof

Using the commutators (1.6) and the claim (1.2) we obtain that the non-zero element $Y=aE_1+bE_2+cE_3+dE_4+eE_5+fE_6 \in \mathfrak {n}_{6,17}(\alpha _i,\beta _j )$ is geodesic if and only if for the real numbers a, b, c, d, e, f with respect to a distinguished orthonormal basis $\{E_1, E_2, E_3, E_4, E_5, E_6\}$ the system of equations

$$\begin{aligned} \left\{ \begin{array}{ll} &{}a f =0, \, a e=0, \\ &{}a\alpha _2d+b\alpha _5f=0,\\ &{}a(\alpha _1c+\beta _1 d)-c\alpha _5f=0,\\ &{}b( \alpha _1 c+ \beta _1 d+ \beta _2e+ \beta _3f) +c(\alpha _2 d+\beta _4 e+\beta _5 f)+d(\alpha _3 e+\beta _6f)+e\alpha _4f =0 \end{array} \right. \nonumber \\ \end{aligned}$$

(4.1)

is satisfied. If $a=0$, then the third and the fourth equations of (4.1) give $bf=0=cf$. In the case $b=c=0=a$, the vector $Y=dE_4+eE_5+fE_6$ is geodesic of $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ if and only if it lies in the set $\overline{C_1}:=\{dE_4+eE_5+fE_6: d(\alpha _3e+\beta _6f)+e\alpha _4f=0 \}$, $d, e, f \in \mathbb R$. For $f=0$ the element $Y=d E_4+ e E_5 \in \overline{C_1}$, $d, e \in \mathbb R$, satisfies the condition $\alpha _3 d e=0$. Since $\alpha _3 \ne 0$ we obtain that either $d=0$ and the element $Y=e E_5$ is in , or $e=0$ and the element $Y=d E_4$ is in . If $d=0$, then for the element $Y=e E_5+ f E_6 \in \overline{C_1}$, $e, f \in \mathbb R$, the condition $\alpha _4 e f=0$ holds. Since $\alpha _4 \ne 0$ we receive that either $e=0$ and the element $Y=f E_6$ is in $\zeta$, or $f=0$ and the element $Y=e E_5$ is in . If $e=0$ and $\beta _6 \ne 0$, then for the element $Y=d E_4+ f E_6 \in \overline{C_1}$, $d, f \in \mathbb R$, the condition $d f=0$ is satisfied. Hence we get either $d=0$ and the element $Y=f E_6$ is in $\zeta$, or $f=0$ and the element $Y=d E_4$ is in . Therefore the conditions for the set $C_1$ in the theorem are proved.

In the case $f=0=a$ the non-zero vector $Y=bE_2+cE_3+dE_4+eE_5$ is geodesic of $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ if and only if it lies in the set $\overline{C_2}:=\{bE_2+cE_3+dE_4+eE_5: b(\alpha _1 c+ \beta _1 d+\beta _2e)+c(\alpha _2 d+\beta _4 e)+d\alpha _3 e =0 \}, b, c, d, e \in \mathbb R$. If $b=c=0$, then the element $Y=d E_4+ e E_5 \in \overline{C_2}$, $d, e \in \mathbb R$, satisfies the condition $\alpha _3 d e=0$. Since $\alpha _3 \ne 0$ we obtain that either $d=0$ and the element $Y=e E_5$ is in , or $e=0$ and the element $Y=d E_4$ is in . If $b=e=0$, then for the element $Y=c E_3+ d E_4 \in \overline{C_2}$, $c, d \in \mathbb R$, the condition $\alpha _2 c d=0$ holds. Since $\alpha _2 \ne 0$ we receive that either $c=0$ and the element $Y=d E_4$ is in , or $d=0$ and the element $Y=c E_3$ is in . If $d=e=0$, then for the element $Y=b E_2+c E_3 \in \overline{C_2}$, $b, c \in \mathbb R$, the condition $\alpha _1 b c=0$ is satisfied. As $\alpha _1 \ne 0$ we get either $b=0$ and the element $Y=c E_3$ is in , or $c=0$ and the element $Y=b E_2$ is in . If $b=d=0$ and $\beta _4 \ne 0$, then for the element $Y=c E_3+ e E_5 \in \overline{C_2}$ the condition $c e=0$ holds. Hence we have either $c=0$ and the element $Y=e E_5$ is in., or $e=0$ and the element $Y=c E_3$ is in . If $c=d=0$ and $\beta _2 \ne 0$, then for the element $Y=b E_2+ e E_5 \in \overline{C_2}$, $b, e \in \mathbb R$ the condition $b e=0$ holds. Therefore we get either $b=0$ and the element $Y=e E_5$ is in , or $e=0$ and the element $Y=b E_2$ is in . Finally, if $c=e=0$ and $\beta _1 \ne 0$, then for the element $Y=b E_2+ d E_4 \in \overline{C_2}$, $b, d \in \mathbb R$ the condition $b d=0$ holds. It follows that either $b=0$ and the element $Y=d E_4$ is in , or $d=0$ and the element $Y=b E_2$ is in . This proves the conditions for the set $C_2$ in the theorem.

If $f=e=0$, then from the third and fourth equations of (4.1) we obtain that $ad=0=ac$. The case $a=0=f=e$ is discussed above. In the case $d=c=0=f=e$ any vector $Y=aE_1+bE_2$, $a, b \in \mathbb R$ is geodesic because it lies in . The intersection $C_1 \cap C_2$ is empty, because for any element of $C_1$ one has $b =c=0$ and any element of $C_2$ one gets $f=0$. Hence the claim above the set of the geodesic vectors of $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ is shown.

In the case of the metric Lie algebra $\left( \mathfrak {n}_{6,18}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ the system (4.1) of equations is satisfied with $\alpha _5=0$. Therefore we obtain that $af=0=ae=ad=ac$. The case $f=e=d=c=0$ is discussed above. In the case $a=0$ the vector $Y=bE_2+cE_3+dE_4+eE_5+fE_6$ is geodesic of $\left( \mathfrak {n}_{6,18}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ precisely if it lies in the set $\overline{C_3}:=\{bE_2+cE_3+dE_4+eE_5+fE_6: b( \alpha _1 c+ \beta _1 d+ \beta _2e+ \beta _3f)+c(\alpha _2 d+\beta _4 e+\beta _5 f)+ d(\alpha _3e+\beta _6f)+e\alpha _4 f =0, b, c, d, e, f \in \mathbb R \}$. If $b=c=d=0$, then the element $Y=e E_5+ f E_6 \in \overline{C_3}$, $e, f \in \mathbb R$, satisfies the condition $\alpha _4 e f=0$. Since $\alpha _4 \ne 0$ we obtain that either $e=0$ and the element $Y=f E_6$ is in $\zeta$, or $f=0$ and the element $Y=e E_5$ is in . If $b=c=f=0$, then for the element $Y=d E_4+ e E_5 \in \overline{C_3}$, $d, e \in \mathbb R$, the condition $\alpha _3 d e=0$ holds. Since $\alpha _3 \ne 0$ we receive that either $d=0$ and the element $Y=e E_5$ is in , or $e=0$ and the element $Y=d E_4$ is in . If $d=e=f=0$, then for the element $Y=b E_2+c E_3 \in \overline{C_3}$, $b, c \in \mathbb R$, the condition $\alpha _1 b c=0$ is satisfied. As $\alpha _1 \ne 0$ we get either $b=0$ and the element $Y=c E_3$ is in , or $c=0$ and the element $Y=b E_2$ is in . In the case $b=c=e=0$ and $\beta _6 \ne 0$, then for the element $Y=d E_4+ f E_6 \in \overline{C_3}$, $d, f \in \mathbb R$, the condition $d f=0$ holds. Hence we have either $d=0$ and the element $Y=f E_6$ is in $\zeta$, or $f=0$ and the element $Y=d E_4$ is in . If $c=d=f=0$ and $\beta _2 \ne 0$, then for the element $Y=b E_2+ e E_5 \in \overline{C_3}$, $b, e \in \mathbb R$, the condition $b e=0$ holds. Therefore we get either $b=0$ and the element $Y=e E_5$ is in , or $e=0$ and the element $Y=b E_2$ is in . Finally, if $c=d=e=0$ and $\beta _3 \ne 0$, then for the element $Y=b E_2+ f E_6 \in \overline{C_3}$, $b, f \in \mathbb R$ the condition $b f=0$ holds. It follows that either $b=0$ and the element $Y=f E_6$ is in $\zeta$, or $f=0$ and the element $Y=b E_2$ is in . This shows the conditions for the set $C_3$ in the theorem. Hence Theorem 4.1 is proved. $\square$

The flat totally geodesic subalgebras of $(\mathfrak {n}_{6,17}(\alpha _i, \beta _j ),\langle .,.\rangle )$ are given in the following theorem.

Theorem 4.2

Let $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.6). The flat totally geodesic subalgebras of dimension $>1$ in the metric Lie algebra $(\mathfrak {n}_{6,17}(\alpha _i,\beta _j ), \langle .,.\rangle )$ are the 2-dimensional subalgebras:

• (1) $\mathfrak {h}_{9}= \text {span}(E_4-\frac{\alpha _3}{\alpha _4}E_6, \quad E_5)$ in the case $\beta _6=0$,

• (2) $\mathfrak {h}_{2}= \text {span}(E_2+k_1E_3-\frac{\beta _2+\beta _4k_1}{\alpha _3}E_4, \quad E_5)$, where $k_1$ is a solution of the equation

  $$\begin{aligned} \alpha _2 \beta _4 k_1^2+(\beta _1 \beta _4+\alpha _2\beta _2-\alpha _1 \alpha _3)k_1+\beta _1 \beta _2=0, \end{aligned}$$

  (4.2)

• (3) $\mathfrak {h}_{4}= \text {span} \left( E_3, E_5 \right)$ in the case $\beta _4=0$,

• (4) $\mathfrak {h}_{1}= \text {span} (E_2-\frac{\beta _1 +\alpha _3k_2}{\alpha _2} E_3+k_2E_5, \quad E_4)$, where $k_2$ is a solution of the equation

  $$\begin{aligned} \alpha _3 \beta _4 k_2^2+(\alpha _1 \alpha _3+\beta _1\beta _4-\alpha _2 \beta _2) k_2+\alpha _1 \beta _1=0, \end{aligned}$$

  (4.3)

• (5) $\mathfrak {h}_{3}= \text {span} (E_3-\frac{\alpha _2}{\alpha _3}E_5, \quad E_4)$ in the case $\beta _4=0$,

• (6) $\mathfrak {h}_{5}= \text {span} \left( E_4, E_6 \right)$ in the case $\beta _6=0$.

Proof

As pointed out in [1, Theorem 1.19], the metric Lie algebra $(\mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle )$ does not have a totally geodesic subalgebra greater than 2. Hence we compute only the 2-dimensional abelian subalgebras in the Lie algebra $\mathfrak {n}_{6,17}(\alpha _i,\beta _j )$. These subalgebras are the following:

$$\begin{aligned}&\mathfrak {h}_{1}= \text {span} \left( E_2+k_1E_3+k_2E_5+k_3E_6, \quad E_4+l_1E_5+l_2E_6 \right) ,\\&\mathfrak {h}_{2}= \text {span} \left( E_2+k_1E_3+k_2E_4+k_3E_6, \quad E_5+l_1E_6 \right) ,\\&\mathfrak {h}_{3}= \text {span} \left( E_3+k_1E_5+k_2E_6, \quad E_4+l_1E_5+l_2E_6 \right) ,\\&\mathfrak {h}_{4}= \text {span} \left( E_3+k_1E_4+k_2E_6, \quad E_5+l_1E_6 \right) ,\\&\mathfrak {h}_{5}= \text {span} \left( E_4+k_1E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{6}= \text {span} \left( E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5, \quad E_6\right) ,\\&\mathfrak {h}_{7}= \text {span} \left( E_2+k_1E_3+k_2E_4+k_3E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{8}= \text {span} \left( E_3+k_1E_4+k_2E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{9}= \text {span} \left( E_4+k_1E_6, \quad E_5+l_1E_6 \right) ,\\&\mathfrak {h}_{10}= \text {span} \left( E_5, E_6 \right) , \end{aligned}$$

where $k_1, k_2, k_3, l_1, l_2 \in \mathbb {R}$.

The subalgebra $\mathfrak {h}_{10}$ is not flat totally geodesic because for the vector $E_5+E_6$ the fifth equation of (4.1) gives the contradiction $\alpha _4=0$. Hence the subalgebra $\mathfrak {h}_{10}$ is excluded. Since for the vector $Y=E_1+k_1E_2+k_2E_3+k_3E_4+k_4E_5+E_6 \in \mathfrak {h}_{6}$ the first equation of the system (4.1) yields the contradiction $1=0$ the subalgebra $\mathfrak {h}_{6}$ is not flat totally geodesic (see Lemma 2.1). Therefore the subalgebra $\mathfrak {h}_{6}$ is excluded. The vector $E_2+k_1E_3+k_2E_4+k_3E_5 +E_6 \in \mathfrak {h}_{7}$ is not geodesic since the third equation of (4.1) gives the contradiction $\alpha _5=0$. Therefore the subalgebra $\mathfrak {h}_{7}$ is not flat totally geodesic. The vector $Y=E_3+k_1E_4+k_2E_5+E_6 \in \mathfrak {h}_{8}$ is not geodesic because the fourth equation of (4.1) leads to the contradiction $\alpha _5=0$. Hence the subalgebra $\mathfrak {h}_{8}$ is not flat totally geodesic.

The vector $E_5+l_1 E_6 \in \mathfrak {h}_{2} \cap \mathfrak {h}_{4} \cap \mathfrak {h}_{9}$ is geodesic if and only if for $a=b=c=d=0$, $e=1, f=l_1$ the system (4.1) of equations is satisfied. From the fifth equation of (4.1) one has $\alpha _4 l_1=0$. As $\alpha _4 \ne 0$ we get $l_1=0$.

Now, we consider the subalgebra $\mathfrak {h}_{9}$. The non-zero vector $E_4+ k_1 E_6 \in \mathfrak {h}_{9}$ is geodesic precisely if for $a=b=c=e=0$, $d=1$, $f=k_1$ the system (4.1) of equations holds. From the fifth equation we obtain

$$\begin{aligned} \beta _6 k_1=0. \end{aligned}$$

(4.4)

Furthermore, the element $E_4+k_1E_6+E_5 \in \mathfrak {h}_{9}$ is geodesic if for $a=b=c=0$, $d=e=1, f=k_1$ the system (4.1) of equations is valid. It follows from the fifth equation of (4.1) that

$$\begin{aligned} \alpha _3+\beta _6 k_1+\alpha _4 k_1=0. \end{aligned}$$

(4.5)

Taking into account (4.4) equation (4.5) reduces to

$$\begin{aligned} \alpha _3+\alpha _4 k_1=0. \end{aligned}$$

(4.6)

The equation (4.6) gives $k_1=-\frac{\alpha _3}{\alpha _4}$. Putting this expression into (4.4) on has $\beta _6=0$. Therefore the case (1) is proved.

Next we deal with the subalgebra $\mathfrak {h}_{2}$. The element $E_2+k_1 E_3+k_2E_4+k_3E_6 \in \mathfrak {h}_{2}$ is geodesic precisely if for $a=e=0$, $b=1, c=k_1, d=k_2, f=k_3$ the system (4.1) of equations holds. From the third equation of (4.1) we obtain $\alpha _5 k_3=0$. Since $\alpha _5 \ne 0$ we receive $k_3=0=f$. Using this the fifth equation of (4.1) gives

$$\begin{aligned} \alpha _1 k_1+\beta _1k_2+k_1\alpha _2 k_2=0. \end{aligned}$$

(4.7)

Moreover, the element $E_2+k_1 E_3+k_2 E_4+E_5 \in \mathfrak {h}_{2}$ is geodesic if and only if for $a=f=0, b=e=1, c=k_1, d=k_2$ the system (4.1) of equations is valid. From the fifth equation of (4.1) we receive

$$\begin{aligned} \alpha _1 k_1+\beta _1 k_2+\beta _2+k_1\alpha _2 k_2+k_1\beta _4 +k_2 \alpha _3=0. \end{aligned}$$

(4.8)

Taking into account (4.7) equation (4.8) reduces to

$$\begin{aligned} \beta _2+\beta _4 k_1+\alpha _3 k_2=0. \end{aligned}$$

(4.9)

From (4.9) one has $k_2=-\frac{\beta _2+\beta _4k_1}{\alpha _3}$. Putting this expression into (4.7) we receive the second order equation (4.2). Therefore the case (2) is proved.

Now we treat the subalgebra $\mathfrak {h}_{4}$. The element $E_3+k_1 E_4+k_2 E_6 \in \mathfrak {h}_{4}$ is geodesic if and only if for $a=b=e=0, c=1, d=k_1, f=k_2$ the system (4.1) of equation is satisfied. From the fourth equation of (4.1) we receive $\alpha _5 k_2=0$. As $\alpha _5 \ne 0$ we get $k_2=f=0$. Using this the fifth equation of (4.1) gives $\alpha _2 k_1=0$. Since $\alpha _2 \ne 0$ we obtain $k_1=0$. The vector $E_3+E_5 \in \mathfrak {h}_{4}$ is geodesic precisely if for $a=b=d=f=0$, $c=e=1$, the system (4.1) of equation holds. From the fifth equation of (4.1) we get $\beta _4=0$. This gives the case (3).

The vector $E_4+l_1 E_5+l_2E_6 \in \mathfrak {h}_{1} \cap \mathfrak {h}_{3}$ is geodesic precisely if for $a=b=c=0, d=1, e=l_1, f=l_2$ the system (4.1) of equation is satisfied. The fifth equation of (4.1) yields

$$\begin{aligned} \alpha _3 l_1+\beta _6 l_2+l_1 \alpha _4 l_2=0. \end{aligned}$$

(4.10)

Let us consider the subalgebra $\mathfrak {h}_{1}$. The element $E_2+k_1 E_3+k_2 E_5+k_3E_6 \in \mathfrak {h}_{1}$ is geodesic if and only if for $a=d=0, b=1, c=k_1, e=k_2, f=k_3$ the system (4.1) of equation is valid. From the third equation of (4.1) one has $\alpha _5 k_3=0$. As $\alpha _5 \ne 0$ we receive $k_3=f=0$. Using this the fifth equation of (4.1) gives

$$\begin{aligned} \alpha _1k_1+\beta _2 k_2+k_1\beta _4 k_2=0. \end{aligned}$$

(4.11)

Additionally, the vector $E_2+E_4+k_1E_3+(k_2+l_1) E_5+l_2 E_6 \in \mathfrak {h}_{1}$ is geodesic if and only if for $a=0, b=d=1, c=k_1, e=k_2+l_1, f=l_2$ the system (4.1) of equation is valid. The third equation of (4.1) gives $\alpha _5 l_2=0$. Since $\alpha _5 \ne 0$ we get $l_2=f=0$. Using this from the equation (4.10) we get $\alpha _3 l_1=0$. As $\alpha _3 \ne 0$ we obtain $l_1=e=0$, and from the fifth equation of (4.1) we obtain

$$\begin{aligned} \alpha _1 k_1+\beta _1+\beta _2 k_2+k_1\alpha _2+k_1 \beta _4 k_2+\alpha _3 k_2=0. \end{aligned}$$

(4.12)

Applying (4.11) equation (4.12) reduces to

$$\begin{aligned} \beta _1+\alpha _2k_1+\alpha _3 k_2=0. \end{aligned}$$

The above equation gives $k_1=-\frac{\beta _1 +\alpha _3k_2}{\alpha _2}$. Putting this expression into (4.11) we receive the second order equation (4.2). Hence the case (4) is proved.

Next we deal with the subalgebra $\mathfrak {h}_{3}$. The vector $E_3+k_1 E_5+k_2E_6 \in \mathfrak {h}_{3}$ is geodesic if for $a=b=d=0, c=1, e=k_1, f=k_2$ the system (4.1) of equation is satisfied. From the fourth equation of (4.1) one obtains $\alpha _5 k_2=0$. As $\alpha _5 \ne 0$ we get $k_2=f=0$. Using this the fifth equation of (4.1) gives

$$\begin{aligned} \beta _4 k_1=0. \end{aligned}$$

(4.13)

In addition, the element $E_3+E_4+(k_1+l_1) E_5+l_2 E_6 \in \mathfrak {h}_{3}$ is geodesic if and only if for $a=b=0, c=d=1, e=k_1+l_1, f=l_2$ the system (4.1) of equation is valid. The fourth equation of (4.1) gives $\alpha _5 l_2=0$. Since $\alpha _5 \ne 0$ one has $l_2=f=0$. Using this from the equation (4.10) we get $\alpha _3 l_1=0$. As $\alpha _3 \ne 0$ we receive $l_1=0$, and from the fifth equation of (4.1) we obtain

$$\begin{aligned} \alpha _2+\beta _4 k_1+\alpha _3 k_1=0. \end{aligned}$$

(4.14)

Applying (4.13) equation (4.14) reduces to

$$\begin{aligned} \alpha _2+\alpha _3 k_1=0. \end{aligned}$$

From the above equation we obtain $k_1=-\frac{\alpha _2}{\alpha _3}$. Putting this expression into (4.13) we receive $\beta _4=0$. This proves the case (5).

Finally we treat the subalgebra $\mathfrak {h}_{5}$. The vector $E_4+k_1 E_5 \in \mathfrak {h}_{5}$ is geodesic if and only if for $a=b=c=f=0$, $d=1$, $e=k_1$ the system (4.1) of equations is valid. It follows from the fifth equation of (4.1) that $\alpha _3 k_1=0$. As $\alpha _3 \ne 0$ we gets $k_1=0$. The vector $E_4+E_6 \in \mathfrak {h}_{5}$ is geodesic if and only if for $a=b=c=e=0$, $d=f=1$ the system (4.1) of equations is valid. The fifth equation gives $\beta _6=0$. Therefore the case (6) is proved. This proves Theorem 4.2. $\square$

Now we determine the flat totally geodesic subalgebras of dimension $>1$ in the standard filiform metric Lie algebra $(\mathfrak {n}_{6,18}(\alpha _i, \beta _j ),\langle .,.\rangle )$.

Theorem 4.3

Let $\left( \mathfrak {n}_{6,18}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$ be the metric Lie algebra defined on $\mathbb {E}^6$ by non-vanishing commutators given by (1.6) with $\alpha _5=0$. The flat totally geodesic subalgebras of dimension $>1$ in the metric Lie algebra $(\mathfrak {n}_{6,18}(\alpha _i, \beta _j ),\langle .,.\rangle )$ are:

• (1) The 4-dimensional subalgebras are the following:

  • (a) $\mathfrak {h}_{1}=\text {span}(E_2- \frac{\alpha _1 \alpha _3}{\alpha _2 \alpha _4} E_6, \quad E_3, \quad E_4-\frac{\alpha _3}{\alpha _4} E_6, \quad E_5)$ in the case $\beta _1=\beta _3=\beta _4=\beta _6=0$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$, $\beta _2= \frac{\alpha _1 \alpha _3}{\alpha _2}$,

  • (b) $\mathfrak {h}_{2}= \text {span} (E_2, \quad E_3-\frac{\alpha _2}{\alpha _3} E_5, \quad E_4, \quad E_6 )$ in the case $\beta _1=\beta _3=\beta _4=\beta _6=0$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$, $\beta _2= \frac{\alpha _1 \alpha _3}{\alpha _2}$.

• (2) the 3-dimensional subalgebras are the following:

  • (a) $\mathfrak {h}_{6}= \text {span} \left( E_2\!+\!k_1E_5\!+\!k_2E_6, \!\quad E_3\!+\!l_1E_5\!+\!l_2E_6, \!\quad E_4\!+\!s_1E_5\!+\!s_2E_6 \right)$ if and only if the following equations

    $$\begin{aligned} \begin{aligned} \beta _2 k_1+ \beta _3 k_2+ k_1 \alpha _4 k_2&=0,\\ \beta _4 l_1+ \beta _5 l_2+ l_1 \alpha _4 l_2&=0, \\ \alpha _3 s_1+ \beta _6 s_2+ s_1 \alpha _4 s_2&=0, \\ \alpha _1 +\beta _2 l_1+\beta _3 l_2+\beta _4 k_1+\beta _5 k_2+k_1 l_2 \alpha _4+l_1 k_2 \alpha _4&=0,\\ \beta _1+\beta _2 s_1+\beta _3 s_2+\alpha _3 k_1+\beta _6 k_2+k_1 s_2 \alpha _4+s_1 k_2 \alpha _4&=0, \\ \alpha _2+\beta _4 s_1+\beta _5 s_2+\alpha _3 l_1+\beta _6 l_2+l_1 s_2 \alpha _4+s_1 l_2 \alpha _4&=0,\\ \beta _2 k_1+\beta _3 k_2+\beta _4 l_1+\beta _5 l_2+\alpha _3 s_1+\beta _6 s_2+k_1 s_2\\ +k_1 k_2 \alpha _4+l_1 l_2 \alpha _4+s_1 s_2 \alpha _4&=0 \end{aligned} \end{aligned}$$

    are satisfied,

  • (b) $\mathfrak {h}_{7}= \text {span} \left( E_2+ k_1 E_4+ k_2 E_6, \quad E_3+l_1 E_4+l_2 E_6, \quad E_5 \right)$ if and only if $\beta _2=0$ the following equations

    $$\begin{aligned} \begin{aligned} \beta _1 k_1+ \beta _3 k_2+ k_1 \beta _6 k_2&=0,\\ \alpha _2 l_1+ \beta _5 l_2+ l_1 \beta _6 l_2&=0, \\ \alpha _1+ \beta _1 l_1+ \beta _3 l_2+ \alpha _2 k_1+ \beta _5 k_2+ k_1 l_2 \beta _6 + l_1 k_2 \beta _6&=0, \\ \beta _2+ \alpha _3 k_1+ \alpha _4 k_2&=0, \\ \beta _4+ \alpha _3 l_1+ \alpha _4 l_2&=0\end{aligned} \end{aligned}$$

    are satisfied,

  • (c) $\mathfrak {h}_{8}= \text {span} \left( E_2+k_1E_4+k_2E_5, \quad E_3+l_1E_4+l_2E_5, \quad E_6 \right)$ if and only if the following equations

    $$\begin{aligned} \begin{aligned} \beta _1 k_1+ \beta _2 k_2+ k_1 \alpha _3 k_2&=0, \\ \alpha _2 l_1+\beta _4 l_2+l_1 l_2 \alpha _3&=0, \\\alpha _1+\beta _1 l_1+\beta _2 l_2+\alpha _2 k_1+\beta _4 k_2+k_1 l_2 \alpha _3+l_1 k_2 \alpha _3&=0, \\\beta _3+\beta _6k_1+\alpha _4 k_2&=0, \\ \beta _5+ l_1\beta _6+ l_2 \alpha _4&=0 \end{aligned} \end{aligned}$$

    hold,

  • (d) $\mathfrak {h}_{9}= \text {span} \left( E_2+ k_1 E_3-\frac{\beta _2+ k_1 \beta _4}{\alpha _4} E_6, \quad E_4- \frac{\alpha _3}{\alpha _4}E_6, \quad E_5 \right)$ such that $k_1$ is a solution of the equation

    $$\begin{aligned} \beta _4 \beta _5 k_1^2+k_1(\beta _2 \beta _5+\beta _3 \beta _4-\alpha _1 \alpha _4)+\beta _2 \beta _3=0, \end{aligned}$$

    (4.15)

    and $\beta _1=\frac{\alpha _3}{\alpha _4} \beta _3$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$,

  • (e) $\mathfrak {h}_{9}= \text {span} \left( E_2+ \frac{\alpha _3 \beta _3-\alpha _4 \beta _1}{\alpha _2 \alpha _4 - \alpha _3 \beta _5} E_3-\frac{\beta _2(\alpha _2 \alpha _4 -\alpha _3 \beta _5)+\beta _4(\beta _3 \alpha _3- \beta _1 \alpha _4)}{\alpha _4 (\alpha _2 \alpha _4- \alpha _3 \beta _5)} E_6, \right. \left. E_4- \frac{\alpha _3}{\alpha _4}E_6, \quad E_5 \right)$ such that $\beta _5 \ne \frac{\alpha _2 \alpha _4}{\alpha _3}$ and the equation

    $$\begin{aligned}{} & {} (\alpha _2 \alpha _4 - \alpha _3 \beta _5)[(\alpha _1 \alpha _4-\beta _3 \beta _4- \beta _2 \beta _5) (\beta _3 \alpha _3- \beta _1 \alpha _4)\nonumber \\{} & {} \qquad -\beta _2 \beta _3(\alpha _2 \alpha _4 - \alpha _3 \beta _5) - \beta _4 \beta _5(\beta _3 \alpha _3- \beta _1 \alpha _4)^2=0 \end{aligned}$$

    (4.16)

    holds,

  • (f) $\mathfrak {h}_{10}= \text {span} \left( E_2+ k_1 E_3-\frac{\beta _3+ k_1 \beta _5}{\alpha _4} E_5, \quad E_4, \quad E_6 \right)$ such that $k_1$ is a solution of the equation (4.15) and $\beta _1=\frac{\alpha _3}{\alpha _4} \beta _3$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$,

  • (g) $\mathfrak {h}_{10}= \text {span} \left( E_2+ \frac{\alpha _3 \beta _3-\alpha _4 \beta _1}{\alpha _2 \alpha _4 - \alpha _3 \beta _5} E_3-\frac{\beta _3(\alpha _2 \alpha _4 -\alpha _3 \beta _5)+\beta _5(\beta _3 \alpha _3- \beta _1 \alpha _4)}{\alpha _4 (\alpha _2 \alpha _4- \alpha _3 \beta _5)} E_6, \quad \right. \left. E_4, \quad E_5 \right)$ such that $\beta _5 \ne \frac{\alpha _2 \alpha _4}{\alpha _3}$ and the equation (4.16) holds.

  • (h) $\mathfrak {h}_{12}= \text {span} \left( E_3, \quad E_4- \frac{\alpha _3}{\alpha _4}E_6, \quad E_5 \right)$ in the case $\beta _4=\beta _6=0$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$,

  • (i) $\mathfrak {h}_{13}= \text {span} \left( E_3- \frac{\alpha _2}{\alpha _3}E_5, \quad E_4, \quad E_6 \right)$ in the case $\beta _4=\beta _6=0$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$.

• (3) the 2-dimensional subalgebras are the following:

  • (a) $\mathfrak {h}_{24}= \text {span}(E_4-\frac{\alpha _3}{\alpha _4}E_6, \quad E_5)$ in the case $\beta _6=0$,

  • (b) $\mathfrak {h}_{25}= \text {span} \left( E_4, E_6 \right)$ in the case $\beta _6=0$,

  • (c) $\mathfrak {h}_{18}= \text {span} \left( E_2+k_1E_3+k_2E_5+k_3E_6, \quad E_4+l_1E_5+l_2E_6 \right)$ if and only if the following equations

    $$\begin{aligned} \begin{aligned} \alpha _3 l_1+\beta _6 l_2+l_1 \alpha _4 l_1&=0, \\ \alpha _1 k_1+\beta _2 k_2+\beta _3 k_3+k_1 \beta _4 k_2+k_1 \beta _5 k_3+k_2 \alpha _4 k_3&=0, \\ \beta _1+\beta _2 l_1+\beta _3 l_2+\alpha _2 k_1+k_1 \beta _4 l_1 + \\+ k_1 \beta _5 l_2+\alpha _3 k_2+\beta _6 k_3+\alpha _4 k_2 l_2+ \alpha _4 l_1 k_3&=0 \end{aligned} \end{aligned}$$

    hold,

  • (d) $\mathfrak {h}_{21}= \text {span} (E_3+k_1E_5+k_2E_6, \quad E_4+l_1E_5+l_2E_6)$ if and only if the following equations

    $$\begin{aligned} \begin{aligned} \beta _4 k_1+ \beta _5 k_2+k_1 \alpha _4 k_2&=0,\\ \alpha _3 l_1+ \beta _6 l_2+l_1 \alpha _4 l_2&=0, \\ \alpha _2+\beta _4 l_1+\beta _5 l_2+ \alpha _3 k_1+\beta _6 k_2+k_1 l_2 \alpha _4+l_1 k_2 \alpha _4&=0 \end{aligned} \end{aligned}$$

    are satisfied,

  • (e) $\mathfrak {h}_{19}= \text {span} (E_2+k_1E_3+k_2E_4-\frac{\beta _2+k_1 \beta _4+k_2 \alpha _3}{\alpha _4 }E_6, \quad E_5)$ such that the equation

    $$\begin{aligned}{} & {} (\alpha _3 k_2+\beta _2 +\beta _4 k_1)(k_1 \beta _5+k_2 \beta _6+\beta _3 )\nonumber \\ {}{} & {} \quad - \alpha _2 \alpha _4 k_1 k_2-\alpha _1 \alpha _4 k_1-\beta _1 \alpha _4 k_2=0 \end{aligned}$$

    (4.17)

    is satisfied,

  • (f) $\mathfrak {h}_{22}= \text {span} (E_3+k_1E_4-\frac{\beta _4+\alpha _3 k_1}{\alpha _4}E_6, \quad E_5 )$, where $k_1$ is a solution of the equation

    $$\begin{aligned} \alpha _3 \beta _6 k_1^2+k_1(\alpha _3 \beta _5+\beta _4 \beta _6-\alpha _2 \alpha _4)+\beta _4 \beta _5=0, \end{aligned}$$

    (4.18)

  • (g) $\mathfrak {h}_{23}= \text {span} (E_3+k_1E_4-\frac{\beta _5+\beta _6 k_1}{\alpha _4}E_5, \quad E_6 )$, where $k_1$ is a solution of the equation (4.18),

  • (h) $\mathfrak {h}_{20}= \text {span} (E_2+k_1E_3+k_2E_4-\frac{\beta _3+ k_1 \beta _5+k_2 \beta _6}{\alpha _4}E_5, \quad E_6 )$ such that the equation (4.17) holds,

  • (i) $\mathfrak {h}_{17}= \text {span} (E_2+k_1E_4+k_2E_5+k_3E_6, \quad E_3+l_1E_4+l_2E_5+l_3E_6)$ if and only if the following equations

    $$\begin{aligned} \begin{aligned} \beta _1 k_1+\beta _2 k_2+\beta _3 k_3+k_1 \alpha _3 k_2+k_1\beta _6 k_3+k_2 \alpha _4 k_3=0, \\ \alpha _2 l_1+\beta _4 l_2+\beta _5 l_3+l_1 \alpha _3 l_2+l_1 \beta _6 l_3+l_2 \alpha _4 l_3=0, \\ \alpha _1+\beta _1 l_1+\beta _2 l_2+\beta _3 l_3+\alpha _2 k_1+\beta _4 k_2+\beta _5 k_3+k_1 l_2 \alpha _3+l_1 k_2 \alpha _3+\\k_1 l_3 \beta _6+l_1 k_3 \beta _6+k_2 l_3 \alpha _4+l_2 l_3 \alpha _4=0 \end{aligned} \end{aligned}$$

    (4.19)

    are satisfied.

Proof

According to Proposition 2.3 b) the dimension of the flat totally geodesic subalgebras of the metric Lie algebra $(\mathfrak {n}_{6,18}(\alpha _i, \beta _j), \langle .,. \rangle )$ is at most 4. Firstly we list the 4-dimensional, the 3-dimensional and the 2-dimensional abelian subalgebras of the Lie algebra $\mathfrak {n}_{6,18}(\alpha _i, \beta _j)$ defined by the commutators (1.6) such that $\alpha _5=0$. The 4-dimensional subalgebras have one of the following forms:

$$\begin{aligned}&\mathfrak {h}_{1}= \text{ span } \left( E_2+k_1E_6, \quad E_3+k_2E_6, \quad E_4+k_3E_6,\quad E_5+k_4E_6 \right) ,\\ {}&\mathfrak {h}_{2}= \text{ span } \left( E_2+k_1E_5, \quad E_3+k_2 E_5, \quad E_4+k_3E_5,\quad E_6 \right) ,\\ {}&\mathfrak {h}_{3}= \text{ span } \left( E_2+k_1E_4, \quad E_3+k_2 E_4, \quad E_5,\quad E_6 \right) ,\\ {}&\mathfrak {h}_{4}= \text{ span }(E_2, E_4, E_5, E_6),\\ {}&\mathfrak {h}_{5}= \text{ span }(E_3, E_4, E_5, E_6), \end{aligned}$$

where $k_1, k_2, k_3, k_4 \in \mathbb {R}$.

The 3-dimensional abelian subalgebras of the Lie algebra $\mathfrak {n}_{6,18}(\alpha _i, \beta _j )$ are:

$$\begin{aligned}&\mathfrak {h}_{6}= \text{ span } \left( E_2+k_1E_5+k_2E_6, \quad E_3+l_1E_5+l_2E_6, \quad E_4+s_1E_5+s_2E_6 \right) ,\\ {}&\mathfrak {h}_{7}= \text{ span } \left( E_2+k_1E_4+k_2E_6, \quad E_3+l_1E_4+l_2E_6, \quad E_5+k_4E_6 \right) ,\\ {}&\mathfrak {h}_{8}= \text{ span } \left( E_2+k_1E_4+k_2E_5, \quad E_3+l_1E_4+l_2E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{9}= \text{ span } \left( E_2+k_1E_3+k_2E_6, \quad E_4+k_3 E_6, \quad E_5+k_4 E_6 \right) ,\\ {}&\mathfrak {h}_{10}= \text{ span } \left( E_2+k_1E_3+k_2E_5, \quad E_4+k_3E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{11}= \text{ span } \left( E_2+k_1E_3+k_2E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{12}= \text{ span } \left( E_3+k_2E_6, \quad E_4+k_3E_6, \quad E_5+k_4 E_6 \right) ,\\ {}&\mathfrak {h}_{13}= \text{ span } \left( E_3+k_2E_5, \quad E_4+k_3E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{14}= \text{ span } \left( E_3+k_2E_4, \quad E_5, \quad E_6 \right) ,\\ {}&\mathfrak {h}_{15}= \text{ span } \left( E_4, E_5, E_6 \right) ,\\ \end{aligned}$$

where $k_1, k_2, k_3, k_4, l_1, l_2, s_1, s_2 \mathbb {R}$.

The 2-dimensional abelian subalgebras of the Lie algebra $\mathfrak {n}_{6,18}(\alpha _i, \beta _j )$ have one of the following forms:

$$\begin{aligned}&\mathfrak {h}_{16}= \text {span} \left( E_1+k_2 E_2+k_3 E_3+k_4 E_4+k_5 E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{17}= \text {span} \left( E_2+k_1E_4+k_2E_5+k_3E_6, \quad E_3+l_1E_4+l_2E_5+l_3E_6 \right) ,\\&\mathfrak {h}_{18}= \text {span} \left( E_2+k_1E_3+k_2E_5+k_3E_6, \quad E_4+l_1E_5+l_2E_6 \right) ,\\&\mathfrak {h}_{19}= \text {span} \left( E_2+k_1E_3+k_2E_4+k_3E_6, \quad E_5+l_1E_6 \right) ,\\&\mathfrak {h}_{20}= \text {span} \left( E_2+k_1E_3+k_2E_4+k_3E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{21}= \text {span} \left( E_3+k_1E_5+k_2E_6, \quad E_4+l_1E_5+l_2E_6 \right) ,\\&\mathfrak {h}_{22}= \text {span} \left( E_3+k_1E_4+k_2E_6, \quad E_5+l_1E_6 \right) ,\\&\mathfrak {h}_{23}= \text {span} \left( E_3+k_1E_4+k_2E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{24}= \text {span} \left( E_4+k_1E_6, \quad E_5+l_1E_6 \right) ,\\&\mathfrak {h}_{25}= \text {span} \left( E_4+k_1E_5, \quad E_6 \right) ,\\&\mathfrak {h}_{26}= \text {span} \left( E_5, E_6 \right) , \end{aligned}$$

where $k_1, k_2, k_3, k_4, k_5, l_1, l_2, l_3 \mathbb {R}$.

The vector $E_1+k_2 E_2+k_3 E_3+k_4 E_4+k_5 E_5+E_6 \in \mathfrak {h}_{16}$ is not geodesic since for $a=f=1$, $b=k_2$, $c=k_3$, $d=k_4$, $e=k_5$, the first equation of (4.1) gives the contradiction $1 =0$. Hence the subalgebra $\mathfrak {h}_{16}$ is not flat totally geodesic.

For all elements of the remaining subalgebras we have $a=0$. Since $\alpha _5=0$ for the metric Lie algebra $(\mathfrak {n}_{6,18} (\alpha _{i},\beta _j ),\langle .,. \rangle )$ the system (4.1) of equations reduces to the equation

$$\begin{aligned} b( \alpha _1 c+ \beta _1 d+ \beta _2e+ \beta _3f) +c(\alpha _2 d+\beta _4 e+\beta _5 f)+d(\alpha _3 e+\beta _6f)+e\alpha _4f =0.\nonumber \\ \end{aligned}$$

(4.20)

The vector $E_5+ E_6 \in \mathfrak {h}_{3} \cap \mathfrak {h}_{4} \cap \mathfrak {h}_{5} \cap \mathfrak {h}_{11} \cap \mathfrak {h}_{14} \cap \mathfrak {h}_{15} \cap \mathfrak {h}_{26}$ is not geodesic because for $b=c=d=0$, $e=f=1$ the equation (4.34.20) gives the contradiction $\alpha _4 =0$. Hence the subalgebras $\mathfrak {h}_{3}$, $\mathfrak {h}_{4}$, $\mathfrak {h}_{5}$, $\mathfrak {h}_{11}$, $\mathfrak {h}_{14}$, $\mathfrak {h}_{15}$, $\mathfrak {h}_{26}$ are not flat totally geodesic.

The subalgebra $\mathfrak { h}_{1}$ is totally geodesic precisely if for all $Y, Z \in \mathfrak { h}_{1}$ and $X \in \mathfrak { h}^\perp _{1} =\text {span}(E_1, E_6- \sum _{i=1}^4 k_i E_{i+1})$ equation (1.1) is satisfied. The subalgebra $\mathfrak { h}_{2}$ is totally geodesic if and only if for all $Y, Z \in \mathfrak { h}_{2}$ and $X \in \mathfrak { h}^\perp _{2} =\text {span}(E_1, E_5- \sum _{i=1}^3 k_i E_{i+1})$ equation (1.1) is satisfied. Since the commutation relations of the element $(E_6- \sum _{i=1}^4 k_i E_{i+1})$ and the elements of $\mathfrak { h}_{1}$ as well as of the element $\left( E_5- \sum _{i=1}^3 k_i E_{i+1}\right)$ and the elements of $\mathfrak { h}_{2}$ are zero, we may assume that $X=E_1$. The element $X=E_1$ lies in the orthogonal complement $\mathfrak { h}^\perp _{i}$ for all $i=3, \cdots , 15$, too. Using the equation (1.1) we receive the following:

(1) For $Y=Z=E_5+ k_4 E_6 \in \mathfrak { h}_{1} \cap \mathfrak { h}_{7} \cap \mathfrak { h}_{9} \cap \mathfrak { h}_{12}$ we have $2 \alpha _4 k_4=0$ and hence $k_4=0$.

(2) Taking the elements $Y=E_4+ k_3 E_6$, $Z=E_5$ in $\mathfrak { h}_{1} \cap \mathfrak { h}_{9} \cap \mathfrak { h}_{12}$ we get $\alpha _3+\alpha _4 k_3=0$ and hence $k_3=-\frac{\alpha _3}{\alpha _4}<0$.

(3) For $Y=Z=E_4+k_3 E_6 \in \mathfrak { h}_{1} \cap \mathfrak { h}_{9} \cap \mathfrak { h}_{12}$ one has $2 \beta _6 k_3= 0$ and hence $\beta _6=0$.

(4) The elements $Y=E_3+ k_2 E_6$, $Z=E_4+k_3 E_6$ in $\mathfrak { h}_{1} \cap \mathfrak { h}_{12}$ yield that $\alpha _2+\beta _5 k_3+\beta _6 k_2= \alpha _2- \beta _5 \frac{ \alpha _3}{\alpha _4}=0$ and hence $\beta _5=\frac{\alpha _2 \alpha _4 }{\alpha _3}>0$.

(5) For $Y=E_3+ k_2 E_6$, $Z=E_5$ in $\mathfrak { h}_{1} \cap \mathfrak { h}_{12}$ we obtain $\beta _4 + \alpha _4 k_2 =0$ and hence $k_2=-\frac{\beta _4}{\alpha _4}$.

(6) For $Y=Z= E_3+k_2 E_6 \in \mathfrak { h}_{1} \cap \mathfrak { h}_{12}$, one has $2 \beta _5 k_2=0$ and hence $k_2=0$ and $\beta _4=0$.

(7) For the elements $Y=E_2+ k_1 E_6$, $Z=E_5$ in $\mathfrak { h}_{1}$ we get $\beta _2 + \alpha _4 k_1 =0$ and therefore $k_1=-\frac{\beta _2}{\alpha _4}$.

(8) For the elements $Y=E_2 +k_1 E_6$, $Z= E_3$ in $\mathfrak { h}_{1}$ we receive $\alpha _1+\beta _5 k_1= \alpha _1-\frac{\alpha _2 }{\alpha _3} {\beta _2}=0$ and hence $\beta _2= \frac{\alpha _1 \alpha _3}{\alpha _2} >0$, moreover $k_1=-\frac{\alpha _1 \alpha _3}{\alpha _4 \alpha _2}<0$.

(9) Taking the elements $Y=Z= E_2+k_1 E_6 \in \mathfrak { h}_{1}$ one has $2 \beta _3 k_1=0$ and hence $\beta _3=0$.

(10) For $Y=E_2+k_1 E_6$, $Z= E_4+k_3 E_6$ of $\mathfrak { h}_{1}$ we obtain $\beta _1+\beta _3 k_3+\beta _6 k_1=\beta _1=0.$

Taking into account (1)–(10) the subalgebra $\mathfrak { h}_{1}$ is flat totally geodesic if and only if $\beta _1=\beta _3=\beta _4=\beta _6=0$,  $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$,  $\beta _2= \frac{\alpha _1 \alpha _3}{\alpha _2}$. Hence the case (a) is proved.

(11) For $Y=Z=E_4+k_3E_5 \in \mathfrak { h}_{2} \cap \mathfrak { h}_{10} \cap \mathfrak { h}_{13}$ we obtain $\alpha _3 k_3=0$ and hence $k_3=0$.

(12) For the elements $Y=E_4, \, Z=E_6$ of $\mathfrak { h}_{2} \cap \mathfrak { h}_{10} \cap \mathfrak { h}_{13}$ we receive $\beta _6=0$.

(13) Taking the elements $Y=E_3+k_2E_5, \, Z=E_4$ in $\mathfrak { h}_{2} \cap \mathfrak { h}_{13}$ one gets $\alpha _2 +\alpha _3 k_2=0$ which implies that $k_2=-\frac{\alpha _2}{\alpha _3}<0$.

(14) For $Y=Z=E_3+k_2E_5 \in \mathfrak { h}_{2} \cap \mathfrak { h}_{13}$ we have $\beta _4 k_2=0$ and hence $\beta _4=0$.

(15) For the elements $Y=E_3+k_2E_5, \, Z=E_6$ of $\mathfrak { h}_{2} \cap \mathfrak { h}_{13}$ one obtains $\beta _5 +\alpha _4 k_2=0$ which yields $\beta _5=\frac{\alpha _4 \alpha _2}{\alpha _3}$.

(16) For $Y=E_2+k_1E_5, \, Z=E_3+k_2E_5$ in $\mathfrak { h}_{2}$ we receive $\alpha _1+\beta _2 k_2+\beta _4 k_1=0$ and hence $\beta _2=\frac{\alpha _1 \alpha _3}{\alpha _2}>0$.

(17) For $Y=Z=E_2+k_1E_5 \in \mathfrak { h}_{2}$ one has $k_1 \beta _2=0$ and hence $k_1=0$.

(18) Taking the elements $Y=E_2, \, Z=E_4$ of $\mathfrak { h}_{2}$ we get $\beta _1=0$.

(19) For the elements $Y=E_2, \, Z=E_6$ in $\mathfrak { h}_{2}$ we receive $\beta _3=0$. According to (11)–(19) the subalgebra $\mathfrak { h}_{2}$ is flat totally geodesic precisely if $\beta _1=\beta _3=\beta _4=\beta _6=0$,  $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$,  $\beta _2= \frac{\alpha _1 \alpha _3}{\alpha _2}$. Hence the assertion (b) follows.

Taking into account (1)–(12) it follows from equation (4.3) that the subalgebra $\mathfrak { h}_{12}$ in (h) as well as the subalgebra $\mathfrak {h}_{13}$ in (i) with the conditions $\beta _4=\beta _6=0$, $\beta _5=\frac{\alpha _2 \alpha _4}{\alpha _3}$ are flat totally geodesic.

Now we consider the subalgebra $\mathfrak { h}_{9}$. Taking into account (1)–(3) we have $k_4=\beta _6=0$, $k_3=-\frac{\alpha _3}{\alpha _4}$. The element $E_2+ k_1 E_3+ k_2 E_6 \in \mathfrak { h}_{9}$ is geodesic if and only if for $b=1$, $c=k_1$, $f=k_2$, $d=e=0$ the equation (4.34.20) is satisfied. This gives the equation

$$\begin{aligned} \alpha _1 k_1+ \beta _3 k_2+ k_1 \beta _5 k_2=0. \end{aligned}$$

(4.21)

The element $E_2+ k_1 E_3+ k_2 E_6 +E_5\in \mathfrak { h}_{9}$ is geodesic if and only if for $b=e=1$, $c=k_1$, $f=k_2$, $d=0$ the equation (4.34.20) is valid. Using equations (4.21) and (4.34.20) we obtain the equation

$$\begin{aligned} \beta _2+ k_1 \beta _4 + \alpha _4 k_2=0. \end{aligned}$$

(4.22)

Since $\alpha _4 \ne 0$ from equation (4.22) we receive

$$\begin{aligned} k_2=- \frac{\beta _2+ k_1 \beta _4 }{\alpha _4}. \end{aligned}$$

(4.23)

The element $E_2+ E_4+ k_1 E_3+ (k_2-\frac{\alpha _3}{\alpha _4}) E_6 \in \mathfrak { h}_{9}$ is geodesic precisely if for $b=d=1$, $c=k_1$, $f=k_2-\frac{\alpha _3}{\alpha _4}$, $e=0$ the equation (4.34.20) holds. Taking into account (4.21) from equation (4.34.20) we get

$$\begin{aligned} \beta _1- \frac{\alpha _3}{\alpha _4} \beta _3 + k_1(\alpha _2 - \beta _5 \frac{\alpha _3}{\alpha _4}). \end{aligned}$$

(4.24)

If $\beta _5= \frac{\alpha _2 \alpha _4}{\alpha _3}$, then from (4.24) we obtain $\beta _1=\frac{\alpha _3}{\alpha _4} \beta _3$ and from (4.21) we get the second order equation (4.15). This proves case (d).

If $\beta _5 \ne \frac{\alpha _2 \alpha _4}{\alpha _3}$, then from (4.24) we have $k_1= \frac{\beta _3 \alpha _3- \beta _1 \alpha _4}{\alpha _2 \alpha _4- \alpha _3 \beta _5}$. Putting this into (3.74.23) we obtain $k_2=- \frac{\beta _2(\alpha _2 \alpha _4- \alpha _3 \beta _5)+ \beta _4(\beta _3 \alpha _3- \beta _1 \alpha _4)}{\alpha _4(\alpha _2 \alpha _4- \alpha _3 \beta _5)}$. Substituting the expression of $k_1$ and $k_2$ into (4.21) we receive equation (4.16). Hence case (e) is shown.

Now we treat the subalgebra $\mathfrak { h}_{10}$. According to (11)–(12) one has $k_3=\beta _6=0$. The element $E_2+ k_1 E_3+ k_2 E_5 \in \mathfrak { h}_{10}$ is geodesic if and only if for $b=1$, $c=k_1$, $e=k_2$, $d=f=0$ the equation (4.34.20) is satisfied. This gives the equation

$$\begin{aligned} \alpha _1 k_1+ \beta _3 k_2+ k_1 \beta _4 k_2=0. \end{aligned}$$

(4.25)

The element $E_2+ k_1 E_3+ k_2 E_5 +E_4\in \mathfrak { h}_{10}$ is geodesic if and only if for $b=d=1$, $c=k_1$, $e=k_2$, $f=0$ the equation (4.34.20) is valid. Using equations (4.34.20) and (4.25) we obtain the equation

$$\begin{aligned} \beta _1+ k_1 \alpha _2 + \alpha _3 k_2=0. \end{aligned}$$

(4.26)

The element $E_2+ k_1 E_3+ k_2 E_5+ E_6 \in \mathfrak { h}_{10}$ is geodesic precisely if for $b=f=1$, $c=k_1$, $e=k_2$, $d=0$ the equation (4.34.20) holds. Taking into account (4.25) from equation (4.34.20) we get

$$\begin{aligned} \beta _3+k_1 \beta _5 + k_2 \alpha _4=0. \end{aligned}$$

(4.27)

Hence we obtain $k_2=- \frac{\beta _3+ k_1 \beta _5}{\alpha _4}$. Putting this expression of $k_2$ into (4.26) we receive

$$\begin{aligned} k_1(\alpha _2 \alpha _4 - \alpha _3 \beta _5)= \alpha _3 \beta _3- \alpha _4 \beta _1. \end{aligned}$$

(4.28)

If $\beta _5= \frac{\alpha _2 \alpha _4}{\alpha _3}$, then from (4.28) we obtain $\beta _1=\frac{\alpha _3}{\alpha _4} \beta _3$ and from (4.25) we get the second order equation (4.15). This proves case (f).

If $\beta _5 \ne \frac{\alpha _2 \alpha _4}{\alpha _3}$, then from (4.28) we have $k_1= \frac{\beta _3 \alpha _3- \beta _1 \alpha _4}{\alpha _2 \alpha _4- \alpha _3 \beta _5}$. Putting this into (4.27) we obtain $k_2=- \frac{\beta _3(\alpha _2 \alpha _4- \alpha _3 \beta _5)+ \beta _5(\beta _3 \alpha _3- \beta _1 \alpha _4)}{\alpha _4(\alpha _2 \alpha _4- \alpha _3 \beta _5)}$. Substituting the expression of $k_1$ and $k_2$ into (4.25) we receive equation (4.16). Hence case (g) is proved.

Here we deal with subalgebra $\mathfrak {h}_{6}$. The element $E_2+ k_1 E_5+ k_2 E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $b=1$, $e=k_1$, $f=k_2$, $c=d=0$ the equation (4.34.20) holds. From this we obtain

$$\begin{aligned} \beta _2 k_1+ \beta _3 k_2+ k_1 \alpha _4 k_2=0. \end{aligned}$$

(4.29)

The vector $E_3+ l_1 E_5+ l_2 E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $c=1$, $e=l_1$, $f=l_2$, $b=d=0$ the equation (4.34.20) is satisfied. This gives

$$\begin{aligned} \beta _4 l_1+ \beta _5 l_2+ l_1 \alpha _4 l_2=0. \end{aligned}$$

(4.30)

The element $E_4+ s_1 E_5+ s_2 E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $d=1$, $e=s_1$, $f=s_2$, $b=c=0$ the equation (4.34.20) holds. This gives

$$\begin{aligned} \alpha _3 s_1+ \beta _6 s_2+ s_1 \alpha _4 s_2=0. \end{aligned}$$

(4.31)

The element $E_2+ E_3+ (k_1+l_1) E_5+ (k_2+l_2) E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $b=c=1$, $e=k_1+l_1$, $f=k_2+l_2$, $d=0$ the equation (4.34.20) is satisfied. Using (4.29), (4.30) we receive

$$\begin{aligned} \alpha _1 +\beta _2 l_1+\beta _3 l_2+\beta _4 k_1+\beta _5 k_2+k_1 l_2 \alpha _4+l_1 k_2 \alpha _4=0. \end{aligned}$$

(4.32)

The element $E_2+ E_4+ (k_1+s_1) E_5+ (k_2+s_2) E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $d=b=1$, $e=k_1+s_1$, $f=k_2+s_2$, $c=0$ the equation (4.34.20) is satisfied. Using (4.29), (4.31) we get

$$\begin{aligned} \beta _1+\beta _2 s_1+\beta _3 s_2+\alpha _3 k_1+\beta _6 k_2+k_1 s_2 \alpha _4+s_1 k_2 \alpha _4=0. \end{aligned}$$

(4.33)

The element $E_3+ E_4+ (l_1+s_1) E_5+ (l_2+s_2) E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $d=c=1$, $e=l_1+s_1$, $f=l_2+s_2$, $b=0$ the equation (4.34.20) is satisfied. Using (4.30), (4.31) we obtain

$$\begin{aligned} \alpha _2+\beta _4 s_1+\beta _5 s_2+\alpha _3 l_1+\beta _6 l_2+l_1 s_2 \alpha _4+s_1 l_2 \alpha _4=0. \end{aligned}$$

(4.34)

The element $E_2+E_3+ E_4+ (k_1+l_1+s_1) E_5+ (k_2+l_2+s_2) E_6 \in \mathfrak {h}_{6}$ is geodesic if and only if for $b=d=c=1$, $e=k_1+l_1+s_1$, $f=k_2+l_2+s_2$, $b=0$ the equation (4.34.20) is satisfied. Using (4.32), (4.33), and (4.34) we receive

$$\begin{aligned}{} & {} \beta _2 k_1+\beta _3 k_2+\beta _4 l_1+\beta _5 l_2+\alpha _3 s_1+\beta _6 s_2\nonumber \\ {}{} & {} \quad +k_1 s_2+k_1 k_2 \alpha _4+l_1 l_2 \alpha _4+s_1 s_2 \alpha _4=0. \end{aligned}$$

(4.35)

This gives the case (a).

Nw we deal with the subalgebra $\mathfrak {h}_{7}$. Taking into account (1) we get $k_4=0$. The element $E_2+ k_1 E_4+ k_2 E_6 \in \mathfrak {h}_{7}$ is geodesic if and only if for $b=1$, $d=k_1$, $f=k_2$, $c=e=0$ the equation (4.34.20) holds. This gives the equation

$$\begin{aligned} \beta _1 k_1+ \beta _3 k_2+ k_1 \beta _6 k_2=0. \end{aligned}$$

(4.36)

The element $E_3+ l_1 E_4+ l_2 E_6 \in \mathfrak {h}_{7}$ is geodesic precisely if for $c=1$, $d=l_1$, $f=l_2$, $b=e=0$ the equation (4.34.20) holds. Hence we obtain the equation

$$\begin{aligned} \alpha _2 l_1+ \beta _5 l_2+ l_1 \beta _6 l_2=0. \end{aligned}$$

(4.37)

The element $E_2+E_3+(k_1+l_1) E_4+ (k_2+l_2) E_6 \in \mathfrak {h}_{7}$ is geodesic if and only if for $b=c=1$, $d=k_1+l_1$, $f=k_2+l_2$, $e=0$ the equation (4.34.20) is valid. Using (4.36) and (4.37) we receive the equation

$$\begin{aligned} \alpha _1+ \beta _1 l_1+ \beta _3 l_2+ \alpha _2 k_1+ \beta _5 k_2+ k_1 l_2 \beta _6 + l_1 k_2 \beta _6=0. \end{aligned}$$

(4.38)

The element $E_2+k_1 E_4+ k_2 E_6+ E_5 \in \mathfrak {h}_{7}$ is geodesic precisely if for $b=e=1$, $d=k_1$, $f=k_2$, $c=0$ the equation (4.34.20) holds. Using (4.36) one gets the equation

$$\begin{aligned} \beta _2+ \alpha _3 k_1+ \alpha _4 k_2=0. \end{aligned}$$

(4.39)

The element $E_3+l_1 E_4+ l_2 E_6+ E_5 \in \mathfrak {h}_{7}$ is geodesic precisely if for $c=e=1$, $d=l_1$, $f=l_2$, $b=0$ the equation (4.34.20) is satisfied. Applying (4.37) one obtains the equation

$$\begin{aligned} \beta _4+ \alpha _3 l_1+ \alpha _4 l_2=0. \end{aligned}$$

(4.40)

Hence, the equations (4.36), (4.37), (4.38), (4.39), (4.40) yield the case (b)

Now we consider the subalgebra $\mathfrak {h}_{8}$. The vector $E_2+ k_1 E_4+ k_2 E_5 \in \mathfrak {h}_{8}$ is geodesic if and only if for $b=1$, $d=k_1$, $e=k_2$, $c=f=0$ the equation (4.34.20) is satisfied. This gives the equation

$$\begin{aligned} \beta _1 k_1+ \beta _2 k_2+ k_1 \alpha _3 k_2=0. \end{aligned}$$

(4.41)

The element $E_3+ l_1 E_4+ l_2 E_5 \in \mathfrak {h}_{8}$ is geodesic if and only if for $c=1$, $d=l_1$, $e=l_2$, $b=f=0$ the equation (4.34.20) holds. Hence we receive

$$\begin{aligned} \alpha _2 l_1+\beta _4 l_2+l_1 l_2 \alpha _3=0. \end{aligned}$$

(4.42)

The vector $E_2+ E_3+ (k_1+l_1) E_4+ (k_2+ l_2) E_5 \in \mathfrak {h}_{8}$ is geodesic if and only if for $b=c=1$, $d=k_1+ l_1$, $e=k_2+ l_2$, $f=0$ the equation (4.34.20) is satisfied. Using (4.41), (4.42) one has the following equation

$$\begin{aligned} \alpha _1+\beta _1 l_1+\beta _2 l_2+\alpha _2 k_1+\beta _4 k_2+k_1 l_2 \alpha _3+l_1 k_2 \alpha _3=0. \end{aligned}$$

(4.43)

The element $E_2+ E_6+ k_1E_4+ k_2E_5 \in \mathfrak {h}_{8}$ is geodesic if and only if for $b=f=1$, $d=k_1$, $e=k_2$, $c=0$ the equation (4.34.20) is satisfied. Applying (4.41) one gets

$$\begin{aligned} \beta _3+\beta _6k_1+\alpha _4 k_2=0. \end{aligned}$$

(4.44)

The element $E_3+ l_1E_4+ l_2E_5+E_6 \in \mathfrak {h}_{8}$ is geodesic if and only if for $c=f=1$, $d=l_1$, $e=l_2$, $b=0$ the equation (4.34.20) holds. Using (4.42) we receive

$$\begin{aligned} \beta _5+ l_1\beta _6+ l_2 \alpha _4=0. \end{aligned}$$

(4.45)

The vector $E_2+E_3+(k_1+l_1)E_4+(k_2+l_2)E_5+E_6 \in \mathfrak {h}_{8}$ is geodesic if and only if for $c=b=f=1$, $d=k_1+l_1$, $e=k_2+ l_2$ the equation (4.34.20) is satisfied. This gives

$$\begin{aligned} \begin{aligned} \alpha _1+\beta _1 k_1+\beta _1 l_1+\beta _2 k_2+\beta _2 l_2+\beta _3+\alpha _2 k_1+\alpha _2 l_1+\beta _4 k_2+\beta _4 l_2+\beta _5\\ \quad +k_1 k_2 \alpha _3+k_1 l_2 \alpha _3+l_1 k_2 \alpha _3+l_1 l_2 \alpha _3 +k_1 \beta _6+l_1 \beta _6+k_2 \alpha _4+l_2 \alpha _4=0. \end{aligned} \nonumber \\ \end{aligned}$$

(4.46)

Using the equations (4.41), (4.42), (4.43), (4.44), and (4.45), the equation (4.46) holds. Therefore the subalgebra $\mathfrak {h}_{8}$ is proved. This gives the case (c)

The subalgebra $\mathfrak {h}_{24}$ coincides with the subalgebra $\mathfrak {h}_{9}$ in $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$. Hence the case (a) is valid. Furthermore, the subalgebra $\mathfrak {h}_{25}$ coincides with the subalgebra $\mathfrak {h}_{5}$ in $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$. Therefore the case (b) is shown.

The vector $E_4+l_1 E_5+l_2E_6 \in \mathfrak {h}_{18} \cap \mathfrak {h}_{21}$ coincides with the element $E_4+l_1 E_5+l_2E_6 \in \mathfrak {h}_{1}$ in the filiform metric Lie algebra $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$. It is geodesic if and only if the equation

$$\begin{aligned} \alpha _3 l_1+\beta _6 l_2+l_1 \alpha _4 l_2=0. \end{aligned}$$

(4.47)

holds. Now we treat the subalgebra $\mathfrak {h}_{18}$. The element $E_2+k_1 E_3+k_2 E_5+k_3E_6 \in \mathfrak {h}_{18}$ is geodesic precisely if for $d=0, b=1, c=k_1, e=k_2, f=k_3$ the equation (4.34.20) is valid. Hence we get

$$\begin{aligned} \alpha _1k_1+\beta _2 k_2+\beta _3 k_3+k_1 \beta _4 k_2+k_1 \beta _5 k_3+k_2 \alpha _4 k_3=0. \end{aligned}$$

(4.48)

In addition, the vector $E_2+E_4+k_1 E_3+ (k_2+l_1) E_5+(k_3+l_2) E_6 \in \mathfrak {h}_{18}$ is geodesic if for $b=d=1, c=k_1, e=k_2+l_1, f=k_3+l_2$ the system (4.34.20) is satisfied. Therefore we obtain

$$\begin{aligned} \begin{aligned}&\alpha _1 k_1+\beta _1+\beta _2 k_2+\beta _2 l_1+\beta _3 k_3+\beta _3l_2+k_1 \alpha _2+k_1 \beta _4 k_2+k_1 \beta _4 l_1\\ {}&\quad +k_1 \beta _5 k_3+k_1 \beta _5 l_2+\alpha _3 k_2+\alpha _3 l_1+\beta _6 k_3+\beta _6 l_2+k_2 k_3 \alpha _4\\&\quad +k_2 l_2 \alpha _4+l_1k_3 \alpha _4+l_1 l_2 \alpha _4=0. \end{aligned} \end{aligned}$$

(4.49)

Due to the equations (4.47) and (4.48), equation (4.49) becomes

$$\begin{aligned}&{} \beta _1+\beta _2 l_1+\beta _3 l_2+\alpha _2 k_1+k_1 \beta _4 l_1\\{}&{} \qquad + k_1 \beta _5 l_2+\alpha _3 k_2+\beta _6 k_3+\alpha _4 k_2 l_2+ \alpha _4 l_1 k_3=0. \end{aligned}$$

This proves case (c).

Here we consider the case $\mathfrak {h}_{21}$. The element $E_3+k_1 E_5+k_2E_6 \in \mathfrak {h}_{21}$ is geodesic precisely if for $b=d=0, c=1, e=k_1, f=k_2$ the equation (4.34.20) is satisfied. From this we obtain

$$\begin{aligned} \beta _4 k_1+\beta _5 k_2+k_1 \alpha _4 k_2=0. \end{aligned}$$

(4.50)

Moreover, the vector $E_3+E_4+ (k_1+l_1) E_5+(k_2+l_2) E_6 \in \mathfrak {h}_{21}$ is geodesic precisely if for $b=0, c=1, d=1, e=k_1+l_1, f=k_2+l_2$ the equation (4.34.20) is valid. This gives

$$\begin{aligned} \begin{aligned} \alpha _2+\beta _4k_1+\beta _4l_1+\beta _5 k_2+\beta _5 l_2+\alpha _3 k_1+\alpha _3 l_1+\beta _6 k_2+ \beta _6 l_2+\\k_1 k_2\alpha _4+k_1 l_2 \alpha _4+l_1 k_2 \alpha _4+l_1 l_2 \alpha _4=0. \end{aligned} \end{aligned}$$

(4.51)

Exploiting equations (4.47) and (4.50), equation (4.51) reduces to

$$\begin{aligned} \alpha _2+\beta _4 l_1+\beta _5 l_2+ \alpha _3 k_1+\beta _6 k_2+k_1 l_2 \alpha _4+l_1 k_2 \alpha _4=0. \end{aligned}$$

This proves the case (d).

The vector $E_5+l_1 E_6 \in \mathfrak {h}_{19} \cap \mathfrak {h}_{22}$ coincides with the element $E_5+l_1 E_6 \in \mathfrak {h}_{2}$ in the filiform metric Lie algebra $\left( \mathfrak {n}_{6,17}(\alpha _i, \beta _j ), \langle .,.\rangle \right)$. It is geodesic if and only if one has $l_1=0$. Now we deal with the subalgebra $\mathfrak {h}_{19}$. The element $E_2+k_1 E_3+k_2 E_4+k_3 E_6 \in \mathfrak {h}_{19}$ is geodesic precisely if for $e=0, b=1, c=k_1, d=k_2, f=k_3$ the equation (4.34.20) is satisfied. This yields

$$\begin{aligned} \alpha _1 k_1+\beta _1 k_2+\beta _3 k_3+k_1 \alpha _2 k_2+k_1 \beta _5 k_3+k_2 \beta _6 k_3=0. \end{aligned}$$

(4.52)

Furthermore, the vector $E_2+k_1E_3+k_2E_4+k_3E_6+E_5 \in \mathfrak {h}_{19}$ is geodesic if and only if for $b=e=1, c=k_1, d=k_2, f=k_3$ the equation (4.34.20) holds. This gives

$$\begin{aligned}{} & {} \alpha _1 k_1+\beta _1 k_2+\beta _2+\beta _3 k_3+k_1 \alpha _2 k_2+k_1 \beta _4\nonumber \\{} & {} \qquad +k_1 \beta _5 k_3+k_2 \alpha _3+k_2 \beta _6 k_3+\alpha _4 k_3=0. \end{aligned}$$

(4.53)

Taking into account (4.52), equation (4.53) can be written as follows

$$\begin{aligned} \beta _2+k_1 \beta _4+k_2 \alpha _3+\alpha _4 k_3=0. \end{aligned}$$

(4.54)

From equation (4.54) we obtain that $k_3=-\frac{\beta _2+k_1 \beta _4+k_2 \alpha _3}{\alpha _4 }$. Putting this expression into equation (4.52) we receive equation (4.17). This proves the case (e).

Next we consider the subalgebra $\mathfrak {h}_{22}$. The vector $E_3+k_1 E_4+k_2 E_6 \in \mathfrak {h}_{22}$ is geodesic precisely if for $b=e=0, c=1, d=k_1, f=k_2$ the equation (4.20) holds. This gives

$$\begin{aligned} \alpha _2 k_1+\beta _5 k_2+k_1 \beta _6 k_2=0. \end{aligned}$$

(4.55)

Additionally, the element $E_3+k_1 E_4+k_2 E_6+E_5 \in \mathfrak {h}_{22}$ is geodesic if and only if for $b=0$, $c=e=1, d=k_1, f=k_2$ the equation 4.3 is satisfied. From this it follows that

$$\begin{aligned} \alpha _2 k_1+\beta _4+\beta _5 k_2+k_1 \alpha _3+k_1 \beta _6 k_2+\alpha _4 k_2=0. \end{aligned}$$

(4.56)

Comparing with equation (4.55), equation (4.56) reduces to

$$\begin{aligned} \beta _4+\alpha _3 k_1+\alpha _4 k_2=0. \end{aligned}$$

(4.57)

From (4.57) one has $k_2=-\frac{\beta _4+\alpha _3 k_1}{\alpha _4}$. Substituting this expression into (4.55) we have the second order equation (4.18) for $k_1$. This gives the case (f).

Now we consider the subalgebra $\mathfrak {h}_{23}$. The element $E_3+k_1 E_4+k_2 E_5 \in \mathfrak {h}_{23}$ is geodesic precisely if for $b=f=0, c=1, d=k_1, e=k_2$ the system (4.1) of equation is satisfied. Hence we receive

$$\begin{aligned} \alpha _2 k_1+\beta _4 k_2+k_1\alpha _3 k_2=0. \end{aligned}$$

(4.58)

Moreover, the vector $E_3+k_1 E_4+k_2 E_5+E_6 \in \mathfrak {h}_{7}$ is geodesic if and only if for $b=0$, $c=f=1, d=k_1, e=k_2$ the equation (4.34.20) holds. This gives

$$\begin{aligned} \alpha _2 k_1+\beta _4 k_2+\beta _5+k_1 \alpha _3 k_2+k_1 \beta _6 +\alpha _4 k_2=0. \end{aligned}$$

(4.59)

Using equation (4.58) equation (4.59) reduces to

$$\begin{aligned} \beta _5+\beta _6 k_1+\alpha _4 k_2=0. \end{aligned}$$

(4.60)

From (4.60) we obtain $k_2=-\frac{\beta _5+\beta _6 k_1}{\alpha _4}$. Putting this expression into (4.58) we have the second order equation (4.18) for $k_1$. Thus, the case (g) is proved.

Now we deal with the subalgebra $\mathfrak {h}_{20}$. The element $E_2+k_1 E_3+k_2 E_4+k_3 E_5 \in \mathfrak {h}_{20}$ is geodesic precisely if for $f=0, b=1, c=k_1, d=k_2, e=k_3$ the equation (4.34.20) is satisfied. From this we get

$$\begin{aligned} \alpha _1 k_1+\beta _1 k_2+\beta _2 k_3+k_1 \alpha _2 k_2+k_1 \beta _4 k_3+k_2 \alpha _3 k_3=0. \end{aligned}$$

(4.61)

Additionally, the vector $E_2+k_1 E_3+k_2 E_4+k_3 E_5+E_6 \in \mathfrak {h}_{4}$ is geodesic if and only if for $b=f=1$, $c=k_1, d=k_2, e=k_3,$ the equation (4.34.20) holds. Hence one obtains

$$\begin{aligned}{} & {} \alpha _1 k_1+\beta _1 k_2+\beta _2 k_3+\beta _3+k_1 \alpha _2 k_2+k_1 \beta _4 k_3\nonumber \\ {}{} & {} \qquad +k_1 \beta _5+k_2 \alpha _3 k_3+k_2 \beta _6+k_3 \alpha _4=0. \end{aligned}$$

(4.62)

Applying (4.61) equation (4.62) reduces to

$$\begin{aligned} \beta _3+ k_1 \beta _5+k_2 \beta _6 +\alpha _4 k_3=0. \end{aligned}$$

(4.63)

From (4.63) we obtain $k_3=-\frac{\beta _3+ k_1 \beta _5+k_2 \beta _6}{\alpha _4}$. Putting this expression into (4.61) we obtain equation (4.19). Hence the case (h) is shown.

Finally, we consider the subalgebra $\mathfrak {h}_{17}$. The vector $E_2+k_1 E_4+k_2 E_5+k_3 E_6 \in \mathfrak {h}_{17}$ is geodesic precisely if for $c=0, b=1, d=k_1, e=k_2, f=k_3$ the equation (4.20) is valid. Hence we get

$$\begin{aligned} \beta _1 k_1+\beta _2 k_2+\beta _3 k_3+k_1 \alpha _3 k_2+k_1 \beta _6 k_3+k_2 \alpha _4 k_3=0. \end{aligned}$$

(4.64)

The element $E_3+l_1 E_4+l_2 E_5+l_3 E_6 \in \mathfrak {h}_{17}$ is geodesic if and only if for $b=0$, $c=1, d=l_1, e=l_2, f=l_3$ the equation (4.3) holds. From this we receive

$$\begin{aligned} \alpha _2 l_1+\beta _4 l_2+\beta _5 l_3+l_1 \alpha _3 l_2+l_1 \beta _6 l_3+l_2 \alpha _4 l_3=0, \end{aligned}$$

(4.65)

Furthermore, the vector $E_2+ E_3+(k_1+l_1 )E_4+(k_2+l_2) E_5+(k_3+l_3)E_6 \in \mathfrak {h}_{17}$ is geodesic precisely if for $b=c=1$, $d=k_1+l_1, e=k_2+l_2, f=k_3+l_3$ the equation (4.20) is satisfied. Therefore one obtains

$$\begin{aligned} \begin{aligned}&\alpha _1+\beta _1 k_1+\beta _1 l_1+\beta _2 k_2+\beta _2 l_2+\beta _3 k_3+\beta _3 l_3+\alpha _2 k_1+\alpha _2 l_1+\beta _4 k_2+\beta _4 l_2\\ {}&\quad +\beta _5k_3+\beta _5 l_3+k_1 k_2 \alpha _3+k_1 l_2\alpha _3+l_1 k_2\alpha _3+l_1 l_2 \alpha _3+k_1 k_3 \beta _6+k_1 l_3 \beta _6\\ {}&\quad +l_1 k_3 \beta _6+l_1 l_3 \beta _6+k_2 k_3 \alpha _4+k_2 l_3 \alpha _4+l_2 k_3 \alpha _4+l_2 l_3 \alpha _4=0. \end{aligned}\nonumber \\ \end{aligned}$$

(4.66)

Taking into account (4.64) and (4.65), equation (4.66) reduces to

$$\begin{aligned} \begin{aligned}&\alpha _1+\beta _1 l_1+\beta _2 l_2+\beta _3 l_3+\alpha _2 k_1+\beta _4 k_2+\beta _5 k_3+k_1 l_2 \alpha _3+l_1 k_2 \alpha _3\\&\quad +k_1 l_3 \beta _6+l_1 k_3 \beta _6+k_2 l_3 \alpha _4+l_2 l_3 \alpha _4=0. \end{aligned} \end{aligned}$$

(4.67)

This gives (i). Hence Theorem 4.3 is proved. $\square$