## 1 Introduction

Let $${\mathfrak {n}}$$ be a real nilpotent Lie algebra and N be the connected 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. Hence $$(N, \langle .,. \rangle _N )$$ becomes a Riemannian manifold. We denote by $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}})$$ the group of orthogonal automorphisms of the Lie algebra $${\mathfrak {n}}$$ consisting of the automorphisms of $${\mathfrak {n}}$$ which preserve the inner product on $${\mathfrak {n}}$$. A connected Riemannian manifold M which admits a transitive nilpotent Lie group of isometries is called a Riemannian nilmanifold. It is pointed out in [10, Theorem 2(4)], that every Riemannian nilmanifold M can be identified with the unique nilpotent Lie subgroup N of the group $${\mathcal {I}}(M)$$ of isometries of M acting simply transitively on M, equipped with a left-invariant metric. Furthermore, $${\mathcal {I}}(N)$$, the group of isometries of $$(N,\langle .,. \rangle _N)$$, is the semi-direct product $$N \rtimes {{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}})$$ of the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}})$$ and the group N itself. From this observation it follows that the determination of the isometry equivalence classes of connected simply connected nilmanifolds and their isometry groups can be carried out by the investigation of the classes of isometrically isomorphic metric nilpotent Lie algebras. Applying this procedure the isometry equivalence classes and the isometry groups of connected simply connected nilmanifolds of dimension at most 5 are established in [5, 7, 9]. In the classes of 6-dimensional nilmanifolds the isometry equivalence classes and the isometry groups on two-step nilpotent Lie groups, respectively on five-step nilpotent Lie groups, this means on filiform Lie groups, are accomplished in , respectively in .

In this paper we deal with 6-dimensional metric Lie algebras having nilpotency class three or four. In [5, Sect. 3], the metric Lie algebras $$({\mathfrak {n}},\langle .,. \rangle )$$ having a decomposition into an orthogonal direct sum of 1-dimensional subspaces such that each orthogonal automorphism of $$({\mathfrak {n}},\langle .,. \rangle )$$ preserves this decomposition play an essential role. We say that these metric Lie algebras have a framing. It turns out in  that there is a strong connection between a special class $${\mathcal {C}}$$ of framed metric Lie algebras and their ideal structures. Namely the framing of a metric Lie algebra in $${\mathcal {C}}$$ can be constructed in a natural way using a descending series of ideals $${\mathfrak {n}}={\mathfrak {n}}^{(0)}\supset {\mathfrak {n}}^{(1)}\supset \dots \supset {\mathfrak {n}}^{(n-1)} \supset {\mathfrak {n}}^{(n)}=\{0\}$$ invariant under all automorphisms of $${\mathfrak {n}}$$ with $$\dim ({\mathfrak {n}}^{(i)})-\dim ({\mathfrak {n}}^{(i+1)})=1$$, $$i=0, \dots , n-1$$. This type of framings we call framing determined by ideals. Every filiform metric Lie algebra of dimension at least four allows a framing determined by ideals (see [5, Theorem 4]).

Applying the classification of 6-dimensional nilpotent Lie algebras given in , Sect. 3 is devoted to the thorough study of the ideal structures of these Lie algebras and to the determination of the 6-dimensional nilpotent metric Lie algebras having a framing determined by ideals. We obtain that 6-dimensional indecomposable nilpotent Lie algebras with the exception of six classes possess a suitable series of ideals (cf. Proposition 3.1).

In Sect. 4 we systematically apply the method of classification of the classes of isometrically isomorphic metric Lie algebras given in . We describe the isometry equivalence classes and determine the group of isometries of connected simply connected nilmanifolds on 6-dimensional indecomposable Lie groups such that their Lie algebras have a framing determined by ideals.

Among the classes of nilmanifolds having nilpotency class $$n>2$$, the geometric properties of filiform nilmanifolds have been considerably improved. In particular the characterization of totally geodesic subalgebras is given in [1, 2, 8]. Our results can be utilized for the enquiry of the totally geodesic subalgebras of 6-dimensional nilmanifolds having nilpotency class $$n \in \{3,4\}$$.

## 2 Preliminaries

The lower central series of a nilpotent Lie algebra $$\ell$$ is $$\ell =S^{0}\ell \supset S^{1}\ell \supset \dots \supset S^{j}\ell \supset S^{j+1}\ell \supset \dots \supset \{0\}$$ such that $$S^{j+1}\ell =[\ell ,S^{j}\ell ], j \in {\mathbb {N}}$$. A Lie algebra $$\ell$$ is called k-step nilpotent if $$S^{k}\ell =\{0\}$$, but $$S^{k-1}\ell \ne \{0\}$$ for some $$k \in {\mathbb {N}}$$. If an n-dimensional Lie algebra $$\ell$$ is $$(n-1)$$-step nilpotent then it is called filiform. The metric Lie algebra is a Lie algebra equipped with an inner product, the automorphisms preserving the inner product are called orthogonal automorphisms.

### Definition 2.1

An orthogonal direct sum decomposition $${\mathfrak {n}} = V_1 \oplus \dots \oplus V_n$$ on one-dimensional subspaces $$V_1$$, ..., $$V_n$$ of a metric Lie algebra $$({\mathfrak {n}},\langle .,. \rangle )$$ is called a framing, if any orthogonal automorphism of $$({\mathfrak {n}},\langle .,. \rangle )$$ preserves this decomposition. An orthonormal basis $$\{G_1,G_2,\dots ,G_n\}$$ of $$({\mathfrak {n}},\langle .,. \rangle )$$ is adapted to the framing $${\mathfrak {n}} = V_1 \oplus \dots \oplus V_n$$ if $$V_i = {\mathbb {R}}\,G_i$$ for $$i=1,\dots ,n$$. The metric Lie algebra $$({\mathfrak {n}},\langle .,. \rangle )$$ is called framed, if it has a framing.

The following concept originates from the assertion in Lemma 3 in .

### Definition 2.2

An n-dimensional metric Lie algebra $$({\mathfrak {n}},\langle .,. \rangle )$$ has a framing determined by ideals, if the Lie algebra $${\mathfrak {n}}=\text {span}(G_1, \dots , G_n)$$ has a descending series of ideals $$n^i=\text {span}(G_i, \dots , G_n)$$, $$i=1, \dots , n$$, which is left invariant under all automorphisms of $${\mathfrak {n}}$$.

In this paper we consider 6-dimensional metric nilpotent Lie algebras having a framing determined by ideals.

It is proved in Section 3.1 in  that the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}})$$ of orthogonal automorphisms of a framed metric nilpotent Lie algebra $$({\mathfrak {n}},\langle .,. \rangle )$$ is a subgroup of the group $${\mathbb {Z}}_2 \times \dots \times {\mathbb {Z}}_2$$, where the number of factors is less than or equal to $$\dim {\mathfrak {n}}$$. Hence the connected component of the isometry group $${\mathcal {I}}(N)$$ of the connected simply connected Riemannian nilmanifold $$(N,\langle .,. \rangle )$$ is isomorphic to the Lie group N.

We often use the following (see [5, Lemma 1]).

### Lemma 2.3

Let $$({\mathfrak {n}},\langle .,. \rangle )$$ and $$({\mathfrak {n}}^*,\langle .,. \rangle ^*)$$ be isometrically isomorphic framed metric Lie algebras of dimension n with framings $${\mathfrak {n}} = {\mathbb {R}}\, G_1 \oplus \dots \oplus {\mathbb {R}}\, G_n$$ and $${\mathfrak {n}}^* = {\mathbb {R}}\, G^*_1 \oplus \dots \oplus {\mathbb {R}}\, G^*_n$$, where $$(G_1, \dots , G_n)$$, respectively $$(G^*_1, \dots , G^*_n)$$ are orthonormal bases. If the commutators [., .] of $${\mathfrak {n}}$$ and $$[.,.]^*$$ of $${\mathfrak {n}}^*$$ are of the form

\begin{aligned}{}[G_i,G_j] = \sum _{k = 1}^n a^k_{i,j} G_k \quad \text {and} \quad [G^*_i,G^*_j]^* = \sum _{k = 1}^n {a^*}^k_{i,j} G^*_k, \quad i, j, k =1,\dots ,n,\end{aligned}

then $$a^k_{i,j} = \pm {a^*}^k_{i,j}$$ for all $$i, j, k =1,\dots ,n$$. Particularly, if $$a^k_{i,j},\,{a^*}^k_{i,j}\ge 0$$ then $$a^k_{i,j} = {a^*}^k_{i,j}$$.

We denote by $${\mathbb {E}}^6$$ a 6-dimensional Euclidean vector space with a distinguished orthonormal basis $${\mathcal {E}} =\{E_1, E_2, E_3, E_4, E_5, E_6\}$$. The classification of metric Lie algebras up to isometric isomorphisms proceeds in the following way given by [5, pp. 371–372]: we apply the Gram–Schmidt process to the ordered basis $$\{ G_6, G_5, G_4,G_3, G_2, G_1 \}$$ in the metric Lie algebra $$({\mathfrak {l}}, \langle .,. \rangle )$$ to get an orthonormal basis $$\{F_1, F_2, F_3, F_4, F_5, F_6\}$$ expressed by $$F_i=\sum _{k=i}^{n} a_{ik}G_k, a_{ik}\in {\mathbb {R}},$$ such that $$a_{ii}\ge 0$$. After this, we define a Lie bracket on $${\mathbb {E}}^6$$ with the same structure coefficients with respect to its distinguished basis $${\mathcal {E}}$$ as that of the metric Lie algebra $$({\mathfrak {l}}, \langle .,. \rangle )$$ with respect to its basis F. The obtained metric Lie algebra $$({\mathfrak {n}}, \langle .,. \rangle )$$ on $${\mathbb {E}}^6$$ is isometrically isomorphic to $$({\mathfrak {l}}, \langle .,. \rangle )$$. Finally, we examine under which conditions on the real parameters of metric Lie algebras on $${\mathbb {E}}^6$$ we receive a one-to-one correspondence between the equivalence classes of isometrically isomorphic metric Lie algebras and a family of metric Lie algebras on $${\mathbb {E}}^6$$.

## 3 Framed metric Lie algebras of dimension 6

In this section we investigate nilpotent Lie algebras of dimension 6 and we wish to determine which Lie algebras in this class have a framing determined by ideals. We deal with Lie algebras which are not direct products of Lie algebras of lower dimensions. According to [4, pp. 646–647], the non-isomorphic Lie algebras in this class are the Lie algebras $${L}_{6,i}, i=10, \dots , 26$$, with respect to a basis $$\{x_1, x_2, \dots , x_6\}$$. The 6-dimensional filiform nilpotent Lie algebras $${L}_{6,14}, \cdots$$, $${L}_{6,18}$$ are treated in , hence we omit these Lie algebras in our consideration. The 6-dimensional 2-step Lie algebras are the Lie algebras $${L}_{6,22}^{\varepsilon }$$ and $${L}_{6,26}$$. The corresponding Lie algebras do not have a framing determined by ideals, because their characteristic ideal is only the centre. The set of their isometric isomorphism classes are studied in . Therefore our list (3.1) doesn’t include these two Lie algebra classes.

For the remaining cases we use the following basis changes:

for $${L}_{6,11}$$, $${L}_{6,12}$$: $$x_1 \mapsto G_1, x_2 \mapsto G_2, x_3 \mapsto G_4, x_4 \mapsto G_5, x_5 \mapsto G_3, x_6 \mapsto G_6,$$ for $${L}_{6,13}$$: $$x_1 \mapsto G_1, x_2 \mapsto G_3, x_3 \mapsto G_4, x_4 \mapsto G_2, x_5 \mapsto G_5, x_6 \mapsto G_6,$$

for $${L}_{6,19}^{\varepsilon }$$: $$x_1 \mapsto G_2, x_2 \mapsto G_1, x_3 \mapsto G_3, x_4 \mapsto G_4, x_5 \mapsto G_5, x_6 \mapsto G_6,$$

for $${L}_{6,23}$$, $${L}_{6,25}$$: $$x_1 \mapsto G_1, x_2 \mapsto G_2, x_3 \mapsto G_4, x_4 \mapsto G_3, x_5 \mapsto G_6, x_6 \mapsto G_5,$$

for all other Lie algebras: $$x_i \mapsto G_i, i=1, \dots 6$$,

to obtain the ordered bases $$(G_6, G_5, G_4, G_3, G_2, G_1)$$ as orthonormal basis adapted to the framing of the corresponding metric Lie algebras. After applying the basis changes we obtain Lie algebras $${\mathfrak {l}}_{6,i}$$, $$i=10, \dots , 13$$, 19, 20, 21,  23, 24, 25, given by the following non-vanishing commutators:

\begin{aligned} \begin{aligned} {\mathfrak {l}}_{6,10}: \ {}&[G_1,G_2]=G_3, [G_1,G_3]=G_6,[G_4,G_5]=G_6;\\ {\mathfrak {l}}_{6,11}: \ {}&[G_1,G_2]=G_4, [G_1,G_4]=G_5,[G_1,G_5]=G_6, [G_2,G_4]=G_6, \\&[G_2,G_3]=G_6; \\ {\mathfrak {l}}_{6,12}: \ {}&[G_1,G_2]=G_4, [G_1,G_4]=G_5,[G_1,G_5]=G_6, [G_2,G_3]=G_6;\\ {\mathfrak {l}}_{6,13}: \ {}&[G_1,G_3]=G_4, [G_1,G_4]=G_5,[G_1,G_5]=G_6, [G_3,G_2]=G_5,\\&[G_4,G_2]=G_6;\\ \end{aligned} \end{aligned}
(3.1)
\begin{aligned} \begin{aligned} {\mathfrak {l}}_{6,19}^{\varepsilon }: \ {}&[G_2,G_1]=G_4,[G_2,G_3]=G_5, [G_1,G_4]=G_6, [G_3,G_5]= \varepsilon G_6;\\ {\mathfrak {l}}_{6,20}: \ {}&[G_1,G_2]=G_4, [G_1,G_3]=G_5, [G_1,G_5]=G_6, [G_2,G_4]=G_6;\\ {\mathfrak {l}}_{6,21}^{\varepsilon }: \ {}&[G_1,G_2]=G_3,[G_1,G_3]=G_4, [G_1,G_4]=G_6, [G_2,G_3]=G_5, \\&[G_2,G_5]=\varepsilon G_6; \\ {\mathfrak {l}}_{6,23}: \ {}&[G_1,G_2]=G_4, [G_1,G_3]=G_5,[G_1,G_4]=G_6, [G_2,G_3]=G_6;\\ {\mathfrak {l}}_{6,24}^{\varepsilon }: \ {}&[G_1,G_2]=G_3, [G_1,G_3]=G_5, [G_1,G_4]=\varepsilon G_6, [G_2,G_3]=G_6, \\&[G_2,G_4]=G_5;\\ {\mathfrak {l}}_{6,25}: \ {}&[G_1,G_2]=G_4, [G_1,G_4]=G_6, [G_1,G_3]=G_5 \end{aligned} \end{aligned}

such that $$\varepsilon \in \{ -1,0,1 \}$$.

### Proposition 3.1

Among the 6-dimensional indecomposable metric Lie algebras the metric Lie algebras $$({\mathfrak {l}}_{6,j},\langle .,. \rangle )$$, $$j=11, \dots , 18, 20, 23, 25$$, $$({\mathfrak {l}}_{6,19}^{\varepsilon =0},\langle .,. \rangle )$$, $$({\mathfrak {l}}_{6,21}^{\varepsilon =0},\langle .,. \rangle )$$ have a framing determined by ideals.

### Proof

According to Theorem 1 in [5, p. 5], the filiform metric Lie algebras $${L}_{6,k}$$ for $$k=14, \dots , 18$$ have a framing determined by ideals.

In the Lie algebras $${\mathfrak {l}}_{6,k}$$, $$k=11, 12, 13$$ the center is $$Z({\mathfrak {l}}_{6,k})=\text {span}(G_6)$$, the commutator subalgebra is $${\mathcal {S}}^1({\mathfrak {l}}_{6,k})=\text {span}(G_4, G_5, G_6)$$, the second member of the lower central series is $${\mathcal {S}}^2({\mathfrak {l}}_{6,k})=\text {span}(G_5,G_6)$$. In the Lie algebras $${\mathfrak {l}}_{6,l}$$, $$l=11, 13$$ the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,l}))$$ is $$\text {span}(G_3,G_4,G_5,G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^2({\mathfrak {l}}_{6,l}))$$ is $$\text {span}(G_2,G_3,G_4,G_5,G_6)$$. For the Lie algebra $${\mathfrak {l}}_{6,12}$$ the preimage $$\pi ^{-1}(Z({\mathfrak {l}}_{6,12}/{\mathcal {S}}^2({\mathfrak {l}}_{6,12})))$$ of the center of the factor algebra $${\mathfrak {l}}_{6,12}/{\mathcal {S}}^2({\mathfrak {l}}_{6,12})$$ in $${\mathfrak {l}}_{6,12}$$ is $$\text {span}(G_3,G_4,G_5,G_6)$$ and the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,12}))$$ is $$\text {span}(G_2,G_3,G_4,G_5,G_6)$$.

In the Lie algebra $${\mathfrak {l}}_{6,19}^{\varepsilon =0}$$ the center is $$Z({\mathfrak {l}}_{6,19}^{\varepsilon =0})=\text {span}(G_5, G_6)$$, the commutator subalgebra is $${\mathcal {S}}^1({\mathfrak {l}}_{6,19}^{\varepsilon =0})=\text {span}(G_4, G_5, G_6)$$, the second member of the lower central series is $${\mathcal {S}}^2({\mathfrak {l}}_{6,19}^{\varepsilon =0})=\text {span}(G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,19}^{\varepsilon =0}))$$ is $$\text {span}(G_2,G_3,G_4,G_5,G_6)$$. The preimage of the center of the factor algebra $${\mathfrak {l}}_{6,19}^{\varepsilon =0}/Z({\mathfrak {l}}_{6,19}^{\varepsilon =0})$$ in $${\mathfrak {l}}_{6,19}^{\varepsilon =0}$$ is $$\pi ^{-1}(Z({\mathfrak {l}}_{6,19}^{\varepsilon =0}/Z({\mathfrak {l}}_{6,19}^{\varepsilon =0})))=\text {span}(G_3,G_4,G_5,G_6)$$.

In the Lie algebra $${\mathfrak {l}}_{6,20}$$ the centre is $$Z({\mathfrak {l}}_{6,20})=\text {span}(G_6)$$, the commutator subalgebra is $${\mathcal {S}}^1({\mathfrak {l}}_{6,20})=\text {span}(G_4, G_5, G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))$$ is $$\text {span}(G_3, G_4, G_5, G_6)$$, the commutator $$[{\mathfrak {l}}_{6,20},{\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))]$$ is $$\text {span}(G_5, G_6)$$. We denote by $$\overline{l_{6,20}}$$ the factor Lie algebra $$l_{6,20}/Z(l_{6,20})=\text {span} (\overline{G_1}, \overline{G_2}, \overline{G_3}, \overline{G_4}, \overline{G_5})$$ with the Lie brackets $$[\overline{G_1}, \overline{G_2}]=\overline{G_4}$$, $$[\overline{G_1}, \overline{G_3}]=\overline{G_5}$$. The factor Lie algebra $$\overline{{\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))}={\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))/ Z(l_{6,20})$$ is the Lie algebra $$\text {span} (\overline{G_3}, \overline{G_4}, \overline{G_5})$$. The centralizer $$\overline{{\mathcal {C}}}(\overline{{\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))})$$ of $$\overline{{\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))}$$ in $$\overline{l_{6,20}}$$ is $$\text {span} (\overline{G_2}, \overline{G_3}, \overline{G_4}, \overline{G_5})$$. The preimage $$\pi ^{-1}(\overline{{\mathcal {C}}}(\overline{{\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,20}))})$$ in $$l_{6,20}$$ is $$\text {span}(G_2, G_3, G_4, G_5, G_6)$$.

In the Lie algebra $${\mathfrak {l}}_{6,21}^{\varepsilon =0}$$ the centre is $$Z({\mathfrak {l}}_{6,21}^{\varepsilon =0})=\text {span}(G_5, G_6)$$, the commutator subalgebra is $${\mathcal {S}}^1({\mathfrak {l}}_{6,21}^{\varepsilon =0})=\text {span}(G_3, G_4, G_5, G_6)$$, the second member of the lower central series is $${\mathcal {S}}^2({\mathfrak {l}}_{6,21}^{\varepsilon =0})=\text {span}(G_4, G_5, G_6)$$, the third member of the lower central series is $${\mathcal {S}}^3({\mathfrak {l}}_{6,21}^{\varepsilon =0})=\text {span}(G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^2({\mathfrak {l}}_{6,21}^{\varepsilon =0}))$$ is $$\text {span}(G_2, G_3, G_4, G_5, G_6)$$.

In the Lie algebras $${\mathfrak {l}}_{6,k}$$, $$k=23, 25$$ the center is $$Z({\mathfrak {l}}_{6,k})=\text {span}(G_5, G_6)$$, the commutator subalgebra is $${\mathcal {S}}^1({\mathfrak {l}}_{6,k})=\text {span}(G_4, G_5, G_6)$$, the second member of the lower central series is $${\mathcal {S}}^2({\mathfrak {l}}_{6,k})=\text {span}(G_6)$$ and the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,k}))$$ is $$\text {span}(G_2,G_3,G_4,G_5,G_6)$$. The preimage $$\pi ^{-1}(Z({\mathfrak {l}}_{6,k}/Z({\mathfrak {l}}_{6,k})))$$ of the center of the factor algebra $${\mathfrak {l}}_{6,k}/Z({\mathfrak {l}}_{6,k})$$ in $${\mathfrak {l}}_{6,k}$$ is $$\text {span}(G_3,G_4,G_5,G_6)$$.

Hence the subspaces $$\text {span}(G_{i}, \cdots , G_6)$$, $$i=1,\dots ,6$$, of the Lie algebras $${\mathfrak {l}}_{6,11}$$, $${\mathfrak {l}}_{6,12}$$, $${\mathfrak {l}}_{6,13}$$, $${\mathfrak {l}}_{6,19}^{\varepsilon =0}$$, $${\mathfrak {l}}_{6,20}$$, $${\mathfrak {l}}_{6,21}^{\varepsilon =0}$$, $${\mathfrak {l}}_{6,23}$$, $${\mathfrak {l}}_{6,25}$$ form a descending series of characteristic ideals. Therefore the metric Lie algebras listed in the proposition have a framing determined by ideals (see Lemma 3 in ).

The metric Lie algebra belonging to $${\mathfrak {l}}_{6,10}$$ does not have a framing determined by ideals, since the characteristic ideals of $${\mathfrak {l}}_{6,10}$$ are the centre $$Z({\mathfrak {l}}_{6,10})=\text {span}(G_6)$$, the commutator subalgebra $${\mathcal {S}}^1({\mathfrak {l}}_{6,10})=\text {span}(G_3, G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,10}))=\text {span}(G_2, G_3, G_4, G_5, G_6)$$ and the preimage $$\pi ^{-1}(Z({\mathfrak {l}}_{6,10}/Z({\mathfrak {l}}_{6,10})))=\text {span}(G_3,G_4,G_5,G_6)$$ of the centre of the factor Lie algebra $${\mathfrak {l}}_{6,10}/Z({\mathfrak {l}}_{6,10})$$ in $${\mathfrak {l}}_{6,10}$$.

A framing determined by ideals does not exist for the metric Lie algebra belonging to $${\mathfrak {l}}_{6,19}^{\varepsilon }$$, $$\varepsilon \in \{ -1,1 \}$$, because the characteristic ideals of $${\mathfrak {l}}_{6,19}^{\varepsilon }$$ are the centre $$Z({\mathfrak {l}}_{6,19}^{\varepsilon })=\text {span}(G_6)$$, the commutator subalgebra $${\mathcal {S}}^1({\mathfrak {l}}_{6,19}^{\varepsilon })=\text {span}(G_4, G_5, G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,19}^{\varepsilon }))=\text {span}(G_2, G_4, G_5, G_6)$$.

The characteristic ideals of $${\mathfrak {l}}_{6,21}^{\varepsilon }$$, $$\varepsilon \in \{ -1,1 \}$$ are the centre $$Z({\mathfrak {l}}_{6,21}^{\varepsilon })=\text {span}(G_6)$$, the commutator subalgebra $${\mathcal {S}}^1({\mathfrak {l}}_{6,21}^{\varepsilon })=\text {span}(G_3, G_4, G_5, G_6)$$, the second member of the lower central series $${\mathcal {S}}^2({\mathfrak {l}}_{6,21}^{\varepsilon })=\text {span}(G_4, G_5, G_6)$$. Hence the metric Lie algebra corresponding to $${\mathfrak {l}}_{6,21}^{\varepsilon }$$, $$\varepsilon \in \{ -1,1 \}$$ does not allow a framing determined by ideals.

The metric Lie algebra belonging to $${\mathfrak {l}}_{6,24}^{\varepsilon }$$, $$\varepsilon \in \{ -1,0,1 \}$$ does not have a framing determined by ideals, because the characteristic ideals of $${\mathfrak {l}}_{6,24}^{\varepsilon }$$ are the centre $$Z({\mathfrak {l}}_{6,24}^{\varepsilon })=\text {span}(G_5, G_6)$$, the commutator subalgebra $${\mathcal {S}}^1({\mathfrak {l}}_{6,24}^{\varepsilon })=\text {span}(G_3, G_5, G_6)$$, the centralizer $${\mathcal {C}}({\mathcal {S}}^1({\mathfrak {l}}_{6,24}^{\varepsilon }))=\text {span}(G_3, G_4, G_5, G_6)$$. $$\square$$

## 4 Isometry classes of metric Lie algebras

Firstly, we consider the 6-dimensional Lie algebras $${\mathfrak {l}}_{6,11}$$ and $${\mathfrak {l}}_{6,12}$$.

### Definition 4.1

Let $$\{E_1, E_2, E_3, E_4, E_5, E_6\}$$ be an orthonormal basis in the Euclidean vector space $${\mathbb {E}}^6$$. We denote by $${\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1,..,4$$, $$j= 1,...,6$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}{} & {} [E_1,E_2]=\alpha _1 E_4+\beta _1 E_5+\beta _2 E_6, \ [E_1,E_4]=\alpha _2 E_5+\beta _5 E_6, \ [E_2,E_3]=\beta _6 E_6, \nonumber \\{} & {} [E_1,E_3]=\beta _3 E_5+\beta _4 E_6, [E_1,E_5]=\alpha _3 E_6, [E_2,E_4]=\alpha _4 E_6. \nonumber \\ \end{aligned}
(4.1)

Let $${\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1, \dots , 4,j=1, \dots , 5$$ with $$\alpha _i \ne 0$$ be the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}&[E_1,E_2]=\alpha _1 E_4 +\beta _1 E_5 +\beta _2 E_6, \ [E_1,E_4]=\alpha _2 E_5 +\beta _5 E_6, \ [E_2,E_3]=\alpha _4 E_6, \nonumber \\&[E_1,E_3]=\beta _3 E_5 +\beta _4 E_6, [E_1,E_5]=\alpha _3 E_6. \end{aligned}
(4.2)

The bracket operations (4.1) and (4.2) satisfy the Jacobi identity.

### Theorem 4.2

Let $$\langle .,. \rangle$$ be an inner product on the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,11}$$, respectively $${\mathfrak {l}}_{6,12}$$.

1. 1.

There is a unique metric Lie algebra $$({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$({\mathfrak {l}}_{6,11}, \langle .,. \rangle )$$ with $$\alpha _i>0$$, $$i=1, \dots , 4$$, and such that one of the following cases is satisfied

1. 1.

at least two of the elements of the set $$\{\beta _1, \beta _3, \beta _4, \beta _5, \beta _6\}$$ are positive with the exception of the pairs $$\{\beta _1, \beta _5\}$$ and $$\{\beta _3, \beta _6\}$$,

2. 2.

$$\beta _1>0$$ or $$\beta _5>0$$, $$\beta _3=\beta _4=\beta _6=0$$,

3. 3.

$$\beta _3>0$$ or $$\beta _6>0$$, $$\beta _1=\beta _4=\beta _5=0$$,

4. 4.

$$\beta _4>0$$, $$\beta _1=\beta _3=\beta _5=\beta _6=0$$,

5. 5.

$$\beta _1=\beta _3=\beta _4=\beta _5=\beta _6=0$$.

There is a unique metric Lie algebra $$({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$({\mathfrak {l}}_{6,12}, \langle .,. \rangle )$$ with $$\alpha _i>0$$, $$i=1, \dots , 4$$ and such that one of the above cases $$1. -5.$$ holds with $$\beta _6=0$$.

2. 2.

The group $${{\mathcal{O}\mathcal{A}}}({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ))$$ of orthogonal automorphisms of the metric Lie algebra $$({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ is the following group:

1. (a)

in case 1. the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ))$$ is trivial,

2. (b)

in case 2. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ))=\{TE_i= E_i,i=1,2,4,5,6, TE_3= \varepsilon E_3, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

3. (c)

in case 3. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ))=\{TE_2= E_2, TE_5= E_5, TE_i= \varepsilon E_i, i=1,3,4,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

4. (d)

in case 4. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ))=\{TE_2= E_2, TE_3= E_3, TE_5=E_5, TE_i=\varepsilon E_i, i=1,4,6, \, \varepsilon =\pm 1\} \simeq {\mathbb {Z}}_2$$,

5. (e)

in case 5. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,11}(\alpha _i,\beta _j ))=\{TE_2= E_2, TE_5=E_5, TE_i=\varepsilon _1 E_i,i=1,4,6, TE_3=\varepsilon _3 E_3, \, \varepsilon _1,\varepsilon _3=\pm 1\} \simeq {\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$.

3. 3.

The group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ))$$ of orthogonal automorphisms of the metric Lie algebra $$({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ is the following group:

1. (a)

in case 1. the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ))$$ is trivial,

2. (b)

in case 2. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_3=E_3, TE_i=\varepsilon E_i,i=2,4,5,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

3. (c)

in case 3. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ))=\{TE_2= E_2, TE_5= E_5, TE_i= \varepsilon E_i, i=1,3,4,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

4. (d)

in case 4. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ))=\{TE_4= E_4, TE_6= E_6, TE_i=\varepsilon E_i, i=1,2,3,5, \, \varepsilon =\pm 1\} \simeq {\mathbb {Z}}_2$$,

5. (e)

in case 5. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,12}(\alpha _i,\beta _j ))=\{TE_i=\varepsilon _1 E_i,i=1,3, TE_j =\varepsilon _2 E_j,j=2,5, TE_k=\varepsilon _1\varepsilon _2 E_k,k=4,6, \, \varepsilon _1,\varepsilon _2=\pm 1\} \simeq {\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$.

### Proof

According to Proposition 3.1 we apply the Gram–Schmidt process to the ordered basis $$\{G_6, G_5, G_4, G_3, G_2, G_1 \}$$ and we obtain an orthonormal basis $$\{ F_1, F_2, F_3, F_4, F_5, F_6 \}$$ of $${\mathfrak {l}}_{6,11}$$ and $${\mathfrak {l}}_{6,12}$$ such that the vector $$F_i$$ is a positive multiple of $$G_i$$ modulo the subspace span $$(G_j; j>i)$$ and orthogonal to span $$(G_j; j>i)$$. Hence the orthogonal direct sum $${\mathbb {R}}F_1\oplus \dots \oplus {\mathbb {R}}F_6$$ is a framing of $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ and $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$. Expressing the vectors of the new basis in the form $$F_i=\sum _{k=i}^{6}a_{ik}G_k$$ with $$a_{ii}>0$$ we receive for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ and $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$

\begin{aligned}{}[F_1,F_2]&=\alpha _1 F_4+\beta _1 F_5+\beta _2 F_6,&[F_1,F_4]&=\alpha _2 F_5+\beta _5 F_6, \nonumber \\ [F_1,F_3]&=\beta _3 F_5+\beta _4 F_6,&[F_1,F_5]&=\alpha _3 F_6, \end{aligned}
(4.3)

and for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$

\begin{aligned}&[F_2,F_3]=\beta _6 F_6, \quad \quad [F_2,F_4]=\alpha _4 F_6, \end{aligned}
(4.4)

for $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$

\begin{aligned}{}[F_2,F_3]=\alpha _4 F_6, \end{aligned}
(4.5)

with $$\alpha _i >0$$, $$i=1,\dots , 4$$, and $$\beta _j \in {\mathbb {R}}$$, $$j=1, \dots , 6$$. Changing the orthonormal basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=-F_3, \tilde{F_4}=-F_4, \tilde{F_5}=F_5, \tilde{F_6}=-F_6$$ we obtain for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ and $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}+\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=\alpha _2 \tilde{F_5}-\beta _5 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=\beta _3 \tilde{F_5}-\beta _4\tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _3 \tilde{F_6}, \end{aligned}

and for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$

\begin{aligned}{}[\tilde{F_2},\tilde{F_3}]=\beta _6 \tilde{F_6}, \quad \quad [\tilde{F_2},\tilde{F_4}]=\alpha _4 \tilde{F_6}, \end{aligned}

for $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$

\begin{aligned} {[}\tilde{F_2},\tilde{F_3}]=\alpha _4 \tilde{F_6}. \end{aligned}

Similarly, for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ the change of the basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=F_3, \tilde{F_4}=-F_4, \tilde{F_5}=F_5, \tilde{F_6}=-F_6$$ yields

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}+\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=\alpha _2 \tilde{F_5}-\beta _5 \tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _3 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=-\beta _3 \tilde{F_5}+\beta _4\tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=-\beta _6 \tilde{F_6},&[\tilde{F_2},\tilde{F_4}]&=\alpha _4 \tilde{F_6}, \end{aligned}

and for $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$ the change of the basis: $$\tilde{F_1}=F_1, \tilde{F_2}=-F_2, \tilde{F_3}=F_3, \tilde{F_4}=-F_4, \tilde{F_5}=-F_5, \tilde{F_6}=-F_6$$ gives

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}+\beta _1 \tilde{F_5}+\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=\alpha _2 \tilde{F_5}+\beta _5 \tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _3 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=-\beta _3 \tilde{F_5}-\beta _4 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Hence there is an orthonormal basis such that in commutators (4.3) and (4.4) as well as (4.3) and (4.5) we have $$\alpha _i>0, i=1, \dots , 4$$ and one of the cases $$1. -5.$$ in assertion 1. is satisfied. This proves the existence of the metric Lie algebras $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ and $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$ having properties as in assertion 1.

Let the linear map $$T:{\mathfrak {n}}_{6,k}(\alpha _i, \beta _j) \rightarrow {\mathfrak {n}}_{6,k}(\alpha _i',\beta _j')$$, $$k=11, 12$$, be an isometric isomorphism. The decomposition $${\mathbb {R}}\, E_1 \oplus {\mathbb {R}}\, E_2 \oplus {\mathbb {R}}\, E_3 \oplus {\mathbb {R}}\, E_4 \oplus {\mathbb {R}}\, E_5 \oplus {\mathbb {R}}\, E_6$$ is a framing of both Lie algebras, where $$\alpha _i, \alpha _i' > 0$$, $$i=1, \dots , 4$$. Hence by Lemma 2.3 we have $$\alpha _i=\alpha _i'$$, $$i=1,\dots , 4$$ and $$|\beta _j'|=\beta _j$$ for $$j=1,\dots , 6.$$ Let $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots ,6$$. Using the commutation relations (4.3), (4.4) and (4.5) we obtain from $$[TE_i,TE_j]' = T[E_i,E_j]$$, $$i, j =1, \dots , 6$$, for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ and $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$ the equations

\begin{aligned}&\varepsilon _1\varepsilon _2 \left( \alpha _1 E_4+\beta _1' E_5+\beta _2' E_6\right) =\alpha _1 \varepsilon _4 E_4+\beta _1 \varepsilon _5 E_5+\beta _2 \varepsilon _6 E_6, \nonumber \\&\varepsilon _1\varepsilon _3\left( \beta _3' E_5+ \beta _4' E_6\right) =\beta _3 \varepsilon _5 E_5+ \beta _4 \varepsilon _6 E_6, \quad \varepsilon _1\varepsilon _5\left( \alpha _3 E_6 \right) =\alpha _3 \varepsilon _6 E_6, \\&\varepsilon _1\varepsilon _4\left( \alpha _2 E_5+\beta _5' E_6 \right) =\alpha _2 \varepsilon _5 E_5+\beta _5 \varepsilon _6 E_6,\nonumber \end{aligned}
(4.6)

and for $$({\mathfrak {l}}_{6,11},\langle .,. \rangle )$$ the equations

\begin{aligned}&\varepsilon _2\varepsilon _3\left( \beta _6' E_6\right) =\beta _6 \varepsilon _6 E_6, \quad \varepsilon _2\varepsilon _4\left( \alpha _4 E_6\right) =\alpha _4 \varepsilon _6 E_6, \end{aligned}
(4.7)

and for $$({\mathfrak {l}}_{6,12},\langle .,. \rangle )$$ the equation

\begin{aligned}&\varepsilon _2\varepsilon _3\left( \alpha _4 E_6\right) =\alpha _4 \varepsilon _6 E_6. \end{aligned}
(4.8)

From (4.6) and (4.7) it follows $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$$$\varepsilon _1\varepsilon _4=\varepsilon _5,\, \varepsilon _1\varepsilon _5=\varepsilon _2 \varepsilon _4=\varepsilon _6$$. Then one has $$\varepsilon _2= \varepsilon _5=1$$, $$\varepsilon _1 = \varepsilon _4= \varepsilon _6$$. Using these relations we have $$\varepsilon _1 \varepsilon _2=\varepsilon _6$$. Therefore one has $$\beta _2'=\beta _2$$.

If $$\beta _1= \beta _1'>0$$ or $$\beta _5= \beta _5'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _5$$ or $$\varepsilon _1\varepsilon _4=\varepsilon _6$$, which yields that $$\varepsilon _i=1, i=1,2,4,5,6$$.

If $$\beta _3= \beta _3'>0$$ or $$\beta _6= \beta _6'>0$$, then we have additionally $$\varepsilon _1\varepsilon _3=\varepsilon _5$$ or $$\varepsilon _2\varepsilon _3=\varepsilon _6$$. Hence one has $$\varepsilon _2=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _6$$.

If $$\beta _4= \beta _4'>0$$, then we get $$\varepsilon _1\varepsilon _3=\varepsilon _6$$, which gives that $$\varepsilon _2=\varepsilon _3=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _4=\varepsilon _6$$.

Using these relations in assertion 1. of the Theorem

in case 1. we obtain $$\varepsilon _i=1, i=1, \dots , 6$$,

in case 2. we have $$\varepsilon _i=1, i=1,2,4,5,6$$,

in case 3. we get $$\varepsilon _2=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _6$$,

in case 4. we have $$\varepsilon _2=\varepsilon _3=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _4=\varepsilon _6$$,

in case 5. we obtain $$\varepsilon _2=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _4=\varepsilon _6$$.

From (4.6) and (4.8) it follows $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$$$\varepsilon _1\varepsilon _4=\varepsilon _5,\, \varepsilon _1\varepsilon _5=\varepsilon _2 \varepsilon _3=\varepsilon _6$$. Then one has $$\varepsilon _1=\varepsilon _3$$, $$\varepsilon _2=\varepsilon _5$$, $$\varepsilon _1\varepsilon _2=\varepsilon _4=\varepsilon _6$$. Using this we have $$\varepsilon _1 \varepsilon _2=\varepsilon _6$$ and hence one has $$\beta _2'=\beta _2$$.

If $$\beta _1= \beta _1'>0$$ or $$\beta _5= \beta _5'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _5$$ or $$\varepsilon _1\varepsilon _4=\varepsilon _6$$, hence in both cases we obtain $$\varepsilon _1=\varepsilon _3=1$$ and $$\varepsilon _2=\varepsilon _4=\varepsilon _5=\varepsilon _6$$.

If $$\beta _3= \beta _3'>0$$, then we have in addition $$\varepsilon _1\varepsilon _3=\varepsilon _5$$, which gives $$\varepsilon _2=\varepsilon _5=1$$, $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _6$$.

If $$\beta _4= \beta _4'>0$$, then we get additionally $$\varepsilon _1\varepsilon _3=\varepsilon _6$$, which yields $$\varepsilon _4=\varepsilon _6=1$$, $$\varepsilon _1=\varepsilon _2=\varepsilon _3=\varepsilon _5$$.

Applying these relations in assertion 1. of the Theorem

in case 1. we obtain $$\varepsilon _i=1, i=1,\dots , 6$$,

in case 2. we get $$\varepsilon _1=\varepsilon _3=1$$ and $$\varepsilon _2=\varepsilon _4=\varepsilon _5=\varepsilon _6$$,

in case 3. we obtain $$\varepsilon _2=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _6$$,

in case 4. we have $$\varepsilon _4=\varepsilon _6=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _3=\varepsilon _5$$,

in case 5. we get $$\varepsilon _1=\varepsilon _3$$, $$\varepsilon _2=\varepsilon _5$$ and $$\varepsilon _1 \varepsilon _2=\varepsilon _4=\varepsilon _6$$.

Hence the system of Eqs. (4.6) and (4.7) as well as (4.6) and (4.8) are satisfied with $$\beta _j'=\beta _j, j=1, \dots , 6$$, in cases $$1.-5.$$ of the Theorem. This proves the uniqueness of the Lie algebras $${\mathfrak {n}}_{6,11}(\alpha _i,\beta _j)$$ and $${\mathfrak {n}}_{6,12}(\alpha _i,\beta _j)$$ in cases $$1.-5$$. This yields assertion 1.

If the map $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1, \dots , 6$$, is an orthogonal automorphism of $${\mathfrak {n}}_{6,11}(\alpha _i,\beta _j)$$, respectively $${\mathfrak {n}}_{6,12}(\alpha _i,\beta _j)$$ then the system of equations given by (4.6) and (4.7), respectively (4.6) and (4.8) is satisfied with $$\alpha _i >0, i=1, \dots , 4, \beta _j'=\beta _j, j=1, \dots , 6$$. Therefore the conditions for $$\varepsilon _i, i=1, \dots , 6$$, are the same as above. Taking this into account the group of orthogonal automorphisms of $${\mathfrak {n}}_{6,11}(\alpha _i,\beta _j)$$ and $${\mathfrak {n}}_{6,12}(\alpha _i,\beta _j)$$ in case 1. is trivial, in cases $$2.-5.$$ is isomorphic to the group given by 2b–2e and 3b–3e. This proves assertions 2 and 3. $$\square$$

### Corollary 4.3

Let $$(\aleph _{6,k}(\alpha _i,\beta _j), \langle .,. \rangle )$$, $$k=11, 12$$, be the connected and simply connected Riemannian nilmanifold corresponding to the metric Lie algebra $$( {\mathfrak {n}}_{6,k}(\alpha _i,\beta _j ), \langle .,. \rangle )$$, $$k=11, 12$$. The isometry group of $$(\aleph _{6,11}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is $${\mathcal {I}} (\aleph _{6,11}(\alpha _i,\beta _j))$$

\begin{aligned} ={\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \ltimes \aleph _{6,11}(\alpha _i,\beta _j) &{} if \ \beta _j=0, j=1,3,4,5,6, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,11}(\alpha _i,\beta _j) &{} if \ \beta _1>0 \ or \ \beta _5>0, \beta _3=\beta _4=\beta _6=0, \\ &{} or \ \beta _3>0 \ or \ \beta _6>0, \beta _1=\beta _4=\beta _5=0, \\ &{} or \ \beta _4>0, \beta _j=0, j=1,3,5,6, \\ \aleph _{6,11}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _3>0, \ or \ \beta _1>0, \beta _4>0, \\ &{} or \ \beta _1>0, \beta _6>0, \ or \ \beta _3>0, \beta _4>0, \\ &{} or \ \beta _3>0, \beta _5>0, \ or \ \beta _4>0, \beta _5>0, \\ &{} or \ \beta _4>0, \beta _6>0, \ or \ \beta _5>0, \beta _6>0. \end{array}\right. } \end{aligned}

The isometry group of $$(\aleph _{6,12}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}}(\aleph _{6,12}(\alpha _i,\beta _j))\nonumber \\{} & {} \quad = {\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \ltimes \aleph _{6,12}(\alpha _i,\beta _j) &{} if \ \beta _1=\beta _3=\beta _4=\beta _5=0, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,12}(\alpha _i,\beta _j) &{} if \ \beta _1>0 \ or \ \beta _5>0, \beta _3=\beta _4=0, \\ &{} or \ \beta _3>0, \beta _1=\beta _4=\beta _5=0, \\ &{} or \ \beta _4>0, \beta _1=\beta _3=\beta _5=0, \\ \aleph _{6,12}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _3>0, \ or \ \beta _1>0, \beta _4>0, \\ &{} or \ \beta _3>0, \beta _4>0, \ or \ \beta _3>0, \beta _5>0, \\ &{} or \ \beta _4>0, \beta _5>0. \end{array}\right. } \end{aligned}

Secondly, we consider the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,13}$$.

### Definition 4.4

Let $$\{E_1, E_2, E_3, E_4, E_5, E_6\}$$ be an orthonormal basis in the Euclidean vector space $${\mathbb {E}}^6$$. Denote by $${\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1, \dots , 4$$, $$j= 1, \dots , 7$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}{}[E_1,E_2]&=\beta _1 E_4+\beta _2 E_5+\beta _3 E_6,&[E_1,E_3]&=\frac{\alpha _2\alpha _3}{\alpha _4} E_4+\beta _4 E_5+\beta _5 E_6, \nonumber \\ [E_1,E_4]&=\alpha _1 E_5+\beta _6 E_6,&[E_1,E_5]&=\alpha _2 E_6, \\ [E_3,E_2]&=\alpha _3 E_5+\beta _7 E_6, \nonumber&[E_4,E_2]&=\alpha _4 E_6. \end{aligned}
(4.9)

The bracket operation (4.9) satisfies the Jacobi identity.

### Theorem 4.5

Let $$\langle .,. \rangle$$ be an inner product on the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,13}$$.

1. 1.

There is a unique metric Lie algebra $$({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$({\mathfrak {l}}_{6,13}, \langle .,. \rangle )$$ with $$\alpha _i>0$$, $$i=1, \dots , 4$$ and such that one of the following cases is satisfied

1. 1.

at least two of the elements of $$\{\beta _1, \beta _2, \beta _3, \beta _4, \beta _6, \beta _7\}$$ are positive with the exception of the pairs $$\{\beta _1, \beta _3\}$$, $$\{\beta _4, \beta _6\}$$, $$\{\beta _4, \beta _7\}$$, $$\{\beta _6, \beta _7\}$$,

2. 2.

$$\beta _1>0$$ or $$\beta _3>0$$, $$\beta _2=\beta _4=\beta _6=\beta _7=0$$,

3. 3.

$$\beta _2>0$$, $$\beta _1=\beta _3=\beta _4=\beta _6=\beta _7=0$$,

4. 4.

$$\beta _4>0$$ or $$\beta _6>0$$ or $$\beta _7>0$$, $$\beta _1=\beta _2=\beta _3=0$$,

5. 5.

$$\beta _1=\beta _2=\beta _3=\beta _4=\beta _6=\beta _7=0$$.

2. 2.

The group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ))$$ of orthogonal automorphisms of the metric Lie algebra $$({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ is the following group:

1. (a)

in case 1. the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ))$$ is trivial,

2. (b)

in case 2. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ))=\{TE_2= E_2, TE_3= E_3, TE_5=E_5, TE_i= \varepsilon E_i, i=1,4,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

3. (c)

in case 3. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ))=\{TE_2=E_2, TE_4= E_4, TE_6= E_6, TE_i=\varepsilon E_i, i=1,3,5, \, \varepsilon =\pm 1\} \simeq {\mathbb {Z}}_2$$,

4. (d)

in case 4. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_2=E_2, TE_i=\varepsilon E_i,i=3,4,5,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

5. (e)

in case 5. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ))=\{TE_2=E_2, TE_1=\varepsilon _1 E_1, TE_i =\varepsilon _3 E_i,i=3,5, TE_j=\varepsilon _1\varepsilon _3 E_j,j=4,6, \, \varepsilon _1,\varepsilon _3=\pm 1\} \simeq {\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$.

### Proof

According to Proposition 3.1 we utilize the Gram–Schmidt process to the ordered basis $$\{G_6, G_5, G_4, G_3, G_2, G_1 \}$$ which yields an orthonormal basis $$\{ F_1, F_2, F_3, F_4, F_5, F_6 \}$$ of $${\mathfrak {l}}_{6,13}$$ such that the vector $$F_i$$ is a positive multiple of $$G_i$$ modulo the subspace span $$(G_j; j>i)$$ and orthogonal to span $$(G_j; j>i)$$. The orthogonal direct sum $${\mathbb {R}}F_1\oplus \dots \oplus {\mathbb {R}}F_6$$ is a framing of $$({\mathfrak {l}}_{6,13},\langle .,. \rangle )$$. Expressing the vectors of the new basis in the form $$F_i=\sum _{k=i}^{6}a_{ik}G_k$$ with $$a_{ii}>0$$ we get

\begin{aligned}{}[F_1,F_2]&=\beta _1 F_4+\beta _2 F_5+\beta _3 F_6,&[F_1,F_3]&=\frac{\alpha _2\alpha _3}{\alpha _4} F_4+\beta _4 F_5+\beta _5 F_6, \nonumber \\ [F_1,F_4]&=\alpha _1 F_5+\beta _6 F_6,&[F_1,F_5]&=\alpha _2 F_6, \\ [F_3,F_2]&=\alpha _3 F_5+\beta _7 F_6,&[F_4,F_2]&=\alpha _4 F_6\nonumber \end{aligned}
(4.10)

with $$\alpha _i >0$$, $$i=1,\dots ,4$$ and $$\beta _j \in {\mathbb {R}}$$, $$j=1, \dots , 7$$. Changing the orthonormal basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=-F_3, \tilde{F_4}=F_4, \tilde{F_5}=-F_5, \tilde{F_6}=F_6$$ we obtain

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=-\beta _1 \tilde{F_4}+\beta _2 \tilde{F_5}-\beta _3 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=\frac{\alpha _2\alpha _3}{\alpha _4} \tilde{F_4}-\beta _4 \tilde{F_5}+\beta _5 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _1 \tilde{F_5}-\beta _6 \tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _2 \tilde{F_6}, \\ [\tilde{F_3},\tilde{F_2}]&=\alpha _3 \tilde{F_5}-\beta _7 \tilde{F_6},&[\tilde{F_4},\tilde{F_2}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Similarly, the change of the basis: $$\tilde{F_1}=F_1, \tilde{F_2}=F_2, \tilde{F_3}=-F_3, \tilde{F_4}=-F_4, \tilde{F_5}=-F_5, \tilde{F_6}=-F_6$$ gives

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=-\beta _1 \tilde{F_4}-\beta _2 \tilde{F_5}-\beta _3 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=\frac{\alpha _2\alpha _3}{\alpha _4} \tilde{F_4}+\beta _4 \tilde{F_5}+\beta _5 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _1 \tilde{F_5}+\beta _6 \tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _2 \tilde{F_6}, \\ [\tilde{F_3},\tilde{F_2}]&=\alpha _3 \tilde{F_5}+\beta _7 \tilde{F_6},&[\tilde{F_4},\tilde{F_2}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Hence there is an orthonormal basis such that in commutators (4.10) we have $$\alpha _i>0, i=1, \dots , 4$$ and one of the cases in assertion 1. is satisfied. This proves the existence of $${\mathfrak {n}}_{6,13}(\alpha _i, \beta _j)$$ with the properties in assertion 1.

Let the linear map $$T:{\mathfrak {n}}_{6,13}(\alpha _i, \beta _j) \rightarrow {\mathfrak {n}}_{6,13}(\alpha _i',\beta _j')$$ be an isometric isomorphism. The decomposition $${\mathbb {R}}\, E_1 \oplus {\mathbb {R}}\, E_2 \oplus {\mathbb {R}}\, E_3 \oplus {\mathbb {R}}\, E_4 \oplus {\mathbb {R}}\, E_5 \oplus {\mathbb {R}}\, E_6$$ is a framing of both Lie algebras, where $$\alpha _i, \alpha _i' > 0$$, $$i=1, \dots , 4$$. According to Lemma 2.3 we have $$\alpha _i=\alpha _i'$$, $$i=1,\dots , 4$$ and $$|\beta _j'|=\beta _j$$, $$j=1, \dots , 7.$$ Let $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1, \dots , 6$$. Using the commutation relations (4.10) we obtain from $$[TE_i,TE_j]' = T[E_i,E_j]$$, $$i, j =1, \dots , 6$$, the equations

\begin{aligned}&\varepsilon _1\varepsilon _2 \left( \beta _1' E_4+\beta _2' E_5+\beta _3' E_6\right) =\beta _1 \varepsilon _4 E_4+\beta _2 \varepsilon _5 E_5+\beta _3 \varepsilon _6 E_6, \nonumber \\&\varepsilon _1\varepsilon _3\left( \frac{\alpha _2\alpha _3}{\alpha _4} E_4+\beta _4' E_5+\beta _5' E_6\right) =\frac{\alpha _2\alpha _3}{\alpha _4} \varepsilon _4 E_4+\beta _4 \varepsilon _5 E_5+\beta _5 \varepsilon _6 E_6,\\&\varepsilon _1\varepsilon _4\left( \alpha _1 E_5+\beta _6' E_6\right) =\alpha _1 \varepsilon _5 E_5+\beta _6 \varepsilon _6 E_6, \quad \varepsilon _1\varepsilon _5\left( \alpha _2 E_6 \right) =\alpha _2 \varepsilon _6 E_6, \nonumber \\&\varepsilon _3\varepsilon _2\left( \alpha _3 E_5+\beta _7' E_6\right) =\alpha _3 \varepsilon _5 E_5+\beta _7 \varepsilon _6 E_6, \quad \varepsilon _4\varepsilon _2\left( \alpha _4 E_6\right) =\alpha _4 \varepsilon _6 E_6. \nonumber \end{aligned}
(4.11)

From (4.11) it follows $$\varepsilon _1 \varepsilon _3=\varepsilon _4$$$$\varepsilon _1\varepsilon _4=\varepsilon _3\varepsilon _2=\varepsilon _5,\, \varepsilon _1\varepsilon _5=\varepsilon _4 \varepsilon _2=\varepsilon _6$$. Hence one has $$\varepsilon _2=1$$, $$\varepsilon _3=\varepsilon _5$$, $$\varepsilon _1\varepsilon _3=\varepsilon _4=\varepsilon _6$$. Using these relations we have $$\varepsilon _1 \varepsilon _3=\varepsilon _6$$. Therefore one has $$\beta _5'=\beta _5$$.

If $$\beta _1= \beta _1'>0$$ or $$\beta _3= \beta _3'>0$$, then we have additionally $$\varepsilon _1\varepsilon _2=\varepsilon _4$$ or $$\varepsilon _1\varepsilon _2=\varepsilon _6$$. Hence one has $$\varepsilon _2=\varepsilon _3=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _4=\varepsilon _6$$.

If $$\beta _2= \beta _2'>0$$, then we get in addition $$\varepsilon _1\varepsilon _2=\varepsilon _5$$, which gives $$\varepsilon _2=\varepsilon _4=\varepsilon _6=1$$, $$\varepsilon _1=\varepsilon _3=\varepsilon _5$$.

If $$\beta _4= \beta _4'>0$$ or $$\beta _6= \beta _6'>0$$ or $$\beta _7= \beta _7'>0$$, then we get additionally $$\varepsilon _1\varepsilon _3=\varepsilon _5$$ or $$\varepsilon _1\varepsilon _4=\varepsilon _6$$ or $$\varepsilon _3\varepsilon _2=\varepsilon _6$$. Hence in these cases we obtain $$\varepsilon _1=\varepsilon _2=1$$ and $$\varepsilon _3=\varepsilon _4=\varepsilon _5=\varepsilon _6$$.

Using these relations in assertion 1. of the Theorem

in case 1. we get $$\varepsilon _i=1, i=1, \dots , 6$$,

in case 2. we obtain $$\varepsilon _2=\varepsilon _3=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _4=\varepsilon _6$$,

in case 3. we have $$\varepsilon _2=\varepsilon _4=\varepsilon _6=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _5$$,

in case 4. we obtain $$\varepsilon _1=\varepsilon _2=1$$ and $$\varepsilon _3=\varepsilon _4=\varepsilon _5=\varepsilon _6$$,

in case 5. we get $$\varepsilon _2=1$$, $$\varepsilon _3=\varepsilon _5$$ and $$\varepsilon _1\varepsilon _3=\varepsilon _4=\varepsilon _6$$.

Hence the system of Eq. (4.11) is satisfied with $$\beta _j'=\beta _j, j=1, \dots , 7$$ in cases $$1.-5.$$ of the Theorem. Therefore the uniqueness of the Lie algebra $${\mathfrak {n}}_{6,13}(\alpha _i,\beta _j)$$ in cases $$1.-5.$$ is proved. This yields assertion 1.

If the map $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1, \dots , 6$$, is an orthogonal automorphism of $${\mathfrak {n}}_{6,13}(\alpha _i,\beta _j)$$, then the system of equations given by (4.11) is satisfied with $$\alpha _i>0, i=1, \dots , 4$$, $$\beta _j'=\beta _j, j=1, \dots , 7$$. Therefore in cases $$1.-5.$$ the conditions for $$\varepsilon _i, i=1, \dots , 6$$, are given above. Taking this into consideration the group of orthogonal automorphisms of $${\mathfrak {n}}_{6,13}(\alpha _i,\beta _j)$$ in case 1. is trivial, in cases $$2.-5.$$ is isomorphic to the group given by 2b–2e. This proves assertion 2. $$\square$$

### Corollary 4.6

Let $$(\aleph _{6,13}(\alpha _i,\beta _j), \langle .,. \rangle )$$ be the connected and simply connected Riemannian nilmanifold belonging to $$( {\mathfrak {n}}_{6,13}(\alpha _i,\beta _j ), \langle .,. \rangle )$$. The group of isometries of $$(\aleph _{6,13}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}} (\aleph _{6,13}(\alpha _i,\beta _j)) \nonumber \\{} & {} ={\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \ltimes \aleph _{6,13}(\alpha _i,\beta _j) &{} if \ \beta _j=0, j=1,2,3,4,6,7, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,13}(\alpha _i,\beta _j) &{} if \ \beta _1>0 \ or \ \beta _3>0, \beta _j=0, j=2,4,6,7, \\ &{} or \ \beta _2>0, \beta _j=0, j=1,3,4,6,7, \\ &{} or \ \beta _4>0 \ or \ \beta _6>0 \ or \ \beta _7>0 \ and \\ &{} \beta _1=\beta _2=\beta _3=0, \\ \aleph _{6,13}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, \ or \ \beta _1>0, \beta _4>0, \\ &{} or \ \beta _1>0, \beta _6>0, \ or \ \beta _1>0, \beta _7>0, \\ &{} or \ \beta _2>0, \beta _3>0, \ or \ \beta _2>0, \beta _4>0, \\ &{} or \ \beta _2>0, \beta _6>0, \ or \ \beta _2>0, \beta _7>0, \\ &{} or \ \beta _3>0, \beta _4>0, \ or \ \beta _3>0, \beta _6>0, \\ &{} or \ \beta _3>0, \beta _7>0. \end{array}\right. } \end{aligned}

We treat the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,19}^{\varepsilon =0}$$.

### Definition 4.7

Let $$\{E_1, E_2, E_3, E_4, E_5, E_6\}$$ be an orthonormal basis in the Euclidean vector space $${\mathbb {E}}^6$$. Denote by $${\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1,2,3, j=1,...,5$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}{}[E_2,E_1]&=\alpha _1 E_4+\beta _1 E_5+\beta _2 E_6,&[E_1,E_3]&=\beta _3 E_5+\beta _4 E_6,\\ [E_1,E_4]&=\alpha _2 E_6, \nonumber&[E_2,E_3]&=\alpha _3 E_5+\beta _5 E_6. \end{aligned}
(4.12)

The bracket operation (4.12) satisfies the Jacobi identity.

### Theorem 4.8

Let $$\langle .,. \rangle$$ be an inner product on the 6-dimensional Lie algebra $$\mathfrak {l}_{6,19}^{\varepsilon =0}$$.

1. 1.

There is a unique metric Lie algebra $$(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$(\mathfrak {l}_{6,19}^{\varepsilon =0}, \langle .,. \rangle )$$ with $$\alpha _i>0$$, $$i=1,2,3$$, and such that one of the following cases is satisfied

1. 1.

at least three of the elements of the set $$\{\beta _1, \beta _2, \beta _3, \beta _4, \beta _5\}$$ are posi- tive with the exception of the triples $$\{\beta _1, \beta _2, \beta _5\}$$ and $$\{\beta _3, \beta _4, \beta _5\}$$,

2. 2.

at least two of the elements of the set $$\{\beta _1,\beta _2,\beta _5\}$$ are positive and $$\beta _3=\beta _4=0$$,

3. 3.

at least two of the elements of the set $$\{\beta _3,\beta _4,\beta _5\}$$ are positive and $$\beta _1=\beta _2=0$$,

4. 4.

$$\beta _1>0$$, $$\beta _3>0$$, $$\beta _2=\beta _4=\beta _5=0$$,

5. 5.

$$\beta _1>0$$, $$\beta _4>0$$, $$\beta _2=\beta _3=\beta _5=0$$,

6. 6.

$$\beta _2>0$$, $$\beta _3>0$$, $$\beta _1=\beta _4=\beta _5=0$$,

7. 7.

$$\beta _2>0$$, $$\beta _4>0$$, $$\beta _1=\beta _3=\beta _5=0$$,

8. 8.

$$\beta _1>0$$, $$\beta _j=0$$, $$j=2,3,4,5$$,

9. 9.

$$\beta _2>0$$, $$\beta _j=0$$, $$j=1,3,4,5$$,

10. 10.

$$\beta _3>0$$, $$\beta _j=0$$, $$j=1,2,4,5$$,

11. 11.

$$\beta _4>0$$, $$\beta _j=0$$, $$j=1,2,3,5$$,

12. 12.

$$\beta _5>0$$, $$\beta _j=0$$, $$j=1,2,3,4$$,

13. 13.

$$\beta _j=0$$, $$j=1,2,3,4,5$$.

2. 2.

The group $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))$$ of orthogonal automorphisms of the metric Lie algebra $$(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ is the following group:

1. (a)

in case 1. the group $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))$$ is trivial,

2. (b)

in case 2. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_3=E_3, TE_i=\varepsilon E_i, i=2,4,5,6, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$$,

3. (c)

in case 3. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_3=E_3, TE_4=E_4, TE_i=\varepsilon E_i, i=1,2,5,6, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$$,

4. (d)

in case 4. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_4=E_4, TE_5=E_5, TE_i=\varepsilon E_i, i=1,2,3,6, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$$,

5. (e)

in case 5. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_2=E_2, TE_6=E_6, TE_i=\varepsilon E_i, i=1,3,4,5, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$$,

6. (f)

in case 6. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_2=E_2, TE_4=E_4, TE_6=E_6, TE_i=\varepsilon E_i, i=3,5, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$$,

7. (g)

in case 7. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_5=E_5, TE_i=\varepsilon E_i, i=2,3,4,6, \, \varepsilon =\pm 1 \}\simeq \mathbb {Z}_2$$,

8. (h)

in case 8. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{ TE_i=\varepsilon _1 E_i, i=1,3, TE_j=\varepsilon _2 E_j, j=2,6, TE_k=\varepsilon _1 \varepsilon _2 E_k, k=4,5, \, \varepsilon _1, \varepsilon _2=\pm 1 \}\simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$,

9. (i)

in case 9. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_i=\varepsilon _2 E_i, i=2,4,6, TE_3=\varepsilon _3 E_3, TE_5=\varepsilon _2 \varepsilon _3 E_5, \, \varepsilon _2, \varepsilon _3=\pm 1 \}\simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$,

10. (j)

in case 10. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_4=E_4, TE_i=\varepsilon _1 E_i, i=1,2,6, TE_3=\varepsilon _3 E_3, TE_5=\varepsilon _1 \varepsilon _3 E_5, \, \varepsilon _1, \varepsilon _3=\pm 1 \}\simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$,

11. (k)

in case 11. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{ TE_i=\varepsilon _1 E_i, i=1,5, TE_j=\varepsilon _2 E_j, j=2,6, TE_k=\varepsilon _1 \varepsilon _2 E_k, k=3,4, \, \varepsilon _1, \varepsilon _2=\pm 1 \}\simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$,

12. (l)

in case 12. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_3=E_3, TE_1=\varepsilon _1 E_1, TE_i=\varepsilon _2 E_i, i=2,5,6, TE_4=\varepsilon _1 \varepsilon _2 E_4, \, \varepsilon _1, \varepsilon _2=\pm 1 \}\simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$,

13. (m)

in case 13. one has $$\mathcal{O}\mathcal{A}(\mathfrak {n}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_1=\varepsilon _1 E_1, TE_i=\varepsilon _2 E_i, i=2,6, TE_3=\varepsilon _3 E_3, TE_4=\varepsilon _1\varepsilon _2 E_4, TE_5=\varepsilon _2\varepsilon _3 E_5,\, \varepsilon _1,\varepsilon _2, \varepsilon _3=\pm 1\} \simeq \mathbb {Z}_2 \times \mathbb {Z}_2 \times \mathbb {Z}_2$$.

### Proof

According to Proposition 3.1 the application of the Gram–Schmidt process to the ordered basis $$\{G_6, G_5, G_4, G_3, G_2, G_1 \}$$ yields an orthonormal basis $$\{ F_1, F_2, F_3, F_4, F_5, F_6 \}$$ of $${\mathfrak {l}}_{6,19}^{\varepsilon =0}$$ such that the vector $$F_i$$ is a positive multiple of $$G_i$$ modulo the subspace span $$(G_j; j>i)$$ and orthogonal to span $$(G_j; j>i)$$. The orthogonal direct sum $${\mathbb {R}}F_1\oplus \dots \oplus {\mathbb {R}}F_6$$ is a framing of $$({\mathfrak {l}}_{6,19}^{\varepsilon =0},\langle .,. \rangle )$$ and the vectors of the new basis can be written into the form $$F_i=\sum _{k=i}^{6}a_{ik}G_k$$ with $$a_{ii}>0$$. Hence we receive

\begin{aligned}{}[F_2,F_1]&=\alpha _1 F_4+\beta _1 F_5+\beta _2 F_6,&[F_1,F_3]&=\beta _3 F_5+\beta _4 F_6,\\ [F_1,F_4]&=\alpha _2 F_6,&[F_2,F_3]&=\alpha _3 F_5+\beta _5 F_6\nonumber \end{aligned}
(4.13)

with $$\alpha _i >0$$, $$i=1,2,3$$ and $$\beta _j \in {\mathbb {R}}$$, $$j=1, \dots , 5$$. The changes of the orthonormal basis:

$$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=F_3, \tilde{F_4}=-F_4, \tilde{F_5}=F_5, \tilde{F_6}=F_6$$, respectively $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=-F_3, \tilde{F_4}=-F_4, \tilde{F_5}=-F_5, \tilde{F_6}=F_6$$, respectively $$\tilde{F_1}=-F_1, \tilde{F_2}=-F_2, \tilde{F_3}=F_3, \tilde{F_4}=F_4, \tilde{F_5}=-F_5, \tilde{F_6}=-F_6$$ give

\begin{aligned}{}[\tilde{F_2},\tilde{F_1}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}-\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=-\beta _3 \tilde{F_5}-\beta _4 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _2 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _3 \tilde{F_5}+\beta _5 \tilde{F_6}, \end{aligned}

respectively

\begin{aligned}{}[\tilde{F_2},\tilde{F_1}]&=\alpha _1 \tilde{F_4}+\beta _1 \tilde{F_5}-\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=-\beta _3 \tilde{F_5}+\beta _4 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _2 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _3 \tilde{F_5}-\beta _5 \tilde{F_6}, \end{aligned}

respectively

\begin{aligned}{}[\tilde{F_2},\tilde{F_1}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}-\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=\beta _3 \tilde{F_5}+\beta _4 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _2 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _3 \tilde{F_5}+\beta _5 \tilde{F_6}. \end{aligned}

Hence there is an orthonormal basis such that in commutators (4.13) one has $$\alpha _i >0$$, $$i=1,2,3$$, and one of the cases in assertion 1. holds. This proves the existence of $${\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i, \beta _j)$$ having properties as in assertion 1.

Let the linear map $$T:{\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i, \beta _j) \rightarrow {\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i',\beta _j')$$ be an isometric isomorphism. The decomposition $${\mathbb {R}}\, E_1 \oplus {\mathbb {R}}\, E_2 \oplus {\mathbb {R}}\, E_3 \oplus {\mathbb {R}}\, E_4 \oplus {\mathbb {R}}\, E_5 \oplus {\mathbb {R}}\, E_6$$ is a framing of both Lie algebras, where $$\alpha _i, \alpha _i'>0$$, $$i=1,2,3$$. Hence by Lemma 2.3 we have $$\alpha _i=\alpha _i'$$, $$i=1,2,3$$ and $$|\beta _j'|=\beta _j$$, $$j=1, \dots , 5.$$ Let $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots ,6$$. Using the commutation relations (4.13) we obtain from $$[TE_i,TE_j]' = T[E_i,E_j]$$, $$i, j =1, \dots , 6$$, the equations

\begin{aligned}&\varepsilon _2\varepsilon _1 \left( \alpha _1 E_4+\beta _1' E_5+\beta _2' E_6\right) =\alpha _1 \varepsilon _4 E_4+\beta _1 \varepsilon _5 E_5+\beta _2 \varepsilon _6 E_6, \nonumber \\&\varepsilon _1\varepsilon _3\left( \beta _3' E_5+\beta _4' E_6\right) =\beta _3 \varepsilon _5 E_5+\beta _4 \varepsilon _6 E_6, \quad \varepsilon _1\varepsilon _4\left( \alpha _2 E_6\right) =\alpha _2 \varepsilon _6 E_6, \\&\varepsilon _2\varepsilon _3\left( \alpha _3 E_5+\beta _5' E_6\right) =\alpha _3 \varepsilon _5 E_5+\beta _5 \varepsilon _6 E_6.\nonumber \end{aligned}
(4.14)

It follows $$\varepsilon _2 \varepsilon _1=\varepsilon _4$$$$\varepsilon _1\varepsilon _4=\varepsilon _6,\, \varepsilon _2\varepsilon _3=\varepsilon _5$$. Hence one has $$\varepsilon _2=\varepsilon _6$$.

If $$\beta _1= \beta _1'>0$$, then we get additionally $$\varepsilon _2\varepsilon _1=\varepsilon _5$$, which yields $$\varepsilon _1=\varepsilon _3$$, $$\varepsilon _2=\varepsilon _6$$ and $$\varepsilon _4=\varepsilon _5$$.

If $$\beta _2= \beta _2'>0$$, then we have additionally $$\varepsilon _2\varepsilon _1=\varepsilon _6$$, which gives $$\varepsilon _1=1$$ and $$\varepsilon _2=\varepsilon _4=\varepsilon _6$$.

If $$\beta _3= \beta _3'>0$$, then one has in addition $$\varepsilon _1\varepsilon _3=\varepsilon _5$$, which yields $$\varepsilon _4=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _6$$.

If $$\beta _4= \beta _4'>0$$, then we get additionally $$\varepsilon _1\varepsilon _3=\varepsilon _6$$. Hence one has $$\varepsilon _1=\varepsilon _5$$, $$\varepsilon _2=\varepsilon _6$$, $$\varepsilon _3=\varepsilon _4$$.

If $$\beta _5= \beta _5'>0$$, then we have additionally $$\varepsilon _2\varepsilon _3=\varepsilon _6$$, which gives $$\varepsilon _3=1$$ and $$\varepsilon _2=\varepsilon _5=\varepsilon _6$$.

Applying these relations in assertion 1. of the Theorem

in case 1. we obtain $$\varepsilon _i=1, i=1,\dots ,6$$,

in case 2. we get $$\varepsilon _1=\varepsilon _3=1$$ and $$\varepsilon _2=\varepsilon _4=\varepsilon _5=\varepsilon _6$$,

in case 3. we have $$\varepsilon _3=\varepsilon _4=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _5=\varepsilon _6$$,

in case 4. we get $$\varepsilon _4=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _3=\varepsilon _6$$,

in case 5. we obtain $$\varepsilon _2=\varepsilon _6=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _5$$,

in case 6. we have $$\varepsilon _1=\varepsilon _2=\varepsilon _4=\varepsilon _6=1$$ and $$\varepsilon _3=\varepsilon _5$$,

in case 7. we receive $$\varepsilon _1=\varepsilon _5=1$$ and $$\varepsilon _2=\varepsilon _3=\varepsilon _4=\varepsilon _6$$,

in case 8. we get $$\varepsilon _1=\varepsilon _3$$, $$\varepsilon _2=\varepsilon _6$$ and $$\varepsilon _1\varepsilon _2=\varepsilon _4=\varepsilon _5$$,

in case 9. we have $$\varepsilon _1=1$$ and $$\varepsilon _2=\varepsilon _4=\varepsilon _6$$, $$\varepsilon _2\varepsilon _3=\varepsilon _5$$,

in case 10. we obtain $$\varepsilon _4=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _6$$, $$\varepsilon _2\varepsilon _3=\varepsilon _1\varepsilon _3=\varepsilon _5$$,

in case 11. we have $$\varepsilon _1=\varepsilon _5$$, $$\varepsilon _2=\varepsilon _6$$ and $$\varepsilon _1\varepsilon _2=\varepsilon _3=\varepsilon _4$$,

in case 12. we get $$\varepsilon _3=1$$ and $$\varepsilon _2=\varepsilon _5=\varepsilon _6$$, $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$,

in case 13. we obtain $$\varepsilon _2=\varepsilon _6$$, $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$, $$\varepsilon _2\varepsilon _3=\varepsilon _5$$.

Hence the system of Eq. (4.14) is satisfied with $$\beta _j'=\beta _j, j=1, \dots , 5$$ in cases $$1.-13.$$ of the Theorem, which proves the uniqueness of the Lie algebra $${\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j)$$ in assertion 1.

If the map $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1, \dots , 6$$, is an orthogonal automorphism of $${\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j)$$, then the system of equations given by (4.14) is satisfied with $$\alpha _i>0, i=1,2,3$$, $$\beta _j'=\beta _j, j=1, \dots , 5$$. Therefore in cases $$1.-13.$$ we obtain the above conditions for $$\varepsilon _i, i=1, \dots , 6$$. Hence the group of orthogonal automorphisms of $${\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j)$$ in case 1. is trivial, in cases $$2.-13.$$ is isomorphic to the group given by 2b–2 m, which proves assertion 2. $$\square$$

### Corollary 4.9

Let $$(\aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j), \langle .,. \rangle )$$ be the connected and simply connected Riemannian nilmanifold corresponding to $$( {\mathfrak {n}}_{6,19}^{\varepsilon =0}(\alpha _i,\beta _j ), \langle .,. \rangle )$$. The isometry group of $$(\aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}}(\aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j)) \\{} & {} = {\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \times {\mathbb {Z}}_2 \ltimes \aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _j=0, j=1,2,3,4,5, \\ {\mathbb {Z}}_2 \times {\mathbb {Z}}_2 \ltimes \aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _j=0, j=2,3,4,5, \\ &{} or \ \beta _2>0, \beta _j=0, j=1,3,4,5, \\ &{} or \ \beta _3>0, \beta _j=0, j=1,2,4,5, \\ &{} or \ \beta _4>0, \beta _j=0, j=1,2,3,5, \\ &{} or \ \beta _5>0, \beta _j=0, j=1,2,3,4, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, \beta _3=\beta _4=0, \\ &{} or \ \beta _1>0, \beta _5>0, \beta _3=\beta _4=0, \\ &{} or \ \beta _2>0, \beta _5>0, \beta _3=\beta _4=0, \\ &{} or \ \beta _3>0, \beta _4>0, \beta _1=\beta _2=0, \\ &{} or \ \beta _3>0, \beta _5>0, \beta _1=\beta _2=0, \\ &{} or \ \beta _4>0, \beta _5>0, \beta _1=\beta _2=0, \\ &{} or \ \beta _1>0, \beta _3>0, \beta _2=\beta _4=\beta _5=0, \\ &{} or \ \beta _1>0, \beta _4>0, \beta _2=\beta _3=\beta _5=0, \\ &{} or \ \beta _2>0, \beta _3>0, \beta _1=\beta _4=\beta _5=0, \\ &{} or \ \beta _2>0, \beta _4>0, \beta _1=\beta _3=\beta _5=0, \\ \aleph _{6,19}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, \beta _3>0, \\ &{} or \ \beta _1>0, \beta _2>0, \beta _4>0, \\ &{} or \ \beta _1>0, \beta _3>0, \beta _4>0, \\ &{} or \ \beta _1>0, \beta _3>0, \beta _5>0, \\ &{} or \ \beta _1>0, \beta _4>0, \beta _5>0, \\ &{} or \ \beta _2>0, \beta _3>0, \beta _4>0, \\ &{} or \ \beta _2>0, \beta _3>0, \beta _5>0, \\ &{} or \ \beta _2>0, \beta _4>0, \beta _5>0. \end{array}\right. } \end{aligned}

We consider the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,20}$$.

### Definition 4.10

Let $$\{E_1, E_2, E_3, E_4, E_5, E_6\}$$ be an orthonormal basis in the Euclidean vector space $${\mathbb {E}}^6$$. Denote by $${\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1, \dots , 4$$, $$j= 1, \dots , 5$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}{}[E_1,E_2]&=\alpha _1 E_4+\beta _1 E_5+\beta _2 E_6, \nonumber&[E_1,E_4]&=\beta _4 E_6,&[E_2,E_3]&=\beta _5 E_6, \\ [E_1,E_3]&=\alpha _2 E_5+\beta _3 E_6,&[E_1,E_5]&=\alpha _3 E_6,&[E_2,E_4]&=\alpha _4 E_6. \end{aligned}
(4.15)

The bracket operation (4.15) satisfies the Jacobi identity.

### Theorem 4.11

Let $$\langle .,. \rangle$$ be an inner product on the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,20}$$.

1. 1.

There is a unique metric Lie algebra $$({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$({\mathfrak {l}}_{6,20}, \langle .,. \rangle )$$ with $$\alpha _i>0$$, $$i=1, \dots , 4$$ and such that one of the following cases is satisfied

1. 1.

at least two of the elements of the set $$\{\beta _1, \beta _2, \beta _3, \beta _4, \beta _5\}$$ are positive with the exception of the pairs $$\{\beta _1, \beta _4\}$$ and $$\{\beta _2, \beta _5\}$$,

2. 2.

$$\beta _1>0$$ or $$\beta _4>0$$, $$\beta _2=\beta _3=\beta _5=0$$,

3. 3.

$$\beta _2>0$$ or $$\beta _5>0$$, $$\beta _1=\beta _3=\beta _4=0$$,

4. 4.

$$\beta _3>0$$, $$\beta _1=\beta _2=\beta _4=\beta _5=0$$,

5. 5.

$$\beta _1=\beta _2=\beta _3=\beta _4=\beta _5=0$$.

2. 2.

The group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ))$$ of orthogonal automorphisms of the metric Lie algebra $$({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ is the following group:

1. (a)

in case 1. the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ))$$ is trivial,

2. (b)

in case 2. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ))=\{TE_4= E_4, TE_5= E_5, TE_i= \varepsilon E_i, i=1,2,3,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

3. (c)

in case 3. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ))=\{TE_2=E_2, TE_5=E_5, TE_i=\varepsilon E_i,i=1,3,4,6, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

4. (d)

in case 4. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ))=\{TE_1=E_1, TE_3=E_3, TE_5=E_5, TE_6=E_6, TE_i=\varepsilon E_i, i=2,4, \, \varepsilon =\pm 1\} \simeq {\mathbb {Z}}_2$$,

5. (e)

in case 5. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ))=\{TE_5=E_5, TE_i=\varepsilon _1 E_i, i=1,3,6, TE_2=\varepsilon _2 E_2, TE_4=\varepsilon _1\varepsilon _2 E_4, \, \varepsilon _1,\varepsilon _2=\pm 1\} \simeq {\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$.

### Proof

According to Proposition 3.1 we apply the Gram–Schmidt process to the ordered basis $$\{G_6, G_5, G_4, G_3, G_2, G_1 \}$$ and obtain an orthonormal basis $$\{ F_1, F_2, F_3, F_4, F_5, F_6 \}$$ of $${\mathfrak {l}}_{6,20}$$ such that the vector $$F_i$$ is a positive multiple of $$G_i$$ modulo the subspace span $$(G_j; j>i)$$ and orthogonal to span $$(G_j; j>i)$$. The orthogonal direct sum $${\mathbb {R}}F_1\oplus \dots \oplus {\mathbb {R}}F_6$$ is a framing of $$({\mathfrak {l}}_{6,20},\langle .,. \rangle )$$. Expressing the vectors of the new basis in the form $$F_i=\sum _{k=i}^{6}a_{ik}G_k$$ with $$a_{ii}>0$$ we get

\begin{aligned}{}[F_1,F_2]&=\alpha _1 F_4+\beta _1 F_5+\beta _2 F_6,&[F_1,F_4]&=\beta _4 F_6,&[F_2,F_3]&=\beta _5 F_6, \\ [F_1,F_3]&=\alpha _2 F_5+\beta _3 F_6, \nonumber&[F_1,F_5]&=\alpha _3 F_6,&[F_2,F_4]&=\alpha _4 F_6 \end{aligned}
(4.16)

with $$\alpha _i >0$$, $$i=1,\dots ,4$$ and $$\beta _j \in {\mathbb {R}}$$, $$j=1, \dots , 5$$. The change of the orthonormal basis: $$\tilde{F_1}=F_1, \tilde{F_2}=-F_2, \tilde{F_3}=F_3, \tilde{F_4}=-F_4, \tilde{F_5}=F_5, \tilde{F_6}=F_6$$ gives

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}-\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=-\beta _4 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=-\beta _5 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=\alpha _2 \tilde{F_5}+\beta _3 \tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _3 \tilde{F_6},&[\tilde{F_2},\tilde{F_4}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Similarly, changing the orthonormal basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=-F_3,$$ $$\tilde{F_4}=-F_4, \tilde{F_5}=F_5, \tilde{F_6}=-F_6$$ we obtain

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}+\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=-\beta _4 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\beta _5 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=\alpha _2 \tilde{F_5}-\beta _3 \tilde{F_6},&[\tilde{F_1},\tilde{F_5}]&=\alpha _3 \tilde{F_6},&[\tilde{F_2},\tilde{F_4}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Hence there exists an orthonormal basis such that in commutators (4.16) we have $$\alpha _i>0, i=1, \dots , 4$$ and one of the cases in assertion 1. is satisfied. This proves the existence of $${\mathfrak {n}}_{6,20}(\alpha _i, \beta _j)$$ with the properties in assertion 1.

Let the linear map $$T:{\mathfrak {n}}_{6,20}(\alpha _i, \beta _j) \rightarrow {\mathfrak {n}}_{6,20}(\alpha _i',\beta _j')$$ be an isometric isomorphism. The decomposition $${\mathbb {R}}\, E_1 \oplus {\mathbb {R}}\, E_2 \oplus {\mathbb {R}}\, E_3 \oplus {\mathbb {R}}\, E_4 \oplus {\mathbb {R}}\, E_5 \oplus {\mathbb {R}}\, E_6$$ is a framing of both Lie algebras, where $$\alpha _i, \alpha _i' > 0$$, $$i=1, \dots , 4$$. Hence by Lemma 2.3 we have $$\alpha _i=\alpha _i'$$, $$i=1,\dots , 4$$ and $$|\beta _j'|=\beta _j$$, $$j=1,\dots ,5.$$ Let $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots ,6$$, then we obtain from $$[TE_i,TE_j]' = T[E_i,E_j]$$, $$i, j =1,\dots ,6$$, using the commutation relations (4.16) the equations

\begin{aligned}&\varepsilon _1\varepsilon _2 \left( \alpha _1 E_4+\beta _1' E_5+\beta _2' E_6\right) =\alpha _1 \varepsilon _4 E_4+\beta _1 \varepsilon _5 E_5+\beta _2 \varepsilon _6 E_6,\nonumber \\&\varepsilon _1\varepsilon _3\left( \alpha _2 E_5+\beta _3' E_6\right) =\alpha _2 \varepsilon _5 E_5+\beta _3 \varepsilon _6 E_6,\quad \varepsilon _1\varepsilon _4\left( \beta _4' E_6\right) =\beta _4 \varepsilon _6 E_6,\\&\varepsilon _1\varepsilon _5\left( \alpha _3 E_6 \right) =\alpha _3 \varepsilon _6 E_6, \quad \varepsilon _2\varepsilon _3\left( \beta _5' E_6\right) =\beta _5 \varepsilon _6 E_6, \quad \varepsilon _2\varepsilon _4\left( \alpha _4 E_6\right) =\alpha _4 \varepsilon _6 E_6. \nonumber \end{aligned}
(4.17)

From (4.17) it follows $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$$$\varepsilon _1\varepsilon _3=\varepsilon _5,\, \varepsilon _1\varepsilon _5=\varepsilon _2\varepsilon _4=\varepsilon _6$$, which yields $$\varepsilon _5=1$$, $$\varepsilon _1=\varepsilon _3=\varepsilon _6$$.

If $$\beta _1= \beta _1'>0$$ or $$\beta _4= \beta _4'>0$$, then we have additionally $$\varepsilon _1\varepsilon _2=\varepsilon _5$$ or $$\varepsilon _1\varepsilon _4=\varepsilon _6$$, which gives that $$\varepsilon _4=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _3=\varepsilon _6$$.

If $$\beta _2= \beta _2'>0$$ or $$\beta _5= \beta _5'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _6$$ or $$\varepsilon _2\varepsilon _3=\varepsilon _6$$. Hence one has $$\varepsilon _2=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _6$$.

If $$\beta _3= \beta _3'>0$$, then we have $$\varepsilon _1\varepsilon _3=\varepsilon _6$$, which yields that $$\varepsilon _1=\varepsilon _3=\varepsilon _5=\varepsilon _6=1$$ and $$\varepsilon _2=\varepsilon _4$$.

Using these relations in assertion 1. of the Theorem

in case 1. we obtain $$\varepsilon _i=1, i=1,\dots ,6$$,

in case 2. we have $$\varepsilon _4=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _3=\varepsilon _6$$,

in case 3. we get $$\varepsilon _2=\varepsilon _5=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _4=\varepsilon _6$$,

in case 4. we obtain $$\varepsilon _1=\varepsilon _3=\varepsilon _5=\varepsilon _6=1$$ and $$\varepsilon _2=\varepsilon _4$$,

in case 5. we get $$\varepsilon _5=1$$, $$\varepsilon _1=\varepsilon _3=\varepsilon _6$$ and $$\varepsilon _1\varepsilon _2=\varepsilon _4$$.

Hence the system of Eq. (4.17) is satisfied with $$\beta _j'=\beta _j, j=1, \dots , 5$$ in cases $$1.-5.$$ of the Theorem, which proves the uniqueness of the Lie algebra $${\mathfrak {n}}_{6,20}(\alpha _i,\beta _j)$$. This shows assertion 1.

If the map $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots ,6$$, is an orthogonal automorphism of $${\mathfrak {n}}_{6,20}(\alpha _i,\beta _j)$$, then the system of equations given by (4.17) is satisfied with $$\alpha _i >0, i=1, \dots , 4$$, $$\beta _j'=\beta _j, j=1, \dots , 5$$. Therefore in cases $$1.-5.$$ the conditions for $$\varepsilon _i, i=1, \dots , 6$$, are given above. Hence the group of orthogonal automorphisms of $${\mathfrak {n}}_{6,20}(\alpha _i,\beta _j)$$ in case 1. is trivial, in cases $$2.-5.$$ it is isomorphic to the group given by 2b–2e and assertion 2 is proved. $$\square$$

### Corollary 4.12

Let $$(\aleph _{6,20}(\alpha _i,\beta _j), \langle .,. \rangle )$$ be the connected and simply connected Riemannian nilmanifold corresponding to $$( {\mathfrak {n}}_{6,20}(\alpha _i,\beta _j ), \langle .,. \rangle )$$. The isometry group of $$(\aleph _{6,20}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}} (\aleph _{6,20}(\alpha _i,\beta _j)) \\{} & {} \quad = {\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \ltimes \aleph _{6,20}(\alpha _i,\beta _j) &{} if \ \beta _j=0, j=1,2,3,4,5, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,20}(\alpha _i,\beta _j) &{} if \ \beta _1>0 \ or \ \beta _4>0, \beta _2=\beta _3=\beta _5=0, \\ &{} or \ \beta _2>0 \ or \ \beta _5>0, \beta _1=\beta _3=\beta _4=0, \\ &{} or \ \beta _3>0, \beta _1=\beta _2=\beta _4=\beta _5=0, \\ \aleph _{6,20}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, \ or \ \beta _1>0, \beta _3>0, \\ &{} or \ \beta _1>0, \beta _5>0, \ or \ \beta _2>0, \beta _3>0, \\ &{} or \ \beta _2>0, \beta _4>0, \ or \ \beta _3>0, \beta _4>0, \\ &{} or \ \beta _3>0, \beta _5>0, \ or \ \beta _4>0, \beta _5>0. \end{array}\right. } \end{aligned}

We consider the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,21}^{\varepsilon =0}$$.

### Definition 4.13

Let $$\{E_1, E_2, E_3, E_4, E_5, E_6\}$$ be an orthonormal basis in the Euclidean vector space $${\mathbb {E}}^6$$. Denote by $${\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1,\dots ,4$$, $$j= 1, \dots ,6$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}{}[E_1,E_2]&=\alpha _1 E_3+\beta _1 E_4+\beta _2 E_5+\beta _3 E_6,&[E_1,E_4]&=\alpha _3 E_6, \\ [E_1,E_3]&=\alpha _2 E_4+\beta _4 E_5+\beta _5 E_6,\nonumber&[E_2,E_3]&=\alpha _4 E_5+\beta _6 E_6. \end{aligned}
(4.18)

The bracket operation (4.18) satisfies the Jacobi identity.

### Theorem 4.14

Let $$\langle .,. \rangle$$ be an inner product on the 6-dimensional Lie algebra $${\mathfrak {l}}_{6,21}^{\varepsilon =0}$$.

1. 1.

There is a unique metric Lie algebra $$({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$({\mathfrak {l}}_{6,21}^{\varepsilon =0}, \langle .,. \rangle )$$ with $$\alpha _i>0$$, $$i=1, \dots , 4$$ and such that one of the following cases is satisfied

1. 1.

at least two of the elements of the set $$\{\beta _1, \beta _2, \beta _4, \beta _5, \beta _6\}$$ are positive with the exception of the pairs $$\{\beta _1, \beta _5\}$$ and $$\{\beta _2, \beta _6\}$$,

2. 2.

$$\beta _1>0$$ or $$\beta _5>0$$, $$\beta _2=\beta _4=\beta _6=0$$,

3. 3.

$$\beta _2>0$$ or $$\beta _6>0$$, $$\beta _1=\beta _4=\beta _5=0$$,

4. 4.

$$\beta _4>0$$, $$\beta _1=\beta _2=\beta _5=\beta _6=0$$,

5. 5.

$$\beta _1=\beta _2=\beta _4=\beta _5=\beta _6=0$$.

2. 2.

The group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ))$$ of orthogonal automorphisms of the metric Lie algebra $$({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ),\langle .,. \rangle )$$ is the following group:

1. (a)

in case 1. the group $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ))$$ is trivial,

2. (b)

in case 2. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{ TE_1= E_1, TE_5= E_5, TE_i=\varepsilon E_i, i=2,3,4,6, \ \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

3. (c)

in case 3. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_2=E_2, TE_4=E_4, TE_i=\varepsilon E_i, i=1,3,5,6, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

4. (d)

in case 4. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_3=E_3, TE_6= E_6, TE_i=\varepsilon E_i, i=1,2,4,5, \, \varepsilon =\pm 1 \}\simeq {\mathbb {Z}}_2$$,

5. (e)

in case 5. one has $${{\mathcal {O}}}{{\mathcal {A}}}({\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ))=\{TE_i=\varepsilon _1 E_i, i=1,5, TE_j =\varepsilon _2 E_j, j=2,4, TE_k=\varepsilon _1 \varepsilon _2 E_k, k=3,6, \varepsilon _1,\varepsilon _2=\pm 1\} \simeq {\mathbb {Z}}_2 \times {\mathbb {Z}}_2$$.

### Proof

Invoking Proposition 3.1, we apply the Gram–Schmidt process to the ordered basis $$\{G_6, G_5, G_4, G_3, G_2, G_1 \}$$ and we receive an orthonormal basis $$\{ F_1, F_2, F_3, F_4, F_5, F_6 \}$$ of $${\mathfrak {l}}_{6,21}^{\varepsilon =0}$$ such that the vector $$F_i$$ is a positive multiple of $$G_i$$ modulo the subspace span $$(G_j; j>i)$$ and orthogonal to span $$(G_j; j>i)$$. The orthogonal direct sum $${\mathbb {R}}F_1\oplus \dots \oplus {\mathbb {R}}F_6$$ is a framing of $$({\mathfrak {l}}_{6,21}^{\varepsilon =0},\langle .,. \rangle )$$ and the vectors of the new basis has the form $$F_i=\sum _{k=i}^{6}a_{ik}G_k$$ with $$a_{ii}>0$$. Using this we have

\begin{aligned}{}[F_1,F_2]&=\alpha _1 F_3+\beta _1 F_4+\beta _2 F_5+\beta _3 F_6,&[F_1,F_4]&=\alpha _3 F_6, \\ [F_1,F_3]&=\alpha _2 F_4+\beta _4 F_5+\beta _5 F_6,\nonumber&[F_2,F_3]&=\alpha _4 F_5+\beta _6 F_6 \end{aligned}
(4.19)

with $$\alpha _i >0$$, $$i=1,\dots ,4$$ and $$\beta _j \in {\mathbb {R}}$$, $$j=1, \dots , 6$$. Changing the orthonormal basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=-F_3, \tilde{F_4}=F_4, \tilde{F_5}=-F_5, \tilde{F_6}=-F_6$$ we obtain

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_3}-\beta _1 \tilde{F_4}+\beta _2 \tilde{F_5}+ \beta _3 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=\alpha _3 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=\alpha _2 \tilde{F_4}-\beta _4 \tilde{F_5}-\beta _5 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _4 \tilde{F_5}+\beta _6 \tilde{F_6}. \end{aligned}

Similarly, the change of the basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=-F_2, \tilde{F_3}=F_3, \tilde{F_4}=-F_4, \tilde{F_5}=-F_5, \tilde{F_6}=F_6$$ gives

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_3}-\beta _1 \tilde{F_4}-\beta _2 \tilde{F_5}+\beta _3 \tilde{F_6},&[\tilde{F_1},\tilde{F_4}]&=\alpha _3 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_3}]&=\alpha _2 \tilde{F_4}+\beta _4 \tilde{F_5}-\beta _5 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _4 \tilde{F_5}-\beta _6 \tilde{F_6}. \end{aligned}

Hence there is an orthonormal basis such that in commutators (4.19) we have $$\alpha _i>0, i=1, \dots , 4$$ and one of the cases in assertion 1. holds. Therefore the existence of $${\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i, \beta _j)$$ with properties given by assertion 1. follows.

Let the linear map $$T:{\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i, \beta _j) \rightarrow {\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i',\beta _j')$$ be an isometric isomorphism. The decomposition $${\mathbb {R}}\, E_1 \oplus {\mathbb {R}}\, E_2 \oplus {\mathbb {R}}\, E_3 \oplus {\mathbb {R}}\, E_4 \oplus {\mathbb {R}}\, E_5 \oplus {\mathbb {R}}\, E_6$$ is a framing of both Lie algebras, where $$\alpha _i, \alpha _i' > 0$$, $$i=1, \dots , 4$$. Hence by Lemma 2.3 we have $$\alpha _i=\alpha _i'$$, $$i=1, \dots , 4$$ and $$|\beta _j'|=\beta _j$$, $$j=1, \dots , 6.$$ Let $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots ,6$$, then we obtain from $$[TE_i,TE_j]' = T[E_i,E_j]$$, $$i, j =1,\dots ,6$$, using the commutation relations (4.19) the equations

\begin{aligned}&\varepsilon _1\varepsilon _2 \left( \alpha _1 E_3+\beta _1' E_4+\beta _2' E_5+\beta _3' E_6\right) =\alpha _1 \varepsilon _3 E_3+\beta _1 \varepsilon _4 E_4+\beta _2 \varepsilon _5 E_5+ \beta _3 \varepsilon _6 E_6, \nonumber \\&\varepsilon _1\varepsilon _3\left( \alpha _2 E_4+\beta _4' E_5+\beta _5' E_6\right) =\alpha _2 \varepsilon _4 E_4+ \beta _4 \varepsilon _5 E_5+\beta _5 \varepsilon _6 E_6, \\&\varepsilon _1\varepsilon _4\left( \alpha _3 E_6\right) =\alpha _3 \varepsilon _6 E_6, \quad \nonumber \varepsilon _2\varepsilon _3\left( \alpha _4 E_5+\beta _6' E_6\right) =\alpha _4 \varepsilon _5 E_5+\beta _6 \varepsilon _6 E_6. \end{aligned}
(4.20)

Hence we obtain $$\varepsilon _1 \varepsilon _2=\varepsilon _3$$$$\varepsilon _1\varepsilon _3=\varepsilon _4$$$$\varepsilon _1\varepsilon _4=\varepsilon _6, \varepsilon _2\varepsilon _3=\varepsilon _5$$, which yields $$\varepsilon _1=\varepsilon _5$$, $$\varepsilon _2=\varepsilon _4$$, $$\varepsilon _1\varepsilon _2=\varepsilon _3=\varepsilon _6$$. Using these relations we have $$\varepsilon _1 \varepsilon _2=\varepsilon _6$$. Therefore one has $$\beta _3'=\beta _3$$.

If $$\beta _1= \beta _1'>0$$ or $$\beta _5= \beta _5'>0$$, then we have additionally $$\varepsilon _1\varepsilon _2=\varepsilon _4$$ or $$\varepsilon _1\varepsilon _3=\varepsilon _6$$, which yields that $$\varepsilon _1=\varepsilon _5=1$$ and $$\varepsilon _2=\varepsilon _3=\varepsilon _4=\varepsilon _6$$.

If $$\beta _2= \beta _2'>0$$ or $$\beta _6= \beta _6'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _5$$ or $$\varepsilon _2\varepsilon _3=\varepsilon _6$$. Hence one has $$\varepsilon _2=\varepsilon _4=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _5=\varepsilon _6$$.

If $$\beta _4= \beta _4'>0$$, then one has in addition $$\varepsilon _1\varepsilon _3=\varepsilon _5$$, which gives $$\varepsilon _3=\varepsilon _6=1$$, $$\varepsilon _1=\varepsilon _2=\varepsilon _4=\varepsilon _5$$.

Applying these relations in assertion 1. of the Theorem

in case 1. we obtain $$\varepsilon _i=1, i=1,\dots ,6$$,

in case 2. we have $$\varepsilon _1=\varepsilon _5=1$$ and $$\varepsilon _2=\varepsilon _3=\varepsilon _4=\varepsilon _6$$,

in case 3. we receive $$\varepsilon _2=\varepsilon _4=1$$ and $$\varepsilon _1=\varepsilon _3=\varepsilon _5=\varepsilon _6$$,

in case 4. we obtain $$\varepsilon _3=\varepsilon _6=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _4=\varepsilon _5$$,

in case 5. we get that $$\varepsilon _1=\varepsilon _5$$, $$\varepsilon _2=\varepsilon _4$$, $$\varepsilon _1\varepsilon _2=\varepsilon _3=\varepsilon _6$$.

Therefore the system of Eq. (4.20) is satisfied with $$\beta _j'=\beta _j, j=1, \dots , 6$$ in cases $$1.-5.$$ of the Theorem and the uniqueness of the metric Lie algebra $${\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j)$$ with properties given by assertion 1. follows. The proof of assertion 1. is done.

If the map $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots ,6$$, is an orthogonal automorphism of $${\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j)$$, then the system of equations given by (4.20) is satisfied with $$\alpha _i>0, i=1, \dots , 4,$$ $$\beta _j'=\beta _j, j=1, \dots , 6$$. Hence for $$\varepsilon _i, i=1, \dots , 6$$, we have the same conditions as above. Taking this into account the group of orthogonal automorphisms of $${\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j)$$ in case 1. is trivial, in cases $$2.-5.$$ it is isomorphic to the group given by 2b-2e. This proves assertion 2. $$\square$$

### Corollary 4.15

Let $$(\aleph _{6,21}^{\varepsilon =0}(\alpha _i,\beta _j), \langle .,. \rangle )$$ be the connected and simply connected Riemannian nilmanifold corresponding to $$( {\mathfrak {n}}_{6,21}^{\varepsilon =0}(\alpha _i,\beta _j ), \langle .,. \rangle )$$. The isometry group of $$(\aleph _{6,21}^{\varepsilon =0}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}} (\aleph _{6,21}^{\varepsilon =0}(\alpha _i,\beta _j)) \\{} & {} \quad = {\left\{ \begin{array}{ll} {\mathbb {Z}}_2 \times {\mathbb {Z}}_2 \ltimes \aleph _{6,21}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _j=0, j=1,2,4,5,6, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,21}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _1>0 \ or \ \beta _5>0, \beta _2=\beta _4=\beta _6=0, \\ &{} or \ \beta _2>0 \ or \ \beta _6>0, \beta _1=\beta _4=\beta _5=0, \\ &{} or \ \beta _4>0, \beta _1=\beta _2=\beta _5=\beta _6=0, \\ \aleph _{6,21}^{\varepsilon =0}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, \ or \ \beta _1>0, \beta _4>0, \\ &{} or \ \beta _1>0, \beta _6>0, \ or \ \beta _2>0, \beta _4>0, \\ &{} or \ \beta _2>0, \beta _5>0, \ or \ \beta _4>0, \beta _5>0, \\ &{} or \ \beta _4>0, \beta _6>0, \ or \ \beta _5>0, \beta _6>0. \end{array}\right. } \end{aligned}

Finally we deal with the 6-dimensional Lie algebras $${\mathfrak {l}}_{6,23}$$ and $${\mathfrak {l}}_{6,25}$$.

### Definition 4.16

Let $$\{E_1, E_2, E_3, E_4, E_5, E_6\}$$ be an orthonormal basis in the Euclidean vector space $${\mathbb {E}}^6$$. We denote by $${\mathfrak {n}}_{6,25}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1,2,3$$, $$j=1,2,3$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by the non-vanishing commutators

\begin{aligned}{}[E_1,E_2]=\alpha _1 E_4+\beta _1 E_5+\beta _2 E_6, \ [E_1,E_3]=\alpha _2 E_5+\beta _3 E_6, \ [E_1,E_4]=\alpha _3 E_6.\nonumber \\ \end{aligned}
(4.21)

Denote by $${\mathfrak {n}}_{6,23}(\alpha _i,\beta _j ), \alpha _i, \beta _j \in {\mathbb {R}}, i=1, \dots , 4,j=1,2,3$$ with $$\alpha _i \ne 0$$ the metric Lie algebra defined on $${\mathbb {E}}^6$$ given by (4.21) and by the additional commutator

\begin{aligned}{}[E_2,E_3]=\alpha _4 E_6. \end{aligned}
(4.22)

The bracket operations (4.21) as well as (4.21) and (4.22) satisfy the Jacobi identity.

### Theorem 4.17

Let $$\langle ., .\rangle$$ be an inner product on the 6-dimensional Lie algebra $$\mathfrak {l}_{6,23}$$, respectively $$\mathfrak {l}_{6,25}$$.

1. 1.

There is a unique metric Lie algebra $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j), \langle ., .\rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$(\mathfrak {l}_{6,25}, \langle ., . \rangle )$$ with $$\alpha _i > 0$$, $$i = 1, 2, 3,$$ such that one of the following cases is satisfied

1. 1.

at least two of the elements of the set $$\{\beta _1, \beta _2, \beta _3\}$$ are positive,

2. 2.

$$\beta _1 > 0, \beta _2 = \beta _3 = 0,$$

3. 3.

$$\beta _2 > 0, \beta _1 = \beta _3 = 0,$$

4. 4.

$$\beta _3 > 0, \beta _1 = \beta _2 = 0,$$

5. 5.

$$\beta _1 = \beta _2 = \beta _3 = 0.$$

There is a unique metric Lie algebra $$(\mathfrak {n}_{6,23}(\alpha _i, \beta _j), \langle ., .\rangle )$$ which is isometrically isomorphic to the metric Lie algebra $$(\mathfrak {l}_{6,23}, \langle ., .\rangle )$$ with $$\alpha _i > 0, i = 1, 2, 3, 4,$$ and such that one of the above cases 1. –5. is satisfied.

1. 2.

The group $$\mathcal{O}\mathcal{A}$$ ($$\mathfrak {n}_{6,23}(\alpha _i, \beta _j))$$ of orthogonal automorphisms of the metric Lie algebra ($$\mathfrak {n}_{6,23}(\alpha _i, \beta _j), \langle ., .\rangle$$) is the following group:

1. (a)

in case 1. the group $$\mathcal{O}\mathcal{A}$$($$\mathfrak {n}_{6,23}\mathrm{(}\alpha _i, \beta _j$$)) is trivial,

2. (b)

in case 2. one has $$\mathcal{O}\mathcal{A}$$($$\mathfrak {n}_{6,23}(\alpha _i, \beta _j)) = \{TE_2 = E_2, TE_3 = E_3, TE_6= E_6, TE_i = \varepsilon E_i, i = 1, 4, 5, \varepsilon = \pm 1\} \simeq \mathbb {Z}_2$$,

3. (c)

in case 3. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,23}(\alpha _i, \beta _j)) = \{TE_1 = E_1, TE_3 = E_3, TE_5= E_5, TE_i = \varepsilon E_i, i = 2, 4, 6, \varepsilon = \pm 1\} \simeq \mathbb {Z}_2$$,

4. (d)

in case 4. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,23}(\alpha _i, \beta _j)) = \{TE_3 = E_3, TE_4 = E_4, TE_i= \varepsilon E_i, i = 1, 2, 5, 6, \varepsilon = \pm 1\} \simeq \mathbb {Z}_2$$,

5. (e)

in case 5. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,23}(\alpha _i, \beta _j)) = \{TE_3 = E_3, TE_i = \varepsilon _1E_i, i = 1, 5, TE_j = \varepsilon _2E_j , j = 2, 6, TE_4 = \varepsilon _1 \varepsilon _2E_4, \varepsilon _1, \varepsilon _2 = \pm 1\}\!\simeq \!\mathbb {Z}_2\!\times \!\mathbb {Z}_2$$.

2. 3.

The group $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j))$$ of orthogonal automorphisms of the metric Lie algebra $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j), \langle ., .\rangle )$$ is the following group:

1. (a)

in case 1. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j)) = \{TE_1=E_1, TE_i = \varepsilon E_i, i = 2, 3, 4, 5, 6, \varepsilon = \pm 1\} \simeq \mathbb {Z}_2$$.

2. (b)

in case 2. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j)) = \{TE_1 = \varepsilon _1E_1, TE_i = \varepsilon _2E_i, \textrm{i}=2,3,6, TE_j = \varepsilon _1 \varepsilon _2E_j , j = 4, 5, \varepsilon _1, \varepsilon _2 = \pm 1\} \simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$.

3. (c)

in case 3. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j)) = \{TE_1 = E_1, TE_i = \varepsilon _2E_i, \textrm{i} = 2, 4, 6, TE_j = \varepsilon _3E_j , j = 3, 5, \varepsilon _2, \varepsilon _3 = \pm 1\} \simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$.

4. (d)

in case 4. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j)) = \{TE_1 = \varepsilon _1E_1, TE_i = \varepsilon _2E_i, \textrm{i} = 2, 5, 6, TE_j = \varepsilon _1\varepsilon _2E_j , j = 3, 4, \varepsilon _1, \varepsilon _2 = \pm 1\} \simeq \mathbb {Z}_2 \times \mathbb {Z}_2$$.

5. (e)

in case 5. one has $$\mathcal{O}\mathcal{A}$$ $$(\mathfrak {n}_{6,25}(\alpha _i, \beta _j)) = \{TE_1 = \varepsilon _1E_1, TE_i = \varepsilon _2E_i, i = 2, 6, TE_3 = \varepsilon _3E_3, TE_4 = \varepsilon _1\varepsilon _2E_4, TE_5=\varepsilon _1 \varepsilon _3 E_5, \varepsilon _1, \varepsilon _2, \varepsilon _3 = \pm 1\} \simeq \mathbb {Z}_2 \times \mathbb {Z}_2 \times \mathbb {Z}_2$$.

### Proof

According to Proposition 3.1 the application of the Gram–Schmidt process to the ordered basis $$\{G_6, G_5, G_4, G_3, G_2, G_1 \}$$ yields an orthonormal basis $$\{ F_1, F_2, F_3, F_4, F_5, F_6 \}$$ of $${\mathfrak {l}}_{6,23}$$, respectively $${\mathfrak {l}}_{6,25}$$ such that the vector $$F_i$$ is a positive multiple of $$G_i$$ modulo the subspace span $$(G_j; j>i)$$ and orthogonal to span $$(G_j; j>i)$$. The orthogonal direct sum $${\mathbb {R}}F_1\oplus \dots \oplus {\mathbb {R}}F_6$$ is a framing of $$({\mathfrak {l}}_{6,23},\langle .,. \rangle )$$, respectively $$({\mathfrak {l}}_{6,25},\langle .,. \rangle )$$. The vectors of the new basis have the form $$F_i=\sum _{k=i}^{6}a_{ik}G_k$$ with $$a_{ii}>0$$. We get for the metric Lie algebras $$({\mathfrak {l}}_{6,23},\langle .,. \rangle )$$ and $$({\mathfrak {l}}_{6,25},\langle .,. \rangle )$$

\begin{aligned}{}[F_1,F_2]=\alpha _1 F_4+\beta _1 F_5+\beta _2 F_6, \ [F_1,F_3]=\alpha _2 F_5+\beta _3 F_6, \ [F_1,F_4]=\alpha _3 F_6, \end{aligned}
(4.23)

and for $$({\mathfrak {l}}_{6,23},\langle .,. \rangle )$$ in addition

\begin{aligned}{}[F_2,F_3]=\alpha _4 F_6, \end{aligned}
(4.24)

where $$\alpha _i >0$$, $$i=1,2,3,4$$ and $$\beta _j \in {\mathbb {R}}$$, $$j=1, 2, 3$$. Changing the orthonormal basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=-F_2, \tilde{F_3}=F_3, \tilde{F_4}=F_4, \tilde{F_5}=-F_5, \tilde{F_6}=-F_6$$ we obtain

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}-\beta _1 \tilde{F_5}-\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=\alpha _2 \tilde{F_5}+\beta _3 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _3 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Similarly, the change of the basis: $$\tilde{F_1}=-F_1, \tilde{F_2}=F_2, \tilde{F_3}=F_3, \tilde{F_4}=-F_4, \tilde{F_5}=-F_5, \tilde{F_6}=F_6$$ yields

\begin{aligned}{}[\tilde{F_1},\tilde{F_2}]&=\alpha _1 \tilde{F_4}+\beta _1 \tilde{F_5}-\beta _2 \tilde{F_6},&[\tilde{F_1},\tilde{F_3}]&=\alpha _2 \tilde{F_5}-\beta _3 \tilde{F_6}, \\ [\tilde{F_1},\tilde{F_4}]&=\alpha _3 \tilde{F_6},&[\tilde{F_2},\tilde{F_3}]&=\alpha _4 \tilde{F_6}. \end{aligned}

Hence there is an orthonormal basis such that in commutators (4.23) and (4.24) we have $$\alpha _i >0$$, $$i=1,\dots ,4$$ and one of the cases in assertion 1. is satisfied. Consequently the existence of $${\mathfrak {n}}_{6,23}(\alpha _i, \beta _j)$$, respectively $${\mathfrak {n}}_{6,25}(\alpha _i, \beta _j)$$ with the properties in assertion 1. is proved.

Let the linear map $$T:{\mathfrak {n}}_{6,k}(\alpha _i, \beta _j) \rightarrow {\mathfrak {n}}_{6,k}(\alpha _i',\beta _j')$$, $$k=23,25$$, be an isometric isomorphism. The decomposition $${\mathbb {R}}\, E_1 \oplus {\mathbb {R}}\, E_2 \oplus {\mathbb {R}}\, E_3 \oplus {\mathbb {R}}\, E_4 \oplus {\mathbb {R}}\, E_5 \oplus {\mathbb {R}}\, E_6$$ is a framing of both Lie algebras, where $$\alpha _i, \alpha _i' > 0$$, $$i=1, \dots , 4$$. Hence by Lemma 2.3 we have $$\alpha _i=\alpha _i'$$, $$i=1,\dots , 4$$ and $$|\beta _j'|=\beta _j$$ for all $$j=1,2,3$$. Let $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots , 6$$. Using the commutation relations (4.23) and (4.24) we obtain from $$[TE_i,TE_j]' = T[E_i,E_j]$$, $$i, j =1,\dots ,6$$, for $${\mathfrak {n}}_{6,25}(\alpha _i, \beta _j)$$ and $${\mathfrak {n}}_{6,23}(\alpha _i, \beta _j)$$ the equations

\begin{aligned}&\varepsilon _1\varepsilon _2 \left( \alpha _1 E_4+\beta _1' E_5+\beta _2' E_6\right) =\alpha _1 \varepsilon _4 E_4+\beta _1 \varepsilon _5 E_5+\beta _2 \varepsilon _6 E_6, \\&\varepsilon _1\varepsilon _3\left( \alpha _2 E_5+\beta _3' E_6\right) =\alpha _2 \varepsilon _5 E_5+\beta _3 \varepsilon _6 E_6, \quad \varepsilon _1\varepsilon _4\left( \alpha _3 E_6\right) =\alpha _3 \varepsilon _6 E_6, \nonumber \end{aligned}
(4.25)

and in addition for $${\mathfrak {n}}_{6,23}(\alpha _i, \beta _j)$$ the equation

\begin{aligned} \varepsilon _2\varepsilon _3\left( \alpha _4 E_6\right) =\alpha _4 \varepsilon _6 E_6. \end{aligned}
(4.26)

From (4.25) and (4.26) for the metric Lie algebra $${\mathfrak {n}}_{6,23}(\alpha _i, \beta _j)$$ we get $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$$$\varepsilon _1\varepsilon _3=\varepsilon _5,\, \varepsilon _1\varepsilon _4=\varepsilon _2\varepsilon _3=\varepsilon _6$$. Then one has $$\varepsilon _3=1$$, $$\varepsilon _1=\varepsilon _5$$, $$\varepsilon _2=\varepsilon _6$$.

If $$\beta _1= \beta _1'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _5$$, which gives $$\varepsilon _2=\varepsilon _3=\varepsilon _6=1$$, $$\varepsilon _1=\varepsilon _4=\varepsilon _5$$.

If $$\beta _2= \beta _2'>0$$, then we obtain $$\varepsilon _1\varepsilon _2=\varepsilon _6$$. Hence one has $$\varepsilon _1=\varepsilon _3=\varepsilon _5=1$$, $$\varepsilon _2=\varepsilon _4=\varepsilon _6$$.

If $$\beta _3= \beta _3'>0$$, then we get additionally $$\varepsilon _1\varepsilon _3=\varepsilon _6$$, which yields $$\varepsilon _3=\varepsilon _4=1$$, $$\varepsilon _1=\varepsilon _2=\varepsilon _5=\varepsilon _6$$.

Using the conditions for $$\beta _j, j=1,2,3$$ given in assertion 1. of the Theorem

in case 1. we get $$\varepsilon _i=1, i=1,\dots , 6$$,

in case 2. we obtain $$\varepsilon _2=\varepsilon _3=\varepsilon _6=1$$ and $$\varepsilon _1=\varepsilon _4=\varepsilon _5$$,

in case 3. we have $$\varepsilon _1=\varepsilon _3=\varepsilon _5=1$$ and $$\varepsilon _2=\varepsilon _4=\varepsilon _6$$,

in case 4. we obtain $$\varepsilon _3=\varepsilon _4=1$$ and $$\varepsilon _1=\varepsilon _2=\varepsilon _5=\varepsilon _6$$,

in case 5. we get $$\varepsilon _3=1$$, $$\varepsilon _1=\varepsilon _5$$, $$\varepsilon _2=\varepsilon _6$$ and $$\varepsilon _1\varepsilon _2=\varepsilon _4$$.

For the metric Lie algebra $${\mathfrak {n}}_{6,25}(\alpha _i, \beta _j)$$ from (4.25) it follows $$\varepsilon _1 \varepsilon _2=\varepsilon _4$$$$\varepsilon _1\varepsilon _3=\varepsilon _5,\, \varepsilon _1\varepsilon _4=\varepsilon _6$$. Then one has $$\varepsilon _2=\varepsilon _6$$.

If $$\beta _1= \beta _1'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _5$$. Hence one has $$\varepsilon _2=\varepsilon _3=\varepsilon _6$$ and $$\varepsilon _4=\varepsilon _5$$.

If $$\beta _2= \beta _2'>0$$, then we get additionally $$\varepsilon _1\varepsilon _2=\varepsilon _6$$, which gives $$\varepsilon _1=1$$, $$\varepsilon _2=\varepsilon _4=\varepsilon _6$$, $$\varepsilon _3=\varepsilon _5$$.

If $$\beta _3= \beta _3'>0$$, then we get additionally $$\varepsilon _1\varepsilon _3=\varepsilon _6$$, which yields that $$\varepsilon _2=\varepsilon _5=\varepsilon _6$$ and $$\varepsilon _3=\varepsilon _4$$.

Applying these relations in assertion 1. of the Theorem

in case 1. we obtain $$\varepsilon _1=1, \varepsilon _2=\varepsilon _3=\varepsilon _4=\varepsilon _5=\varepsilon _6$$,

in case 2. we get $$\varepsilon _2=\varepsilon _3=\varepsilon _6$$, $$\varepsilon _4=\varepsilon _5$$ and $$\varepsilon _1\varepsilon _2=\varepsilon _4$$,

in case 3. we have $$\varepsilon _1=1$$, $$\varepsilon _2=\varepsilon _4=\varepsilon _6$$ and $$\varepsilon _3=\varepsilon _5$$,

in case 4. we get $$\varepsilon _2=\varepsilon _5=\varepsilon _6$$, $$\varepsilon _3=\varepsilon _4$$ and $$\varepsilon _1\varepsilon _2=\varepsilon _4$$,

in case 5. we obtain $$\varepsilon _2=\varepsilon _6$$, $$\varepsilon _1\varepsilon _2=\varepsilon _4$$ and $$\varepsilon _1\varepsilon _3=\varepsilon _5$$.

Hence in both metric Lie algebras the system of Eq. (4.25) is satisfied with $$\beta _j'=\beta _j, i=1,2,3$$, in cases $$1.-5.$$ This proves the uniqueness of the Lie algebra $${\mathfrak {n}}_{6,23}(\alpha _i,\beta _j)$$, respectively $${\mathfrak {n}}_{6,25}(\alpha _i,\beta _j)$$ in cases $$1.-5.$$, which gives assertion 1.

If the map $$T(E_i)=\varepsilon _iE_i$$, $$\varepsilon _i = \pm 1$$, $$i=1,\dots , 6$$, is an orthogonal automorphism of $${\mathfrak {n}}_{6,25}(\alpha _i,\beta _j)$$, respectively $${\mathfrak {n}}_{6,23}(\alpha _i,\beta _j)$$, then the system of equations given by (4.25), respectively (4.25) and (4.26) is satisfied with $$\beta _j'=\beta _j, j=1,2,3$$. Therefore in cases $$1.-5.$$ for $$\varepsilon _i, i=1, \dots , 6$$ we have the conditions as above. Hence the group of orthogonal automorphisms of $${\mathfrak {n}}_{6,23}(\alpha _i,\beta _j)$$, respectively $${\mathfrak {n}}_{6,25}(\alpha _i,\beta _j)$$ in cases $$1.-5.$$ is isomorphic to the group given by 2a–2e, respectively 3a–3e. This proves assertions 2 and 3. $$\square$$

### Corollary 4.18

Let $$(\aleph _{6,k}(\alpha _i,\beta _j), \langle .,. \rangle )$$, $$k=23, 25$$, be the connected and simply connected Riemannian nilmanifold corresponding to the metric Lie algebra $$({\mathfrak {n}}_{6,k}(\alpha _i,\beta _j ), \langle .,. \rangle )$$, $$k=23, 25$$. The isometry group of $$(\aleph _{6,23}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}} (\aleph _{6,23}(\alpha _i,\beta _j))\\{} & {} = {\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \ltimes \aleph _{6,23}(\alpha _i,\beta _j) &{} if \ \beta _1=\beta _2=\beta _3=0, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,23}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2=\beta _3=0, \\ &{} or \ \beta _2>0, \beta _1=\beta _3=0, \\ &{} or \ \beta _3>0, \beta _1=\beta _2=0, \\ \aleph _{6,23}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, or \ \beta _1>0, \beta _3>0, \\ &{} or \ \beta _2>0, \beta _3>0. \end{array}\right. } \end{aligned}

The isometry group of $$(\aleph _{6,25}(\alpha _i,\beta _j), \langle .,. \rangle )$$ is

\begin{aligned}{} & {} {\mathcal {I}}(\aleph _{6,25}(\alpha _i,\beta _j)) \\{} & {} = {\left\{ \begin{array}{ll} {\mathbb {Z}}_2\times {\mathbb {Z}}_2 \times {\mathbb {Z}}_2 \ltimes \aleph _{6,25}(\alpha _i,\beta _j) &{} if \ \beta _1=\beta _2=\beta _3=0, \\ {\mathbb {Z}}_2 \times {\mathbb {Z}}_2 \ltimes \aleph _{6,25}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2=\beta _3=0, \\ &{} or \ \beta _2>0, \beta _1=\beta _3=0, \\ &{} or \ \beta _3>0, \beta _1=\beta _2=0, \\ {\mathbb {Z}}_2 \ltimes \aleph _{6,25}(\alpha _i,\beta _j) &{} if \ \beta _1>0, \beta _2>0, or \ \beta _1>0, \beta _3>0, \\ &{} or \ \beta _2>0, \beta _3>0. \\ \end{array}\right. } \end{aligned}