(R)
1. Introduction
The main goal of this paper is to address the following problem.
Determine the full isometry group of each left-invariant Riemannian metric on the simply connected unimodular nilpotent or solvable
$(R)$
-type Lie groups of dimension four.
There are exactly seven such
$4$
-dimensional Lie groups which are [Reference Van Thuong21].
\begin{equation*}\operatorname {Nil}^3 \times \mathbb{R} \qquad \operatorname {Nil}^4 \qquad \operatorname {Sol_{m, n}^4} \qquad \operatorname {Sol}^3 \times \mathbb{R} \qquad \operatorname {Sol}_0^4 \qquad \operatorname {Sol}_0^{' 4} \qquad \operatorname {Sol}_1^4\end{equation*}
where the two Lie groups
$\operatorname {Nil}^3 \times \mathbb{R}$
and
$\operatorname {Nil}^4$
are nilpotent and the others are solvable
$(R)$
-type Lie groups. To understand why the Lie groups
$\operatorname {Sol_{m, n}^4}, \operatorname {Sol}^3 \times \mathbb{R}, \operatorname {Sol}_0^4, \operatorname {Sol}_0^{' 4}, \operatorname {Sol}_1^4$
are of type
$(R)$
, see the abstract and Theorems 4.1, 4.2, 4.3, 4.4, 5.1 of the paper [Reference Van Thuong21]. The classification of left-invariant Riemannian metrics on these Lie groups is due to Van Thuong [Reference Van Thuong21]. There is a complete study of the two problems of classifying left-invariant Riemannian metrics, and determining their isometry groups, in the three-dimensional Lie groups given by [Reference Ha and Lee10, Reference Ha and Lee11, Reference Shin19] in the unimodular case and by [Reference Cosgaya and Reggiani6] in the non-unimodular case. See also [Reference Biggs3] for other general remarks.
Let
$G$
be a connected and simply connected Lie group with Lie algebra
$\mathfrak{g}$
, and let
$g$
be a left-invariant Riemannian metric on
$G$
. The group of isometries of
$G$
with respect to the left-invariant Riemannian metric
$g$
denoted
$\operatorname {Isom}\!(G, g)$
is the group of all diffeomorphisms
$\theta$
of
$G$
such that the pullback of
$g$
by
$\theta$
denoted
$\theta ^{\ast }g$
is equal to
$g$
. This means that
\begin{equation*}g(u, v) = g(\theta _{\ast }^{-1}u, \theta _{\ast }^{-1}v) \quad \forall u, v \in \mathfrak{g}, (\text{see}\;[11])\end{equation*}
where
$\theta _{\ast }$
is the differential of
$\theta$
at the identity element
$e$
of
$G$
. This definition is used because to describe the group of isometries, it suffices to examine the isotropy subgroup at the identity element, as will be explained below. The isometry group
$\operatorname {Isom}\!(G, g)$
is a Lie group under the compact-open topology and acts on
$G$
transitively [Reference Myers and Steenrod17]. The isotropy subgroup at the identity element
$e$
of
$G$
is denoted by
$\operatorname {Isom}\!(G, g)_{e}$
, it consists of isometries of
$G$
fixing the identity element
$e$
. Therefore, one simply gets that
\begin{equation*}\operatorname {Isom}\!(G, g) = L(G) \cdot \operatorname {Isom}\!(G, g)_{e} \cong G \cdot \operatorname {Isom}\!(G, g)_{e}\end{equation*}
where
$L(G)$
is the subgroup of left translations which can be identified with
$G$
. In fact, if
$\theta$
is an element of
$\operatorname {Isom}\!(G, g)$
such that
$\theta (e) = p \in G$
, then
$\theta$
decomposes as
\begin{equation*}\theta = L_{p} \circ (L_{p^{-1}} \circ \theta ) \quad \text{where}\,\, L_{p^{-1}} \circ \theta \in \operatorname {Isom}\!(G, g)_{e}.\end{equation*}
The product
$L(G) \cdot \operatorname {Isom}\!(G, g)_{e}$
is not, in general, a semidirect product; it depends on whether
$L(G)$
is normal in
$\operatorname {Isom}\!(G, g)$
. In [Reference Wilson22, Reference Wolf23], it was proved that if
$G$
is nilpotent then
\begin{equation} \operatorname {Isom}\!(G, g) \cong G \rtimes \operatorname {Aut}\!(G)_{g}. \end{equation}
where
$\operatorname {Aut}\!(G)_{g}$
is the group of isometric automorphisms of
$G$
which is described below. The isometry groups of
$4$
-dimensional nilpotent Lie groups with respect to a family of Lorentzian metrics are studied in [Reference Šukilović20].
Since
$G$
is simply connected, then
$\operatorname {Aut}\!(G) \backsimeq \operatorname {Aut}\!(\mathfrak{g})$
, and we have an action of this group on the set
$\mathscr{L}$
of all left-invariant Riemannian metrics on
$G$
given by
\begin{equation*} \begin{array}{rcl} \operatorname {Aut}\!(\mathfrak{g}) \times \mathscr{L}&\longrightarrow & \mathscr{L}\\[5pt] (\theta , g) &\longmapsto & \theta ^{\ast }g = g_{\theta } \end{array} \end{equation*}
where
$g_{\theta }(u, v) = g(\theta ^{-1}u, \theta ^{-1}v) \,\, \forall u, v\in \mathfrak{g}$
. See [Reference Ha and Lee11] for more detail.
We denote the isotropy subgroup of
$\operatorname {Aut}\!(\mathfrak{g})$
at
$g$
by
$\operatorname {Aut}\!(\mathfrak{g})_{g} = \left \lbrace \theta \in \operatorname {Aut}\!(\mathfrak{g}) / \theta ^{\ast }g = g \right \rbrace$
.
The Lie group
$G$
is said to be of type
$(R)$
(solvable of type
$(R)$
is also called completely solvable), if for every
$x \in \mathfrak{g}$
, the endomorphism
$\operatorname {ad} x \;:\; \mathfrak{g} \longrightarrow \mathfrak{g}$
has only real eigenvalues. Completely solvable Lie groups are in standard position inside their isometry group; consequently, a unimodular Lie group of this kind is normal in the isometry group of any left-invariant Riemannian metric
$g$
on it. In this case, the isometry group is given by [Reference Gordon and Wilson9, Reference Ha and Lee11]
\begin{equation} \operatorname {Isom}\!(G, g) \cong G \rtimes \operatorname {Aut}\!(G)_g. \end{equation}
The result in [Reference Ha and Lee11] is a consequence of two results given by Gordon with Wilson and Helgason in [Reference Gordon and Wilson9, Reference Helgason12]. In fact, Corollary (5.3) in [Reference Gordon and Wilson9], which is key to obtaining the result above, is itself a consequence of the work by Alekseevskii in [Reference Alekseevskii2]. In the case of the four-dimensional oscillator group, which is unimodular and solvable, but not of type
$(R)$
, the problem of determining the group of isometries is resolved in the following paper [Reference Ait Ben Haddou and Ayad1]. To apply the previous two results, we need the automorphism group of our seven simply connected unimodular
$4$
-dimensional Lie groups. The automorphism group of four-dimensional Lie groups has been studied by several authors, including [Reference Biggs and Remsing4, Reference Christodoulakis, Papadopoulos and Dimakis5, Reference Fisher, Gray and Hydon7, Reference Van Thuong21]. From the description of the automorphism group of the simply connected unimodular Lie groups of dimension four, given in [Reference Van Thuong21], we provide the groups
$\operatorname {Aut}\!(G)_g$
and then describe the full group of isometries of our four-dimensional Lie groups. In particular, we obtain a generalization of some of the results presented in [Reference Lee, Lee, Shin and Yi15].
2. Nilpotent and solvable
$(R)$
-type Lie groups of dimension 4
In this section, we describe the Lie bracket structures of the Lie algebras corresponding to simply connected nilpotent or solvable
$(R)$
-type Lie groups of dimension four. We also revisit the structures of such simply connected unimodular Lie groups.
The Lie group
$\operatorname {Nil}^3$
is referred to as the standard three-dimensional Heisenberg Lie group.
$\operatorname {Nil}^3 \times \mathbb{R} (U1[(2, 1)], \mu = 0 \;\text{in}\;$
[Reference MacCallum16]): The Lie algebra of
$\operatorname {Nil}^3 \times \mathbb{R}$
has a basis
$(e_i)$
such that [Reference Van Thuong21]
\begin{eqnarray*} & [e_i, e_j] &= 0 \quad \text{for}\quad 1 \leq i \leq j \leq 3\\[5pt] & [e_i, e_4] &= 0 \quad \text{for}\quad i = 1, 2\\[5pt] & [e_3, e_4] &= e_1. \end{eqnarray*}
Nil4 (U1[3] in [Reference MacCallum16]): The Lie algebra of
$\operatorname {Nil}^4$
has a basis
$(e_i)$
such that [Reference Van Thuong21]
\begin{eqnarray*} & [e_i, e_j] =& 0 \quad \text{for}\quad 1 \leq i \leq j \leq 3\\[5pt] & [e_1, e_4] =& 0, \quad [e_2, e_4] = e_1, \quad [e_3, e_4] = e_2. \end{eqnarray*}
The Lie algebras of the five solvable Lie groups of type
$(R)$
and of dimension four are described in [Reference Van Thuong21] as follows
\begin{equation*} \begin{aligned} \operatorname {Sol}_{m,n}^{4} &\quad (U1[1,1,1]\ \text{in}\ [16]),\\[5pt] \operatorname {Sol}^{3}\times \mathbb{R} &\quad (U1[1,1,1],\ \nu =0\ \text{in}\ [16]),\\[5pt] \operatorname {Sol}_{0}^{4} &\quad (U1[1,1,1],\ \lambda =\mu =1,\ \nu =-2\ \text{in}\ [16]),\\[5pt] \operatorname {Sol}_{0}^{'4} &\quad (U1[2,1]\ \text{in}\ [16]),\\[5pt] \operatorname {Sol}_{1}^{4} &\quad (U3I0\ \text{in}\ [16]). \end{aligned} \end{equation*}
More precisely, in the case of
$\operatorname {Sol}_1^4$
(U3I0 in [Reference MacCallum16]), the basis
$(e_i)$
of the Lie algebra satisfies the following relations [Reference Van Thuong21]
\begin{eqnarray*} & [e_2, e_3] =& e_1, \quad [e_1, e_i] = 0 \quad \text{for all}\; i\\[5pt] & [e_4, e_2] =& e_2, \\[5pt] & [e_4, e_3] =& -e_3. \end{eqnarray*}
We now describe the matrix realizations of the simply connected unimodular Lie groups of dimension four. This problem was addressed by Biggs and Remsing in [Reference Biggs and Remsing4]. Our approach follows Table 3 in [Reference Van Thuong21].
The universal covering group
$\operatorname {Nil}^3 \times \mathbb{R}$
, is the universal covering group for the Lie algebra
$\mathfrak{g}_{3.1} \oplus \mathfrak{g}_1$
in [Reference Biggs and Remsing4]. It is given by
\begin{equation*}\operatorname {Nil}^3 \times \mathbb{R} = \left \lbrace \begin{bmatrix} 1 & & & x & & & w & & & 0\\[5pt] 0 & & & 1 & & & y & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & e^{z} \end{bmatrix} = p(w, x, y, z) \;:\; w, x, y, z \in \mathbb{R}\right \rbrace .\end{equation*}
The universal covering group
$\operatorname {Nil}^4$
is the universal covering group for the Lie algebra
$\mathfrak{g}_{4.1}$
in [Reference Biggs and Remsing4]. It is given by
\begin{equation*}\operatorname {Nil}^4 = \left \lbrace \begin{bmatrix} 1 & & & z & & & \dfrac {z^2}{2} & & & w\\[7pt] 0 & & & 1 & & & z & & & -x + z\\[5pt] 0 & & & 0 & & & 1 & & & y\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = p(w, x, y, z) \;:\; w, x, y, z \in \mathbb{R}\right \rbrace .\end{equation*}
The universal covering group
$\operatorname {Sol_{m, n}^4}$
is the universal covering group for the Lie algebra
$\mathfrak{g}_{4.5}^{\alpha , \beta }$
where
$\alpha = -1 - \beta$
(see Table 3 in [Reference Van Thuong21]) in [Reference Biggs and Remsing4]. It is given by
\begin{equation*}\operatorname {Sol_{m, n}^4} = G_{4.5}^{\alpha , \beta } = \left \lbrace \begin{bmatrix} e^{-z} & & & 0 & & & 0 & & & w\\[5pt] 0 & & & e^{-\alpha z} & & & 0 & & & y\\[5pt] 0 & & & 0 & & & e^{(1 + \alpha )z} & & & x\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \;:\; w, x, y, z \in \mathbb{R}\right \rbrace .\end{equation*}
The universal covering group
$\operatorname {Sol}^3 \times \mathbb{R}$
is the universal covering group for the Lie algebra
$\mathfrak{g}_{3.4}^0 \oplus \mathfrak{g}_1$
in [Reference Biggs and Remsing4]. It is isomorphic to
$G_{3.4}^0 \times \mathbb{R}$
where
\begin{equation*}G_{3.4}^0 = \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0\\[5pt] x & & & \cosh z & & & -\sinh z\\[5pt] y & & & -\sinh z & & & \cosh z \end{bmatrix} \;:\; x, y, z \in \mathbb{R}\right \rbrace = \operatorname {SE}(1, 1).\end{equation*}
The universal covering group
$\operatorname {Sol}_0^4$
is the universal covering group for the Lie algebra
$\mathfrak{g}_{4.5}^{\alpha , \beta }$
where
$\alpha = \beta = \frac {-1}{2}$
(see Table 3 in [Reference Van Thuong21]) in [Reference Biggs and Remsing4]. It is given by
\begin{equation*}\operatorname {Sol}_0^4 = G_{4.5}^{\alpha , \beta } = \left \lbrace \begin{bmatrix} e^{-z} & & & 0 & & & 0 & & & w\\[5pt] 0 & & & e^{\frac {1}{2} z} & & & 0 & & & y\\[5pt] 0 & & & 0 & & & e^{\frac {1}{2} z} & & & x\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \;:\; w, x, y, z \in \mathbb{R}\right \rbrace .\end{equation*}
The universal covering group
$\operatorname {Sol}_0^{' 4}$
is the universal covering group for the Lie algebra
$\mathfrak{g}_{4.2}^{\alpha }$
where
$\alpha = -2$
(see Table 3 in [Reference Van Thuong21]) in [Reference Biggs and Remsing4]. It is given by
\begin{equation*}\operatorname {Sol}_0^{' 4} = G_{4.2}^{\alpha } = \left \lbrace \begin{bmatrix} e^{2z} & & & 0 & & & 0 & & & w\\[5pt] 0 & & & e^{-z} & & & 2ze^{-z} & & & -2x\\[5pt] 0 & & & 0 & & & e^{-z} & & & y\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \;:\; w, x, y, z \in \mathbb{R}\right \rbrace .\end{equation*}
The universal covering group
$\operatorname {Sol}_1^{4}$
is the universal covering group for the Lie algebra
$\mathfrak{g}_{4.8}^{-1}$
in [Reference Biggs and Remsing4]. It is given by
\begin{equation*}\operatorname {Sol}_1^{4} = G_{4.8}^{-1} = \left \lbrace \begin{bmatrix} 1 & & & x & & & w\\[5pt] 0 & & & e^z & & & y\\[5pt] 0 & & & 0 & & & 1 \end{bmatrix} = p(w, x, y, z) \;:\; w, x, y, z \in \mathbb{R}\right \rbrace .\end{equation*}
3. Metrics on 4-dimensional unimodular Lie groups
A left-invariant Riemannian metric on a Lie group
$G$
is simply an inner product
$\langle \cdot ,\cdot \rangle$
on the tangent space of
$G$
, which varies smoothly across
$G$
and such that the left translations are isometries.
It is well known that there is a bijective correspondence between left-invariant Riemannian metrics on a Lie group
$G$
and inner products on its Lie algebra
$\mathfrak{g}$
. Therefore, the classification of left-invariant Riemannian metrics on
$G$
is equivalent to the classification of inner products on
$\mathfrak{g}$
.
Proposition 3.1 [Reference Van Thuong21]. The set of inner products on
$\mathfrak{g}$
is
$(n^2 + n)/2$
-dimensional and can be identified with the set of upper triangular matrices with positive diagonal entries.
Proof. This fact was already proven in [Reference Van Thuong21]. We proceed with an alternative classical proof. The space of inner products on
$\mathfrak{g}$
is the space of symmetric, positive definite matrices denoted
$\operatorname {S}_n$
. We adopt the following notations.
$\operatorname {Tsup}_n$
: = the group consisting of upper triangular matrices with positive diagonal entries.
$\operatorname {Tinf}_n$
: = the group consisting of lower triangular matrices with positive diagonal entries.
$\operatorname {D}_n^{+}$
: = the group consisting of diagonal matrices with positive elements.
Consider the following map
\begin{equation*} \begin{array}{rcl} \varphi\; : \;\operatorname {Tsup}_{n}&\longrightarrow & \operatorname {S}_{n}\\[5pt] B &\longmapsto & (B^{-1})^t(B^{-1}) \end{array} \end{equation*}
It is clear that the map
$\varphi$
is well defined. For the injectivity, let
$B, C \in \operatorname {Tsup}_{n}$
such that
$\varphi (B) = \varphi (C)$
. This is equivalent to the following equality
$B^{-1}C = (C^{-1}B)^{t}$
. Put
$M = C^{-1}B$
, then
$B^{-1}C = (C^{-1}B)^{t} \Leftrightarrow M^{-1} = M^{t}$
. Since
$\operatorname {Tsup}_{n}$
is a group, then
$M, M^{-1} \in \operatorname {Tsup}_{n}$
. Since
$M^{-1} = M^{t} \in \operatorname {Tinf}_n$
, then
$M \in \operatorname {Tsup}_n \cap \operatorname {Tinf}_n = \operatorname {D}_n^{+}$
. Hence, we have
$M^{-1} = M^{t} = M$
, then
$M = I_n$
. This implies that
$B = C$
and
$\varphi$
is injective.
For the surjectivity, let
$A \in \operatorname {S}_{n}$
and consider the inner product on
$\mathbb{R}^{n}$
defined by
\begin{equation*}\langle u, v\rangle = u^{t}Av, \forall u, v \in \mathbb{R}^{n}.\end{equation*}
Let
$\mathcal{B} = \left \lbrace e_1, \ldots , e_n\right \rbrace$
be the canonical basis of
$\mathbb{R}^{n}$
. The Gram-Schmidt procedure produces an orthonormal basis
$\mathscr{B} = \left \lbrace v_1, \ldots , v_n\right \rbrace$
of the Euclidean space
$\left ( \mathbb{R}^{n}, \langle \cdot ,\cdot \rangle \right )$
such that
\begin{equation*}[v_1, \ldots , v_n] = X \in Tsup_n.\end{equation*}
This means that
$Xe_i = v_i$
. Since
$\mathscr{B} = \{v_1, \ldots , v_n\}$
is an orthonormal basis of
$(\mathbb{R}^n, \langle \cdot ,\cdot \rangle )$
, then
\begin{equation*}\operatorname {Mat}\left (\langle \cdot ,\cdot \rangle , \mathscr{B}\right ) =X^tAX = I_n.\end{equation*}
Thus,
$A = (X^{-1})^{t}(X^{-1}) = \varphi (X)$
. Hence,
$\varphi$
is surjective.
Therefore,
$\varphi$
is a bijection.
Remark 3.1 [Reference Van Thuong21]. The classification of inner products (symmetric positive definite matrices) on Lie algebras is equivalent to the classification of upper triangular matrices with positive diagonal entries.
Now, following [Reference Van Thuong21], we describe all left-invariant Riemannian metrics on simply connected unimodular nilpotent or solvable
$(R)$
-type 4-dimensional Lie groups, up to isometry. These metrics are represented by upper triangular matrices with positive diagonal entries. However, we will use the bijection
$\varphi$
, as introduced above, to describe the associated symmetric positive definite matrices.
Remark 3.2. In each unimodular Lie group of dimension four, we will investigate different metrics that yield different isometry groups. This means that for any other metric, its isometry group is isomorphic to that of one of the investigated metrics.
3.1. Left-invariant Riemannian metrics on
$\operatorname {Nil}^3 \times \mathbb{R}$
By Theorem 3.1 in [Reference Van Thuong21], any metric on
$\operatorname {Nil}^3 \times \mathbb{R}$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} \alpha & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \quad \alpha \gt 0.\end{equation*}
According to the bijection
$\varphi$
, the left-invariant Riemannian metric (symmetric positive definite matrix) associated with
$M$
is
\begin{equation*}\langle \cdot ,\cdot \rangle = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[7pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}.\end{equation*}
3.2. Left-invariant Riemannian metrics on
$\operatorname {Nil}^4$
By Theorem 3.2 in [Reference Van Thuong21], any metric on
$\operatorname {Nil}^4$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} \alpha & & & \beta & & & 0 & & & 0\\[5pt] 0 & & & \gamma & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \quad \alpha , \gamma \gt 0 \quad \beta \geq 0.\end{equation*}
We distinguish two cases associated to this metric
-
1. If
$\beta = 0$
, then the left-invariant Riemannian metric associated with
$M$
is
\begin{equation*}\langle \cdot ,\cdot \rangle _1 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[8pt] 0 & & & \dfrac {1}{\gamma ^2} & & & 0 & & & 0\\[8pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\end{equation*}
-
2. If
$\beta \gt 0$
, then our left-invariant Riemannian metric is
\begin{equation*}\langle \cdot ,\cdot \rangle _2 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & \dfrac {-\beta }{\alpha ^2\gamma } & & & 0 & & & 0\\[8pt] \dfrac {-\beta }{\alpha ^2\gamma } & & & \dfrac {1}{\gamma ^2} + \dfrac {\beta ^2}{\alpha ^2\gamma ^2} & & & 0 & & & 0\\[8pt] 0 & & & 0 & & & 1 & & & 0\\[8pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}.\end{equation*}
3.3. Left-invariant Riemannian metrics on
$\operatorname {Sol_{m, n}^4}$
By Theorem 4.3 in [Reference Van Thuong21], any metric on
$\operatorname {Sol_{m, n}^4}$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} 1 & & & \alpha & & & \beta & & & 0\\[5pt] 0 & & & 1 & & & \gamma & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \mu \end{bmatrix} \quad \mu \gt 0 \quad \alpha , \gamma \geq 0 \quad \beta \in \mathbb{R}.\end{equation*}
We distinguish three cases associated with this metric
-
1. If
$\alpha = \beta = \gamma = 0$
, then the left-invariant Riemannian metric associated with
$M$
is
\begin{equation*}\langle \cdot ,\cdot \rangle _1 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\end{equation*}
-
2. If
$\alpha \gt 0$
and
$\beta = \gamma = 0$
, then our left-invariant Riemannian metric is
\begin{equation*}\langle \cdot ,\cdot \rangle _2 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} 1 & & & -\alpha & & & 0 & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\end{equation*}
-
3. If
$\alpha \gt 0$
,
$\beta \neq 0$
and
$\gamma \geq 0$
, then our left-invariant Riemannian metric isThe remaining cases for the parameters
\begin{equation*}\langle \cdot ,\cdot \rangle _3 = \begin{bmatrix} 1 & & & -\alpha & & & \alpha \gamma - \beta & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & -\alpha (\alpha \gamma - \beta ) - \gamma & & & 0\\[5pt] \alpha \gamma - \beta & & & -\alpha (\alpha \gamma - \beta ) - \gamma & & & (\alpha \gamma - \beta )^2 + \gamma ^2 + 1 & & & 0\\[7pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}.\end{equation*}
$(\alpha ,\beta ,\gamma ,\mu )$
do not give rise to additional isometry groups.
3.4. Left-invariant Riemannian metrics on
$\operatorname {Sol}^3 \times \mathbb{R}$
By Theorem 4.4 in [Reference Van Thuong21], any metric on
$\operatorname {Sol}^3 \times \mathbb{R}$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} 1 & & & \alpha & & & \beta & & & 0\\[5pt] 0 & & & 1 & & & \gamma & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \mu \end{bmatrix} \quad \mu \gt 0 \quad \alpha , \gamma \geq 0 \quad \beta \in \mathbb{R}.\end{equation*}
Hence, this case is similar to the precedent one, we distinguish the following left-invariant Riemannian metrics
\begin{equation*}\langle \cdot ,\cdot \rangle _1 = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix} \qquad \langle \cdot ,\cdot \rangle _2 = \begin{bmatrix} 1 & & & -\alpha & & & 0 & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\end{equation*}
\begin{equation*}\langle \cdot ,\cdot \rangle _3 = \begin{bmatrix} 1 & & & -\alpha & & & \alpha \gamma - \beta & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & -\alpha (\alpha \gamma - \beta ) - \gamma & & & 0\\[5pt] \alpha \gamma - \beta & & & -\alpha (\alpha \gamma - \beta ) - \gamma & & & (\alpha \gamma - \beta )^2 + \gamma ^2 + 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}.\end{equation*}
The remaining cases for the parameters
$(\alpha ,\beta ,\gamma ,\mu )$
do not give rise to additional isometry groups.
3.5. Left-invariant Riemannian metrics on
$\operatorname {Sol}_0^4$
By Theorem 4.1 in [Reference Van Thuong21], any metric on
$\operatorname {Sol}_0^4$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} 1 & & & 0 & & & \alpha & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \beta \end{bmatrix} \quad \alpha \geq 0 \quad \beta \gt 0.\end{equation*}
We distinguish two cases associated with this metric
-
1. If
$\alpha = 0$
, then the left-invariant Riemannian metric associated with
$M$
is
\begin{equation*}\langle \cdot ,\cdot \rangle _1 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}\end{equation*}
-
2. If
$\alpha \gt 0$
, then our left-invariant Riemannian metric is
\begin{equation*}\langle \cdot ,\cdot \rangle _2 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} 1 & & & 0 & & & -\alpha & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] -\alpha & & & 0 & & & 1 + \alpha ^2 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}.\end{equation*}
3.6. Left-invariant Riemannian metrics on
$\operatorname {Sol}_0^{' 4}$
By Theorem 4.2 in [Reference Van Thuong21], any metric on
$\operatorname {Sol}_0^{' 4}$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} 1 & & & 0 & & & \beta & & & 0\\[5pt] 0 & & & \alpha & & & \gamma & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \mu \end{bmatrix} \quad \alpha , \mu \gt 0 \quad \beta \geq 0 \quad \gamma \in \mathbb{R}.\end{equation*}
We distinguish two cases associated with this metric
-
1. If
$\beta = \gamma = 0$
, then the left-invariant Riemannian metric associated with
$M$
is
\begin{equation*}\langle \cdot ,\cdot \rangle _1 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \dfrac {1}{\alpha ^2} & & & 0 & & & 0\\[7pt] 0 & & & 0 & & & 1 & & & 0\\[6pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\end{equation*}
-
2. If
$\beta \neq 0$
and
$\gamma = 0$
, then our left-invariant Riemannian metric is
\begin{equation*}\langle \cdot ,\cdot \rangle _2 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} 1 & & & 0 & & & -\beta & & & 0\\[5pt] 0 & & & \dfrac {1}{\alpha ^2} & & & 0 & & & 0\\[7pt] -\beta & & & 0 & & & 1 + \beta ^2 & & & 0\\[6pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}.\end{equation*}
The remaining cases for the parameters
$(\alpha ,\beta ,\gamma ,\mu )$
do not give rise to additional isometry groups.
3.7. Left-invariant Riemannian metrics on
$\operatorname {Sol}_1^4$
By Theorem 5.1 in [Reference Van Thuong21], any metric on
$\operatorname {Sol}_1^4$
is equivalent, up to isometry, to the following metric
\begin{equation*}M = \begin{bmatrix} \alpha & & & \beta & & & \gamma & & & 0\\[5pt] 0 & & & 1 & & & \mu & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \nu \end{bmatrix} \quad \alpha , \nu \gt 0 \quad \beta , \mu \geq 0 \quad \gamma \in \mathbb{R}.\end{equation*}
We distinguish three cases associated with this metric
-
1. If
$\beta = \gamma = \mu = 0$
, then the left-invariant Riemannian metric associated with
$M$
is
\begin{equation*}\langle \cdot ,\cdot \rangle _1 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[7pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\end{equation*}
-
2. If
$\beta \gt 0$
and
$\gamma = \mu = 0$
, then our left-invariant Riemannian metric is
\begin{equation*}\langle \cdot ,\cdot \rangle _2 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & \dfrac {-\beta }{\alpha ^2} & & & 0 & & & 0\\[7pt] \dfrac {-\beta }{\alpha ^2} & & & 1 + \bigg(\dfrac {\beta }{\alpha }\bigg)^2 & & & 0 & & & 0\\[7pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\end{equation*}
-
3. If
$\beta , \mu \gt 0$
and
$\gamma = 0$
, then our left-invariant Riemannian metric is
\begin{equation*}\langle \cdot ,\cdot \rangle _3 = \left ( M^{-1}\right )^t\left ( M^{-1}\right ) = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & \dfrac {-\beta }{\alpha ^2} & & & \dfrac {\beta \mu }{\alpha ^2} & & & 0\\[7pt] \dfrac {-\beta }{\alpha ^2} & & & 1 + \bigg(\dfrac {\beta }{\alpha }\bigg)^2 & & & \dfrac {-\beta ^2\mu }{\alpha ^2} - \mu & & & 0\\[7pt] \dfrac {\beta \mu }{\alpha ^2} & & & \dfrac {-\beta ^2\mu }{\alpha ^2} - \mu & & & \bigg(\dfrac {\beta \mu }{\alpha }\bigg)^2 + \mu ^2 + 1 & & & 0\\[7pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}.\end{equation*}
The remaining cases for the parameters
$(\alpha ,\beta ,\gamma ,\mu , \nu )$
do not give rise to additional isometry groups.
4. The isometry groups
Let us recall some well-known and important results about the isometry groups of left-invariant Riemannian metrics on Lie groups.
Assume that our Lie group
$G$
is compact, then for any left-invariant Riemannian metric on
$G$
. The isometry group
$\operatorname {Isom}\!(G, g)$
is also compact [Reference Myers and Steenrod17].
If
$G$
is compact simple and if
$g$
is a left-invariant Riemannian metric on
$G$
. Then
$\operatorname {Isom}_0(G, g) \subset L(G)R(G)$
[Reference Ochiai and Takahashi18]. This means that any isometry in the connected component of the identity
$\operatorname {Isom}_0(G, g)$
decomposes into a left translation and a right translation.
Let
$\theta$
be an automorphism of
$G$
and let
$g$
be a left-invariant Riemannian metric on
$G$
. The pullback
$\theta ^{\ast }g$
of
$g$
by
$\theta$
is defined by:
$\theta ^{\ast }g(u, v) = g\left (\theta _{\ast }^{-1}(u), \theta _{\ast }^{-1}(v)\right ) \; \forall u, v \in \mathfrak{g}$
where
$\theta _{\ast }$
is the differential of
$\theta$
at the identity element
$e$
of
$G$
. This equation is equivalent to the following matrix equation
$[g] = [\theta _{\ast }]^t[\theta ^{\ast }g][\theta _{\ast }]$
. Hence, in term of matrix calculation, the group of isometric automorphisms of
$G$
is given by [Reference Ha and Lee11]
\begin{equation} \operatorname {Aut}\!(G)_{g} = \left \lbrace \theta \in \operatorname {Aut}\!(G) \mid [g] = [\theta _{\ast }]^t[g][\theta _{\ast }] \right \rbrace . \end{equation}
Remark 4.1.
It is a standard fact that if two left-invariant Riemannian metrics on a Lie group
$G$
are equivalent up to automorphism, then their isometry groups are conjugate.
According to this remark, one only needs to describe the isometry groups of representative metrics of the non-equivalent classes of left-invariant Riemannian metrics on the Lie group.
Remark 4.2. The computations of the groups of isometric automorphisms determined by the matrix equation
\begin{equation*}A^t\langle \cdot ,\cdot \rangle _i A=\langle \cdot ,\cdot \rangle _i\end{equation*}
were carried out using MAPLE. The detailed calculations are presented in each case.
Note that determining the isometry groups of each Lie group enables us to answer Question4.3.
Question 4.3. Does every nontrivial compact subgroup of the automorphism group arise as the isotropy subgroup of some left-invariant Riemannian metric?
4.1. Nilpotent case
Using the result (1), we give the isometry group of our two nilpotent four-dimensional Lie groups.
4.1.1. Isometry group of
$\operatorname {Nil}^3 \times \mathbb{R}$
Theorem 4.4.
The isometry group of
$\operatorname {Nil}^3 \times \mathbb{R}$
is given by
\begin{equation*}\operatorname {Isom}\!\left (\operatorname {Nil}^3 \times \mathbb{R}, \langle \cdot ,\cdot \rangle \right ) \cong \left (\operatorname {Nil}^3 \times \mathbb{R}\right ) \rtimes \left (\operatorname {diag}\left \lbrace \pm 1, \pm 1, \operatorname {O}(2) \right \rbrace \right )\end{equation*}
Proof. The automorphism group
$\operatorname {Aut}\!\left (\operatorname {Nil}^3 \times \mathbb{R}\right )$
consists of elements of the form [Reference Van Thuong21]
\begin{equation*}\begin{bmatrix} ad - bc & & & x & & & y & & & u\\[5pt] 0 & & & e & & & z & & & v\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & c & & & d \end{bmatrix} \quad \begin{bmatrix} a & & & b\\[5pt] c & & & d \end{bmatrix} \in \operatorname {GL}\!(2, \mathbb{R}) \quad e \in \mathbb{R}^{\ast } \quad x, y, z, u, v \in \mathbb{R}.\end{equation*}
Let
$A = \begin{bmatrix} ad - bc & & & x & & & y & & & u\\[3pt] 0 & & & e & & & z & & & v\\[3pt] 0 & & & 0 & & & a & & & b\\[3pt] 0 & & & 0 & & & c & & & d \end{bmatrix} \in \operatorname {Aut}\!\left (\operatorname {Nil}^3 \times \mathbb{R}\right )$
, then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!\left (\operatorname {Nil}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle }\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle A = \langle \cdot ,\cdot \rangle \\[5pt] \Leftrightarrow & \begin{bmatrix} ad - bc & & & 0 & & & 0 & & & 0\\[5pt] x & & & e & & & 0 & & & 0\\[5pt] y & & & z & & & a & & & c\\[5pt] u & & & v & & & b & & & d \end{bmatrix}\begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\begin{bmatrix} ad - bc & & & x & & & y & & & u\\[5pt] 0 & & & e & & & z & & & v\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & c & & & d \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} \dfrac {(ad - bc)^2}{\alpha ^2} & & & \dfrac {(ad - bc)x}{\alpha ^2} & & & \dfrac {(ad - bc)y}{\alpha ^2} & & & \dfrac {(ad - bc)u}{\alpha ^2}\\[7pt] \dfrac {(ad - bc)x}{\alpha ^2} & & & \dfrac {x^2}{\alpha ^2} + e^2 & & & \dfrac {xy}{\alpha ^2} + ez & & & \dfrac {xu}{\alpha ^2} + ev\\[7pt] \dfrac {(ad - bc)y}{\alpha ^2} & & & \dfrac {xy}{\alpha ^2} + ez & & & \dfrac {y^2}{\alpha ^2} + z^2 + a^2 + c^2 & & & \dfrac {yu}{\alpha ^2} + zv + ab + cd\\[7pt] \dfrac {(ad - bc)u}{\alpha ^2} & & & \dfrac {xu}{\alpha ^2} + ev & & & \dfrac {yu}{\alpha ^2} + zv + ab + cd & & & \dfrac {u^2}{\alpha ^2} + v^2 + b^2 + d^2 \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[7pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}. \end{eqnarray*}
By equating the
$(1,2)$
,
$(1,3)$
, and
$(1,4)$
entries of the two matrices above, we obtain
$x = y = u = 0$
. Then, equating the
$(2,3)$
and
$(2,4)$
entries yields
$z = v = 0$
. Hence, the matrix equality reduces to
\begin{eqnarray*} & \begin{bmatrix} \dfrac {(ad - bc)^2}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[7pt] 0 & & & e^2 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & a^2 + c^2 & & & ab + cd\\[3pt] 0 & & & 0 & & & ab + cd & & & b^2 + d^2 \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[7pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\\[3pt] \Leftrightarrow & \begin{bmatrix} a & & & c\\[2pt] b & & & d \end{bmatrix}\begin{bmatrix} a & & & b\\[2pt] c & & & d \end{bmatrix} = I_2, \quad e = \pm 1, \quad ad - bc = \pm 1\\[2pt] \Leftrightarrow & \begin{bmatrix} a & & & b\\[2pt] c & & & d \end{bmatrix} \in \operatorname {O}(2), \quad e = \pm 1. \end{eqnarray*}
Hence,
\begin{equation*}\operatorname {Isom}\!\left (\operatorname {Nil}^3 \times \mathbb{R}, \langle \cdot ,\cdot \rangle \right ) \cong \left (\operatorname {Nil}^3 \times \mathbb{R}\right ) \rtimes \left (\operatorname {diag}\left \lbrace \pm 1, \pm 1, \operatorname {O}(2)\right \rbrace \right ).\end{equation*}
Answer. The answer to Question4.3 is negative. For example, the compact subgroup
$(\mathbb{Z}_2)^2$
of
$\operatorname {Aut}\!\left (\operatorname {Nil}^3 \times \mathbb{R}\right )$
cannot occur as the isotropy subgroup of any left-invariant Riemannian metric.
4.1.2. Isometry group of
$\operatorname {Nil}^4$
Theorem 4.5.
The isometry group of
$\operatorname {Nil}^4$
is given by
\begin{eqnarray*} \operatorname {Isom}\!\left (\operatorname {Nil}^4, \langle \cdot ,\cdot \rangle _1\right ) &\cong & \operatorname {Nil}^4 \rtimes \left (\mathbb{Z}_2\right )^2\\[2pt] \operatorname {Isom}\!\left (\operatorname {Nil}^4, \langle \cdot ,\cdot \rangle _2\right ) &\cong & \operatorname {Nil}^4 \rtimes (\mathbb{Z}_2). \end{eqnarray*}
Proof. The automorphism group
$\operatorname {Aut}\!\left (\operatorname {Nil}^4\right )$
consists of elements of the form [Reference Van Thuong21]
\begin{equation*}\begin{bmatrix} ad^2 & & & ed & & & x & & & y\\[5pt] 0 & & & ad & & & e & & & z\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & 0 & & & d \end{bmatrix} \quad a, d \in \mathbb{R}^{\ast } \quad b, e, x, y, z \in \mathbb{R}.\end{equation*}
Let
$A = \begin{bmatrix} ad^2 & & & ed & & & x & & & y\\[5pt] 0 & & & ad & & & e & & & z\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & 0 & & & d \end{bmatrix} \in \operatorname {Aut}\!\left (\operatorname {Nil}^4\right )$
, then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!\left (\operatorname {Nil}^4\right )_{\langle \cdot ,\cdot \rangle _1}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[5pt] \Leftrightarrow & \begin{bmatrix} ad^2 & & & 0 & & & 0 & & & 0\\[5pt] ed & & & ad & & & 0 & & & 0\\[5pt] x & & & e & & & a & & & 0\\[5pt] y & & & z & & & b & & & d \end{bmatrix}\begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \dfrac {1}{\gamma ^2} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\begin{bmatrix} ad^2 & & & ed & & & x & & & y\\[5pt] 0 & & & ad & & & e & & & z\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & 0 & & & d \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \dfrac {1}{\gamma ^2} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} \dfrac {a^{2} d^{4}}{\alpha ^{2}} & & & \dfrac {a \,d^{3} e}{\alpha ^{2}} & & & \dfrac {a \,d^{2} x}{\alpha ^{2}} & & & \dfrac {a \,d^{2} y}{\alpha ^{2}}\\[5pt] \dfrac {a \,d^{3} e}{\alpha ^{2}} & & & \dfrac {e^{2} d^{2}}{\alpha ^{2}}+\dfrac {a^{2} d^{2}}{\gamma ^{2}} & & & \dfrac {e d x}{\alpha ^{2}}+\dfrac {a d e}{\gamma ^{2}} & & & \dfrac {e d y}{\alpha ^{2}}+\dfrac {a d z}{\gamma ^{2}}\\[5pt] \dfrac {a \,d^{2} x}{\alpha ^{2}} & & & \dfrac {e d x}{\alpha ^{2}}+\dfrac {a d e}{\gamma ^{2}} & & & \dfrac {x^{2}}{\alpha ^{2}}+\dfrac {e^{2}}{\gamma ^{2}}+a^{2} & & & \dfrac {x y}{\alpha ^{2}}+\dfrac {e z}{\gamma ^{2}}+a b\\[5pt] \dfrac {a \,d^{2} y}{\alpha ^{2}} & & & \dfrac {e d y}{\alpha ^{2}}+\dfrac {a d z}{\gamma ^{2}} & & & \dfrac {x y}{\alpha ^{2}}+\dfrac {e z}{\gamma ^{2}}+a b & & & \dfrac {y^{2}}{\alpha ^{2}}+\dfrac {z^{2}}{\gamma ^{2}}+b^{2}+d^{2} \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \dfrac {1}{\gamma ^2} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}. \end{eqnarray*}
By equating the
$(1,2)$
,
$(1,3)$
, and
$(1,4)$
entries of the two matrices above, we obtain
$e = x = y = 0$
. Then, equating the
$(2,4)$
and
$(3,4)$
entries yields
$z = b = 0$
. Hence, the matrix equality reduces to
\begin{equation*} \begin{bmatrix} \dfrac {a^{2} d^{4}}{\alpha ^{2}} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \dfrac {a^{2} d^{2}}{\gamma ^{2}} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & a^{2} & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & d^{2} \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \dfrac {1}{\gamma ^2} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \Leftrightarrow a = \pm 1, \quad d = \pm 1.\end{equation*}
Therefore,
\begin{eqnarray*} \operatorname {Aut}\!\left (\operatorname {Nil}^4\right )_{\langle \cdot ,\cdot \rangle _1} &=& \left \lbrace I_4, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace \\[5pt] &\cong & \left (\mathbb{Z}_2\right )^2. \end{eqnarray*}
Hence,
\begin{equation*}\operatorname {Isom}\!\left (\operatorname {Nil}^4, \langle \cdot ,\cdot \rangle _1\right ) \cong \operatorname {Nil}^4 \rtimes \left (\mathbb{Z}_2\right )^2.\end{equation*}
For the metric
$\langle \cdot ,\cdot \rangle _2$
, we have by (3):
\begin{eqnarray*} & A \in \operatorname {Aut}\!\left (\operatorname {Nil}^4\right )_{\langle \cdot ,\cdot \rangle _2}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _2A = \langle \cdot ,\cdot \rangle _2\\[5pt] \Leftrightarrow & \begin{bmatrix} ad^2 & & & 0 & & & 0 & & & 0\\[5pt] ed & & & ad & & & 0 & & & 0\\[5pt] x & & & e & & & a & & & 0\\[5pt] y & & & z & & & b & & & d \end{bmatrix}\begin{bmatrix} \dfrac {1}{\alpha ^2} & & & \dfrac {-\beta }{\alpha ^2\gamma } & & & 0 & & & 0\\[5pt] \dfrac {-\beta }{\alpha ^2\gamma } & & & \dfrac {1}{\gamma ^2} + \dfrac {\beta ^2}{\alpha ^2\gamma ^2} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\begin{bmatrix} ad^2 & & & ed & & & x & & & y\\[5pt] 0 & & & ad & & & e & & & z\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & 0 & & & d \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & \dfrac {-\beta }{\alpha ^2\gamma } & & & 0 & & & 0\\[5pt] \dfrac {-\beta }{\alpha ^2\gamma } & & & \dfrac {1}{\gamma ^2} + \dfrac {\beta ^2}{\alpha ^2\gamma ^2} & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \end{eqnarray*}
We first note that
$\operatorname {Aut}\!\left (\operatorname {Nil}^4\right )_{\langle \cdot ,\cdot \rangle _2} \subset \operatorname {O}(4)$
and:
\begin{eqnarray*} & \begin{bmatrix} ad^2 & & & 0 & & & 0 & & & 0\\[5pt] ed & & & ad & & & 0 & & & 0\\[5pt] x & & & e & & & a & & & 0\\[5pt] y & & & z & & & b & & & d \end{bmatrix}\begin{bmatrix} ad^2 & & & ed & & & x & & & y\\[5pt] 0 & & & ad & & & e & & & z\\[5pt] 0 & & & 0 & & & a & & & b\\[5pt] 0 & & & 0 & & & 0 & & & d \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} a^{2} d^{4} & & & a \,d^{3} e & & & a \,d^{2} x & & & a \,d^{2} y\\[5pt] a \,d^{3} e & & & a^{2} d^{2}+e^{2} d^{2} & & & a d e+e d x & & & a d z+e d y\\[5pt] a \,d^{2} x & & & a d e+e d x & & & a^{2}+e^{2}+x^{2} & & & a b+e z+x y\\[5pt] a \,d^{2} y & & & a d z+e d y & & & a b+e z+x y & & & b^{2}+d^{2}+y^{2}+z^{2} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\\[5pt] \Leftrightarrow & e = x = y = z = b = 0, \quad a = \pm 1, \quad d = \pm 1. \end{eqnarray*}
Now since the form of
$A$
is restricted, one simply gets that (it suffices to check which elements of
$\left (\mathbb{Z}_2\right )^2$
preserve the metric
$\langle \cdot , \cdot \rangle _2$
)
\begin{eqnarray*} A \in \operatorname {Aut}\!\left (\operatorname {Nil}^4\right )_{\langle \cdot ,\cdot \rangle _2} &\Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _2A = \langle \cdot ,\cdot \rangle _2\\[5pt] &\Leftrightarrow & A \in \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}\right \rbrace \cong \mathbb{Z}_2. \end{eqnarray*}
Hence,
$\operatorname {Aut}\!\left (\operatorname {Nil}^4\right )_{\langle \cdot ,\cdot \rangle _2} \cong \mathbb{Z}_2$
. Therefore,
$\operatorname {Isom}\!\left (\operatorname {Nil}^4, \langle \cdot ,\cdot \rangle _2\right ) \cong \operatorname {Nil}^4 \rtimes \mathbb{Z}_2.$
Answer. The answer to Question4.3 is affirmative. Note that
$(\mathbb{Z}_2)^2$
has three distinct subgroups of order
$2$
; however, these are different subgroups inside
$(\mathbb{Z}_2)^2$
, even though they are all isomorphic to
$\mathbb{Z}_2$
. We emphasize that it is these abstract group types that occur as isotropy groups of left-invariant Riemannian metrics.
Remark 4.6. Note that when we find the group of isometric automorphisms of the first class of metrics, the problem becomes easy, because the group of isometric automorphisms of the first class of metrics is the maximal one. This fact has a significant impact on the rest of the article.
A major development in the study of left-invariant metrics on solvable Lie groups was the discovery that distinguished metrics have maximal isometry groups. More precisely, Jablonski and Jablonski–Gordon established this property for Einstein and Ricci soliton metrics in [Reference Gordon and Jablonski8, Reference Jablonski13, Reference Jablonski14].
4.2. Solvable
$(R)$
-type case
Using the result (2), we give the isomerty groups of our five solvable Lie groups of type
$(R)$
.
4.2.1. Isometry group of
$\operatorname {Sol_{m, n}^4}$
The automorphism group of
$\operatorname {Sol}_{m, n}^4$
consists of elements of the form [Reference Van Thuong21]
\begin{equation*}\begin{bmatrix} a & & & 0 & & & 0 & & & x\\[5pt] 0 & & & b & & & 0 & & & y\\[5pt] 0 & & & 0 & & & c & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \qquad a, b, c \in \mathbb{R}^{\ast }, \quad x, y, z \in \mathbb{R}. \end{equation*}
Theorem 4.7.
The group of isometric automorphisms of
$\operatorname {Sol}_{m, n}^4$
is given by
\begin{eqnarray*} \operatorname {Aut}\!\big(\!\operatorname {Sol}_{m, n}^4\big)_{\langle \cdot ,\cdot \rangle _1} &\cong & (\mathbb{Z}_2)^3\\[5pt] \operatorname {Aut}\big(\!\operatorname {Sol}_{m, n}^4\big)_{\langle \cdot ,\cdot \rangle _2} &\cong & (\mathbb{Z}_2)^2\\[5pt] \operatorname {Aut}\big(\!\operatorname {Sol}_{m, n}^4\big)_{\langle \cdot ,\cdot \rangle _3} &\cong & \mathbb{Z}_2. \end{eqnarray*}
Proof. Let
$A = \begin{bmatrix} a & & & 0 & & & 0 & & & x\\[5pt] 0 & & & b & & & 0 & & & y\\[5pt] 0 & & & 0 & & & c & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol_{m, n}^4}\!)$
. Then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!\big(\!\operatorname {Sol_{m,n}^4}\big)_{\langle \cdot ,\cdot \rangle _1}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[5pt] \Leftrightarrow & \begin{bmatrix} a & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & b & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & c & & & 0\\[5pt] x & & & y & & & z & & & 1 \end{bmatrix}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\begin{bmatrix} a & & & 0 & & & 0 & & & x\\[5pt] 0 & & & b & & & 0 & & & y\\[5pt] 0 & & & 0 & & & c & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} a^{2} & & & 0 & & & 0 & & & a x\\[5pt] 0 & & & b^{2} & & & 0 & & & b y\\[5pt] 0 & & & 0 & & & c^{2} & & & c z\\[5pt] a x & & & b y & & & c z & & & x^{2}+y^{2}+z^{2}+\dfrac {1}{\mu ^{2}} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & x = y = z = 0, \quad a = \pm 1, \quad b = \pm 1, \quad c = \pm 1. \end{eqnarray*}
Therefore,
\begin{equation*}\operatorname {Aut}\!(\!\operatorname {Sol_{m,n}^4}\!)_{\langle \cdot ,\cdot \rangle _1} = \left \lbrace \begin{bmatrix} \pm 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \pm 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & \pm 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace \cong (\mathbb{Z}_2)^3. \end{equation*}
For the metric
$\langle \cdot ,\cdot \rangle _2$
, by using (3), one gets that
\begin{eqnarray*} & A \in \operatorname {Aut}\!\big(\!\operatorname {Sol_{m,n}^4}\big)_{\langle \cdot ,\cdot \rangle _2}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _2A = \langle \cdot ,\cdot \rangle _2\\[5pt] \Leftrightarrow & \begin{bmatrix} a & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & b & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & c & & & 0\\[5pt] x & & & y & & & z & & & 1 \end{bmatrix}\begin{bmatrix} 1 & & & -\alpha & & & 0 & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\begin{bmatrix} a & & & 0 & & & 0 & & & x\\[5pt] 0 & & & b & & & 0 & & & y\\[5pt] 0 & & & 0 & & & c & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & & & -\alpha & & & 0 & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} a^{2} & & & -a \alpha b & & & 0 & & & -a \alpha y+a x\\[5pt] -a \alpha b & & & b^{2} (\alpha ^{2}+1) & & & 0 & & & -b \alpha x+b (\alpha ^{2}+1) y\\[5pt] 0 & & & 0 & & & c^{2} & & & c z\\[5pt] (\!-\!y \alpha +x) a & & & (\!-\!x \alpha +y (\alpha ^{2}+1)) b & & & c z & & & (\!-\!y \alpha +x) x+(\!-\!x \alpha +y (\alpha ^{2}+1)) y+z^{2}+\dfrac {1}{\mu ^{2}} \end{bmatrix}\\[5pt] & = \begin{bmatrix} 1 & & & -\alpha & & & 0 & & & 0\\[5pt] -\alpha & & & 1 + \alpha ^2 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & z = 0, \quad a = \pm 1, \quad b = \pm 1, \quad ab = 1, \quad c = \pm 1\\[5pt] \Leftrightarrow & z = 0, \quad a = b = \pm 1, \quad c = \pm 1. \end{eqnarray*}
By equating the
$(1,4)$
and
$(2,4)$
entries of the two matrices above, we see that
$x, y$
verify the following system
\begin{equation*}\left \lbrace \begin{array}{l@{\quad}l@{\quad}l} x - \alpha y = 0\\[5pt] -\alpha x + \left (1 + \alpha ^2\right )y = 0 \end{array}\right .\end{equation*}
This implies that
$x = y = 0$
. Hence,
\begin{eqnarray*} \operatorname {Aut}\!\big(\!\operatorname {Sol_{m,n}^4}\big)_{\langle \cdot ,\cdot \rangle _2} &=& \left \lbrace I_4, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace \\[5pt] &\cong & (\mathbb{Z}_2)^2. \end{eqnarray*}
Finally, using the fact that
$(\mathbb{Z}_2)^3$
as above is the maximal group of isometric automorphisms of
$\operatorname {Sol_{m, n}^4}$
, a straightforward verification shows that
\begin{eqnarray*} A \in \operatorname {Aut}\!\big(\!\operatorname {Sol_{m,n}^4}\big)_{\langle \cdot ,\cdot \rangle _3} &\Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _3A = \langle \cdot ,\cdot \rangle _3\\[5pt] &\Leftrightarrow & A \in \left \lbrace I_4, \operatorname {diag}\left \lbrace -1, -1, -1, 1\right \rbrace \right \rbrace \cong \mathbb{Z}_2. \end{eqnarray*}
Corollary 4.8.
The isometry group of
$\operatorname {Sol}_{m, n}^4$
is given by
\begin{eqnarray*} \operatorname {Isom}\!\left (\operatorname {Sol}_{m, n}^4, \langle \cdot ,\cdot \rangle _1\right ) &\cong & \operatorname {Sol}_{m, n}^4 \rtimes (\mathbb{Z}_2)^3\\[5pt] \operatorname {Isom}\!\left (\operatorname {Sol}_{m, n}^4, \langle \cdot ,\cdot \rangle _2\right ) &\cong & \operatorname {Sol}_{m, n}^4 \rtimes (\mathbb{Z}_2)^2\\[5pt] \operatorname {Isom}\!\left (\operatorname {Sol}_{m, n}^4, \langle \cdot ,\cdot \rangle _3\right ) &\cong & \operatorname {Sol}_{m, n}^4 \rtimes \mathbb{Z}_2. \end{eqnarray*}
Answer. The answer to Question4.3 is affirmative. Note that
$(\mathbb{Z}_2)^3$
has seven distinct subgroups of order
$2$
and seven distinct subgroups of order
$4$
; however, these are different subgroups inside
$(\mathbb{Z}_2)^3$
, even though they are all isomorphic to
$\mathbb{Z}_2$
and
$(\mathbb{Z}_2)^2$
, respectively. We emphasize that it is these abstract group types that occur as isotropy groups of left-invariant Riemannian metrics.
4.2.2. Isometry group of
$\operatorname {Sol}^3 \times \mathbb{R}$
We denote by
$\mathbf{D}(4)$
the dihedral group of order
$8$
.
Theorem 4.9.
The group of isometric automorphisms of
$\operatorname {Sol}^3 \times \mathbb{R}$
is given by
\begin{eqnarray*} \operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _1} &\cong & \mathbf{D}(4) \times \mathbb{Z}_{2}\\[5pt] \operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _2} &\cong & (\mathbb{Z}_2)^2\\[5pt] \operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _3} &\cong & \mathbb{Z}_2. \end{eqnarray*}
Proof. The group of automorphisms of
$\operatorname {Sol}^3 \times \mathbb{R}$
consists of all elements of
$\operatorname {GL}\!(4, \mathbb{R})$
of the form [Reference Van Thuong21]
\begin{equation*}\begin{bmatrix} e & & & 0 & & & 0 & & & x\\[3pt] 0 & & & a & & & 0 & & & y\\[3pt] 0 & & & 0 & & & d & & & z\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \quad \text{or}\quad \begin{bmatrix} e & & & 0 & & & 0 & & & x\\[3pt] 0 & & & 0 & & & b & & & y\\[3pt] 0 & & & c & & & 0 & & & z\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \quad a, b, c, d, e \in \mathbb{R}^{\ast }, x, y, z \in \mathbb{R}.\end{equation*}
Let
$A = \begin{bmatrix} e & & & 0 & & & 0 & & & x\\[3pt] 0 & & & a & & & 0 & & & y\\[3pt] 0 & & & 0 & & & d & & & z\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol}^3 \times \mathbb{R})$
. Then by (3)
\begin{align*} & \;\,\qquad\qquad\qquad\qquad\quad A \in \operatorname {Aut}\!(\!\operatorname {Sol}^3 \times \mathbb{R})_{\langle \cdot ,\cdot \rangle _1}\\[3pt] \Leftrightarrow & \;\, \qquad\qquad\qquad\qquad\qquad A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[3pt] \Leftrightarrow & \quad \begin{bmatrix} e & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & a & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & d & & & 0\\[3pt] x & & & y & & & z & & & 1 \end{bmatrix}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\begin{bmatrix} e & & & 0 & & & 0 & & & x\\[3pt] 0 & & & a & & & 0 & & & y\\[3pt] 0 & & & 0 & & & d & & & z\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix} \end{align*}
\begin{align*} \Leftrightarrow & \qquad\qquad\qquad \begin{bmatrix} e^{2} & & & 0 & & & 0 & & & e x\\[5pt] 0 & & & a^{2} & & & 0 & & & a y\\[5pt] 0 & & & 0 & & & d^{2} & & & d z\\[5pt] e x & & & a y & & & d z & & & x^{2}+y^{2}+z^{2}+\dfrac {1}{\mu ^{2}} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & \qquad\qquad\qquad\quad \;\,x = y = z = 0, \quad e = \pm 1, \quad a = \pm 1, \quad d = \pm 1\\[5pt] \Leftrightarrow & \qquad\qquad\qquad\qquad\quad \;\,A \in \left \lbrace \begin{bmatrix} \pm 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & \pm 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & \pm 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace \cong (\mathbb{Z}_2)^3. \end{align*}
Let
$A = \begin{bmatrix} e & & & 0 & & & 0 & & & x\\[5pt] 0 & & & 0 & & & b & & & y\\[5pt] 0 & & & c & & & 0 & & & z\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol}^3 \times \mathbb{R})$
. Then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!(\!\operatorname {Sol}^3 \times \mathbb{R})_{\langle \cdot ,\cdot \rangle _1}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[5pt] \Leftrightarrow & \begin{bmatrix} e & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & c & & & 0\\[5pt] 0 & & & b & & & 0 & & & 0\\[5pt] x & & & y & & & z & & & -1 \end{bmatrix}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\begin{bmatrix} e & & & 0 & & & 0 & & & x\\[5pt] 0 & & & 0 & & & b & & & y\\[5pt] 0 & & & c & & & 0 & & & z\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} e^{2} & & & 0 & & & 0 & & & e x\\[5pt] 0 & & & c^{2} & & & 0 & & & c z\\[5pt] 0 & & & 0 & & & b^{2} & & & b y\\[5pt] e x & & & c z & & & b y & & & x^{2}+y^{2}+z^{2}+\dfrac {1}{\mu ^{2}} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & x = y = z = 0, \quad e = \pm 1, \quad b = \pm 1, \quad c = \pm 1\\[5pt] \Leftrightarrow & A \in \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace \\[5pt] & \cup \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace \end{eqnarray*}
Hence,
\begin{eqnarray*} & \operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _1} \\[2pt] &= (\mathbb{Z}_2)^3 \cup \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace \\[2pt] & \qquad \cup \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace \end{eqnarray*}
Therefore,
$\operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _1}$
is a group of order 16 such that:
The following
$11$
elements are of order
$2$
\begin{equation*}\hspace {1cm} \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \end{equation*}
\begin{equation*}\hspace {1.3cm}\begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \end{equation*}
\begin{equation*}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}.\end{equation*}
The following
$4$
elements are of order
$4$
\begin{equation*}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 1 & & & 0\\[2pt] 0 & & & -1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & -1 & & & 0\\[2pt] 0 & & & 1 & & & 0 & & & 0\\[2pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}.\end{equation*}
In the classification of groups of order 16, the group that contains 11 elements of order 2 and 4 elements of order 4 is exactly
$\mathbf{D}_4 \times \mathbb{Z}_2$
. Thus,
\begin{equation*}\operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _1} \cong \mathbf{D}(4) \times \mathbb{Z}_2.\end{equation*}
The maximal group of isometric automorphisms of
$\operatorname {Sol}^3 \times \mathbb{R}$
is
$\mathbf{D}(4) \times \mathbb{Z}_2$
, whose elements are listed above. To determine the group of isometric automorphisms corresponding to the metric
$\langle \cdot ,\cdot \rangle _2$
, it suffices to check which elements of
$\mathbf{D}(4) \times \mathbb{Z}_2$
preserve this metric. A straightforward verification shows that
\begin{equation*}\operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _2} = \operatorname {Aut}\!(\!\operatorname {Sol}_{m, n}^4\!)_{\langle \cdot ,\cdot \rangle _2} \cong (\mathbb{Z}_2)^2.\end{equation*}
The same for the third metric, we have
\begin{equation*}\operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )_{\langle \cdot ,\cdot \rangle _3} = \operatorname {Aut}\!(\!\operatorname {Sol}_{m, n}^4\!)_{\langle \cdot ,\cdot \rangle _3} \cong \mathbb{Z}_2. \end{equation*}
Corollary 4.10.
The isometry group of
$\operatorname {Sol}^3 \times \mathbb{R}$
is given by
\begin{eqnarray*} \operatorname {Isom}\!\left (\operatorname {Sol}^3 \times \mathbb{R}, \langle \cdot ,\cdot \rangle _1\right ) &\cong & \left (\operatorname {Sol}^3 \times \mathbb{R}\right ) \rtimes \left (\mathbf{D}(4) \times \mathbb{Z}_{2}\right )\\[3pt] \operatorname {Isom}\!\left (\operatorname {Sol}^3 \times \mathbb{R}, \langle \cdot ,\cdot \rangle _2\right ) &\cong & \left (\operatorname {Sol}^3 \times \mathbb{R}\right ) \rtimes (\mathbb{Z}_{2})^2\\[3pt] \operatorname {Isom}\!\left (\operatorname {Sol}^3 \times \mathbb{R}, \langle \cdot ,\cdot \rangle _3\right ) &\cong & \left (\operatorname {Sol}^3 \times \mathbb{R}\right ) \rtimes \mathbb{Z}_{2}. \end{eqnarray*}
Answer. The answer to Question4.3 is negative. For example, the compact subgroup
$\left (\mathbb{Z}_2\right )^3$
of
$\operatorname {Aut}\!\left (\operatorname {Sol}^3 \times \mathbb{R}\right )$
cannot occur as the isotropy subgroup of any left-invariant Riemannian metric.
4.2.3. Isometry group of
$\operatorname {Sol}_0^4$
Theorem 4.11.
The isometry group of
$\operatorname {Sol}_0^4$
is given by
\begin{eqnarray*} \operatorname {Isom}\!\left (\operatorname {Sol}_0^4, \langle \cdot ,\cdot \rangle _1\right ) &\cong & \operatorname {Sol}_0^4 \rtimes \left (\operatorname {diag}\left \lbrace \operatorname {O}(2), \pm 1, 1\right \rbrace \right )\\[3pt] \operatorname {Isom}\!\left (\operatorname {Sol}_0^4, \langle \cdot ,\cdot \rangle _2\right ) &\cong & \operatorname {Sol}_0^4 \rtimes (\mathbb{Z}_2)^2. \end{eqnarray*}
Proof. The automorphism group
$\operatorname {Aut}\!(\!\operatorname {Sol_0^4}\!)$
consists of elements of the form [Reference Van Thuong21]
\begin{equation*}\begin{bmatrix} a & & & b & & & 0 & & & x\\[5pt] c & & & d & & & 0 & & & y\\[5pt] 0 & & & 0 & & & e & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \quad \begin{bmatrix} a & & & b\\[5pt] c & & & d \end{bmatrix} \in \operatorname {GL}\!(2, \mathbb{R}) \quad e \in \mathbb{R}^{\ast } \quad x, y, z \in \mathbb{R}.\end{equation*}
Let
$A = \begin{bmatrix} a & & & b & & & 0 & & & x\\[5pt] c & & & d & & & 0 & & & y\\[5pt] 0 & & & 0 & & & e & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol_0^4}\!)$
, then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!\left (\operatorname {Sol_0^4}\right )_{\langle \cdot ,\cdot \rangle _1}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1 A = \langle \cdot ,\cdot \rangle _1\\[5pt] \Leftrightarrow & \begin{bmatrix} a & & & c & & & 0 & & & 0\\[5pt] b & & & d & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & e & & & 0\\[5pt] x & & & y & & & z & & & 1 \end{bmatrix}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}\begin{bmatrix} a & & & b & & & 0 & & & x\\[5pt] c & & & d & & & 0 & & & y\\[5pt] 0 & & & 0 & & & e & & & z\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} a^{2}+c^{2} & & & ab+cd & & & 0 & & & ax+cy\\[5pt] ab+cd & & & b^{2}+d^{2} & & & 0 & & & bx+dy\\[5pt] 0 & & & 0 & & & e^{2} & & & ez\\[5pt] ax+cy & & & bx+dy & & & ez & & & x^{2}+y^{2}+z^{2}+\dfrac {1}{\beta ^{2}} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}. \end{eqnarray*}
By equating the
$(3,4)$
and
$(4,4)$
entries of the two matrices above, we obtain
\begin{equation*}z = x = y = 0.\end{equation*}
Hence, the matrix equality reduces to
\begin{eqnarray*} & \begin{bmatrix} a^{2}+c^{2} & & & ab+cd & & & 0 & & & 0\\[3pt] ab+cd & & & b^{2}+d^{2} & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & e^{2} & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^{2}} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}\\[3pt] \Leftrightarrow & \begin{bmatrix} a & & & c\\[3pt] b & & & d \end{bmatrix}\begin{bmatrix} a & & & b\\[3pt] c & & & d \end{bmatrix} = I_2, \quad e = \pm 1\\[3pt] \Leftrightarrow & \begin{bmatrix} a & & & b\\[3pt] c & & & d \end{bmatrix} \in \operatorname {O}(2), \quad e = \pm 1. \end{eqnarray*}
Hence
\begin{equation*}\operatorname {Isom}\!\left (\operatorname {Sol}_0^4, \langle \cdot ,\cdot \rangle _1\right ) \cong \operatorname {Sol}_0^4 \rtimes \left (\operatorname {diag}\left \lbrace \operatorname {O}(2), \pm 1, 1\right \rbrace \right ).\end{equation*}
For the metric
$\langle \cdot ,\cdot \rangle _2$
, using (3), one gets that
\begin{eqnarray*} & A \in \operatorname {Aut}\!(\!\operatorname {Sol_0^4}\!)_{\langle \cdot ,\cdot \rangle _2}\\[3pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _2A = \langle \cdot ,\cdot \rangle _2\\[3pt] \Leftrightarrow & \begin{bmatrix} a & & & c & & & 0 & & & 0\\[3pt] b & & & d & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & e & & & 0\\[3pt] x & & & y & & & z & & & 1 \end{bmatrix}\begin{bmatrix} 1 & & & 0 & & & -\alpha & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] -\alpha & & & 0 & & & 1 + \alpha ^2 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}\begin{bmatrix} a & & & b & & & 0 & & & x\\[3pt] c & & & d & & & 0 & & & y\\[3pt] 0 & & & 0 & & & e & & & z\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & -\alpha & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] -\alpha & & & 0 & & & 1 + \alpha ^2 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix}\\[3pt] \Leftrightarrow & {\begin{bmatrix} a^{2}+c^{2} & & & ab+cd & & & -a\alpha e & & & -a\alpha z+ax+cy\\[3pt] ab+cd & & & b^{2}+d^{2} & & & -b\alpha e & & & -b\alpha z+bx+dy\\[3pt] -a\alpha e & & & -b\alpha e & & & e^{2}(\alpha ^{2}+1) & & & -e\alpha x+e(\alpha ^{2}+1)z\\[3pt] (\!-\!z\alpha +x)a+cy & & & (\!-\!z\alpha +x)b+dy & & & (\!-\!x\alpha +z(\alpha ^{2}+1))e & & & (\!-\!z\alpha +x)x+y^{2}+(\!-\!x\alpha +z(\alpha ^{2}+1))z+\dfrac {1}{\beta ^{2}} \end{bmatrix}}\\[3pt] & = \begin{bmatrix} 1 & & & 0 & & & -\alpha & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] -\alpha & & & 0 & & & 1 + \alpha ^2 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\beta ^2} \end{bmatrix} \end{eqnarray*}
By equating the
$(3,3)$
and
$(3,1)$
entries of the two matrices above, we obtain
$e = \pm 1$
and
$ae = 1$
. Since
$\alpha \gt 0$
and
$e \neq 0$
, equating the
$(3,2)$
entries yields
$b = 0$
. Next, the
$(2,2)$
entries imply that
$d = \pm 1$
. The
$(2,1)$
entries then imply that
$c = 0$
, while the
$(4,2)$
entries give
$y = 0$
. Finally, the
$(4,1)$
and
$(4,3)$
entries show that
$x$
and
$z$
satisfy the following system:
\begin{equation*}\left \lbrace \begin{array}{l@{\quad}l@{\quad}l} x - \alpha z = 0\\[3pt] -\alpha x + \left (1 + \alpha ^2\right )z = 0 \end{array}\right .\end{equation*}
This implies that
$x = z = 0$
. Hence,
\begin{eqnarray*} \operatorname {Aut}\!(\!\operatorname {Sol_0^4}\!)_{\langle ., .\rangle _2} &=& \left \lbrace I_4, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace \\[3pt] &\cong & (\mathbb{Z}_2)^2. \end{eqnarray*}
Consequently,
\begin{equation*}\operatorname {Isom}\!\left (\operatorname {Sol}_0^4, \langle \cdot ,\cdot \rangle _2\right ) \cong \operatorname {Sol}_0^4 \rtimes (\mathbb{Z}_2)^2. \end{equation*}
Answer. The answer to Question4.3 is negative. For example, the compact subgroup
$\mathbb{Z}_2$
of
$\operatorname {Aut}\!(\!\operatorname {Sol}_0^4)$
cannot occur as the isotropy subgroup of any left-invariant Riemannian metric.
4.2.4. Isometry group of
$\operatorname {Sol}_0^{' 4}$
Theorem 4.12.
The isometry group of
$\operatorname {Sol}_0^{' 4}$
is given by
\begin{eqnarray*} \operatorname {Isom}\!\big(\!\operatorname {Sol}_0^{' 4}, \langle \cdot ,\cdot \rangle _1\big) &\cong & \operatorname {Sol}_0^{' 4} \rtimes (\mathbb{Z}_2)^2\\[3pt] \operatorname {Isom}\!\big(\!\operatorname {Sol}_0^{' 4}, \langle \cdot ,\cdot \rangle _2\big) &\cong & \operatorname {Sol}_0^{' 4} \rtimes \mathbb{Z}_2. \end{eqnarray*}
Proof. Let
$A = \begin{bmatrix} a & & & b & & & 0 & & & x\\[3pt] 0 & & & a & & & 0 & & & y\\[3pt] 0 & & & 0 & & & e & & & z\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol}_0^{' 4}\!)$
(see [Reference Van Thuong21]). Then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!\big(\!\operatorname {Sol}_0^{' 4}\!\big)_{\langle \cdot ,\cdot \rangle _1}\\[3pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[3pt] \Leftrightarrow & \begin{bmatrix} a & & & 0 & & & 0 & & & 0\\[3pt] b & & & a & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & e & & & 0\\[3pt] x & & & y & & & z & & & 1 \end{bmatrix}\begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & \dfrac {1}{\alpha ^2} & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\begin{bmatrix} a & & & b & & & 0 & & & x\\[3pt] 0 & & & a & & & 0 & & & y\\[3pt] 0 & & & 0 & & & e & & & z\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & \dfrac {1}{\alpha ^2} & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[3pt] \Leftrightarrow & \begin{bmatrix} a^{2} & & & ab & & & 0 & & & ax\\[3pt] ab & & & b^{2}+\dfrac {a^{2}}{\alpha ^{2}} & & & 0 & & & bx+\dfrac {ay}{\alpha ^{2}}\\[3pt] 0 & & & 0 & & & e^{2} & & & ez\\[3pt] ax & & & bx+\dfrac {ay}{\alpha ^{2}} & & & ez & & & x^{2}+\dfrac {y^{2}}{\alpha ^{2}}+z^{2}+\dfrac {1}{\mu ^{2}} \end{bmatrix} = \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & \dfrac {1}{\alpha ^2} & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\mu ^2} \end{bmatrix}\\[3pt] \Leftrightarrow & x = y = z = b = 0, \quad e = \pm 1, \quad a = \pm 1. \end{eqnarray*}
Thus,
\begin{equation*}\operatorname {Aut}\!(\!\operatorname {Sol}_0^{' 4})_{\langle ., .\rangle _1} = \left \lbrace I_4, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace .\end{equation*}
This is exactly the Klein 4-group, thus
$\operatorname {Aut}\!(\!\operatorname {Sol}_0^{' 4}\!)_{\langle \cdot ,\cdot \rangle _1} \cong (\mathbb{Z}_2)^2.$
Hence,
\begin{equation*}\operatorname {Isom}\!\big(\!\operatorname {Sol}_0^{' 4}, \langle \cdot ,\cdot \rangle _1\big) \cong \operatorname {Sol}_0^{' 4} \rtimes (\mathbb{Z}_2)^2.\end{equation*}
For the metric
$\langle \cdot ,\cdot \rangle _2$
, it suffices to determine which of the elements of
$(\mathbb{Z}_2)^2$
listed above preserve this metric. We obtain that
\begin{eqnarray*} A \in \operatorname {Aut}\!\big(\!\operatorname {Sol}_0^{' 4}\big)_{\langle \cdot ,\cdot \rangle _2} &\Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _2A = \langle \cdot ,\cdot \rangle _2\\[3pt] &\Leftrightarrow & A \in \left \lbrace I_4, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace \cong \mathbb{Z}_2. \end{eqnarray*}
Therefore,
\begin{equation*}\operatorname {Isom}\!\big(\!\operatorname {Sol}_0^{' 4}, \langle \cdot ,\cdot \rangle _2\big) \cong \operatorname {Sol}_0^{' 4} \rtimes \mathbb{Z}_2.\end{equation*}
Answer. The answer to Question4.3 is affirmative. Note that
$(\mathbb{Z}_2)^2$
has three distinct subgroups of order
$2$
; however, these are different subgroups inside
$(\mathbb{Z}_2)^2$
, even though they are all isomorphic to
$\mathbb{Z}_2$
. We emphasize that it is these abstract group types that occur as isotropy groups of left-invariant Riemannian metrics.
4.2.5. Isometry group of
$\operatorname {Sol}_1^4$
Theorem 4.13.
The group of isometric automorphisms of
$\operatorname {Sol}_1^4$
is given by
\begin{eqnarray*} \operatorname {Aut}\!\big(\!\operatorname {Sol}_1^4\big)_{\langle \cdot ,\cdot \rangle _1} &\cong & \mathbf{D}(4)\\[3pt] \operatorname {Aut}\!\big(\!\operatorname {Sol}_1^4\big)_{\langle \cdot ,\cdot \rangle _2} &\cong & \mathbb{Z}_2\\[3pt] \operatorname {Aut}\!\big(\!\operatorname {Sol}_1^4\big)_{\langle \cdot ,\cdot \rangle _3} &=& \{I_4\}. \end{eqnarray*}
Proof. The group of automorphisms of
$\operatorname {Sol}_1^4$
consists of all elements of
$\operatorname {GL}\!(4, \mathbb{R})$
of the form [Reference Van Thuong21]
\begin{equation*}\begin{bmatrix} ad & & & -aq & & & -dp & & & x\\[3pt] 0 & & & a & & & 0 & & & p\\[3pt] 0 & & & 0 & & & d & & & q\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \quad \text{or}\quad \begin{bmatrix} -bc & & & cp & & & bq & & & x\\[3pt] 0 & & & 0 & & & b & & & p\\[3pt] 0 & & & c & & & 0 & & & q\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \quad a, b, c, d \in \mathbb{R}^{\ast }, x, p, q \in \mathbb{R}.\end{equation*}
Let
$A = \begin{bmatrix} ad & & & -aq & & & -dp & & & x\\[3pt] 0 & & & a & & & 0 & & & p\\[3pt] 0 & & & 0 & & & d & & & q\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol}_1^4)$
. Then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!(\!\operatorname {Sol}_1^4)_{\langle \cdot ,\cdot \rangle _1}\\[3pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[3pt] \Leftrightarrow & \begin{bmatrix} ad & & & 0 & & & 0 & & & 0\\[3pt] -aq & & & a & & & 0 & & & 0\\[3pt] -dp & & & 0 & & & d & & & 0\\[3pt] x & & & p & & & q & & & 1 \end{bmatrix}\begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\begin{bmatrix} ad & & & -aq & & & -dp & & & x\\[3pt] 0 & & & a & & & 0 & & & p\\[3pt] 0 & & & 0 & & & d & & & q\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix} \end{eqnarray*}
\begin{eqnarray*} \Leftrightarrow & \begin{bmatrix} \dfrac {a^{2}d^{2}}{\alpha ^{2}} & & & -\dfrac {a^{2}dq}{\alpha ^{2}} & & & -\dfrac {a\,d^{2}p}{\alpha ^{2}} & & & \dfrac {adx}{\alpha ^{2}}\\[7pt] -\dfrac {a^{2}dq}{\alpha ^{2}} & & & \dfrac {a^{2}q^{2}}{\alpha ^{2}}+a^{2} & & & \dfrac {aqdp}{\alpha ^{2}} & & & -\dfrac {aqx}{\alpha ^{2}}+ap\\[7pt] -\dfrac {a\,d^{2}p}{\alpha ^{2}} & & & \dfrac {aqdp}{\alpha ^{2}} & & & \dfrac {d^{2}p^{2}}{\alpha ^{2}}+d^{2} & & & -\dfrac {dpx}{\alpha ^{2}}+dq\\[7pt] \dfrac {adx}{\alpha ^{2}} & & & -\dfrac {aqx}{\alpha ^{2}}+ap & & & -\dfrac {dpx}{\alpha ^{2}}+dq & & & \dfrac {x^{2}}{\alpha ^{2}}+p^{2}+q^{2}+\dfrac {1}{\nu ^{2}} \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[7pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & x = p = q = 0, \quad a = \pm 1, \quad d = \pm 1\\[5pt] \Leftrightarrow & A \in \left \lbrace I_4, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace . \end{eqnarray*}
Let
$A = \begin{bmatrix} -bc & & & cp & & & bq & & & x\\[5pt] 0 & & & 0 & & & b & & & p\\[5pt] 0 & & & c & & & 0 & & & q\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \in \operatorname {Aut}\!(\!\operatorname {Sol}_1^4)$
. Then by (3)
\begin{eqnarray*} & A \in \operatorname {Aut}\!(\!\operatorname {Sol}_1^4)_{\langle \cdot ,\cdot \rangle _1}\\[5pt] \Leftrightarrow & A^t\langle \cdot ,\cdot \rangle _1A = \langle \cdot ,\cdot \rangle _1\\[5pt] \Leftrightarrow & \begin{bmatrix} -bc & & & 0 & & & 0 & & & 0\\[5pt] cp & & & 0 & & & c & & & 0\\[5pt] bq & & & b & & & 0 & & & 0\\[5pt] x & & & p & & & q & & & -1 \end{bmatrix}\begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\begin{bmatrix} -bc & & & cp & & & bq & & & x\\[5pt] 0 & & & 0 & & & b & & & p\\[5pt] 0 & & & c & & & 0 & & & q\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & \begin{bmatrix} \dfrac {b^{2}c^{2}}{\alpha ^{2}} & & & -\dfrac {b\,c^{2}p}{\alpha ^{2}} & & & -\dfrac {b^{2}cq}{\alpha ^{2}} & & & -\dfrac {bcx}{\alpha ^{2}}\\[7pt] -\dfrac {b\,c^{2}p}{\alpha ^{2}} & & & \dfrac {c^{2}p^{2}}{\alpha ^{2}}+c^{2} & & & \dfrac {cpbq}{\alpha ^{2}} & & & \dfrac {cpx}{\alpha ^{2}}+cq\\[7pt] -\dfrac {b^{2}cq}{\alpha ^{2}} & & & \dfrac {cpbq}{\alpha ^{2}} & & & \dfrac {b^{2}q^{2}}{\alpha ^{2}}+b^{2} & & & \dfrac {bqx}{\alpha ^{2}}+bp\\[7pt] -\dfrac {bcx}{\alpha ^{2}} & & & \dfrac {cpx}{\alpha ^{2}}+cq & & & \dfrac {bqx}{\alpha ^{2}}+bp & & & \dfrac {x^{2}}{\alpha ^{2}}+p^{2}+q^{2}+\dfrac {1}{\nu ^{2}} \end{bmatrix} = \begin{bmatrix} \dfrac {1}{\alpha ^2} & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & \dfrac {1}{\nu ^2} \end{bmatrix}\\[5pt] \Leftrightarrow & x = p = q = 0, \quad b = \pm 1, \quad c = \pm 1\\[5pt] \Leftrightarrow & A \in \left \lbrace \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & 1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 1 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & -1 & & & 0\\[5pt] 0 & & & -1 & & & 0 & & & 0\\[5pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace . \end{eqnarray*}
Consequently:
\begin{eqnarray*} & \operatorname {Aut}\!\big(\!\operatorname {Sol}_1^4\big)_{\langle \cdot ,\cdot \rangle _1}\\[3pt] = & \left \lbrace \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix} \right \rbrace \\[3pt] & \cup \left \lbrace \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 1 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix}, \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & -1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right \rbrace \\[3pt] = & \left\langle \begin{bmatrix} -1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & 1 \end{bmatrix}, \begin{bmatrix} 1 & & & 0 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & -1 & & & 0\\[3pt] 0 & & & 1 & & & 0 & & & 0\\[3pt] 0 & & & 0 & & & 0 & & & -1 \end{bmatrix} \right\rangle \cong \mathbf{D}(4). \end{eqnarray*}
Therefore,
$\operatorname {Aut}\!(\!\operatorname {Sol}_1^4)_{\langle \cdot ,\cdot \rangle _1} \cong \mathbf{D}(4)$
.
The maximal group of isometric automorphisms of
$\operatorname {Sol}_1^4$
is
$\mathbf{D}(4)$
, whose elements are listed above. To determine the group of isometric automorphisms corresponding to the metric
$\langle \cdot ,\cdot \rangle _2$
, it suffices to check which elements of
$\mathbf{D}(4)$
preserve this metric. A straightforward verification shows that
\begin{equation*}\operatorname {Aut}\!\big(\!\operatorname {Sol}_1^4\big)_{\langle \cdot ,\cdot \rangle _2} = \left \lbrace I_4, \operatorname {diag}\{-1, -1, 1, 1\}\right \rbrace \cong \mathbb{Z}_2.\end{equation*}
The same for the third metric, we obtain
$\operatorname {Aut}\!(\!\operatorname {Sol}_1^4)_{\langle \cdot ,\cdot \rangle _3} = \{I_4\}.$
Corollary 4.14.
The isometry group of
$\operatorname {Sol}_1^4$
is given by
\begin{eqnarray*} \operatorname {Isom}\!\big(\!\operatorname {Sol}_1^4, \langle \cdot ,\cdot \rangle _1\big) &\cong & \operatorname {Sol}_1^4 \rtimes \mathbf{D}(4)\\[3pt] \operatorname {Isom}\!\big(\!\operatorname {Sol}_1^4, \langle \cdot ,\cdot \rangle _2\big) &\cong & \operatorname {Sol}_1^4 \rtimes \mathbb{Z}_{2}\\[3pt] \operatorname {Isom}\!\big(\!\operatorname {Sol}_1^4, \langle \cdot ,\cdot \rangle _3\big) &\cong & \operatorname {Sol}_1^4. \end{eqnarray*}
Answer. The answer to Question4.3 is negative. For example, the compact subgroup
$\left (\mathbb{Z}_2\right )^2$
of
$\operatorname {Aut}\!(\!\operatorname {Sol}_1^4)$
cannot occur as the isotropy subgroup of any left-invariant Riemannian metric.
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