On the Classification of Totally Geodesic and Parallel Hypersurfaces of the Lie Group Nil⁴

Abstract

This work establishes a complete algebraic classification of hypersurfaces with totally symmetric cubic form, including the Codazzi, parallel, and totally geodesic cases, on the 4-dimensional 3-step nilpotent Lie group $Ni{l}^4$ endowed with six left-invariant Lorentzian metrics. Combined with prior results, we achieve a complete classification of such hypersurfaces on 4-dimensional nilpotent Lie groups. The core of our approach lies in the explicit derivation and solution of the Codazzi tensor equations, which directly leads to the construction of these hypersurfaces and provides their explicit parametrizations. Our main results establish the existence of Codazzi hypersurfaces on $Ni{l}^4$, demonstrate the non-existence of totally geodesic hypersurfaces, specify the algebraic condition for a Codazzi hypersurface to become parallel, and provide their explicit parametrizations. This observation highlights fundamental differences between Lorentzian and Riemannian settings within hypersurface theory. This work thus clarifies the distinct geometric properties inherent to the Lorentzian cases on nilpotent Lie groups.

1. Introduction

Parallel submanifolds represent a fundamental class of submanifolds in differential geometry. They are natural generalizations of totally geodesic submanifolds and are of significant interest because their extrinsic invariants do not vary from point to point [1]. The study of these structures is not only central to geometric analysis but also finds relevance in physical theories.

While the classification of parallel submanifolds is well-established in Riemannian symmetric spaces [2,3], the situation becomes more complex and less explored for general homogeneous spaces, particularly in the pseudo-Riemannian setting. Some examples of parallel hypersurfaces in the pseudo-Riemannian setting can be found in [4,5,6,7]. The extension of our study to 4-dimensional Lorentzian manifolds is motivated by their direct relevance to theoretical physics. Furthermore, there are some foundational works including characterizations of parallel hypersurfaces in specific four-manifolds [8,9,10,11,12,13,14,15].

As shown by Magnin [16], there are precisely two non-Abelian, nilpotent Lie groups of dimension 4: $Ni{l}^4$ and $H_3\times \mathbb{R}$. There are many studies on geometric properties of the 4-dimensional nilpotent Lie groups, such as Ricci solitons [17], generalized Ricci solitons [18], homogeneous Lorentzian structure [19,20], Douglas $(\alpha ,\beta )$-metrics [21], and isometry groups [22]. In this paper, we consider $Ni{l}^4$ equipped with its left-invariant Lorentzian metrics. Our focus on this group is motivated by its algebraic structure, which is further from being Abelian compared to 2-step groups, presenting a richer and more challenging classification problem. Together with the recently solved case of $H_3\times \mathbb{R}$ [13], the investigation of $Ni{l}^4$ in this work thus establishes a full classification of parallel hypersurfaces for all four-dimensional nilpotent Lie groups. Our main results are shown below.

Based on the data presented in

Table 1. The main result.

Group	Normal Vector Field	Parallel	In Total, Geodesic
$(Ni{l}^4,g_A)$	$\xi =\pm e_1$	✓	×
$(Ni{l}^4,g_A^{\pm })$	$\xi =\pm e_1$	✓	×
$(Ni{l}^4,g_1^{\lambda })$	$\xi =\pm e_1+a_4{e}_4$	✓	×
$(Ni{l}^4,g_1^{\lambda })$	$\xi =\pm {\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$	✓	×
$(Ni{l}^4,g_2^{\lambda })$	$\xi =\pm e_2+a_4{e}_4$	✓	×
$(Ni{l}^4,g_2^{\lambda })$	$\xi =\pm {\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$	✓	×
$(Ni{l}^4,g_3^{\lambda })$	×	×	×
$(Ni{l}^4,g_4^{\lambda })$	$\xi =\pm e_1$	✓	×

regarding parallel hypersurfaces of four-dimensional $Ni{l}^4$, we formulate the following theorem.

Theorem 1.

Let $F:M\to (Ni{l}^4,g)$ be a parallel hypersurface of 4-dimensional 3-step nilpotent Lie group $Ni{l}^4$, where the metric g corresponds to one of the metrics in (2). Consider the coordinate (1) on $Ni{l}^4$. Then, there exist local coordinates $(u_1,u_2,u_3)$ on M such that, the immersion takes one of the following expression under these coordinates.

1.

$F(u_1,u_2,u_3)=(0,u_1,u_2,u_3)$;

2.

$F(u_1,u_2,u_3)=(H_1^{\prime }(u_1),-u_1,H_1(u_1)+u_2,u_3)$;

3.

$F(u_1,u_2,u_3)=(u_1,\sqrt{\lambda }{u}_2,H_1(u_2)+u_1{u}_2,u_3)$;

4.

$F(u_1,u_2,u_3)=(u_1,{\displaystyle \frac{1}{6}}{u}_1^3-{\displaystyle \frac{1}{2}}C{u}_1^2-D{u}_1,u_2,u_3)$;

5.

$F(u_1,u_2,u_3)=(\sqrt{\lambda }{u}_2,u_1,\sqrt{\lambda }{u}_1{u}_2+E(u_2),u_3)$,

where in case (2), $H_1(u_1)=a_{1}cos(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1)+a_{2}sin(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1)$ and $a_1,a_2\in \mathbb{R}$; in case (3), $G_1(u_2)={\displaystyle \frac{k_1}{1+\lambda }}(k_{1}sin(1+\lambda )u_2-k_{2}cos(1+\lambda )u_2)$, and $k_1,k_2\in \mathbb{R}$; in case (4), $C,D\in \mathbb{R}$; in case (5), $E(u_2)={\displaystyle \frac{1}{\lambda }}(k_1{e}^{\lambda u_2}-k_2{e}^{-\lambda u_2})$, and $k_1,k_2\in \mathbb{R}$.

The paper structure is outlined as follows. Section 2 reviews necessary concepts and definitions concerning 4-dimensional 3-step nilpotent Lie group $Ni{l}^4$ and these hypersurface types. Section 3 presents a complete algebraic classification of hypersurfaces with totally symmetric cubic forms for every left-invariant Lorentzian metric. Section 4 provides an explicit expression of the immersion of the parallel hypersurfaces for the $Ni{l}^4$.

2. Preliminary

2.1. On the Geometry Property of $Ni{l}^4$

As both a canonical nilpotent Lie group and one of Thurston’s model geometries, $Ni{l}^4$ is structured as the semidirect product ${\mathbb{R}}^3{⋉}_U\mathbb{R}$, where the one-parameter subgroup $U(x)=exp(xL)$ is generated by a specific nilpotent matrix L:

$$L=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right),exp(tL)=I_3+xL+{\displaystyle \frac{x^2}{2}}{L}^2=\left(\begin{array}{ccc}1& x& {\displaystyle \frac{x^2}{2}}\\ 0& 1& x\\ 0& 0& 1\end{array}\right)$$

The Lie algebra ${\mathfrak{nil}}^4$ associated with the four-dimensional nilpotent Lie group $Ni{l}^4$ is a 3-step nilpotent algebra of dimension four [16]. Its basis $\{E_1,E_2,E_3,E_4\}$ satisfies non-zero brackets:

$$[E_1,E_2]=E_3,[E_1,E_3]=E_4.$$

This structure exhibits a one-dimensional center $Z({\mathfrak{nil}}^4)=span\left\{E_4\right\}$. The left-invariant vector fields $\{e_1,e_2,e_3,e_4\}$ of $Ni{l}^4$ corresponding to the basis elements. In global coordinate system $(x,y,z,w)$, they admit the explicit realizations:

$$e_1={\displaystyle \frac{\partial }{\partial x}},e_2={\displaystyle \frac{\partial }{\partial y}}+x{\displaystyle \frac{\partial }{\partial z}}+{\displaystyle \frac{x^2}{2}}{\displaystyle \frac{\partial }{\partial w}},e_3={\displaystyle \frac{\partial }{\partial z}}+x{\displaystyle \frac{\partial }{\partial w}},e_4={\displaystyle \frac{\partial }{\partial w}}.$$

(1)

The coframe ${\left\{{\omega }_i\right\}}_{i=1}^4$ corresponding to these vector fields can be expressed as the differential forms:

$${\omega }_1=dx,{\omega }_2=dy,{\omega }_3=dz-xdy,{\omega }_4={\displaystyle \frac{x^2}{2}}dy-xdz+dw.$$

Bokan, Šukilović, and Vukmirović classified left-invariant Lorentzian metrics on $Ni{l}^4$ in [23]. The result is summarized as follows.

Theorem 2

([23]). Left-invariant Lorentzian metrics on $Ni{l}^4$ belong to one of six types. These metrics are defined as follows:

$$\begin{array}{cc}\hfill g_A^{\pm }& =\pm {\omega }_1\otimes {\omega }_1\mp {\omega }_2\otimes {\omega }_2+a{\omega }_3\otimes {\omega }_3+2b{\omega }_3\otimes {\omega }_4+c{\omega }_4\otimes {\omega }_4,\hfill \\ \hfill g_A& ={\omega }_1\otimes {\omega }_1+{\omega }_2\otimes {\omega }_2+a{\omega }_3\otimes {\omega }_3+2b{\omega }_3\otimes {\omega }_4+c{\omega }_4\otimes {\omega }_4,\hfill \\ \hfill g_1^{\lambda }& ={\omega }_1\otimes {\omega }_1+2{\omega }_2\otimes {\omega }_4+\lambda {\omega }_3\otimes {\omega }_3,\hfill \\ \hfill g_2^{\lambda }& ={\omega }_2\otimes {\omega }_2+2{\omega }_1\otimes {\omega }_4+\lambda {\omega }_3\otimes {\omega }_3,\hfill \\ \hfill g_3^{\lambda }& ={\omega }_2\otimes {\omega }_2+2{\omega }_1\otimes {\omega }_3+\lambda {\omega }_4\otimes {\omega }_4,\hfill \\ \hfill g_4^{\lambda }& ={\omega }_1\otimes {\omega }_1+2{\omega }_2\otimes {\omega }_3+\lambda {\omega }_4\otimes {\omega }_4.\hfill \end{array}$$

(2)

Here, $a,b,c$ denote constants with $b\ge 0$ and $\lambda >0$. For the metrics $g_A^{\pm }$, $ac-b^2>0$, whereas for $g_A$, $ac-b^2<0$.

2.2. Introduction to in Total, Geodesic and Parallel Hypersurface

We consider an isometric immersion $F:M\to N$ of pseudo-Riemannian manifolds. The Levi-Civita connections on M and N are denoted ${\nabla }^M$ and ∇, respectively. Let $\xi$ be a unit normal vector field on the hypersurface M. The shape operator S is defined by $S(X)=-{\nabla }_X\xi$ for tangent vector fields X on M. For tangent vector fields X, Y on M, ${\nabla }_{X}Y$ tangent to M is identified as ${\nabla }_X^{M}Y$. Then, the well-known Gauss formula decomposes the ambient connection as:

$${\nabla }_{X}Y={\nabla }_X^{M}Y+h(X,Y)\xi ,$$

(3)

where h is the second fundamental form, which is a symmetric $(0,2)$-tensor field on M measuring the deviation of M from being totally geodesic. A hypersurface is totally geodesic if h vanishes identically, meaning every geodesic in M is also a geodesic in N. Let $R^M$ and R denote the Riemann-Christoffel curvature tensors of M and N, respectively. The Codazzi equation is:

$$\langle R(X,Y)Z,\xi \rangle =({\nabla }^{M}h)(X,Y,Z)-({\nabla }^{M}h)(Y,X,Z),$$

(4)

for tangent vectors fields $X,Y,Z$ on M. Here, ${\nabla }^{M}h$ is defined as:

$$({\nabla }^{M}h)(X,Y,Z)=X(h(Y,Z))-h({\nabla }_X^{M}Y,Z)-h(Y,{\nabla }_X^{M}Z).$$

A hypersurface is called a Codazzi hypersurface if its second fundamental form h satisfies the Codazzi equation: $({\nabla }_X^{M}h)(Y,Z)=({\nabla }_Y^{M}h)(X,Z)$ for all tangent vectors $X,Y,Z$. This is equivalent to the covariant derivative $\nabla h$ being totally symmetric in its arguments. In total, geodesic hypersurfaces are characterized by the vanishing of the second fundamental form ($h\equiv 0$). We conclude the above relationship as follows:

$$\mathrm{In}\mathrm{total},\mathrm{geodesic}(h\equiv 0)⊊\mathrm{Parallel}(\nabla h=0)⊊\mathrm{Codazzi}(\nabla h\mathrm{symmetric}).$$

3. Hypersurfaces of ${\mathit{Nil}}^{\mathbf{4}}$ with in Total, Symmetry Form

Consider an immersion $F:M\to (Ni{l}^4,g)$ where g is one of the metrics in (2). Let $\xi$ be a unit normal vector field along M. Since parallel and totally geodesic hypersurfaces are special cases of Codazzi hypersurfaces (having totally symmetric cubic form), we apply the Codazzi Equation (4) to derive the following Theorems. These Theorems impose necessary algebraic conditions on the components of $\xi$ for hypersurfaces to admit such symmetric cubic forms.

Theorem 3.

Consider a hypersurface immersion $F:M\to (Ni{l}^4,g_A)$ with a totally symmetric form. Suppose ξ is a unit timelike or spacelike normal vector field on M. For each point in M, there exists a neighborhood $U\subseteq M$ such that $\xi =\pm {\tilde{e}}_1$.

Proof. 

With respect to the basis in (2), we construct an orthonormal frame for $Ni{l}^4$ equipped with the $g_A$ in (2). The orthonormal vector fields with respect to the coordinates in (1) are defined as follows:

$${\tilde{e}}_1=e_1,{\tilde{e}}_2=e_2,{\tilde{e}}_3={\displaystyle \frac{e_3}{\sqrt{a}}},{\tilde{e}}_4={\displaystyle \frac{(e_4-\frac{b}{a}{e}_3)}{\sqrt{ac-b^2}}}.$$

This frame satisfies the non-vanishing Lie bracket relations:

$$[{\tilde{e}}_1,{\tilde{e}}_2]=\sqrt{a}{\tilde{e}}_3,[{\tilde{e}}_1,{\tilde{e}}_3]={\displaystyle \frac{\lambda }{a}}{\tilde{e}}_4+{\displaystyle \frac{b}{a}}{\tilde{e}}_3,[{\tilde{e}}_1,{\tilde{e}}_4]=-{\displaystyle \frac{b}{a}}({\tilde{e}}_4+{\displaystyle \frac{b}{\lambda }}{\tilde{e}}_3),$$

where $\lambda =\sqrt{ac-b^2}$. The non-vanishing components of Levi-Civita connection ∇ are given by:

$${\nabla }_{{\tilde{e}}_i}{\tilde{e}}_j=\left(\begin{array}{cccc}0& {\displaystyle \frac{\sqrt{a}}{2}}{\tilde{e}}_3& {\displaystyle \frac{\sqrt{a}}{2}}{\tilde{e}}_2+{\displaystyle \frac{A}{2}}{\tilde{e}}_4& -{\displaystyle \frac{A}{2}}{\tilde{e}}_3\\ -{\displaystyle \frac{\sqrt{a}}{2}}{\tilde{e}}_3& 0& {\displaystyle \frac{\sqrt{a}}{2}}{\tilde{e}}_1& 0\\ {\displaystyle \frac{\sqrt{a}}{2}}{\tilde{e}}_2-{\displaystyle \frac{b}{a}}{\tilde{e}}_3-{\displaystyle \frac{B}{2}}{\tilde{e}}_4& {\displaystyle \frac{\sqrt{a}}{2}}{\tilde{e}}_1& {\displaystyle \frac{b}{a}}{\tilde{e}}_1& {\displaystyle \frac{B}{2}}{\tilde{e}}_1\\ -{\displaystyle \frac{B}{2}}{\tilde{e}}_3+{\displaystyle \frac{b}{a}}{\tilde{e}}_4& 0& {\displaystyle \frac{B}{2}}{\tilde{e}}_1& -{\displaystyle \frac{b}{a}}{\tilde{e}}_1\end{array}\right),$$

with $A={\displaystyle \frac{\lambda }{a}}+{\displaystyle \frac{b^2}{\lambda a}}$ and $B={\displaystyle \frac{\lambda }{a}}-{\displaystyle \frac{b^2}{\lambda a}}$, with $\lambda =\sqrt{ac-b^2}$. The curvature R exhibits non-vanishing components determined by the following structural constants:

$$\begin{array}{c}R({\tilde{e}}_1,{\tilde{e}}_2){\tilde{e}}_1=-{\displaystyle \frac{3a}{4}}{\tilde{e}}_2+{\displaystyle \frac{\sqrt{a}b}{a}}{\tilde{e}}_3+({\displaystyle \frac{\sqrt{a}}{2}}B-{\displaystyle \frac{\sqrt{a}}{4}}A){\tilde{e}}_4,R({\tilde{e}}_1,{\tilde{e}}_2){\tilde{e}}_2=-{\displaystyle \frac{3a}{4}}{\tilde{e}}_1,\hfill \\ R({\tilde{e}}_1,{\tilde{e}}_2){\tilde{e}}_3=-{\displaystyle \frac{\sqrt{a}b}{a}}{\tilde{e}}_1,R({\tilde{e}}_1,{\tilde{e}}_2){\tilde{e}}_4=-({\displaystyle \frac{\sqrt{a}}{2}}B-{\displaystyle \frac{\sqrt{a}}{4}}A){\tilde{e}}_1,\hfill \\ R({\tilde{e}}_1,{\tilde{e}}_3){\tilde{e}}_1=-{\displaystyle \frac{\sqrt{a}b}{a}}{\tilde{e}}_2+({\displaystyle \frac{a}{4}}+{\displaystyle \frac{AB}{4}}+{\displaystyle \frac{\lambda B}{2a}}+{\displaystyle \frac{b^2}{a^2}}){\tilde{e}}_3-{\displaystyle \frac{b}{a}}A{\tilde{e}}_4,R({\tilde{e}}_1,{\tilde{e}}_3){\tilde{e}}_2=-{\displaystyle \frac{\sqrt{a}b}{a}}{\tilde{e}}_1,\hfill \\ R({\tilde{e}}_1,{\tilde{e}}_3){\tilde{e}}_3=-({\displaystyle \frac{a}{4}}+{\displaystyle \frac{AB}{4}}+{\displaystyle \frac{\lambda B}{2a}}+{\displaystyle \frac{b^2}{a^2}}){\tilde{e}}_1,R({\tilde{e}}_1,{\tilde{e}}_3){\tilde{e}}_4={\displaystyle \frac{b}{a}}A{\tilde{e}}_1,\hfill \\ R({\tilde{e}}_1,{\tilde{e}}_4){\tilde{e}}_1=-{\displaystyle \frac{\sqrt{a}}{2}}({\displaystyle \frac{B}{2}}-{\displaystyle \frac{b^2}{\lambda a}}){\tilde{e}}_2-{\displaystyle \frac{b}{a}}A{\tilde{e}}_3+({\displaystyle \frac{b^2}{a^2}}-{\displaystyle \frac{AB}{4}}-{\displaystyle \frac{b^2}{2\lambda a}}){\tilde{e}}_4,\hfill \end{array}$$

$$\begin{array}{c}R({\tilde{e}}_1,{\tilde{e}}_4){\tilde{e}}_2=-{\displaystyle \frac{\sqrt{a}}{2}}({\displaystyle \frac{B}{2}}-{\displaystyle \frac{b^2}{\lambda a}}){\tilde{e}}_1,R({\tilde{e}}_1,{\tilde{e}}_4){\tilde{e}}_3={\displaystyle \frac{b}{a}}A{\tilde{e}}_1,\hfill \\ R({\tilde{e}}_1,{\tilde{e}}_4){\tilde{e}}_4=-({\displaystyle \frac{b^2}{a^2}}-{\displaystyle \frac{AB}{4}}-{\displaystyle \frac{b^2}{2\lambda a}}){\tilde{e}}_1,R({\tilde{e}}_2,{\tilde{e}}_3){\tilde{e}}_2=-{\displaystyle \frac{a}{4}}{\tilde{e}}_3,\hfill \\ R({\tilde{e}}_2,{\tilde{e}}_3){\tilde{e}}_3=-{\displaystyle \frac{a}{4}}{\tilde{e}}_2+{\displaystyle \frac{\sqrt{a}}{4}}B{\tilde{e}}_4,R({\tilde{e}}_2,{\tilde{e}}_3){\tilde{e}}_4=-{\displaystyle \frac{\sqrt{a}}{4}}B{\tilde{e}}_3,\hfill \\ R({\tilde{e}}_2,{\tilde{e}}_4){\tilde{e}}_3=-{\displaystyle \frac{\sqrt{a}b}{2a}}{\tilde{e}}_4,R({\tilde{e}}_2,{\tilde{e}}_4){\tilde{e}}_4={\displaystyle \frac{\sqrt{a}b}{2a}}{\tilde{e}}_3,\hfill \\ R({\tilde{e}}_3,{\tilde{e}}_4){\tilde{e}}_2={\displaystyle \frac{\sqrt{a}}{4}}B{\tilde{e}}_3-{\displaystyle \frac{\sqrt{a}b}{2a}}{\tilde{e}}_4,R({\tilde{e}}_3,{\tilde{e}}_4){\tilde{e}}_3={\displaystyle \frac{\sqrt{a}}{4}}B{\tilde{e}}_2-{\displaystyle \frac{A^2}{4}}{\tilde{e}}_4,\hfill \\ R({\tilde{e}}_3,{\tilde{e}}_4){\tilde{e}}_4=-{\displaystyle \frac{\sqrt{a}b}{2a}}{\tilde{e}}_2+{\displaystyle \frac{A^2}{4}}{\tilde{e}}_3.\hfill \end{array}$$

Assume that $\xi$ on M has the form $\xi ={\sum }_{i=1}^4{a}_i{\tilde{e}}_i$. The coefficients $a_i:U\subseteq M\to \mathbb{R}$ ($i=1,2,3,4$) are smooth functions. The tangent space to M is spanned by the vector fields:

$$\begin{array}{c}\hfill X_1=a_2{\tilde{e}}_1+a_1{\tilde{e}}_2,X_2=a_3{\tilde{e}}_1-a_1{\tilde{e}}_3,\\ \hfill X_3=a_4{\tilde{e}}_1-a_1{\tilde{e}}_4,X_4=a_3{\tilde{e}}_2+a_2{\tilde{e}}_3,\\ \hfill X_5=a_4{\tilde{e}}_2+a_2{\tilde{e}}_4,X_6=a_4{\tilde{e}}_3-a_3{\tilde{e}}_4.\end{array}$$

Under the condition that $\nabla h$ is totally symmetric, the Codazzi Equation (4) implies the constraint $R(X_i,X_j)\xi =0$ for all $i,j\in \{1,\dots ,6\}$. We analyze constraints on $a_i$ by categorizing into distinct cases:

Case 1: All $a_i\ne 0$. Examining components of ${\tilde{e}}_2$ in $R(X_1,X_2)\xi =0$ and $R(X_4,X_5)\xi =0$, ${\tilde{e}}_3$ in $R(X_1,X_3)\xi =0$, we derive:

$$\left\{\begin{array}{c}{\displaystyle \frac{\sqrt{a}b}{a}}{a}_2+a{a}_3=0,\hfill \\ ({\displaystyle \frac{b}{a}}A{a}_2-{\displaystyle \frac{\sqrt{a}b}{a}}{a}_4)a_1-{\displaystyle \frac{\sqrt{a}b}{2a}}{a}_1{a}_4=0,\hfill \\ ({\displaystyle \frac{a}{4}}{a}_4+{\displaystyle \frac{\sqrt{a}}{4}}B{a}_2)a_3-{\displaystyle \frac{\sqrt{a}b}{2a}}{a}_2{a}_4=0.\hfill \end{array}\right.$$

Substituting into the third equation yields $2A+B=0$, which contradicts $\lambda \ne 0$ by definitions of A and B. Hence, no solutions exist with all $a_i\ne 0$.

Case 2: Exactly one $a_i=0$.

• Subcase 2a: $a_1=0$. From $R(X_1,X_4)\xi =0$ and analyzing the ${\tilde{e}}_2,{\tilde{e}}_4$-component of $R(X_4,X_5)\xi$ we derive:

  $$\left\{\begin{array}{c}(a+{\displaystyle \frac{\lambda B}{2a}}+{\displaystyle \frac{b^2}{a^2}}-{\displaystyle \frac{A^2}{2}}+{\displaystyle \frac{AB}{2}})a_3-{\displaystyle \frac{3b}{2a}}A{a}_4=0,\hfill \\ (a+{\displaystyle \frac{\lambda B}{2a}}+{\displaystyle \frac{b^2}{a^2}}-{\displaystyle \frac{A^2}{2}}-{\displaystyle \frac{AB}{4}})a_2-{\displaystyle \frac{3\sqrt{a}}{4}}A{a}_4=0.\hfill \end{array}\right.$$

  (5)
  From Equation (5), it is observed that the coefficients $a_2$, $a_3$, and $a_4$ exhibit proportionality. We introduce real constants $k_1$ and $k_2$ such that $a_3=k_1{a}_2$ and $a_4=k_2{a}_2$. Through normalization enforced by the condition $\langle \xi ,\xi \rangle =\pm 1$, which translates to $-a_2^2+a_3^2+a_4^2=\pm 1$, we deduce that $a_2$, $a_3$, and $a_4$ must be constant functions. To investigate the integrability of the distribution $\mathrm{span}\left\{{\xi }^{\perp }\right\}=\mathrm{span}\{{\tilde{e}}_1,a_3{\tilde{e}}_2+a_2{\tilde{e}}_3,a_4{\tilde{e}}_2+a_2{\tilde{e}}_4\}$, the Frobenius theorem imposes the following necessary conditions:

  $$\langle [{\tilde{e}}_1,a_3{\tilde{e}}_2+a_2{\tilde{e}}_3],\xi \rangle =0and\langle [{\tilde{e}}_1,a_4{\tilde{e}}_2+a_2{\tilde{e}}_4],\xi \rangle =0.$$
  Further analysis of the ${\tilde{e}}_2$-component within the curvature tensor expression $R(X_4,X_5)\xi =0$ yields additional constraints that refine these conditions, ensuring the structural consistency required for the distribution’s integrability.
• Subcase 2b–2d: $a_2=0$, $a_3=0$, or $a_4=0$. Each leads to contradictions. For example, if $a_2=0$, the ${\tilde{e}}_2$-component of $R(X_1,X_3)\xi$ force ${\displaystyle \frac{3a}{4}}{a}_1{a}_4=0$, contradicting $a_1{a}_4\ne 0$.

Case 3: Exactly two $a_i=0$. All subcases fail. For instance, if $a_1=a_4=0$, the $X_4$-component of $R(X_4,X_5)\xi$ gives ${\displaystyle \frac{\sqrt{a}}{4}}B{a}_2=0$. Since $B\ne 0$ (as $\lambda \ne 0$), this implies $a_2=0$, which is a contradiction.

Case 4: Exactly three $a_i=0$.

• Subcase 4a: Only $a_1\ne 0$. Directly yields $\xi =\pm {\tilde{e}}_1$, consistent with the unit normal condition.
• Subcase 4b–4d: Other single non-zero $a_i$. Each leads to contradictions. For example, if $a_2\ne 0$, the curvature component $R(X_2,X_4)\xi =0$ forces ${\displaystyle \frac{\sqrt{a}b}{a}}{a}_2^2=0$, contradicting $a_2\ne 0$.

The only valid Codazzi normal vector fields are $\xi =\pm e_1$. Since totally geodesic and parallel hypersurfaces are special cases of Codazzi hypersurfaces (having totally symmetric cubic form), the above classification of admissible unit normal vectors applies to them. Therefore, for the metric $g_A$, any totally geodesic or parallel hypersurface must have a unit normal vector field $\xi =\pm e_1$. □

Remark 1.

The classification results for the metrics $g_A^{\pm }$ are identical to those for $g_A$ derived above. The explicit expressions for the normal vector fields and the resulting hypersurfaces can be obtained by an analogous derivation process, and are therefore omitted here.

Theorem 4.

Consider a hypersurface immersion $F:M\to (Ni{l}^4,g_1^{\lambda })$ with a totally symmetric form. Suppose ξ is a unit timelike or spacelike normal vector field on M. For each point in M, there exists a neighborhood $U\subseteq M$ such that $\xi =e_1+a_4{e}_4$, or $\xi ={\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$, where $a_4:U\to \mathbb{R}$ is a function.

Proof. 

The Levi-Civita connection ∇ for the left-invariant Lorentzian metric $g_1^{\lambda }$ on $Ni{l}^4$ with respect to the basis in (2) has non-zero components:

$${\nabla }_{e_1}{e}_2=-{\nabla }_{e_2}{e}_1={\displaystyle \frac{1}{2}}{e}_3,{\nabla }_{e_1}{e}_3={\nabla }_{e_3}{e}_1=-{\displaystyle \frac{1}{2}}{e}_4,{\nabla }_{e_2}{e}_3={\nabla }_{e_3}{e}_2={\displaystyle \frac{1}{2}}{e}_1.$$

The curvature tensor R has non-zero components:

$$R(e_1,e_2)e_1={\displaystyle \frac{3}{4}}{e}_4,R(e_1,e_2)e_2=-{\displaystyle \frac{3}{4}}{e}_1,R(e_2,e_3)e_2=-{\displaystyle \frac{1}{4}}{e}_3,R(e_2,e_3)e_3={\displaystyle \frac{1}{4}}{e}_4.$$

Assume that $\xi$ on M has the form $\xi ={\sum }_{i=1}^4{a}_i{e}_i$. The coefficients $a_i:U\subseteq M\to \mathbb{R}$ ($i=1,2,3,4$) are smooth functions. The tangent space to M is spanned by the vector fields:

$$\begin{array}{c}\hfill X_1=a_4{e}_1-a_1{e}_2,X_2=a_3{e}_1-a_1{e}_3,\\ \hfill X_3=a_2{e}_1-a_1{e}_4,X_4=a_3{e}_2-a_4{e}_3,\\ \hfill X_5=a_2{e}_2-a_4{e}_4,X_6=a_2{e}_3-a_3{e}_4.\end{array}$$

Imposing total symmetry of $\nabla h$ via the Codazzi equation forces $R(X_i,X_j)\xi =0$ yields critical conditions:

$$\begin{array}{cc}\hfill & 0=R(X_3,X_5)\xi =-{\displaystyle \frac{3}{4}}{a}_2^2{X}_3,\hfill \\ \hfill & 0=R(X_5,X_6)\xi =-{\displaystyle \frac{1}{4}}{a}_2^2{X}_6,\hfill \\ \hfill & 0=R(X_2,X_4)\xi =-{\displaystyle \frac{3}{4}}{a}_3^2{X}_3-{\displaystyle \frac{1}{4}}{a}_1{a}_3{X}_6.\hfill \end{array}$$

From these, we deduce: $a_2=0$, and $a_1=0$ or $a_3=0$,

Case 1: $a_3=0$. The normal vector simplifies to $\xi =e_1+a_4{e}_4$. This aligns $\xi$ with the $e_1-e_4$ plane, consistent with the metric’s Lorentzian signature.

Case 2: $a_1=0$. The normal vector simplifies to $\xi =e_3+a_4{e}_4$. □

Theorem 5.

Consider a hypersurface immersion $F:M\to (Ni{l}^4,g_2^{\lambda })$ with a totally symmetric form. Suppose ξ is a unit timelike or spacelike normal vector field on M. For each point in M, there exists a neighborhood $U\subseteq M$ such that $\xi =e_2+a_4{e}_4$, or $\xi ={\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$, where $a_4:U\to \mathbb{R}$ is a function.

Proof. 

The Levi-Civita connection ∇ for the left-invariant Lorentzian metric $g_2^{\lambda }$ on $Ni{l}^4$ with respect to the basis in (2) has non-zero components:

$$\begin{array}{cc}\hfill & {\nabla }_{e_1}{e}_1=-{\displaystyle \frac{1}{\lambda }}{e}_3,{\nabla }_{e_1}{e}_2={\displaystyle \frac{1}{2}}{e}_3,{\nabla }_{e_1}{e}_3=-{\displaystyle \frac{\lambda }{2}}{e}_2+e_4,\hfill \\ \hfill & {\nabla }_{e_2}{e}_1=-{\displaystyle \frac{1}{2}}{e}_3,{\nabla }_{e_2}{e}_3={\displaystyle \frac{\lambda }{2}}{e}_4,{\nabla }_{e_3}{e}_1=-{\displaystyle \frac{\lambda }{2}}{e}_2,{\nabla }_{e_3}{e}_2={\displaystyle \frac{\lambda }{2}}{e}_4.\hfill \end{array}$$

The curvature tensor R has non-zero components:

$$R(e_1,e_2)e_1={\displaystyle \frac{3\lambda }{4}}{e}_2,R(e_1,e_2)e_2=-{\displaystyle \frac{3\lambda }{4}}{e}_4,R(e_1,e_3)e_1=-{\displaystyle \frac{\lambda }{4}}{e}_3,R(e_1,e_3)e_3={\displaystyle \frac{{\lambda }^2}{4}}{e}_4.$$

Assume that $\xi$ on M has the form $\xi ={\sum }_{i=1}^4{a}_i{e}_i$. The coefficients $a_i:U\subseteq M\to \mathbb{R}$ ($i=1,2,3,4$) are smooth functions. The tangent space to M is spanned by the vector fields:

$$\begin{array}{c}\hfill X_1=a_2{e}_1-a_4{e}_2,X_2=\lambda a_3{e}_1-a_4{e}_3,\\ \hfill X_3=a_1{e}_1-a_4{e}_4,X_4=\lambda a_3{e}_2-a_2{e}_3,\\ \hfill X_5=a_1{e}_2-a_2{e}_4,X_6=a_1{e}_3-\lambda a_3{e}_4.\end{array}$$

Imposing total symmetry of $\nabla h$ via the Codazzi equation forces $R(X_i,X_j)\xi =0$ yields critical conditions:

$$\begin{array}{cc}\hfill & 0=R(X_3,X_5)\xi ={\displaystyle \frac{3\lambda }{4}}{a}_1^2{X}_5,\hfill \\ \hfill & 0=R(X_3,X_6)\xi =-{\displaystyle \frac{\lambda }{4}}{a}_1^2{X}_6,\hfill \\ \hfill & 0=R(X_1,X_4)\xi ={\displaystyle \frac{3{\lambda }^2}{4}}{a}_2{a}_3{X}_5+{\displaystyle \frac{\lambda }{4}}{a}_2^2{X}_6.\hfill \end{array}$$

Thus, we obtain $a_1=0$, and from $R(X_1,X_4)\xi =0$, it follows that ${\displaystyle \frac{{\lambda }^2}{2}}{a}_2{a}_3{e}_4=0$. Which split into: $a_2=0$ or $a_3=0$.

Case 1: $a_2=0$. The normal vector simplifies to $\xi =a_3{e}_3+a_4{e}_4$. Normalizing gives $\xi ={\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$.

Case 2: $a_3=0$. The normal vector simplifies to $\xi =a_2{e}_2+a_4{e}_4$. Normalizing gives $\xi =e_2+a_4{e}_4$. □

Theorem 6.

Consider a hypersurface immersion $F:M\to (Ni{l}^4,g_3^{\lambda })$ with a totally symmetric form. Suppose ξ is a unit timelike or spacelike normal vector field on M. Then, there is no exist unit normal vector field ξ on M.

Proof. 

The Levi-Civita connection ∇ for the left-invariant Lorentzian metric $g_3^{\lambda }$ on $Ni{l}^4$ with respect to the basis in (2) has non-zero components:

$$\begin{array}{cc}\hfill & {\nabla }_{e_1}{e}_1=-e_2,{\nabla }_{e_1}{e}_2=e_3,{\nabla }_{e_1}{e}_3={\displaystyle \frac{1}{2}}{e}_4,{\nabla }_{e_1}{e}_4=-{\displaystyle \frac{\lambda }{2}}{e}_1,\hfill \\ \hfill & {\nabla }_{e_3}{e}_1=-{\displaystyle \frac{1}{2}}{e}_4,{\nabla }_{e_3}{e}_4={\displaystyle \frac{\lambda }{2}}{e}_3,{\nabla }_{e_4}{e}_1=-{\displaystyle \frac{\lambda }{2}}{e}_1,{\nabla }_{e_4}{e}_3={\displaystyle \frac{\lambda }{2}}{e}_3.\hfill \end{array}$$

The curvature tensor R has non-zero components:

$$\begin{array}{cc}\hfill & R(e_1,e_2)e_1={\displaystyle \frac{1}{2}}{e}_4,R(e_1,e_2)e_4=-{\displaystyle \frac{\lambda }{2}}{e}_3,R(e_1,e_3)e_1={\displaystyle \frac{3\lambda }{4}}{e}_1,R(e_1,e_3)e_3=-{\displaystyle \frac{3\lambda }{4}}{e}_3,\hfill \\ \hfill & R(e_1,e_4)e_1={\displaystyle \frac{\lambda }{2}}{e}_2,R(e_1,e_4)e_2=-{\displaystyle \frac{\lambda }{2}}{e}_3,R(e_1,e_4)e_3={\displaystyle \frac{\lambda }{4}}{e}_4,R(e_1,e_4)e_4=-{\displaystyle \frac{{\lambda }^2}{4}}{e}_1,\hfill \\ \hfill & R(e_3,e_4)e_1={\displaystyle \frac{\lambda }{4}}{e}_4,R(e_3,e_4)e_4=-{\displaystyle \frac{{\lambda }^2}{4}}{e}_3.\hfill \end{array}$$

Assume that $\xi$ on M has the form $\xi ={\sum }_{i=1}^4{a}_i{e}_i$. The coefficients $a_i:U\subseteq M\to \mathbb{R}$ ($i=1,2,3,4$) are smooth functions. The tangent space to M is spanned by the vector fields:

$$\begin{array}{cccc}\hfill & X_1=a_2{e}_1-a_3{e}_2,\hfill & \hfill & X_2=a_1{e}_1-a_3{e}_3,\hfill \\ \hfill & X_3=\lambda a_4{e}_1-a_3{e}_4,\hfill & \hfill & X_4=a_1{e}_2-a_2{e}_3,\hfill \\ \hfill & X_5=\lambda a_4{e}_2-a_2{e}_4,\hfill & \hfill & X_6=\lambda a_4{e}_3-a_1{e}_4.\hfill \end{array}$$

Under the condition that $\nabla h$ is totally symmetric, the Codazzi Equation (4) implies the constraint $R(X_i,X_j)\xi =0$ for all $i,j\in \{1,\dots ,6\}$. In particular, $R(X_4,X_5)\xi =-{\displaystyle \frac{\lambda }{4}}{a}_2^2{X}_6=0$ splits into two cases: $a_2=0$ or $a_1=a_4=0$.

Case 1: $a_2=0$. From $e_4$-component of $R(X_2,X_4)\xi =0$ yields $a_1=0$. Analogously, considering the $e_3$-component of $R(X_3,X_5)\xi =0$ forces $a_4=0$. This reduces the normal vector to $\xi =\pm e_3$. However, this contradicts the normalization condition.

Case 2: $a_1=a_4=0$. Substituting into $R(X_1,X_5)\xi =0$ and isolating the $e_3$-component yields $a_2=0$. Consequently, the normal vector becomes $\xi =\pm e_3$. This again contradicts the normalization condition.

Since all possible cases lead to a contradiction, there exists no unit normal vector field $\xi$ that satisfies the Codazzi equation for the metric $g_3^{\lambda }$. No non-trivial Codazzi hypersurfaces exist in $Ni{l}^4$ under the metric $g_3^{\lambda }$. □

Theorem 7.

Consider a hypersurface immersion $F:M\to (Ni{l}^4,g_4^{\lambda })$ with a totally symmetric form. Suppose ξ is a unit timelike or spacelike normal vector field on M. For each point in M, there exists a neighborhood $U\subseteq M$ such that $\xi =\pm e_1$.

Proof. 

The Levi-Civita connection ∇ for the left-invariant Lorentzian metric $g_4^{\lambda }$ on $Ni{l}^4$ with respect to the basis in (2) has non-zero components:

$$\begin{array}{cc}\hfill & {\nabla }_{e_1}{e}_3={\displaystyle \frac{1}{2}}{e}_4,{\nabla }_{e_1}{e}_4=-{\displaystyle \frac{\lambda }{2}}{e}_2,{\nabla }_{e_2}{e}_1=-e_3,{\nabla }_{e_2}{e}_2=e_1,\hfill \\ \hfill & {\nabla }_{e_3}{e}_1=-{\displaystyle \frac{1}{2}}{e}_4,{\nabla }_{e_3}{e}_4={\displaystyle \frac{\lambda }{2}}{e}_1,{\nabla }_{e_4}{e}_1=-{\displaystyle \frac{\lambda }{2}}{e}_2,{\nabla }_{e_4}{e}_3={\displaystyle \frac{\lambda }{2}}{e}_1.\hfill \end{array}$$

The curvature tensor R has non-zero components:

$$\begin{array}{cc}\hfill & R(e_1,e_3)e_1={\displaystyle \frac{3\lambda }{4}}{e}_2,R(e_1,e_3)e_3=-{\displaystyle \frac{3\lambda }{4}}{e}_1,R(e_2,e_3)e_2={\displaystyle \frac{1}{2}}{e}_4,R(e_2,e_3)e_4=-{\displaystyle \frac{\lambda }{2}}{e}_3,\hfill \\ \hfill & R(e_2,e_4)e_2={\displaystyle \frac{\lambda }{2}}{e}_2,R(e_2,e_4)e_3=-{\displaystyle \frac{\lambda }{2}}{e}_3,R(e_3,e_4)e_3=-{\displaystyle \frac{\lambda }{4}}{e}_4,R(e_3,e_4)e_4={\displaystyle \frac{{\lambda }^2}{4}}{e}_2.\hfill \end{array}$$

Assume that $\xi$ on M has the form $\xi ={\sum }_{i=1}^4{a}_i{e}_i$. The coefficients $a_i:U\subseteq M\to \mathbb{R}$ ($i=1,2,3,4$) are smooth functions. The tangent space to M is spanned by the vector fields:

$$\begin{array}{cccc}\hfill & X_1=a_3{e}_1-a_1{e}_2,\hfill & \hfill & X_2=a_2{e}_1-a_1{e}_3,\hfill \\ \hfill & X_3=\lambda a_4{e}_1-a_1{e}_4,\hfill & \hfill & X_4=a_2{e}_2-a_3{e}_3,\hfill \\ \hfill & X_5=\lambda a_4{e}_2-a_3{e}_4,\hfill & \hfill & X_6=\lambda a_4{e}_3-a_2{e}_4.\hfill \end{array}$$

Under the condition that $\nabla h$ is totally symmetric, the Codazzi Equation (4) implies the constraint $R(X_i,X_j)\xi =0$ for all $i,j\in \{1,\dots ,6\}$. In particular, $R(X_1,X_3)\xi ={\displaystyle \frac{\lambda }{2}}{a}_1^2{X}_4=0$ splits into two cases: $a_1=0$ and $a_2=a_3=0$.

Case 1: $a_1=0$. The $R(X_1,X_4)\xi ={\displaystyle \frac{3\lambda }{4}}{a}_3^2{X}_1=0$ forces $a_3=0$. Analogously, considering the $e_3$-component of $R(X_1,X_4)\xi =0$ forces $a_4=0$. This reduces the normal vector to $\xi =\pm e_2$. However, this contradicts the normalization $\langle \xi ,\xi \rangle =\pm 1$.

Case 2: $a_2=a_3=0$. Substituting into $R(X_5,X_6)\xi =0$ and extracting the $e_3$-component, we obtain the constraint $a_4=0$. This result simplifies the normal vector field to $\xi =\pm e_1$. which directly contradicts the normalization condition. □

4. Parallel Hypersurfaces of ${\mathit{Nil}}^{\mathbf{4}}$

Based on the preceding analysis and utilizing the explicit form of the normal vector field $\xi$ within the coordinate defined by (1), we can now give the proof of Theorem 1, which provide a complete and explicit characterization of the corresponding hypersurfaces.

Proof of Theorem 1.

For the ambient space $(Ni{l}^4,g_A)$: Consider $\xi =\pm e_1$. The tangent space to M is spanned by:

$$Y_1={\tilde{e}}_2,Y_2={\tilde{e}}_3,Y_3={\tilde{e}}_4.$$

Using the metric (2), the Levi-Civita connection in terms of $\left\{Y_i\right\}$ on M is flat(${\nabla }_{Y_i}^M{Y}_j=0$), with non-zero components:

$${\nabla }_{Y_1}{Y}_2={\nabla }_{Y_2}{Y}_1={\displaystyle \frac{\sqrt{a}}{2}}\xi ,{\nabla }_{Y_2}{Y}_3={\nabla }_{Y_3}{Y}_2={\displaystyle \frac{B}{2}}\xi ,{\nabla }_{Y_2}{Y}_2=-{\nabla }_{Y_3}{Y}_3={\displaystyle \frac{b}{a}}\xi .$$

By using the Gauss formula (3), the non-zero components of second fundamental form h is determined by:

$$\begin{array}{cc}\hfill & h(Y_1,Y_2)=h(Y_2,Y_1)={\displaystyle \frac{\sqrt{a}}{2}},h(Y_2,Y_2)={\displaystyle \frac{b}{a}},\hfill \\ \hfill & h(Y_2,Y_3)=h(Y_3,Y_2)={\displaystyle \frac{B}{2}},h(Y_3,Y_3)=-{\displaystyle \frac{b}{a}}.\hfill \end{array}$$

The covariant derivative $\nabla h$ of h with respect to the flat connection vanishes identically (${\nabla }^{M}h=0$). This confirms that M is a parallel hypersurface in $(Ni{l}^4,g_A)$, maintaining constant shape under ambient translations. Since M is flat, the vector fields $Y_i={\partial }_{u_i}$ ($i=1,2,3$) can serve as coordinate vector fields on M.

Let $F:M\to (Ni{l}^4,g_A)$ be the immersion defined by local coordinates $(u_1,u_2,u_3)$ as $F=(F_1(u_1,u_2,u_3),\dots ,F_4(u_1,u_2,u_3))$. Then, we can compute the derivatives of F:

$${\partial }_{u_1}F=(0,1,0,0),{\partial }_{u_2}F=(0,0,1,0),{\partial }_{u_3}F=(0,0,0,1).$$

We obtain after reparametrization the simplified immersion F:

$$F_1=c_1,F_2=u_1+c_2,F_3=u_2+c_3,F_4=u_3+c_4.$$

for constants $c_i\in \mathbb{R}$. This linear parametrization reflects the flatness of M, aligning with Case (1) of Theorem 1.

For the ambient space $(Ni{l}^4,g_1^{\lambda })$: Consider $\xi =e_1+a_4{e}_4$. The tangent space to M is spanned by the global frame:

$$Y_1=a_4{e}_1-e_2,Y_2=e_3,Y_3=e_4.$$

Using the metric (2), the non-vanishing components of the Levi-Civita connection in terms of $\left\{Y_i\right\}$ are:

$$\begin{array}{cc}\hfill & {\nabla }_{Y_1}{Y}_1=Y_1(a_4)\xi +{\displaystyle \frac{1}{\lambda }}{a}_4{Y}_2-Y_1(a_4)a_4{Y}_3,{\nabla }_{Y_1}{Y}_2=-{\displaystyle \frac{\lambda +1}{2}}\xi +a_4{Y}_3,\hfill \\ \hfill & {\nabla }_{Y_2}{Y}_1=(Y_2(a_4)-{\displaystyle \frac{1+\lambda }{2}})\xi -Y_2(a_4)a_4{Y}_3,{\nabla }_{Y_3}{Y}_1=Y_3(a_4)\xi -Y_3(a_4)a_4{Y}_3.\hfill \end{array}$$

By the Gauss formula (3), the non-zero components of second fundamental form h is determined by:

$$\begin{array}{cc}\hfill & h(Y_1,Y_1)=Y_1(a_4),h(Y_1,Y_2)=-{\displaystyle \frac{\lambda +1}{2}},h(Y_1,Y_3)=0,\hfill \\ \hfill & h(Y_2,Y_1)=Y_2(a_4)-{\displaystyle \frac{1+\lambda }{2}},h(Y_3,Y_1)=Y_3(a_4).\hfill \end{array}$$

The symmetry of h implies $Y_2(a_4)=Y_3(a_4)=0$. The induced connection on M satisfies:

$${\nabla }_{Y_1}^M{Y}_1=-{\displaystyle \frac{1}{\lambda }}{a}_4{Y}_2-Y_1(a_4)a_4{Y}_3,{\nabla }_{Y_1}^M{Y}_2=a_4{Y}_3.$$

For parallel second fundamental form ${\nabla }^{M}h\equiv 0$, we impose:

$$({\nabla }_{Y_1}^{M}h)(Y_1,Y_1)=Y_1(Y_1·a_4)-{\displaystyle \frac{\lambda +1}{\lambda }}{a}_4=0.$$

(6)

We construct explicit coordinates on M using a local frame ${\left\{u_i\right\}}_{i=1}^3$. The coordinate vector fields are defined as:

$${\partial }_{u_1}=Y_1,{\partial }_{u_2}=\alpha Y_2+\beta Y_3,{\partial }_{u_3}=Y_3,$$

where $\alpha ,\beta$ are smooth functions on M. Solving Equation (6) yields:

$$a_4=a_{1}cos(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1)+a_{2}sin(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1),$$

with $a_1,a_2\in \mathbb{R}$. Note that $h(Y_1,Y_2)=-{\displaystyle \frac{\lambda +1}{2}}$ where $\lambda >0$, M is not totally geodesic. The commutativity conditions for the local frame yields:

$$\left\{\begin{array}{c}\hfill Y_1(\alpha )=0,\\ \hfill Y_1(\beta )+\alpha a_4=0.\end{array}\right.$$

Observe that we only need one solution for $\alpha$ and $\beta$ in the system above in order to find a coordinate system $(u_1,u_2,u_3)$ on the surface M. So, we take $\alpha =1$ and $\beta =-{\int }_c^{u_1}{a}_{4}ds$.

Let $F:M\to (Ni{l}^4,g_A)$ be the immersion defined by local coordinates $(u_1,u_2,u_3)$ as $F=(F_1(u_1,u_2,u_3),\dots ,F_4(u_1,u_2,u_3))$. Then, we can compute the derivatives of F:

$${\partial }_{u_1}F=(a_4,-1,-F_1,-{\displaystyle \frac{F_1^2}{2}}),{\partial }_{u_2}F=(0,0,1,0),{\partial }_{u_3}F=(0,0,0,1).$$

Integrating the coordinate derivatives yields the explicit parametrization:

$$\begin{array}{cc}\hfill & F_1=\sqrt{{\displaystyle \frac{\lambda }{1+\lambda }}}(a_{1}sin(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1)-a_{2}cos(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1))+c_1,\hfill \\ \hfill & F_2=-u_1+c_2,\hfill \\ \hfill & F_3=a_{1}cos(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1)+a_{2}sin(\sqrt{{\displaystyle \frac{\lambda +1}{\lambda }}}{u}_1)+u_2+c_3,\hfill \\ \hfill & F_4=u_3+c_4.\hfill \end{array}$$

This parametrization matches Case (2) of Theorem 1. confirming that M is a non-totally geodesic parallel hypersurface in $Ni{l}^4$.

For the ambient space $(Ni{l}^4,g_1^{\lambda })$: Consider the unit normal vector field $\xi ={\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$. The tangent space to M is spanned by the global frame:

$$Y_1=e_1,Y_2=\sqrt{\lambda }{e}_2-a_4{e}_3,Y_3=e_4.$$

By the Frobenius integrability condition, the Lie brackets satisfy:

$$\langle [e_1,\sqrt{\lambda }{e}_2-a_4{e}_3],\xi \rangle =\langle [e_4,\sqrt{\lambda }{e}_2-a_4{e}_3],\xi \rangle =0,$$

forcing ${\partial }_{e_1}(a_4)-\sqrt{\lambda }={\partial }_{e_4}(a_4)=0$. Using the metric (2), the non-vanishing components of the Levi-Civita connection in terms of $\left\{Y_i\right\}$ are:

$$\begin{array}{cc}\hfill & {\nabla }_{Y_1}{Y}_2=\lambda a_4{Y}_3-{\displaystyle \frac{1+\lambda }{2}}\xi ,{\nabla }_{Y_2}{Y}_1=(1+\lambda )a_4{Y}_3-{\displaystyle \frac{1+\lambda }{2}}\xi ,\hfill \\ \hfill & {\nabla }_{Y_2}{Y}_2=-\sqrt{\lambda }(1+\lambda )a_4{Y}_1+\sqrt{\lambda }{Y}_2(a_4)a_4{Y}_3-\sqrt{\lambda }{Y}_2(a_4)\xi .\hfill \end{array}$$

By using the Gauss formula (3), the second fundamental form h is determined by:

$$h(Y_1,Y_2)=h(Y_2,Y_1)=-{\displaystyle \frac{1+\lambda }{2}},h(Y_2,Y_2)=-\sqrt{\lambda }{Y}_2(a_4).$$

The induced connection on M are given by:

$$\begin{array}{cc}\hfill & {\nabla }_{Y_1}^M{Y}_2=\lambda a_4{Y}_3,{\nabla }_{Y_2}^M{Y}_1=(1+\lambda )a_4{Y}_3,\hfill \\ \hfill & {\nabla }_{Y_2}^M{Y}_2=-\sqrt{\lambda }(1+\lambda )a_4{Y}_1+\sqrt{\lambda }{Y}_2(a_4)a_4{Y}_3.\hfill \end{array}$$

Note that $h(Y_1,Y_2)=h(Y_2,Y_1)=-{\displaystyle \frac{1+\lambda }{2}}$, M is not totally geodesic. For parallel second fundamental form ${\nabla }^{M}h\equiv 0$, we impose:

$$({\nabla }_{Y_2}^{M}h)(Y_2,Y_2)=-Y_2(Y_2·a_4)-{(1+\lambda )}^2{a}_4=0.$$

(7)

We construct explicit coordinates on M using a local frame ${\left\{u_i\right\}}_{i=1}^3$. The coordinate vector fields are defined as:

$${\partial }_{u_1}=Y_1,{\partial }_{u_2}=\alpha Y_2+\beta Y_3,{\partial }_{u_3}=Y_3,$$

for some smooth functions $\alpha ,\beta$ on M. Solving the function (7) gives:

$$a_4=k_{1}cos(1+\lambda )u_2+k_{2}sin(1+\lambda )u_2,$$

where $k_1,k_2\in \mathbb{R}$. Note that $h(Y_1,Y_2)=-{\displaystyle \frac{\lambda +1}{2}}$ where $\lambda >0$, M is not totally geodesic. The commutativity conditions for the local frame yields:

$$\left\{\begin{array}{c}\hfill Y_1(\alpha )=0,\\ \hfill Y_1(\beta )-\alpha a_4=0.\end{array}\right.$$

A single solution for $\alpha$ and $\beta$ is sufficient to define the coordinate system $(u_1,u_2,u_3)$ on M rather than the general solution. So, we set $\alpha =1$, and $\beta =-{\int }_c^{u_1}{a}_{4}ds$.

Let $F:M\to (Ni{l}^4,g_A)$ be the immersion defined by local coordinates $(u_1,u_2,u_3)$ as $F=(F_1(u_1,u_2,u_3),\dots ,F_4(u_1,u_2,u_3))$. Then, we can compute the derivatives of F:

$$\begin{array}{cc}\hfill & {\partial }_{u_1}F=(1,0,0,0),\hfill \\ \hfill & {\partial }_{u_2}F=(0,\sqrt{\lambda },\sqrt{\lambda }{F}_1-a_4,{\displaystyle \frac{\sqrt{\lambda }{F}_1^2}{2}}-F_1{a}_4+\beta ),\hfill \\ & {\partial }_{u_3}F=(0,0,0,1).\hfill \end{array}$$

Integrating the coordinate derivatives yields the explicit parametrization:

$$\begin{array}{cc}\hfill & F_1=u_1+c_1,\hfill \\ \hfill & F_2=\sqrt{\lambda }{u}_2+c_2,\hfill \\ \hfill & F_3={\displaystyle \frac{k_1}{1+\lambda }}(k_{1}sin(1+\lambda )u_2-k_{2}cos(1+\lambda )u_2)+(u_1+c_1)u_2+c_3,\hfill \\ \hfill & F_4=u_3+c_4.\hfill \end{array}$$

This parametrization matches case (3) of Theorem 1.

For the ambient space $(Ni{l}^4,g_2^{\lambda })$: Consider the unit normal vector field $\xi =e_2+a_4{e}_4$. The tangent space to M is spanned by the global frame:

$$Y_1=e_1-a_4{e}_2,Y_2=e_3,Y_3=e_4.$$

Using the metric (2), the non-vanishing components of the Levi-Civita connection in terms of $\left\{Y_i\right\}$ are:

$$\begin{array}{cc}\hfill & {\nabla }_{Y_1}{Y}_1=-Y_1(a_4)\xi -{\displaystyle \frac{1}{\lambda }}{Y}_2+Y_1(a_4)a_4{Y}_3,{\nabla }_{Y_1}{Y}_2=-{\displaystyle \frac{\lambda }{2}}\xi +Y_3,\hfill \\ \hfill & {\nabla }_{Y_2}{Y}_1=-({\displaystyle \frac{\lambda }{2}}+Y_2(a_4))\xi +Y_2(a_4)a_4{Y}_3,{\nabla }_{Y_3}{Y}_1=-Y_3(a_4)\xi +Y_3(a_4)a_4{Y}_3.\hfill \end{array}$$

Using the Gauss formula (3), the non-zero components of second fundamental form h is determined by:

$$\begin{array}{cc}\hfill & h(Y_1,Y_1)=-Y_1(a_4),h(Y_1,Y_2)=-{\displaystyle \frac{\lambda }{2}},\hfill \\ \hfill & h(Y_2,Y_1)=-({\displaystyle \frac{\lambda }{2}}+Y_2(a_4)),h(Y_3,Y_1)=-Y_3(a_4).\hfill \end{array}$$

The symmetry condition of h implies $Y_2(a_4)=Y_3(a_4)=0$. The induced connection on M is fully determined by:

$${\nabla }_{Y_1}^M{Y}_1=-{\displaystyle \frac{1}{\lambda }}{Y}_2+Y_1(a_4)a_4{Y}_3,{\nabla }_{Y_1}^M{Y}_2=Y_3.$$

For parallel second fundamental form ${\nabla }^{M}h\equiv 0$, we impose:

$$({\nabla }_{Y_1}^{M}h)(Y_1,Y_1)=-Y_1(Y_1·a_4)-1=0.$$

(8)

We construct explicit coordinates on M using a local frame ${\left\{u_i\right\}}_{i=1}^3$. The coordinate vector fields are defined as:

$${\partial }_{u_1}=Y_1,{\partial }_{u_2}=\alpha Y_2+\beta Y_3,{\partial }_{u_3}=Y_3,$$

for some smooth functions $\alpha ,\beta$ on M. Solving the function (8) gives:

$$a_4=-{\displaystyle \frac{1}{2}}{u}_1^2+C{u}_1+D,$$

where $C,D\in \mathbb{R}$. Note that, $h(Y_1,Y_2)=-{\displaystyle \frac{\lambda }{2}}$, M is not totally geodesic. The commutativity conditions for the local frame yields:

$$\left\{\begin{array}{c}\hfill Y_1(\alpha )=0,\\ \hfill Y_1(\beta )+\alpha =0.\end{array}\right.$$

We choose $\alpha =1$ and $\beta =-u_1$ as solutions to the above system.

Let $F:M\to (Ni{l}^4,g_A)$ be the immersion defined by local coordinates $(u_1,u_2,u_3)$ as $F=(F_1(u_1,u_2,u_3),\dots ,F_4(u_1,u_2,u_3))$. Then, we can compute the derivatives of F:

$${\partial }_{u_1}F=(1,-a_4,-a_4{F}_1,-{\displaystyle \frac{a_4{F}_1^2}{2}}),{\partial }_{u_2}F=(0,0,1,0),{\partial }_{u_3}F=(0,0,0,1).$$

Integrating these, and using a suitable reparametrization, we obtain the simplified immersion F:

$$F_1=u_1+c_1,F_2={\displaystyle \frac{1}{6}}{u}_1^3-{\displaystyle \frac{1}{2}}C{u}_1^2-D{u}_1+c_2,F_3=u_2+c_3,F_4=u_3+c_4.$$

This parametrization matches case (4) of Theorem 1. Confirming M as a non-totally geodesic parallel hypersurface in $Ni{l}^4$.

For the ambient space $(Ni{l}^4,g_2^{\lambda })$: Consider the unit normal vector field $\xi ={\displaystyle \frac{1}{\sqrt{\lambda }}}{e}_3+a_4{e}_4$. The tangent space to M is spanned by the global frame:

$$Y_1=e_2,Y_2=\sqrt{\lambda }{e}_1-a_4{e}_3,Y_3=e_4.$$

The Frobenius integrability condition imposes:

$$\langle [e_2,\sqrt{\lambda }{e}_1-a_4{e}_3],\xi \rangle =\langle [e_4,\sqrt{\lambda }{e}_1-a_4{e}_3],\xi \rangle =0,$$

force ${\partial }_{e_2}(a_4)+\sqrt{\lambda }={\partial }_{e_4}(a_4)=0$. Using the metric (2), the non-vanishing components of the Levi-Civita connection in terms of $\left\{Y_i\right\}$ are:

$$\begin{array}{cc}\hfill & {\nabla }_{Y_1}{Y}_2={\nabla }_{Y_2}{Y}_1=-\lambda a_4{Y}_3+{\displaystyle \frac{\lambda }{2}}\xi ,\hfill \\ \hfill & {\nabla }_{Y_2}{Y}_2={\lambda }^{{\displaystyle \frac{3}{2}}}{a}_4{Y}_1+\sqrt{\lambda }{a}_4(Y_2(a_4-1))Y_3-\sqrt{\lambda }{Y}_2(a_4)\xi .\hfill \end{array}$$

By the Gauss formula (3), the non-zero components of second fundamental form h is determined by:

$$h(Y_1,Y_2)=h(Y_2,Y_1)={\displaystyle \frac{\lambda }{2}},h(Y_2,Y_2)=\sqrt{\lambda }{Y}_2(a_4).$$

The induced connection on M is given as follows:

$${\nabla }_{Y_1}^M{Y}_2={\nabla }_{Y_2}^M{Y}_1=-\lambda a_4{Y}_3,{\nabla }_{Y_2}^M{Y}_2={\lambda }^{{\displaystyle \frac{3}{2}}}{a}_4{Y}_1+\sqrt{\lambda }{a}_4(Y_2(a_4-1))Y_3.$$

Note that $h(Y_1,Y_2)=h(Y_2,Y_1)={\displaystyle \frac{\lambda }{2}}$, M is not totally geodesic. For parallel second fundamental form ${\nabla }^{M}h\equiv 0$, we impose:

$$({\nabla }_{Y_2}^{M}h)(Y_2,Y_2)=Y_2(Y_2·a_4)-{\lambda }^2{a}_4=0.$$

(9)

We construct explicit coordinates on M using a local frame ${\left\{u_i\right\}}_{i=1}^3$. It follows that the vector fields $Y_i={\partial }_{u_i}$$(i=1,2,3)$ are precisely the coordinate vector fields M. Solving the function (9) gives:

$$a_4=k_1{e}^{\lambda u_2}+k_2{e}^{-\lambda u_2}.$$

Let $F:M\to (Ni{l}^4,g_A)$ be the immersion defined by local coordinates $(u_1,u_2,u_3)$ as $F=(F_1(u_1,u_2,u_3),\dots ,F_4(u_1,u_2,u_3))$. Then, we can compute the derivatives of F:

$${\partial }_{u_1}F=(0,1,F_1,{\displaystyle \frac{F_1^2}{2}}),{\partial }_{u_2}F=(\sqrt{\lambda },0,-a_4,-a_4{F}_1),{\partial }_{u_3}F=(0,0,0,1).$$

Solving these and applying an appropriate reparameterization, we derive the simplified expression for the immersion F:

$$\begin{array}{cc}\hfill & F_1=\sqrt{\lambda }{u}_2+c_1,\hfill \\ \hfill & F_2=u_1+c_2,\hfill \\ \hfill & F_3={\displaystyle \frac{1}{\lambda }}(k_1{e}^{\lambda u_2}-k_2{e}^{-\lambda u_2})+\sqrt{\lambda }{u}_1(u_2+c_1)+c_3,\hfill \\ \hfill & F_4=u_3+c_4.\hfill \end{array}$$

This parametrization matches Case (5) of Theorem 1.

For the ambient space $(Ni{l}^4,g_4^{\lambda })$: Consider the unit normal vector field $\xi =\pm e_1$. The tangent space to M is spanned by the global frame:

$$Y_1=e_2,Y_2=e_3,Y_3=e_4.$$

Using the metric (2), the non-vanishing components of the Levi-Civita connection in terms of $\left\{Y_i\right\}$ are:

$${\nabla }_{Y_1}{Y}_1=\xi ,{\nabla }_{Y_2}{Y}_3={\nabla }_{Y_3}{Y}_2={\displaystyle \frac{\lambda }{2}}\xi .$$

By the Gauss formula (3), the Levi-Civita connection in terms of $\left\{Y_i\right\}$ on M is flat (${\nabla }_{Y_i}^M{Y}_j=0$). Thus M is flat and the vector fields $Y_i={\partial }_{u_i}$ ($i=1,2,3$), are precisely the coordinate vector fields M.

Let $F:M\to (Ni{l}^4,g_A)$ be the immersion defined by local coordinates $(u_1,u_2,u_3)$ as $F=(F_1(u_1,u_2,u_3),\dots ,F_4(u_1,u_2,u_3))$. Then, we can compute the derivatives of F:

$${\partial }_{u_1}F=(0,1,0,0),{\partial }_{u_2}F=(0,0,1,0),{\partial }_{u_3}F=(0,0,0,1).$$

Integrating these, and using a suitable reparametrization, we obtain the simplified immersion F:

$$F_1=c_1,F_2=u_1+c_2,F_3=u_2+c_3,F_4=u_3+c_4.$$

This coincides with case (1) of Theorem 1.

Again from the above equations and the Gauss formula, we deduce that the second fundamental form is determined by:

$$\begin{array}{c}\hfill h(Y_1,Y_1)=1,h(Y_2,Y_3)={\displaystyle \frac{\lambda }{2}},h(Y_3,Y_2)={\displaystyle \frac{\lambda }{2}}.\end{array}$$

This confirms M is a parallel hypersurface, maintaining constant shape under ambient translations. However, since $h\ne 0$, M is not totally geodesic. □

5. Conclusions

This study investigates the geometry of four-dimensional nilpotent Lie groups, denoted as $Ni{l}^4$, equipped with non-flat left-invariant Lorentzian metrics. Our results demonstrate that, except for the specific case of $(Ni{l}^4,g_3^{\lambda })$, all other configurations admit parallel hypersurfaces. However, none of these cases exhibit totally geodesic hypersurfaces. Together with the recently solved case of $H_3\times \mathbb{R}$ [13], the investigation of $Ni{l}^4$ in this work thus establishes a full classification of parallel hypersurfaces for all four-dimensional nilpotent Lie groups. For future research, we intend to explore other geometrically significant models and extend our investigation to broader pseudo-Riemannian settings, with the aim of constructing further examples of parallel and totally geodesic hypersurfaces.