Generalized Ricci Solitons of Three-Dimensional Lorentzian Lie Groups Associated Canonical Connections and Kobayashi-Nomizu Connections

Abstract

In this paper, we study the affine generalized Ricci solitons on three-dimensional Lorentzian Lie groups associated canonical connections and Kobayashi-Nomizu connections and we classifying these left-invariant affine generalized Ricci solitons with some product structure.

1 Introduction

The notion of generalized Ricci soliton or Einstein-type manifolds is introduced by Catino et al. as a generalization of Einstein spaces [5]. Study of the generalization Ricci soliton, over different geometric spaces is one of interesting topics in geometry and normalized physics. A pseudo-Riemannian manifold (M, g) is called an generalized Ricci soliton if there exists a vector field $X\in \mathcal {X}(M)$ and a smooth function $\lambda$ on M such that

$$\begin{aligned} \alpha Ric+\frac{\beta }{2}\mathcal {L}_{X}g+\mu X^{\flat }\otimes X^{\flat }=(\rho S+\lambda )g, \end{aligned}$$

(1)

for some constants $\alpha ,\beta , \mu ,\rho \in \mathbb {R}$, with $(\alpha ,\beta , \mu )\ne (0,0,0)$, where $\mathcal {L}_{X}$ denotes the Lie derivative in the direction of X, $X^{\flat }$ denotes a 1-form such that $X^{\flat }(Y)=g(X,Y)$, S is the scalar curvature, and Ric is the Ricci tensor. The generalized Ricci soliton becomes

• (i) the homothetic vector field equation when $\alpha =\mu =\rho =0$ and $\beta \ne 0$,

• (ii) the Ricci soliton equation when $\alpha =1$, $\mu =0$, and $\rho =0$,

• (iii) the Ricci-Bourguignon soliton ( or $\rho$-Einstein soliton equation when $\alpha =1$ and $\mu =0$.

In the special case that (M, g) is a Lie group and g is a left-invariant metric, we say that g is a left-invariant generalized Ricci soliton on M if the Eq. (1) holds.

In [11, 14, 16, 17, 21, 22], Einstein manifolds associated to affine connections were studied and affine Ricci solitons had been studied in [7, 10, 12, 13, 15]. In [4], Calvaruso studied the Eq. (1) for $\rho =0$ on three-dimensional generalized Lie groups. Also, in [20] Wang classified affine Ricci solitons associated to canonical connections and Kobayashi-Nomizu connections on three-dimensional Lorentzian Lie groups. In [8], Etayo and Santamaria investigated the canonical connection and the Kobayashi-Nomizu connection for a product structure. Motivated by [1, 19, 23, 24], we consider the distribution $V=span\{e_{1},e_{2}\}$ and $V^{\perp }=span\{e_{3}\}$ for the three dimensional Lorentzian Lie group $G_{i}$, $i = 1,.....,7$, with product structure J such that $Je_{1}=e_{1},\,\,Je_{2}=e_{2}$, and $Je_{3}=-e_{3}$. Then we obtain affine generalized Ricci solitons associated to the canonical connection and the Kobayashi-Nomizu connection.

The paper is organaized as follows. In Sect. 2 we review some necessary concepts on three-dimensional Lie groups which be used throughout this paper. In the Sect. 3 we state the main results and their proof.

2 Three-Dimensional Lorentzian Lie Groups

In the following we give a brief description of all three-dimensional unimodular and non-unimodular Lie groups. Complete and simply connected three-dimensional Lorentzian homogeneous manifolds are either symmetric or a Lie group with left-invariant Lorentzian metric [3].

2.1 Unimodular Lie Groups

Let $\{ e_{1}, e_{2},e_{3}\}$ be an orthonormal basis of signature $(+\,+\,-)$. We denote the Lorentzian vector product on $\mathbb {R}_{1}^{3}$ induced by the product of the para-quaternions by $\times$ i.e.,

$$\begin{aligned} e_{1}\times e_{2}=-e_{3},\,\,\,\,e_{2}\times e_{3}=-e_{1},\,\,\,e_{3}\times e_{1}=-e_{2}. \end{aligned}$$

Then the Lie bracket $[\, ,\,]$ defines the corresponding Lie algebra $\mathfrak {g}$, which is unimodular if and only if the endomorphism L defined by $[Z,Y]=L(Z\times Y)$ is self-adjoint and non-unimodular if L is not self-adjoint [18]. By assuming the different types of L, we get the following four classes of unimodular three-dimensional Lie algebra [9].

$\mathfrak {g}_{1}$::

If L is diagonalizable with eigenvalues $\{a, b, c\}$ with respect to an orthonormal basis $\{ e_{1}, e_{2},e_{3}\}$ of signature $(+\,+\,-)$, then the corresponding Lie algebra is given by

$$\begin{aligned} \qquad [e_{1}, e_{2}]=-c e_{3},\,\,\,\,[e_{1}, e_{3}]=-b e_{2},\,\,\,[e_{2}, e_{3}]=a e_{1}. \end{aligned}$$

$\mathfrak {g}_{2}$::

Assume L has a complex eigenvalues. Then, with respect to an orthonormal basis $\{ e_{1}, e_{2},e_{3}\}$ of signature $(+\,+\,-)$, one has

$$\begin{aligned} L=\left( \begin{array}{ccc} a &{}0 &{} 0 \\ 0 &{} c &{}-b \\ 0 &{}b&{} c \\ \end{array} \right) ,\qquad \quad b\ne 0, \end{aligned}$$

then the corresponding Lie algebra is given by

$$\begin{aligned} \qquad [e_{1}, e_{2}]=b e_{2}-c e_{3},\,\,\,\,[e_{1}, e_{3}]=-c e_{2}-b e_{3},\,\,\,[e_{2}, e_{3}]=a e_{1}. \end{aligned}$$

$\mathfrak {g}_{3}$::

Assume L has a triple root of its minimal polynomial. Then, with respect to an orthonormal basis $\{ e_{1}, e_{2},e_{3}\}$ of signature $(+\,+\,-)$, the corresponding Lie algebra is given by

$$\begin{aligned}{}[e_{1}, e_{2}]=a e_{1}-b e_{3},\,\,\,\,[e_{1}, e_{3}]=-a e_{1}-b e_{2},\,\,\,\, [e_{2}, e_{3}]=b e_{1}+a e_{2}+a e_{3},\,\,\,a\ne 0. \end{aligned}$$

$\mathfrak {g}_{4}$::

Assume L has a double root of its minimal polynomial. Then, with respect to an orthonormal basis $\{ e_{1}, e_{2},e_{3}\}$ of signature $(+\,+\,-)$, the corresponding Lie algebra is given by

$$\begin{aligned} \qquad [e_{1}, e_{2}]=- e_{2}-(2d-b) e_{3},\,\,\,\,[e_{1}, e_{3}]=-b e_{2}+ e_{3},\,\,\,[e_{2}, e_{3}]=a e_{1},\,\,\,\,d=\pm 1. \end{aligned}$$

2.2 Non-unimodular Lie Groups

Next we treat the non-unimodular case. Let $\mathfrak {G}$ denotes a special class of the solvable Lie algebra $\mathfrak {g}$ such that [x, y] is a linear combination of x and y for any $x,y\in \mathfrak {g}$. From [6], the non-unimodular Lorentzian Lie algebras of non-constant sectional curvature not belonging to class $\mathfrak {G}$ with respect to a pseudo-orthonormal basis $\{e_{1},e_{2}, e_{3}\}$ with $e_{3}$ time-like are one of the following:

$\mathfrak {g}_{5}$::

$$\begin{aligned} &{}[e_{1}, e_{2}]=0,\,\,\,\,[e_{1}, e_{3}]=a e_{1}+b e_{2},\,\,\,\\ &[e_{2}, e_{3}]=c e_{1}+d e_{2},\,\,\,\,a+d\ne 0,\,\,\,\,ac+bd=0. \end{aligned}$$

$\mathfrak {g}_{6}$::

$$\begin{aligned} &{}[e_{1}, e_{2}]=a e_{2}+b e_{3},\,\,\,\,[e_{1}, e_{3}]=c e_{2}+d e_{3},\,\,\, \\ & [e_{2}, e_{3}]=0,\,\,\,\,a+d\ne 0,\,\,\,\,ac-bd=0. \end{aligned}$$

$\mathfrak {g}_{7}$::

$$\begin{aligned} &[e_{1}, e_{2}]=- ae_{1}-be_{2}-b e_{3},\,\,\,\,[e_{1}, e_{3}]=ae_{1}+b e_{2}+ be_{3},\\&[e_{2}, e_{3}]=c e_{1}+de_{2}+de_{3},\,\,\,\,a+d\ne 0,\,\,\,\,ac=0. \end{aligned}$$

Throughout this paper, we assume that $G_{i} ,{\mkern 1mu} i = 1,2,.....,7$ are the connected, simply connected three-dimensional Lie group equipped with a left-invariant Lorentzian metric g and having Lie algebra $g_{i} ,{\mkern 1mu} i = 1,2,.....,7$, respectively. Let $\nabla$ be the Levi-Civita connection of $G_{i}$ and $R(X,Y)Z=[\nabla _{X},\nabla _{Y}]Z-\nabla _{[X,Y]}Z$ be its curvature tensor. The Ricci tensor of $(G_{i},g)$ with respect to orthonormal basis $\{ e_{1}, e_{2},e_{3}\}$ of signature $(+\,+\,-)$ is defined by

$$\begin{aligned} Ric(X,Y)=-g(R(X,e_{1})Y,e_{1})-g(R(X,e_{2})Y,e_{2})+g(R(X,e_{3})Y,e_{3}). \end{aligned}$$

We consider a product structure J on $G_{i}$ by $Je_{1}=e_{1}, \, Je_{2}=e_{2},\, Je_{3}=-e_{3}$. Similar [8], we consider the canonical connection and the Kobayashi-Nomizu connection as

$$\begin{aligned} \nabla _{X}^{0}Y=\nabla _{X}Y-\frac{1}{2}(\nabla _{X}J)JY,\qquad \nabla _{X}^{1}Y=\nabla _{X}^{0}Y-\frac{1}{4}[(\nabla _{Y}J)JX-(\nabla _{JY}J)X], \end{aligned}$$

respectively. We define

$$\begin{aligned} R^{i}(X,Y)Z=[\nabla _{X}^{i},\nabla _{Y}^{i}]Z-\nabla _{[X,Y]}^{i}Z,\,\,\,i=0,1, \end{aligned}$$

and the Ricci tensors of $(G_{i},g)$ associated to the canonical connection and the Kobayashi-Nomizu connection are defined by

$$\begin{aligned} Ric^{i}(X,Y)=-g(R^{i}(X,e_{1})Y,e_{1})-g(R^{i}(X,e_{2})Y,e_{2})+g(R^{i}(X,e_{3})Y,e_{3}),\,\,\,i=0,1. \end{aligned}$$

Let

$$\begin{aligned} \widetilde{Ric}^{i}(X,Y)=\frac{Ric^{i}(X,Y)+Ric^{i}(Y,X)}{2},\,\,\,i=0,1. \end{aligned}$$

Similar to definition of $(\mathcal {L}_{V}g)$ where $(\mathcal {L}_{X}g)(Y,Z)=g(\nabla _{Y}V,Z)+g(Y,\nabla _{Z}V)$, we define

$$\begin{aligned} (\mathcal {L}_{V}^{i}g)(Y,Z):=g(\nabla _{Y}^{i}V,Z)+g(Y,\nabla _{Z}^{i}V),\,\,\,i=0,1. \end{aligned}$$

Definition 1

The Lie group (G, g, J) is called the affine generalized Ricci soliton associated to the connection $\nabla ^{i},\,i=0,1$ if it satisfies

$$\begin{aligned} \alpha \widetilde{Ric}^{i}(Y,Z)+\frac{\beta }{2}\mathcal {L}_{X}^{i}g(Y,Z)+\mu X^{\flat }\otimes X^{\flat }(Y,Z)=(\rho \widetilde{S}^{i}+\lambda )g(Y,Z),\,\,\,i=0,1, \end{aligned}$$

(2)

where $\widetilde{S}^{i}=g^{jk}\widetilde{Ric}_{jk}^{i}$.

Throughout this paper for prove of our results we use the results of [19, 20].

3 Lorentzian Affine Generalized Ricci Solitons on 3D Lorentzian Lie Groups

In this section, we investigate the existence of left-invariant solutions to Eq. (2) on the Lorentzian Lie groups discussed in Sect. 2. We completely solve the corresponding equations and obtain a complete description of all left-invariant affine generalized Ricci solitons.

Theorem 1

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{1},g,J,X)$ are the following:

• (i) $\mu =\lambda =0$, $a+b-c=0$, and for all $x_{1},x_{2},x_{3},\alpha , \beta , \rho$ such that $(\alpha ,\beta , \mu )\ne (0,0,0)$,

• (ii) $\mu =0$, $a+b-c\ne 0$, $\alpha =0$, $\beta \ne 0$, $x_{1}=x_{2}=0$, $\lambda =\rho c(a+b-c)$, and for all $x_{3}, \rho$,

• (iii) $\mu =0$, $a+b-c\ne 0$, $\alpha \ne 0$, $c=\beta =\lambda =0$, and for all $x_{1},x_{2},x_{3}, \rho$,

• (iv) $\mu =0$, $a+b-c\ne 0$, $\alpha \ne 0$, $c=\lambda =0$,$\beta \ne 0$, $x_{1}=x_{2}=0$, and for all $x_{3}, \rho$,

• (v) $\mu \ne 0$, $x_{1}=x_{2}=0$, $\lambda =(\rho -\frac{1}{2}\alpha ) c(a+b-c)$, $x_{3}^{2}=\frac{\rho c(a+b-c)-\lambda }{\mu }$, and for all $x_{3},\alpha , \beta , \rho ,a,b,c$ such that $\frac{\rho c(a+b-c)-\lambda }{\mu }\ge 0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{0}=\left( \begin{array}{ccc} -\frac{1}{2}c(a+b-c)&{}0 &{}0 \\ 0 &{} -\frac{1}{2}c(a+b-c) &{}0 \\ 0 &{}0 &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} 0&{}0 &{}-\frac{1}{2}x_{2}(a+b-c) \\ 0 &{} 0&{} \frac{1}{2}x_{1}(a+b-c)\\ -\frac{1}{2}x_{2}(a+b-c) &{}\frac{1}{2}x_{1}(a+b-c)&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}=-c(a+b-c)$ and $X^{\flat }\otimes X^{\flat }(e_{i},e_{j})=\epsilon _{i}\epsilon _{j}x_{i}x_{j}$ where $(\epsilon _{1},\epsilon _{2},\epsilon _{3})=(1,1,-1)$. Hence, by Eq. (2) there exists a affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ if and only if the following system of equations is satisfied

$$\begin{aligned} {\left\{ \begin{array}{ll} -\frac{1}{2}\alpha c(a+b-c)+\mu x_{1}^{2}=-\rho c(a+b-c)+\lambda ,\\ \mu x_{1}x_{2}=0,\\ -\frac{\beta }{4}x_{2}(a+b-c)-\mu x_{1}x_{3}=0,\\ -\frac{1}{2}\alpha c(a+b-c)+\mu x_{2}^{2}=-\rho c(a+b-c)+\lambda ,\\ \frac{\beta }{4}x_{1}(a+b-c)-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho c(a+b-c)-\lambda . \end{array}\right. } \end{aligned}$$

(3)

Using the first and fourth equations of the system Eq. (3) we have $\mu (x_{1}^{2}-x_{2}^{2})=0$. From the third and fiveth equations of the system Eq. (3) we get

$$\begin{aligned} \frac{\beta }{4}(x_{1}-x_{2})(a+b-c)-\mu x_{3}( x_{1}+x_{2})=0. \end{aligned}$$

Multiplying both sides of last equality by $(x_{1}-x_{2})$ we conclude

$$\begin{aligned} \beta (x_{1}-x_{2})^{2}(a+b-c)=0. \end{aligned}$$

(4)

The second equation of the system Eq. (3) implies that $\mu =0$, or $x_{1}=0$ or $x_{2}=0$. Suppose that $\mu =0$. In this case, the system Eq. (3) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha c(a+b-c)=0,\\ \beta x_{2}(a+b-c)=0,\\ \beta x_{1}(a+b-c)=0,\\ \rho c(a+b-c)=\lambda . \end{array}\right. } \end{aligned}$$

(5)

If $a+b-c=0$ then the system Eq. (5) holds for any $x_{1},x_{2}$, and $x_{3}$. If $a+b-c\ne 0$ for the cases (ii)–(iv) the sytem Eq. (5) holds. Now we assume that $\mu \ne 0$ and $x_{1}=0$, then $x_{2}=0$ and the system Eq. (3) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\frac{1}{2}\alpha c(a+b-c)=-\rho c(a+b-c)+\lambda ,\\ \mu x_{3}^{2}=\rho c(a+b-c)-\lambda . \end{array}\right. } \end{aligned}$$

(6)

This shows that the case (v) holds. $\square$

Theorem 2

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{1},g,J,X)$ are the following:

• (i) $\mu =0$, $c=0$, $\lambda =0$, $\beta =0$, and for all $a,b, x_{1},x_{2},x_{3},\alpha ,\rho$ such that $\alpha \ne 0$,

• (ii) $\mu =0$, $c=0$, $\lambda =0$, $\beta \ne 0$, $a=b=0$, and for all $x_{1},x_{2},x_{3},\alpha ,\rho$,

• (iii) $\mu =0$, $c=0$, $\lambda =0$, $\beta \ne 0$, $a=x_{1}=0$, and for all $b,x_{2},x_{3},\alpha ,\rho$,

• (iv) $\mu =0$, $c=0$, $\lambda =0$, $\beta \ne 0$, $a\ne 0$, $x_{2}=b=0$, and for all $x_{1},x_{3},\alpha ,\rho$,

• (v) $\mu =0$, $c=0$, $\lambda =0$, $\beta \ne 0$, $a\ne 0$, $x_{2}=x_{1}=0$, and for all $b,x_{3},\alpha ,\rho$,

• (vi) $\mu =0$, $c\ne 0$, $\lambda =\rho c(a+b)$, $b=0$, $\beta =a=0$, and for all $x_{1},x_{2},x_{3},\alpha ,\rho$ such that $\alpha \ne 0$,

• (vii) $\mu =0$, $c\ne 0$, $\lambda =\rho c(a+b)$, $b=0$, $\beta \ne 0$, $a=0$, and for all $x_{1},x_{2},x_{3},\alpha ,\rho$,

• (viii) $\mu =0$, $c\ne 0$, $\lambda =\rho c(a+b)$, $b=0$, $\beta \ne 0$, $a\ne 0$, $x_{2}=0$, and for all $x_{1},x_{3},\alpha ,\rho$,

• (ix) $\mu =0$, $c\ne 0$, $\lambda =\rho c(a+b)$, $b\ne 0$, $\alpha =0$, $a=x_{1}=0$, and for all $x_{2},x_{3},\beta ,\rho$, such that $\beta \ne 0$,

• (x) $\mu =0$, $c\ne 0$, $\lambda =\rho c(a+b)$, $b\ne 0$, $\alpha =0$, $a\ne 0$, $x_{2}=x_{1}=0$, and for all $x_{3},\beta ,\rho$, such that $\beta \ne 0$,

• (xi) $\mu \ne 0$, $x_{1}=0$, $x_{3}=0$, $c=0$, $\lambda =x_{2}=0$, for all $a,b, \alpha , \beta , \rho$,

• (xii) $\mu \ne 0$, $x_{1}=0$, $x_{3}=0$, $c\ne 0$, $\alpha =0$, $\lambda =\rho c(a+b)$, $x_{2}=0$, for all $a,b, \beta , \rho$,

• (xiii) $\mu \ne 0$, $x_{1}=0$, $x_{3}=0$, $c\ne 0$, $\alpha \ne 0$, $b=0$, $\lambda =x_{2}=a=0$, for all $\beta , \rho$,

• (xiv) $\mu \ne 0$, $x_{1}=0$, $x_{3}=0$, $c\ne 0$, $\alpha \ne 0$, $b=0$, $x_{2}\ne 0$, $\lambda =\rho ca$, $\beta =0$, $x_{2}^{2}=\frac{a\alpha c}{\mu }$ for all $a, \rho$,

• (xv) $\mu \ne 0$, $x_{1}=0$, $x_{3}\ne 0$, $x_{2}=0$, $x_{3}^{2}=\frac{ac\alpha }{\mu }>0$, for all $a,b,c, \alpha , \beta , \rho , \lambda$ such that $bc\alpha =ac\alpha =\rho c (a+b)-\lambda$,

• (xvi) $\mu \ne 0$, $x_{1}\ne 0$, $x_{2}=x_{3}=0$, $\lambda =\rho c(a+b)$, for all $a,b,c,\alpha ,\rho , \beta$ such that $\beta b=ac\alpha =0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{1}=\left( \begin{array}{ccc} -bc&{}0 &{}0 \\ 0 &{} -ac &{}0 \\ 0 &{}0 &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} 0&{}0 &{}-ax_{2} \\ 0 &{} 0&{} bx_{1}\\ -ax_{2} &{}bx_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}=-c(a+b)$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -bc\alpha +\mu x_{1}^{2}=-\rho c(a+b)+\lambda ,\\ \mu x_{1}x_{2}=0,\\ -\frac{\beta }{2}ax_{2}-\mu x_{1}x_{3}=0,\\ -ac\alpha +\mu x_{2}^{2}=-\rho c(a+b)+\lambda ,\\ \frac{\beta }{2}bx_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho c(a+b)-\lambda . \end{array}\right. } \end{aligned}$$

(7)

The second equation of the system Eq. (7) implies that $\mu =0$ or $x_{1}=0$ or $x_{2}=0$. We consider $\mu =0$, then the first equation yields $bc\alpha =0$. If $c=0$ then we get $\lambda =0$ and the cases (i)-(v) hold. If we assume that $c\ne 0$ and $\lambda =\rho c(a+b)$ and in this we obtain the cases (vi)-(x). Now, we consider the case $\mu \ne 0$ and $x_{1}=0$. In this case the system Eq. (7) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} -bc\alpha =-\rho c(a+b)+\lambda ,\\ \beta ax_{2}=0,\\ -ac\alpha +\mu x_{2}^{2}=-\rho c(a+b)+\lambda ,\\ x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho c(a+b)-\lambda . \end{array}\right. } \end{aligned}$$

(8)

The fourth equation of the system Eq. (8) implies that $x_{2}=0$ or $x_{3}=0$. If $x_{3}=0$ then we obtain the cases (xi)-(xiv). If $x_{3}\ne 0$ and $x_{2}=0$ then the case (xv) holds. Also, if we consider $\mu \ne 0$ and $x_{1}\ne 0$ then $x_{2}=0$ and the case (xvi) is true. $\square$

Theorem 3

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{2},g,J,X)$ are the following:

• (i) $\mu =0$, $\beta \ne 0$, $x_{1}=x_{2}=\alpha =0$, $\lambda =\rho (2b^{2}+ac)$, for all $a,b,c, x_{3}, \rho$ such that $b\ne 0$,

• (ii) $\mu \ne 0$, $x_{2}=0$, $x_{1}=0$, $\alpha =0$, $\lambda =\rho (2b^{2}+ac)$, $x_{3}=0$, for all $\beta , \rho , a,b,c$ such that $b\ne 0$.

• (iii) $\mu \ne 0$, $x_{2}=0$, $x_{1}=0$, $\alpha \ne 0$, $a=2c$, $\lambda =(2\rho -\alpha )(b^{2}+c^{2})$, $x_{3}^{2}=\frac{\alpha }{\mu }(b^{2}+c^{2})$, for all $\beta , \rho , b$ such that $b\ne 0$ and $\alpha \mu \ge 0$,

• (iv) $\mu \ne 0$, $x_{2}=0$, $x_{1}=-\frac{\beta b}{\mu }\ne 0$, $x_{3}=0$, $\lambda =\rho (2b^{2}+ac)$, for all $a,b,c, \alpha , \beta ,\rho$ such that $\beta \ne 0$, $\alpha \mu (2b^{2}+ac)+\beta ^{2}b^{2}=0$, and $\alpha \mu (2c+a)-\beta ^{2}a=0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{0}=\left( \begin{array}{ccc} -(b^{2}+\frac{ac}{2})&{}0 &{}0 \\ 0 &{} -(b^{2}+\frac{ac}{2}) &{}\frac{bc}{2}-\frac{ab}{4} \\ 0 &{}\frac{bc}{2}-\frac{ab}{4} &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} 0&{}bx_{2} &{}-\frac{a}{2}x_{2} \\ bx_{2} &{} -2bx_{1}&{} \frac{a}{2}x_{1}\\ -\frac{a}{2}x_{2} &{} \frac{a}{2}x_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Then $\widetilde{S}= -(2b^{2}+ac)$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+\frac{ac}{2})+\mu x_{1}^{2}=-\rho (2b^{2}+ac)+\lambda ,\\ \frac{\beta }{2}bx_{2}+ \mu x_{1}x_{2}=0,\\ -\frac{\beta a}{4}x_{2}-\mu x_{1}x_{3}=0,\\ -\alpha (b^{2}+\frac{ac}{2})-\beta b x_{1}+\mu x_{2}^{2}=-\rho (2b^{2}+ac)+\lambda ,\\ \alpha (\frac{bc}{2}-\frac{ab}{4})+\frac{\beta a}{4}x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2b^{2}+ac)-\lambda . \end{array}\right. } \end{aligned}$$

(9)

At the first we assume $\mu =0$. In this case, the system Eq. (9) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+\frac{ac}{2})=0,\\ \beta x_{2}=0,\\ \beta x_{1}=0,\\ \alpha (\frac{bc}{2}-\frac{ab}{4})+\frac{\beta a}{4}x_{1}=0,\\ \rho (2b^{2}+ac)=\lambda . \end{array}\right. } \end{aligned}$$

(10)

The second equation of Eq. (10) implies that $\beta =0$ or $x_{2}=0$. If $\beta =0$ then $\alpha \ne 0$ and the fourth equation of the system Eq. (10) yields $a=2c$ and replacing it in the first equation we obtain $b^{2}+c^{2}=0$ which is a contradiction. Thus $\beta \ne 0$ and $x_{1}=x_{2}=\alpha =0$.

Now we consider $\mu \ne 0$. Using the first and fourth equations of Eq. (9) we obtain

$$\begin{aligned} \mu x_{1}^{2}+\beta b x_{1}=\mu x_{2}^{2}. \end{aligned}$$

(11)

The second equation of the system Eq. (9) implies that $x_{2}=0$ or $x_{1}=-\frac{\beta b}{2\mu }$. If $x_{2}\ne 0$ then $x_{1}=-\frac{\beta b}{2\mu }$ and plugging it in Eq. (11) we get $x_{2}^{2}+\frac{\beta ^{2}b^{2}}{4\mu ^{2}}=0$ which is a contradiction. Therefore $x_{2}=0$ and in this case we have

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+\frac{ac}{2})+\mu x_{1}^{2}=-\rho (2b^{2}+ac)+\lambda ,\\ \mu x_{1}^{2}+\beta b x_{1}=0,\\ x_{1}x_{3}=0,\\ -\alpha (b^{2}+\frac{ac}{2})-\beta b x_{1}=-\rho (2b^{2}+ac)+\lambda ,\\ \alpha (\frac{bc}{2}-\frac{ab}{4})+\frac{\beta a}{4}x_{1}=0,\\ \mu x_{3}^{2}=\rho (2b^{2}+ac)-\lambda . \end{array}\right. } \end{aligned}$$

(12)

The third equation of the system Eq. (12) implies that $x_{1}=0$ or $x_{3}=0$. If $x_{1}=0$ then $\alpha (2c-a)=0$. Thus $\alpha =0$ or $a=2c$. In the case $\alpha =0$ we have $\lambda =\rho (2b^{2}+ac)$ and $x_{3}=0$. In the case $\alpha \ne 0$ and $a=2c$ we get $\lambda =(2\rho -\alpha )(b^{2}+c^{2})$ and $x_{3}^{2}=\frac{\alpha }{\mu }(b^{2}+c^{2})$. Now we assume that $\mu \ne 0$, $x_{2}=0$, $x_{1}\ne 0$, and $x_{3}=0$. In this case we have (iv). $\square$

Theorem 4

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{2},g,J,X)$ are the following:

• (i) $\mu =0$, $\alpha =0$, $\beta \ne 0$, $x_{1}=x_{2}=x_{3}=0$, $\lambda =\rho (2b^{2}+c^{2}+ac)$, for all $\rho , a,b,c$ such that $b\ne 0$,

• (ii) $\mu \ne 0$, $x_{2}=x_{3}=0$, $\beta =0$, $\alpha =0$, $x_{1}=0$, $\lambda =\rho (2b^{2}+c^{2}+ac)$, for all $\rho , a,b,c$ such that $b\ne 0$,

• (iii) $\mu \ne 0$, $x_{2}=x_{3}=0$, $\beta \ne 0$, $c=0$, $\alpha =0$, $x_{1}=0$, $\lambda =\rho (2b^{2})$, for all $\rho , a,b,c$ such that $b\ne 0$,

• (iv) $\mu \ne 0$, $x_{2}=x_{3}=0$, $\beta \ne 0$, $c=0$, $\alpha \ne 0$, $a=0$, $x_{1}=-\frac{\alpha b}{\beta }$, $\lambda =\rho (2b^{2}+c^{2})$, for all $\rho , a,b,c$ such that $b\ne 0$, $\mu \alpha b^{2}=\beta ^{2}(b^{2}+c^{2})$,

• (v) $\mu \ne 0$, $x_{2}=x_{3}=0$, $\beta \ne 0$, $c\ne 0$, $x_{1}=\frac{\alpha ab}{\beta c}$, $\lambda =\rho (2b^{2}+c^{2}+ac)$, for all $\alpha , \rho , a,b,c$ such that $b\ne 0$, $\alpha a b^{2}=-\alpha c(b^{2}+ac)$, $\mu (\alpha ab)^{2}=\alpha \beta ^{2} c^{2}(b^{2}+c^{2})$,

• (vi) $\mu \ne 0$, $x_{2}=0$, $x_{3}^{2}=\frac{\alpha }{\mu }(b^{2}+c^{2})-\frac{\beta ^{2}b^{2}}{4\mu ^{2}}\ne 0$, $x_{1}=\frac{\beta b}{2\mu }$, $\lambda = \rho (2b^{2}+c^{2}+ac)-\alpha (b^{2}+c^{2})-\frac{\beta ^{2}b^{2}}{2\mu }$, for all $\alpha , \beta , \rho , a,b,c$ such that $b\ne 0$, $2a\alpha \mu =\beta ^{2}c$, $3\beta ^{2}b^{2}-4\mu \alpha c(c-a)=0$,

• (vii) $\mu \ne 0$, $x_{2}^{2}=-\frac{\alpha }{\mu }c(c-a)\ne 0$, $x_{1}=0$, $\beta =0$, $x_{3}=\frac{\alpha }{\mu }(b^{2}+c^{2})$, $\lambda =-\alpha (b^{2}+c^{2})+\rho (2b^{2}+c^{2}+ac)$ for all $\alpha , \rho , a,b,c$ such that $b\ne 0$, $-4c(c-a)(b^{2}+c^{2})=a^{2}b^{2}$, $\frac{\alpha }{\mu }\ge 0$, $-c(c-a)\ge 0$,

• (viii) $\mu \ne 0$, $x_{2}\ne 0$, $x_{1}=-\frac{\beta b}{2\mu }$, $\beta \ne 0$, $x_{3}=\frac{a}{2b}x_{2}=-\frac{2\alpha \mu ab+c\beta ^{2}b}{4\mu ^{2}}$, $\lambda =-\alpha (b^{2}+c^{2})+\frac{\beta ^{2}b^{2}}{4\mu }+\rho (2b^{2}+c^{2}+ac)$, for all $\alpha , \rho , a,b,c$ such that $x_{2}^{2}=-(\frac{\beta b}{2\mu })^{2}-\frac{\alpha }{\mu }c(c-a)$, $x_{3}^{2}=\frac{\alpha }{\mu } (b^{2}+c^{2})-\frac{\beta ^{2}b^{2}}{4\mu ^{2}}>0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{1}=\left( \begin{array}{ccc} -(b^{2}+c^{2})&{}0 &{}0 \\ 0 &{} -(b^{2}+ac) &{}-\frac{ab}{2} \\ 0 &{}-\frac{ab}{2} &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} 0&{}bx_{2} &{}-ax_{2}+bx_{3} \\ bx_{2} &{} -2bx_{1}&{} cx_{1}\\ ax_{2}+bx_{3} &{}cx_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}= -(2b^{2}+c^{2}+ac)$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+c^{2})+\mu x_{1}^{2}=-\rho (2b^{2}+c^{2}+ac)+\lambda ,\\ \frac{\beta }{2} bx_{2}+\mu x_{1}x_{2}=0,\\ \frac{\beta }{2}(-ax_{2}+bx_{3})-\mu x_{1}x_{3}=0,\\ -\alpha (b^{2}+ac)-\beta bx_{1}+\mu x_{2}^{2}=-\rho (2b^{2}+c^{2}+ac)+\lambda ,\\ -\alpha \frac{ab}{2}+\frac{\beta }{2}cx_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2b^{2}+c^{2}+ac)-\lambda . \end{array}\right. } \end{aligned}$$

(13)

We first consider $\mu =0$. In this case, the system Eq. (13) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+c^{2})=0,\\ \beta bx_{2}=0,\\ \beta (-ax_{2}+bx_{3})=0,\\ -\alpha (b^{2}+ac)-\beta bx_{1}=0,\\ -\alpha ab+\beta cx_{1}=0,\\ \rho (2b^{2}+c^{2}+ac)=\lambda . \end{array}\right. } \end{aligned}$$

(14)

Since $b\ne 0$, the first equation of Eq. (14) implies that $\alpha =0$. Due to $(\alpha , \beta , \mu )\ne (0,0,0)$ we conclude $\beta \ne 0$. Then the second equation of the system Eq. (14) yields $x_{2}=0$. Using, the third and fourth equations of Eq. (14) we obtain $x_{1}=x_{3}=0$.

Now we consider $\mu \ne 0$. The second equation of the system Eq. (13) implies that $x_{2}=0$ or $x_{1}=-\frac{\beta b}{2\mu }$. If $x_{2}=0$ then we get

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+c^{2})+\mu x_{1}^{2}+\mu x_{3}^{2}=0,\\ \frac{\beta }{2}bx_{3}-\mu x_{1}x_{3}=0,\\ -\alpha (b^{2}+ac)-\beta bx_{1}+\mu x_{3}^{2}=0,\\ -\alpha \frac{ab}{2}+\frac{\beta }{2}cx_{1}=0,\\ \mu x_{3}^{2}=\rho (2b^{2}+c^{2}+ac)-\lambda . \end{array}\right. } \end{aligned}$$

(15)

From the second equation of the system Eq. (15) we obtain $x_{3}=0$ or $x_{1}=\frac{\beta b}{2\mu }$. If $x_{3}=0$ then the cases (ii)-(v) hold. If $x_{3}\ne 0$ and $x_{1}=\frac{\beta b}{2\mu }$ then the case (vi) holds. Now we assume that $\mu \ne 0$, $x_{2}\ne 0$ and $x_{1}=-\frac{\beta b}{2\mu }$. In this cases, the system Eq. (13) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (b^{2}+c^{2})+\frac{\beta ^{2}b^{2}}{4\mu }=-\rho (2b^{2}+c^{2}+ac)+\lambda ,\\ \frac{\beta }{2} bx_{2}+\mu x_{1}x_{2}=0,\\ \beta (-ax_{2}+2bx_{3})=0,\\ -\alpha (b^{2}+ac)+\frac{\beta ^{2}b^{2}}{2\mu }+\mu x_{2}^{2}=-\rho (2b^{2}+c^{2}+ac)+\lambda ,\\ -\alpha \frac{ab}{2}-\frac{c\beta ^{2}b}{4\mu }-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2b^{2}+c^{2}+ac)-\lambda . \end{array}\right. } \end{aligned}$$

(16)

Thus the cases (vii)-(viii) are true. $\square$

Theorem 5

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{3},g,J,X)$ are the following:

• (i) $\mu =0$, $\alpha =0$, $\beta \ne 0$, $x_{1}=x_{2}=0$, for all $\rho , a,b,c, x_{3}$ such that $a\ne 0$,

• (ii) $\mu \ne 0$, $x_{1}=0$, $x_{2}=0$, $x_{3}=0$, $\alpha =0$, $\lambda =\rho (2a^{2}+b^{2})$, for all $\beta , a,b,c$ such that $a\ne 0$,

• (iii) $\mu \ne 0$, $x_{1}=0$, $x_{2}=\frac{\beta a}{\mu }\ne 0$, $x_{3}=\frac{a\alpha }{2\beta }$, $\lambda = (2\rho -\alpha )(a^{2}+\frac{b^{2}}{2})+\frac{\beta ^{2}a^{2}}{\mu }$, for all $\alpha , \beta , a,b,c, \rho$ such that $a\ne 0$, $\mu \alpha b=\beta ^{2} b$, $\frac{\alpha ^{2}a^{2}}{4\beta ^{2}}=\frac{\alpha }{\mu }(a^{2}+\frac{b^{2}}{2})-\frac{\beta ^{2}a^{2}}{\mu ^{2}}$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{0}=\left( \begin{array}{ccc} -(a^{2}+\frac{b^{2}}{2})&{}0 &{}\frac{ab}{4} \\ 0 &{} -(a^{2}+\frac{b^{2}}{2}) &{}\frac{a^{2}}{2} \\ \frac{ab}{4}&{}\frac{a^{2}}{2} &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} 2ax_{2}&{}-ax_{1} &{}-\frac{b}{2}x_{2} \\ -ax_{1} &{} 0&{} \frac{b}{2}x_{1}\\ -\frac{b}{2}x_{2} &{} \frac{b}{2}x_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Then $\widetilde{S}= -(2a^{2}+b^{2})$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+\frac{b^{2}}{2})+\beta a x_{2}+\mu x_{1}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ -\frac{\beta }{2}ax_{1}+ \mu x_{1}x_{2}=0,\\ \frac{\alpha ab}{4}-\frac{\beta b}{4}x_{2}-\mu x_{1}x_{3}=0,\\ -\alpha (a^{2}+\frac{b^{2}}{2})+\mu x_{2}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \alpha \frac{a^{2}}{2}+\frac{\beta b}{4}x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+b^{2})-\lambda . \end{array}\right. } \end{aligned}$$

(17)

Let $\mu =0$. In this case, the system Eq. (17) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} \beta x_{2}=0,\\ \beta x_{1}=0,\\ \alpha =0,\\ \lambda =\rho (2a^{2}+b^{2}). \end{array}\right. } \end{aligned}$$

(18)

Since $(\alpha , \beta , \mu )\ne (0,0,0)$ we get $\beta \ne 0$ and $x_{1}=x_{2}=0$. Thus the case (i) holds. Using of the first and fourth equations of the system Eq. (17) we get

$$\begin{aligned} \beta a x_{2}+\mu x_{1}^{2}=\mu x_{2}^{2}. \end{aligned}$$

(19)

Now, we consider $\mu \ne 0$, in this case, the second equation of the system Eq. (17) implies that $x_{1}=0$ or $x_{2}=\frac{\beta a}{2\mu }$. If $x_{1}\ne 0$ then $x_{2}=\frac{\beta a}{2\mu }$. Substutiting it in Eq. (19) we have $(\frac{\beta a}{2\mu })^{2}+x_{1}^{2}=0$ which is a contradiction. Hence $x_{1}=0$ and the system Eq. (17) and Eq. (19) become

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+\frac{b^{2}}{2})+\beta a x_{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \frac{\alpha ab}{4}-\frac{\beta b}{4}x_{2}=0,\\ -\alpha (a^{2}+\frac{b^{2}}{2})+\mu x_{2}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \alpha \frac{a^{2}}{2}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+b^{2})-\lambda ,\\ \beta a x_{2}=\mu x_{2}^{2}. \end{array}\right. } \end{aligned}$$

(20)

The sixth equation of Eq. (20) yields $x_{2}=0$ or $x_{2}=\frac{\beta a}{\mu }$. If $x_{2}=0$ then the case (ii) is true. If $x_{2}\ne 0$ and $x_{2}=\frac{\beta a}{\mu }$ then the case (iii) holds. $\square$

Theorem 6

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{3},g,J,X)$ are the following:

• (i) $\mu =0$, $\alpha =0$, $\beta \ne 0$, $x_{1}=x_{2}=x_{3}=0$, $\lambda =2\rho (a^{2}+b^{2})$, for all $\rho , a,b,c$ such that $a\ne 0$,

• (ii) $\mu \ne 0$, $\alpha b=0$, $x_{1}=\beta =x_{2}=\alpha =x^{3}=0$, $\lambda =2\rho (a^{2}+b^{2})$,

• (iii) $\mu \ne 0$, $\alpha b=0$, $x_{1}=0$, $\beta =0$, $x_{2}=0$, $x_{3}=-\frac{\alpha a}{\beta }$, $\lambda =(2\rho -\alpha )(a^{2}+b^{2})$, $\alpha ^{2}a^{2}\mu =\beta ^{2}\alpha (a^{2}+b^{2})$,

• (iv) $\mu \ne 0$, $\alpha b=0$, $x_{1}=0$, $\beta = 0$, $x_{2}=\frac{\beta a}{\mu }$, $b=0$, $\alpha \mu =-\beta ^{2}$, $x_{3}=\beta ^{2}a^{2}\frac{\mu -1}{\mu ^{2}}$, $\lambda =(2\rho -\alpha )(a^{2}+b^{2})+\frac{\beta ^{2}a^{2}}{\mu }$,

• (v) $\mu \ne 0$, $\alpha b\ne 0$,

  $$\begin{aligned} x_{1}=\epsilon _{1}\sqrt{\frac{-\beta ^{2}a^{2}+\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}{8\mu ^{2}}},\,\,\,\,\,\epsilon _{1}=\pm 1, \end{aligned}$$
  $$\begin{aligned} x_{2}=\frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu },\,\,\,\,\,\epsilon _{2}=\pm 1, \end{aligned}$$
  $$\begin{aligned} \lambda =(2\rho -\alpha ) (a^{2}+b^{2})+\mu \left( \frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu }\right) ^{2}, \end{aligned}$$

  and

  $$\begin{aligned} x_{3}=\epsilon _{3}\sqrt{\frac{\alpha }{\mu } (a^{2}+b^{2})-\left( \frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu }\right) ^{2} }, \end{aligned}$$

  where $\epsilon _{3}=\pm 1$, $\alpha ab= \epsilon _{1}\epsilon _{2}\Vert \alpha ab\Vert$,

  $$\begin{aligned}&\frac{\alpha ab +\beta b \left( \frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu } \right) }{\epsilon _{1}\sqrt{\frac{-\beta ^{2}a^{2}+\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}{8\mu ^{2}}}}-\beta a \\&= 2\mu \epsilon _{3}\sqrt{\frac{\alpha }{\mu } (a^{2}+b^{2})-\left( \frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu }\right) ^{2}}, \end{aligned}$$

  and

  $$\begin{aligned}&\frac{2\mu \alpha a^{2}-2\mu \beta b \epsilon _{1}\sqrt{\frac{-\beta ^{2}a^{2}+\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}{8\mu ^{2}}}}{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}\\&+ \frac{2 \mu a\beta \epsilon _{3}\sqrt{\frac{\alpha }{\mu } (a^{2}+b^{2})-\left( \frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu }\right) ^{2}}}{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}} -a\beta \\&=2\mu \epsilon _{3}\sqrt{\frac{\alpha }{\mu } (a^{2}+b^{2})-\left( \frac{-\beta a +\epsilon _{2}\sqrt{\frac{1}{2}\beta ^{2}a^{2}+\frac{1}{2}\sqrt{\beta ^{4}a^{4}+64\mu ^{2}\alpha ^{2} b^{2}a^{2} }}}{-2\mu }\right) ^{2}}. \end{aligned}$$

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{1}=\left( \begin{array}{ccc} -(a^{2}+b^{2})&{}ab &{}-\frac{ab}{2} \\ ab &{} -(a^{2}+b^{2}) &{}\frac{a^{2}}{2} \\ -\frac{ab}{2} &{}\frac{a^{2}}{2} &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} 2ax_{2}&{}-ax_{1} &{}ax_{1}-bx_{2} \\ -ax_{1} &{} 0&{} bx_{1}-ax_{2}-ax_{3}\\ ax_{1}-bx_{2}&{} bx_{1}-ax_{2}-ax_{3}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}= -2(a^{2}+b^{2})$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+b^{2})+\beta ax_{2}+\mu x_{1}^{2}=-2\rho (a^{2}+b^{2})+\lambda ,\\ ab\alpha -\frac{\beta }{2} ax_{1}+\mu x_{1}x_{2}=0,\\ -\frac{\alpha ab}{2}+\frac{\beta }{2}(ax_{1}-bx_{2})-\mu x_{1}x_{3}=0,\\ -\alpha (a^{2}+b^{2})+\mu x_{2}^{2}=-2\rho (a^{2}+b^{2})+\lambda ,\\ -\alpha \frac{a^{2}}{2}+\frac{\beta }{2}(bx_{1}-ax_{2}-ax_{3})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=2\rho (a^{2}+b^{2})-\lambda . \end{array}\right. } \end{aligned}$$

(21)

Let $\mu =0$ then we have $\alpha =0$ and $\beta \ne 0$. Thus $x_{1}=x_{2}=x_{3}=0$ and the case (i) holds. Now, we assume that $\mu \ne 0$. The first and fourth equations of the system Eq. (21) imply that

$$\begin{aligned} \beta ax_{2}+\mu x_{1}^{2}-\mu x_{2}^{2}=0, \end{aligned}$$

(22)

and the fourth and sixth equations imply that

$$\begin{aligned} x_{2}^{2}+ x_{3}^{2}=\frac{\alpha }{\mu }(a^{2}+b^{2}). \end{aligned}$$

(23)

From the second equation we have

$$\begin{aligned} (2\alpha ba)^{2}=x_{1}^{2}(\beta a-2\mu x_{2})^{2}=x_{1}^{2}(\beta ^{2}a^{2}+4\mu (\mu x_{2}^{2}-\beta a x_{2})). \end{aligned}$$

Plugging Eq. (22) into last equality we get

$$\begin{aligned} 4\mu ^{2}x_{1}^{4}+\beta ^{2}a^{2}x_{1}^{2}-(2\alpha ba)^{2}=0. \end{aligned}$$

(24)

If $\alpha b=0$ then $x_{1}=0$ and we obtain three cases (ii)-(iv). If $\alpha b\ne 0$, then $x_{1}\ne 0$ and the case (v) is true. $\square$

Theorem 7

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{4},g,J,X)$ are the following:

• (i) $\mu =0$, $\alpha =0$, $\beta \ne 0$, $x_{1}=x_{2}=0$, $\lambda =-\rho ((2d-b)(a+2d)-2)$ for all $a,b,c,\rho , x_{3}$ such that $d=\pm 1$.

• (ii) $\mu =0$, $\alpha \ne 0$, $d=b$, $a=0$ $\beta =0$, $\lambda =0$, for all $x_{1},x_{2},x_{3},\rho$ such that $d=\pm 1$,

• (iii) $\mu =0$, $\alpha \ne 0$, $d=b$, $a=0$ $\beta \ne 0$, $x_{1}=x_{2}=0$, $\lambda =0$, for all $x_{3},\rho$ such that $d=\pm 1$,

• (iv) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $x_{1}=0$, $\alpha =0$, $\lambda =-\rho ((2d-b)(a+2d)-2)$, for all $a,b,\rho ,\beta$, $d=\pm 1$,

• (v) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $x_{1}=0$, $\alpha \ne 0$, $b=d$, $a=0$, $\lambda =0$, for all $\rho ,\beta$, $d=\pm 1$,

• (vi) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $x_{1}=\frac{\beta }{\mu }\ne 0$, $\lambda =-\rho ((2d-b)(a+2d)-2)$ for all $a,b,\rho , \alpha$ such that $d=\pm 1$, $\beta ^{2}=-\alpha \mu \big ( (2d-b)(\frac{a}{2}+d)-1\big )>0$, $\alpha \big (a+b-(2d-b)(\frac{a}{2}+d)^{2} \big )=0$,

• (vii) $\mu \ne 0$, $x_{2}=0$, $x_{3}\ne 0$, $x_{1}=0$, $x_{3}^{2}=-\frac{\alpha }{\mu }\big ( (2d-b)(\frac{a}{2}+d)-1\big )>0$, $\lambda =(\alpha -2\rho )\big ( (2d-b)(\frac{a}{2}+d)-1\big )$, for all $\rho , \beta$ such that $d=\pm 1$, $\alpha (a+2b-2d)=0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{0}=\left( \begin{array}{ccc} (2d-b)(\frac{a}{2}+d)-1&{}0 &{}0 \\ 0 &{} (2d-b)(\frac{a}{2}+d)-1&{}\frac{a}{4}+\frac{d}{2}-\frac{b}{2} \\ 0&{}\frac{a}{4}+\frac{d}{2}-\frac{b}{2} &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} 0&{}-x_{2} &{}-(\frac{a}{2}+d)x_{2} \\ -x_{2} &{} 2x_{1}&{}(\frac{a}{2}+d)x_{1}\\ -(\frac{a}{2}+d)x_{2} &{}(\frac{a}{2}+d)x_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Then $\widetilde{S}= (2d-b)(a+2d)-2$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( (2d-b)(\frac{a}{2}+d)-1\big )+\mu x_{1}^{2}=\rho ((2d-b)(a+2d)-2)+\lambda ,\\ -\frac{\beta }{2} x_{2}+ \mu x_{1}x_{2}=0,\\ -\frac{\beta }{2}(\frac{a}{2}+d)x_{2}-\mu x_{1}x_{3}=0,\\ \alpha \big ( (2d-b)(\frac{a}{2}+d)-1\big )+\beta x_{1}+\mu x_{2}^{2}=\rho ((2d-b)(a+2d)-2)+\lambda ,\\ \alpha \big (\frac{a}{4}+\frac{d}{2}-\frac{b}{2}\big )+\frac{\beta }{2}(\frac{a}{2}+d)x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=-\rho ((2d-b)(a+2d)-2)-\lambda . \end{array}\right. } \end{aligned}$$

(25)

We consider $\mu =0$, then the system Eq. (25) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( (2d-b)(\frac{a}{2}+d)-1\big )=0,\\ \beta x_{2}=0,\\ \beta x_{1}=0,\\ \alpha \big (\frac{a}{4}+\frac{d}{2}-\frac{b}{2}\big )=0,\\ \lambda =-\rho ((2d-b)(a+2d)-2). \end{array}\right. } \end{aligned}$$

(26)

If $\alpha =0$ then $\beta \ne 0$ and $x_{1}=x_{2}=0$. Thus the case (i) holds. If $\alpha \ne 0$ then $d=b$, $a=0$ and the cases (ii)-(iii) hold. Now we consider $\mu \ne 0$. The first and third equations of the system Eq. (25) yield

$$\begin{aligned} \mu x_{1}^{2}=\beta x_{1}+\mu x_{2}^{2}. \end{aligned}$$

(27)

In this case the second eqution of the system Eq. (25) implies that $x_{2}=0$ or $x_{1}=\frac{\beta }{2\mu }$. If $x_{2}\ne 0$ then $x_{1}=\frac{\beta }{2\mu }$ and substutiting it in Eq. (27) we get $x_{2}^{2}+\frac{\beta ^{2}}{4\mu ^{2}}=0$ which is a cotradiction. Hence, $x_{2}=0$ and the system Eq. (25) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( (2d-b)(\frac{a}{2}+d)-1\big )+\mu x_{1}^{2}=\rho ((2d-b)(a+2d)-2)+\lambda ,\\ x_{1}x_{3}=0,\\ \alpha \big ( (2d-b)(\frac{a}{2}+d)-1\big )+\beta x_{1}=\rho ((2d-b)(a+2d)-2)+\lambda ,\\ \alpha \big (\frac{a}{4}+\frac{d}{2}-\frac{b}{2}\big )+\frac{\beta }{2}(\frac{a}{2}+d)x_{1}=0,\\ \mu x_{3}^{2}=-\rho ((2d-b)(a+2d)-2)-\lambda . \end{array}\right. } \end{aligned}$$

(28)

In this cases (iv)-(vi) hold. $\square$

Theorem 8

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{4},g,J,X)$ are the following:

• (i) $\mu =0$, $\alpha =0$, $x_{1}=x_{2}=x_{3}=0$, $\lambda =2\rho [1+(b-2d)b]$, for all $a,b,\rho$, $d=\pm 1$,

• (ii) $\mu =0$, $\alpha \ne 0$, $\beta \ne 0$, $x_{2}=x_{3}=0$, $b=d$, $a=2b$, $x_{1}=\frac{\alpha }{2\beta }$, $\lambda =0$, for all $\rho$, $d=\pm 1$,

• (iii) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $\alpha =0$, $x_{1}=0$, $\lambda =2\rho [1+(b-2d)b]$, for all $a,b, \rho ,\beta$ such that $d=\pm 1$,

• (iv) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $\alpha \ne 0$, $\beta \ne 0$, $b\ne 0$, $x_{1}=\frac{\alpha a}{\beta b}$, $\lambda =2\rho [1+(b-2d)b]$, for all $a,b, \rho$ such that $d=\pm 1$, $b [1+(b-2d)a]=a$,

  $$\begin{aligned} -\beta ^{2} [1+(b-2d)b]+\mu \alpha [1+(b-2d)a]^{2}=0, \end{aligned}$$

• (v) $\mu \ne 0$, $x_{2}=0$, $x_{3}^{2}=\frac{\alpha }{\mu }[1+(b-2d)b]-\frac{\beta ^{2}}{4\mu ^{2}}>0$, $x_{1}=-\frac{\beta }{2\mu }$, $\lambda =(2\rho -\alpha )[1+(b-2d)b] +\frac{\beta ^{2}}{4\mu }$, for all $a,b,\rho , \alpha$ such that

  $$\begin{aligned} d=\pm 1,\,\,\,\,\, \alpha (b-2d)(a-b)=-\frac{3\beta ^{2}}{4\mu },\,\,\,\,2\mu \alpha a=-b \beta ^{2}, \end{aligned}$$

• (vi) $\mu \ne 0$, $x_{2}\ne 0$, $x_{1}=\frac{\beta }{2\mu }$, $\beta =0$, $\lambda =(2\rho -\alpha ) [1+(b-2d)b]$, $x_{2}=\epsilon _{1}\sqrt{-\frac{\alpha }{\mu } [(b-2d)(b-a)]}$, $x_{3}=\epsilon _{2}\sqrt{\frac{\alpha }{\mu } [1+(b-2d)b]}$, for all $\rho$ such that $d=\pm 1$, $\frac{\alpha }{\mu } [(b-2d)(b-a)]\le 0$, $\frac{\alpha }{\mu } [1+(b-2d)b]\ge 0$, $\epsilon _{1}=\pm 1$, $\epsilon _{2}=\pm 1$,

  $$\begin{aligned} -\frac{a\alpha }{\mu }=\epsilon _{1}\epsilon _{2}\sqrt{-\frac{\alpha }{\mu } [(b-2d)(b-a)]}\sqrt{\frac{\alpha }{\mu } [1+(b-2d)b]} \end{aligned}$$

• (vii) $\mu \ne 0$, $x_{2}\ne 0$, $x_{1}=\frac{\beta }{2\mu }$, $\beta \ne 0$, $x_{3}= -\frac{a}{2}x_{2}$, $\lambda =(2\rho -\alpha ) [1+(b-2d)b]+\frac{\beta ^{2}}{4\mu }$, $x_{2}^{2}=\frac{-2\mu \alpha [1+(b-2d)a]+\beta ^{2}}{-2\mu ^{2}(1+\frac{a^{2}}{4})}>0$,

  $$\begin{aligned} d=\pm 1,\,\,\,\, -a\alpha +\frac{\beta ^{2}}{2\mu }b+\frac{a\mu }{2}\frac{-2\mu \alpha [1+(b-2d)a]+\beta ^{2}}{-2\mu ^{2}(1+\frac{a^{2}}{4})}=0, \end{aligned}$$
  $$\begin{aligned} -\alpha [1+(b-2d)b]+\frac{\beta ^{2}}{4\mu }=-\mu \frac{a^{2}}{4}\frac{-2\mu \alpha [1+(b-2d)a]+\beta ^{2}}{-2\mu ^{2}(1+\frac{a^{2}}{4})}. \end{aligned}$$

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{1}=\left( \begin{array}{ccc} -[1+(b-2d)b]&{}0 &{}0 \\ 0 &{} -[1+(b-2d)a] &{}\frac{a}{2} \\ 0 &{}\frac{a}{2} &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} 0&{}-x_{2} &{}-ax_{2}-x_{3} \\ -x_{2} &{} 2x_{1}&{} bx_{1}\\ -ax_{2}-x_{3}&{} bx_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}= -2[1+(b-2d)b]$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha [1+(b-2d)b]+\mu x_{1}^{2}=-2\rho [1+(b-2d)b]+\lambda ,\\ -\frac{\beta }{2} x_{2}+\mu x_{1}x_{2}=0,\\ \frac{\beta }{2}(-ax_{2}-x_{3} )-\mu x_{1}x_{3}=0,\\ -\alpha [1+(b-2d)a]+\beta x_{1}+\mu x_{2}^{2}=-2\rho [1+(b-2d)b]+\lambda ,\\ -\alpha \frac{a}{2}+\frac{\beta }{2}(bx_{1})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=2\rho [1+(b-2d)b]-\lambda . \end{array}\right. } \end{aligned}$$

(29)

Let $\mu =0$, then the system Eq. (29) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha [1+(b-2d)b]=0,\\ \beta x_{2}=0,\\ \beta x_{3}=0,\\ -\alpha [1+(b-2d)a]+\beta x_{1}=0,\\ -a\alpha +\beta bx_{1}=0,\\ \lambda =2\rho [1+(b-2d)b]. \end{array}\right. } \end{aligned}$$

(30)

and the cases (i)-(ii) holds. Now we consider $\mu \ne 0$. In this case the second equation of the system Eq. (29) implies that $x_{2}=0$ or $x_{1}=\frac{\beta }{2\mu }$. If $x_{2}=0$ then the system Eq. (29) gives

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha [1+(b-2d)b]+\mu x_{1}^{2}=-2\rho [1+(b-2d)b]+\lambda ,\\ \beta x_{3}+2\mu x_{1}x_{3}=0,\\ -\alpha [1+(b-2d)a]+\beta x_{1}=-2\rho [1+(b-2d)b]+\lambda ,\\ -\alpha a+\beta (bx_{1})=0,\\ \mu x_{3}^{2}=2\rho [1+(b-2d)b]-\lambda . \end{array}\right. } \end{aligned}$$

(31)

The second equation of the system Eq. (31) implies that $x_{3}=0$ or $x_{1}=-\frac{\beta }{2\mu }$. We assume that $x_{3}=0$, thus

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha [1+(b-2d)b]+\mu x_{1}^{2}=0,\\ -\alpha [1+(b-2d)a]+\beta x_{1}=0,\\ -\alpha a+\beta bx_{1}=0,\\ \lambda =2\rho [1+(b-2d)b], \end{array}\right. } \end{aligned}$$

and the cases (iii)-(iv) are true. If $x_{3}\ne 0$ and $x_{1}=-\frac{\beta }{2\mu }$ then the case (v) is true. Now, we consider $x_{2}\ne 0$ and $x_{1}=\frac{\beta }{2\mu }$. In this case the system Eq. (29) yields

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha [1+(b-2d)b]+\frac{\beta ^{2}}{4\mu }=-2\rho [1+(b-2d)b]+\lambda ,\\ \beta ax_{2}+2\beta x_{3} =0,\\ -\alpha [1+(b-2d)a]+\frac{\beta ^{2}}{2\mu }+\mu x_{2}^{2}=-2\rho [1+(b-2d)b]+\lambda ,\\ -a\alpha +\frac{\beta ^{2}}{2\mu }b-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=2\rho [1+(b-2d)b]-\lambda . \end{array}\right. } \end{aligned}$$

(32)

The second equation of Eq. (32) implies that $\beta =0$ or $x_{3}=-\frac{a}{2}x_{2}$. If $\beta =0$ then we obtain the case (vi). If $\beta \ne 0$ then $x_{3}=-\frac{a}{2}x_{2}$ and the case (vii) holds.

$\square$

Theorem 9

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{5},g,J,X)$ are the following:

• (i) $\mu =\beta =\lambda =0$ and for all $\alpha , \rho , x_{1},x_{2},x_{3},a,b,c,d$ such that $a+d\ne 0$ and $ac+bd=0$,

• (ii) $\mu =\lambda =0$, $\beta \ne 0$, $b=c$, and for all $\alpha , \rho , x_{1},x_{2},x_{3},a,d$ such that $a+d\ne 0$ and $ac+bd=0$,

• (iii) $\mu \ne 0$, $x_{1}=x_{2}=x_{3}=\lambda =0$ and for all $\alpha , \rho , a,b,c,d$ such that $a+d\ne 0$ and $ac+bd=0$.

Proof

From [19, 20], we have $\widetilde{Ric}^{0}=0$ and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} 0&{}0 &{}\frac{b-c}{2}x_{2} \\ 0 &{}0&{}-\frac{b-c}{2}x_{1}\\ \frac{b-c}{2}x_{2} &{}-\frac{b-c}{2}x_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Then $\widetilde{S}= 0$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \mu x_{1}^{2}=\lambda ,\\ \mu x_{1}x_{2}=0,\\ \frac{\beta }{2}\frac{b-c}{2}x_{2}-\mu x_{1}x_{3}=0,\\ \mu x_{2}^{2}=\lambda ,\\ -\frac{\beta }{2}\frac{b-c}{2}x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=-\lambda . \end{array}\right. } \end{aligned}$$

(33)

The first, fourth and sixth equations of system Eq. (33) imply that

$$\begin{aligned} \mu ( x_{1}^{2}+ x_{3}^{2})=\mu ( x_{2}^{2}+ x_{3}^{2})=0. \end{aligned}$$

We consider $\mu =0$, then $\lambda =0$. If $\beta =0$ or $b=c$ then the system Eq. (33) holds for any $x_{1},x_{2}$, and $x_{3}$. Now, if $\mu \ne 0$ then $x_{1}=x_{2}=x_{3}=\lambda =0$. $\square$

Theorem 10

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{5},g,J,X)$ are the following:

• (i) $\mu =\beta =\lambda =0$ and for all $\alpha , \rho , x_{1},x_{2},x_{3},a,b,c,d$ such that $a+d\ne 0$ and $ac+bd=0$,

• (ii) $\mu =\lambda =0$, $\beta \ne 0$, $a=b=x_{2}=0$, and for all $\alpha , \rho , x_{1},x_{3},c,d$ such that $d\ne 0$,

• (iii) $\mu =\lambda =0$, $\beta \ne 0$, $a\ne 0$, $x_{1}=x_{2}=0$, and for all $\alpha , \rho , x_{3},c,d$ such that $d\ne 0$,

• (iv) $\mu =\lambda =0$, $\beta \ne 0$, $a\ne 0$, $x_{1}=c=d=0$, $x_{2}\ne 0$, and for all $\alpha , \rho , x_{3}$,

• (v) $\mu \ne 0$, $x_{1}=x_{2}=x_{3}=\lambda =0$ and for all $\alpha , \rho ,a,b,c,d$ such that $a+d\ne 0$ and $ac+bd=0$.

Proof

From [19, 20], we have $\widetilde{Ric}^{1}=0$ and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} 0&{}0 &{}-ax_{1}-cx_{2} \\ -0 &{} 0&{} -bx_{1}-dx_{2}\\ -ax_{1}-cx_{2}&{}-bx_{1}-dx_{2}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}= 0$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \mu x_{1}^{2}=\lambda ,\\ \mu x_{1}x_{2}=0,\\ \frac{\beta }{2}(-ax_{1}-cx_{2} )-\mu x_{1}x_{3}=0,\\ \mu x_{2}^{2}=\lambda ,\\ \frac{\beta }{2}( -bx_{1}-dx_{2})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=-\lambda . \end{array}\right. } \end{aligned}$$

(34)

The first, fourth and sixth equations of system Eq. (34) imply that

$$\begin{aligned} \mu ( x_{1}^{2}+ x_{3}^{2})=\mu ( x_{2}^{2}+ x_{3}^{2})=0. \end{aligned}$$

We consider $\mu =0$, then $\lambda =0$. Let $\beta =0$, then the system Eq. (34) holds for any $x_{1},x_{2}$, and $x_{3}$. If $\beta \ne 0$ then the third and fiveth equations of Eq. (34) given

$$\begin{aligned} {\left\{ \begin{array}{ll} ax_{1}+cx_{2}=0,\\ bx_{1}+dx_{2}=0. \end{array}\right. } \end{aligned}$$

Since $ac+bd=0$ we get $(a^{2}+b^{2})x_{1}=0$ and $(c^{2}+d^{2})x_{2}=0$. We consider $a=0$. In this case we obtain $d\ne 0$ and $b=x_{2}=0$. if $a\ne 0$ then $x_{1}=0$ and $cx_{2}=dx_{2}=0$. For case $x_{2}\ne 0$ we have $c=d=0$. Now, we assume that $\mu \ne 0$, then $x_{1}=x_{2}=x_{3}=\lambda =0$. $\square$

Theorem 11

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{6},g,J,X)$ are the following:

• (i) $\mu =0$, $a=0$, $d\ne 0$, $b=0$, $c=0$, $\lambda = 0$, for all $\alpha , \beta , \rho , x_{1},x_{2}, x_{3}$, such that $( \alpha , \beta )\ne (0,0)$,

• (ii) $\mu =0$, $a=0$, $d\ne 0$, $b=0$, $c\ne 0$, $\beta \ne 0$, $\lambda = 0$, $x_{2}=0$, $x_{1}=-\frac{\alpha d}{\beta }$ for all $\alpha , \rho , x_{3}$,

• (iii) $\mu =0$, $a\ne 0$, $\beta =0$, $\alpha \ne 0$, $\lambda =0$, $c=\frac{bd}{a}$, for all $b,d, \rho , x_{1}, x_{2}, x_{3}$, such that $b^{2}(a-d)=2a^{3}$, $d(a-d)=2\alpha$,

• (iv) $\mu =0$, $a\ne 0$, $\beta \ne 0$, $x_{1}=x_{2}=0$, $\alpha =0$, $c=\frac{bd}{a}$, $\lambda =-\rho (b(b-c)-2a^{2})$ for all $b,d, x_{3},\rho$ such that $a+d\ne 0$,

• (v) $\mu \ne 0$, $x_{2}=0$, $x_{1}=0$, $x_{3}^{2}=-\frac{\alpha }{\mu }\big ( \frac{1}{2}b^{2}-a^{2}\big )\ge 0$, $\lambda =(-2\rho +\alpha )\big ( \frac{1}{2}b(b-c)-a^{2}\big )$, for all $a,b,c,d, \rho , \beta , \alpha$ such that $\alpha c=0, ac-bd=0, a+d\ne 0$,

• (vi) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $x_{1}=-\frac{\beta a}{\mu }\ne 0$, $\lambda =-\rho (b(b-c)-2a^{2})$, for all $a,b,c,d, \rho , \beta , \alpha$ such that

  $$\begin{aligned} \alpha \big ( \frac{1}{2}b(b-c)-a^{2}\big )+\frac{\beta ^{2}a^{2}}{\mu }=0,\,\,\,\,\alpha \frac{1}{2}[-ac+\frac{1}{2}d(b-c)]-\frac{\beta ^{2}a}{4\mu }(b-c)=0, \end{aligned}$$
  $$\begin{aligned} ac-bd=0,\,\,\,\, a+d\ne 0. \end{aligned}$$

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{0}=\left( \begin{array}{ccc} \frac{1}{2}b(b-c)-a^{2}&{}0 &{}0 \\ 0 &{} \frac{1}{2}b(b-c)-a^{2}&{}\frac{1}{2}[-ac+\frac{1}{2}d(b-c)] \\ 0&{}\frac{1}{2}[-ac+\frac{1}{2}d(b-c)] &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} 0&{}ax_{2} &{}\frac{c-b}{2}x_{2} \\ ax_{2} &{} -2ax_{1}&{}\frac{b-c}{2}x_{1}\\ \frac{c-b}{2}x_{2} &{}\frac{b-c}{2}x_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Then $\widetilde{S}=b(b-c)-2a^{2}$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( \frac{1}{2}b(b-c)-a^{2}\big )+\mu x_{1}^{2}=\rho (b(b-c)-2a^{2})+\lambda ,\\ \frac{\beta }{2} ax_{2}+ \mu x_{1}x_{2}=0,\\ \frac{\beta }{4}(c-b)x_{2}-\mu x_{1}x_{3}=0,\\ \alpha \big ( \frac{1}{2}b(b-c)-a^{2}\big )-\beta ax_{1}+\mu x_{2}^{2}=\rho (b(b-c)-2a^{2})+\lambda ,\\ \alpha \frac{1}{2}[-ac+\frac{1}{2}d(b-c)]+\frac{\beta }{4}(b-c)x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=-\rho (b(b-c)-2a^{2})-\lambda . \end{array}\right. } \end{aligned}$$

(35)

Let $\mu =0$, then the system Eq. (35) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( \frac{1}{2}b(b-c)-a^{2}\big )=0,\\ \beta ax_{2}=0,\\ \beta (c-b)x_{2}=0,\\ \beta ax_{1}=0,\\ 2\alpha [-ac+\frac{1}{2}d(b-c)]+\beta (b-c)x_{1}=0,\\ \lambda =-\rho (b(b-c)-2a^{2}). \end{array}\right. } \end{aligned}$$

(36)

If $a=0$ then $d\ne 0$, $b=0$, and $\beta c x_{2}=0$. If $c=0$ then the case (i) holds. Now, if $c\ne 0$ then $\beta \ne 0$ and the case (ii) holds. For $a\ne 0$ the cases (iii)-(iv) hold.

Now we assume that $\mu \ne 0$. The first and fourth equations of the system Eq. (35) give

$$\begin{aligned} \mu x_{1}^{2}=-\beta ax_{1}+\mu x_{2}^{2}. \end{aligned}$$

(37)

The second equation of the system Eq. (35) yields $x_{2}=0$ or $x_{1}=-\frac{\beta a}{2\mu }$. The Eq. (36) implies that $x_{1}\ne -\frac{\beta a}{2\mu }$ thus $x_{2}= 0$. The third equation of the system Eq. (35) implies that $x_{1}=0$ or $x_{3}=0$. If $x_{1}=0$ then we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( \frac{1}{2}b(b-c)-a^{2}\big )=\rho (b(b-c)-2a^{2})+\lambda ,\\ ac-bd=0,\\ \alpha [-ac+\frac{1}{2}d(b-c)]=0,\\ \mu x_{3}^{2}=-\rho (b(b-c)-2a^{2})-\lambda . \end{array}\right. } \end{aligned}$$

(38)

Hence, the case (v) holds. If $x_{1}\ne 0$ and $x_{3}=0$ then the Eq. (37) gives $x_{1}=-\frac{\beta a}{\mu }$ and we get

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha \big ( \frac{1}{2}b(b-c)-a^{2}\big )+\frac{\beta ^{2}a^{2}}{\mu }=0,\\ \alpha \frac{1}{2}[-ac+\frac{1}{2}d(b-c)]-\frac{\beta ^{2}a}{4\mu }(b-c)=0,\\ \lambda =-\rho (b(b-c)-2a^{2}). \end{array}\right. } \end{aligned}$$

(39)

Therefore the case (vi) holds. $\square$

Theorem 12

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{6},g,J,X)$ are the following:

• (i) $\mu =0$, $a=0$, $b=0$, $d\ne 0$, $\beta =0$, $\lambda =0$, for all $c,\rho , \alpha , x_{1},x_{2},x_{3}$ such that $\alpha \ne 0$,

• (ii) $\mu =0$, $a=0$, $b=0$, $d\ne 0$, $\beta \ne 0$, $\lambda =0$, $x_{3}=0$, for all $c,\rho , \alpha , x_{1},x_{2}$ such that $cx_{1}=0$,

• (iii) $\mu =0$, $a\ne 0$, $\beta \ne 0$, $x_{2}=0$, $\alpha =0$, $x_{1}=0$, $\lambda =\rho (2a^{2}+bc)$, for all $x_{3},b,c,d,\rho$ such that $a+d\ne 0$, $c=\frac{bd}{a}$,

• (iv) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $a=0$, $b=0$, $d\ne 0$, $x_{1}=0$, $\lambda =0$, for all $c,d,\rho , \alpha , \beta$ such that $(\alpha ,\beta )\ne (0,0)$, $d\ne 0$,

• (v) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $a\ne 0$, $c=\frac{bd}{a}$, $d\ne 0$, $\alpha =0$, $x_{1}=0$, $\lambda =\rho (2a^{2}+bc)$, for all $d,\rho , \alpha , \beta$ such that $(\alpha ,\beta )\ne (0,0)$,

• (vi) $\mu \ne 0$, $x_{2}=0$, $x_{3}=0$, $a\ne 0$, $d\ne 0$, $\alpha \ne 0$, $c=0$, $x_{1}=-\frac{\beta a}{\mu }$, $b=0$, $\lambda =2\rho a^{2}$, for all $\rho ,\beta$, such that $a+d\ne 0$,

• (vii) $\mu \ne 0$, $x_{2}=0$, $x_{3}^{2}=\frac{\alpha a^{2}}{\mu }+\frac{\beta ^{2} a d}{2\mu ^{2}}>0$, $x_{1}=\frac{\beta d}{2\mu }$, $c=\frac{bd}{a}$, $\lambda =-\alpha a^{2}-\frac{\beta ^{2} a d}{2\mu }+\rho (2a^{2}+bc)$, for all $a,b, \rho , \alpha , \beta$ such that $a+d\ne 0$, $\beta ^{2} cd=0$, $\beta ^{2}d^{2}=-2\beta ^{2} a d$,

• (viii) $\mu \ne 0$, $x_{2}^{2}=\frac{\alpha }{\mu }a^{2}-\frac{\beta ^{2} a^{2}}{2\mu ^{2}}>0$, $x_{1}=-\frac{\beta a}{2\mu }$, $x_{3}=0$, $\lambda =\rho (2a^{2}+bc)$, $c=\frac{bd}{a}$, for all $b,d,a,\rho , \alpha , \beta$ such that $a+d\ne 0$, $\beta ac=0$, $4\mu \alpha (a^{2}+bc)+\beta ^{2}a^{2}=0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{1}=\left( \begin{array}{ccc} -(a^{2}+bc)&{}0 &{}0 \\ 0 &{} -a^{2}&{}0 \\ 0 &{}0 &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} 0&{}ax_{2} &{}dx_{3} \\ ax_{2} &{}- 2ax_{1}&{} -cx_{1}\\ dx_{3}&{} -cx_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}= -(2a^{2}+bc)$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+bc)+\mu x_{1}^{2}=-\rho (2a^{2}+bc)+\lambda ,\\ \frac{\beta }{2}a x_{2}+\mu x_{1}x_{2}=0,\\ \frac{\beta }{2}dx_{3} -\mu x_{1}x_{3}=0,\\ -\alpha a^{2}-\beta ax_{1}+\mu x_{2}^{2}=-\rho (2a^{2}+bc)+\lambda ,\\ -\frac{\beta }{2}cx_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(40)

Let $\mu =0$, then

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha (a^{2}+bc)=0,\\ a\beta x_{2}=0,\\ d\beta x_{3} =0,\\ -\alpha a^{2}-\beta ax_{1}=0,\\ \beta cx_{1}=0,\\ \lambda =\rho (2a^{2}+bc). \end{array}\right. } \end{aligned}$$

(41)

If we assume that $a=0$ then the cases (i)-(ii) hold. If we consider $a\ne 0$ then $\beta \ne 0$ and $x_{2}=0$, $\alpha =0$ and the case (iii) holds.

Now we consider $\mu \ne 0$. The second equation of the system Eq. (40) implies that $x_{2}=0$ or $x_{1}=-\frac{\beta a}{2\mu }$. If $x_{2}=0$ then the system Eq. (40) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+bc)+\mu x_{1}^{2}=-\rho (2a^{2}+bc)+\lambda ,\\ \frac{\beta }{2}dx_{3} -\mu x_{1}x_{3}=0,\\ -\alpha a^{2}-\beta ax_{1}=-\rho (2a^{2}+bc)+\lambda ,\\ \beta cx_{1}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(42)

The second equation of the system Eq. (42) yields $x_{3}=0$ or $x_{1}=\frac{\beta d}{2\mu }$. We consider $x_{3}=0$, then $\lambda = \rho (2a^{2}+bc)$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+bc)+\mu x_{1}^{2}=0,\\ -\alpha a^{2}-\beta ax_{1}=0,\\ \beta cx_{1}=0. \end{array}\right. } \end{aligned}$$

(43)

Thus, the cases (iv)-(vi) hold. If $x_{3}\ne 0$ then $x_{1}=\frac{\beta d}{2\mu }$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+bc)+\frac{\beta ^{2}d^{2}}{4\mu }=-\rho (2a^{2}+bc)+\lambda ,\\ -\alpha a^{2}-\frac{\beta ^{2} a d}{2\mu }=-\rho (2a^{2}+bc)+\lambda ,\\ \beta ^{2} cd=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(44)

Hence, the case (vii) holds. Now we assume that $x_{2}\ne 0$, then $x_{1}=-\frac{\beta a}{2\mu }$ and we have

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+bc)+\frac{\beta ^{2} a^{2}}{4\mu }=-\rho (2a^{2}+bc)+\lambda ,\\ \beta x_{3} =0,\\ -\alpha a^{2}+\frac{\beta ^{2} a^{2}}{2\mu }+\mu x_{2}^{2}=-\rho (2a^{2}+bc)+\lambda ,\\ \frac{\beta ^{2} ac}{4\mu }-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(45)

Therefore $x_{3}=0$ and the case (viii) holds.

$\square$

Theorem 13

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{0}$ on the Lie group $(G_{7},g,J,X)$ are the following:

• (i) $\mu =0$, $\beta =0$, $\alpha \ne 0$, $a=0$, $d\ne 0$, $c=0$, $\lambda =0$, for all $\rho , x_{1},x_{2},x_{3}$,

• (ii) $\mu =0$, $\beta \ne 0$, $a=0$, $d\ne 0$, $b=0$, $c=0$, $\lambda =0$, for all $\rho , \alpha , x_{1},x_{2},x_{3}$,

• (iii) $\mu =0$, $\beta \ne 0$, $a=0$, $d\ne 0$, $b=0$, $c\ne 0$, $\lambda =0$, $x_{1}=0$, $x_{2}=-\frac{\alpha dc}{\beta c}$ for all $\rho , \alpha , x_{3}$,

• (iv) $\mu =0$, $\beta \ne 0$, $a=0$, $d\ne 0$, $b\ne 0$, $x_{1}=x_{2}=0$, $\lambda =\rho bc$, for all $c, \rho , \alpha , x_{3}$, such that $\alpha b=\alpha c=0$,

• (v) $\mu =0$, $\beta \ne 0$, $a\ne 0$, $c=0$, $\alpha =x_{1}=x_{2}=0$, $\lambda =2\rho a^{2}$, for all $\rho , x_{3}, b,d$ such that $a+d\ne 0$,

• (vi) $\mu \ne 0$, $a=0$, $d\ne 0$, $x_{2}=0$, $x_{1}=0$, $\lambda =\rho bc$, $x_{3}=0$, for all $b,c,\rho , \alpha , \beta$ such that $\alpha c=0$,

• (vii) $\mu \ne 0$, $a=0$, $d\ne 0$, $x_{2}=0$, $x_{1}=\frac{\beta b}{\mu }\ne 0$, $x_{3}=-\frac{\alpha dc}{4\beta \mu }$, $\lambda =-\alpha \frac{bc}{2}+\frac{\beta ^{2}b^{2}}{\mu }+\rho bc$, for all $\rho , c,\alpha$ such that $\mu \alpha c+\beta ^{2}(c-2b)=0$, $\frac{\alpha ^{2} d^{2}c^{2}}{16\beta ^{2} \mu }=\frac{\alpha bc}{2}-\frac{\beta ^{2}b^{2}}{\mu }$,

• (viii) $\mu \ne 0$, $a\ne 0$, $c=0$, $x_{1}=x_{2}=x_{3}=0$, $\alpha =0$, $\lambda =2\rho a^{2}$, for all $\rho , \beta , d$ such that $a+d\ne 0$,

• (ix) $\mu \ne 0$, $a\ne 0$, $c=0$, $x_{3}=x_{2}=-\frac{\alpha a}{2\beta }\ne 0$, $x_{1}=0$, $\lambda = (2\rho -\frac{\alpha }{2})a^{2}$, for all $b,d,\rho , \alpha , \beta$ such that $2\alpha \beta ^{2}=\alpha ^{2}\mu$, $a+d\ne 0$.

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{0}=\left( \begin{array}{ccc} -(a^{2}+\frac{bc}{2})&{}0 &{}-\frac{1}{2}(ac+\frac{dc}{2}) \\ 0 &{} -(a^{2}+\frac{bc}{2})&{}\frac{1}{2}(a^{2}+\frac{bc}{2})\\ -\frac{1}{2}(ac+\frac{dc}{2})&{}\frac{1}{2}(a^{2}+\frac{bc}{2}) &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{0}g)=\left( \begin{array}{ccc} -2ax_{2}&{}ax_{1}-bx_{2} &{}(b-\frac{c}{2})x_{2} \\ ax_{1}-bx_{2} &{} 2bx_{1}&{}(\frac{c}{2}-b)x_{1}\\ (b-\frac{c}{2})x_{2} &{}(\frac{c}{2}-b)x_{1}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Then $\widetilde{S}=-(2a^{2}+bc)$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+\frac{bc}{2})-\beta ax_{2}+\mu x_{1}^{2}=-\rho (2a^{2}+bc)+\lambda ,\\ \frac{\beta }{2} (ax_{1}-bx_{2})+ \mu x_{1}x_{2}=0,\\ -\frac{\alpha }{2}(ac+\frac{dc}{2})+\frac{\beta }{2}(b-\frac{c}{2})x_{2}-\mu x_{1}x_{3}=0,\\ -\alpha (a^{2}+\frac{bc}{2})+\beta bx_{1}+\mu x_{2}^{2}=-\rho (2a^{2}+bc)+\lambda ,\\ \frac{\alpha }{2}(a^{2}+\frac{bc}{2})+\frac{\beta }{2}(\frac{c}{2}-b)x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(46)

Let $\mu =0$ then the system Eq. (46) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha (a^{2}+\frac{bc}{2})-\beta ax_{2}=0,\\ \frac{\beta }{2} (ax_{1}-bx_{2})=0,\\ -\frac{\alpha }{2}(ac+\frac{dc}{2})+\frac{\beta }{2}(b-\frac{c}{2})x_{2}=0,\\ -\alpha (a^{2}+\frac{bc}{2})+\beta bx_{1}=0,\\ \frac{\alpha }{2}(a^{2}+\frac{bc}{2})+\frac{\beta }{2}(\frac{c}{2}-b)x_{1}=0,\\ \lambda =\rho (2a^{2}+bc). \end{array}\right. } \end{aligned}$$

(47)

If $\beta =0$ then $\alpha \ne 0$ and the case (i) is true. If $\beta \ne 0$ and $a=0$ then $d\ne 0$ and we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha bc=0,\\ bx_{2}=0,\\ \alpha dc+\beta c x_{2}=0,\\ bx_{1}=0,\\ cx_{1}=0,\\ \lambda =\rho bc. \end{array}\right. } \end{aligned}$$

(48)

Hence the cases (ii)-(iv) hold.

For the case $\beta \ne 0$ and $a\ne 0$ we have $c=0$ and $\alpha =x_{1}=x_{2}=0$. Therefore the case (v) holds.

Now, we assume that $\mu \ne 0$. The first and the fourth equations of the system Eq. (46) imply

$$\begin{aligned} -\beta ax_{2}+\mu x_{1}^{2}=\beta bx_{1}+\mu x_{2}^{2}. \end{aligned}$$

(49)

Since $ac=0$ then $a=0$ or $c=0$. If $a=0$ then the system Eq. (46) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha \frac{bc}{2}+\mu x_{1}^{2}=-\rho bc+\lambda ,\\ -\frac{\beta }{2} bx_{2}+ \mu x_{1}x_{2}=0,\\ -\frac{\alpha }{2}\frac{dc}{2}+\frac{\beta }{2}(b-\frac{c}{2})x_{2}-\mu x_{1}x_{3}=0,\\ -\alpha \frac{bc}{2}+\beta bx_{1}+\mu x_{2}^{2}=-\rho bc+\lambda ,\\ \frac{\alpha }{2}\frac{bc}{2}+\frac{\beta }{2}(\frac{c}{2}-b)x_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho bc-\lambda . \end{array}\right. } \end{aligned}$$

(50)

The second equation of Eq. (50) yields $x_{2}=0$ or $x_{1}=\frac{\beta b}{2\mu }$. If $x_{2}\ne 0$ then $x_{1}=\frac{\beta b}{2\mu }$ and substituting it in Eq. (49) we get $\frac{\beta ^{2} b^{2}}{4\mu ^{2}}+x_{2}^{2}=0$ and this is a contradiction. Thus $x_{2}=0$ and from the Eq. (49) we have $x_{1}=0$, or $x_{1}=\frac{\beta b}{\mu }$. If $x_{1}=0$ then $\lambda =\rho bc$, $\alpha c=0$, $x_{3}=0$, and the case (vi) is true. Also, if $x_{1}=\frac{\beta b}{\mu }\ne 0$ then

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha \frac{bc}{2}+\frac{\beta ^{2}b^{2}}{\mu }=-\rho bc+\lambda ,\\ -\frac{\alpha }{2}\frac{dc}{2}-\beta b x_{3}=0,\\ \frac{\alpha }{2}\frac{bc}{2}+\frac{\beta ^{2} b}{2\mu }(\frac{c}{2}-b)=0,\\ \mu x_{3}^{2}=\rho bc-\lambda . \end{array}\right. } \end{aligned}$$

(51)

Thus, the case (vii) holds. Now for case $\mu \ne 0$ and $a\ne 0$ we have $c=0$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha a^{2}-\beta ax_{2}+\mu x_{1}^{2}=-2\rho a^{2}+\lambda ,\\ \frac{\beta }{2} (ax_{1}-bx_{2})+ \mu x_{1}x_{2}=0,\\ \frac{\beta }{2}bx_{2}-\mu x_{1}x_{3}=0,\\ -\alpha a^{2}+\beta bx_{1}+\mu x_{2}^{2}=-2\rho a^{2}+\lambda ,\\ \frac{\alpha }{2}a^{2}-\frac{\beta }{2}bx_{1}-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=2\rho a^{2}-\lambda . \end{array}\right. } \end{aligned}$$

(52)

The fourth, fiveth and sixth equations of Eq. (52) imply that $x_{2}=x_{3}$. The second and third equations imply that $\beta x_{1}=0$. Then

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha a^{2}-\beta ax_{2}+\mu x_{1}^{2}=-2\rho a^{2}+\lambda ,\\ -\frac{\beta }{2}bx_{2}+ \mu x_{1}x_{2}=0,\\ -\alpha a^{2}+\mu x_{2}^{2}=-2\rho a^{2}+\lambda ,\\ \frac{\alpha }{2}a^{2}-\mu x_{2}^{2}=0,\\ \mu x_{2}^{2}=2\rho a^{2}-\lambda . \end{array}\right. } \end{aligned}$$

(53)

Using the second equation of Eq. (53) we have $x_{2}=0$ or $x_{1}=\frac{\beta b}{2\mu }$. If $x_{2}=0$ then $\alpha =0$ and the case (viii) holds. If $x_{2}\ne 0$ then $x_{1}=\frac{\beta b}{2\mu }=0$ and

$$\begin{aligned} {\left\{ \begin{array}{ll} -\beta ax_{2}=\frac{\alpha }{2}a^{2},\\ \frac{\alpha }{2}a^{2}-\mu x_{2}^{2}=0,\\ \frac{\alpha }{2}a^{2}=2\rho a^{2}-\lambda . \end{array}\right. } \end{aligned}$$

(54)

Thus the case (ix) is true. $\square$

Theorem 14

The left-invariant affine generalized Ricci soliton associated to the connection $\nabla ^{1}$ on the Lie group $(G_{7},g,J,X)$ are the following:

• (i) $\mu =0$, $a=0$, $d\ne 0$, $\beta \ne 0$, $b=0$, $x_{2}=0$, $x_{3}=\frac{2\alpha d}{\beta }$, $\lambda =0$, for all $c,\rho , x_{1}$,

• (ii) $\mu =0$, $a=0$, $d\ne 0$, $\beta \ne 0$, $b=0$, $x_{2}\ne 0$, $c=0$,$x_{2}+x_{3}=\frac{2\alpha d}{\beta }$, $\lambda =0$, for all $\rho , x_{1}$,

• (iii) $\mu =0$, $a=0$, $d\ne 0$, $\beta \ne 0$, $b\ne 0$, $x_{1}=\frac{\alpha (b+c)}{\beta }$, $x_{2}=\frac{\alpha bd }{\beta b}$, $x_{3}=\frac{2\alpha d}{\beta }-\frac{c\alpha d}{\beta b}$, $\lambda =\rho (b^{2}+bc)$, for all $c, \rho ,\alpha$ such that $c\alpha d^{2}-b^{3}\alpha -\alpha bd^{2}=0$,

• (iv) $\mu =0$, $a\ne 0$, $c=0$, $\beta \ne 0$, $b=0$, $\alpha =0$, $x_{1}=x_{2}=0$, $\lambda = 2\rho a^{2}$, for all $\rho , d$ such that $a+d\ne 0$, $dx_{3}=0$,

• (v) $\mu =0$, $a\ne 0$, $c=0$, $\beta \ne 0$, $b\ne 0$, $\alpha =0$, $x_{1}=x_{2}=x_{3}=0$, $\lambda = \rho (2a^{2}+b^{2})$, for all $\rho , d$ such that $a+d\ne 0$,

• (vi) $\mu \ne 0$, $a=0$, $d\ne 0$, $x_{1}=x_{3}=0$, $\lambda =\rho (b^{2}+bc)$, $\beta =0$, $\alpha =0$, $x_{2}=0$, for all $\rho , b,c$,

• (vii) $\mu \ne 0$, $a=0$, $d\ne 0$, $x_{1}=x_{3}=0$, $\lambda =0$, $\beta \ne 0$, $x_{2}=\alpha =0$, for all $\rho , b,c$,

• (viii) $\mu \ne 0$, $a\ne 0$, $c=0$, $\beta = 0$, $x_{1}=\epsilon _{1}\sqrt{\frac{\alpha a^{2}-\rho (2a^{2}+b^{2})+\lambda }{\mu }}$, $x_{2}=\epsilon _{2}\sqrt{\frac{\alpha (a^{2}+b^{2})-\rho (2a^{2}+b^{2})+\lambda }{\mu }}$, $x_{3}=\epsilon _{3}\sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}$, for all $b,d,\rho ,\lambda , \alpha$ such that $a+d\ne 0$,

  $$\begin{aligned}&\frac{\alpha a^{2}-\rho (2a^{2}+b^{2})+\lambda }{\mu }\ge 0,\,\,\,\,\,\,\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }\ge 0,\\&\frac{\alpha }{2}(bd-ab) +\mu \epsilon _{1}\epsilon _{2}\sqrt{\frac{\alpha a^{2}-\rho (2a^{2}+b^{2})+\lambda }{\mu }}\sqrt{\frac{\alpha (a^{2}+b^{2})-\rho (2a^{2}+b^{2})+\lambda }{\mu }}=0,\\&\alpha b(a+d)-\mu \epsilon _{1}\epsilon _{3}\sqrt{\frac{\alpha a^{2}-\rho (2a^{2}+b^{2})+\lambda }{\mu }}\sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}=0,\\&\frac{\alpha }{2}(ad+2d^{2})-\mu \epsilon _{2}\epsilon _{3}\sqrt{\frac{\alpha (a^{2}+b^{2})-\rho (2a^{2}+b^{2})+\lambda }{\mu }}\sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}=0. \end{aligned}$$

• (ix) $\mu \ne 0$, $a\ne 0$, $c=0$, $\beta \ne 0$, $x_{1}=F$, $x_{2}=\frac{\rho (2a^{2}+b^{2})-\lambda -\alpha a^{2}+\mu F^{2}}{\beta a}$, $x_{3}=\epsilon \sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}$, for all $b,d, \rho , \alpha$ such that $a+d\ne 0$, $\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }\ge 0$,

  $$\begin{aligned}&\alpha b(a+d)+\frac{\beta }{2}\left( -aF-b\epsilon \sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}\right) -\mu F\epsilon \sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}=0,\\&-\alpha (a^{2}+b^{2})+\beta bF+\mu \Big (\frac{\rho (2a^{2}+b^{2})-\lambda -\alpha a^{2}+\mu F^{2}}{\beta a}\Big )^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\&\frac{\alpha }{2}(ad+2d^{2})+\frac{\beta }{2}\left( -bF-d\frac{\rho (2a^{2}+b^{2})-\lambda -\alpha a^{2}+\mu F^{2}}{\beta a}-d\epsilon \sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}\right) \\&-\mu \epsilon \frac{\rho (2a^{2}+b^{2})-\lambda -\alpha a^{2}+\mu F^{2}}{\beta a}\sqrt{\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }}=0,\\&\frac{\rho (2a^{2}+b^{2})-\lambda }{\mu }\ge 0,\\&27C^{2}+(4A^{3}-18AB)C+4B^{3}-A^{2}B^{2}\ge 0,\\ \end{aligned}$$

  where $\epsilon =\pm 1$,

  $$\begin{aligned}&F=E-\frac{\frac{B}{3}-\frac{A^{2}}{9}}{E}-\frac{A}{3},\\&E=\left( \frac{\sqrt{27C^{2}+(4A^{3}-18AB)C+4B^{3}-A^{2}B^{2}}}{2(3^{\frac{3}{2}})}+\frac{AB-3C}{6}-\frac{A^{3}}{27}\right) ^{\frac{1}{3}}\ne 0,\\&A=-\frac{b\beta }{2\mu },\\&B=\frac{\beta ^{2}a^{2}}{2\mu ^{2}}+\frac{1}{\mu }\Big (\rho (2a^{2}+b^{2})-\lambda -\alpha a^{2} \Big ),\\&C=\frac{\beta a}{2\mu ^{2}}\left( \alpha (bd-ab)-\frac{b}{a}\Big (\rho (2a^{2}+b^{2})-\lambda -\alpha a^{2} \Big ) \right) . \end{aligned}$$

Proof

From [19, 20], we have

$$\begin{aligned} \widetilde{Ric}^{1}=\left( \begin{array}{ccc} -a^{2}&{}\frac{1}{2}(bd-ab) &{}b(a+d) \\ \frac{1}{2}(bd-ab)&{} -(a^{2}+b^{2}+bc) &{}\frac{1}{2}(bc+ad+2d^{2}) \\ b(a+d) &{}\frac{1}{2}(bc+ad+2d^{2}) &{} 0 \\ \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} (\mathcal {L}_{X}^{1}g)=\left( \begin{array}{ccc} -2a x_{2}&{}ax_{1}-bx_{2} &{}-ax_{1}-cx_{2}-bx_{3} \\ ax_{1}-bx_{2} &{} 2bx_{1}&{} -bx_{1}-dx_{2}-dx_{3}\\ -ax_{1}-cx_{2}-bx_{3} &{} -bx_{1}-dx_{2}-dx_{3}&{} 0 \\ \end{array} \right) \end{aligned}$$

with respect to the basis $\{e_{1},e_{2},e_{3}\}$. Therefore $\widetilde{S}= -(2a^{2}+b^{2}+bc)$ and the Eq. (2) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha a^{2}-\beta ax_{2}+\mu x_{1}^{2}=-\rho (2a^{2}+b^{2}+bc)+\lambda ,\\ \frac{\alpha }{2}(bd-ab) +\frac{\beta }{2}(ax_{1}-bx_{2})+\mu x_{1}x_{2}=0,\\ \alpha b(a+d)+\frac{\beta }{2}(-ax_{1}-cx_{2}-bx_{3} )-\mu x_{1}x_{3}=0,\\ -\alpha (a^{2}+b^{2}+bc)+\beta bx_{1}+\mu x_{2}^{2}=-\rho (2a^{2}+b^{2}+bc)+\lambda ,\\ \frac{\alpha }{2}(bc+ad+2d^{2})+\frac{\beta }{2}(-bx_{1}-dx_{2}-dx_{3})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+b^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(55)

Let $\mu =0$, then the system Eq. (55) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha a^{2}-\beta ax_{2}=0,\\ \frac{\alpha }{2}(bd-ab) +\frac{\beta }{2}(ax_{1}-bx_{2})=0,\\ \alpha b(a+d)+\frac{\beta }{2}(-ax_{1}-cx_{2}-bx_{3} )=0,\\ -\alpha (a^{2}+b^{2}+bc)+\beta bx_{1}=0,\\ \frac{\alpha }{2}(bc+ad+2d^{2})+\frac{\beta }{2}(-bx_{1}-dx_{2}-dx_{3})=0,\\ \lambda =\rho (2a^{2}+b^{2}+bc). \end{array}\right. } \end{aligned}$$

(56)

Since $ac=0$ we get $a=0$ or $c=0$. If $a=0$ then $a+d\ne 0$ implies that $d\ne 0$ and we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha bd - \beta bx_{2}=0,\\ \alpha bd+\frac{\beta }{2}(-cx_{2}-bx_{3} )=0,\\ -\alpha (b^{2}+bc)+\beta bx_{1}=0,\\ \frac{\alpha }{2}(bc+2d^{2})+\frac{\beta }{2}(-bx_{1}-dx_{2}-dx_{3})=0,\\ \lambda =\rho (b^{2}+bc). \end{array}\right. } \end{aligned}$$

(57)

If $\beta =0$ then $(\alpha , \beta ,\mu )\ne (0,0,0)$ yields $\alpha \ne 0$. Also, the first equation of Eq. (57) gives $b=0$ and the fourth equation of Eq. (57) implies that $\alpha d^{2}=0$ which is a contradiction. Hence, $\beta \ne 0$ and the cases (i)–(iii) hold.

Now, we consider $\mu =0$ and $a\ne 0$, then $c=0$ and we get

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha a+\beta x_{2}=0,\\ \alpha (bd-ab) +\beta (ax_{1}-bx_{2})=0,\\ \alpha b(a+d)+\frac{\beta }{2}(-ax_{1}-bx_{3} )=0,\\ -\alpha (a^{2}+b^{2})+\beta bx_{1}=0,\\ \frac{\alpha }{2}(ad+2d^{2})+\frac{\beta }{2}(-bx_{1}-dx_{2}-dx_{3})=0,\\ \lambda =\rho (2a^{2}+b^{2}). \end{array}\right. } \end{aligned}$$

(58)

If $\beta =0$ then the first equation of the system Eq. (58) yields $\alpha =0$ which is a contradiction, then $\beta \ne 0$. If $b=0$ then the case (iv) holds. If $b\ne 0$ then from the first three equations of the system Eq. (58) we obtain $x_{1}=-\frac{\alpha bd}{\beta a}$, $x_{2}=-\frac{\alpha a}{\beta }$, $x_{3}=\frac{2\alpha a+3\alpha d}{\beta }$. Hence using the fourth and fiveth equations of the system Eq. (58) we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha (a^{2}+b^{2}+\frac{b^{2}d}{a})=0,\\ \alpha (-d^{2}+\frac{b^{2}d}{a})=0. \end{array}\right. } \end{aligned}$$

(59)

If $\alpha \ne 0$ then $a^{2}+b^{2}+d^{2}=0$ which is a contradiction, then $\alpha =0$, $x_{1}=x_{2}=x_{3}=0$ and the case (v) is true. Now, we consider $\mu \ne 0$. If $a=0$ then $d\ne 0$ and the system Eq. (55) gives

$$\begin{aligned} {\left\{ \begin{array}{ll} \mu x_{1}^{2}=-\rho (b^{2}+bc)+\lambda ,\\ \alpha bd -\beta bx_{2}+2\mu x_{1}x_{2}=0,\\ \alpha bd+\frac{\beta }{2}(-cx_{2}-bx_{3} )-\mu x_{1}x_{3}=0,\\ -\alpha (b^{2}+bc)+\beta bx_{1}+\mu x_{2}^{2}=-\rho (b^{2}+bc)+\lambda ,\\ \frac{\alpha }{2}(bc+2d^{2})+\frac{\beta }{2}(-bx_{1}-dx_{2}-dx_{3})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (b^{2}+bc)-\lambda . \end{array}\right. } \end{aligned}$$

(60)

The first and sixth equations of the system Eq. (60) imply that $x_{1}=x_{3}=0$ and $\lambda =\rho (b^{2}+bc)$. Thus the system Eq. (60) gives

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha bd -\beta bx_{2}=0,\\ 2\alpha bd-\beta cx_{2}=0,\\ -\alpha (b^{2}+bc)+\mu x_{2}^{2}=0,\\ \alpha (bc+2d^{2})-\beta dx_{2}=0.\\ \end{array}\right. } \end{aligned}$$

(61)

If $\beta =0$ then $\alpha =0$, $x_{2}=0$ and the case (vi) holds. If $\beta \ne 0$ and $b=0$ then $x_{2}=\alpha =0$ and the case (vii) is true. Notice if $b\ne 0$ and $\alpha \ne 0$ then from the first two equations of the system Eq. (61) we infer $c=2b$ and Replacing it with $x_{2}=\frac{\alpha d}{\beta }$ in the fourth equation of the system Eq. (61) we obtain $2b^{2}+d^{2}=0$ which is a contradiction. Let $\mu \ne 0$ and $a\ne 0$ then $c=0$ and the system Eq. (55) gives

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha a^{2}-\beta ax_{2}+\mu x_{1}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \frac{\alpha }{2}(bd-ab) +\frac{\beta }{2}(ax_{1}-bx_{2})+\mu x_{1}x_{2}=0,\\ \alpha b(a+d)+\frac{\beta }{2}(-ax_{1}-bx_{3} )-\mu x_{1}x_{3}=0,\\ -\alpha (a^{2}+b^{2})+\beta bx_{1}+\mu x_{2}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \frac{\alpha }{2}(ad+2d^{2})+\frac{\beta }{2}(-bx_{1}-dx_{2}-dx_{3})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+b^{2})-\lambda . \end{array}\right. } \end{aligned}$$

(62)

If $\beta =0$ then the system Eq. (62) reduces to

$$\begin{aligned} {\left\{ \begin{array}{ll} -\alpha a^{2}+\mu x_{1}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \frac{\alpha }{2}(bd-ab) +\mu x_{1}x_{2}=0,\\ \alpha b(a+d)-\mu x_{1}x_{3}=0,\\ -\alpha (a^{2}+b^{2})+\mu x_{2}^{2}=-\rho (2a^{2}+b^{2})+\lambda ,\\ \frac{\alpha }{2}(ad+2d^{2})-\mu x_{2}x_{3}=0,\\ \mu x_{3}^{2}=\rho (2a^{2}+b^{2})-\lambda . \end{array}\right. } \end{aligned}$$

(63)

Then the case (viii) is true. If $\beta \ne 0$, then the first and second equations of the system Eq. (62) imply that

$$\begin{aligned} x_{1}^{3}+A x_{1}^{2}+Bx_{1}+C=0. \end{aligned}$$

(64)

Thus the case (ix) holds.

$\square$