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

Journal of Nonlinear Mathematical Physics https://doi.org/10.1007/s44198-022-00069-2 RESEARCH ARTICLE Generalized Ricci Solitons of Three‑Dimensional Lorentzian Lie Groups Associated Canonical Connections and Kobayashi‑Nomizu Connections Shahroud Azami1 Received: 21 April 2022 / Accepted: 8 June 2022 © The Author(s) 2022 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. Keywords Generalized Ricci soliton · Lie group · Three-dimensional Lorentzian Mathematics Subject Classification 58C40 · 53E20 · 53C21 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 ∈ X(M) and a smooth function 𝜆 on M such that 𝛼Ric + 𝛽 L g + 𝜇X ♭ ⊗ X ♭ = (𝜌S + 𝜆)g, 2 X (1) for some constants 𝛼, 𝛽, 𝜇, 𝜌 ∈ ℝ, with (𝛼, 𝛽, 𝜇) ≠ (0, 0, 0), where LX denotes the Lie derivative in the direction of X, X ♭ denotes a 1-form such that X ♭ (Y) = g(X, Y), S is the scalar curvature, and Ric is the Ricci tensor. The generalized Ricci soliton becomes * Shahroud Azami azami@sci.ikiu.ac.ir 1 Department of Pure Mathematics, Imam Khomeini International University, Persian Gulf, Qazvin, Qazvin 34148-96818, Iran 13 Vol.:(0123456789) Journal of Nonlinear Mathematical Physics (i) the homothetic vector field equation when 𝛼 = 𝜇 = 𝜌 = 0 and 𝛽 ≠ 0, (ii) the Ricci soliton equation when 𝛼 = 1, 𝜇 = 0, and 𝜌 = 0, (iii) the Ricci-Bourguignon soliton ( or 𝜌-Einstein soliton equation when 𝛼 = 1 and 𝜇 = 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 𝜌 = 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{e1 , e2 } and V ⟂ = span{e3 } for the three dimensional Lorentzian Lie group Gi , i = 1, ....., 7, with product structure J such that Je1 = e1 , Je2 = e2, and Je3 = −e3. 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 leftinvariant Lorentzian metric [3]. 2.1 Unimodular Lie Groups Let {e1 , e2 , e3 } be an orthonormal basis of signature (+ + −). We denote the Lorentzian vector product on ℝ31 induced by the product of the para-quaternions by × i.e., e1 × e2 = −e3 , e2 × e3 = −e1 , e3 × e1 = −e2 . Then the Lie bracket [ , ] defines the corresponding Lie algebra 𝔤, which is unimodular if and only if the endomorphism L defined by [Z, Y] = L(Z × 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]. 𝔤1: If L is diagonalizable with eigenvalues {a, b, c} with respect to an orthonormal basis {e1 , e2 , e3 } of signature (+ + −), then the corresponding Lie algebra is given by 13 Journal of Nonlinear Mathematical Physics [e1 , e2 ] = −ce3 , [e1 , e3 ] = −be2 , [e2 , e3 ] = ae1 . 𝔤2: Assume L has a complex eigenvalues. Then, with respect to an orthonormal basis {e1 , e2 , e3 } of signature (+ + −), one has ⎛a 0 0 ⎞ L = ⎜ 0 c − b ⎟, ⎜ ⎟ ⎝0 b c ⎠ b ≠ 0, then the corresponding Lie algebra is given by [e1 , e2 ] = be2 − ce3 , [e1 , e3 ] = −ce2 − be3 , [e2 , e3 ] = ae1 . 𝔤3: Assume L has a triple root of its minimal polynomial. Then, with respect to an orthonormal basis {e1 , e2 , e3 } of signature (+ + −), the corresponding Lie algebra is given by [e1 , e2 ] = ae1 − be3 , [e1 , e3 ] = −ae1 − be2 , [e2 , e3 ] = be1 + ae2 + ae3 , a ≠ 0. 𝔤4: Assume L has a double root of its minimal polynomial. Then, with respect to an orthonormal basis {e1 , e2 , e3 } of signature (+ + −), the corresponding Lie algebra is given by [e1 , e2 ] = −e2 − (2d − b)e3 , [e1 , e3 ] = −be2 + e3 , [e2 , e3 ] = ae1 , d = ±1. 2.2 Non-unimodular Lie Groups Next we treat the non-unimodular case. Let 𝔊 denotes a special class of the solvable Lie algebra 𝔤 such that [x, y] is a linear combination of x and y for any x, y ∈ 𝔤 . From [6], the non-unimodular Lorentzian Lie algebras of non-constant sectional curvature not belonging to class 𝔊 with respect to a pseudo-orthonormal basis {e1 , e2 , e3 } with e3 time-like are one of the following: 𝔤5: 𝔤6: [e1 , e2 ] = 0, [e1 , e3 ] = ae1 + be2 , [e2 , e3 ] = ce1 + de2 , a + d ≠ 0, ac + bd = 0. [e1 , e2 ] = ae2 + be3 , [e1 , e3 ] = ce2 + de3 , [e2 , e3 ] = 0, a + d ≠ 0, ac − bd = 0. 13 Journal of Nonlinear Mathematical Physics [e1 , e2 ] = −ae1 − be2 − be3 , [e1 , e3 ] = ae1 + be2 + be3 , [e2 , e3 ] = ce1 + de2 + de3 , a + d ≠ 0, ac = 0. 𝔤7: Throughout this paper, we assume that Gi , 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 gi , i = 1, 2, ....., 7 , respectively. Let ∇ be the Levi-Civita connection of Gi and R(X, Y)Z = [∇X , ∇Y ]Z − ∇[X,Y] Z be its curvature tensor. The Ricci tensor of (Gi , g) with respect to orthonormal basis {e1 , e2 , e3 } of signature (+ + −) is defined by Ric(X, Y) = −g(R(X, e1 )Y, e1 ) − g(R(X, e2 )Y, e2 ) + g(R(X, e3 )Y, e3 ). We consider a product structure J on Gi by Je1 = e1 , Je2 = e2 , Je3 = −e3. Similar [8], we consider the canonical connection and the Kobayashi-Nomizu connection as 1 ∇0X Y = ∇X Y − (∇X J)JY, 2 1 ∇1X Y = ∇0X Y − [(∇Y J)JX − (∇JY J)X], 4 respectively. We define Ri (X, Y)Z = [∇iX , ∇iY ]Z − ∇i[X,Y] Z, i = 0, 1, and the Ricci tensors of (Gi , g) associated to the canonical connection and the Kobayashi-Nomizu connection are defined by Rici (X, Y) = −g(Ri (X, e1 )Y, e1 ) − g(Ri (X, e2 )Y, e2 ) + g(Ri (X, e3 )Y, e3 ), i = 0, 1. Let i i i ̃ (X, Y) = Ric (X, Y) + Ric (Y, X) , i = 0, 1. Ric 2 Similar to definition of (LV g) where (LX g)(Y, Z) = g(∇Y V, Z) + g(Y, ∇Z V), we define (LiV g)(Y, Z) ∶= g(∇iY V, Z) + g(Y, ∇iZ V), i = 0, 1. Definition 1 The Lie group (G, g, J) is called the affine generalized Ricci soliton associated to the connection ∇i , i = 0, 1 if it satisfies i � (Y, Z) + 𝛼 Ric 𝛽 i L g(Y, Z) + 𝜇X ♭ ⊗ X ♭ (Y, Z) = (𝜌� Si + 𝜆)g(Y, Z), i = 0, 1, (2) 2 X i ̃ . where ̃ Si = gjk Ric jk Throughout this paper for prove of our results we use the results of [19, 20]. 13 Journal of Nonlinear Mathematical Physics 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 ∇0 on the Lie group (G1 , g, J, X) are the following: (i) 𝜇 = 𝜆 = 0, a + b − c = 0, and for all x1 , x2 , x3 , 𝛼, 𝛽, 𝜌 such that (𝛼, 𝛽, 𝜇) ≠ (0, 0, 0) , (ii) 𝜇 = 0 , a + b − c ≠ 0 , 𝛼 = 0 , 𝛽 ≠ 0 , x1 = x2 = 0 , 𝜆 = 𝜌c(a + b − c), and for all x3 , 𝜌, (iii) 𝜇 = 0, a + b − c ≠ 0, 𝛼 ≠ 0, c = 𝛽 = 𝜆 = 0, and for all x1 , x2 , x3 , 𝜌, (iv) 𝜇 = 0, a + b − c ≠ 0, 𝛼 ≠ 0, c = 𝜆 = 0,𝛽 ≠ 0, x1 = x2 = 0, and for all x3 , 𝜌, (v) 𝜇 ≠ 0 , x1 = x2 = 0 , 𝜆 = (𝜌 − 12 𝛼)c(a + b − c) , x32 = 𝜌c(a+b−c)−𝜆 , and for all 𝜇 ≥ 0. x3 , 𝛼, 𝛽, 𝜌, a, b, c such that 𝜌c(a+b−c)−𝜆 𝜇 Proof From [19, 20], we have 1 0 0⎞ ⎛ − 2 c(a + b − c) 1 ⎜ ̃ Ric = 0 − 2 c(a + b − c) 0 ⎟ ⎟ ⎜ ⎝ 0 0 0⎠ 0 and ⎛ 0 0 − 12 x2 (a + b − c) ⎞ ⎜ ⎟ 1 x (a + b − c) ⎟ 0 0 (L0X g) = ⎜ 2 1 ⎜ − 1 x (a + b − c) 1 x (a + b − c) ⎟ 0 ⎝ 2 2 ⎠ 2 1 S = −c(a + b − c) and with respect to the basis {e1 , e2 , e3 }. Therefore ̃ X ♭ ⊗ X ♭ (ei , ej ) = 𝜖i 𝜖j xi xj where (𝜖1 , 𝜖2 , 𝜖3 ) = (1, 1, −1). Hence, by Eq. (2) there exists a affine generalized Ricci soliton associated to the connection ∇0 if and only if the following system of equations is satisfied 13 Journal of Nonlinear Mathematical Physics 1 2 ⎧ − 2 𝛼c(a + b − c) + 𝜇x1 = −𝜌c(a + b − c) + 𝜆, ⎪ ⎪ 𝜇x x = 0, ⎪ 1 2 ⎪ ⎪ − 𝛽 x (a + b − c) − 𝜇x x = 0, 1 3 ⎪ 4 2 ⎨ ⎪ − 1 𝛼c(a + b − c) + 𝜇x2 = −𝜌c(a + b − c) + 𝜆, 2 ⎪ 2 ⎪𝛽 ⎪ x1 (a + b − c) − 𝜇x2 x3 = 0, ⎪4 ⎪ 2 ⎩ 𝜇x3 = 𝜌c(a + b − c) − 𝜆. (3) Using the first and fourth equations of the system Eq. (3) we have 𝜇(x12 − x22 ) = 0. From the third and fiveth equations of the system Eq. (3) we get 𝛽 (x − x2 )(a + b − c) − 𝜇x3 (x1 + x2 ) = 0. 4 1 Multiplying both sides of last equality by (x1 − x2 ) we conclude 𝛽(x1 − x2 )2 (a + b − c) = 0. (4) The second equation of the system Eq. (3) implies that 𝜇 = 0, or x1 = 0 or x2 = 0. Suppose that 𝜇 = 0. In this case, the system Eq. (3) reduces to ⎧ 𝛼c(a + b − c) = 0, ⎪ 𝛽x2 (a + b − c) = 0, ⎨ 𝛽x (a + b − c) = 0, ⎪ 1 ⎩ 𝜌c(a + b − c) = 𝜆. (5) If a + b − c = 0 then the system Eq. (5) holds for any x1 , x2, and x3. If a + b − c ≠ 0 for the cases (ii)–(iv) the sytem Eq. (5) holds. Now we assume that 𝜇 ≠ 0 and x1 = 0, then x2 = 0 and the system Eq. (3) becomes { 1 − 2 𝛼c(a + b − c) = −𝜌c(a + b − c) + 𝜆, (6) 𝜇x32 = 𝜌c(a + b − c) − 𝜆. This shows that the case (v) holds. ◻ Theorem 2 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G1 , g, J, X) are the following: (i) (ii) (iii) (iv) 𝜇 𝜇 𝜇 𝜇 = 0, c = 0, 𝜆 = 0, 𝛽 = 0, c = 0, 𝜆 = 0, 𝛽 = 0, c = 0, 𝜆 = 0, 𝛽 = 0, c = 0, 𝜆 = 0, 𝛽 13 = 0, and for all a, b, x1 , x2 , x3 , 𝛼, 𝜌 such that 𝛼 ≠ 0, ≠ 0, a = b = 0, and for all x1 , x2 , x3 , 𝛼, 𝜌, ≠ 0, a = x1 = 0, and for all b, x2 , x3 , 𝛼, 𝜌, ≠ 0, a ≠ 0, x2 = b = 0, and for all x1 , x3 , 𝛼, 𝜌, Journal of Nonlinear Mathematical Physics (v) (vi) (vii) (viii) (ix) (x) (xi) (xii) (xiii) (xiv) (xv) (xvi) 𝜇 = 0, c = 0, 𝜆 = 0, 𝛽 ≠ 0, a ≠ 0, x2 = x1 = 0, and for all b, x3 , 𝛼, 𝜌, 𝜇 = 0, c ≠ 0, 𝜆 = 𝜌c(a + b), b = 0, 𝛽 = a = 0, and for all x1 , x2 , x3 , 𝛼, 𝜌 such that 𝛼 ≠ 0, 𝜇 = 0, c ≠ 0, 𝜆 = 𝜌c(a + b), b = 0, 𝛽 ≠ 0, a = 0, and for all x1 , x2 , x3 , 𝛼, 𝜌, 𝜇 = 0, c ≠ 0, 𝜆 = 𝜌c(a + b), b = 0, 𝛽 ≠ 0, a ≠ 0, x2 = 0, and for all x1 , x3 , 𝛼, 𝜌, 𝜇 = 0, c ≠ 0, 𝜆 = 𝜌c(a + b), b ≠ 0, 𝛼 = 0, a = x1 = 0, and for all x2 , x3 , 𝛽, 𝜌, such that 𝛽 ≠ 0, 𝜇 = 0 , c ≠ 0 , 𝜆 = 𝜌c(a + b) , b ≠ 0 , 𝛼 = 0 , a ≠ 0 , x2 = x1 = 0 , and for all x3 , 𝛽, 𝜌, such that 𝛽 ≠ 0, 𝜇 ≠ 0, x1 = 0, x3 = 0, c = 0, 𝜆 = x2 = 0, for all a, b, 𝛼, 𝛽, 𝜌, 𝜇 ≠ 0, x1 = 0, x3 = 0, c ≠ 0, 𝛼 = 0, 𝜆 = 𝜌c(a + b), x2 = 0, for all a, b, 𝛽, 𝜌, 𝜇 ≠ 0, x1 = 0, x3 = 0, c ≠ 0, 𝛼 ≠ 0, b = 0, 𝜆 = x2 = a = 0, for all 𝛽, 𝜌, 𝜇 ≠ 0, x1 = 0, x3 = 0, c ≠ 0, 𝛼 ≠ 0, b = 0, x2 ≠ 0, 𝜆 = 𝜌ca, 𝛽 = 0, x22 = a𝛼c 𝜇 for all a, 𝜌, > 0, for all a, b, c, 𝛼, 𝛽, 𝜌, 𝜆 such that 𝜇 ≠ 0, x1 = 0, x3 ≠ 0, x2 = 0, x32 = ac𝛼 𝜇 bc𝛼 = ac𝛼 = 𝜌c(a + b) − 𝜆, 𝜇 ≠ 0 , x1 ≠ 0 , x2 = x3 = 0 , 𝜆 = 𝜌c(a + b) , for all a, b, c, 𝛼, 𝜌, 𝛽 such that 𝛽b = ac𝛼 = 0. Proof From [19, 20], we have ⎛ −bc 0 0 ⎞ 1 ̃ = ⎜ 0 − ac 0 ⎟ Ric ⎜ ⎟ 0 0⎠ ⎝ 0 and ⎛ 0 0 − ax2 ⎞ 0 bx1 ⎟ (L1X g) = ⎜ 0 ⎜ ⎟ ⎝ −ax2 bx1 0 ⎠ S = −c(a + b) and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Therefore ̃ becomes ⎧ −bc𝛼 + 𝜇x2 = −𝜌c(a + b) + 𝜆, ⎪ 𝜇x x = 0, 1 ⎪ 𝛽1 2 ⎪ − 2 ax2 − 𝜇x1 x3 = 0, ⎨ −ac𝛼 + 𝜇x2 = −𝜌c(a + b) + 𝜆, 2 ⎪𝛽 ⎪ 2 bx1 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌c(a + b) − 𝜆. ⎩ 3 (7) The second equation of the system Eq. (7) implies that 𝜇 = 0 or x1 = 0 or x2 = 0. We consider 𝜇 = 0, then the first equation yields bc𝛼 = 0. If c = 0 then we get 𝜆 = 0 and the cases (i)-(v) hold. If we assume that c ≠ 0 and 𝜆 = 𝜌c(a + b) and in this we 13 Journal of Nonlinear Mathematical Physics obtain the cases (vi)-(x). Now, we consider the case 𝜇 ≠ 0 and x1 = 0. In this case the system Eq. (7) reduces to ⎧ −bc𝛼 = −𝜌c(a + b) + 𝜆, ⎪ 𝛽ax = 0, 2 ⎪ 2 ⎨ −ac𝛼 + 𝜇x2 = −𝜌c(a + b) + 𝜆, ⎪ x2 x3 = 0, ⎪ 𝜇x2 = 𝜌c(a + b) − 𝜆. ⎩ 3 (8) The fourth equation of the system Eq. (8) implies that x2 = 0 or x3 = 0. If x3 = 0 then we obtain the cases (xi)-(xiv). If x3 ≠ 0 and x2 = 0 then the case (xv) holds. Also, if we consider 𝜇 ≠ 0 and x1 ≠ 0 then x2 = 0 and the case (xvi) is true. ◻ Theorem 3 The left-invariant affine generalized Ricci soliton associated to the connection ∇0 on the Lie group (G2 , g, J, X) are the following: (i) 𝜇 = 0, 𝛽 ≠ 0, x1 = x2 = 𝛼 = 0, 𝜆 = 𝜌(2b2 + ac), for all a, b, c, x3 , 𝜌 such that b ≠ 0, (ii) 𝜇 ≠ 0, x2 = 0, x1 = 0, 𝛼 = 0, 𝜆 = 𝜌(2b2 + ac), x3 = 0, for all 𝛽, 𝜌, a, b, c such that b ≠ 0. (iii) 𝜇 ≠ 0, x2 = 0, x1 = 0, 𝛼 ≠ 0, a = 2c , 𝜆 = (2𝜌 − 𝛼)(b2 + c2 ), x32 = 𝜇𝛼 (b2 + c2 ), for all 𝛽, 𝜌, b such that b ≠ 0 and 𝛼𝜇 ≥ 0, ≠ 0 , x3 = 0 , 𝜆 = 𝜌(2b2 + ac) , for all a, b, c, 𝛼, 𝛽, 𝜌 (iv) 𝜇 ≠ 0 , x2 = 0 , x1 = − 𝛽b 𝜇 such that 𝛽 ≠ 0, 𝛼𝜇(2b2 + ac) + 𝛽 2 b2 = 0, and 𝛼𝜇(2c + a) − 𝛽 2 a = 0. Proof From [19, 20], we have ⎛ −(b2 + ̃ = ⎜⎜ 0 Ric ⎜ 0 ⎝ 0 ac ) 2 − 0 ) + ac 2 ab − 4 (b2 bc 2 bc 2 0 − 0 ab 4 ⎞ ⎟ ⎟ ⎟ ⎠ and bx2 − a2 x2 ⎞ ⎛ 0 a (L0X g) = ⎜ bx2 − 2bx1 2 x1 ⎟ ⎜ a ⎟ a 0 ⎠ ⎝ − 2 x2 2 x1 S = −(2b2 + ac) and the Eq. (2) becomes with respect to the basis {e1 , e2 , e3 }. Then ̃ 13 Journal of Nonlinear Mathematical Physics ) + 𝜇x12 = −𝜌(2b2 + ac) + 𝜆, ⎧ −𝛼(b2 + ac ⎪ 𝛽 bx + 𝜇x2 x = 0, 1 2 ⎪2 2 ⎪ − 𝛽a x2 − 𝜇x1 x3 = 0, 4 ⎨ −𝛼(b 2 + ac ) − 𝛽bx1 + 𝜇x22 = −𝜌(2b2 + ac) + 𝜆, ⎪ 2 ⎪ 𝛼( bc − ab ) + 𝛽a x1 − 𝜇x2 x3 = 0, 4 4 ⎪ 22 2 ⎩ 𝜇x3 = 𝜌(2b + ac) − 𝜆. (9) At the first we assume 𝜇 = 0. In this case, the system Eq. (9) reduces to ) = 0, ⎧ −𝛼(b2 + ac ⎪ 𝛽x = 0, 2 ⎪ 2 ⎨ 𝛽x1 = 0, ⎪ 𝛼( bc − ab ) + 𝛽a x1 = 0, 4 ⎪ 22 4 ⎩ 𝜌(2b + ac) = 𝜆. (10) The second equation of Eq. (10) implies that 𝛽 = 0 or x2 = 0. If 𝛽 = 0 then 𝛼 ≠ 0 and the fourth equation of the system Eq. (10) yields a = 2c and replacing it in the first equation we obtain b2 + c2 = 0 which is a contradiction. Thus 𝛽 ≠ 0 and x1 = x2 = 𝛼 = 0. Now we consider 𝜇 ≠ 0. Using the first and fourth equations of Eq. (9) we obtain 𝜇x12 + 𝛽bx1 = 𝜇x22 . (11) 𝛽b The second equation of the system Eq. (9) implies that x2 = 0 or x1 = − 2𝜇 . If x2 ≠ 0 𝛽b then x1 = − 2𝜇 and plugging it in Eq. (11) we get x22 + tion. Therefore x2 = 0 and in this case we have 𝛽 2 b2 4𝜇2 = 0 which is a contradic- ⎧ −𝛼(b2 + ac ) + 𝜇x2 = −𝜌(2b2 + ac) + 𝜆, 1 2 ⎪ 2 ⎪ 𝜇x1 + 𝛽bx1 = 0, ⎪ x1 x3 = 0, ⎨ −𝛼(b2 + ac ) − 𝛽bx = −𝜌(2b2 + ac) + 𝜆, ⎪ bc ab 2 𝛽a 1 ⎪ 𝛼( 2 − 4 ) + 4 x1 = 0, ⎪ 𝜇x2 = 𝜌(2b2 + ac) − 𝜆. ⎩ 3 (12) The third equation of the system Eq. (12) implies that x1 = 0 or x3 = 0. If x1 = 0 then 𝛼(2c − a) = 0. Thus 𝛼 = 0 or a = 2c. In the case 𝛼 = 0 we have 𝜆 = 𝜌(2b2 + ac) and x3 = 0. In the case 𝛼 ≠ 0 and a = 2c we get 𝜆 = (2𝜌 − 𝛼)(b2 + c2 ) and x32 = 𝜇𝛼 (b2 + c2 ). Now we assume that 𝜇 ≠ 0, x2 = 0, x1 ≠ 0, and x3 = 0. In this case we have (iv). ◻ Theorem 4 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G2 , g, J, X) are the following: 13 Journal of Nonlinear Mathematical Physics 𝜇 = 0, 𝛼 = 0, 𝛽 ≠ 0, x1 = x2 = x3 = 0, 𝜆 = 𝜌(2b2 + c2 + ac), for all 𝜌, a, b, c such that b ≠ 0, (ii) 𝜇 ≠ 0, x2 = x3 = 0, 𝛽 = 0, 𝛼 = 0, x1 = 0, 𝜆 = 𝜌(2b2 + c2 + ac), for all 𝜌, a, b, c such that b ≠ 0, (iii) 𝜇 ≠ 0 , x2 = x3 = 0 , 𝛽 ≠ 0 , c = 0 , 𝛼 = 0 , x1 = 0 , 𝜆 = 𝜌(2b2 ), for all 𝜌, a, b, c such that b ≠ 0, (iv) 𝜇 ≠ 0, x2 = x3 = 0, 𝛽 ≠ 0, c = 0, 𝛼 ≠ 0, a = 0, x1 = − 𝛼b , 𝜆 = 𝜌(2b2 + c2 ), for 𝛽 (i) all 𝜌, a, b, c such that b ≠ 0, 𝜇𝛼b2 = 𝛽 2 (b2 + c2 ), , 𝜆 = 𝜌(2b2 + c2 + ac) , for all (v) 𝜇 ≠ 0 , x2 = x3 = 0 , 𝛽 ≠ 0 , c ≠ 0 , x1 = 𝛼ab 𝛽c 𝛼, 𝜌, a, b, c such that b ≠ 0, 𝛼ab2 = −𝛼c(b2 + ac), 𝜇(𝛼ab)2 = 𝛼𝛽 2 c2 (b2 + c2 ), 2 2 𝛽b (vi) , 𝜆 = 𝜌(2b2 + c2 + ac) 𝜇 ≠ 0 , x2 = 0 , x32 = 𝜇𝛼 (b2 + c2 ) − 𝛽4𝜇b2 ≠ 0 , x1 = 2𝜇 𝛽 2 b2 , 2𝜇 −𝛼(b2 + c2 ) − for all 𝛼, 𝛽, 𝜌, a, b, c such that b ≠ 0 , 2a𝛼𝜇 = 𝛽 2 c , 3𝛽 2 b2 − 4𝜇𝛼c(c − a) = 0, (vii) 𝜇 ≠ 0 , x22 = − 𝜇𝛼 c(c − a) ≠ 0 , x1 = 0 , 𝛽 = 0 , x3 = 𝜇𝛼 (b2 + c2 ) , 𝜆 = −𝛼(b2 + c2 ) + 𝜌(2b2 + c2 + ac) for all 𝛼, 𝜌, a, b, c such that b ≠ 0 , −4c(c − a)(b2 + c2 ) = a2 b2 , 𝜇𝛼 ≥ 0, −c(c − a) ≥ 0, (viii) 𝜇≠0 , 2 x2 ≠ 0 , 2 𝜆 = −𝛼(b + c ) + x22 = 𝛽b 2 ) −( 2𝜇 − 𝛽 2 b2 4𝜇 𝛼 c(c 𝜇 𝛽b x1 = − 2𝜇 , 𝛽≠0 , x3 = a x 2b 2 = − 2𝛼𝜇ab+c𝛽 4𝜇2 2b , + 𝜌(2b + c + ac) , for all 𝛼, 𝜌, a, b, c such that 2 2 − a), x32 = 𝜇𝛼 (b2 + c2 ) − 𝛽 2 b2 4𝜇2 > 0. Proof From [19, 20], we have 2 2 0 0 ⎛ −(b + c ) ̃ =⎜ 0 − (b2 + ac) − ab Ric 2 ⎜ 0 0 − ab ⎝ 2 1 ⎞ ⎟ ⎟ ⎠ and ⎛ 0 bx2 − ax2 + bx3 ⎞ ⎟ − 2bx1 cx1 (L1X g) = ⎜ bx2 ⎜ ⎟ 0 ⎝ ax2 + bx3 cx1 ⎠ S = −(2b2 + c2 + ac) and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Therefore ̃ becomes ⎧ −𝛼(b2 + c2 ) + 𝜇x12 = −𝜌(2b2 + c2 + ac) + 𝜆, ⎪ 𝛽 bx + 𝜇x x = 0, 1 2 ⎪2 2 ⎪ 𝛽 (−ax2 + bx3 ) − 𝜇x1 x3 = 0, 2 ⎨ −𝛼(b 2 + ac) − 𝛽bx1 + 𝜇x22 = −𝜌(2b2 + c2 + ac) + 𝜆, ⎪ ab ⎪ −𝛼 2 + 𝛽2 cx1 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2b2 + c2 + ac) − 𝜆. ⎩ 3 13 (13) Journal of Nonlinear Mathematical Physics We first consider 𝜇 = 0. In this case, the system Eq. (13) becomes ⎧ −𝛼(b2 + c2 ) = 0, ⎪ ⎪ 𝛽bx2 = 0, ⎪ 𝛽(−ax2 + bx3 ) = 0, ⎨ −𝛼(b2 + ac) − 𝛽bx = 0, 1 ⎪ ⎪ −𝛼ab + 𝛽cx1 = 0, ⎪ 𝜌(2b2 + c2 + ac) = 𝜆. ⎩ (14) Since b ≠ 0, the first equation of Eq. (14) implies that 𝛼 = 0. Due to (𝛼, 𝛽, 𝜇) ≠ (0, 0, 0) we conclude 𝛽 ≠ 0. Then the second equation of the system Eq. (14) yields x2 = 0. Using, the third and fourth equations of Eq. (14) we obtain x1 = x3 = 0. Now we consider 𝜇 ≠ 0. The second equation of the system Eq. (13) implies that 𝛽b . If x2 = 0 then we get x2 = 0 or x1 = − 2𝜇 ⎧ −𝛼(b2 + c2 ) + 𝜇x12 + 𝜇x32 = 0, ⎪ 𝛽 bx − 𝜇x x = 0, 1 3 ⎪ 2 32 2 ⎨ −𝛼(b + ac) − 𝛽bx1 + 𝜇x3 = 0, 𝛽 ab ⎪ −𝛼 + cx1 = 0, 2 ⎪ 22 ⎩ 𝜇x3 = 𝜌(2b2 + c2 + ac) − 𝜆. (15) From the second equation of the system Eq. (15) we obtain x3 = 0 or x1 = x3 = 0 then the cases (ii)-(v) hold. If x3 ≠ 0 and x1 = Now we assume that 𝜇 ≠ 0, x2 ≠ 0 and x1 = reduces to 𝛽b . − 2𝜇 𝛽b 2𝜇 If then the case (vi) holds. In this cases, the system Eq. (13) ⎧ 𝛽 2 b2 2 2 2 2 ⎪ −𝛼(b + c ) + 4𝜇 = −𝜌(2b + c + ac) + 𝜆, ⎪ 𝛽 bx + 𝜇x x = 0, 1 2 ⎪2 2 ⎪ 𝛽(−ax2 + 2bx3 ) = 0, ⎨ −𝛼(b2 + ac) + 𝛽 2 b2 + 𝜇x2 = −𝜌(2b2 + c2 + ac) + 𝜆, 2 ⎪ 2𝜇 ⎪ −𝛼 ab − c𝛽 2 b − 𝜇x x = 0, 2 3 2 4𝜇 ⎪ ⎪ 𝜇x32 = 𝜌(2b2 + c2 + ac) − 𝜆. ⎩ Thus the cases (vii)-(viii) are true. 𝛽b . 2𝜇 (16) ◻ Theorem 5 The left-invariant affine generalized Ricci soliton associated to the connection ∇0 on the Lie group (G3 , g, J, X) are the following: (i) (ii) 𝜇 = 0, 𝛼 = 0, 𝛽 ≠ 0, x1 = x2 = 0, for all 𝜌, a, b, c, x3 such that a ≠ 0, 𝜇 ≠ 0 , x1 = 0 , x2 = 0 , x3 = 0 , 𝛼 = 0 , 𝜆 = 𝜌(2a2 + b2 ), for all 𝛽, a, b, c such that a ≠ 0, 13 Journal of Nonlinear Mathematical Physics (iii) 𝜇 ≠ 0 , x1 = 0 , x2 = 𝛽a 𝜇 ≠ 0 , x3 = a𝛼 , 2𝛽 2 𝜆 = (2𝜌 − 𝛼)(a2 + 𝛼, 𝛽, a, b, c, 𝜌 such that a ≠ 0, 𝜇𝛼b = 𝛽 b, 𝛼 2 a2 4𝛽 2 = 𝛼 2 (a 𝜇 + b2 ) 2 b2 ) 2 − 𝛽 2 a2 , 𝜇 𝛽 2 a2 . 𝜇2 + for all Proof From [19, 20], we have ⎛ −(a2 + 0 ̃ = ⎜⎜ Ric 0 ab ⎜ ⎝ 4 b2 ) 2 ⎞ ⎟ + 2) ⎟ a2 0 ⎟⎠ 2 0 − (a2 b2 ab 4 a2 2 and ⎛ 2ax2 − ax1 − b x2 ⎞ 2 ⎜ ⎟ b x (L0X g) = ⎜ −ax1 0 2 1 ⎟ ⎜ −bx bx 0 ⎟⎠ ⎝ 2 2 2 1 S = −(2a2 + b2 ) and the Eq. (2) becomes with respect to the basis {e1 , e2 , e3 }. Then ̃ 2 b ⎧ −𝛼(a2 + 2 ) + 𝛽ax2 + 𝜇x12 = −𝜌(2a2 + b2 ) + 𝜆, ⎪ − 𝛽 ax + 𝜇x x = 0, 1 2 ⎪ 𝛼ab2 1𝛽b ⎪ − x − 𝜇x1 x3 = 0, 2 ⎨ 4 2 4 b2 2 2 2 ⎪ −𝛼(a + 2 ) + 𝜇x2 = −𝜌(2a + b ) + 𝜆, 2 𝛽b a ⎪ 𝛼 + x − 𝜇x x = 0, 2 3 ⎪ 22 4 1 2 ⎩ 𝜇x3 = 𝜌(2a + b2 ) − 𝜆. (17) Let 𝜇 = 0. In this case, the system Eq. (17) reduces to ⎧ 𝛽x2 = 0, ⎪ 𝛽x1 = 0, ⎨ 𝛼 = 0, ⎪ ⎩ 𝜆 = 𝜌(2a2 + b2 ). (18) Since (𝛼, 𝛽, 𝜇) ≠ (0, 0, 0) we get 𝛽 ≠ 0 and x1 = x2 = 0. Thus the case (i) holds. Using of the first and fourth equations of the system Eq. (17) we get 𝛽ax2 + 𝜇x12 = 𝜇x22 . (19) Now, we consider 𝜇 ≠ 0, in this case, the second equation of the system Eq. (17) 𝛽a 𝛽a implies that x1 = 0 or x2 = 2𝜇 . If x1 ≠ 0 then x2 = 2𝜇 . Substutiting it in Eq. (19) we 𝛽a 2 ) + x12 = 0 which is a contradiction. Hence x1 = 0 and the system Eq. (17) have ( 2𝜇 and Eq. (19) become 13 Journal of Nonlinear Mathematical Physics 2 ⎧ −𝛼(a2 + b ) + 𝛽ax2 = −𝜌(2a2 + b2 ) + 𝜆, ⎪ 𝛼ab 𝛽b 2 ⎪ 4 − 4 x22 = 0, ⎪ −𝛼(a2 + b ) + 𝜇x2 = −𝜌(2a2 + b2 ) + 𝜆, 2 2 ⎨ a2 ⎪ 𝛼 2 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + b2 ) − 𝜆, ⎪ 3 ⎩ 𝛽ax2 = 𝜇x22 . The sixth equation of Eq. (20) yields x2 = 0 or x2 = true. If x2 ≠ 0 and x2 = 𝛽a 𝜇 𝛽a . 𝜇 (20) If x2 = 0 then the case (ii) is then the case (iii) holds. ◻ Theorem 6 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G3 , g, J, X) are the following: (i) 𝜇 = 0, 𝛼 = 0, 𝛽 ≠ 0, x1 = x2 = x3 = 0, 𝜆 = 2𝜌(a2 + b2 ), for all 𝜌, a, b, c such that a ≠ 0, (ii) 𝜇 ≠ 0, 𝛼b = 0, x1 = 𝛽 = x2 = 𝛼 = x3 = 0, 𝜆 = 2𝜌(a2 + b2 ), (iii) 𝜇 ≠ 0, 𝛼b = 0, x1 = 0, 𝛽 = 0, x2 = 0, x3 = − 𝛼a , 𝜆 = (2𝜌 − 𝛼)(a2 + b2 ), 𝛽 𝛼 2 a2 𝜇 = 𝛽 2 𝛼(a2 + b2 ), (iv) 𝜇 ≠ 0, 𝛼b = 0, x1 = 0, 𝛽 = 0, x2 = 𝛽a , 𝜇 b = 0, 𝛼𝜇 = −𝛽 2 , x3 = 𝛽 2 a2 𝜇−1 , 𝜇2 2 2 𝜆 = (2𝜌 − 𝛼)(a2 + b2 ) + 𝛽 𝜇a , (v) 𝜇 ≠ 0, 𝛼b ≠ 0, � √ −𝛽 2 a2 + 𝛽 4 a4 + 64𝜇2 𝛼 2 b2 a2 x1 = 𝜖1 , 𝜖1 = ±1, 8𝜇2 −𝛽a + 𝜖2 x2 = � 1 2 2 𝛽 a 2 + 1 2 √ 𝛽 4 a4 + 64𝜇2 𝛼 2 b2 a2 −2𝜇 ⎛ ⎜ −𝛽a + 𝜖2 2 2 𝜆 = (2𝜌 − 𝛼)(a + b ) + 𝜇⎜ ⎜ ⎝ and � 1 2 2 𝛽 a 2 , 𝜖2 = ±1, 2 √ ⎞ 𝛽 4 a4 + 64𝜇2 𝛼 2 b2 a2 ⎟ ⎟ , −2𝜇 ⎟ ⎠ + 1 2 13 Journal of Nonlinear Mathematical Physics � � 2 � � √ ⎛ � 1 2 2 1 4 a4 + 64𝜇 2 𝛼 2 b2 a2 ⎞ � −𝛽a + 𝜖 𝛽 𝛽 a + 2 ⎟ ⎜ 2 2 �𝛼 x3 = 𝜖3 � (a2 + b2 ) − ⎜ ⎟, �𝜇 −2𝜇 ⎟ ⎜ ⎠ ⎝ where 𝜖3 = ±1, 𝛼ab = 𝜖1 𝜖2 ‖𝛼ab‖, � √1 √ 1 𝛼ab + 𝛽b −𝛽a+𝜖2 2 𝛽 2 a2 + 2 � 𝛽 4 a4 +64𝜇2 𝛼 2 b2 a2 −2𝜇 � √ −𝛽 2 a2 + 𝛽 4 a4 +64𝜇2 𝛼 2 b2 a2 8𝜇2 − 𝛽a 𝜖1 � � 2 � � √ ⎛ � 1 1 2 2 4 a4 + 64𝜇 2 𝛼 2 b2 a2 ⎞ � 𝛽 a + −𝛽a + 𝜖 𝛽 2 ⎟ ⎜ 2 2 �𝛼 = 2𝜇𝜖3 � (a2 + b2 ) − ⎜ ⎟, �𝜇 −2𝜇 ⎟ ⎜ ⎝ ⎠ and � 2 √ −𝛽 2 a2 + 𝛽 4 a4 +64𝜇2 𝛼 2 b2 a2 2𝜇𝛼a − 2𝜇𝛽b𝜖1 8𝜇2 � √ −𝛽a + 𝜖2 12 𝛽 2 a2 + 12 𝛽 4 a4 + 64𝜇2 𝛼 2 b2 a2 � �2 � � √1 √ � 2 a2 + 1 4 a4 +64𝜇 2 𝛼 2 b2 a2 −𝛽a+𝜖2 𝛽 𝛽 𝛼 2 � 2 2 2𝜇a𝛽𝜖3 𝜇 (a + b2 ) − −2𝜇 − a𝛽 � √ −𝛽a + 𝜖2 12 𝛽 2 a2 + 12 𝛽 4 a4 + 64𝜇2 𝛼 2 b2 a2 � � 2 � � √ ⎛ � 1 1 2 2 4 a4 + 64𝜇 2 𝛼 2 b2 a2 ⎞ � 𝛽 a + −𝛽a + 𝜖 𝛽 2 ⎜ ⎟ 2 2 �𝛼 = 2𝜇𝜖3 � (a2 + b2 ) − ⎜ ⎟. �𝜇 −2𝜇 ⎜ ⎟ ⎝ ⎠ + Proof From [19, 20], we have ⎛ −(a2 + b2 ) ab − ab 2 2 ⎜ ̃ =⎜ ab − (a2 + b2 ) a2 Ric a2 ⎜ 0 − ab ⎝ 2 2 1 and 13 ⎞ ⎟ ⎟ ⎟ ⎠ Journal of Nonlinear Mathematical Physics ⎛ 2ax2 − ax1 ax1 − bx2 ⎞ 0 bx1 − ax2 − ax3 ⎟ (L1X g) = ⎜ −ax1 ⎜ ⎟ 0 ⎝ ax1 − bx2 bx1 − ax2 − ax3 ⎠ S = −2(a2 + b2 ) and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Therefore ̃ becomes ⎧ −𝛼(a2 + b2 ) + 𝛽ax2 + 𝜇x12 = −2𝜌(a2 + b2 ) + 𝜆, ⎪ ab𝛼 − 𝛽 ax + 𝜇x x = 0, 1 2 ⎪ 𝛼ab 2 𝛽 1 ⎪ − 2 + 2 (ax1 − bx2 ) − 𝜇x1 x3 = 0, ⎨ −𝛼(a2 + b2 ) + 𝜇x2 = −2𝜌(a2 + b2 ) + 𝜆, 2 ⎪ 2 ⎪ −𝛼 a + 𝛽 (bx1 − ax2 − ax3 ) − 𝜇x2 x3 = 0, 2 ⎪ 𝜇x2 2= 2𝜌(a 2 + b2 ) − 𝜆. ⎩ 3 (21) Let 𝜇 = 0 then we have 𝛼 = 0 and 𝛽 ≠ 0. Thus x1 = x2 = x3 = 0 and the case (i) holds. Now, we assume that 𝜇 ≠ 0. The first and fourth equations of the system Eq. (21) imply that 𝛽ax2 + 𝜇x12 − 𝜇x22 = 0, (22) and the fourth and sixth equations imply that x22 + x32 = 𝛼 2 (a + b2 ). 𝜇 (23) From the second equation we have (2𝛼ba)2 = x12 (𝛽a − 2𝜇x2 )2 = x12 (𝛽 2 a2 + 4𝜇(𝜇x22 − 𝛽ax2 )). Plugging Eq. (22) into last equality we get 4𝜇2 x14 + 𝛽 2 a2 x12 − (2𝛼ba)2 = 0. (24) If 𝛼b = 0 then x1 = 0 and we obtain three cases (ii)-(iv). If 𝛼b ≠ 0, then x1 ≠ 0 and the case (v) is true. ◻ Theorem 7 The left-invariant affine generalized Ricci soliton associated to the connection ∇0 on the Lie group (G4 , g, J, X) are the following: (i) 𝜇 = 0 , 𝛼 = 0 , 𝛽 ≠ 0 , x1 = x2 = 0 , 𝜆 = −𝜌((2d − b)(a + 2d) − 2) for all a, b, c, 𝜌, x3 such that d = ±1. (ii) 𝜇 = 0, 𝛼 ≠ 0, d = b, a = 0 𝛽 = 0, 𝜆 = 0, for all x1 , x2 , x3 , 𝜌 such that d = ±1, (iii) 𝜇 = 0 , 𝛼 ≠ 0 , d = b , a = 0 𝛽 ≠ 0 , x1 = x2 = 0 , 𝜆 = 0 , for all x3 , 𝜌 such that d = ±1, 13 Journal of Nonlinear Mathematical Physics (iv) 𝜇 ≠ 0 , x2 = 0 , x3 = 0 , x1 = 0 , 𝛼 = 0 , 𝜆 = −𝜌((2d − b)(a + 2d) − 2) , for all a, b, 𝜌, 𝛽 , d = ±1, (v) 𝜇 ≠ 0, x2 = 0, x3 = 0, x1 = 0, 𝛼 ≠ 0, b = d , a = 0, 𝜆 = 0, for all 𝜌, 𝛽 , d = ±1, (vi) 𝜇 ≠ 0 , x2 = 0 , x3 = 0 , x1 = 𝜇𝛽 ≠ 0 , 𝜆 = −𝜌((2d − b)(a + 2d) − 2) for all ) ( a, b, 𝜌, 𝛼 s u c h t h a t d = ±1 , 𝛽 2 = −𝛼𝜇 (2d − b)( 2a + d) − 1 > 0 , ) ( 𝛼 a + b − (2d − b)( a2 + d)2 = 0, ( ) (vii) 𝜇 ≠ 0 , x2 = 0 , x3 ≠ 0 , x1 = 0 , x32 = − 𝜇𝛼 (2d − b)( a2 + d) − 1 > 0 , ) ( 𝜆 = (𝛼 − 2𝜌) (2d − b)( a2 + d) − 1 , for all 𝜌, 𝛽 such that d = ±1 , 𝛼(a + 2b − 2d) = 0. Proof From [19, 20], we have ⎛ (2d − b)( a + d) − 1 0 2 0 ̃ = ⎜⎜ 0 (2d − b)( 2a + d) − 1 Ric a ⎜ + d2 − 2b 0 ⎝ 4 0 a 4 + d 2 0 − b 2 ⎞ ⎟ ⎟ ⎟ ⎠ and 0 − x2 − ( a2 + d)x2 ⎞ ⎛ −x 2x1 ( a2 + d)x1 ⎟ (L0X g) = ⎜ ⎜ a 2 ⎟ a 0 ⎝ −( 2 + d)x2 ( 2 + d)x1 ⎠ S = (2d − b)(a + 2d) − 2 and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Then ̃ becomes � � ⎧ 𝛼 (2d − b)( 2a + d) − 1 + 𝜇x12 = 𝜌((2d − b)(a + 2d) − 2) + 𝜆, ⎪ − 𝛽 x + 𝜇x x = 0, 1 2 ⎪ 2 2 ⎪ − 𝛽 ( a + d)x2 − 𝜇x1 x3 = 0, � (25) ⎨ �2 2 a 2 ⎪ 𝛼 �(2d − b)( 2�+ d) − 1 + 𝛽x1 + 𝜇x2 = 𝜌((2d − b)(a + 2d) − 2) + 𝜆, ⎪ 𝛼 a + d − b + 𝛽 ( a + d)x1 − 𝜇x2 x3 = 0, 2 2 ⎪ 24 2 2 ⎩ 𝜇x3 = −𝜌((2d − b)(a + 2d) − 2) − 𝜆. We consider 𝜇 = 0, then the system Eq. (25) reduces to � � ⎧ 𝛼 (2d − b)( a2 + d) − 1 = 0, ⎪ 𝛽x = 0, ⎪ 2 ⎨ 𝛽x�1 = 0, � ⎪ 𝛼 a + d − b = 0, 4 2 2 ⎪ ⎩ 𝜆 = −𝜌((2d − b)(a + 2d) − 2). (26) If 𝛼 = 0 then 𝛽 ≠ 0 and x1 = x2 = 0. Thus the case (i) holds. If 𝛼 ≠ 0 then d = b, a = 0 and the cases (ii)-(iii) hold. Now we consider 𝜇 ≠ 0. The first and third equations of the system Eq. (25) yield 13 Journal of Nonlinear Mathematical Physics 𝜇x12 = 𝛽x1 + 𝜇x22 . (27) In this case the second eqution of the system Eq. (25) implies that x2 = 0 or x1 = 2 𝛽 . 2𝜇 𝛽 𝛽 = 0 which is a and substutiting it in Eq. (27) we get x22 + 4𝜇 If x2 ≠ 0 then x1 = 2𝜇 2 cotradiction. Hence, x2 = 0 and the system Eq. (25) becomes � � a ⎧ 𝛼 (2d − b)( 2 + d) − 1 + 𝜇x12 = 𝜌((2d − b)(a + 2d) − 2) + 𝜆, ⎪ x x = 0, � ⎪ 1� 3 a (28) ⎨ 𝛼 �(2d − b)( 2�+ d) − 1 + 𝛽x1 = 𝜌((2d − b)(a + 2d) − 2) + 𝜆, ⎪ 𝛼 a + d − b + 𝛽 ( a + d)x1 = 0, 2 2 ⎪ 24 2 2 ⎩ 𝜇x3 = −𝜌((2d − b)(a + 2d) − 2) − 𝜆. In this cases (iv)-(vi) hold. ◻ Theorem 8 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G4 , g, J, X) are the following: (i) 𝜇 = 0, 𝛼 = 0, x1 = x2 = x3 = 0, 𝜆 = 2𝜌[1 + (b − 2d)b], for all a, b, 𝜌, d = ±1, 𝛼 (ii) 𝜇 = 0 , 𝛼 ≠ 0 , 𝛽 ≠ 0 , x2 = x3 = 0 , b = d , a = 2b , x1 = 2𝛽 , 𝜆 = 0 , for all 𝜌 , d = ±1, (iii) 𝜇 ≠ 0 , x2 = 0 , x3 = 0 , 𝛼 = 0 , x1 = 0 , 𝜆 = 2𝜌[1 + (b − 2d)b], for all a, b, 𝜌, 𝛽 such that d = ±1, (iv) 𝜇 ≠ 0, x2 = 0, x3 = 0, 𝛼 ≠ 0, 𝛽 ≠ 0, b ≠ 0, x1 = 𝛼a , 𝜆 = 2𝜌[1 + (b − 2d)b], for 𝛽b all a, b, 𝜌 such that d = ±1, b[1 + (b − 2d)a] = a, −𝛽 2 [1 + (b − 2d)b] + 𝜇𝛼[1 + (b − 2d)a]2 = 0, (v) 𝜇 ≠ 0 , x32 = 𝜇𝛼 [1 + (b − 2d)b] − x2 = 0 , 𝜆 = (2𝜌 − 𝛼)[1 + (b − 2d)b] + 𝛽2 , 4𝜇 d = ±1, 𝛼(b − 2d)(a − b) = − 𝛽2 4𝜇2 >0 , 𝛽 x1 = − 2𝜇 , for all a, b, 𝜌, 𝛼 such that 3𝛽 2 , 2𝜇𝛼a = −b𝛽 2 , 4𝜇 𝛽 (vi) 𝜇 ≠ 0 , x1 = 2𝜇 , 𝛽 = 0 , 𝜆 = (2𝜌 − 𝛼)[1 + (b − 2d)b] , x2 ≠ 0 , √ √ 𝛼 x2 = 𝜖1 − 𝜇 [(b − 2d)(b − a)] , x3 = 𝜖2 𝜇𝛼 [1 + (b − 2d)b] , for all 𝜌 such that d = ±1, 𝜇𝛼 [(b − 2d)(b − a)] ≤ 0, 𝜇𝛼 [1 + (b − 2d)b] ≥ 0, 𝜖1 = ±1, 𝜖2 = ±1, √ √ a𝛼 𝛼 𝛼 = 𝜖1 𝜖2 − [(b − 2d)(b − a)] [1 + (b − 2d)b] − 𝜇 𝜇 𝜇 𝛽 , 2𝜇 −2𝜇𝛼[1+(b−2d)a]+𝛽 2 (vii) 𝜇 ≠ 0 , x2 ≠ 0 , x1 = x22 = 2 −2𝜇2 (1+ a4 ) 𝛽 ≠ 0 , x3 = − a2 x2 , 𝜆 = (2𝜌 − 𝛼)[1 + (b − 2d)b] + 𝛽2 , 4𝜇 > 0, 13 Journal of Nonlinear Mathematical Physics d = ±1, −a𝛼 + a𝜇 −2𝜇𝛼[1 + (b − 2d)a] + 𝛽 2 𝛽2 b+ = 0, 2 2𝜇 2 −2𝜇2 (1 + a ) 4 −𝛼[1 + (b − 2d)b] + 𝛽2 a2 −2𝜇𝛼[1 + (b − 2d)a] + 𝛽 2 = −𝜇 . 2 4𝜇 4 −2𝜇2 (1 + a ) 4 Proof From [19, 20], we have 0 0⎞ ⎛ −[1 + (b − 2d)b] 1 a ⎟ ⎜ ̃ 0 − [1 + (b − 2d)a] Ric = 2 ⎟ ⎜ a 0 0⎠ ⎝ 2 and ⎛ 0 − x2 − ax2 − x3 ⎞ ⎟ 2x1 bx1 (L1X g) = ⎜ −x2 ⎜ ⎟ 0 ⎝ −ax2 − x3 bx1 ⎠ S = −2[1 + (b − 2d)b] and the Eq. with respect to the basis {e1 , e2 , e3 }. Therefore ̃ (2) becomes ⎧ −𝛼[1 + (b − 2d)b] + 𝜇x12 = −2𝜌[1 + (b − 2d)b] + 𝜆, ⎪ − 𝛽 x + 𝜇x x = 0, 1 2 ⎪ 2 2 ⎪ 𝛽 (−ax2 − x3 ) − 𝜇x1 x3 = 0, 2 ⎨ −𝛼[1 + (b − 2d)a] + 𝛽x1 + 𝜇x22 = −2𝜌[1 + (b − 2d)b] + 𝜆, ⎪ a ⎪ −𝛼 2 + 𝛽2 (bx1 ) − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 2𝜌[1 + (b − 2d)b] − 𝜆. ⎩ 3 (29) Let 𝜇 = 0, then the system Eq. (29) becomes ⎧ 𝛼[1 + (b − 2d)b] = 0, ⎪ ⎪ 𝛽x2 = 0, ⎪ 𝛽x3 = 0, ⎨ −𝛼[1 + (b − 2d)a] + 𝛽x = 0, 1 ⎪ ⎪ −a𝛼 + 𝛽bx1 = 0, ⎪ 𝜆 = 2𝜌[1 + (b − 2d)b]. ⎩ (30) and the cases (i)-(ii) holds. Now we consider 𝜇 ≠ 0. In this case the second equation 𝛽 of the system Eq. (29) implies that x2 = 0 or x1 = 2𝜇 . If x2 = 0 then the system Eq. (29) gives 13 Journal of Nonlinear Mathematical Physics ⎧ −𝛼[1 + (b − 2d)b] + 𝜇x2 = −2𝜌[1 + (b − 2d)b] + 𝜆, 1 ⎪ 𝛽x + 2𝜇x x = 0, 1 3 ⎪ 3 ⎨ −𝛼[1 + (b − 2d)a] + 𝛽x1 = −2𝜌[1 + (b − 2d)b] + 𝜆, ⎪ −𝛼a + 𝛽(bx1 ) = 0, ⎪ 𝜇x2 = 2𝜌[1 + (b − 2d)b] − 𝜆. ⎩ 3 (31) 𝛽 The second equation of the system Eq. (31) implies that x3 = 0 or x1 = − 2𝜇 . We assume that x3 = 0, thus 2 ⎧ −𝛼[1 + (b − 2d)b] + 𝜇x1 = 0, ⎪ −𝛼[1 + (b − 2d)a] + 𝛽x1 = 0, ⎨ −𝛼a + 𝛽bx = 0, 1 ⎪ ⎩ 𝜆 = 2𝜌[1 + (b − 2d)b], 𝛽 and the cases (iii)-(iv) are true. If x3 ≠ 0 and x1 = − 2𝜇 then the case (v) is true. Now, we consider x2 ≠ 0 and x1 = 𝛽 . 2𝜇 In this case the system Eq. (29) yields ⎧ 𝛽2 ⎪ −𝛼[1 + (b − 2d)b] + 4𝜇 = −2𝜌[1 + (b − 2d)b] + 𝜆, ⎪ 𝛽ax2 + 2𝛽x3 = 0, ⎪ 𝛽2 2 ⎨ −𝛼[1 + (b − 2d)a] + 2𝜇 + 𝜇x2 = −2𝜌[1 + (b − 2d)b] + 𝜆, 2 ⎪ 𝛽 ⎪ −a𝛼 + 2𝜇 b − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 2𝜌[1 + (b − 2d)b] − 𝜆. ⎩ 3 (32) The second equation of Eq. (32) implies that 𝛽 = 0 or x3 = − a2 x2. If 𝛽 = 0 then we obtain the case (vi). If 𝛽 ≠ 0 then x3 = − a2 x2 and the case (vii) holds. ◻ Theorem 9 The left-invariant affine generalized Ricci soliton associated to the connection ∇0 on the Lie group (G5 , g, J, X) are the following: (i) 𝜇 = 𝛽 = 𝜆 = 0 and for all 𝛼, 𝜌, x1 , x2 , x3 , a, b, c, d such that a + d ≠ 0 and ac + bd = 0, (ii) 𝜇 = 𝜆 = 0, 𝛽 ≠ 0, b = c, and for all 𝛼, 𝜌, x1 , x2 , x3 , a, d such that a + d ≠ 0 and ac + bd = 0, (iii) 𝜇 ≠ 0, x1 = x2 = x3 = 𝜆 = 0 and for all 𝛼, 𝜌, a, b, c, d such that a + d ≠ 0 and ac + bd = 0. 13 Journal of Nonlinear Mathematical Physics 0 ̃ = 0 and Proof From [19, 20], we have Ric ⎛ ⎜ (L0X g) = ⎜ ⎜ ⎝ 0 0 b−c x 2 2 0 0 − b−c x 2 1 b−c x 2 2 b−c − 2 x1 0 ⎞ ⎟ ⎟ ⎟ ⎠ with respect to the basis {e1 , e2 , e3 }. Then ̃ S = 0 and the Eq. (2) becomes ⎧ 𝜇x2 = 𝜆, ⎪ 𝜇x1 x = 0, 1 2 ⎪ 𝛽 b−c ⎪ 2 2 x2 − 𝜇x1 x3 = 0, ⎨ 𝜇x2 = 𝜆, ⎪ 𝛽2 b−c ⎪ − 2 2 x1 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = −𝜆. ⎩ 3 (33) The first, fourth and sixth equations of system Eq. (33) imply that 𝜇(x12 + x32 ) = 𝜇(x22 + x32 ) = 0. We consider 𝜇 = 0, then 𝜆 = 0. If 𝛽 = 0 or b = c then the system Eq. (33) holds for any x1 , x2, and x3. Now, if 𝜇 ≠ 0 then x1 = x2 = x3 = 𝜆 = 0. ◻ Theorem 10 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G5 , g, J, X) are the following: (i)𝜇 = 𝛽 = 𝜆 = 0 and for all 𝛼, 𝜌, x1 , x2 , x3 , a, b, c, d such that a + d ≠ 0 and ac + bd = 0, (ii)𝜇 = 𝜆 = 0, 𝛽 ≠ 0, a = b = x2 = 0, and for all 𝛼, 𝜌, x1 , x3 , c, d such that d ≠ 0, (iii)𝜇 = 𝜆 = 0, 𝛽 ≠ 0, a ≠ 0, x1 = x2 = 0, and for all 𝛼, 𝜌, x3 , c, d such that d ≠ 0, (iv)𝜇 = 𝜆 = 0, 𝛽 ≠ 0, a ≠ 0, x1 = c = d = 0, x2 ≠ 0, and for all 𝛼, 𝜌, x3, (v)𝜇 ≠ 0 , x1 = x2 = x3 = 𝜆 = 0 and for all 𝛼, 𝜌, a, b, c, d such that a + d ≠ 0 and ac + bd = 0. 1 ̃ = 0 and Proof From [19, 20], we have Ric ⎛ 0 0 − ax1 − cx2 ⎞ −0 0 − bx1 − dx2 ⎟ (L1X g) = ⎜ ⎜ ⎟ 0 ⎝ −ax1 − cx2 − bx1 − dx2 ⎠ with respect to the basis {e1 , e2 , e3 }. Therefore ̃ S = 0 and the Eq. (2) becomes 13 Journal of Nonlinear Mathematical Physics ⎧ 𝜇x2 = 𝜆, ⎪ 𝜇x1 x = 0, ⎪𝛽 1 2 ⎪ 2 (−ax1 − cx2 ) − 𝜇x1 x3 = 0, ⎨ 𝜇x2 = 𝜆, ⎪𝛽 2 ⎪ 2 (−bx1 − dx2 ) − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = −𝜆. ⎩ 3 (34) The first, fourth and sixth equations of system Eq. (34) imply that 𝜇(x12 + x32 ) = 𝜇(x22 + x32 ) = 0. We consider 𝜇 = 0, then 𝜆 = 0. Let 𝛽 = 0, then the system Eq. (34) holds for any x1 , x2, and x3. If 𝛽 ≠ 0 then the third and fiveth equations of Eq. (34) given { ax1 + cx2 = 0, bx1 + dx2 = 0. Since ac + bd = 0 we get (a2 + b2 )x1 = 0 and (c2 + d2 )x2 = 0. We consider a = 0. In this case we obtain d ≠ 0 and b = x2 = 0. if a ≠ 0 then x1 = 0 and cx2 = dx2 = 0. For case x2 ≠ 0 we have c = d = 0. Now, we assume that 𝜇 ≠ 0, then x1 = x2 = x3 = 𝜆 = 0. ◻ Theorem 11 The left-invariant affine generalized Ricci soliton associated to the connection ∇0 on the Lie group (G6 , g, J, X) are the following: (i) 𝜇 = 0 , a = 0 , d ≠ 0 , b = 0 , c = 0 , 𝜆 = 0 , for all 𝛼, 𝛽, 𝜌, x1 , x2 , x3 , such that (𝛼, 𝛽) ≠ (0, 0), (ii) 𝜇 = 0, a = 0, d ≠ 0, b = 0, c ≠ 0, 𝛽 ≠ 0, 𝜆 = 0, x2 = 0, x1 = − 𝛼d for all 𝛼, 𝜌, x3, 𝛽 (iii) 𝜇 = 0 , a ≠ 0 , 𝛽 = 0 , 𝛼 ≠ 0 , 𝜆 = 0 , c = bd , for all b, d, 𝜌, x1 , x2 , x3, such that a b2 (a − d) = 2a3, d(a − d) = 2𝛼, (iv) 𝜇 = 0, a ≠ 0, 𝛽 ≠ 0, x1 = x2 = 0, 𝛼 = 0, c = bd , 𝜆 = −𝜌(b(b − c) − 2a2 ) for all a b, d, x3 , 𝜌 such that a + d ≠ 0,( ) ) ( (v) 𝜇 ≠ 0, x2 = 0, x1 = 0, x32 = − 𝜇𝛼 12 b2 − a2 ≥ 0, 𝜆 = (−2𝜌 + 𝛼) 21 b(b − c) − a2 , for all a, b, c, d, 𝜌, 𝛽, 𝛼 such that 𝛼c = 0, ac − bd = 0, a + d ≠ 0, ≠ 0 , 𝜆 = −𝜌(b(b − c) − 2a2 ) , for all (vi) 𝜇 ≠ 0 , x2 = 0 , x3 = 0 , x1 = − 𝛽a 𝜇 a, b, c, d, 𝜌, 𝛽, 𝛼 such that 𝛼 ) 𝛽 2 a2 (1 𝛽 2a 1 1 = 0, 𝛼 [−ac + d(b − c)] − (b − c) = 0, b(b − c) − a2 + 2 𝜇 2 2 4𝜇 ac − bd = 0, a + d ≠ 0. 13 Journal of Nonlinear Mathematical Physics Proof From [19, 20], we have ⎛ 1 b(b − c) − a2 2 ̃ = ⎜⎜ 0 Ric ⎜ 0 ⎝ 0 ⎞ 0 0 ⎟ 1 1 − c) − a2 [−ac + 2 d(b − c)] ⎟ 2 1 ⎟ [−ac + 12 d(b − c)] 0 ⎠ 2 1 b(b 2 and ⎛ 0 ax2 ⎜ 0 (LX g) = ⎜ ax2 − 2ax1 ⎜ c−b x b−c x ⎝ 2 2 2 1 c−b x 2 2 b−c x 2 1 0 ⎞ ⎟ ⎟ ⎟ ⎠ S = b(b − c) − 2a2 and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Then ̃ becomes � � ⎧ 𝛼 12 b(b − c) − a2 + 𝜇x12 = 𝜌(b(b − c) − 2a2 ) + 𝜆, ⎪ 𝛽 ax + 𝜇x x = 0, 1 2 ⎪2 2 ⎪ 𝛽 (c − b)x2 − 𝜇x1 x3 = 0, � (35) ⎨ 4� 1 2 2 2 ⎪ 𝛼 2 b(b − c) − a − 𝛽ax1 + 𝜇x2 = 𝜌(b(b − c) − 2a ) + 𝜆, ⎪ 𝛼 1 [−ac + 1 d(b − c)] + 𝛽 (b − c)x1 − 𝜇x2 x3 = 0, 2 4 ⎪ 22 ⎩ 𝜇x3 = −𝜌(b(b − c) − 2a2 ) − 𝜆. Let 𝜇 = 0, then the system Eq. (35) becomes ⎧ 𝛼 � 1 b(b − c) − a2 � = 0, ⎪ 2 ⎪ 𝛽ax2 = 0, ⎪ 𝛽(c − b)x2 = 0, ⎨ 𝛽ax = 0, 1 ⎪ 1 ⎪ 2𝛼[−ac + 2 d(b − c)] + 𝛽(b − c)x1 = 0, ⎪ 𝜆 = −𝜌(b(b − c) − 2a2 ). ⎩ (36) If a = 0 then d ≠ 0, b = 0, and 𝛽cx2 = 0. If c = 0 then the case (i) holds. Now, if c ≠ 0 then 𝛽 ≠ 0 and the case (ii) holds. For a ≠ 0 the cases (iii)-(iv) hold. Now we assume that 𝜇 ≠ 0. The first and fourth equations of the system Eq. (35) give 𝜇x12 = −𝛽ax1 + 𝜇x22 . (37) 𝛽a The second equation of the system Eq. (35) yields x2 = 0 or x1 = − 2𝜇 . The Eq. (36) 𝛽a thus x2 = 0. The third equation of the system Eq. (35) implies implies that x1 ≠ − 2𝜇 that x1 = 0 or x3 = 0. If x1 = 0 then we have 13 Journal of Nonlinear Mathematical Physics � ⎧ �1 2 2 ⎪ 𝛼 2 b(b − c) − a = 𝜌(b(b − c) − 2a ) + 𝜆, ⎪ ac − bd = 0, ⎨ 𝛼[−ac + 1 d(b − c)] = 0, ⎪ 2 ⎪ 𝜇x32 = −𝜌(b(b − c) − 2a2 ) − 𝜆. ⎩ (38) Hence, the case (v) holds. If x1 ≠ 0 and x3 = 0 then the Eq. (37) gives x1 = − 𝛽a and 𝜇 we get � � 2 2 ⎧ 𝛼 1 b(b − c) − a2 + 𝛽 a = 0, 2 𝜇 ⎪ 1 2 ⎨ 𝛼 2 [−ac + 21 d(b − c)] − 𝛽4𝜇a (b − c) = 0, ⎪ 2 ⎩ 𝜆 = −𝜌(b(b − c) − 2a ). Therefore the case (vi) holds. (39) ◻ Theorem 12 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G6 , g, J, X) are the following: (i) 𝜇 = 0, a = 0, b = 0, d ≠ 0, 𝛽 = 0, 𝜆 = 0, for all c, 𝜌, 𝛼, x1 , x2 , x3 such that 𝛼 ≠ 0 , (ii) 𝜇 = 0, a = 0, b = 0, d ≠ 0, 𝛽 ≠ 0, 𝜆 = 0, x3 = 0, for all c, 𝜌, 𝛼, x1 , x2 such that cx1 = 0, (iii) 𝜇 = 0, a ≠ 0, 𝛽 ≠ 0, x2 = 0, 𝛼 = 0, x1 = 0, 𝜆 = 𝜌(2a2 + bc), for all x3 , b, c, d, 𝜌 such that a + d ≠ 0, c = bd , a (iv) 𝜇 ≠ 0 , x2 = 0 , x3 = 0 , a = 0 , b = 0 , d ≠ 0 , x1 = 0 , 𝜆 = 0 , for all c, d, 𝜌, 𝛼, 𝛽 such that (𝛼, 𝛽) ≠ (0, 0), d ≠ 0, (v) 𝜇 ≠ 0 , x2 = 0 , x3 = 0 , a ≠ 0 , c = bd , d ≠ 0 , 𝛼 = 0 , x1 = 0 , 𝜆 = 𝜌(2a2 + bc), a for all d, 𝜌, 𝛼, 𝛽 such that (𝛼, 𝛽) ≠ (0, 0), (vi) 𝜇 ≠ 0, x2 = 0, x3 = 0, a ≠ 0, d ≠ 0, 𝛼 ≠ 0, c = 0, x1 = − 𝛽a , b = 0, 𝜆 = 2𝜌a2 , 𝜇 for all 𝜌, 𝛽 , such that a + d ≠ 0, 2 2 𝛽d c = bd x32 = 𝛼a𝜇 + 𝛽2𝜇ad2 > 0 , (vii) 𝜇 ≠ 0 , x1 = 2𝜇 , , x2 = 0 , a 𝛽 2 ad 2𝜇 2 2 𝜆 = −𝛼a2 − + 𝜌(2a2 + bc) , for all a, b, 𝜌, 𝛼, 𝛽 such that a + d ≠ 0 , 𝛽 2 cd = 0, 𝛽 d = −2𝛽 2 ad , 2 2 𝛽a (viii) 𝜇 ≠ 0, x22 = 𝜇𝛼 a2 − 𝛽2𝜇a2 > 0, x1 = − 2𝜇 , x3 = 0, 𝜆 = 𝜌(2a2 + bc), c = bd , for all a b, d, a, 𝜌, 𝛼, 𝛽 such that a + d ≠ 0, 𝛽ac = 0, 4𝜇𝛼(a2 + bc) + 𝛽 2 a2 = 0. Proof From [19, 20], we have ⎛ −(a2 + bc) 0 0 ⎞ 1 ̃ 0 − a2 0 ⎟ Ric = ⎜ ⎜ ⎟ 0 0 0⎠ ⎝ 13 Journal of Nonlinear Mathematical Physics and ⎛ 0 ax2 dx3 ⎞ (L1X g) = ⎜ ax2 − 2ax1 − cx1 ⎟ ⎜ ⎟ 0 ⎠ ⎝ dx3 − cx1 S = −(2a2 + bc) and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Therefore ̃ becomes ⎧ −𝛼(a2 + bc) + 𝜇x12 = −𝜌(2a2 + bc) + 𝜆, ⎪ 𝛽 ax + 𝜇x x = 0, 1 2 ⎪2 2 ⎪ 𝛽 dx3 − 𝜇x1 x3 = 0, 2 ⎨ −𝛼a 2 2 2 ⎪ 𝛽 − 𝛽ax1 + 𝜇x2 = −𝜌(2a + bc) + 𝜆, ⎪ − 2 cx1 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + bc) − 𝜆. ⎩ 3 (40) ⎧ 𝛼(a2 + bc) = 0, ⎪ ⎪ a𝛽x2 = 0, ⎪ d𝛽x3 = 0, ⎨ −𝛼a2 − 𝛽ax = 0, 1 ⎪ ⎪ 𝛽cx1 = 0, ⎪ 𝜆 = 𝜌(2a2 + bc). ⎩ (41) Let 𝜇 = 0, then If we assume that a = 0 then the cases (i)-(ii) hold. If we consider a ≠ 0 then 𝛽 ≠ 0 and x2 = 0, 𝛼 = 0 and the case (iii) holds. Now we consider 𝜇 ≠ 0. The second equation of the system Eq. (40) implies that 𝛽a . If x2 = 0 then the system Eq. (40) becomes x2 = 0 or x1 = − 2𝜇 ⎧ −𝛼(a2 + bc) + 𝜇x12 = −𝜌(2a2 + bc) + 𝜆, ⎪ 𝛽 dx − 𝜇x x = 0, 1 3 ⎪ 2 32 ⎨ −𝛼a − 𝛽ax1 = −𝜌(2a2 + bc) + 𝜆, ⎪ 𝛽cx1 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + bc) − 𝜆. ⎩ 3 The second equation of the system Eq. (42) yields x3 = 0 or x1 = x3 = 0, then 𝜆 = 𝜌(2a + bc) and (42) 𝛽d . 2𝜇 We consider 2 ⎧ −𝛼(a2 + bc) + 𝜇x2 = 0, 1 ⎪ 2 ⎨ −𝛼a − 𝛽ax1 = 0, ⎪ 𝛽cx1 = 0. ⎩ Thus, the cases (iv)-(vi) hold. If x3 ≠ 0 then x1 = 13 𝛽d 2𝜇 and (43) Journal of Nonlinear Mathematical Physics 2 2 ⎧ 2 + 𝛽4𝜇d = −𝜌(2a2 + bc) + 𝜆, ⎪ −𝛼(a + bc) ⎪ −𝛼a2 − 𝛽 2 ad = −𝜌(2a2 + bc) + 𝜆, 2𝜇 ⎨ 2 ⎪ 𝛽 cd = 0, ⎪ 𝜇x2 = 𝜌(2a2 + bc) − 𝜆. ⎩ 3 (44) 𝛽a Hence, the case (vii) holds. Now we assume that x2 ≠ 0, then x1 = − 2𝜇 and we have ⎧ 𝛽 2 a2 2 2 ⎪ −𝛼(a + bc) + 4𝜇 = −𝜌(2a + bc) + 𝜆, ⎪ 𝛽x3 = 0, ⎪ 𝛽 2 a2 2 2 2 ⎨ −𝛼a + 2𝜇 + 𝜇x2 = −𝜌(2a + bc) + 𝜆, ⎪ 𝛽 2 ac ⎪ 4𝜇 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + bc) − 𝜆. ⎩ 3 (45) Therefore x3 = 0 and the case (viii) holds. ◻ Theorem 13 The left-invariant affine generalized Ricci soliton associated to the connection ∇0 on the Lie group (G7 , g, J, X) are the following: (i) 𝜇 = 0, 𝛽 = 0, 𝛼 ≠ 0, a = 0, d ≠ 0, c = 0, 𝜆 = 0, for all 𝜌, x1 , x2 , x3, (ii) 𝜇 = 0, 𝛽 ≠ 0, a = 0, d ≠ 0, b = 0, c = 0, 𝜆 = 0, for all 𝜌, 𝛼, x1 , x2 , x3, (iii) 𝜇 = 0 , 𝛽 ≠ 0 , a = 0 , d ≠ 0 , b = 0 , c ≠ 0 , 𝜆 = 0 , x1 = 0 , x2 = − 𝛼dc for all 𝛽c 𝜌, 𝛼, x3, (iv) 𝜇 = 0, 𝛽 ≠ 0, a = 0, d ≠ 0, b ≠ 0, x1 = x2 = 0, 𝜆 = 𝜌bc, for all c, 𝜌, 𝛼, x3, such that 𝛼b = 𝛼c = 0, (v) 𝜇 = 0, 𝛽 ≠ 0, a ≠ 0, c = 0, 𝛼 = x1 = x2 = 0, 𝜆 = 2𝜌a2 , for all 𝜌, x3 , b, d such that a + d ≠ 0, (vi) 𝜇 ≠ 0, a = 0, d ≠ 0, x2 = 0, x1 = 0, 𝜆 = 𝜌bc , x3 = 0, for all b, c, 𝜌, 𝛼, 𝛽 such that 𝛼c = 0, 2 2 𝛼dc ≠ 0, x3 = − 4𝛽𝜇 + 𝛽 𝜇b + 𝜌bc, (vii) 𝜇 ≠ 0, a = 0, d ≠ 0, x2 = 0, x1 = 𝛽b , 𝜆 = −𝛼 bc 𝜇 2 2 2 2 for all 𝜌, c, 𝛼 such that 𝜇𝛼c + 𝛽 2 (c − 2b) = 0, 𝛼16𝛽d 2c𝜇 = 𝛼bc 2 − 𝛽 2 b2 , 𝜇 (viii) 𝜇 ≠ 0, a ≠ 0, c = 0, x1 = x2 = x3 = 0, 𝛼 = 0, 𝜆 = 2𝜌a2, for all 𝜌, 𝛽, d such that a + d ≠ 0, ≠ 0 , x1 = 0 , 𝜆 = (2𝜌 − 𝛼2 )a2 , for all (ix) 𝜇 ≠ 0 , a ≠ 0 , c = 0 , x3 = x2 = − 𝛼a 2𝛽 b, d, 𝜌, 𝛼, 𝛽 such that 2𝛼𝛽 2 = 𝛼 2 𝜇, a + d ≠ 0. Proof From [19, 20], we have 13 Journal of Nonlinear Mathematical Physics ⎛ −(a2 + bc ) 0 − 21 (ac + dc )⎞ 2 2 ⎟ ⎜ 1 2 bc bc 2 ̃ 0 − (a + 2 ) 2 (a + 2 ) ⎟ Ric = ⎜ ⎟ ⎜ − 1 (ac + dc ) 1 (a2 + bc ) 0 ⎠ ⎝ 2 2 2 2 0 and (L0X g) c ⎛ −2ax2 ax1 − bx2 (b − 2 )x2 ⎞ c = ⎜ ax1 − bx2 2bx1 ( 2 − b)x1 ⎟ ⎜ ⎟ c c 0 ⎝ (b − 2 )x2 ( 2 − b)x1 ⎠ S = −(2a2 + bc) and the Eq. (2) becomes with respect to the basis {e1 , e2 , e3 }. Then ̃ ) − 𝛽ax2 + 𝜇x12 = −𝜌(2a2 + bc) + 𝜆, ⎧ −𝛼(a2 + bc ⎪ 𝛽 (ax − bx2 ) + 𝜇x x = 0, 2 1 2 ⎪2 1 ⎪ − 𝛼 (ac + dc ) + 𝛽 (b − c )x2 − 𝜇x1 x3 = 0, 2 2 ⎨ 2 2 bc2 2 2 ⎪ −𝛼(a + 2 ) + 𝛽bx1 + 𝜇x2 = −𝜌(2a + bc) + 𝜆, bc 𝛽 c 𝛼 2 ⎪ (a + ) + ( − b)x1 − 𝜇x2 x3 = 0, 2 2 2 ⎪2 2 ⎩ 𝜇x3 = 𝜌(2a2 + bc) − 𝜆. (46) Let 𝜇 = 0 then the system Eq. (46) becomes ⎧ −𝛼(a2 + bc ) − 𝛽ax2 = 0, ⎪ 𝛽 (ax − bx2 ) = 0, 2 ⎪2 1 ⎪ − 𝛼 (ac + dc ) + 𝛽 (b − c )x2 = 0, 2 2 ⎨ 2 2 bc2 ⎪ −𝛼(a + 2 ) + 𝛽bx1 = 0, ⎪ 𝛼 (a2 + bc ) + 𝛽 ( c − b)x1 = 0, 2 2 2 ⎪2 ⎩ 𝜆 = 𝜌(2a2 + bc). (47) If 𝛽 = 0 then 𝛼 ≠ 0 and the case (i) is true. If 𝛽 ≠ 0 and a = 0 then d ≠ 0 and we have ⎧ 𝛼bc = 0, ⎪ ⎪ bx2 = 0, ⎪ 𝛼dc + 𝛽cx2 = 0, ⎨ bx = 0, ⎪ 1 ⎪ cx1 = 0, ⎪ 𝜆 = 𝜌bc. ⎩ (48) Hence the cases (ii)-(iv) hold. For the case 𝛽 ≠ 0 and a ≠ 0 we have c = 0 and 𝛼 = x1 = x2 = 0. Therefore the case (v) holds. Now, we assume that 𝜇 ≠ 0. The first and the fourth equations of the system Eq. (46) imply 13 Journal of Nonlinear Mathematical Physics −𝛽ax2 + 𝜇x12 = 𝛽bx1 + 𝜇x22 . (49) Since ac = 0 then a = 0 or c = 0. If a = 0 then the system Eq. (46) becomes + 𝜇x12 = −𝜌bc + 𝜆, ⎧ −𝛼 bc 2 ⎪ − 𝛽 bx + 𝜇x x = 0, 1 2 ⎪ 2 2 ⎪ − 𝛼 dc + 𝛽 (b − c )x2 − 𝜇x1 x3 = 0, 2 2 ⎨ 2 bc2 2 ⎪ −𝛼 2 + 𝛽bx1 + 𝜇x2 = −𝜌bc + 𝜆, bc 𝛽 c 𝛼 ⎪ + ( − b)x1 − 𝜇x2 x3 = 0, ⎪ 2 22 2 2 ⎩ 𝜇x3 = 𝜌bc − 𝜆. The second equation of Eq. (50) yields x2 = 0 or x1 = 𝛽 2 b2 4𝜇2 𝛽b . 2𝜇 (50) If x2 ≠ 0 then x1 = 𝛽b 2𝜇 and + = 0 and this is a contradiction. Thus substituting it in Eq. (49) we get . If x1 = 0 then 𝜆 = 𝜌bc, x2 = 0 and from the Eq. (49) we have x1 = 0, or x1 = 𝛽b 𝜇 x22 𝛼c = 0, x3 = 0, and the case (vi) is true. Also, if x1 = 𝛽b 𝜇 ≠ 0 then ⎧ −𝛼 bc + 𝛽 2 b2 = −𝜌bc + 𝜆, 2 𝜇 ⎪ ⎪ − 𝛼 dc − 𝛽bx3 = 0, 2 2 ⎨ 𝛼 bc 𝛽2b c ⎪ 2 2 + 2𝜇 ( 2 − b) = 0, ⎪ 𝜇x2 = 𝜌bc − 𝜆. ⎩ 3 (51) Thus, the case (vii) holds. Now for case 𝜇 ≠ 0 and a ≠ 0 we have c = 0 and ⎧ −𝛼a2 − 𝛽ax2 + 𝜇x12 = −2𝜌a2 + 𝜆, ⎪ 𝛽 (ax − bx ) + 𝜇x x = 0, 2 1 2 ⎪2 1 ⎪ 𝛽 bx2 − 𝜇x1 x3 = 0, 2 ⎨ −𝛼a 2 + 𝛽bx1 + 𝜇x22 = −2𝜌a2 + 𝜆, ⎪𝛼 2 ⎪ 2 a − 𝛽2 bx1 − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 2𝜌a2 − 𝜆. ⎩ 3 (52) The fourth, fiveth and sixth equations of Eq. (52) imply that x2 = x3. The second and third equations imply that 𝛽x1 = 0. Then ⎧ −𝛼a2 − 𝛽ax2 + 𝜇x12 = −2𝜌a2 + 𝜆, ⎪ − 𝛽 bx + 𝜇x x = 0, 1 2 ⎪ 2 22 2 2 ⎨ −𝛼a + 𝜇x2 = −2𝜌a + 𝜆, 𝛼 2 2 ⎪ a − 𝜇x = 0, 2 ⎪2 2 2 ⎩ 𝜇x2 = 2𝜌a − 𝜆. Using the second equation of Eq. (53) we have x2 = 0 or x1 = 𝛼 = 0 and the case (viii) holds. If x2 ≠ 0 then x1 = 𝛽b 2𝜇 (53) 𝛽b . 2𝜇 If x2 = 0 then = 0 and 13 Journal of Nonlinear Mathematical Physics ⎧ −𝛽ax = 𝛼 a2 , ⎪ 𝛼 2 2 22 ⎨ 2 a − 𝜇x2 = 0, ⎪ 𝛼 a2 = 2𝜌a2 − 𝜆. ⎩2 (54) Thus the case (ix) is true. ◻ Theorem 14 The left-invariant affine generalized Ricci soliton associated to the connection ∇1 on the Lie group (G7 , g, J, X) are the following: (i) 𝜇 = 0, a = 0, d ≠ 0, 𝛽 ≠ 0, b = 0, x2 = 0, x3 = 2𝛼d ,𝜆 𝛽 = 0, for all c, 𝜌, x1, , 𝜆 = 0, for all (ii) 𝜇 = 0, a = 0, d ≠ 0, 𝛽 ≠ 0, b = 0, x2 ≠ 0, c = 0, x2 + x3 = 2𝛼d 𝛽 𝜌, x1, − c𝛼d (iii) 𝜇 = 0 , a = 0 , d ≠ 0 , 𝛽 ≠ 0 , b ≠ 0 , x1 = 𝛼(b+c) , x2 = 𝛼bd , x3 = 2𝛼d , 𝛽 𝛽b 𝛽 𝛽b (iv) (v) (vi) (vii) 𝜆 = 𝜌(b2 + bc), for all c, 𝜌, 𝛼 such that c𝛼d2 − b3 𝛼 − 𝛼bd2 = 0, 𝜇 = 0 , a ≠ 0 , c = 0 , 𝛽 ≠ 0 , b = 0 , 𝛼 = 0 , x1 = x2 = 0 , 𝜆 = 2𝜌a2 , for all 𝜌, d such that a + d ≠ 0, dx3 = 0, 𝜇 = 0 , a ≠ 0 , c = 0 , 𝛽 ≠ 0 , b ≠ 0 , 𝛼 = 0 , x1 = x2 = x3 = 0 , 𝜆 = 𝜌(2a2 + b2 ), for all 𝜌, d such that a + d ≠ 0, 𝜇 ≠ 0, a = 0, d ≠ 0, x1 = x3 = 0, 𝜆 = 𝜌(b2 + bc), 𝛽 = 0, 𝛼 = 0, x2 = 0, for all 𝜌, b, c, , for all 𝜌, b, c, 𝜇 ≠ 0, a = 0, d ≠ 0, x1 = x3 = 0, 𝜆 = 0, 𝛽 ≠ 0, x2 = 𝛼 = 0√ 2 2 2 c=0 , (viii) 𝜇 ≠ 0 , a≠0 , x1 = 𝜖1 𝛼a −𝜌(2a𝜇 +b )+𝜆 , 𝛽=0 , √ 2 2 √ 2 +b2 )+𝜆 2 2 )−𝜆 , x3 = 𝜖3 𝜌(2a +b , for all b, d, 𝜌, 𝜆, 𝛼 such that x2 = 𝜖2 𝛼(a +b )−𝜌(2a 𝜇 𝜇 a + d ≠ 0, 𝜌(2a2 + b2 ) − 𝜆 𝛼a2 − 𝜌(2a2 + b2 ) + 𝜆 ≥ 0, ≥ 0, 𝜇 𝜇 √ √ 𝛼a2 − 𝜌(2a2 + b2 ) + 𝜆 𝛼(a2 + b2 ) − 𝜌(2a2 + b2 ) + 𝜆 𝛼 (bd − ab) + 𝜇𝜖1 𝜖2 = 0, 2 𝜇 𝜇 √ √ 𝛼a2 − 𝜌(2a2 + b2 ) + 𝜆 𝜌(2a2 + b2 ) − 𝜆 = 0, 𝛼b(a + d) − 𝜇𝜖1 𝜖3 𝜇 𝜇 √ √ 2 + b2 ) − 𝜌(2a2 + b2 ) + 𝜆 𝛼(a 𝜌(2a2 + b2 ) − 𝜆 𝛼 (ad + 2d2 ) − 𝜇𝜖2 𝜖3 = 0. 2 𝜇 𝜇 𝜌(2a2 +b2 )−𝜆−𝛼a2 +𝜇F 2 , x3 𝛽a 𝜌(2a2 +b2 )−𝜆 ≥ 0, 𝜇 (ix) 𝜇 ≠ 0, a ≠ 0, c = 0, 𝛽 ≠ 0, x1 = F, x2 = for all b, d, 𝜌, 𝛼 such that a + d ≠ 0, 13 =𝜖 √ 𝜌(2a2 +b2 )−𝜆 , 𝜇 Journal of Nonlinear Mathematical Physics � � ⎛ ⎞ 𝛽⎜ 𝜌(2a2 + b2 ) − 𝜆 ⎟ 𝜌(2a2 + b2 ) − 𝜆 − 𝜇F𝜖 = 0, 𝛼b(a + d) + −aF − b𝜖 ⎟ 2⎜ 𝜇 𝜇 ⎝ ⎠ � 𝜌(2a2 + b2 ) − 𝜆 − 𝛼a2 + 𝜇F 2 �2 − 𝛼(a2 + b2 ) + 𝛽bF + 𝜇 = −𝜌(2a2 + b2 ) + 𝜆, 𝛽a � ⎞ ⎛ 𝜌(2a2 + b2 ) − 𝜆 − 𝛼a2 + 𝜇F 2 𝛽⎜ 𝜌(2a2 + b2 ) − 𝜆 ⎟ 𝛼 2 (ad + 2d ) + −bF − d − d𝜖 ⎟ 2 2⎜ 𝛽a 𝜇 ⎠ ⎝ � 2 2 2 2 𝜌(2a + b ) − 𝜆 − 𝛼a + 𝜇F 𝜌(2a2 + b2 ) − 𝜆 − 𝜇𝜖 = 0, 𝛽a 𝜇 𝜌(2a2 + b2 ) − 𝜆 ≥ 0, 𝜇 27C2 + (4A3 − 18AB)C + 4B3 − A2 B2 ≥ 0, where 𝜖 = ±1, F=E− E= B 3 − A2 9 − E A , 3 �√ 27C2 + (4A3 − 18AB)C + 4B3 − A2 B2 3 2 2(3 ) AB − 3C A3 − + 6 27 �1 3 ≠ 0, b𝛽 , 2𝜇 � 𝛽 2 a2 1 � 2 2 2 , B= 𝜌(2a + b ) − 𝜆 − 𝛼a + 𝜇 2𝜇2 � �� 𝛽a � b 2 2 2 𝛼(bd − ab) − C= 𝜌(2a + b ) − 𝜆 − 𝛼a . a 2𝜇2 A=− Proof From [19, 20], we have 1 ⎞ ⎛ −a2 (bd − ab) b(a + d) 2 ⎜ 1 1 2 2 2 ̃ = ⎜ (bd − ab) − (a + b + bc) (bc + ad + 2d ) ⎟⎟ Ric 2 2 ⎟ ⎜ b(a + d) 1 (bc + ad + 2d2 ) 0 ⎠ ⎝ 2 1 and ⎛ −2ax2 ax1 − bx2 − ax1 − cx2 − bx3 ⎞ ax1 − bx2 2bx1 − bx1 − dx2 − dx3 ⎟ (L1X g) = ⎜ ⎜ ⎟ 0 ⎝ −ax1 − cx2 − bx3 − bx1 − dx2 − dx3 ⎠ S = −(2a2 + b2 + bc) and the Eq. (2) with respect to the basis {e1 , e2 , e3 }. Therefore ̃ becomes 13 Journal of Nonlinear Mathematical Physics ⎧ −𝛼a2 − 𝛽ax2 + 𝜇x12 = −𝜌(2a2 + b2 + bc) + 𝜆, ⎪ 𝛼 (bd − ab) + 𝛽 (ax − bx ) + 𝜇x x = 0, 1 2 1 2 ⎪2 2 ⎪ 𝛼b(a + d) + 𝛽 (−ax1 − cx2 − bx3 ) − 𝜇x1 x3 = 0, ⎨ −𝛼(a2 + b2 +2 bc) + 𝛽bx + 𝜇x2 = −𝜌(2a2 + b2 + bc) + 𝜆, 1 2 ⎪𝛼 ⎪ 2 (bc + ad + 2d2 ) + 𝛽2 (−bx1 − dx2 − dx3 ) − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + b2 + bc) − 𝜆. ⎩ 3 (55) Let 𝜇 = 0, then the system Eq. (55) becomes ⎧ −𝛼a2 − 𝛽ax2 = 0, ⎪ 𝛼 (bd − ab) + 𝛽 (ax − bx ) = 0, 1 2 ⎪2 2 ⎪ 𝛼b(a + d) + 𝛽 (−ax1 − cx2 − bx3 ) = 0, ⎨ −𝛼(a2 + b2 +2 bc) + 𝛽bx = 0, 1 ⎪𝛼 ⎪ 2 (bc + ad + 2d2 ) + 𝛽2 (−bx1 − dx2 − dx3 ) = 0, ⎪ 𝜆 = 𝜌(2a2 + b2 + bc). ⎩ (56) Since ac = 0 we get a = 0 or c = 0. If a = 0 then a + d ≠ 0 implies that d ≠ 0 and we have ⎧ 𝛼bd − 𝛽bx2 = 0, ⎪ 𝛼bd + 𝛽 (−cx − bx ) = 0, 2 3 2 ⎪ 2 + bc) + 𝛽bx = 0, −𝛼(b ⎨ 1 ⎪ 𝛼 (bc + 2d2 ) + 𝛽 (−bx1 − dx2 − dx3 ) = 0, 2 ⎪2 ⎩ 𝜆 = 𝜌(b2 + bc). (57) If 𝛽 = 0 then (𝛼, 𝛽, 𝜇) ≠ (0, 0, 0) yields 𝛼 ≠ 0. Also, the first equation of Eq. (57) gives b = 0 and the fourth equation of Eq. (57) implies that 𝛼d2 = 0 which is a contradiction. Hence, 𝛽 ≠ 0 and the cases (i)–(iii) hold. Now, we consider 𝜇 = 0 and a ≠ 0, then c = 0 and we get ⎧ 𝛼a + 𝛽x2 = 0, ⎪ 𝛼(bd − ab) + 𝛽(ax − bx ) = 0, 1 2 ⎪ 𝛽 ⎪ 𝛼b(a + d) + 2 (−ax1 − bx3 ) = 0, ⎨ −𝛼(a2 + b2 ) + 𝛽bx = 0, 1 ⎪𝛼 𝛽 ⎪ 2 (ad + 2d2 ) + 2 (−bx1 − dx2 − dx3 ) = 0, ⎪ 𝜆 = 𝜌(2a2 + b2 ). ⎩ (58) If 𝛽 = 0 then the first equation of the system Eq. (58) yields 𝛼 = 0 which is a contradiction, then 𝛽 ≠ 0. If b = 0 then the case (iv) holds. If b ≠ 0 then from the first three equations of the system Eq. (58) we obtain x1 = − 𝛼bd , x2 = − 𝛼a , x3 = 2𝛼a+3𝛼d . 𝛽a 𝛽 𝛽 Hence using the fourth and fiveth equations of the system Eq. (58) we have 13 Journal of Nonlinear Mathematical Physics { 𝛼(a2 + b2 + 𝛼(−d2 + b2 d ) a b2 d ) a = 0, = 0. (59) If 𝛼 ≠ 0 then a2 + b2 + d2 = 0 which is a contradiction, then 𝛼 = 0, x1 = x2 = x3 = 0 and the case (v) is true. Now, we consider 𝜇 ≠ 0. If a = 0 then d ≠ 0 and the system Eq. (55) gives ⎧ 𝜇x2 = −𝜌(b2 + bc) + 𝜆, 1 ⎪ 𝛼bd − 𝛽bx2 + 2𝜇x1 x2 = 0, ⎪ 𝛽 ⎪ 𝛼bd + 2 (−cx2 − bx3 ) − 𝜇x1 x3 = 0, ⎨ −𝛼(b2 + bc) + 𝛽bx + 𝜇x2 = −𝜌(b2 + bc) + 𝜆, 1 2 ⎪𝛼 𝛽 ⎪ 2 (bc + 2d2 ) + 2 (−bx1 − dx2 − dx3 ) − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(b2 + bc) − 𝜆. ⎩ 3 (60) The first and sixth equations of the system Eq. (60) imply that x1 = x3 = 0 and 𝜆 = 𝜌(b2 + bc). Thus the system Eq. (60) gives ⎧ 𝛼bd − 𝛽bx2 = 0, ⎪ 2𝛼bd − 𝛽cx2 = 0, ⎨ −𝛼(b2 + bc) + 𝜇x2 = 0, 2 ⎪ ⎩ 𝛼(bc + 2d2 ) − 𝛽dx2 = 0. (61) If 𝛽 = 0 then 𝛼 = 0, x2 = 0 and the case (vi) holds. If 𝛽 ≠ 0 and b = 0 then x2 = 𝛼 = 0 and the case (vii) is true. Notice if b ≠ 0 and 𝛼 ≠ 0 then from the first two equations of the system Eq. (61) we infer c = 2b and Replacing it with x2 = 𝛼d 𝛽 in the fourth equation of the system Eq. (61) we obtain 2b2 + d2 = 0 which is a contradiction. Let 𝜇 ≠ 0 and a ≠ 0 then c = 0 and the system Eq. (55) gives ⎧ −𝛼a2 − 𝛽ax2 + 𝜇x12 = −𝜌(2a2 + b2 ) + 𝜆, ⎪ 𝛼 (bd − ab) + 𝛽 (ax − bx ) + 𝜇x x = 0, 1 2 1 2 ⎪2 2 ⎪ 𝛼b(a + d) + 𝛽 (−ax1 − bx3 ) − 𝜇x1 x3 = 0, 2 ⎨ −𝛼(a2 + b2 ) + 𝛽bx1 + 𝜇x22 = −𝜌(2a2 + b2 ) + 𝜆, ⎪𝛼 ⎪ 2 (ad + 2d2 ) + 𝛽2 (−bx1 − dx2 − dx3 ) − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + b2 ) − 𝜆. ⎩ 3 (62) If 𝛽 = 0 then the system Eq. (62) reduces to ⎧ −𝛼a2 + 𝜇x2 = −𝜌(2a2 + b2 ) + 𝜆, 1 ⎪𝛼 ⎪ 2 (bd − ab) + 𝜇x1 x2 = 0, ⎪ 𝛼b(a + d) − 𝜇x1 x3 = 0, ⎨ −𝛼(a2 + b2 ) + 𝜇x2 = −𝜌(2a2 + b2 ) + 𝜆, 2 ⎪𝛼 2 ⎪ 2 (ad + 2d ) − 𝜇x2 x3 = 0, ⎪ 𝜇x2 = 𝜌(2a2 + b2 ) − 𝜆. ⎩ 3 (63) 13 Journal of Nonlinear Mathematical Physics Then the case (viii) is true. If 𝛽 ≠ 0, then the first and second equations of the system Eq. (62) imply that x13 + Ax12 + Bx1 + C = 0. (64) Thus the case (ix) holds. ◻ Author Contributions Not applicable. Funding Not applicable. Data Availability All data generated or analysed during this study are included in this published article. Code Availability Not applicable. Declarations Conflict of Interest We declare that we do not have any commercial or associative interest that represents a confict of interest in connection with the work submitted. Shahroud Azami, April 21, 2022 Ethical Approval Not applicable Consent to Participate Not applicable Consent for Publication Not applicable Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. 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