Rigity results on ρ-Einstein solitons with zero scalar curvature
aa r X i v : . [ m a t h . DG ] F e b RIGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZEROSCALAR CURVATURE ROMILDO PINA, ILTON MENEZES, AND LUCYJANE SILVA
Abstract.
In this paper we show that a ρ -Einstein solitons conformal to apseudo-Euclidean space, invariant under the action of the pseudo-orthogonalgroup with zero scalar curvature is stady and consequently flat. How applica-tion of the results obtained we present an explicit example for a the questionproposed by Kazdan in [17]. Introduction and main statements
In this paper, we study two related problems. The first problem is on the exis-tence of ρ –Einstein sotiton with scalar curvature K ¯ g = 0. Besides that we presentsome rigity results.The second problem consists in find all metrics that are conformal to the pseudoEuclidean metrics, with zero scalar curvature, which are invariant under the ac-tion of the pseudo-orthogonal group. This provides explicit solutions to Yamabe’sproblem in the non-compact case. In the Riemannian case under some additionalassumptions, all metrics obtained are complete. As application of this results weobtain a family of complete metrics in R n \ { } with scalar curvature positive, neg-ative and zero, presenting an explicit example for a question proposed by Kazdanin [17].In 1982, R. Hamilton introduced a nonlinear evolution equation for Riemannianmetrics with the aim of finding canonical metrics on manifolds (see [1] or [16]). Thisevolution equation is known as the Ricci flow, and it has since been used widely andwith great success, most notably in Perelman’s solution of the Poincar´e conjecture.Furthermore, several convergence theorems have been established. One importantaspect in the treatment of the Ricci flow is the study of Ricci solitons, whichgenerate self-similar solutions to the flow and often arise as singularity models.Given a semi-Riemannian manifold ( M n , g ), n ≥
3, we say that (
M, g ) is agradient Ricci soliton if there exists a differentiable function h : M −→ R (calledthe potential function) such that(1.1) Ric g + Hess g ( h ) = λg, λ ∈ R , where Ric g is the Ricci tensor, Hess g ( h ) is the Hessian of h with respect to themetric g , and λ is a real number. We say that a gradient Ricci soliton is shrinking,steady, or expanding if λ > λ = 0, or λ <
0, respectively. Bryant [10] proved thatthere exists a complete, steady, gradient Ricci soliton that is spherically symmetricfor any n ≥
3, which is known as Bryant’s soliton. In the bi-dimensional case an
Mathematics Subject Classification.
Primary 53A30, 53C21.
Key words and phrases.
Conformal metric, ρ –Einstein solitons, rigidity ρ –Einstein solitons,Scalar curvature. analogous nontrivial rotationally symmetric solution was obtained explicitly, and isknown as the Hamilton cigar. Recently Cao-Chen [11] showed that any complete,steady, gradient Ricci soliton, locally conformally flat, up to homothety, is eitherflat or isometric to the Bryant’s soliton. The results obtained in [11] were extendedto bach – flat gradient steady Ricci solitons (see [4]). Complete, conformally flatshrinking gradient solitons have been characterized as being quotients of R n , S n or R × S n − (see [13]). In the case of steady gradient Ricci solitons, [2] provides allsuch solutions when the metric is conformal to an n-dimensional pseudo-Euclideanspace and invariant under the action of an ( n − − dimensional translation group.Motivated by the notion of Ricci solitons on a semi-Riemannian manifold ( M n , g ), n ≥
3, it is natural to consider geometric flows of the following type:(1.2) ∂∂t g ( t ) = − Ric − ρRg )for ρ ∈ R , ρ = 0, as in [3] and [6]. We call these the Ricci-Bourguignon flows.We notice that short time existence for the geometric flows described in (1.2) isprovided in ([5]). Associated to the flows, we have the following notion of gradient ρ -Einstein solitons, which generate self-similar solutions: Definition 1.1.
Let ( M n , g ) , n ≥
3, be a Riemannian manifold and let ρ ∈ R , ρ =0. We say that ( M n , g ) is a gradient ρ − Einstein soliton if there exists a smoothfunction h : M −→ R , such that the metric g satisfies the equations Ric g + Hess g h = ρK g g + λg (1.3)for some constant λ ∈ R , where K g is the scalar curvature of the metric g .A ρ -Einstein soliton is said to be shrinking, steady, or expanding if λ > λ = 0,or λ <
0, respectively. Furthermore, a ρ -Einstein solitons is said to be a gradientEinstein soliton, gradient traceless Ricci soliton, and gradient Schouten soliton if ρ = , ρ = n , and ρ = n − , respectively.The gradient ρ − Einstein solitons equation (1.3) links geometric informationabout the curvature of the manifold through the Ricci tensor and the geometryof the level sets of the potential function by means of their second fundamentalform. Hence, classifying gradient ρ − Einstein solitons under some curvature condi-tions is a natural problem. The ρ − Einstein solitons were investigated by Catinoand Mazzieri in [6], they obtained important rigidity results, proving that everycompact gradient Einstein, Schouten, or traceless Ricci soliton is trivial. In addi-tion, they proved that every complete gradient steady Schouten soliton is trivial,hence Ricci flat.Gradient Ricci solitons with constant scalar curvature were investigated by Pe-tersen and Wylie in [15], they proved that: If a non-steady gradient Ricci soliton hasconstant scalar curvature K g , then it is bounded as 0 ≤ K g ≤ nλ in the shrinkingcase, and nλ ≤ K g ≤ K g , then K g must be a multiple of λ .In [9] the authors considered a ρ –Einstein solitons that are conformal to a pseudoEuclidean space and invariant under the action of the pseudo-orthogonal group.They provide all the solutions for the gradient Schouten soliton case. Moreover,they proved that if a gradient Schouten soliton is both complete, conformal to aEuclidean metric, and rotationally symmetric, then it is isometric to R × S n − . IGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 3 In [12] the autors used the variational method to study the existence problemof metrics with constant scalar curvature on complete non-compact Riemannianmanifolds. The assumptions of results are motivated from question in the work ofKazdan [17]. The question is that if M has complete metrics g + and g − with positive(respectively negative) scalar curvature, is there one with zero scalar curvature?With several additional hypotheses o autor provide an answer to the question posedby Kazdan, more details see [14].We studied the equation (1.3) in semi-Riemannian manifolds with scalar curva-ture constante. We consider gradient ρ -Einstein solitons conformal to a pseudo-Euclidean space, which are invariant under the action of the pseudo-orthogonalgroup. More precisely, let ( R n , g ) be the standard pseudo-Euclidean space withmetric g and coordinates ( x , ..., x n ), with g ij = δ ij ε i , 1 ≤ i, j ≤ n , where δ ij isthe Kronecker delta, and ε i = ±
1. Let r = P ni =1 ε i x i be a basic invariant foran ( n − − dimensional pseudo-orthogonal group. The main goal of this paper isto present in the Riemannian case a family of complete metrics and some resultsof rigidity on a large class of noncompact semi-Riemannian manifolds in the casewhere the scalar curvature is zero. In the Riemannian case the same results hold.We initially find a system of differential equations, such that the functions h and ψ must satisfy, so that the metric ¯ g = g/ψ satisfies (1.3) (see Theorem 1.2).Note that if the solutions are invariant under the action of the pseudo-orthogonalgroup, the system of partial differential equations given in Theorem 1.2 can betransformed into a system of ordinary differential equations (see Corollary 1.3). Inthe Theorem 1.4 we found all metrics that are conformal to the pseudo Euclideanmetrics, with zero scalar curvature, which are invariant under the action of thepseudo-orthogonal group. As a consequence of the Theorem 1.4, we obtain theCorollary 1.6 we constuct a family of complete metrics with zero scalar curvature.We present results of rigidity on gradient ρ - Einstein soliton with scalar curvaturezero ( Theorem1.7, Corollary 1.8 and Corollary 1.9).In the Proposition 1.1 we construct a family of complete metrics with zero scalarcurvature. In the Corollary 1.12 we constuct an explicit example for Kazdan’squestion.In what follows, we state our main results. We denote the second order derivativeof ψ and h by ψ ,x i x j and h ,x i x j , respectively, with respect to x i x j . Theorem 1.2.
Let ( R n , g ) , n ≥ , be a pseudo-Euclidean space with coordinates x = ( x , ..., x n ) and metric components g ij = δ ij ε i , ≤ i, j ≤ n , where ε i = ± .Consider a smooth function h : R n −→ R . There exists a metric ¯ g = ψ g such that ( R n , ¯ g ) is a gradient ρ -Einstein soliton with h as a potential function if, and onlyif, the functions ψ , and h satisfy ( n − ψ ,x i x j + ψh ,x i x j + ψ ,x i h ,x j + ψ ,x j h ,x i = 0 , i = j, (1.4) and ψ [( n − ψ ,x i x i + ψh ,x i x i + 2 ψ ,x i h ,x i ](1.5)+ ε i n X k =1 ε k (cid:2) ( n − (cid:0) ρnψ ,x k − ρψψ ,x k x k − ψ ,x k (cid:1) − ψψ ,x k h ,x k + ψψ ,x k x k (cid:3) = λε i , i = j. ROMILDO PINA, ILTON MENEZES, AND LUCYJANE SILVA
Our objective is to determine solutions of the system (1.4), (1.5) of the form ψ ( r ) and h ( r ), where r = P ni =1 ε i x i . The following theorem reduces the system ofpartial differential equations (1.4) and (1.5) into an system of ordinary differentialequations. Corollary 1.3.
Let ( R n , g ) , n ≥ , be a pseudo-Euclidean space with coordinates x = ( x , ..., x n ) and metric components g ij = δ ij ε i , ≤ i, j ≤ n , where ε i = ± .Consider smooth functions ψ ( r ) and h ( r ) with r = P nk =1 ε k x k . Then there existsa metric ¯ g = ψ g such that ( R n , ¯ g ) is a gradient ρ -Einstein soliton with h as apotential function if, and only if, the functions ψ and h satisfy (1.6) ( n − ψ ′′ + ψh ′′ + 2 ψ ′ h ′ = 0 , and (1.7) 2 ψ [( n − ψ ′ + ψh ′ ] + 2 n [1 − n − ρ ] ψψ ′ +4 r (cid:8) ( n − (cid:2) ( ρn −
1) ( ψ ′ ) − ρψψ ′′ (cid:3) − ψψ ′ h ′ + ψψ ′′ (cid:9) = λ. The next we found all metrics that are conformal to the pseudo Euclidean met-rics, with zero scalar curvature, which are invariant under the action of the pseudo-orthogonal group.
Theorem 1.4.
Let ( R n , g ) be a pseudo-Euclidean space, n ≥ , with coordinates x = ( x , · · · , x n ) and g ij = δ ij ε i , ≤ i, j ≤ n , where ε i = ± . Consider ¯ g = ψ ( r ) g where r = n P k =1 ε k x k . Then ¯ g have scalar curvature K ¯ g = 0 , if and only if (1.8) ψ ( r ) = k r (cid:16) Ar n − (cid:17) n − , where A, k ∈ R with k > . If A ≥ the metric ¯ g is defined in R n \ { } . If A < the set of singularity points of ¯ g consist of the origin and a sphere ( n − –dimensional, with center at the origin and radius R = q ( − A ) n − .Remark . If ( R n , g ) is the Euclidean space, then we find in the Theorem 1.4all metrics conformal to g and spherically symmetrical with zero scalar curvature.This provides explicit solutions to Yamabe’s problem in the non-compact case.In [7], the authors showed that { R n \ { } , ¯ g = ϕ g , ϕ ( r ) = √ r } is a completeRiemannian manifold and isometric at S n − × R . As a consequence of Theorem 1.4together with this fact, we obtain the following result: Corollary 1.6.
Let ( R n , g ) be a Euclidean space, n ≥ , with coordinates x =( x , · · · , x n ) and ( g ) ij = δ ij , ≤ i, j ≤ n . Consider ¯ g = ψ ( r ) g where r = n P k =1 x k .The metrics obtained in the Theorem 1.4 are complete whenever A > . As an consequence of the Theorem 1.4, we get the following rigidity results.
IGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 5 Theorem 1.7.
Let ( R n , g ) be a pseudo-Euclidean space, n ≥ , with coordinates x = ( x , · · · , x n ) and g ij = δ ij ε i , ≤ i, j ≤ n , where ε i = ± . Consider ( R n , ¯ g ) , ¯ g = ψ g a ρ –Einstein sotiton with scalar curvature K ¯ g = 0 , where ψ ( r ) and h ( r ) smooth functions, r = n P k =1 ε k x k and h as a potential function. Then λ = 0 , that is ( R n , ¯ g ) is steady. Corollary 1.8.
Let ( R n , g ) be a pseudo-Euclidean space, n ≥ , with coordinates x = ( x , · · · , x n ) and g ij = δ ij ε i , ≤ i, j ≤ n , where ε i = ± . Then ( R n , ¯ g ) , ¯ g = ψ g is a ρ –Einstein sotiton steady with scalar curvature K ¯ g = 0 , where ψ ( r ) and h ( r ) smooth functions, r = n P k =1 ε k x k and h as a potential function, if and onlyif, ( R n , ¯ g ) is flat. As a consequence of the previous results, we have the following result in theRiemannian case.
Corollary 1.9.
Let ( M n , ¯ g ) be n ≥ a ρ –Einstein sotiton, Riemannian,locallyconformally flat and rotationally symmetric with zero scalar curvature.Then ( M n , ¯ g ) is necessarily steady. Besides that ( M n , ¯ g ) is flat.Remark . These results hold for ρ = 0 and therefore they are extended to theRicci solitons gradients, proving that a Ricci soliton gradient, conformal to the Eu-clidean space and spherically symmetrical with zero scalar curvature is necessarilysteady and consequently flat. Remark . As a consequently of the results obtained let’s make an aplicationgiving a positive answer for a question proposed by Kazdan in [17], as follows:If M has complete metrics g + and g − with positive (respectively, negative) scalarcurvature, is there one with zero scalar curvature? kazdan mostrou em [18], que nocaso compacto the answer is ”yes”.We built in ( R n \ { } ) complete metrics with positive, negative and zero scalarcurvature, respectively. Proposition 1.1.
Let ( R n , g ) be a Euclidean space, n ≥ , with coordinates x =( x , · · · , x n ) and ( g ) ij = δ ij , ≤ i, j ≤ n . Consider g = ϕ ( r ) g where r = n P k =1 x k . If ϕ ( r ) = re − (cid:18) r n − (cid:19) n − , then the metric g on R n is complete withscalar curvature negative given by K g = h ( r ) (cid:20) ( n − (cid:16) r n − (cid:17) n − + 2( n − r − n + ( n + 2) (cid:21) , where h ( r ) = − n − r n − (cid:18) r n − (cid:19) − n ) n − e r n − ! n − . In the next result we construct an explicit example in Riemannian manifolds forthe question left by Kazdan [17].
ROMILDO PINA, ILTON MENEZES, AND LUCYJANE SILVA
Corollary 1.12.
Note that { R n \{ } , ¯ g = ϕ g , ϕ ( r ) = √ r } is a complete Riemann-ian manifold with scalar curvature positive and R n \ { } , ¯ g = ϕ g , ϕ ( r ) = re − (cid:18) r n − (cid:19) n − is a complete Riemannian manifold with scalar curvature negative, exists a completemetric of scalar curvature zero. Proofs of the main results
Proof.
Proof of Theorem 1.2. It is well known (see, e.g., [2]) that if ¯ g = gψ , then Ric ¯ g = 1 ψ { ( n − ψHess g ( ψ ) + [ ψ ∆ g ψ − ( n − |∇ g ψ ] g } and ¯ R = ( n − (cid:0) ψ ∆ g ψ − n |∇ g ψ | (cid:1) . Hence, the equation
Ric ¯ g + Hess ¯ g ( h ) = ρ ¯ R ¯ g + λ ¯ g, is equivalent to1 ψ { ( n − ψHess g ( ψ ) ij + [ ψ ∆ g ψ − ( n − |∇ g ψ | ] δ ij ε i } + Hess ¯ g ( h ) ij (2.1) = (cid:2) ρ ( n − ψ ∆ g ψ − n |∇ g ψ | ) + λ (cid:3) ψ δ ij ε i . Recall that,
Hess ¯ g ( h ) ij = h ,x i x j − n X k =1 ¯Γ kij h ,x k where ¯Γ kij are the Christoffel symbols of the metric ¯ g . For a distinct i, j, k , we have¯Γ kij = 0 , ¯Γ iij = − ψ ,x j ψ , ¯Γ kii = ε i ε k ψ ,x k ψ , ¯Γ iii = − ψ ,x i ψ , therefore,(2.2) Hess ¯ g ( h ) ij = h ,x i x j + ψ ,x j h ,x i ψ + ψ ,x i h ,x j ψ , i = j. Similarly, by considering i = j , we have(2.3) Hess ¯ g ( h ) ii = h ,,x i x i + 2 ψ ,x i h ,x i ψ − ε i n X k =1 ε k ψ ,x k h ,x k ψ . However, we note that |∇ g ψ | = n X k =1 ε k (cid:18) ∂ψ∂x k (cid:19) , ∆ g ψ = n X k =1 ε k ψ ,x k x k , Hess g ( ψ ) ij = ψ ,x i x j . (2.4)If i = j in (2.1), we obtain( n − Hess g ( ψ ) ij ψ + Hess ¯ g ( h ) ij = 0 . (2.5) IGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 7 Substituting the expressions found in (2.2), and (2.4) into (2.5), we obtain( n − ψ ,x i x j + ψh ,x i x j + ψ ,x i h ,x j + ψ ,x j h ,x i = 0 , i = j. Similarly, if i = j in (2.1), we have( n − ψHess g ( ψ ) ii + ψ ∆ g ψε i − ( n − |∇ g ψ | ε i + ψ Hess ¯ g ( h ) ii (2.6) = 2( n − ρ ∆ g ψε i − n ( n − ρ |∇ g ψ | ε i + λε i . Substituting the expressions found in (2.3), and (2.4) into (2.6), we obtain ψ [( n − ψ ,x i x i + ψh ,x i x i + 2 ψ ,x i h ,x i ]+ ε i n X k =1 ε k (cid:2) ( n − (cid:0) ρnψ ,x k − ρψψ ,x k x k − ψ ,x k (cid:1) − ψψ ,x k h ,x k + ψψ ,x k x k (cid:3) = λε i . This concludes the proof of Theorem 1.2. (cid:3)
Proof.
Proof of Corollary 1.3. Let ¯ g = ψ − g be a conformal metric of g . We areassuming that ψ ( r ) and h ( r ) are functions of r , where r = P nk =1 ε k x k . Hence, wehave ψ ,x i = 2 ε i x i ψ ′ , ψ ,x i x i = 4 x i ψ ′′ + 2 ε i ψ ′ , ψ ,x i x j = 4 ε i ε j x i x j ψ ′′ and h ,x i = 2 ε i x i h ′ , h ,x i x i = 4 x i h ′′ + 2 ε i h ′ , h ,x i x j = 4 ε i ε j x i x j h ′′ . Substituting these expressions into (1.4), we obtain4 ε i ε j ( n − x i x j ψ ′′ + 4 ε i ε j x i x j ψh ′′ + (2 ε i x i ψ ′ ) . (2 ε j x j h ′ ) + (2 ε j x j ψ ′ ) . (2 ε i x i f ′ ) = 0 , which is equivalent to4 ε i ε j [( n − ψ ′′ + ψh ′′ + 2 ψ ′ h ′ ] x i x j = 0 . Since there exist i = j , such that x i x j = 0, we have( n − ψ ′′ + h ′′ ψ + 2 ψ ′ h ′ = 0 . Similarly, considering the equation (1.5), we obtain4 ψ [( n − ψ ′′ + ψh ′′ + 2 ψ ′ h ′ ] x i + 2 ψ [( n − ψ ′ + ψh ′ ] ε i + 2 ε i n [1 − n − ρ ] ψψ ′ +4 ε i n X k =1 ε k x k (cid:8) ( n − (cid:2) ( ρn −
1) ( ψ ′ ) − ρψψ ′′ (cid:3) − ψψ ′ h ′ + ψψ ′′ (cid:9) = λε i . Note that ( n − ψ ′′ + ψh ′′ + 2 ψ ′ h ′ = 0 and r = P nk =1 ε k x k . Therefore, we obtain2 ψ [( n − ψ ′ + ψh ′ ] + 2 n [1 − n − ρ ] ψψ ′ +4 r (cid:8) ( n − (cid:2) ( ρn −
1) ( ψ ′ ) − ρψψ ′′ (cid:3) − ψψ ′ h ′ + ψψ ′′ (cid:9) = λ. ROMILDO PINA, ILTON MENEZES, AND LUCYJANE SILVA
This concludes the proof of Corollary 1.3. (cid:3)
Proof.
Proof of the Theorem 1.4 It is well known (see, e.g., [2] or [9]) that if ¯ g = gψ ,then K ¯ g = ( n − (cid:0) ψ ∆ g ψ − n |∇ g ψ | (cid:1) . How we are assuming that ψ ( r ) is a functions of r , where r = P nk =1 ε k x k , thenwe have that K ¯ g = 0 if, and only, if − nψψ ′ − rψψ ′′ + nr ( ψ ′ ) = 0 , which is equivalent to − n r ψ ′ ψ + n (cid:18) ψ ′ ψ (cid:19) − ψ ′′ ψ = 0 . By equality ψ ′′ ψ = (cid:16) ψ ′ ψ (cid:17) ′ + (cid:16) ψ ′ ψ (cid:17) , follows that − n r ψ ′ ψ + n (cid:18) ψ ′ ψ (cid:19) − (cid:18) ψ ′ ψ (cid:19) ′ − (cid:18) ψ ′ ψ (cid:19) = 0 . Taking y = ψ ′ ψ , the previous equation becomes(2.7) y ′ = − n r y + n − y . Note that equation (2.7) is an ordinary differential equation of Bernoulli. There-fore, you can determine all your solutions, whose general solution is given by(2.8) y − = Ce F − ( n − e F Z e − F dr, where F ( r ) = n Z r dr = ln r n , C is an arbitrary constant (for more details see [19]). Thus y − = Cr n − ( n − r n Z r − n dr equivalently, y − = (cid:18) C − ( n − k (cid:19) r n + r, where k is a real number. Implies that y − = Ar n + r, where A = C − ( n − k . It follow that y = r − n A + r − n , how y = ψ ′ ψ , we get(2.9) ψ ( r ) = exp (cid:26)Z r − n A + r − n dr + ln k (cid:27) IGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 9 where k ∈ R ∗ + . Note that(2.10) Z r − n A + r − n dr = ln (cid:16) A + r − n (cid:17) − , combining the equations (2.9) and (2.10), the following that(2.11) ψ ( r ) = k B − n , where B = A + r − n . How n ≥ ψ ( r ) = k r (cid:16) Ar n − (cid:17) n − , (cid:3) Proof.
Proof of the Corollary 1.6 If K ¯ g = 0, by Theorem 1.4 we get ψ ( r ) = k r (cid:18) Ar n − (cid:19) n − . We will show that ¯ g = g ψ is complete.Consider the manifolds M = (cid:16) R n \ { } , ¯ g = g ψ (cid:17) , where ψ ( r ) = k r (cid:18) Ar n − (cid:19) n − , k ∈ R ∗ + and N = (cid:16) R n \ { } , g = g ϕ (cid:17) , where ϕ ( r ) = √ r , and g is a Euclideanmetric. Note that | v | ¯ g = 1 ψ | v | g and | v | g = 1 ϕ | v | g By other hand, we get | v | ¯ g = (cid:16) Ar n − (cid:17) n − k r | v | g = (cid:16) Ar n − (cid:17) n − k √ r √ r | v | g , thus, | v | ¯ g = f ( r ) | v | g , where f ( r ) = (cid:18) Ar n − (cid:19) n − k √ r .To find c > | v | ¯ g ≥ c | v | g , just solve the following problemmin r ∈ R ∗ + f ( r )The first derivative of f takes us f ′ ( r ) = r n − (cid:16) Ar n − (cid:17) − nn − A n − r n − − r − (cid:16) Ar n − (cid:17) n − k r , equaivalently, f ′ ( r ) = (cid:16) Ar n − (cid:17) n − k r (cid:20)(cid:16) Ar n − (cid:17) − r n − − r (cid:21) . Therefore, f ′ ( r ) = (cid:16) Ar n − (cid:17) − nn − k r (cid:16) Ar n − − (cid:17) . Given f is a real function, we have that r is a critical point if, and only if, f ′ ( r ) = 0. How r >
0, the minimum point candidate is given by r = 1 A n − . Let’s calculate the second derivative of f and evaluate at this point, this is, f ′′ ( r ) = 12 k (cid:20) − nn − (cid:16) Ar n − (cid:17) − nn − A n − r n − (cid:16) Ar n − − r − (cid:17)(cid:21) + 12 k (cid:20)(cid:16) Ar n − (cid:17) − nn − (cid:18) n − Ar n − + 32 r − (cid:19)(cid:21) equivalently, f ′′ ( r ) = 12 k (cid:16) Ar n − (cid:17) − nn − (cid:18) − n A r n − − (4 − n ) A r n − (cid:19) + 12 k (cid:16) Ar n − (cid:17) − nn − (cid:18) n − Ar n − + 3 A r − (cid:19) implies that f ′′ ( r ) = − (cid:16) Ar n − (cid:17) − nn − k r (cid:16) A r n − + 6 Ar n − − Anr n − − (cid:17) . Now let’s evaluate the second derivative at the point r = A n − , that is, f ′′ (cid:18) A n − (cid:19) = − A (cid:18) A n − (cid:19) n − ! − nn − k (cid:18) A n − (cid:19) A (cid:18) A n − (cid:19) n − + 6 A (cid:18) A n − (cid:19) n − − An (cid:18) A n − (cid:19) n − − ! , equivalently, f ′′ (cid:18) A n − (cid:19) = − − nn − k A n − (1 + 6 − n − , implies that f ′′ (cid:18) A n − (cid:19) = − − nn − k A n − (4 − n ) . Therefore, f ′′ (cid:18) A n − (cid:19) = 2 − nn − k A n − ( n − . How n ≥ A, k ∈ R ∗ + , we get f ′′ (cid:18) A n − (cid:19) >
0, consequently r = A n − it’s aminimum point. Therefore,(2.13) f (cid:18) A n − (cid:19) = A (cid:18) A n − (cid:19) n − ! n − k s(cid:18) A n − (cid:19) = (4 A ) n − k ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 11 Just take c = (4 A ) n − k , we get f ( r ) ≥ c , ∀ r . Thus | v | ¯ g ≥ c | v | g , how N = (cid:16) R n \ { } , g = g ϕ (cid:17) is complete, implies that M = (cid:16) R n \ { } , ¯ g = g ψ (cid:17) is complete.Therefore, the proof is done. (cid:3) Proof.
Proof of the Theorem 1.7 How ( R n , ¯ g ) is a ρ –Einstein sotiton, with zeroscalar curvature follows by Theorem 1.4 that(2.14) ψ ( r ) = k B − n , where B = A + r − n . Consequently,(2.15) ψ ′ ( r ) = k B − n r − n , ψ ′′ ( r ) = k n (cid:16) B − n r − n − B − n r − ( n +2)2 (cid:17) . Replacing the expressions found in (2.14) and (2.15) in (1.7), we have2 ( n −
1) (1 − nρ ) k B − n B n − n r − n + 2 ( n −
1) ( nρ − k rB n − n r − n +2 (1 − n − ρ ) rk B − n k n (cid:16) B n − − n r − n − B n − n r − ( n +2)2 (cid:17) + k B − n (cid:16) B − n − B n − n r − n (cid:17) h ′ = λ , equivalently,2 ( n −
1) (1 − nρ ) k B − n B n − n r − n + 2 (1 − n − ρ ) k n B − n B n − − n r − n − − n − ρ ) k n B − n B n − n r − n + 2 ( n −
1) ( nρ − k B n − − n B − n r − n + k B − n (cid:16) B − n − B n − n r − n (cid:17) h ′ = λ . How B = 0, we get2 ( n −
1) (1 − nρ ) B n − n r − n +(1 − n − ρ ) nB n − − n r − n − (1 − n − ρ ) nB n − n r − n − n −
1) (1 − nρ ) B n − − n r − n + k B − n (cid:16) B − n − B n − n r − n (cid:17) h ′ = λ k B − n . Consequently,( n − (cid:16) B n − n r − n − B n − − n r − n (cid:17) + (cid:16) B − n − B n − n r − n (cid:17) h ′ = λ k B − n equivalently,2 k B − n ( n − (cid:16) B n − n r − n − B n − − n r − n (cid:17) +2 k B − n (cid:16) B − n − B n − n r − n (cid:17) h ′ = λ. Note that B − n − B n − n r − n = 0, otherwise B = 2 r − n and Consequently B = 2 A .Thus, h ′ ( r ) = λ k B − n (cid:16) B − n − B n − n r − n (cid:17) + (2 − n ) B n − n r − n − B n − − n r − n B − n − B n − n r − n ! . Making ϕ ( r ) = λ k B − n (cid:16) B − n − B n − n r − n (cid:17) , and w ( r ) = (2 − n ) B n − n r − n − B n − − n r − n B − n − B n − n r − n ! , the first derivative of this equations leads us to(2.16) ϕ ′ ( r ) = − λ k (cid:0) nB − r − n − ( n + 2) B − r − n (cid:1)(cid:16) B − n − B n − n r − n (cid:17) , and(2.17) w ′ ( r ) = (2 − n )2 (cid:16) nB n − n r − n − n − B n − − n r − n + 2( n − B n − − n r − n ) − nB n +22 − n r − ( n +2)2 (cid:17)(cid:16) B − n − B n − n r − n (cid:17) Replacing the functions in the equation (1.6), that is,( n − ψ ′′ + ψh ′′ + 2 ψ ′ h ′ = 0 , we obtain( n − k n (cid:16) B n − − n r − n − B n − n r − n (cid:17) (cid:16) B − n − B n − n r − n + 4 B n − n r − n (cid:17) + (2 − n )2 k B − n (cid:16) nB n − n r − n − n − B − n − n r − n + 2 ( n − B n − − n r − n ) − nB n +22 − n r − n +22 (cid:17) − n ) k B n − n r − n (cid:16) B n − n r − n − B n − − n r − n (cid:17) (cid:16) B − n − B n − n r − n (cid:17) − λ k (cid:16) nB n − n r − n − ( n + 2) B n − − n r − n (cid:17) = 0 , Which is equivalent to B n +1)2 − n r − n − B n − n r − n + B n − − n r − n ) = λ ( n − k (cid:16) nB n − n r − n − ( n + 2) B n − − n r − n (cid:17) . Note that nB n − n r − n − ( n +2) B n − − n r − n = 0, because otherwise B = n +2 n r − n .On the other hand, B n +1)2 − n r − n − B n − n r − n + B n − − n r − n ) = 0therefore, (cid:16) B n +12 − n r − n − B n − − n r − n (cid:17) = 0 , and consequently B = r − n , but this is a contradiction, because B = n +2 n r − n .Therefore,(2.18) λ = ( n − k (cid:16) B n +12 − n r − n − B n − − n r − n (cid:17) (cid:16) nB n − n r − n − ( n + 2) B n − − n r − n (cid:17) . How λ is constant, we have that IGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 13 (2.19) ddr (cid:16) B n +12 − n r − n − B n − − n r − n (cid:17) (cid:16) nB n − n r − n − ( n + 2) B n − − n r − n (cid:17) = 0 . if, and only if,(4 n − B n − n r − n − (4 n − n +2) B n − − n r − n − n B n +22 − n r − n +22 +( n +2) nB n − n r − n +(2 − n ) nB n − − n r − n − (2 − n )( n + 2) B n − − n r − n + (2 n − nB n − − n r − n − (cid:0) n + 2 n − (cid:1) B n − n r − n − (2 n − n +2) B n − − n r − n +( n + n − B n − − n r − n + n B n +22 − n r − n +22 + (3 n + 2 n −
4) + n B n − − n r − − n − n + n − B n − − n r − n − n B n − n r − n − (cid:0) n + 2 n − (cid:1) B n − − n r − n + n B n − − n r − − n +( n + n − B n − − n r − − n = 0 , if, and only if,(2.20) A h − n r n − A + ( n + 4) A + 4(1 − n ) r − n i = 0 . We will prove that equation (2.20) is satisfied if, and only, if A = 0. For this,consider f ( r ) = − n r n − A + ( n + 4) A + 4(1 − n ) r − n . If f ( r ) = 0 ∀ r > A = 0, then its derivative also is zero. Since there is a singlevalue of r such that f ′ ( r ) = 0 given by r = (cid:16) n − n A (cid:17) n − , we get a contradiction.Therefore the equation (2.20) is satisfied if, and only, if A = 0. In this case B = r − n . Substituindo B = r − n in (2.18) we obtain that λ = 0. Therefore theproof is done. (cid:3) Proof.
Proof of the Theorem 1.8 Follows by Theorem 1.7 that ( R n , ¯ g ), ¯ g = ψ g is a ρ –Einstein sotiton with scalar curvature K ¯ g = 0 is steady if, and only if, A = 0. Besides that h ( r ) is constant. How A = 0 we obtain from Lemma 1.4, that ψ ( r ) = k r . Follow of the [14] that ( R n , ¯ g ) have sectional curvature zero. Therefore,we conclude que ( R n , ¯ g ) is flat. The reciprocal is automatically satisfied. (cid:3) Proof.
Proof of the Corollary 1.9 How ( M n , ¯ g ) is locally conformally flat and rota-tionally symmetric, then locally the metric ¯ g is given by ¯ g = ψ g where ψ = ψ ( r )and r = n P k =1 x k and g is the euclidean metric.Therefore the results obtained inTheorem 1.7 and Corollary 1.8 are satisfied. (cid:3) Proof.
Proof of the Proposition 1.1 Consider the manifolds M = (cid:16) R n \ { } , g = g ϕ (cid:17) ,where ϕ ( r ) = re − (cid:18) r n − (cid:19) n − , k ∈ R ∗ + and N = (cid:16) R n \ { } , ¯ g = g ψ (cid:17) , where ψ ( r ) = k r (cid:18) r n − (cid:19) n − , and g is a Euclidean metric. Note that | v | g = 1 ϕ | v | g and | v | ¯ g = 1 ψ | v | g By other hand, we get | v | g = e (cid:18) r n − (cid:19) n − r | v | g = e (cid:18) r n − (cid:19) n − (cid:16) r n − (cid:17) n − (cid:16) r n − (cid:17) n − k r | v | g . Thus, | v | g = h ( r ) | v | ¯ g , where, h ( r ) = e r n − ! n − (cid:18) r n − (cid:19) n − .To find c > | v | ¯ g ≥ c | v | g , just solve the following problemmin r ∈ R ∗ + h ( r )The first derivative of h takes us h ′ ( r ) = e (cid:18) r n − (cid:19) n − (cid:16) r n − (cid:17) − nn − r n − (cid:20)(cid:16) r n − (cid:17) n − − (cid:21) . How r >
0, we get h ′ ( r ) > ∀ r , so the h function is strictly increasing. There-fore,(2.21) min r ∈ R ∗ + h ( r ) = lim r −→ e (cid:18) r n − (cid:19) n − (cid:16) r n − (cid:17) n − = e. Just take c = e , we get h ( r ) ≥ c , ∀ r . Thus | v | g ≥ c | v | ¯ g , how M = (cid:16) R n \ { } , ¯ g = g ψ (cid:17) is complete, implies that N = (cid:16) R n \ { } , g = g ϕ (cid:17) is complete.We will show that ( R n , ¯ g ) has negative scalar curvature. Indeed, note that(2.22) ϕ ′ ( r ) = 1 − r n − (cid:16) r n − (cid:17) − nn − e (cid:18) r n − (cid:19) n − , and ϕ ′′ ( r ) = − e (cid:18) r n − (cid:19) n − (cid:20) (4 − n ) r n − (cid:16) r n − (cid:17) − nn − + ( n − r n − (cid:16) r n − (cid:17) − nn − − r n − (cid:16) r n − (cid:17) − nn − − r n − (cid:16) r n − (cid:17) (4 − n ) n − , implies that,(2.23) ϕ ′′ ( r ) = − " nr n − (cid:16) r n − (cid:17) − nn − + (4 − n ) r n − (cid:16) r n − (cid:17) − nn − − r n − (cid:16) r n − (cid:17) (4 − n ) n − e (cid:18) r n − (cid:19) n − . IGIDITY RESULTS ON ρ –EINSTEIN SOLITONS WITH ZERO SCALAR CURVATURE 15 It is well known (see, e.g., [2] or [9]) that if ¯ g = gψ , then(2.24) K ¯ g = 4 r h n − ϕϕ ′′ − n ( n −
1) ( ϕ ′ ) i + 4 n ( n − ϕϕ ′ . Substituting the expressions found in (2.22) and (2.23) into (2.24), we K ¯ g = 4( n − e (cid:18) r n − (cid:19) n − " (2 − n ) r n − (cid:16) r n − (cid:17) − n ) n − + 2(1 − n ) r n (cid:16) r n − (cid:17) − nn − + ( n − r n − (cid:16) r n − (cid:17) − n ) n − equivalently, K ¯ g = 4( n − (cid:16) r n − (cid:17) − nn − e (cid:18) r n − (cid:19) n − (cid:20) (2 − n ) r n − (cid:16) r n − (cid:17) − nn − + 2(1 − n ) r n + ( n − r n − (cid:16) r n − (cid:17) − (cid:21) implies that, K ¯ g = 4( n − (cid:16) r n − (cid:17) − n ) n − e (cid:18) r n − (cid:19) n − (cid:20) (2 − n ) r n − (cid:16) r n − (cid:17) n − + 2(1 − n ) r n (cid:16) r n − (cid:17) − (4 − n ) r n − (cid:3) . Therefore, K ¯ g = − n − (cid:16) r n − (cid:17) − n ) n − e (cid:18) r n − (cid:19) n − (cid:20) ( n − (cid:16) r n − (cid:17) n − + 2( n − r − n + ( n + 2) (cid:21) r n − . (cid:3) Proof.
Proof of Corollary 1.12 Follow immediately from Corollary 1.6. (cid:3)
The authors would like to thank the referee for his careful reading, relevantremarks, and valuable suggestion.
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