Spherical Symmetric 4d the Ricci Flow Solution with Constant Scalar Curvature Space as Initial Geometry

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1 International Mathematical Forum, Vol. 6, 011, no. 7, Spherical Symmetric 4d the Ricci Flow Solution with Constant Scalar Curvature Space as Initial Geometry Fiki T. Akbar and Bobby E. Gunara 1, 1 Indonesian Center for Theoretical and Mathematical Physics ICTMP) and Theoretical Physics Laboratory Theoretical High Energy Physics and Instrumentation Research Group Faculty of Mathematics and Natural Sciences Institut Teknologi Bandung Jl. Ganesha no. 10 Bandung, Indonesia, 4013 ft akbar@students.itb.ac.id bobby@fi.itb.ac.id Abstract In this paper, we show the existence a solution of the Ricci flow equation in four dimensional spherical symmetric spacetime with a constant scalar curvature as initial geometry. Our strategy is that away from singularities the metric can be expanded perturbatively with respect to a real parameter. Then, the lowest order of the soliton is chosen to be Ricci flat or Einstein. Moreover, we simplify the problem using the separation variable method and carry out up to the second order computation. Keywords: Ricci flow, Ricci flat, Einstein metrics 1 Introduction and Main Results The Ricci flow equation firstly introduced by R. Hamilton in is an equation describing the evolution of a Riemannian metric g = R, 1.1) τ where R is the Ricci tensor of metric g. The Ricci flow is a nonlinear second order partial differential equations on the metric which can be viewed as a nonlinear heat equation of a metric and contains at most the second derivative of the metric. The

2 350 F. T. Akbar and B. E. Gunara solution of this equation is a one-parameter family of metrics g τ) parameterized by a real parameter τ in the interval 0 τ T on smooth manifold M. In general, there exists an initial point τ = 0 describing the initial geometry with initial metric g 0). Moreover, the solution of 1.1) might end up to a point at finite τ which is the singular point. In serial papers by G. Perelman, he modified Hamilton works on Ricci flow to prove the celebrated three dimensional puzzles called Thurston s geometrization conjecture 3. The conjecture is about the classification of three dimensional smooth manifolds in which it includes the 100 years old Poincaré conjecture. Some excellent reviews of the Ricci flow equation and also its developments can further read for example in 4, 5, 6, 7. Apart from the above application, in this paper we study a simple class of solutions of 1.1) on four dimensional manifolds. The solution admits spherical symmetries and the manifold is chosen to be pseudo Riemannian or we call it spacetime. This is motivated by the standard solution in general relativity textbook, namely the Schwarzschild-anti) de Sitter metric. Hence, our particular result is the following statement: Theorem 1. Let M τ,gτ)) be a four dimensional spherical symmetric solution of 1.1), then away from singularities there exists perturbatively a solution whose initial geometry has a constant scalar curvature. The proof of the theorem can roughly be mentioned as follows. First, far away from singularities we assume that there exists a real small parameter ε which is constant and positive, and the solution of 1.1) can be regarded as a series expansion of four dimensional spherical symmetric metrics. Or in other words, This is nothing but the perturbative solution of 1.1) 1. Then, it turns out that the Ricci tensor can be expanded with respect to same parameter see section ). Inserting into 1.1), we then get an equation for each order of ε and the resulting equations are recursive. Furthermore, we simply take two background metric or the lowest order solution of 1.1) has to be a Ricci flat or an Einstein metrics. Then, for higher order we solve the resulting nonlinear equations by using separation variables which splits a function of τ and a function of a radial coordinate r. Finally, the solution can be easily obtained and we particularly employ the computation up to the second order. The structure of this paper can be mentioned as follow. In Section, we construct some nonlinear equations coming from a spherical symmetric metric ansatz of the Ricci flow using series expansion. Then, in Section 3 we solve the equations in the case where the zeroth order is Ricci flat, whereas the Einstein metric case is in Section 4. Both of them are computed up to the second order. Finally, the proof of Theorem 1 we put in Section 5. 1 Such a method has been used by R. Hirota 8 to obtain a soliton of nonlinear differential equations which is analog to the perturbation method, but the expansion parameter is not necessary have a small value.

3 Spherical symmetric 4d Ricci flow solution 351 Spherical Symmetric Equation of the Ricci Flow This section is devoted to construct some dynamical equations that admit spherical symmetries which are useful for our analysis throughout the paper. It is worth mentioning that it is sufficient to prove Theorem 1 by employing the second order computation. Our starting point is to consider the spherical symmetric metric ansatz which in local coordinates t, r, θ, φ) given by ) g = diag e ν, e λ, fr, fr sin θ,.1) where ν ν r, τ ), λ λ r, τ ), f f τ). Firstly, away from singular region we could expand the functions ν, λ and f as power series with respect to a real parameter ε, ν r, τ ) = ε n ν n) = ν 0) + εν 1) + ε ν ) +, λ r, τ ) = f τ) = n=0 ε n λ n) = λ 0) + ελ 1) + ε λ ) +, n=0 ε n f n) = f 0) + εf 1) + ε f ) +..) n=0 Or in other words, the metric components.1) can be written down as g = ε n g μν n) = g μν 0) + εg 1) + ε g ) +,.3) n=0 with g 0) = diag e ν0), e λ0), f 0) r, f 0) r sin θ,.4) g 1) = diag e ν0) ν 1), e λ0) λ 1), f 1) r, f 1) r sin θ,.5) ) ) g ) ν = diag e 1) ) ν0) ν ) λ 1) ), e λ0) λ ), f ) r, f ) r sin θ.6). The non vanishing components of Ricci tensor of the metric.1) are then R 00 = 1 4 eν λ ν + ν ν λ ) + 4 r ν, R 11 = 1 ν + ν ν λ ) 4r 4 λ, R = 1+fe λ 1+ r ν λ ), R 33 = R sin θ,

4 35 F. T. Akbar and B. E. Gunara where the prime represents the derivative with respect to the radial coordinate r. Using expansion in.), we can also write down the components of Ricci tensor as where R = n=0 ε n R n) = R0) + εr1) + ε R ) +..7) R 0) 00 = 1 4 eν0) λ 0) F 1, R 0) 11 = 1 4 F, R 0) = 1+f 0) e λ0) F 3, R 0) 33 = R 0) sin θ,.8) R 1) 00 = 1 4 eν0) λ 0) F 4 + F 1 ν 1) λ 1)), R 1) 11 = 1 4 F 5, R 1) = e λ0) f 0) F 6 + f 1) λ 1) f 0)) F 3 R 1) 33 = R 1) sin θ,.9) R ) 00 = 1 λ 0) 4 eν0) F 7 + F 4 ν 1) λ 1)) 1 + F 1 ν 1) λ 1)) ) + ν ) λ ), R ) 11 = 1 4 F 8, ) R ) = e f λ0) 0) F 9 + f 1) λ 1) f 0)) F 6 + F 3 f ) f 1) λ 1) + f 0) λ 1) ) f 0) λ ), R ) 33 = R ) sin θ,.10) where F 1 = ν 0) + ν 0) ν 0) λ 0) ) + 4 r ν0), F = ν 0) + ν 0) ν 0) λ 0) ) 4 r λ0), F 3 = 1+ r ν 0) λ 0) ),, F 4 = ν 1) + 4 r ν1) + ν 0) ν 1) λ 1) ) + ν 1) λ 0), F 5 = ν 1) 4 r λ1) + ν 0) ν 1) λ 1) ) + ν 1) λ 0), F 6 = r ν 1) λ 1) ),

5 Spherical symmetric 4d Ricci flow solution 353 F 7 = ν ) + 4 r ν) + ν 0) ν ) λ ) ) + ν 1) ν 1) λ 1) ) + ν ) ν 0) λ 0) ), F 8 = ν ) 4 r λ) + ν 0) ν ) λ ) ) + ν 1) ν 1) λ 1) ) + ν ) ν 0) λ 0) ), F 9 = r ν ) λ ) ). Inserting the above equations into 1.1), we then obtain ) ) g 0) τ + εg1) + ε g ) + = R 0) + εr1) + ε R ) + Thus, for each order of ε we then have..11) g 0) τ g 1) τ g ) τ = R 0),.1) = R 1),.13) = R ),.14) describing the zeroth, the first and the second order equations of 1.1), respectively. Note that.1)-.14) are recursive, therefore we need to know the lowest order solution in order to obtain the higher order solution. In the rest of the paper we consider two cases for zeroth order solution, namely the Ricci flat and the Einstein metric cases which will be discussed in the next sections. 3 Ricci Flat Cases In the Ricci flat case, we choose the lowest order of the Ricci tensor satisfying the condition R 0) =0. 3.1) Hence, from.1) and.) the zeroth order functions are independent of τ, ν 0) ν 0) r), λ 0) λ 0) r), f 0) 1, 3.) which gives the Schwarzschild solution g 0) μν = diag 1 m r, 1 m r ) 1, r, r sin θ). 3.3)

6 354 F. T. Akbar and B. E. Gunara 3.1 The First Order Solution Let us next consider the first order solution. Inserting.9) into.13), we get equation the first order equations e λ0) ν 1) = ν 1) λ 0) 3ν 1) λ 1) ) + 4r ν1), 3.4) λ 1) = ν e λ0) 1) λ 0) 3ν 1) λ 1) ) 4r λ1), 3.5) f 1) = r f 1) λ 1)) + e r λ0) ν 1) λ 1) ), 3.6) where the dot represents the derivative with respect to τ. Equations 3.4)-3.6) are the decoupled partial differential equations. For the sake of simplicity, we apply the separation variables method to these equations as ν 1) r, τ ) = Θ 1) τ)+ρ 1) r), λ 1) r, τ ) = Π 1) τ)+χ 1) r), such that equations 3.4) and 3.5) simply become Θ 1) = e λ0) Π 1) = e λ0) ρ 1) λ 0) 3ρ 1) χ 1) ) + 4r ρ1) ρ 1) λ 0) 3ρ 1) χ 1) ) 4r χ1) The left hand sides of the above equations depend only on τ, while the other sides depend only on r. Therefore, we must have Θ 1) = K 1, 3.7) ρ 1) λ 0) 3ρ 1) χ 1) ) + 4r ρ1) = K 1 e λ0), 3.8) Π 1) = K 1, 3.9) ρ 1) λ 0) 3ρ 1) χ 1) ) 4r χ1) = K 1 e λ0), 3.10) where K 1 is constant. Then, after some straightforward computations, 3.8) and 3.10) can be written down as ρ 1) + χ 1) = 0, 3.11) 4ρ 1) λ 0) 3ρ 1) χ 1) ) + 4 ρ 1) χ 1) ) = 4K 1 e λ0). 3.1) r,. From 3.11) and 3.1), one can obtain the equation for ρ 1) as ρ 1) + r m ρ1) = K 1r r m, 3.13)

7 Spherical symmetric 4d Ricci flow solution 355 which gives ρ 1) r) = C 1 r m + C + K 1 6 r 3 4m r +8m 3 ) r m, 3.14) where C 1 and C are constants. It follows from equation 3.11) that χ 1) r) = C C 1 r m K 1 6 r 3 4m r +8m 3 ) r m. 3.15) Next, with direct integration, we have the solutions of 3.7) and 3.9), namely Θ 1) τ) = K 1 τ + C 3, 3.16) Π 1) τ) = K 1 τ + C ) where C 3 and C 4 are constants. Thus, the complete solutions for ν 1) and λ 1) are given by ν 1) r, τ ) = K 1 τ + C 5 + C 1 r m + K 1 r 3 4m r +8m 3 ), 3.18) 6 r m λ 1) r, τ ) = K 1 τ + C 5 C 1 r m K 1 r 3 4m r +8m 3 ). 3.19) 6 r m Now let us look at 3.6). Using equations 3.18) and 3.19), we have f 1) = r f 1) K 1 τ C 4 + K 1 r which results in 3. The Second Order Solution f 1) τ) =K 1 τ + C 4 3.0) Now we can turn to look at the second order solution. Inserting.10) into.14), we get a set of second order equations ) K1 τ + C 5K 1 + ν ) = 1 r m r ν ) + r m r r m ν ) m 1 r r m λ) + gr), 3.1) ) K1 τ + C 5K λ) 1 = 1 r m r ν ) + 6m 1 r r m ν) r 3m r r m λ ) + gr), 3.) where f ) = e λ0) F ) r + f 1) λ 1)) ) F 1) + e λ0) f ) f 1) λ 1) + λ1) ) λ ), 3.3) gr) = K1 r r 3m)+6C 1 36r m) 3 + K 1 ρ 1) r).

8 356 F. T. Akbar and B. E. Gunara Applying the separation variable method ν ) = Θ ) τ)+ρ ) r), λ ) = Π ) τ)+χ ) r), into 3.1) and 3.), they can be simplified as K1 τ + C 5K 1 + Θ ) = 1 r m r K 1τ + C 5 K 1 + Π ) = 1 r m r ρ ) + ) r m ν ) m r r m r ρ ) + 6m 1 r r m ρ) r 1 r m λ) + gr), ) χ ) + gr). r 3m r m The left hand sides of the above equations depends only on τ, while the other sides depend only on r. So, it must be ρ ) + r ρ ) + 6m r ) r m ν ) m r m r 1 r m ρ) r K1τ + C 5 K 1 + Θ ) = K 3.4) 1 r m λ) + gr) = K r 3.5) r m K1 τ + C 5K 1 + Π ) = K 3.6) ) χ ) + gr) = K r 3.7) r m r 3m r m where K is constant. After some computations, we can write equation 3.5) and 3.7) as ρ ) + χ ) = 0, 3.8) 4ρ ) + 4 ) r + m ρ ) 4 ) r m χ ) +gr) = 4K r r r m r r m r m, 3.9) and then we have the equation for ρ ) r) as, which gives ρ ) + ρ ) r) = C 6 + C 7 r m + K 6 and it follows K 1 7 χ ) r) = C 6 C 7 r m K 6 + K 1 7 r m ρ) = r r m K gr)), 3.30) r 3 4m r +8m 3 C 1 K1 r8m 4mr + r )+6C 1 r m 1 r m) r 6 8m r 4 +16m 3 r 3 48m 4 r + 18m 5 r 64m 6 r m), 3.31) r 3 4m r +8m 3 + C 1 K1 r8m 4mr + r )+6C 1 r m 1 r m) r 6 8m r 4 +16m 3 r 3 48m 4 r + 18m 5 r 64m 6 r m) 3.3)

9 Spherical symmetric 4d Ricci flow solution 357 On the other hand, the solutions of 3.4) and 3.6) can be easily obtained as Θ ) τ) = 1 K 1τ C 5 K 1 K )τ + C 8, 3.33) Π ) τ) = 1 K 1τ C 5 K 1 K )τ + C 9, 3.34) where C 8 and C 9 are constants. So, the complete solutions of ν ) and λ ) are respectively ν ) r, τ ) = C 10 + C 7 C 1 K1 r8m 4mr+r )+6C 1 1 r m) K 1 7 λ ) r, τ ) = C 10 C 7 K 1 7 r m + K r 3 4m r+8m 3 6 r m r 6 8m r 4 +16m 3 r 3 48m 4 r +18m 5 r 64m 6 r m K r 3 4m r+8m 3 6 r m 1 r m) K 1 τ C 5 K 1 K )τ, 3.35) + C 1 K1 r8m 4mr+r )+6C 1 1 r m) r 6 8m r 4 +16m 3 r 3 48m 4 r +18m 5 r 64m 6 r m) 1 K 1 τ C 5 K 1 K )τ,3.36) together with the solution of f ) as ) f ) τ) = K + K 1 m τ + K 1C 5 m m 6 K m K 1 m3 3 K 1 C 1 ) 3.37) after using 3.35) and 3.36). 4 Einstein Metrics Cases In the Einstein metric case, we choose such that g 0) satisfies R 0) =Λg0) The solution of this equation is, see for example in 4, 0). 4.1) where g 0) τ) =u τ) g0) μν 0), 4.) u τ) =1 Λτ, and g 0) 0) is nothing but the Schwarzschild-de Sitter metric g 0) r, 0) = diag { p r), p 1 r), r, r sin θ }, p r) = 1 m r Λ 3 r. So, we find e ν0) = u τ) p r), 4.3) e λ0) = u τ) p 1 r), 4.4) f 0) = u τ). 4.5)

10 358 F. T. Akbar and B. E. Gunara 4.1 The First Order Solution The next step is to consider the first order solution. Inserting.9) into.13) and using 4.3) and 4.4), we get then τ τ uν 1)) = Λ ν 1) λ 1)) + p ν 1) + p 3ν 1) λ 1) ) + 4 p r ν1), 4.6) ν 1) + p 3ν 1) λ 1) ) 4 p r λ1), 4.7) f 1) = p ν 1) λ 1) ) ) 1 Λr + τ r ur f 1) uλ 1)). 4.8) uλ 1)) = p We apply the separation variables method into 4.6)-4.8), then it results in ν 1) r, τ ) = Θ 1) τ)+ρ 1) r), λ 1) r, τ ) = Π 1) τ)+χ 1) r), uθ 1)) +Λ Θ 1) Π 1)) = Λχ 1) + p ρ 1) + p τ p uπ 1)) = Λχ 1) + p τ ρ 1) + p p 3ρ 1) χ 1) ) + 4 r ρ1),4.9) 3ρ 1) χ 1) ) 4 r χ1) 4.10). So, we have here that the left hand sides on 4.9) and 4.10) are only τ dependent, whereas the other sides depend only on r. Therefore, we must have uθ 1)) +Λ Θ 1) Π 1)) = K 1, 4.11) τ Λχ 1) + p ρ 1) + p 3ρ 1) χ 1) ) + 4 p r ρ1) = K 1, 4.1) uπ 1)) = K 1, 4.13) τ Λχ 1) + p ρ 1) + p 3ρ 1) χ 1) ) 4 p r χ1) = K ) After some straightforward computations, 4.1) and 4.14) can be written down as ρ 1) + χ 1) = 0, 4.15) 4Λχ 1) + p 4ρ 1) + p 3ρ 1) χ 1) ) + 4 ρ 1) χ 1) ) = K ) p r From 4.15) and 4.16), we can obtain the equation of χ 1) as χ 1) Λr ) 3r 6m Λr 3 χ1) 6Λr 3K 1 r 3r 6m Λr 3 χ1) = 3r 6m Λr 3, 4.17)

11 Spherical symmetric 4d Ricci flow solution 359 which results in χ 1) 3C 1 r) = 3r 6m Λr 3 K 1 r 3 4m r +8m 3 ) 3r 6m Λr 3, 4.18) and then from 4.15), we get ρ 1) r) = 3C 1 3r 6m Λr 3 + K 1 r 3 4m r +8m 3 ) 3r 6m Λr ) The equations 4.11) and 4.13) can be solved Θ 1) τ) =Π 1) τ) = K 1τ + C 5 u τ). 4.0) Thus, the complete solutions of ν 1) and λ 1) are given by ν 1) r, τ ) = C 5 + K 1 τ u τ) λ 1) r, τ ) = C 5 + K 1 τ u τ) + 3C 1 3r 6m Λr 3 + K 1 3C 1 3r 6m Λr 3 K 1 r 3 4m r +8m 3 ) 3r 6m Λr 3 r 3 4m r +8m 3 ) 3r 6m Λr 3, 4.1), 4.) where u τ) =1 Λτ). In addition, we have 4. The Second Order Solution f 1) = K 1 τ + C 5 + 4K 1m 1 Λτ). 4.3) 6 Finally, we work out the second order equations. Inserting.9) into.14) and using 4.3) and 4.4), we then get ) Θ 1) u + ν ) = Λν ) λ ) )+ p ν ) + p 3ν ) λ ) ) + 4 τ p r ν) + gr), 4.4) ) Π 1) u + λ ) = p ν ) + p 3ν ) λ ) ) + 4 τ p r λ) + hr), 4.5) ) αe λ0) f ) = r f 0) F 0) + F 0) f 1) λ 1) f 0) )+F 0) f ) f 1) λ 1) + f 0) λ 1) f 0) λ ) 4.6) where gr) = p ρ 1) ρ 1) χ 1) ) + F 1) 00 ρ1) χ 1) ) Λρ 1) χ 1) ) K 1 ρ 1) +Λρ 1), hr) = p ρ1) ρ 1) χ 1) ) + K 1 χ 1) +Λχ 1).,

12 360 F. T. Akbar and B. E. Gunara We then apply the separation of variables method ν ) = Θ ) τ)+ρ ) r), λ ) = Π ) τ)+χ ) r), into 4.4) and 4.5) which simplify to u Θ1) + uπ ) + ΛΘ ) Π ) ) = Λχ ) + p ρ ) + p τ p u Π1) + uπ ) = Λχ ) + p τ 3ρ ) χ ) ) + 4 r ρ) + gr), ρ ) + p 3ρ ) χ ) ) + 4 p r χ) + hr). Again, the left hand sides of the above equations depends only on τ, while the other sides depend only on r. Therefore, we must have u Θ1) + uπ ) + ΛΘ ) Π ) ) = K, 4.7) τ Λχ ) + p ρ ) + p 3ρ ) χ ) ) + 4 p r ρ) + gr) = K, 4.8) u Π1) + uπ ) = K, 4.9) τ Λχ ) + p ρ ) + p 3ρ ) χ ) ) + 4 p r χ) + hr) = K. 4.30) After some straightforward computations, 4.8) and 4.30) can be written down as 4Λχ ) + p 4ρ ) + p p Also, we have the equation of ρ ) r) ρ ) Λr ) 3r 6m Λr 3 ρ) whose solution has the form ρ ) r) = and then it follows χ ) r) = 3C 7 3r 6m+Λr + K 3 3ρ ) χ ) ) + 4 r ρ) χ ) ) 6Λr 3r 6m Λr 3 ρ) = 3r ρ ) χ ) = 0, 4.31) = K gr)+hr))4.3). 1 3r 6m Λr 3 K gr)+hr)) r 3 4m r+8m 3 3r 6m+Λr 3C 3 1 K1 r8m 4mr+r )+6C 1 4 3r 6m+Λr 3 ) K 1 r 6 8m r 4 +16m 3 r 3 48m 4 r +18m 5 r 64m 6 8 3r 6m+Λr 3 ) + 4mK 1Λ 1 3C 7 3r 6m+Λr K 3 r 3 4m r+8m 3 3r 6m+Λr 3 + 3C K 1 r 6 8m r 4 +16m 3 r 3 48m 4 r +18m 5 r 64m 6 8 4mK 1Λ 3r 6m+Λr 3 ) 1 r 3 m r)8k 1 m 3 +6C 1 ) 3r 6m+Λr 3 ) K1 r8m 4mr+r )+6C 1 3r 6m+Λr 3 ) r 3 m r)8k 1 m 3 +6C 1 ) 3r 6m+Λr 3 ) 4.33) 4.34), 4.35).

13 Spherical symmetric 4d Ricci flow solution 361 The solutions of equations 4.7) and 4.9) are given by Θ ) τ) = K 1 u τ C 5K 1 K u u τ C 5 u, 4.36) Π ) τ) = K 1 u τ C 5K 1 K u u τ C 5 u. 4.37) Thus, the complete forms of ν ) and λ ) are ν ) r, τ ) = K 1 τ C 5K 1 K u τ C u u 5 + u K 1 r 6 8m r 4 +16m 3 r 3 48m 4 r +18m 5 r 64m 6 8 3r 6m+Λr 3 ) λ ) r, τ ) = K 1 u τ C 5K 1 K u u τ C 5 3C 9 K 3r 6m+Λr 3 r 3 4m r+8m 3 + 3C 3r 6m+Λr 3 1 K1 r8m 4mr+r ) 6C 1 4 3r 6m+Λr 3 ) + 4mK 1Λ r 3 m r)8k 1 m 3 6C 1 ) 1, 4.38) 3r 6m+Λr 3 ) 3C 9 + K u 3r 6m+Λr 3 r 3 4m r+8m 3 3C 3r 6m+Λr 3 1 K1 r8m 4mr+r ) 6C 1 4 3r 6m+Λr 3 ) + K 1 r 6 8m r 4 +16m 3 r 3 48m 4 r +18m 5 r 64m 6 8 4mK 1Λ 3r 6m+Λr 3 ) 1 together with the solution of f ) as r 3 m r)8k 1 m 3 +3C 1 α), 4.39) 3r 6m+Λr 3 ) ) f ) τ) = K + K 1 m τ + K 1C 5 m m 3 K m K 1 m3 3 K 1 C 1 ) 1 Λτ). 4.40) 5 The Proof of Theorem 1 In the previous sections, we have constructed perturbative solutions of 1.1) whose backgrounds are Ricci flat and Einstein. To be precise, away from singularities, the metric can be expanded into a series with respect to a real parameter ε. In section 3 and 4, we computed the solution up to second order for Ricci flat and Einstein metric backgrounds. So in this section we just provide the some final steps of the proof of Theorem 1. Let us consider the Einstein backgrounds since the Ricci flat case just setting Λ = 0. The scalar curvature of the metric.1) up to the second order has the form Rτ) = 4Λ 1 τλ εk 1 +C 5 Λ) 1 Λτ) + ε C 5 + K 1 τ)k 1 +C 5 Λ) 1 Λτ)K 1 Λτ) ) which is not depend on the radial coordinate r. Since the parameter ε is constant, then at τ = 0 the scalar curvature 5.1) becomes R0) = 4Λ εk 1 +C 5 Λ) + ε C 5 K 1 K +C 5Λ) 5.) which is constant. This completes the proof of Theorem 1. 6 Acknowledgement We acknowledge H. Alatas, A. A. Iskandar, and Triyanta for useful discussions. The work of B. E. G is supported by Hibah Kompetensi DIKTI 010 no. 7/SPH/PP/DPM/III/010.

14 36 F. T. Akbar and B. E. Gunara References 1 R. S. Hamilton, Three-manifolds with positive Ricci curvature. J. Differential Geom ) 55. G. Perelman, The entropy formula for the Ricci flow and its geometric applications, math.dg/011159; Ricci flow with surgery on three-manifolds, math.dg/ ; Finite extinction time for the solutions to the Ricci flow on certain three-manifolds, math.dg/ W. P. Thurston, Three-dimensional manifolds, Kleinian groups and hyperbolic geometry. Bull. Amer. Math. Soc. 6, 198) J. Morgan dan G. Tian, Ricci Flow and the Poincare Conjecture, arxiv:math DG/ v. 5 P. Topping, Lectures on the Ricci Flow, Cambrigde University Press, H.-D. Cao and X.-P. Zhu, A Complete Proof of the Poincare and Geometrization Conjectures. Application of the Hamilton-Perelman Theory of The Ricci Flow 006. Asian J. Math ), B. Chow and D. Knopf. The Ricci Flow: An Introduction 004). Mathematical Surveys and Monographs, Vol. 110, American Mathematical Society. 8 R. Hirota, The Direct Method in Soliton Theory 004). Cambridge University Press. Received: August, 010