aa r X i v : . [ m a t h . DG ] F e b ON COMPLETE GRADIENT SCHOUTEN SOLITONS
VALTER BORGES
Abstract.
In this paper we prove an optimal inequality between the potentialfunction of a complete Schouten soliton and the norm of its gradient. We alsoprove that these metrics have bounded scalar curvature of defined sign. Asan application, we prove that the potential function of a Shrinking Schoutensoliton grows linearly and provide optimal estimates for the growth of thevolume of geodesic balls, without any further assumption. Introduction and main results
We say that a Riemannian manifold ( M n , g ) is a gradient Schouten soliton ifthere are f ∈ C ∞ ( M ), called the potential function , and λ ∈ R satisfying Ric + ∇∇ f = (cid:18) R n −
1) + λ (cid:19) g. (1.1)Here ∇∇ f is the Hessian of f , R is the scalar curvature and Ric is the Ricci tensorof M . The soliton is called shrinking , steady or expanding , provided λ is positive,zero or negative, respectively. In this case we use the notation ( M n , g, f, λ ).More generally, one can define a gradient ρ -Einstein manifold, ρ ∈ R , as aRiemannian manifold ( M, g ) for which a similar equation holds, namely when
Ric + ∇∇ f = ( ρR + λ ) g (1.2)is satisfied for f ∈ C ∞ ( M ), also called potential function , and λ ∈ R . Withthis terminology, Schouten solitons are 1 / n − ρ = 0, for that leads to the Ricci solitons, which areimportant since they can be realized as self-similar solutions of the Ricci flow, andthe latter played a crucial rule in the proof of the Poincar´e conjecture. Examples of ρ -Einstein manifolds were given in [2] by Catino and Mazzieri, where this conceptwas introduced.Similar to Ricci solitons, it was shown in [3] that for each ρ these metrics arerealized as self-similar solutions of a geometric flow known as Ricci-Bourguignonflow (associated to the fixed ρ ). Such flow was shown to have short time solutionfor any initial metric whenever ρ < / n −
1) and M is compact (see [4] and [7]for more details). On the other hand, if ρ > / n −
1) solutions are not likely toexist for any initial metric, even when M is compact [4]. Consequently the value1 / n −
1) is where the parameter ρ bifurcates and this makes it special among the Date : February 11, 2021.2000
Mathematics Subject Classification.
Primary 35Q51, 53B20, 53C20, 53C25; Secondary34A40.
Key words and phrases.
Schouten Soliton, Schouten Flow, Gradient Estimate, VolumeEstimate. other values. At this stage, it is worth saying that the following problem is stillunsolved:
Question 1 (Short time existence for the Schouten flow, [4]) . Let ( M n , g ) bea compact Riemannian manifold. Are there T > and a family of Riemannianmetrics g ( t ) , t ∈ [0 , T ) , for which the IVP ∂∂t g ( t ) = − (cid:18) Ric ( t ) − R ( t )2( n − g ( t ) (cid:19) (1.3) g (0) = g holds true? Here R ( t ) and Ric ( t ) are the scalar and Ricci curvatures of g ( t ) . Thus, besides detecting which geometries allow equation (1 .
1) to have solution,another motivation to investigate Schouten solitons is that, as in the Ricci flowsetting, it is expected that a solid knowledge on self-similar solutions of (1 .
3) mayhelp to understand a general solution of it and, perhaps, help to solve Question 1.The simplest examples of Schouten solitons are Einstein manifolds. Other ex-amples are obtained as follows. Given n ≥ k ≤ n and λ ∈ R , consider anEinstein manifold ( N k , g ) of dimension k ≤ n and scalar curvature R N = n − kλ n − − k ,and the potential function f ( x, p ) = (cid:0) R N n − + λ (cid:1) k x k (Euclidean norm), ( x, p ) ∈ R n − k × N k . Then ( R n − k × Γ N k , g, f, λ ) is an n dimensional Schouten soliton, where g = h , i + g N and Γ acts freely on N and by orthogonal transformations on R n − k .A Schouten soliton is called rigid if it is isometric to one of those described above[8]. In [2] the following classes of Schouten solitons were proven to be rigid:(1) compact ;(2) complete noncompact with λ = 0 and n ≥ complete noncompact with λ > and n = 3;(4) complete warped products B × h N n − (including those with rotational sym-metry), where B is a one dimensional manifold, h : B → R is a positivesmooth function and N n − a space form. The goal of this paper is to investigate expanding and shrinking complete non-compact Schouten solitons. By the rigidity results mentioned above this is the bestscenario to consider. Our main result is the following
Theorem 1.1.
Let ( M n , g, f, λ ) , λ = 0 , be a complete noncompact Schouten solitonwith f nonconstant. If λ > λ < , respectively ) , then the potential function f attains a global minimum ( maximum, respectively ) and is unbounded above ( below,respectively ) . Furthermore, ≤ λR ≤ n − λ , (1.4) 2 λ ( f − f ) ≤ |∇ f | ≤ λ ( f − f ) , (1.5) with f = min p ∈ M f ( p ) , if λ > , ( f = max p ∈ M f ( p ) , if λ < , respectively ) . The theorem above can be seen as the analog of Hamilton’s identity for Riccisolitons, namely R + |∇ f | − λf = c , (1.6) N COMPLETE GRADIENT SCHOUTEN SOLITONS 3 for a c ∈ R . The equation above is central in obtaining many results concerningclassification of Ricci solitons as well as general geometric properties of these man-ifolds, as one can see for example in [1, 6, 8] and references therein. Its proof relieson a computation using (1 . ρ = 0, with the fixed direction ∇ f , the secondcontracted Bianchi identity and the Bochner formula [6]. Such approach does notgive a similar result when ρ = 0 and much effort has to be done to proof Theorem1.1 (see sections 2 and 3). We first find an ordinary differential inequality satisfiedby |∇ f | along suitable curves (see Proposition 2.2). Then we describe the behaviorof solutions of such inequality, obtaining various bounds (see Propositions 2.6, 2.8,2.9 and 3.3) which are used to prove Theorem 1.1.When λ >
0, the left hand side of (1 .
4) was first proved in [3] using a parabolicMaximum Principle. We also point out that (1 .
5) has appeared in [2] and [3] underadditional assumptions for other values of ρ . Remark . If one replaces the factor ρR + λ by a general smooth function˜ λ ∈ C ∞ ( M ), one obtains the concept of gradient Ricci almost solitons. Thiswas introduced in [9] as a generalization of Ricci solitons. Notice that this con-cept encloses all ρ -Einstein manifolds as particular cases. One says that the Riccialmost solitons is shrinking, steady, expanding or undefined provided the function˜ λ is positive, null, negative or changes sign. It follows from Theorem 1.1 that ashrinking (resp., expanding) Schouten soliton is automatically a shrinking (resp.,expanding) Ricci almost soliton.Now we state results concerning the growth of the potential function and volumeof shrinking Schouten solitons. These results are based in Theorem 1.1 in the sameway their counterparts are based in (1 . .
5) and the secondvariation of arc length to provide optimal lower and upper growth estimates on thepotential function of a shrinking Schouten soliton. More precisely:
Theorem 1.3.
Let ( M n , g, f, λ ) be a complete noncompact shrinking Schouten soli-ton with f nonconstant, f = min p ∈ M f ( p ) and q ∈ M . Then λ d ( p ) − A ) + f ≤ f ( p ) ≤ λ ( d ( p ) + A ) + f , (1.7) where d ( p ) = d ( p, q ) > , A and A is a positive constant depending on λ and thegeometry of the soliton on the unity ball B q (1) . It is worth mentioning that the rigid examples mentioned above show that (1 . .
7) is not true ingeneral for expanding Schouten solitons.Next we provide volume estimates for geodesic balls on complete Schouten soli-tons. Related results can be obtained in [1] and [5].
Theorem 1.4.
Let ( M n , g, f, λ ) be a complete noncompact shrinking Schouten soli-ton with f nonconstant, q ∈ M , δ = inf p ∈ M R ( p ) and θ = sup p ∈ M R ( p ) . Then there arepositive constants C , C and r , depending only on n and λ , so that C r n − ( n − θ n − λ ≤ vol ( B r ( q )) ≤ C r n − ( n − δ n − λ . (1.8) for any r > r . In particular V ol ( M ) = + ∞ . VALTER BORGES
By Theorem 1.1 (item (1 . δ, θ ∈ [0 , n − λ ]. Therefore, (1 .
8) im-plies that the volume growth of geodesic balls in a complete noncompact shrinkingSchouten soliton is at least linear and at most Euclidean. Also, both estimates areoptimal, as one can verify the equality for the rigid Schouten solitons describedearlier.Although the volume of M is infinite, we have the following information on theweighted volume of M . Theorem 1.5.
Let ( M n , g, f, λ ) be a complete noncompact shrinking Schouten soli-ton with f nonconstant. Then, V ol f ( M ) = Z M e − f dV < + ∞ . (1.9)In other words, the corresponding measure metric space ( M n , e − f dV ) has finitevolume. The theorem above is a consequence of the arguments in [5]. See alsoSection 4 for a sketch of the proof.This paper is organized in the following way. In Section 2, after setting notation,we provide the basic tools for the development of the paper. In this section we proveProposition 2.2 and Lemma 2.7, which play a major role in our approach. Theformer provides our claimed differential inequality for |∇ f | . The latter shows that acertain geometric quantity is monotone along suitable curves, which is fundamentalin obtaining certain estimates for |∇ f | . In Section 3 we give the proof of Theorem1.1 and in Section 4 we prove Theorem 1.3, Theorem 1.4 and Theorem 1.5.2. Preliminary Results
We start this section with some important identities on Schouten solitons. Forits proof see [2].
Proposition 2.1 ([2]) . If ( M n , g, f, λ ) is a gradient Schouten soliton, then ∆ f = nλ − n − n − R, (2.1) Ric ( ∇ f, X ) = 0 , ∀ X ∈ X ( M ) , (2.2) h∇ f, ∇ R i + (cid:18) Rn − λ (cid:19) R = 2 | Ric | . (2.3)Now let p ∈ M be a regular point of f and α p : ( ω ( p ) , ω ( p )) → M the maximalintegral curve of ∇ f |∇ f | through p . When the dependence on p is irrelevant to thecontext we write α : ( ω , ω ) → M . Below we recall some facts about α .(1) Since ( f ◦ α ) ′ ( s ) = 1 , ∀ s ∈ ( ω , ω ), for any [ s , s ] ⊂ ( ω , ω ) we have( f ◦ α )( s ) − ( f ◦ α )( s ) = s − s , (2.4) that is, f ◦ α is a linear function of s .(2) Fix s ∈ ( ω , ω ) and consider the diffeomorphism t ( s ) = Z ss dξ |∇ f ( α ( ξ )) | , s ∈ ( ω , ω ) . (2.5) N COMPLETE GRADIENT SCHOUTEN SOLITONS 5
Observe that for each s > s , t ( s ) is the length of α restricted to [ s , s ].Consequently, for any [ s , s ] ⊂ ( ω , ω ) we have d ( α ( s ) , α ( s )) ≤ Z s s dξ |∇ f ( α p ( ξ )) | . (2.6) Denote by s ( t ) the inverse of t ( s ). Then the curve( α ◦ s )( t ) = α ( s ( t ))(2.7) is an integral curve of ∇ f |∇ f | . Furthermore, a standard computation showsthat ( α ◦ s )( t ) is a geodesic for each t .The next result, although simple to proof, is of major importance, for it allowsto estimate |∇ f | along α ( s ), as we will see in the subsequent results. Proposition 2.2.
Let ( M n , g, f, λ ) , λ = 0 , be a Schouten soliton with f non-constant and α ( s ) , s ∈ ( ω , ω ) , a maximal integral curve of ∇ f |∇ f | . The function b : ( ω , ω ) → R , defined by b ( s ) = |∇ f ( α ( s )) | . (2.8) satisfies the differential inequality (2.9) bb ′′ − ( b ′ ) + 6 λb ′ − λ ≥ , where b ′ and b ′′ are the first and the second derivative of b with respect to s .Proof. Consider the smooth function a : ( ω , ω ) → R given by a ( s ) = R ( α ( s )).From d ( |∇ f | )( X ) = 2 ∇∇ f ( X, ∇ f ) and equation (1 .
1) one has b ′ ( s ) = d ( |∇ f | )( α ′ ( s ))= (cid:18) R ( α ( s )) n − λ (cid:19) df ( α ′ ( s ))= a ( s ) n − λ, (2.10)which after differentiating gives a ′ ( s ) = ( n − b ′′ ( s ). Consequently, h∇ f ( α ( s )) , ∇ R ( α ( s )) i = |∇ f ( α ( s )) | dR ( α ′ ( s ))= b ( s ) a ′ ( s )= ( n − b ( s ) b ′′ ( s ) . (2.11)Putting (2 . .
10) and (2 .
11) together we have( n − b ( s ) b ′′ ( s ) + b ′ ( s )( b ′ ( s ) − λ )) = 2 | Ric | ( α ( s )) ≥ n − b ′ ( s ) − λ ) , where in the second line we have used the inequality ( n − | Ric | ≥ R , which isa consequence of (2 . .
10) once again. Consequently,(2.12) bb ′′ ≥ ( b ′ − λ )( b ′ − λ ) , This finishes the proof, since inequalities (2 .
12) and (2 .
9) are equivalent. (cid:3) If p ∈ M is a given regular point of f , the solution of (2 . associated to p is theone obtained as in (2 .
8) using the curve α p : ( ω ( p ) , ω ( p )) → M through p . VALTER BORGES
Remark . If b ( s ), s ∈ ( s , s ), is a solution of (2 .
9) with λ <
0, then φ ( z ) = b ( − z ), z ∈ ( − s , − s ), satisfies φφ zz − ( φ z ) + 6 µφ z − µ ≥ , with µ = − λ >
0. Here φ z is the derivative of φ with respect to r . Remark . Let ( R n − k × Γ N k , g, f, λ ) be the Schouten soliton described earlier.Then it has constant scalar curvature R = n − kλ n − − k and, if k ≤ n −
1, its potentialfunction f is not constant and |∇ f | = (cid:16) Rn − + 2 λ (cid:17) f . After a linear change ofcoordinates using (2 .
4) with the condition f ( α (0)) = 0 we can replace f by s obtaining a function b ( s ) = (cid:16) Rn − + 2 λ (cid:17) s . A simple computation shows that this b ( s ) is a solution of (2 . . ω , ω ) ⊂ R any fixed interval, possibly unbounded, and let b : ( ω , ω ) → R be a positive smooth solution of (2 . λ = 0. Note that the latter conditionprevents constant functions as solutions to such inequality. Also, in view of Remark2.3, their proofs may be presented only for the case where λ > . Lemma 2.5. If b ( s ) , s ∈ ( ω , ω ) , is a positive solution of (2 . , then it has thefollowing properties: (1) b ( s ) has at most one critical point, which must be a minimum. In particular, b ′ changes sign in ( ω , ω ) at most once; (2) If there is s ∈ ( ω , ω ) so that b ′ ( s ) > max { λ, λ } ( b ′ ( s ) < min { λ, λ } ,respectively ) , then b ′ ( s ) ≥ b ′ ( s ) , ∀ s ∈ [ s , ω ) ( b ′ ( s ) ≤ b ′ ( s ) , ∀ s ∈ ( ω , s ] ,respectively ) .Proof. To prove (1) assume by contradiction that s , s ∈ ( ω , ω ) are two distinctcritical points of b with s < s . Let i ∈ { , } . Using equation (2 .
9) we obtain b ( s i ) b ′′ ( s i ) ≥ λ >
0, and then s i is a local minimum of b , in virtue of b ( s i ) > s ∈ ( s , s ), that is b ′′ ( s ) ≤ b ′ ( s ) = 0.As before, the latter equality implies b ′′ ( s ) >
0, what is a contradiction.Now we will prove (2). If b ′ ( s ) > max { λ, λ } then equation (2 .
9) implies that b ( s ) b ′′ ( s ) ≥ ( b ′ ( s ) − λ )( b ′ ( s ) − λ ) > . (2.13)Since b ( s ) >
0, we conclude that b ′ is increasing around s , which implies that b ′ ( s ) ≥ b ′ ( s ) > max { λ, λ } , (2.14)for some s ≥ s . Consequently (2 .
13) is true for such s , which implies that (2 . s ∈ [ s , ω ). The case where b ′ ( s ) < min { λ, λ } is analogous. (cid:3) Proposition 2.6.
Let b : ( ω , ω ) → R be a positive smooth solution of (2 . and ˜ s ∈ ( ω , ω ) . If c = min n b (˜ s ) , b (˜ s ) e b ′ (˜ s )6 λ o > , then (1) b ( s ) ≥ c, ∀ s ∈ [˜ s, ω ) , if λ > ; (2) b ( s ) ≥ c, ∀ s ∈ ( ω , ˜ s ] , if λ < . N COMPLETE GRADIENT SCHOUTEN SOLITONS 7
Proof.
We will prove the proposition in the case where λ >
0. The proof of theremaining case is similar.Suppose that b ′ ( s ) ≤ , ∀ s ∈ [˜ s, ω ). Note that any solution b ( s ) of (2 .
9) satisfies( b ′ ( s ) + 6 λ ln( b ( s ))) ′ ≥ (( b ′ ( s )) + 8 λ ) b ( s ) > , what for each s ∈ [˜ s, ω ) implies e b ′ (˜ s ) ( b (˜ s )) λ ≤ e b ′ ( s ) ( b ( s )) λ . As a consequence,for each s ∈ [˜ s, ω ) we have b ( s ) ≥ b ( s ) e b ′ ( s )6 λ ≥ b (˜ s ) e b ′ (˜ s )6 λ ≥ c .Suppose that b ′ ( s ) ≥ , ∀ s ∈ [˜ s, ω ). Then b ( s ) ≥ b (˜ s ) ≥ c .Suppose that b ′ ( s ) changes sign on [˜ s, ω ). For Lemma 2.5 it happens exactlyonce at a point of minimum, which we denote by s ∈ (˜ s, ω ). Then b ′ ( s ) < ∀ s ∈ [˜ s, s ), b ′ ( s ) > ∀ s ∈ ( s , ω ), and b ′ ( s ) = 0. Repeating the argument ofthe first part we have b ( s ) ≥ c , ∀ s ∈ [˜ s, s ). In particular b ( s ) ≥ c . On the otherhand, since b is nondecreasing in [ s , ω ) we get b ( s ) ≥ b ( s ) ≥ c in this interval.This finishes the proof. (cid:3) A solution of (2 .
9) gives rise to a quantity which turns out to be monotone inmany situations, as the lemma below shows.
Lemma 2.7.
Let b ( s ) be a smooth solution of (2 . and consider the function σ ( s ) = ( b ′ ( s ) − λ ) b ( s )( b ′ ( s ) − λ ) , (2.15) defined whenever b ( s )( b ′ ( s ) − λ ) = 0 . If in addition b ′ ( s )( b ′ ( s ) − λ ) = 0 , then σ ′ ( s ) b ′ ( s )( b ′ ( s ) − λ ) ≥ . (2.16) Proof.
A straightforward computation involving (2 .
9) gives: σ ′ ( s ) b ′ ( s )( b ′ ( s ) − λ ) = 2( b ′ − λ ) bb ′′ − ( b ′ − λ )( b ′ ( b ′ − λ ) + bb ′′ ) b ′ b ( b ′ − λ ) = bb ′′ − ( b ′ − λ )( b ′ − λ ) b ( b ′ − λ ) ≥ , for each s such that b ( s ) b ′ ( s )( b ′ ( s ) − λ )( b ′ ( s ) − λ ) = 0. (cid:3) In presence of additional assumptions we can improve Proposition 2.6 by usingLemma 2.7 in the following way.
Proposition 2.8.
Let b : ( ω , ω ) → R be a positive smooth solution of (2 . . Ifthere is ˜ s ∈ ( ω , ω ) such that λ ( b ′ (˜ s ) − λ ) < , then there is a constant c > such that b ( s ) ≥ c , ∀ s ∈ ( ω , ω ) .Proof. Because λ ( b ′ (˜ s ) − λ ) < b (˜ s ) > c = min (cid:26) b (˜ s ) , − λσ (˜ s ) (cid:27) , (2.17)where σ is the function defined in (2 . b ( s ) ≥ c , ∀ s ∈ ( ω , ˜ s ], if λ > b ( s ) ≥ c , ∀ s ∈ [˜ s, ω ), if λ < VALTER BORGES
If we take c = min { c, c } , where c is the constant given by Proposition 2.6, then b ( s ) ≥ c , ∀ s ∈ ( ω , ω ), and the proposition is proved. In view of Remark 2.3, wemay assume that λ > b ′ ( s ) ≤ s ∈ ( ω , ˜ s ]. Then b ( s ) ≥ b (˜ s ) ≥ c , ∀ s ∈ ( ω, ˜ s ].Suppose that b ′ ( s ) ≥ s ∈ ( ω , ˜ s ]. Consider σ ( s ) , s ∈ ( ω , ˜ s ], defined asin (2 . σ ( s ) < σ ′ ( s ) ≤
0. Therefore, σ (˜ s ) ≤ σ ( s ) <
0. Equivalently, 0 ≤ ( b ′ ( s )) ≤ ( b ′ ( s ) − λ )( σ (˜ s ) b ( s ) + 8 λ ). On theother hand, item (2) of Lemma 2.5 implies that b ′ ( s ) < λ < λ , ∀ s ∈ ( ω , ˜ s ], whichgives σ (˜ s ) b ( s ) + 8 λ ≤
0, and then b ( s ) ≥ − λσ (˜ s ) ≥ c , ∀ s ∈ ( ω, ˜ s ].Suppose that b ′ ( s ) changes sign on ( ω , ˜ s ]. Let s ∈ ( ω , ˜ s ) be the only pointwhere b ( s ) takes its minimum, given by Lemma 2.5. Then b ′ ( s ) < ∀ s ∈ ( ω , s ),and b ′ ( s ) ≥ ∀ s ∈ [ s , ˜ s ]. Proceeding as in the end of the proof of Proposition 2.6we get b ( s ) ≥ − λσ (˜ s ) ≥ c > , ∀ s ∈ ( ω, ˜ s ]. (cid:3) The next result shows that certain solutions of (2 .
9) grow exponentially. Later,in Proposition 3.3, we show that those solutions cannot be constructed as in Propo-sition 2.2.
Proposition 2.9.
Let b : ( ω , ω ) → R be a positive smooth solution of (2 . . Ifthere is ˜ s ∈ ( ω , ω ) satisfying ( b ′ (˜ s ) − λ )( b ′ (˜ s ) − λ ) > , (2.18) and ( b ′ (˜ s ) − λ ) b ′ (˜ s ) > , (2.19) then there are positive constants κ and K so that (1) b ( s ) ≥ K e κs , ∀ s ∈ [˜ s, ω ) , if b ′ (˜ s ) > max { λ, λ } ; (2) b ( s ) ≥ K e − κs , ∀ s ∈ ( ω , ˜ s ] , if b ′ (˜ s ) < min { λ, λ } .Proof. In view of Remark 2.3 we assume that λ >
0. Inequality (2 .
18) gives rise totwo cases:
Case 1. b ′ (˜ s ) > λ > λ . It follows from item (2) of Lemma 2.5 that b ′ ( s ) > λ > λ, ∀ s ∈ [˜ s, ω ). If σ ( s )is defined as in (2 . σ ( s ) > σ ′ ( s ) ≥
0, andthen σ ( s ) ≥ ˜ σ = σ (˜ s ) > , ∀ s ∈ [˜ s, ω ). Equivalently, for each s ∈ [˜ s, ω ) we have( b ′ ( s )) − (˜ σb ( s ) + 8 λ ) b ′ ( s ) + 2 λ (˜ σb ( s ) + 8 λ ) ≥ , which gives B + ( s ) B − ( s ) ≥ , (2.20)with B + ( s ) = b ′ ( s ) − (cid:16) ˜ σb ( s ) + 8 λ + p ˜ σb ( s )(˜ σb ( s ) + 8 λ ) (cid:17) ,B − ( s ) = b ′ ( s ) − (cid:16) ˜ σb ( s ) + 8 λ − p ˜ σb ( s )(˜ σb ( s ) + 8 λ ) (cid:17) . (2.21)Once b ′ ( s ) > λ , we cannot have B − ( s ) <
0, which by (2 .
20) implies that B + ( s ) ≥
0, and then b ′ ( s ) ≥
12 (˜ σb ( s ) + 8 λ ) , ∀ s ∈ [˜ s, ω ) . N COMPLETE GRADIENT SCHOUTEN SOLITONS 9
Integrating the inequality above after dropping its last term we get b ( s ) ≥ K e κs , ∀ s ∈ [˜ s, ω ), with K = b (˜ s ) e − κ ˜ s > κ = σ (˜ s )4 >
0, which is (1).
Case 2. b ′ (˜ s ) < λ < λ . It follows from (2 .
19) that b ′ (˜ s ) < b ′ ( s ) < ∀ s ∈ ( ω , ˜ s ]. Consider φ ( z ) = b ( − z ) , z ∈ [˜ z, − ω ), with ˜ z = − ˜ s and µ = − λ . Then φ ( z ) satisfies (2 .
9) (see Remark 2.3) and φ z ( z ) >
0. Consider σ ( z )defined as in (2 .
15) using φ ( z ) and µ . It follows from Lemma 2.7 that σ ( z ) > σ z ( z ) ≥ ∀ z ∈ [˜ z, − ω ). Considering ˜ σ = σ (˜ z ) > σφ ( z ) + 8 µ > , ∀ z ∈ [˜ z, − ω ) . (2.22)To see this first notice that by the definition of ˜ σ we have (˜ σφ (˜ z )+8 µ )( φ z (˜ z ) − µ ) =( φ z (˜ z )) . This proves (2 .
22) at ˜ z . Using that φ is increasing in [˜ z, − ω ), (2 .
22) isproved in this interval. Considering functions as in (2 .
21) and proceeding as before,we get φ z ( z ) ≥ ˜ σφ ( z ) + 8 µ, ∀ z ∈ [˜ z, − ω ). Returning to the variable s we get b ′ ( s ) ≤ − ˜ σb ( s ) + 8 λ, ∀ s ∈ ( ω , ˜ s ] , which after integrating gives b ( s ) ≥ K e − κs , ∀ s ∈ ( ω , ˜ s ], with κ = ˜ σ > K = (cid:0) b (˜ s ) − λ ˜ σ (cid:1) e ˜ σ ˜ s = (cid:0) φ (˜ z ) + µ ˜ σ (cid:1) e − ˜ σ ˜ z >
0, by (2 . (cid:3) Estimates on the Gradient of f In this section we use the estimates obtained in the previous one to prove The-orem 1.1 and Theorem 1.3. Our first step is to use Proposition 2.6 to show that f is unbounded and has at most one critical level. Proposition 3.1.
Let ( M n , g, f, λ ) , λ = 0 , be a complete Schouten soliton with f nonconstant. If p ∈ M is a critical point of f , then (1) f ( p ) = inf M f , if λ > ; (2) f ( p ) = sup M f , if λ < .Proof. Assume that λ > p is not a global minimum of f . Asthe set of regular values of f is dense in f ( M ), there is a regular value f of f suchthat f < f ( p ). Since f − ( f ) is a closed submanifold of M , there are p ∈ f − ( f )and a normalized geodesic γ : [0 , l ] → M so that γ (0) = p and γ ( l ) = p . Since γ is orthogonal to f − ( f ) at p , the uniqueness property of geodesics asserts that γ ( t ) = ( α ◦ s )( t ) as long as γ does not hit a critical point of f , where α ( s ) and s ( t )are given as in (2 .
5) and (2 . t ∗ ∈ (0 , l ] be the first parameter satisfying ∇ f ( γ ( t ∗ )) = 0. Using (2 .
5) wewrite α ( s ) = γ ( t ( s )) , s ∈ [˜ s, s ∗ ), where [˜ s, s ∗ ) is the image of [0 , t ∗ ) by t ( s ) and α (˜ s ) = p . By Proposition 2.2, the function b ( s ) = |∇ f ( α ( s )) | , s ∈ [ s , s ), is asolution of (2 .
9) and then, by Proposition 2.6, there is a constant c > b ( s ) ≥ c , for all s ∈ [˜ s, s ∗ ). Therefore, |∇ f ( γ ( t )) | ≥ c, ∀ t ∈ [0 , t ∗ ), what is acontradiction, in view of |∇ f ( γ ( t ∗ )) | = 0. This proves (1).If λ < . − f )( p ) = min M ( − f ), which finishes the proof of (2). (cid:3) Proposition 3.2.
Let ( M n , g, f, λ ) , λ = 0 , be a complete Schouten soliton and p ∈ M a regular point of f . If b : ( ω ( p ) , ω ( p )) → R is defined as in (2 . , then (1) ω ( p ) = + ∞ and sup M f = + ∞ , if λ > ; (2) ω ( p ) = −∞ and inf M f = −∞ , if λ < .Proof. Assume that λ >
0, otherwise we use Remark (2 . p ∈ M such that ω ( p ) < + ∞ . Let α p : ( ω ( p ) , ω ( p )) → M be the maximal integral curve of ∇ f / |∇ f | so that α p (˜ s ) = p , for some ˜ s ∈ ( ω ( p ) , ω ( p )). If b : [˜ s, ω ( p )) → R is defined as in(2 .
8) using α p , Lemma 2.6 gives a constant c > |∇ f ( α p ( s )) | = p b ( s ) ≥ √ c, (3.1)for s ∈ [˜ s, ω ( p )). Let ( s k ) k be a monotone sequence in [˜ s, ω ( p )) converging to ω ( p ). By (2 .
6) we have d ( α p ( s k ) , α p ( s l )) ≤ √ c | s k − s l | , and then ( α p ( s k )) k converges to a point p ∈ M . By continuity, (3 .
1) implies that |∇ f ( p ) | ≥ √ c , and then p is a regular point. This is a contradiction, since theintegral curve α p of ∇ f / |∇ f | through p extends α p past ω ( p ), contradictingthe maximality of ω ( p ). Then ω ( p ) = + ∞ , for any regular point p ∈ M of f .To see that f is unbounded above, use (2 .
4) and the first part of this proof. (cid:3)
The next proposition gives a necessary condition for a solution of (2 .
9) to beconstructed as in (2 . Proposition 3.3.
Let ( M n , g, f, λ ) , λ = 0 , be a complete Schouten soliton with f nonconstant and b : ( ω , ω ) → R defined as in (2 . . Then ( b ′ ( s ) − λ )( b ′ ( s ) − λ ) ≤ , ∀ s ∈ ( ω , ω ) . (3.2) Proof.
Suppose by contradiction that (3 .
2) is not true and let ˜ s ∈ ( ω , ω ) satisfying(2 . λ >
0, otherwise use Remark 2.3.Assume that b ′ (˜ s ) > λ > λ . Let γ ( t ) be the reparametrization of α ( s ) givenby (2 . t, + ∞ ), where γ (˜ t ) = α (˜ s ). Denote f ( γ ( t )) by f ( t ), t ∈ [˜ t, + ∞ ), andnote that its derivative with respect to t is ˙ f ( t ) = |∇ f ( γ ( t )) | . Using Proposition2.9, (1), we can find positive constants K and κ so that˙ f ( t ) ≥ K e κ f ( t ) , t ∈ [˜ t, + ∞ ) , where we have used that s = f ( t ) + ˜ s − f (˜ t ). Integrating the above inequality from0 to t we have t − ˜ t ≤ κ K ( e − κ f (˜ t ) − e − κ f ( t ) ) ≤ e − κ f (˜ t ) κ K , what gives a contradiction when t → + ∞ .Assume that b ′ (˜ s ) < λ < λ . We will first prove the following claim. Claim 1. ( ω , ω ) = R and there is s < ˜ s satisfying (2 . and (2 . .Proof. Notice that according to Proposition 2.8 there is a positive constant c sothat b ( s ) ≥ c , ∀ s ∈ ( ω , ω ). Arguing as in the proof of Proposition 3.2, thisimplies that ( ω , ω ) = R . If we show the existence of s < ˜ s so that b ′ ( s ) < N COMPLETE GRADIENT SCHOUTEN SOLITONS 11 then the proof of the second part is finished. Assume by contradiction that b isnondecreasing on ( −∞ , ˜ s ]. Then (2 .
12) implies that b ′′ ( s ) ≥ ˜ a, ∀ s ∈ ( −∞ , ˜ s ], where˜ a = (2 λ − b ′ (˜ s ))(4 λ − b ′ (˜ s )) b (˜ s ) > , once b ′ ( s ) < λ and b ′′ ( s ) >
0. But then 0 ≤ b ′ ( s ) ≤ ˜ as + b ′ (˜ s ) − ˜ a ˜ s , which providesa contradiction when s → −∞ . (cid:3) Let s be given as in Claim 1. As ω = −∞ , it follows that γ ( t ) and f ( t ), definedas before, are defined for t ∈ ( −∞ , t ] where γ ( t ) = α ( s ). Using Proposition 2.9,item (2), we have ˙ f ( t ) ≥ K e − κ f ( t ) , t ∈ ( −∞ , t ] , for some constants K and κ . Integrating the above inequality from t to t implies t − t ≤ κ K ( e κ f ( t ) − e κ f ( t ) ) ≤ e κ f ( t ) κ K , what gives a contradiction when t → −∞ . (cid:3) Remark . Inequality (3 .
2) is optimal. To see that consider the function b ( s ) ofRemark 2.4. Then b ′ ( s ) = n − λ n − − k , with k ∈ { , , · · · , n − } , which clearly satisfy(3 . k = 0 or k = n − Proof of Theorem 1.1.
Assume that λ >
0, otherwise we use Remark 2.3 to goback to this case. • Proof of (1.4).
Assume that p be a regular point of f . Let b ( s ) be the function definedas (2 .
8) and s ∈ ( ω ( p ) , ω ( p )) with α ( s ) = p . Then b ′ ( s ) = R ( p ) n − λ ,and using (3 .
2) we have0 ≤ R ( p ) ≤ n − λ, (3.3) as desired. Now assume that p is a critical point of f . If we can approximate p with a sequence of regular points of f , the result follows by continuity.Then we will show that this is always the case. In order to do that, assumeby contradiction that the set C of critical points of f has nonempty interior.Let q be a point in the boundary of the interior of C . By approximating q by a sequence in the interior of C we get R ( q ) = 2( n − nλn − , (3.4) where we have used (2 .
1) and the fact that ∆ f vanishes in the interior of C . On the other hand, by approximating q by a sequence lying outside C and using (3 .
3) we get R ( q ) ≤ n − λ. (3.5) If we put (3 .
4) and (3 .
5) together we get a contradiction. • Proof of f ( M ) = [ f , + ∞ ) . By Proposition 3.1 and Proposition 3.2 we only need to show that f has acritical point. Let p ∈ M be a regular point of f and b : ( ω , + ∞ ) → R thecorresponding solution of (2 . λ ≤ b ′ ( s ) ≤ λ, ∀ s ∈ ( ω , + ∞ ), which gives2 λ ( s − s ) ≤ b ( s ) − b ( s ) ≤ λ ( s − s ) , (3.6) with s ≤ s in ( ω , + ∞ ). We claim that ω > −∞ . Suppose by contra-diction that ω = −∞ . Using (3 .
6) we have for s ∈ ( −∞ , s ] b ( s ) ≤ λs + b ( s ) − λs , showing that b ( s ) must become negative somewhere below s , which is acontradiction.Now let ( s k ) k be a sequence so that lim k → + ∞ s k = ω . It follows from (3 . b ( s k )) k is a Cauchy sequence, and hence it converges to b ∈ [0 , + ∞ ).It is not hard to see that b does not depend on the choice of the sequence( s k ) k . Letting s → ω in (3 .
6) we obtain2 λ ( s − ω ) + b ≤ b ( s ) ≤ λ ( s − ω ) + b . (3.7) On the other hand, if s l < s k it follows from (2 .
6) and (3 . d ( α ( s k ) , α ( s l )) ≤ Z s k s l ds p λ ( s − ω ) + b = r λ (cid:16)p s k + C − p s l + C (cid:17) with C = b λ − ω . This shows that ( α ( s k )) k is a Cauchy sequence andthen it converges to a certain p ∈ M . Again, p does not depend on thechoice of the sequence. Then, lim s → ω |∇ f ( α ( s )) | = |∇ f ( p ) | = b . Now,since ω > −∞ and ( ω , + ∞ ) is the maximal interval of definition of b ( s ),we must have |∇ f ( p ) | = b = 0, and then f ( p ) = f = min M f . • Proof of (1.5).
Let p ∈ M be a regular point of f , b : ( ω , + ∞ ) → R the correspondingsolution of (2 .
9) and p ∈ M a critical point of f as constructed above. Let s ∈ ( ω , + ∞ ) so that α ( s ) = p . Using (2 .
4) and b ( s ) = |∇ f ( α ( s )) | in(3 .
6) and making s → −∞ we have the desired inequality at p , namely2 λ ( f ( p ) − f ) ≤ |∇ f ( p ) | ≤ λ ( f ( p ) − f ) . If p ∈ M is a critical point of f , then (1 .
5) is trivially satisfied since, aswe have seen, f ( p ) = f and ∇ f ( p ) = 0. (cid:3) Growth Estimates
In this section we show that (1 .
4) and (1 .
5) can be used to investigate Schoutensolitons in a similar way (1 .
6) is used to understand Ricci solitons. This is doneby proving Theorem 1.3 and Theorem 1.4, whose versions to Ricci solitons wereproved in [1].
N COMPLETE GRADIENT SCHOUTEN SOLITONS 13
Proof of Theorem 1.3.
Let q ∈ M and d ( q ) = d ( p, q ). We follow closely theproof of Proposition 1.2 in [1]. It follows from the second inequality of (1 .
5) that (cid:12)(cid:12)(cid:12) ∇ p f − f (cid:12)(cid:12)(cid:12) ≤ p | λ | . (4.1)This implies that √ f − f is Lipschitz, and then f ( p ) ≤ | λ | d ( p ) + f ( q ) − f p | λ | ! + f . Assume that λ >
0. Fix q ∈ M , let p ∈ M be a point so that d ( p, q ) = l > γ : [0 , l ] → M be a minimizing geodesic, with γ (0) = q and γ ( l ) = p . Denoting f ( γ ( s )) by f ( s ) and its derivative by ˙ f ( s ) we get˙ f ( l − − ˙ f (1) = Z l − ∇∇ f ( ˙ γ ( s ) , ˙ γ ( s )) ds = Z l − (cid:18) R n −
1) + λ − Ric ( ˙ γ ( s ) , ˙ γ ( s )) (cid:19) ds ≥ λ ( l − − Z l − Ric ( ˙ γ ( s ) , ˙ γ ( s )) ds. (4.2)In order to estimate the last term in the inequality above we use the secondvariational formula. Since γ : [0 , l ] → M is minimizing, for any continuous piecewiseregular function ϕ : [0 , l ] → R satisfying ϕ (0) = ϕ ( l ) = 0 one has Z l ( ϕ ( s )) Ric ( ˙ γ ( s ) , ˙ γ ( s )) ds ≤ ( n − Z l | ˙ ϕ ( s ) | ds. (4.3)Consider the function ϕ ( s ) = s, if s ∈ [0 , , , if s ∈ [1 , l − ,l − s, if s ∈ [ l − , l ] . (4.4)Then Z l − Ric ( ˙ γ, ˙ γ ) ds = Z l ϕ Ric ( ˙ γ, ˙ γ ) ds − Z ϕ Ric ( ˙ γ, ˙ γ ) ds − Z ll − ϕ Ric ( ˙ γ, ˙ γ ) ds ≤ ( n − Z l | ˙ ϕ | ds + max B q (1) | Ric | − Z ll − ϕ Ric ( ˙ γ, ˙ γ ) ds = 2( n −
1) + max B q (1) | Ric | − Z ll − ϕ Ric ( ˙ γ, ˙ γ ) ds On the other hand, the last term above can be estimated in the following way Z ll − ϕ Ric ( ˙ γ, ˙ γ ) ds ≥ λ − Z ll − ( ϕ ( s )) ∇∇ f ( ˙ γ ( s ) , ˙ γ ( s )) ds = λ − ( ϕ ( l )) ˙ f ( l ) + ( ϕ ( l − ˙ f ( l − − Z ll − ϕ ( s ) ˙ f ( s ) ds ≥ λ f ( l − − max l − ≤ s ≤ l | ˙ f ( s ) | , which gives Z l − Ric ( ˙ γ, ˙ γ ) ds ≤ n − − λ − ˙ f ( l −
1) + max B q (1) | Ric | + max l − ≤ s ≤ l | ˙ f ( s ) | and by (4 . l − ≤ s ≤ l | ˙ f ( s ) | ≥ λl − n − − λ f (1) − max B q (1) | Ric | . (4.5)For each s ∈ [0 , l ], using (4 . (cid:12)(cid:12)(cid:12)p f ( s ) − f − p f ( l ) − f (cid:12)(cid:12)(cid:12) ≤ √ λ ( l − s ) ≤ √ λ, and then p f ( s ) − f ≤ p f ( l ) − f + √ λ , which for (1 .
5) givesmax l − ≤ s ≤ l | ˙ f ( s ) | ≤ max l − ≤ s ≤ l p λ ( f ( s ) − f ) ≤ p λ ( f ( l ) − f ) + 2 λ. (4.6)Analogously | ˙ f (1) | ≤ max B q (1) |∇ f | ≤ max B q (1) p λ ( f − f ) + 2 λ. (4.7)Putting (4 . .
6) and (4 .
7) together we conclude that f ( p ) ≥ λ d ( p, q ) − A ) + f , where A = 1 λ (cid:18) n −
1) + 17 λ B q (1) np λ ( f − f ) + | Ric | o(cid:19) (cid:3) Now we will prove the growth estimate for the volume of geodesic balls of com-plete shrinking Schouten solitons, announced in Theorem 1.4.Fix q ∈ M , consider r ( p ) = d ( p, q ) and ρ ( p ) = 2 p f ( p ) − f . According to (4 . ∇ ρ = 1 √ f − f ∇ ( f − f ) , (4.8)and if r is large enough (1 .
7) imply that √ λ ( r ( p ) − A ) ≤ ρ ( p ) ≤ √ λ ( r ( p ) + A ) , (4.9)where A = max { A , A } . According to Theorem 1.3, the set D ( r ) = { p ∈ M ; ρ ( p ) < r } is precompact and then has finite volume, that is V ( r ) = Z D ( r ) dV = Z r Z ∂D ( s ) dS |∇ ρ | ! ds < + ∞ , (4.10)where in the second equality we used the Coarea Formula.In the next result we present estimates for V ( r ). This is going to be importantin the proof of Theorem 1.4 N COMPLETE GRADIENT SCHOUTEN SOLITONS 15
Proposition 4.1.
Let ( M, g, f, λ ) be a complete noncompact shrinking Schoutensoliton. Consider the real numbers θ = sup p ∈ M R ( p ) , δ = inf p ∈ M R ( p ) and r > . Then V ( r ) r n − ( n − θ n − λ r n − ( n − θ n − λ ≤ V ( r ) ≤ V ( r ) r n − ( n − δ n − λ r n − ( n − δ n − λ , (4.11) for any r > r .Proof. Integrating (2 .
1) on the set D ( r ) and using the Divergence Theorem gives Z ∂D ( r ) |∇ ( f − f ) | dS = nλV ( r ) − n − n − Z D ( r ) RdV. (4.12)On the other hand, (4 .
10) gives dVdr ( r ) = Z ∂D ( r ) dS |∇ ρ | Taking (4 .
8) into account, the first inequality of (1 .
5) gives Z ∂D ( r ) |∇ ( f − f ) | dS ≥ λ Z ∂D ( r ) ( f − f ) dS |∇ ( f − f ) | = λr Z ∂D ( r ) dS |∇ ρ | = λr dVdr , and the second inequality of (1 . Z ∂D ( r ) |∇ ( f − f ) | dS ≤ λr dVdr . Plugging all this with (4 .
12) implies λr dVdr ≤ nλV − n − n − Z D ( r ) RdV ≤ (cid:18) nλ − ( n − δ n − (cid:19) V, and 2 λr dVdr ≥ nλV − n − n − Z D ( r ) RdV ≥ (cid:18) nλ − ( n − θ n − (cid:19) V. Integrating these inequalities from r > r > r we get (4 . (cid:3) As a consequence of Proposition 4.1 we have
Proof of Theorem 1.4.
It follows from (4 .
9) that D ( √ λ ( r − A )) ⊂ B q ( r ) ⊂ D (2 √ λ ( r + A )) for r sufficiently large. Taking r > C and C so that C r n − ( n − δ n − λ ≤ V ( √ λ ( r − A ) ≤ V ol ( B q ( r )) ≤ V (2 √ λ ( r + A )) ≤ C r n − n − n − λ θ , for every r > r . In particular, since δ ≤ n − λ , we have V ol ( B q ( r )) ≥ C r, for all r ≥ r . This completes the prove. (cid:3) We end this paper proving that the weighted volume of M with respect to thefunction f − f is finite. This opens new possibilities of investigation, once weightedmanifolds have been a topic of constant development in recent years. The proofof this result follows closely the proof of Theorem 2.1 of [5]. We observe thatafter normalization issues we could directly apply the main result of [5]. Since wedecided do not make any normalization, we present a sketch of the proof, for theconvenience of the reader. For more details see the referred article. Proposition 4.2 ([5]) . Let ( M, g, f, λ ) be a complete noncompact shrinking Schoutensoliton. Then (4.13) Z M e − ( f − f ) dV < + ∞ , and, for any r > , (4.14) V ( r ) ≤ (cid:18) e Z M e − ( f − f ) dV (cid:19) r n − n − n − λ θ . Proof.
Consider the function F ( t ) = 1 t k Z D ( r ) e − t ( f − f ) dV, for k = n − n − n − λ δ , where δ = inf p ∈ M R ( p ) ∈ [0 , n − λ ]. Then dFdt = t − k − Z D ( r ) (cid:18) f − f t − k (cid:19) dV. On the other hand, using items (1 .
4) and (1 .
5) of Theorem 1.1, we have Z D ( r ) div ( e − t ( f − f ) ∇ ( f − f )) dV ≤ − λ Z D ( r ) (cid:18) f − f t − k (cid:19) dV = t k +1 dFdt Using the Divergence Theorem the inequality above gives dFdt ( t ) ≤ − t − k − λ Z ∂D ( r ) e − t ( f − f ) |∇ ( f − f ) | dS ≤ . Integrating this inequality from 1 to r > e − V ( r ) ≤ Z D ( r ) e − r ( f − f ) dV ≤ r k Z D ( r ) e − ( f − f ) dV, (4.15)where the first inequality comes from the fact that on the set D ( r ) we have r ≥ f − f ). Using the inequality above it is possible to show (see [5] for details) that Z D ( r + m ) e − ( f − f ) dV ≤ − e − r Z D ( r + m − e − ( f − f ) dV, for any integer m ≥
1. Iterating the last inequality we have (4 .
13) which, togetherwith (4 . . (cid:3) N COMPLETE GRADIENT SCHOUTEN SOLITONS 17
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