Notes 11: Shrinking GRS and quadratic curvature decay

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1 Notes 11: Shrinking GRS and quadratic curvature decay In this section, by applications of the maximum principle, we prove a pair of quadratic decay estimates for the curvature. The upper bound requires the assumption that the curvature tends to zero at infinity. We begin by considering a lower bound for the scalar curvature R. We have We compute that f R = R 2 Rc 2 R. (1) f (f 1 ) = f 1 f 2 ( n 2 2 f 2 f f (f 2 ) = 2f 2 f 3 ( n 6 f 2 f ), (2) ). (3) Assume that our shrinking GRS is non-gaussian, so that R > 0. We shall show that the scalar curvature decays at most inverse quadratically in distance. Using the above functions to construct a lower barrier for R, we first compute for c R +, ( ) f R cf 1 = R cf 1 + cf 2 ( n 2 2 f 2 ) 2 Rc 2. (4) f To correct for the (bad) positive term nc 2 f 2 on the rhs, we consider the quantity ϕ R c (f 1 + nf 2 ). We compute that ( ) f f ϕ = 2 Rc 2 + ϕ cnf 3 2 n cf 4 (2f + 6n) f 2 (5) ( ) f ϕ cnf 3 2 n. Remarkably, each term on the rhs is good. Since R > 0, by choosing c > 0 sufficiently small, we may assume that ϕ > 0 in B 8n (o), where o is a minimum point of f. Suppose that inf M B8n (o) ϕ < 0. Then, since lim inf x ϕ(x) 0, a negative minimum of ϕ is attained at some point x 0, where x 0 M B 8n (o). By the maximum principle, f ϕ(x 0 ) 0. Hence (5) implies f(x 0 ) n < 0. (6) 2 On the other hand, since d(x 0, o) 8n, we have f(x 0 ) 9 4 n2, yielding a contradiction. We conclude that R cf 1 ϕ 0 on all of M. (7) Summarizing, we have proved a scalar curvature gap theorem due to the work of Lu, Yang, and the author combined with an earlier version due to Ni and Wilking. Theorem 1 For any non-gaussian shrinking GRS (M n, g, f) there exists c > 0 such that R(x) c (d(x, o) + 1) 2. (8) Therefore any asymptotic Riemannian cone of (M, g) has positive scalar curvature and hence is nonflat. 1

2 Definition 2 Let (M n, g) be a complete noncompact Riemannian manifold. Its asymptotic scalar curvature ratio is ASCR (g) lim sup R (x) d (x, o) 2, (9) d(x,o) where o M. This definition is independent of the choice of o. Next, we consider an upper bound for Rm assuming that Rm 0. The idea of the proof is the same as Theorem 1, except that some signs are flipped and one has to be more careful with the choice of constants in the barrier function. We have p R pjkl = l R kj k R lj = R lkjm m f. Hence, by applying i to this equation and using the commutator formula for an r-tensor β, we obtain Now i j β k1 k r j i β k1 k r i R lkjm m f + R lkjm i m f = r h=1 p=1 n R p ijk h β k1 k h 1 p k h + 1 k r (10) = i p R pjkl = p i R pjkl R ippq R qjkl R ipjq R pqkl R ipkq R pjql R iplq R pjkq. p i R pjkl = p p R ijkl + p j R pikl = R ijkl + j p R pikl R pjpq R qikl R pjiq R pqkl R pjkq R piql R pjlq R pikq (here, R ijkl ( Rm) ijkl denote the components of the rough Laplacian of the Riemann curvature tensor) and j p R pikl = j (R lkim m f). Hence which we may rewrite as R ijkl = i R lkjm m f + R lkjm i m f + R ippq R qjkl + R ipjq R pqkl + R ipkq R pjql + R iplq R pjkq j R lkim m f R lkim j m f + R pjpq R qikl + R pjiq R pqkl + R pjkq R piql + R pjlq R pikq = m f m R lkji + R ijkl + R ipjq R pqkl + R ipkq R pjql + R iplq R pjkq + R pjiq R pqkl + R pjkq R piql + R pjlq R pikq, f R ijkl R ijkl = 2 (R ipjq R pqkl + R ipkq R pjql + R iplq R pjkq ), where f-laplacian f f acts on tensors. We conclude on a shrinking GRS that f Rm = Rm + Rm Rm. (11) 2

3 Here, denotes some product of tensors, including traces. We write formulas in this way when we do not care what the specific form of the product is. From this we compute that f Rm 2 = 2 Rm f Rm Rm (12) = 2 Rm Rm 2 + Rm Rm Rm. On the other hand, Rm = Since Rm Rm by Kato s inequality Rm 2 2 Rm Rm 2. (13) Rm for any tensor T, we obtain T 2 = 1 T 2 T, T 2 T 2 (14) f Rm Rm C Rm 2, (15) where C is a universal constant. The technique used to prove the scalar curvature gap theorem can be applied to prove a quadratic curvature decay estimate provided the curvature tends to zero at infinity. In abstract, we have the following result of Munteanu and Wang. Theorem 3 Suppose on a noncompact shrinking GRS with n 4 that w 0 satisfies and w(x) 0 as x. Then there exists a constant C such that f w w C 0 w 2 (16) w C f. (17) Hence w(x) C (1 + d(x, o)) 2. (18) The asymptotic curvature ratio (ACR) of a complete noncompact Riemannian manifold (M n, g) is defined by ACR (g) = lim sup Rm (x) d (x, p) 2. (19) x It is easy to see that this value is independent of the choice of p M. By taking w = Rm in Theorem 3, we immediately obtain the following result of Munteanu and Wang. Theorem 4 If a noncompact shrinking GRS with n 4 satisfies Rm 0 as x, then the Riemann curvature decays quadratically: In other words, ACR (g) <. Of course, AVR(g) > 0. Rm (x) C (1 + d(x, o)) 2. (20) 3

4 By (20) and Shi s local derivative of curvature estimates, there exists an asymptotic cone of (M, g) which is a Riemannian cone. Proof of Theorem 3. Since n 4, (2) implies that From (3) we have f (f 2 ) 2f 2 nf 3. Hence, if f 2n, then f (f 1 ) f 1. (21) Therefore, if f 2n and a 0, then Note that if f > max{a, 2n}, then f 1 af 2 > 0 and f (f 2 ) 3 2 f 2. (22) f (f 1 af 2 ) f af 2. (23) f (f 1 af 2 ) f 1 af 2 a 2 (f 1 af 2 ) 2. (24) Let ψ = b(f 1 af 2 ), where b > 0. Then ψ bf 1 and Since w(x) 0 as x, there exists C 1 > 4n such that Choose a and b so that a 2b = C 0. Then (16) and (25) imply f ψ ψ a 2b ψ2. (25) w(x) 1 10C 0 if f(x) C 1. (26) f (w ψ) w ψ C 0 w 2 + C 0 ψ 2 (27) = (1 C 0 (w + ψ)) (w ψ). Let a = 1 2 C 1. If f C 1, then Thus, if f C 1, then C 0 ψ C 0b C 1 = a 2C 1 = 1 4. C 0 (w + ψ) (28) Now, if f = C 1 = 2a, then Therefore, if f = C 1 = 2a, then ψ = b(f 1 af 2 ) = b 4a = 1 8C 0. On the other hand, by (27) and (28), on {f C 1 } we have w ψ 1 10C 0 1 8C 0 < 0. (29) f (w ψ) (1 C 0 (w + ψ)) (w ψ), (30) 4

5 where 1 C 0 (w + ψ) (31) Since w ψ < 0 on {f = C 1 } and since (w ψ)(x) 0 as x, if w ψ is positive somewhere on M {f < C 1 } = {f C 1 } Ω, then a positive maximum of w ψ occurs at some point x 0 Ω. Then 0 f (w ψ)(x 0 ) > 0 (32) by (30) and (31). Since this is a contradiction, we conclude that w(x) ψ(x) = b(f 1 af 2 )(x) b f(x) (33) for all x {f C 1 }. This completes the proof of Theorem 3. The following is due to Munteanu and Wang; when n = 3 this was originally due to the works of Perelman, Naber, Ni and Wallach, and Cao, Chen, and Zhu. Theorem 5 Any shrinking GRS with sect 0 and Rc > 0 must be compact. Proof. Since sect 0, we have R kijl R kl 0 (nonnegative semidefinite). Indeed, at any point p M we may write Rc = n a=1 λ av a V a, where λ a 0 and where {V a } n a=1 is an orthonormal basis of Tp M; hence, for any W T p M we have R kijl R kl W i W j = n a=1 λ a R kijl V a k V a l W i W j 0. Therefore f Rc Rc. (34) Suppose that M is noncompact. Seeing the analogy with (1), we can apply the same idea as in the proof of Theorem 1. Define the symmetric 2-tensor S Rc c ( f 1 + nf 2) g, (35) where c > 0. From (34), (2), and (3), we compute analogously to (5) that f S S cnf 3 ( f 2 n ) g. (36) Since Rc > 0, there exists c > 0 such that S 0 in the set {f 2n}. Since S tends to nonnegative at infinity, if S has a negative eigenvalue somewhere, then by applying the tensor maximum principle to (36) we conclude that there exists a point x 0 {f > 2n} and unit tangent vector U 0 at x 0 such that ( ) ( ) f f 0 S(U 0, U 0 ) cnf 3 2 n < cnf 3 2 n, (37) which implies f(x 0) 2 n < 0, a contradiction. Therefore there exists c > 0 such that Rc c ( f 1 + nf 2) g cf 1 g on all of M. That is, the Ricci curvature decays at most quadratically. 5

6 Let β : R M be an integral curve of f. We have d R(β(t)) = 2 Rc( f, f) 2c f 2 dt f = 2c (1 Rf ) (β(t)). (38) Let x M and choose β so that β(0) = x. If we have R(β(t)) 1 for all t [ n, 0], then (38) f 2 c implies d dt R(β(t)) c for n c t 0. Integrating this on [ n, 0] yields c R(x) n + R(β( n )) > n. c Otherwise, there exists t [ n, 0] such that y = β(t) satisfies R(y) 1, so that c f 2 R(x) R (y) 1 2 f (y) 1 2 e n c f(x), where the last inequality follows from d dt f(β(t)) = f 2 f(β(t)). We conclude that R(x) max{n, 1 2 e n c f(x)} for all x M. This contradicts R dµ n Vol {f < t} for t sufficient large {f<t} 2 since g has infinite volume and f is proper (f grows quadratically). 6