RFB 5 The Li - Yau Estimate for Curvature in the Ricci Flow

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Clyde Morris
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1 RFB 5 The Li - Yau Estimate for Curvature in the Ricci Flow I) The Li - Yau Estimate Suppose we have a solution of the Ricci Flow We define the tensors M ab and P abc by M ab = DRc ab D2 R ab + 2 Rm acbd R cd - R ac R bc P abc = D a Rc bc - D b Rc ac and form the quadratic Z(W, U) by Z HW, UL = M ac W a W c + 2 P abc U ab W c + Rm abcd U ab U cd where W is a one-form and U is a two-form We say the solution has positive curvature operator if Rm(U) = Rm abcd U ab U cd > 0 for all two-forms U! 0 This implies positive Ricci curvature Rc(W) = R ab W a W b > 0 if W! 0 When the solution has positive curvature operator we can define the Li - Yau quadratic H(W) = H ac W a W c = inf Z(W, U) U I1 Theorem : If a solution to the Ricci Flow on a compact manifold! for 0 t T has positive curvature operator then H(W) t Rc(W) > 0 for all W! 0 The proof is below II) Computation The proof ot the Li - Yau Estimate will follow from the following computation II 1 Theorem : If Rm > 0 and H ac W a W c = Inf U 8 M ac W a W c + 2 P abc U ab W c + Rm abcd U ab U cd < then D t H ac DH ac + 2 Rc -1 bd H ab H cd + 2 Rm abcd H bd Proof : Given some point and time and some W and U, we can extend W and U to sections locally with D t W a = DW a D a W b = 0 D t U ab = DU ab

2 2 RFB 5 The Li - Yau Estimate for Curvaturenb t ab ab D a U bc = 1 2 HRc ab W c - Rc ac W b L Hg ab Y c - g ac Y b L where we will choose Y a later to give a strong term II2 Lemma : At the chosen point where and and D t Z = DZ + S + N + Q S = P abc W c + Rm abcd U cd 2 N = -@4 HM ac W a + 2 P abc U ab L Y c + 2 Rc ac Y a Y c D Q = 2 Rm abcd M ac W b W d - 2 P cab P dba W c W d + 8 P cab Rm adbe U cd W e + 4 Rm acbd Rm aebf U ce U df Proof : The computation is tedious but straightforward Next we interpret these terms Suppose at a given point and time and for a given W we choose the minimal U II3 The minimal U ab satisfies and P abc W c + Rm abcd U cd = 0 H ac W a = M ac W a + P abc U ab Consequently S = P abc W c + Rm abcd U cd 2 = 0 Proof : We use the definition H ac W a W c = Inf U 8 M ac W a W c + 2 P abc U ab W c + Rm abcd U ab U cd < Take a variation U è ab in U ab The variation H è ac in H ac is H è ac W a W c = 2(P abc W c + Rm abcd U cd ) U è ab and for the minimal U this will always vanish for all U è ab; so P abc W c + Rm abcd U cd = 0 This proves the first identity Next write the quadratic Z as a symmetric bilinear function of two pairs (W 1, U 1 ) and (W 2, U 2 ), so that

3 RFB 5 The Li - Yau Estimate for Curvaturenb 3 and (W, U ), so that Z = M ac W 1 a W 2 c + P abc U 1 ab W 2 c + P abc U 2 ab W 1 c + Rm abcd U 1 ab U 2 cd If we choose U 1 as the minimizer for W 1 and U 2 as the minimizer for W 2 then H ac W 1 2 a W c = M ac W 1 a W 2 c + P abc U 1 ab W 2 c + P abc U 2 ab W 1 c + Rm abcd U 1 ab U 2 cd Factor this as H ac W 1 2 a W c = ( M ac W 1 a + P abc U 1 ab M W 2 c + I P abc W 1 c + Rm abcd U 1 cd )U 2 ab By the first identity P abc W 1 1 c + Rm abcd U cd = 0 Then H ac W 1 2 a W c = ( M ac W 1 a + P abc U 1 2 ab M W c for all W c 2, so H ac W 1 a = M ac W 1 1 a + P abc U ab which is the second identity II4 Lemma : If we choose Y to solve then H ac W a Y c + Rc ac Y a Y c = 0 N = 2 Rc -1 bd H ab H cd W a W c Proof : From Lemma II3, N = -H4 H ac W a Y c + 2 Rc ac Y a Y c L We choose Y a to solve H ac W a Y c + Rc ac Y a Y c = 0 and this makes N = 2 Rc -1 bd H ab H cd W a W c which is a strong new term We define a tensor A abc anti -symmetric in {a,b} by solving P abe + Rm abcd A cde = 0 On a soliton P abc = Rm abcd D d f, so on a soliton the solution is A abc = 1 2 Hg bc D a f - g ac D b fl Let F be the trace-free part of A F abc = A abc - 1 n-1 HA dbd g ac - A dad g bc L Then on a soliton F abc = 0 II5 Lemma : The minimal U ab is given by

4 4 RFB 5 The Li - Yau Estimate for Curvaturenb U ab = A abc W c Proof : By Lemma II3 the minimal U is the solution of P abc W c + Rm abcd U cd = 0 By the definition of A P abe W e + Rm abcd A cde W e = 0 so U cd = A cde W e solves the equation II6 Corollary : H ac = M ac + A dea P dec Proof : By Lemmas II3 and IIV H ce W e = H ac W a = M ac W a + P abc U ab = (M ae + P abc A abe L W e Since this is true for all W e H ce = M ae + P abc A abe Next we use the fact that a positive quadratic function can be written as a sum of the squares of each eigenvector times the square root of its positive eigenvalue When Rm(U) = Rm abcd U ab U cd > 0 for al U! 0, we can find two-forms {X p = X p ab, 1 p n (n-1) / 2} such that Rm acbd = X p ab X p cd II7 Lemma : Q = 2 Rm acbd H cd +! where! = X p ab X q cd A cdb W a -X q ab X p cd A cdb W a + 2 X p bd X q cd A bca W a 2 Proof : Recall Q = 2 Rm abcd M ac W b W d - 2 P cab P dba W c W d + 8 P cab Rm adbe U cd W e + 4 Rm acbd Rm aebf U ce U df Substitute M ac = H ac - A dea P dec from Lemma IIVII to get Q = 2 Rm acbd H cd +! where! = - 2 Rm abcd A dea P dec W b W d - 2 P cab P dba W c W d + 8 P cab Rm adbe U cd W e + 4 Rm acbd Rm aebf U ce U df Now use the definition of A as the solution of P abe + Rm abcd A cde = 0 to replace each P by an A: P dec = - Rm defg A fgc P cab = - Rm caef A efb P dba = - Rm dbh A gha to get! = 2 Rm abcd A dea Rm defg A fgc W b W d - 2 Rm caef A efb Rm dbh A gha W c W d + 8 Rm caef A efb Rm adbe U cd W e + 4 Rm acbd Rm aebf U ce U df

5 D t H è ac D H è ac + 2 Rc -1 H è ab H è cd + 2 Rm abcd H è bd - 1 H è ac! = - 2 Rm abcd A dea P dec W b W d - 2 P cab P dba W c W d + 8 P cab Rm adbe U cd W e + 4 Rm acbd Rm aebf U ce U df Now use the definition of A as the solution of P abe + Rm abcd A cde = 0 to replace each P by an A: P dec = - Rm defg A fgc P cab = - Rm caef A efb P dba = - Rm dbh A gha to get! = 2 Rm abcd A dea Rm defg A fgc W b W d - 2 Rm caef A efb Rm dbh A gha W c W d + 8 Rm caef A efb Rm adbe U cd W e + 4 Rm acbd Rm aebf U ce U df Next replace each U by Lemma IIV U cd = A cdf W f and U ce = A ceg W cg and U df = A dfh W h to get RFB 5 The Li - Yau Estimate for Curvaturenb 5! = 2 Rm abcd A dea Rm defg A fgc W b W d - 2 Rm caef A efb Rm dbh A gha W c W d + 8 Rm caef A efb Rm adbe A cdf W f W e + 4 Rm acbd Rm aebf A ceg W cg A dfh W h Rearranging to make it pretty! = 2 Rm acbd Rm efgh A efa A ghb W c W d - 2 Rm acef Rm bdgh A efb A gha W c W d - 8 Rm eafg Rm ahbc A fgb A ehd W c W d + 4 Rm aebf Rm agbh A gec A hfd W c W d Replace each Rm by a sum of products of two X's using Rm acbd = X p ab X p cd and factor to get! = X p ab X q cd A cdb W a -X q ab X p cd A cdb W a + 2 X p bd X q cd A bca W 2 a If Rm > 0 then! 0 Moreover we can check that in the above expression the trace part of A abc cancels so that in fact! = X p ab X q cd F cdb W a - X q ab X p cd F cdb W a + 2 X p bd X q cd F bca W a 2 and F = 0 on a soliton, so on a soliton! = 0 as we would expect Now we can finish the proof of Theorem II1 From Lemma II2 D t Z = DZ + S + N + Q At the chosen point where U minimizes Z = H ac W a W c while we saw S = 0 there, while from Lemma II4 N = 2 Rc -1 bd H ab H cd W a W c and from Lemma II7 Q = 2 Rm acbd H cd +! where! = X p ab X q cd A cdb W a -X q ab X p cd A cdb W a + 2 X p bd X q cd A bca W 2 a 0 Thus D t H ac W a W c DH ac W a W c + 2 Rc -1 bd H ab H cd W a W c + 2 Rm abcd H bd This proves Theorem II1 One a compact manifold it is easy using this to prove Theorem I1 Let H è ac = H ac t R ac Then we easily compute

6 Q = 2 Rm acbd H cd +! where! = X p ab X q cd A cdb W a -X q ab X p cd A cdb W a + 2 X p bd X q cd A bca W 2 a 0 Thus D t H ac W a W c DH ac W a W c + 2 Rc -1 bd H ab H cd W a W c + 2 Rm abcd H bd This proves Theorem II1 6 RFB 5 The Li - Yau Estimate for Curvaturenb One a compact manifold it is easy using this to prove Theorem I1 Let H è ac = H ac t D t H è ac R ac Then we easily compute D H è ac + 2 Rc -1 bd H è ab H è cd + 2 Rm abcd H è bd - 1 Hè t ac and then H è ac 0 follows from the maximum principle for systems

arxiv: v3 [math.dg] 13 Mar 2011

arxiv: v3 [math.dg] 13 Mar 2011 GENERALIZED QUASI EINSTEIN MANIFOLDS WITH HARMONIC WEYL TENSOR GIOVANNI CATINO arxiv:02.5405v3 [math.dg] 3 Mar 20 Abstract. In this paper we introduce the notion of generalized quasi Einstein manifold,