A complex geometric proof of Tian-Yau-Zelditch expansion

Transcription

1 A complex geometric proof of Tian-Yau-Zelditch expansion Zhiqin Lu Department of Mathematics, UC Irvine, Irvine CA October 21, 2010 Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 1/40

2 Let (M, L) be a polarized Kähler manifold. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 2/40

3 Let (M, L) be a polarized Kähler manifold. Let h be a Hermitian metric on L such that c 1 (L) = ω, the Kähler metric. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 2/40

4 Let (M, L) be a polarized Kähler manifold. Let h be a Hermitian metric on L such that c 1 (L) = ω, the Kähler metric. Let E be a Hermitian vector bundle. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 2/40

5 H 0 (M, L m E) and H 0 (M, L m ) are Hermitian inner product spaces with respect to the L 2 metrics. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 3/40

6 When L is ample, we know that L m is very ample for m big enough. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 4/40

7 When L is ample, we know that L m is very ample for m big enough. Let S 0,, S d be an orthonormal basis of H 0 (M, L m ). Then the map x [S 0,, S d ] gives an embedding f m of M to CP d. The metric 1 m f m(ω F S ) is called the Bergman metric of M. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 4/40

8 Theorem (Tian) as m. 1 m f m(ω F S ) ω 0 Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 5/40

9 Theorem (Tian) as m. 1 m f m(ω F S ) ω 0 (In fact, Tian proved the convergence rate to be m 1 2, and Ruan improved it to 1/m.) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 5/40

10 Let {S m 1,, S m d m } be any orthonormal basis of H 0 (M, L m ). Then B(x) = B m (x) = S j 2 is called the Bergman kernel of the polarized Kähler manifold. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 6/40

11 Let {S m 1,, S m d m } be any orthonormal basis of H 0 (M, L m ). Then B(x) = B m (x) = S j 2 is called the Bergman kernel of the polarized Kähler manifold. Important observation 1 m f m(ω F S ) ω = 1 m logb(x). Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 6/40

12 Theorem (Catlin, Zelditch) There is an asymptotic expansion: B a 0 (x)m n + a 1 (x)m n 1 + a 2 (x)m n 2 + for certain smooth coefficients a j (x) Hom(E, E) with a 0 = I. More precisely, for any k B(x) N a j (x)m n k C µ C N,µ m n N 1, k=0 where C N,µ depends on N, µ and the manifold M and the bundles L, E. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 7/40

13 Result of BBS Robert Berman and Bo Berndtsson and Johannes Sjöstrand A direct approach to Bergman Kernel Asymptotics for Positive Line Bundles Arkiv Math, 46(2): , 2008 Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 8/40

14 In the paper, they proved the existence of the Bergman kernel expansion directly, without using the deep result of Fefferman and Boutet de Monvel-Johannes Sjöstrand on Bergman kernel. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 9/40

15 In the paper, they proved the existence of the Bergman kernel expansion directly, without using the deep result of Fefferman and Boutet de Monvel-Johannes Sjöstrand on Bergman kernel. Their methods also provide an effective way of computing all the coefficients. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 9/40

16 In the paper, they proved the existence of the Bergman kernel expansion directly, without using the deep result of Fefferman and Boutet de Monvel-Johannes Sjöstrand on Bergman kernel. Their methods also provide an effective way of computing all the coefficients. Their methods can be generalized to the twisted case as well. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 9/40

17 Result of Ross-Thomas Julius Ross and Richard Thomas Weighted Bergman kernels on Orbifolds ArXiv: v2 Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 10/40

18 Result of Ross-Thomas Julius Ross and Richard Thomas Weighted Bergman kernels on Orbifolds ArXiv: v2 C convergence on orbifolds. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 10/40

19 Motivations of seeking a complex geometric proof. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

20 Motivations of seeking a complex geometric proof. 1 Give a purely complex geometric proof of the existence of the expansion; Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

21 Motivations of seeking a complex geometric proof. 1 Give a purely complex geometric proof of the existence of the expansion; 2 Give an explicit algorithm of the coefficients; more importantly, give the effective error estimates; Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

22 Motivations of seeking a complex geometric proof. 1 Give a purely complex geometric proof of the existence of the expansion; 2 Give an explicit algorithm of the coefficients; more importantly, give the effective error estimates; Bergman kernels on family of manifolds? Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

23 Motivations of seeking a complex geometric proof. 1 Give a purely complex geometric proof of the existence of the expansion; 2 Give an explicit algorithm of the coefficients; more importantly, give the effective error estimates; Bergman kernels on family of manifolds? 3 The expansion when the metrics are real analytic; Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

24 Motivations of seeking a complex geometric proof. 1 Give a purely complex geometric proof of the existence of the expansion; 2 Give an explicit algorithm of the coefficients; more importantly, give the effective error estimates; Bergman kernels on family of manifolds? 3 The expansion when the metrics are real analytic; 4 Orbiford Cases? Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

25 Motivations of seeking a complex geometric proof. 1 Give a purely complex geometric proof of the existence of the expansion; 2 Give an explicit algorithm of the coefficients; more importantly, give the effective error estimates; Bergman kernels on family of manifolds? 3 The expansion when the metrics are real analytic; 4 Orbiford Cases? 5 (less important but more difficult) A recursive formula for all coefficients? Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 11/40

26 Technical Difficulties 1 The C convergence (C 0 convergence is proved in my paper On the lower order terms of the asymptotic expansion of Tian-Yau-Zelditch expansion, AJM 122, 2000.) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 12/40

27 Technical Difficulties 1 The C convergence (C 0 convergence is proved in my paper On the lower order terms of the asymptotic expansion of Tian-Yau-Zelditch expansion, AJM 122, 2000.) 2 Effective error estimate. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 12/40

28 Technical Difficulties 1 The C convergence (C 0 convergence is proved in my paper On the lower order terms of the asymptotic expansion of Tian-Yau-Zelditch expansion, AJM 122, 2000.) 2 Effective error estimate. 3 Hom(E, E)-valued objects. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 12/40

29 Definition of Bergman kernel (vector bundle case) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 13/40

30 Definition of Bergman kernel (vector bundle case) Let S 1,, S d be any basis of H 0 (M, E). Let F = (F ij ) = (S i, S j ). Let P be the matrix such that P F P = I. If we write S i = r b ij e j, j=1 then the Bergman kernel can be represented by B = HB F 1 B, where B = (b ij ) is a d r matrix. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 13/40

31 Definition of Bergman kernel (vector bundle case) Let S 1,, S d be any basis of H 0 (M, E). Let F = (F ij ) = (S i, S j ). Let P be the matrix such that P F P = I. If we write S i = r b ij e j, j=1 then the Bergman kernel can be represented by B = HB F 1 B, where B = (b ij ) is a d r matrix. B is an r r matrix, where r is the rank of E. H is the Hermitian metric of E. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 13/40

32 Important remark For any linearly independent sections S 1,, S t (subspace of H 0 (M, E)), Bergman kernels are defined. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 14/40

33 Reduction of the problem For Bergman kernel of H 0 (M, L m E) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 15/40

34 Reduction of the problem For Bergman kernel of H 0 (M, L m E) 1 the C asymptotic expansion of B F 1 B Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 15/40

35 Reduction of the problem For Bergman kernel of H 0 (M, L m E) 1 the C asymptotic expansion of B F 1 B 2 the C asymptotic expansion at one point Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 15/40

36 Reduction of the problem For Bergman kernel of H 0 (M, L m E) 1 the C asymptotic expansion of B F 1 B 2 the C asymptotic expansion at one point 3 the C 0 asymptotic expansion at one point (F 1 P Q zp z Q ). Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 15/40

37 Reduction of the problem For Bergman kernel of H 0 (M, L m E) 1 the C asymptotic expansion of B F 1 B 2 the C asymptotic expansion at one point 3 the C 0 asymptotic expansion at one point (F 1 P Q zp z Q ). 4 the expansion of the inverse of the metric matrix Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 15/40

38 Reduction of the problem For Bergman kernel of H 0 (M, L m E) 1 the C asymptotic expansion of B F 1 B 2 the C asymptotic expansion at one point 3 the C 0 asymptotic expansion at one point (F 1 P Q zp z Q ). 4 the expansion of the inverse of the metric matrix 5 Effective version of Ruan s lemma. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 15/40

39 Definition We say a sequence of functions f m (x) has a C µ asymptotic expansion, if there exist matrix-valued functions a 0 (x),, a s (x), such that for any s, µ, ( f m(x) m n a 0 (x) + a 1(x) m + + a s(x) m s where C is a constant independent to m. ) C µ C m s+1, Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 16/40

40 Definition We say a sequence of functions f m (x) has a C µ asymptotic expansion at the point x 0, if there exist matrix-valued functions a 0 (x),, a s (x), in a neighborhood of x 0 such that for any s, we have ( ( Dµ f m (x) m n a 0 (x) + a 1(x) m + + a )) s(x) (x m s 0 ) where C is a constant independent to m and x 0. The derivative is taken with respect to a K-coordinate system. C m s+1, Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 17/40

41 Definition In particular, we say f m (x) has a C 0 asymptotic expansion at the point x 0, if there exists matrices a 0,, a s, such that f m (x) m n ( a 0 + a 1 m + + a s m s ) (x0 ) where C is independent to m. C m s+1, If a sequence of functions has a C µ asymptotic expansion, then it has the C µ asymptotic expansion at any point x 0. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 18/40

42 Lemma A sequence of functions f m has a C µ asymptotic expansion if and only if for every µ 0 and at each point x 0 M, f m has a C µ asymptotic expansion at x 0. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 19/40

43 Definition of K-coordinates and K-frames Definition Let p > 0 be any positive integer. Let x 0 M be a point. Let (z 1,, z n ) be a holomorphic coordinate system centered at x 0. Let (g α β) be the Kähler metric matrix. If g α β(x 0 ) = δ αβ, p 1+ +p n g α β z p (x 1 1 zn pn 0 ) = 0 for α, β = 1,, n and any nonnegative integers (p 1,, p n ) with p > p p n 0. Then we call the coordinate system a K-coordinate system of order p. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 20/40

44 Definition Let e L be a local holomorphic frame of L at x 0. If for p > 0, the local representation function a of the Hermitian metric h L satisfies p 1+ +p n a a(x 0 ) = 1, z p (x 1 1 zn pn 0 ) = 0 (1) for any nonnegative integers (p 1,, p n ) with p > p p n 0. Then we call e L is a K-frame of order p. If a is analytic, then again we can take p = +. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 21/40

45 Under K-coordinates, the coefficients of the Taylor expansions of the metrics only contain the curvatures and their derivatives. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 22/40

46 Let P = (p 1,, p n ) be a multiple index and let 1 j r. Define the lexicographical order on the set of (P, j) s. That is, (P, j) < (Q, k) if 1 pi < q i, or 2 p 1 = q 1,, p l = q l but p l+1 < q l+1 for some 0 l n 1, or 3 j < k. Such an order gives rise to the function P = P (j). For example, P (1) = ((0,, 0), 1), P (2r + 2) = ((0, 1,, 0), 2), etc. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 23/40

47 Definition Let S 1,, S k, S k+1,, S d be a basis of H 0 (M, L m E). We say that it is a regular basis at x 0 of order µ, if under the local K-coordinates at x 0 1 for 1 j k, S j (z) = z P (j) + o( z µ ); 2 for j > k, S j (z) = o( z µ ). Moreover, the (i, j)-th entry of F 1 has a C 0 asymptotic expansion at x 0. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 24/40

48 Lemma If a regular basis exists, then the Catlin-Zelditch s result is valid. Proof. The Taylor expansion for the smooth vector-valued function H gives the asymptotic expansion. Thus in order to prove the result, we only need to prove the existence of the C µ expansion of B F 1 B. It is not hard to see that if P + Q B z P z Q has the C 0 asymptotic expansion at x 0 for all P + Q µ. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 25/40

49 Peak Sections 1, z, z 2,, z p are the peak functions on C with respect to the norm e z 2. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 26/40

50 Peak Sections 1, z, z 2,, z p are the peak functions on C with respect to the norm e z 2. Graphs of f i (x) = x i e x for i = 1,, 4. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 26/40

51 Peak Sections 1, z, z 2,, z p are the peak functions on C with respect to the norm e z 2. Graphs of f i (x) = x i e x for i = 1,, 4. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 26/40

52 Peak Sections 1, z, z 2,, z p are the peak functions on C with respect to the norm e z 2. Graphs of f i (x) = x i e x for i = 1,, 4. x i e x goes to zero but not uniformly. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 26/40

53 In C n, z α 1 1 z αn n are peak sections for the norm e z 2. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 27/40

54 In C n, z α 1 1 z αn n are peak sections for the norm e z 2. On the polarized manifold, the fact that c 1 (L) = ω implies that in a neighborhood of x 0, the metric is close to e z 2. ( log e z 2 = j dz j d z j ) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 27/40

55 In C n, z α 1 1 z αn n are peak sections for the norm e z 2. On the polarized manifold, the fact that c 1 (L) = ω implies that in a neighborhood of x 0, the metric is close to e z 2. ( log e z 2 = j dz j d z j ) The metric on L m is close to e m z 2. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 27/40

56 In C n, z α 1 1 z αn n are peak sections for the norm e z 2. On the polarized manifold, the fact that c 1 (L) = ω implies that in a neighborhood of x 0, the metric is close to e z 2. ( log e z 2 = j dz j d z j ) The metric on L m is close to e m z 2. After rescaling { z log m m }, e m z 2 { z log m}, e z 2. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 27/40

57 Theorem (Peak Section Theorem) We can find holomorphic sections S H 0 (M, L m ) such that 1 in a neighborhood of x 0, S is close to z α 1 1 z αn n 2 Outside the neighborhood, S is very small. Here α j < ε log m. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 28/40

58 joint with Chiung-ju Liu Theorem Let S 1,, S k be peak sections k = [ε log m]. Let B k peak be the Bergman kernel with respect to the peak sections S 1,, S k. Then 1 B k peak has an C asymptotic expansion 2 The expansion stables to the TYZ expansion: B(x) B k peak C µ C m ε 0 log k. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 29/40

59 W.D. Ruan s Lemma Lemma Let S P be peak sections. Let T be another section of L m. Near x 0, T = fe m L for a holomorphic function f. When we say T s Taylor expansion at x 0, we mean the Taylor expansion of f at x 0 under the coordinate system (z 1,, z n ). 1 If z P is not in T s Taylor expansion at 0, then ( ) 1 (S P, T ) = O S T. m 2 If T contains terms z Q for Q P + σ in the Taylor expansion, then ( ) 1 (S P, T ) = O S T. m 1+ σ 2 Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 30/40

60 We worked out the effective version of Ruan s Lemma. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 31/40

61 Let S k+1,, S d be an orthonormal basis of the space V s+1 = {T H 0 (M, L m E) T vanishes at x 0 of order at least s + 1}. Let S 1,, S k be peak sections. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 32/40

62 Define the matrix A ij to be δ ij (S i, S j ) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 33/40

63 Define the matrix A ij to be δ ij (S i, S j ) Then we have F 1 = I A. By our effective Ruan s Lemma, we have A α,β C α β 1+ m 2. (This is a little over-simplified version) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 33/40

64 We represent F 1 1 = (C αβ ) as a block matrix using the same partition as in the matrix A. Using the expansion for any fixed (α 0, β 0 ), we have We have C αβ I = k=s+1 F 1 = I + A + A 2 +, k=1 i 1,,i k 1 A α0 i 1 A i1 i 2 A ik 1 β 0. A α0 i 1 A i1 i 2 A ik 1 β 0 i 1,,i k 1 C m s+1. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 34/40

65 Similarly, we consider the terms s k=1 some i j =s+1 If some i j = s + 1, we must have A α0 i 1 A i1 i 2 A ik 1 β 0. α 0 i 1 + i 1 i i k 1 β 0 2s + 2 α 0 β 0. Thus we have s k=1 some i j =s+1 A α0 i 1 A i1 i 2 A ik 1 β 0 C m s (α 0+β 0 ). Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 35/40

66 Real analytic case Theorem (Liu-L) Assume that the metrics are real analytic. Then 1 The TYZ expansion is convergent for m large. j=0 a j m j 2 B(x) m n j=0 a j m m)1/n j C µ Cm ε(log. (Fefferman-Boutet de Monvel-Sjöstrand s method possible) Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 36/40

67 The technical heart is that, if we work harder, we can have B(x) m n N j=0 a j m j C µ CN m N+1, where N is up to ε log m and C is independent to N. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 37/40

68 The technical heart is that, if we work harder, we can have B(x) m n N j=0 a j m j C µ CN m N+1, where N is up to ε log m and C is independent to N. By setting N = [ε log m], we can get the result. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 37/40

69 The orbifold case Assume M is an orbifold with only one orbifold singularity point. Assume that the local group is Z 2. Such an orbifold may not exist. Just use it as an example. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 38/40

70 The orbifold case Assume M is an orbifold with only one orbifold singularity point. Assume that the local group is Z 2. Such an orbifold may not exist. Just use it as an example. It is not possible to have the C convergence of the Begman kernel at the orbifold point. What Ross-Thomas did was to average several Bergman kernels. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 38/40

71 The orbifold case Assume M is an orbifold with only one orbifold singularity point. Assume that the local group is Z 2. Such an orbifold may not exist. Just use it as an example. It is not possible to have the C convergence of the Begman kernel at the orbifold point. What Ross-Thomas did was to average several Bergman kernels. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 38/40

72 In terms of peak sections, this is clear. The set of peak sections can be decomposed into two parts: {z 2k+1 } and {z 2k } and these two sets are perpendicular to each other. We can make the Begman kernel for each of them. The sum of the Bergman kernels is equal to the Bergman kernel of the local uniformization. Since the Bergman kernel expansion can be localized, using the ordinary method on manifold, we can recover the result of Ross-Thomas. Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 39/40

73 Thank you! Zhiqin Lu, Dept. Math, UCI A complex geometric proof of TYZ expansion 40/40