Bergman Kernel and Kobayashi Pseudodistance in Convex Domains

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1 Bergman Kernel and Kobayashi Pseudodistance in Convex Domains Zbigniew B locki Uniwersytet Jagielloński, Kraków, Poland blocki (Joint work with W lodzimierz Zwonek) NORDAN Reykjavík, April 26, 2015

2 Ω C n, w Ω K Ω (w) = sup{ f (w) 2 : f O(Ω), f 2 dλ 1} Ω (Bergman kernel on the diagonal) G w (z) = G Ω (z, w) = sup{u(z) : u PSH (Ω) : lim z w ( u(z) log z w ) < } (pluricomplex Green function) Theorem 0 Assume Ω is pseudoconvex in C n. Then for w Ω and t 0 K Ω (w) 1 e 2nt λ({g w < t}). Optimal constant: = if Ω = B(w, r).

3 Proof 1 Using Donnelly-Fefferman s estimate for one can prove where c(n, t) = K Ω (w) ( c(n, t)λ({g w < t}), (1) ) 2 C, Ei(a) = Ei( nt) a ds se s (B. 2005). Now use the tensor power trick: Ω = Ω Ω C nm, w = (w,..., w) for m 0. Then K Ω( w) = (K Ω (w)) m, λ({g w < t}) = (λ({g w < t})) m, and by (1) for Ω But K Ω (w) lim c(nm, m t)1/m = e 2nt. 1 c(nm, t) 1/m λ({g w < t}).

4 Proof 2 (Lempert) By Berndtsson s result on log-(pluri)subharmonicity of the Bergman kernel for sections of a pseudoconvex domain it follows that log K {Gw <t}(w) is convex for t (, 0]. Therefore t 2nt + log K {Gw <t}(w) is convex and bounded, hence non-decreasing. It follows that K Ω (w) e 2nt K {Gw <t}(w) e 2nt λ({g w < t}). Berndtsson-Lempert: This method can be improved to show the Ohsawa-Takegoshi extension theorem with optimal constant.

5 Theorem 0 Assume Ω is pseudoconvex in C n. Then for w Ω and t 0 K Ω (w) 1 e 2nt λ({g w < t}). What happens when t? For n = 1 Theorem 0 immediately gives: Theorem (Suita conjecture) For a domain Ω C one has K Ω (w) c Ω (w) 2 /π, w Ω, (2) where c Ω (w) = exp ( lim z w (G Ω (z, w) log z w ) ) (logarithmic capacity of C \ Ω w.r.t. w). Theorem (Guan-Zhou) Equality holds in (2) iff Ω \ F, where is the unit disk and F a closed polar subset.

6 πk Ω c 2 Ω for Ω = {e 5 < z < 1} as a function of 2 log w

7 What happens with e 2nt λ({g w < t}) as t for arbitrary n? For convex Ω using Lempert s theory one can get Proposition If Ω is bounded, smooth and strongly convex in C n then for w Ω lim t e 2nt λ({g w < t}) = λ(i K Ω (w)), where I K Ω (w) = {ϕ (0) : ϕ O(, Ω), ϕ(0) = w} (Kobayashi indicatrix). Corollary If Ω C n is convex then For general Ω one can prove 1 K Ω (w) λ(iω K w Ω. (w)), Theorem If Ω is bounded and hyperconvex in C n and w Ω then lim t e 2nt λ({g w < t}) = λ(iω A (w)), where I A Ω (w) = {X Cn : lim ζ 0 ( Gw (w + ζx ) log ζ ) 0} (Azukawa indicatrix)

8 Corollary (SCV version of the Suita conjecture) If Ω C n is pseudoconvex and w Ω then K Ω (w) 1 λ(i A Ω (w)). Remark 1. For n = 1 one has λ(i A Ω (w)) = π/c Ω(w) If Ω is convex then I A Ω (w) = I K Ω (w). Conjecture For Ω pseudoconvex and w Ω the function is non-decreasing in t. t e 2nt λ({g w < t}) It would easily follow if we knew that the function t log λ({g w < t}) is convex on (, 0]. Fornæss however constructed a counterexample to this (already for n = 1).

9 Theorem The conjecture is true for n = 1. Proof It is be enough to prove that f (t) 0 where f (t) := log λ({g w < t}) 2t and t is a regular value of G w. By the co-area formula t dσ λ({g w < t}) = G w ds and therefore f (t) = {G w =t} {G w =s} dσ G w λ({g w < t}) By the Schwarz inequality dσ {G w =t} G w (σ({g w = t})) 2 G w dσ {G w =t} 2. = (σ({g w = t})) 2. 2π

10 The isoperimetric inequality gives and we obtain f (t) 0. (σ({g w = t})) 2 4πλ({G w < t}) The conjecture for arbitrary n is equivalent to the following pluricomplex isoperimetric inequality for smooth strongly pseudoconvex Ω dσ G w 2nλ(Ω). Ω The conjecture also turns out to be closely related to the problem of symmetrization of the complex Monge-Ampère equation.

11 What about the corresponding upper bound in the Suita conjecture? Not true in general: Proposition Let Ω = {r < z < 1}. Then K Ω ( r) (c Ω ( 2 log r r)) 2 π 3. It would be interesting to find un upper bound of the Bergman kernel for domains in C in terms of logarithmic capacity which would in particular imply the part in the well known equivalence (due to Carleson) K Ω > 0 c Ω > 0 (c 2 Ω πk Ω being a quantitative version of ).

12 The upper bound for the Bergman kernel holds for convex domains: Theorem For a convex Ω and w Ω set Then F Ω (w) 4. F Ω (w) := ( K Ω (w)λ(i K Ω (w)) ) 1/n. Sketch of proof Denote I := int IΩ K (w) and assume that w = 0. One can show that I 2 Ω. Then K Ω (0)λ(I ) K I /2 (0)λ(I ) = λ(i ) λ(i /2) = 4n. If Ω is in addition symmetric w.r.t. w then F Ω (w) 16/π 2 = Remark The proof of the optimal lower bound F Ω 1 used. The proof of the (probably) non-optimal upper bound F Ω 4 is much more elementary!

13 For convex domains F Ω (w) = ( λ(i Ω (w))k Ω (w)) ) 1/n is a biholomorphically invariant function satisfying 1 F Ω 4. Find an example with F Ω 1. What are the properties of the function w λ(i Ω (w))? What is the optimal upper bound for F Ω?

14 Formulas for some convex complex ellipsoids in C 2 E(p, q) = {z C 2 : z 1 2p + z 2 2q < 1}, p, q 1/2. Blank-Fan-Klein-Krantz-Ma-Pang (1992) found implicit formulas for the Kobayashi function of E(m, 1). They can be made explicit for m = 1/2. Using this one can prove Theorem For Ω = { z 1 + z 2 2 < 1} and b [0, 1) one has and λ(i Ω ((b, 0))) = π2 3 (1 b)3 (1 + 3b + 3b 2 b 3 ) λ(i Ω ((b, 0)))K Ω ((b, 0)) = 1 + (1 b)3 b 2 3(1 + b) 3.

15 F Ω ((b, 0)) for Ω = { z 1 + z 2 2 < 1}

16 Although the Kobayashi function of E(m, 1) is given by implicit formulas, it turns out that the volume of the Kobayashi indicatrix can be computed explicitly: Theorem For Ω = { z 1 2m + z 2 2 < 1}, m 1/2, and b [0, 1) one has λ(i Ω ((b, 0))) = π [ 2 m 1 3(m 1) 2m(3m 2)(3m 1) b6m+2 2m(m 2)(m + 1) b2m+2 For m = 2/3 m + 2(m 2)(3m 2) b6 + 3m 3m 1 b4 4m 1 2m b2 + m m + 1 ( ) λ(i Ω ((b, 0))) = π2 65b b 6 log b + 160b 4 27b 10/3 100b , 80 and m = 2 λ(i Ω ((b, 0))) = π2 ( 3b 14 25b 6 120b 6 log b + 288b 4 420b ). 240 ].

17 About the proof Main tool: Jarnicki-Pflug-Zeinstra (1993) formula for geodesics in convex complex ellipsoids. If C U z (f (z), g(z)) I is a parametrization of an S 1 -invariant portion of I then the volume of the corresponding part of I is given by π H(z) dλ(z), (3) 2 where U H = f 2 ( g z 2 g z 2 ) + g 2 ( f z 2 f z 2 ) + 2Re ( f ḡ(f z g z f z g z ) ). Both H and the integral (3) are computed with the help of Mathematica. The same method is used for computations in other ellipsoids.

18 For Ω = { z 1 2m + z 2 2 < 1} the formula for the Bergman kernel is well known: K Ω (w) = 1 π 2 (1 w 2 2 ) 1/m 2 (1/m + 1)(1 w 2 2 ) 1/m + (1/m 1) w 1 2 ( (1 w2 2 ) 1/m w 1 2) 3, so that K Ω ((b, 0)) = m (1 m)b2 π 2 m(1 b 2 ) 3. Since for t R and a the mapping Ω z (e it (1 a 2 ) 1/2m (1 āz 2 ) z 1, z ) 2 a 1/m 1 āz 2 is a holomorphic automorphism of Ω, F Ω ((b, 0)) for b [0, 1) attains all values of F Ω in Ω.

19 F Ω ((b, 0)) in Ω = { z 1 2m + z 2 2 < 1} for m = 1/2, 4, 8, 16, 32, 64, 128 sup F Ω ((b, 0)) as m 0<b<1 (highest value of F Ω obtained so far in arbitrary dimension)

20 Theorem For Ω = { z 1 + z 2 < 1} and b [0, 1) one has and λ(i Ω ((b, 0))) = π2 6 (1 b)4( (1 b) 4 + 8b ) 2 (1 b)4 λ(i Ω ((b, 0)))K Ω ((b, 0)) = 1 + b (1 + b) 4. The Bergman kernel for this ellipsoid was found by Hahn-Pflug (1988): K Ω (w) = 2 π 2 3(1 w 2 ) 2 (1 + w 2 ) + 4 w 1 2 w 2 2 (5 3 w 2 ) ( (1 w 2 ) 2 4 w 1 2 w 2 2) 3, so that K Ω ((b, 0)) = 6(1 + b2 ) π 2 (1 b 2 ) 4. In all examples so far the function w λ(i Ω (w)) is analytic. Is it true in general?

21 Theorem For Ω = { z 1 + z 2 < 1} and b [0, 1/4] one has λ(i Ω ((b, b))) = π2 ( 30b 8 64b b 6 80b b 4 16b 3 8b ). 6

22 Since K Ω ((b, b)) = 2(3 6b2 + 8b 4 ) π 2 (1 4b 2 ) 3, we get the following picture: F Ω ((b, b)) in Ω = { z 1 + z 2 < 1} for b [0, 1/4]

23 Since K Ω ((b, b)) = 2(3 6b2 + 8b 4 ) π 2 (1 4b 2 ) 3, we get the following picture: F Ω ((b, b)) in Ω = { z 1 + z 2 < 1} for b [0, 1/4] By either of the estimates 1 F Ω 4, the function b F Ω ((b, b)) cannot be analytic on (0, 1/2)!

24 Theorem For Ω = { z 1 + z 2 < 1} and b [0, 1/4] one has λ(i Ω ((b, b))) = π2 ( 30b 8 64b b 6 80b b 4 16b 3 8b ). 6 For b [1/4, 1/2) ( ) 2π 2 b(1 2b) 3 2b 3 + 3b 2 6b + 4 λ(i Ω ((b, b))) = 3(1 b) 2 ( ) π 30b b b 8 176b 7 260b b 5 76b 4 144b b 2 18b (1 b) 2 ( arccos 1 + 4b 1 ) 2b 2 ( ) π(1 2b) 180b b 6 554b b b b 2 305b b 1 ( 72(1 b) ) 4πb(1 2b) 4 7b 2 + 2b 2 + 3(1 b) 2 arctan 4b 1 + 4πb2 (1 2b) 4 (2 b) 1 3b (1 b) 2 arctan (1 b) 4b 1.

25 By χ (b), resp. χ + (b), denote λ(i Ω ((b, b))) for b 1/4, resp. b 1/4. Then at b = 1/4 but χ = χ + = π2, χ = χ + = π2, χ = χ + = π2, χ (3) = χ (3) + = π2, χ (4) = π2, χ (4) + =. Corollary For Ω = { z 1 + z 2 < 1} the function w λ(i Ω (w)) is not C 3,1 at w = (1/4, 1/4).

26 F Ω ((b, b)) in Ω = { z 1 + z 2 < 1} for b [0, 1/2)

27 Mahler Conjecture K - convex symmetric body in R n Mahler volume := λ(k)λ(k ) K := {y R n : x y 1 for every x K} Mahler volume is an invariant of the Banach space defined by K: it is independent of linear transformations and of the choice of inner product. Blaschke-Santaló Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes True for n = 2: Hansen-Lima bodies: starting from an interval they are produced by taking products of lower dimensional HL bodies and their duals. n = 2 n = 3

28 Equivalent SCV formulation (Nazarov, 2012) For u L 2 (K ) we have û(0) 2 = u dλ K with equality for u = χ K. Therefore 2 λ(k ) u 2 L 2 (K ) = (2π) n λ(k ) û 2 L 2 (R n ) λ(k ) = (2π) n sup f P f (0) 2 f 2, L 2 (R n ) where P = {û : u L 2 (K )} O(C n ). By the Paley-Wiener thm P = {f O(C n ): f (z) Ce C z, f (iy) Ce q K (y) }, where q K is the Minkowski function for K. Therefore the Mahler conjecture is equivalent to finding f P with f (0) = 1 and ( π ) n f (x) 2 dλ(x) n! λ(k). R 2 n

29 Bourgain-Milman Inequality Bourgain-Milman (1987) There exists c > 0 such that Mahler Conjecture: c = 1 G. Kuperberg (2006) c = π/4 λ(k)λ(k ) c n 4n n!. Nazarov (2012) SCV proof using Hörmander s estimate (c = (π/4) 3 ) Consider the tube domain T K := intk + ir n C n. Then ( π 4 ) 2n 1 (λ n (K)) 2 K T K (0) n! π n λ n (K ) λ n (K). Therefore λ n (K)λ n (K ) ( π 4 ) 3n 4 n n!.

30 The upper bound K TK (0) n! λ n (K ) π n easily follows from Rothaus λ n (K) formula (1968): K TK (0) = (2π) n dλ, J K where J K (y) = K R n e 2x y dλ(x). ( π ) 2n 1 To show the lower bound K TK (0) we can use the 4 (λ n (K)) 2 estimate: and K TK (0) Proposition I TK (0) 4 (K + ik) π ( π ) n 1 Conjecture K TK (0) 4 (λ n (K)) 2 1 λ 2n (I TK (0)) This would be optimal, since we have equality for cubes.

31 However, one can check that for K = { x 1 + x 2 + x 3 1} we have ( π ) 3 1 K TK (0) > 4 (λ 3 (K)) 2. This shows that Nazarov s proof of the Bourgain-Milman inequality cannot give the Mahler conjecture directly.

32 Thank you!

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