Bergman Kernel and Pluripotential Theory

Transcription

1 Bergman Kernel and Pluripotential Theory Zbigniew B locki Uniwersytet Jagielloński, Kraków, Poland blocki İstanbul Analysis Seminar March 21, 2014

2 Bergman Completeness

3 Bergman Completeness Ω bounded domain in C n H 2 (Ω) = O(Ω) L 2 (Ω)

4 Bergman Completeness Ω bounded domain in C n H 2 (Ω) = O(Ω) L 2 (Ω) K Ω (, ) Bergman kernel f (w) = f K Ω (, w) dλ, Ω w Ω, f H 2 (Ω) K Ω (w) = K Ω (w, w) = sup{ f (w) 2 : f H 2 (Ω), f 1}

5 Bergman Completeness Ω bounded domain in C n H 2 (Ω) = O(Ω) L 2 (Ω) K Ω (, ) Bergman kernel f (w) = f K Ω (, w) dλ, Ω w Ω, f H 2 (Ω) K Ω (w) = K Ω (w, w) = sup{ f (w) 2 : f H 2 (Ω), f 1} Ω is called Bergman complete if it is complete w.r.t. the Bergman metric B Ω = i log K Ω

6 Kobayashi Criterion (1959) If f (w) 2 lim w Ω K Ω (w) = 0, f H2 (Ω), then Ω is Bergman complete.

7 Kobayashi Criterion (1959) If f (w) 2 lim w Ω K Ω (w) = 0, f H2 (Ω), then Ω is Bergman complete. The opposite is not true even for n = 1 (Zwonek, 2001).

8 Kobayashi Criterion (1959) If f (w) 2 lim w Ω K Ω (w) = 0, f H2 (Ω), then Ω is Bergman complete. The opposite is not true even for n = 1 (Zwonek, 2001). Kobayashi Criterion easily follows using the embedding ι : Ω w [K Ω (, w)] P(H 2 (Ω)) and the fact that ι ω FS = B Ω.

9 Kobayashi Criterion (1959) If f (w) 2 lim w Ω K Ω (w) = 0, f H2 (Ω), then Ω is Bergman complete. The opposite is not true even for n = 1 (Zwonek, 2001). Kobayashi Criterion easily follows using the embedding ι : Ω w [K Ω (, w)] P(H 2 (Ω)) and the fact that ι ω FS = B Ω. Since ι is distance decreasing, distω B K Ω (z, w) (z, w) arccos KΩ (z)k Ω (w).

10 Some Pluripotential Theory

11 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω).

12 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex.

13 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex. Pluricomplex Green function For a pole w Ω we set G Ω (, w) = G w = sup{v PSH (Ω) : v log w + C}

14 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex. Pluricomplex Green function For a pole w Ω we set G Ω (, w) = G w = sup{v PSH (Ω) : v log w + C} Lempert (1981) Ω convex G Ω symmetric

15 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex. Pluricomplex Green function For a pole w Ω we set G Ω (, w) = G w = sup{v PSH (Ω) : v log w + C} Lempert (1981) Ω convex G Ω symmetric D ly (1985) Ω hyperconvex e G Ω C( Ω Ω)

16 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex. Pluricomplex Green function For a pole w Ω we set G Ω (, w) = G w = sup{v PSH (Ω) : v log w + C} Lempert (1981) Ω convex G Ω symmetric D ly (1985) Ω hyperconvex e G Ω C( Ω Ω) Open Problem e G Ω C( Ω Ω \ Ω )

17 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex. Pluricomplex Green function For a pole w Ω we set G Ω (, w) = G w = sup{v PSH (Ω) : v log w + C} Lempert (1981) Ω convex G Ω symmetric D ly (1985) Ω hyperconvex e G Ω C( Ω Ω) Open Problem e G Ω C( Ω Ω \ Ω ) Equivalently: G(, w k ) 0 loc. uniformly as w k Ω?

18 Some Pluripotential Theory Ω is called hyperconvex if it admits a negative plurisubharmonic (psh) exhaustion function (u PSH (Ω) s.th. u = 0 on Ω). D ly (1985) If Ω is pseudoconvex with Lipschitz boundary then it is hyperconvex. Pluricomplex Green function For a pole w Ω we set G Ω (, w) = G w = sup{v PSH (Ω) : v log w + C} Lempert (1981) Ω convex G Ω symmetric D ly (1985) Ω hyperconvex e G Ω C( Ω Ω) Open Problem e G Ω C( Ω Ω \ Ω ) Equivalently: G(, w k ) 0 loc. uniformly as w k Ω? True if Ω C 2 (Herbort, 2000)

19 D ly (1985) If Ω is hyperconvex then G w = G Ω (, w) is the unique solution to u PSH(Ω) C( Ω \ {w}) (dd c u) n = (2π) n δ w u = 0 on Ω u log w + C

20 D ly (1985) If Ω is hyperconvex then G w = G Ω (, w) is the unique solution to u PSH(Ω) C( Ω \ {w}) (dd c u) n = (2π) n δ w u = 0 on Ω u log w + C B. (2000) If Ω is smooth and strongly pseudoconvex then G w C 1,1 ( Ω \ {w}).

21 D ly (1985) If Ω is hyperconvex then G w = G Ω (, w) is the unique solution to u PSH(Ω) C( Ω \ {w}) (dd c u) n = (2π) n δ w u = 0 on Ω u log w + C B. (2000) If Ω is smooth and strongly pseudoconvex then G w C 1,1 ( Ω \ {w}). B. (1995) If Ω is hyperconvex then! u = u Ω s.th. u PSH(Ω) C( Ω) (dd c u) n = 1 dλ u = 0 on Ω.

22 B.-Y. Chen, Pflug - B. / Herbort (1998) Hyperconvex domains are Bergman complete

23 B.-Y. Chen, Pflug - B. / Herbort (1998) Hyperconvex domains are Bergman complete Herbort If Ω is pseudoconvex then f (w) 2 K Ω (w) c n {G w < 1} f 2 dλ, w Ω, f H 2 (Ω).

24 B.-Y. Chen, Pflug - B. / Herbort (1998) Hyperconvex domains are Bergman complete Herbort If Ω is pseudoconvex then Corollary f (w) 2 K Ω (w) c n {G w < 1} f 2 dλ, w Ω, f H 2 (Ω). lim λ({g w < 1})=0 Ω is Bergman complete w Ω

25 B.-Y. Chen, Pflug - B. / Herbort (1998) Hyperconvex domains are Bergman complete Herbort If Ω is pseudoconvex then f (w) 2 K Ω (w) c n {G w < 1} f 2 dλ, w Ω, f H 2 (Ω). Corollary lim λ({g w < 1})=0 Ω is Bergman complete w Ω Proposition If Ω is hyperconvex then lim G w L w Ω n (Ω) = 0.

26 B.-Y. Chen, Pflug - B. / Herbort (1998) Hyperconvex domains are Bergman complete Herbort If Ω is pseudoconvex then f (w) 2 K Ω (w) c n {G w < 1} f 2 dλ, w Ω, f H 2 (Ω). Corollary lim λ({g w < 1})=0 Ω is Bergman complete w Ω Proposition If Ω is hyperconvex then lim G w L w Ω n (Ω) = 0. Sketch of proof G w n n = Ω G w n (dd c u Ω ) n n! u Ω n 1 u Ω (dd c G w ) n C(n, λ(ω)) u Ω (w) Ω

27 Lower Bound for the Bergman Distance

28 Lower Bound for the Bergman Distance Diederich-Ohsawa (1994), B. (2005) If Ω is pseudoconvex with C 2 boundary then distω B log (, w) where δ Ω (z) = dist Ω (z, Ω). δ 1 Ω C log log δ 1 Ω,

29 Lower Bound for the Bergman Distance Diederich-Ohsawa (1994), B. (2005) If Ω is pseudoconvex with C 2 boundary then distω B log (, w) where δ Ω (z) = dist Ω (z, Ω). δ 1 Ω C log log δ 1 Ω Pluripotential theory is the main tool in proving this estimate, in particular we have the following:,

30 Lower Bound for the Bergman Distance Diederich-Ohsawa (1994), B. (2005) If Ω is pseudoconvex with C 2 boundary then distω B log (, w) where δ Ω (z) = dist Ω (z, Ω). δ 1 Ω C log log δ 1 Ω Pluripotential theory is the main tool in proving this estimate, in particular we have the following: B. (2005) If Ω is pseudoconvex and z, w Ω are such that {G z < 1} {G w < 1} =, then dist B Ω (z, w) c n > 0.

31 Lower Bound for the Bergman Distance Diederich-Ohsawa (1994), B. (2005) If Ω is pseudoconvex with C 2 boundary then distω B log (, w) where δ Ω (z) = dist Ω (z, Ω). δ 1 Ω C log log δ 1 Ω Pluripotential theory is the main tool in proving this estimate, in particular we have the following: B. (2005) If Ω is pseudoconvex and z, w Ω are such that {G z < 1} {G w < 1} =, then dist B Ω (z, w) c n > 0. Open Problem dist B Ω (, w) 1 C log δ 1 Ω

32 From Herbort s estimate f (w) 2 K Ω (w) c n for f 1 we get {G w < 1} K Ω (w) f 2 dλ, w Ω, f H 2 (Ω), 1 c n λ({g w < 1}).

33 From Herbort s estimate f (w) 2 K Ω (w) c n for f 1 we get {G w < 1} K Ω (w) f 2 dλ, w Ω, f H 2 (Ω), 1 c n λ({g w < 1}). To find the optimal constant c n here turns out to have very interesting consequences!

34 From Herbort s estimate f (w) 2 K Ω (w) c n for f 1 we get {G w < 1} K Ω (w) f 2 dλ, w Ω, f H 2 (Ω), 1 c n λ({g w < 1}). To find the optimal constant c n here turns out to have very interesting consequences! Herbort (1999) c n = 1 + 4e 4n+3+R2, where Ω B(z 0, R) (Main tool: Hörmander s estimate for )

35 From Herbort s estimate f (w) 2 K Ω (w) c n for f 1 we get {G w < 1} K Ω (w) f 2 dλ, w Ω, f H 2 (Ω), 1 c n λ({g w < 1}). To find the optimal constant c n here turns out to have very interesting consequences! Herbort (1999) c n = 1 + 4e 4n+3+R2, where Ω B(z 0, R) (Main tool: Hörmander s estimate for ) B. (2005) c n = (1 + 4/Ei(n)) 2 ds, where Ei(t) = t se s (Main tool: Donnelly-Fefferman s estimate for )

36 Suita Conjecture

37 Suita Conjecture D bounded domain in C

38 Suita Conjecture D bounded domain in C c D (z) := exp lim ζ z (G D (ζ, z) log ζ z ) (logarithmic capacity of C \ D w.r.t. z)

39 Suita Conjecture D bounded domain in C c D (z) := exp lim ζ z (G D (ζ, z) log ζ z ) (logarithmic capacity of C \ D w.r.t. z) c D dz is an invariant metric (Suita metric)

40 Suita Conjecture D bounded domain in C c D (z) := exp lim ζ z (G D (ζ, z) log ζ z ) (logarithmic capacity of C \ D w.r.t. z) c D dz is an invariant metric (Suita metric) Curv cd dz = (log c D) z z c 2 D

41 Suita Conjecture D bounded domain in C c D (z) := exp lim ζ z (G D (ζ, z) log ζ z ) (logarithmic capacity of C \ D w.r.t. z) c D dz is an invariant metric (Suita metric) Curv cd dz = (log c D) z z c 2 D Suita Conjecture (1972): Curv cd dz 1 = if D is simply connected < if D is an annulus (Suita) Enough to prove for D with smooth boundary = on D if D has smooth boundary

42 Curv cd dz for D = {e 5 < z < 1} as a function of log z

43 Curv (log KD ) z z dz 2 for D = {e 5 < z < 1} as a function of log z

44 2 z z (log c D) = πk D (Suita)

45 2 z z (log c D) = πk D (Suita) Therefore the Suita conjecture is equivalent to c 2 D πk D.

46 2 z z (log c D) = πk D (Suita) Therefore the Suita conjecture is equivalent to c 2 D πk D. Ohsawa (1995) observed that it is really an extension problem: for z D find holomorphic f in D such that f (z) = 1 and f 2 π dλ (c D (z)) 2. D

47 2 z z (log c D) = πk D (Suita) Therefore the Suita conjecture is equivalent to c 2 D πk D. Ohsawa (1995) observed that it is really an extension problem: for z D find holomorphic f in D such that f (z) = 1 and f 2 π dλ (c D (z)) 2. D Using the methods of the Ohsawa-Takegoshi extension theorem he showed the estimate c 2 D CπK D with C = 750.

48 2 z z (log c D) = πk D (Suita) Therefore the Suita conjecture is equivalent to c 2 D πk D. Ohsawa (1995) observed that it is really an extension problem: for z D find holomorphic f in D such that f (z) = 1 and f 2 π dλ (c D (z)) 2. D Using the methods of the Ohsawa-Takegoshi extension theorem he showed the estimate c 2 D CπK D with C = 750. C = 2 (B., 2007) C = (Guan-Zhou-Zhu, 2011)

49 Ohsawa-Takegoshi extension theorem (1987)

50 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013)

51 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013) 0 D C, Ω C n 1 D, Ω pseudoconvex,

52 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013) 0 D C, Ω C n 1 D, Ω pseudoconvex, ϕ PSH(Ω) f holomorphic in Ω := Ω {z n = 0}

53 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013) 0 D C, Ω C n 1 D, Ω pseudoconvex, ϕ PSH(Ω) f holomorphic in Ω := Ω {z n = 0} Then there exists a holomorphic extension F of f to Ω such that Ω F 2 e ϕ dλ π c D (0) 2 Ω f 2 e ϕ dλ.

54 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013) 0 D C, Ω C n 1 D, Ω pseudoconvex, ϕ PSH(Ω) f holomorphic in Ω := Ω {z n = 0} Then there exists a holomorphic extension F of f to Ω such that Ω F 2 e ϕ dλ π c D (0) 2 Ω f 2 e ϕ dλ. For n = 1 and ϕ 0 we get the Suita conjecture.

55 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013) 0 D C, Ω C n 1 D, Ω pseudoconvex, ϕ PSH(Ω) f holomorphic in Ω := Ω {z n = 0} Then there exists a holomorphic extension F of f to Ω such that Ω F 2 e ϕ dλ π c D (0) 2 Ω f 2 e ϕ dλ. For n = 1 and ϕ 0 we get the Suita conjecture. Main tool: Hörmander s estimate for

56 Ohsawa-Takegoshi extension theorem (1987) with optimal constant (B., 2013) 0 D C, Ω C n 1 D, Ω pseudoconvex, ϕ PSH(Ω) f holomorphic in Ω := Ω {z n = 0} Then there exists a holomorphic extension F of f to Ω such that Ω F 2 e ϕ dλ π c D (0) 2 Ω f 2 e ϕ dλ. For n = 1 and ϕ 0 we get the Suita conjecture. Main tool: Hörmander s estimate for B.-Y. Chen (2011) proved that the Ohsawa-Takegoshi theorem (without optimal constant) follows form Hörmander s estimate.

57 Tensor Power Trick

58 Tensor Power Trick We have K Ω (w) where c n = (1 + 4/Ei(n)) 2. 1 c n λ({g w < 1})

59 Tensor Power Trick We have K Ω (w) where c n = (1 + 4/Ei(n)) 2. 1 c n λ({g w < 1}) Take m 0 and set Ω := Ω m C nm, w := (w,..., w).

60 Tensor Power Trick We have K Ω (w) 1 c n λ({g w < 1}) where c n = (1 + 4/Ei(n)) 2. Take m 0 and set Ω := Ω m C nm, w := (w,..., w). Then K Ω( w) = (K Ω (w)) m, λ 2nm ({G w < 1}) = (λ 2n ({G w < 1}) m. Therefore K Ω (w) 1 c 1/m nm λ({g w < 1})

61 Tensor Power Trick We have K Ω (w) 1 c n λ({g w < 1}) where c n = (1 + 4/Ei(n)) 2. Take m 0 and set Ω := Ω m C nm, w := (w,..., w). Then K Ω( w) = (K Ω (w)) m, λ 2nm ({G w < 1}) = (λ 2n ({G w < 1}) m. Therefore but K Ω (w) 1 c 1/m nm λ({g w < 1}) lim m c1/m nm = e 2n.

62 Repeating this argument for any sublevel set we get Theorem 1 Assume Ω is pseudoconvex in C n. Then for t 0 and w Ω 1 K Ω (w) e 2nt λ({g w < t}).

63 Repeating this argument for any sublevel set we get Theorem 1 Assume Ω is pseudoconvex in C n. Then for t 0 and w Ω 1 K Ω (w) e 2nt λ({g w < t}). Lempert recently noticed that this estimate can also be proved using Berndtsson s result on positivity of direct image bundles.

64 Repeating this argument for any sublevel set we get Theorem 1 Assume Ω is pseudoconvex in C n. Then for t 0 and w Ω 1 K Ω (w) e 2nt λ({g w < t}). Lempert recently noticed that this estimate can also be proved using Berndtsson s result on positivity of direct image bundles. What happens when t?

65 Repeating this argument for any sublevel set we get Theorem 1 Assume Ω is pseudoconvex in C n. Then for t 0 and w Ω 1 K Ω (w) e 2nt λ({g w < t}). Lempert recently noticed that this estimate can also be proved using Berndtsson s result on positivity of direct image bundles. What happens when t?

66 Repeating this argument for any sublevel set we get Theorem 1 Assume Ω is pseudoconvex in C n. Then for t 0 and w Ω 1 K Ω (w) e 2nt λ({g w < t}). Lempert recently noticed that this estimate can also be proved using Berndtsson s result on positivity of direct image bundles. What happens when t? For n = 1 we get K Ω cω 2 /π (another proof of Suita Conjecture).

67 Repeating this argument for any sublevel set we get Theorem 1 Assume Ω is pseudoconvex in C n. Then for t 0 and w Ω 1 K Ω (w) e 2nt λ({g w < t}). Lempert recently noticed that this estimate can also be proved using Berndtsson s result on positivity of direct image bundles. What happens when t? For n = 1 we get K Ω cω 2 /π (another proof of Suita Conjecture). Theorem 2 If Ω is a convex domain in C n then for w Ω K Ω (w) 1 λ(i Ω (w)), I Ω (w) = {ϕ (0) : ϕ O(, Ω), ϕ(0) = w} (Kobayashi indicatrix).

68 Mahler Conjecture

69 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(k )

70 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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.

71 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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. Santaló Inequality (1949) Mahler volume is maximized by balls

72 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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. Santaló Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes

73 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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. Santaló Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes Hansen-Lima bodies: starting from an interval they are produced by taking products of lower dimensional HL bodies and their duals.

74 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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. Santaló Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes Hansen-Lima bodies: starting from an interval they are produced by taking products of lower dimensional HL bodies and their duals. n = 2: square

75 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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. Santaló Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes Hansen-Lima bodies: starting from an interval they are produced by taking products of lower dimensional HL bodies and their duals. n = 2: square n = 3: cube & octahedron

76 Mahler Conjecture K - convex symmetric body in R n K := {y R n : x y 1 for every x K} Mahler volume := λ(k)λ(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. Santaló Inequality (1949) Mahler volume is maximized by balls Mahler Conjecture (1938) Mahler volume is minimized by cubes Hansen-Lima bodies: starting from an interval they are produced by taking products of lower dimensional HL bodies and their duals. n = 2: square n = 3: cube & octahedron n = 4:...

77 Bourgain-Milman (1987) There exists c > 0 such that λ(k)λ(k ) c n 4n n!.

78 Bourgain-Milman (1987) There exists c > 0 such that Mahler Conjecture: c = 1 λ(k)λ(k ) c n 4n n!.

79 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!.

80 Bourgain-Milman (1987) There exists c > 0 such that Mahler Conjecture: c = 1 G. Kuperberg (2006) c = π/4 Nazarov (2012) λ(k)λ(k ) c n 4n n!. equivalent SCV formulation of the Mahler Conjecture via the Fourier transform and the Paley-Wiener theorem

81 Bourgain-Milman (1987) There exists c > 0 such that Mahler Conjecture: c = 1 G. Kuperberg (2006) c = π/4 Nazarov (2012) λ(k)λ(k ) c n 4n n!. equivalent SCV formulation of the Mahler Conjecture via the Fourier transform and the Paley-Wiener theorem proof of the Bourgain-Milman Inequality (c = (π/4) 3 ) using Hörmander s estimate for

82 K - convex symmetric body in R n Nazarov: consider T K := intk + ir n C n.

83 K - convex symmetric body in R n Nazarov: consider 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).

84 K - convex symmetric body in R n Nazarov: consider 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!.

85 K - convex symmetric body in R n Nazarov: consider 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!. To show the lower bound we can use Theorem 2: K TK (0) 1 λ 2n (I TK (0)).

86 K - convex symmetric body in R n Nazarov: consider 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!. To show the lower bound we can use Theorem 2: K TK (0) 1 λ 2n (I TK (0)). Proposition I TK (0) 4 (K + ik) π

87 K - convex symmetric body in R n Nazarov: consider 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!. To show the lower bound we can use Theorem 2: K TK (0) 1 λ 2n (I TK (0)). Proposition I TK (0) 4 (K + ik) π ( ) 4 2n In particular, λ 2n (I TK (0)) (λ n (K)) 2 π

88 K - convex symmetric body in R n Nazarov: consider 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!. To show the lower bound we can use Theorem 2: K TK (0) 1 λ 2n (I TK (0)). Proposition I TK (0) 4 (K + ik) π ( ) 4 2n In particular, λ 2n (I TK (0)) (λ n (K)) 2 π ( ) 4 n Conjecture λ 2n (I TK (0)) (λ n (K)) 2 π

89 Lempert Theory (1981)

90 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary

91 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary ϕ O(, Ω) C(, Ω) is a geodesic if and only if ϕ( ) Ω and there exists h O(, C n ) C(, C n ) s.th. the vector e it h(e it ) is outer normal to Ω at ϕ(e it ) for every t R.

92 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary ϕ O(, Ω) C(, Ω) is a geodesic if and only if ϕ( ) Ω and there exists h O(, C n ) C(, C n ) s.th. the vector e it h(e it ) is outer normal to Ω at ϕ(e it ) for every t R. F O(Ω, ) a left-inverse to ϕ (i.e. F ϕ = id ) s.th. (z ϕ(f (z))) h(f (z)) = 0, z Ω.

93 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary ϕ O(, Ω) C(, Ω) is a geodesic if and only if ϕ( ) Ω and there exists h O(, C n ) C(, C n ) s.th. the vector e it h(e it ) is outer normal to Ω at ϕ(e it ) for every t R. F O(Ω, ) a left-inverse to ϕ (i.e. F ϕ = id ) s.th. (z ϕ(f (z))) h(f (z)) = 0, z Ω. Lempert s Theory for Tube Domains (S. Zaj ac, 2013)

94 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary ϕ O(, Ω) C(, Ω) is a geodesic if and only if ϕ( ) Ω and there exists h O(, C n ) C(, C n ) s.th. the vector e it h(e it ) is outer normal to Ω at ϕ(e it ) for every t R. F O(Ω, ) a left-inverse to ϕ (i.e. F ϕ = id ) s.th. (z ϕ(f (z))) h(f (z)) = 0, z Ω. Lempert s Theory for Tube Domains (S. Zaj ac, 2013) Ω = T K, where K is smooth and strongly convex in R n

95 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary ϕ O(, Ω) C(, Ω) is a geodesic if and only if ϕ( ) Ω and there exists h O(, C n ) C(, C n ) s.th. the vector e it h(e it ) is outer normal to Ω at ϕ(e it ) for every t R. F O(Ω, ) a left-inverse to ϕ (i.e. F ϕ = id ) s.th. (z ϕ(f (z))) h(f (z)) = 0, z Ω. Lempert s Theory for Tube Domains (S. Zaj ac, 2013) Ω = T K, where K is smooth and strongly convex in R n Since Im (e it h(e it )) = 0, h must be of the form for some w C n and b R n. h(ζ) = w + ζb + ζ 2 w

96 Lempert Theory (1981) Ω - bounded strongly convex domain in C n with smooth boundary ϕ O(, Ω) C(, Ω) is a geodesic if and only if ϕ( ) Ω and there exists h O(, C n ) C(, C n ) s.th. the vector e it h(e it ) is outer normal to Ω at ϕ(e it ) for every t R. F O(Ω, ) a left-inverse to ϕ (i.e. F ϕ = id ) s.th. (z ϕ(f (z))) h(f (z)) = 0, z Ω. Lempert s Theory for Tube Domains (S. Zaj ac, 2013) Ω = T K, where K is smooth and strongly convex in R n Since Im (e it h(e it )) = 0, h must be of the form h(ζ) = w + ζb + ζ 2 w for some w C n and b R n. Therefore ( b + 2Re (e Re ϕ(e it ) = ν 1 it ) w) b + 2Re (e it, w) where ν : K S n 1 is the Gauss map.

97 By the Schwarz formula ϕ(ζ) = 1 2π e it ( + ζ b + 2Re (e it ) w) 2π 0 e it ζ ν 1 b + 2Re (e it dt + iim ϕ(0). w)

98 By the Schwarz formula ϕ(ζ) = 1 2π e it ( + ζ b + 2Re (e it ) w) 2π 0 e it ζ ν 1 b + 2Re (e it dt + iim ϕ(0). w) If K is in addition symmetric then all geodesics in T K with ϕ(0) = 0 are of the form for some w (C n ). ϕ(ζ) = 1 2π e it ( + ζ Re (e it ) w) 2π 0 e it ζ ν 1 Re (e it dt w)

99 By the Schwarz formula ϕ(ζ) = 1 2π e it ( + ζ b + 2Re (e it ) w) 2π 0 e it ζ ν 1 b + 2Re (e it dt + iim ϕ(0). w) If K is in addition symmetric then all geodesics in T K with ϕ(0) = 0 are of the form ϕ(ζ) = 1 2π e it ( + ζ Re (e it ) w) 2π 0 e it ζ ν 1 Re (e it dt w) for some w (C n ). Then ϕ (0) = 1 π 2π parametrizes I TK (0) for w S 2n 1. 0 ( ) Re (e e it ν 1 it w) Re (e it dt w)

100 By the Schwarz formula ϕ(ζ) = 1 2π e it ( + ζ b + 2Re (e it ) w) 2π 0 e it ζ ν 1 b + 2Re (e it dt + iim ϕ(0). w) If K is in addition symmetric then all geodesics in T K with ϕ(0) = 0 are of the form ϕ(ζ) = 1 2π e it ( + ζ Re (e it ) w) 2π 0 e it ζ ν 1 Re (e it dt w) for some w (C n ). Then ϕ (0) = 1 π 2π 0 ( ) Re (e e it ν 1 it w) Re (e it dt w) parametrizes I TK (0) for w S 2n 1. ( ) 4 n Conjecture λ 2n (I TK (0)) (λ n (K)) 2 π