Bergman Kernel in the Annulus

Bergman Kernel in the Annulus

Zbigniew Błocki

(Jagiellonian University, Kraków, Poland) http://gamma.im.uj.edu.pl/ eblocki

Tambara Workshop

on the Bergman Kernel and Related Topics

October 24-28, 2011

Formulas for the Bergman kernel in P = {r < |z| < 1}

I. Since (z ^j ) _j∈Z is an orthogonal system in P , we have K _P (z, w) = h(λ)

πλ , where λ = z ¯ w and

h(λ) = 1

2 log(1/r) + X

j∈Z

jλ ^j 1 − r ^2j . Zeros of K

: If r < e ⁻⁴ then h has a zero in S := {|λ| = r} ∪ {λ ∈ R : r ² < |λ| < 1}

(Skwarczyński, 1969).

Proof: h(λ) ∈ R for λ ∈ S, h > 0 in S ∩ R + and

h < 0 in S ∩ R − near −1.

II. Weierstrass elliptic function P:

ω 1 = − log r, ω 2 = πi, Λ := {2jω 1 +2kω 2 : (j, k) ∈ Z ² } P(z) := 1

z ² + X

ω∈Λ

 1

(z − ω) ² − 1 ω ²

 .

h(λ) = P(log λ) + η ₁ ω 1

, (Zarankiewicz, 1934) where η ₁ = ζ(ω ₁ ) and the Weierstrass elliptic function ζ is determined by

ζ ⁰ = −P, ζ(z) = 1

z + O(|z|).

Zeros of K

: h has two zeros in {r ² < |λ| < 1} (for every r) (Rosenthal, 1969).

Proof: P attains every value of ¯ C twice in

{2tω ₁ + 2sω ₂ : s, t ∈ [0, 1)}.

III. For any bounded Ω ⊂ C K _Ω = 2

π

∂ ² G _Ω

∂z∂ ¯ w , (Schiffer), where G Ω (·, w) is the (negative) Green function.

If p : ∆ → Ω is a covering then for any λ 0 ∈ p ⁻¹ (w) G _Ω (z, w) = X

µ∈p

(z)

G _∆ (λ ₀ , µ) (Myrberg, 1933).

If U is a neighb. of w and V j are s.th. p ⁻¹ (U ) = S V _j and p| V

→ U are biholomorphic, then for ϕ _j := (p| V

) ⁻¹

G Ω (z, w) = X

j

log     

ϕ j (z) − ϕ 0 (w) 1 − ϕ j (z)ϕ 0 (w)

     .

K Ω (z, w) = π X

j

ϕ ⁰ _j (z)ϕ ⁰ ₀ (w)

(1 − ϕ _j (z)ϕ ₀ (w)) ² .

For Ω = P

p(ζ) = exp  log r πi Log

 i 1 + ζ

1 − ζ



,

ϕ j (z) = e πi(Log z+2jπi)/ log r − i

e πi(Log z+2jπi)/ log r + i , j ∈ Z.

Therefore

h(λ) = − π ² log ² r

X

j∈Z

f j (λ) (1 − f j (λ)) ² , where

f _j (z) = exp πi(Log z + 2jπi)

log r .

We will get

h(−r) = π ² log ² r

X

j∈Z

q ^2j+1

(1 + q ^2j+1 ) ² > 0, h(−r ² ) = h(−1) = − π ²

log ² r X

j∈Z

q ^2j+1

(1 − q ^2j+1 ) ² < 0 where q = e ^π

^{/ log r} < 1.

Theorem. h has exactly two zeros in {r ² < |λ| < 1}: one

on the interval (−1, −r) and one on (−r, −r ² ).

Suita Conjecture (1972): For Ω ⊂⊂ C we have e ^2ψ(z) ≤ πK _Ω (z, z),

where ψ(w) := lim

z→w (G _Ω (z, w) − log |z − w|) (Robin f.) Another formulation: since

K _Ω (z, z) = 1

π ψ _{z ¯ z} (Suita, 1972), we have

Suita Conjecture ⇔ e ^2ψ ≤ ψ _{z ¯ z} ⇔ K _e

|dz| ≤ −1.

Slightly more general statement would be:

K _e

|dz| satisfies the maximum principle.

Ohsawa, 1995: e ^2ψ ≤ 750ψ _{z ¯ z} B., 2007: e ^2ψ ≤ 2ψ _{z ¯ z}

Guan-Zhou-Zhu, 2011: e ^2ψ ≤ 1.954 . . . ψ _{z ¯ z}

Suita, 1972: e ^2ψ < ψ z ¯ z for Ω = P Sketch of proof:

One can show that ψ(z) = γ(t), where t = −2 log |z|, γ(t) = c

2 t ² + t

2 − log σ(t),

c = η 1 /ω 1 , and the Weierstrass elliptic function σ is determined by σ ⁰ /σ = ζ, σ(z) = z + O(|z| ² ). Then

ψ _{z ¯ z} (z) = e ^t γ ⁰⁰ (t) = e ^t (P(t) + c).

Set

F := log(−K _e

|dz| ) = log ψ _{z ¯ z} e ^2ψ

= log(P + c) + 2 log σ − ct ² .

Have to show that F > 0 on (0, 2ω 1 ).

F = log(P + c) + 2 log σ − ct ² .

We have F (2ω ₁ − t) = F (t), F (0) = F ⁰ (0) = F ⁰ (ω ₁ ) = 0 and F (ω 1 ) > 0. Key: differential equation for P

(P ⁰ ) ² = 4P ³ − g ₂ P − g ₃ , where

g 2 = 60 X

ω∈Λ

1

ω ⁴ , g 3 = 140 X

ω∈Λ

1 ω ⁶ . Therefore

F ⁰⁰ = b(P + c) − a (P + c) ² , where

a = −4c ³ + cg ₂ − g ₃ > 0, b = g ₂

2 − 6c ² > 0,

and F ⁰⁰ vanishes exactly once in (0, ω ₁ ) and thus F > 0.

2 4 6 8 10

-7 -6 -5 -4 -3 -2 -1

K _e

|dz| for r = e ⁻⁵

Curvature of the “Bergman” metric K

(z, z)|dz|

R 1 = 2K _ψ

_|dz|

= − (log ψ z ¯ z ) z ¯ z

ψ z ¯ z

= − Q ◦ P (P + c) ³ , where

Q(x) = 2(x + c) ³ + b(x + c) − a.

Therefore

R ⁰ ₁ = 2b(P + c) − 3a (P + c) ⁴ P ⁰ .

5 10 15 20

-3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5

r = e ⁻¹⁰

Curvature of the Bergman metric (log K

(z, z))

|dz|

R ₂ := 2 K _{(log ψ}

₎

_|dz|

= − (log(log ψ _{z ¯ z} ) _{z ¯ z} ) _{z ¯ z} (log ψ _{z ¯ z} ) _{z ¯ z} Since ψ z ¯ z = (P + c)e ^t , we can compute that

R ₂ = 2 − (P + c) ³ S ◦ P (Q ◦ P) ³ , where

S(y − c) = 24y ⁶ + 60by ⁴ − 96(a + bc)y ³ + 6(36ac − b ² )y ² + 24aby − 12a ² + b ³ + 12abc.

Then

R ⁰ ₂ = (P + c) ³ W ◦ P

(Q ◦ P) ⁴ P ⁰ ,

where W is a polynomial of degree 7.

W (y − c) = − 36a ³ + 3ab ³ + 36a ² bc + 96a ² by − 54ab ² y ² + 1080a ² cy ² − 720a ² y ³ + 24b ³ y ³ − 864abcy ³ + 948aby ⁴ + 288b ² cy ⁴ − 288b ² y ⁵ + 1728acy ⁵

− 360ay ⁶ − 576bcy ⁶ + 96by ⁷ Proposition. W (P(ω ₁ )) > 0

Conjecture. W (P(ω ₁ /2)) < 0

R ⁰ ₂ = (P + c) ³ W ◦ P (Q ◦ P) ⁴ P ⁰ ,

1 2 3 4 5

-6 -5 -4 -3 -2 -1

r = e ^−2.5