Suita Conjecture for Some Convex Ellipsoids in \mathbb{C}^{2}

# On the Suita Conjecture for Some Convex Ellipsoids in C2

Zbigniew Błocki, Włodzimierz Zwonek Uniwersytet Jagielloński
a   Instytut Matematyki
a   Łojasiewicza 6
a   30-348 Kraków
a   Poland
a   Zbigniew.Blocki@im.uj.edu.pl
a   Wlodzimierz.Zwonek@im.uj.edu.pl
###### Abstract.

It has been recently shown that for a convex domain in and the function , where is the Bergman kernel on the diagonal and the Kobayashi indicatrix, satisfies . While the lower bound is optimal, not much more is known about the upper bound. In general it is quite difficult to compute even numerically and the highest value of it obtained so far is In this paper we present precise, although rather complicated formulas for the ellipsoids (with ) and all , as well as for and on the diagonal. The Bergman kernel for those ellipsoids had been known, the main point is to compute the volume of the Kobayashi indicatrix. It turns out that in the second case the function is not .

The first named author was supported by the Ideas Plus grant 0001/ID3/2014/63 of the Polish Ministry Of Science and Higher Education and the second named author by the Polish National Science Centre grant 2011/03/B/ST1/04758

## Introduction

For a convex domain in and the following estimates have been recently established:

 (1) 1λ(IΩ(w))≤KΩ(w)≤4nλ(IΩ(w)).

Here

 KΩ(w)=sup{|f(w)|2:f∈O(Ω), ∫Ω|f|2dλ≤1}

is the Bergman kernel on the diagonal and

 IΩ(w)={φ′(0):φ∈O(Δ,Ω), φ(0)=w}

is the Kobayashi indicatrix, where denotes the unit disc. The first inequality in (1) was shown in [3], the proof uses -estimates for and Lempert’s theory [9]. It is optimal, for example if is balanced with respect to (that is every intersection of with a complex line containing is a disc) then we have equality. It can be viewed as a multi-dimensional version of the Suita conjecture [11] proved in [2] (see also [5] for the precise characterization when equality holds).

The second equality in (1) was proved in [4] using rather elementary methods. It was also shown that the constant can be replaced by if is in addition symmetric with respect to . We can write (1) as

 1≤FΩ(w)≤4,

where is a biholomorphically invariant function in . It is not clear what the optimal upper bound should be. It was in fact quite difficult to prove that one can at all have . It was done in [4] for ellipsoids of the form , where and . The function was also computed numerically for the ellipsoid , , based on an implicit formula for the Kobayashi function from [1]. Our first result is the precise formula in this case:

###### Theorem 1.

For define

 Ωm={z∈C2:|z1|2m+|z2|2<1}.

Then for , and with , we have

 λ(IΩm((b,0)))=π2 [−m−12m(3m−2)(3m−1)b6m+2−3(m−1)2m(m−2)(m+1)b2m+2 +m2(m−2)(3m−2)b6+3m3m−1b4−4m−12mb2+mm+1].

For and one has

 λ(IΩ2/3((b,0)))=π280(−65b6+40b6logb+160b4−27b10/3−100b2+32),
 λ(IΩ2((b,0)))=π2240(−3b14−25b6−120b6logb+288b4−420b2+160).

The general formula for the Kobayashi function for is known, see [1], but it is implicit in the sense that it requires solving a nonlinear equation which is polynomial of degree if it is an integer. It turns out however that the volume of the Kobayashi indicatrix for , that is the set where the Kobayashi function is not bigger than 1, can be found explicitly. It would be interesting to check whether Theorem 1 also holds in the non-convex case, that is when (see [10] for computations of the Kobayashi metric in this case).

The formula for the Bergman kernel for this ellipsoid is well known (see e.g. [7], Example 6.1.6):

 KΩm(w)=1π2(1−|w2|2)1/m−2(1/m+1)(1−|w2|2)1/m+(1/m−1)|w1|2((1−|w2|2)1/m−|w1|2)3,

so that

 KΩm((b,0))=m+1+(1−m)b2π2m(1−b2)3,

and we can obtain the following graphs of for example for , 8, 16, 32, 64 and 128:

They are consistent with the graphs from [4] obtained numerically using the implicit formula from [1]. Note that for and the mapping

 Ωm∋z⟼(eit(1−|a|2)1/2m(1−¯az2)1/mz1,z2−a1−¯az2)

is a holomorphic automorphism of and therefore where attains all values of in . One can show numerically that

 supm≥1/2supΩmFΩm=1.010182…

which was already noticed in [4]. This is the highest value of (in arbitrary dimension) obtained so far.

In [4] it was also shown that for and with one has

 λ(IΩ((b,0))=π26(1−b)4((1−b)4+8b),

so that in particular similarly as in Theorem 1 it is an analytic function on this part of . This raises a question whether is smooth in general. In [4] it was also predicted that the highest value of for convex in should be attained for for on the diagonal. The following result will answer both of these questions in the negative:

###### Theorem 2.

Let . Then for with we have

 (2) λ(IΩ((b,b)))=π26(30b8−64b7+80b6−80b5+76b4−16b3−8b2+1)

and when

 (3) λ(IΩ +π(30b10−124b9+238b8−176b7−260b6+424b5−76b4−144b3+89b2−18b+1)6(1−b)2 ×arccos(−1+4b−12b2) +π(1−2b)(−180b7+444b6−554b5+754b4−1214b3+922b2−305b+37)72(1−b)√4b−1 +4πb(1−2b)4(7b2+2b−2)3(1−b)2arctan√4b−1 +4πb2(1−2b)4(2−b)(1−b)2arctan1−3b(1−b)√4b−1.

The function

 b⟼λ(IΩ((b,b)))

is on the interval but not at .

Again, the formula for the Bergman metric for this ellipsoid is known, see [6] or [7], Example 6.1.9:

 KΩ(w)=2π2⋅3(1−|w|2)2(1+|w|2)+4|w1|2|w2|2(5−3|w|2)((1−|w|2)2−4|w1|2|w2|2)3,

so that

 (4) KΩ((b,b))=2(3−6b2+8b4)π2(1−4b2)3.

The first part of Theorem 2, formula (2) on the interval , is easier to prove than the second one. Combining it with (4) one obtains the following graph of for :

One can show that its analytic continuation to attains values below 1 and thus it follows already from (1) that cannot be analytic. To conclude that it is in fact not one has to prove much harder formula (3). Here is the full picture on the interval , the analytic continuation of from and the actual graph of :

One can check that the maximal value of for is

All pictures and numerical computations in this paper, as well as a lot of formal ones in the proofs of Theorems 1 and 2 have been done using Mathematica.

## 1. General formula for geodesics in convex complex ellipsoids

Boundary of the Kobayashi indicatrix of a convex domain at consists of the vectors where is a geodesic of satisfying . Theorems 1 and 2 will be proved using a general formula for geodesics in convex complex ellipsoids from [8] based on Lempert’s theory [9] describing geodesics of smooth strongly convex domains.

For with set

 E(p)={z∈Cn:|z1|2p1+⋯+|zn|2pn<1}

and define

 φj(ζ)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ajζ−αj1−¯αjζ(1−¯αjζ1−¯α0ζ)1/pj,j∈Aaj(1−¯αjζ1−¯α0ζ)1/pj,j∉A,

where , for , for ,

 (5) α0=|a1|2p1α1+⋯+|an|2pnαn,

and

 (6) 1+|α0|2=|a1|2p1(1+|α1|2)+⋯+|an|2pn(1+|αn|2).

A component has a zero in if and only if . We have

 (7) φj(0)={−ajαj,j∈Aaj,j∉A,

and

 (8) φ′j(0)=⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩aj(1+(1pj−1)|αj|2−αj¯α0pj),j∈Aaj¯α0−¯αjpj,j∉A.

For the set of vectors where forms a subset of of a full measure. The geodesics in are uniquely determined: for a given and there exists unique geodesic such that and .

## 2. Proof of Theorem 1

First note that the formulas for and easily follow from the first one by approximation. For and there are two possibilities for a geodesic : either crosses the axis or it does not. By and denote the respective parts of . In the first case must be of the form

 φ(ζ)=(a1ζ−α11−¯α1ζ(1−¯α1ζ1−¯α0ζ)1/m,a2ζ−α21−¯α0ζ),

where and satisfy (5), (6). By (7) and since we have , and by (5) . By (6)

 1+b4m|α1|2−4m=b2m|α1|−2m(1+|α1|2)+|a2|2,

that is

 (9) |a2|2=(1−b2m|α1|−2m)(1−b2m|α1|2−2m).

Since , it follows that . Write , , then by (8) and (9), with ,

 φ′(0) =((br+b(1m−1)r−b2m+1r1−2mm)eit,√(1−b2mr−2m)(1−b2mr2−2m)eis) =:(γ1(r)eit,γ2(r)eis).

The mapping

 (10) Δ×[0,2π)×(b,1)∋(ζ,t,r)⟼ζ(γ1(r)eit,γ2(r))

parametrizes . We will need a lemma.

###### Lemma 3.

Let be a function of two complex variables, where and are . Then the real Jacobian of is equal to , where

 H=|f|2(|g¯z|2−|gz|2)+|g|2(|f¯z|2−|fz|2)+2\rm Re(f¯g(¯¯¯¯¯fzgz−¯¯¯¯¯f¯zg¯z)).

The proof is left to the reader. For the mapping (10) we can compute that

 H=γ1γ2(γ1γ′2−γ′1γ2)=−b2m2r−6m−3[b2m(−mr2+m−1)+r2m][r2m((m−1)r2+m)−(2m−1)r2b2m]        ×[r2b2m+r2m((m−1)r2−m)].

Since

 (11) ∫Δ|ζ|2dλ(ζ)=π2,

we obtain

 (12) λ(I12) =π2∫1b|H|dr =π2((1−2m)2m2(3m−1)(3m−2)b6m+2−3m2(m+1)(m−2)b2m+2−32m2b4m+2 +m2(m−2)(3m−2)b6+3m3m−1b4−4m2−m+12m2b2+mm+1).

To compute the volume of we consider geodesics of the form

 φ(ζ)=(a1(1−¯α1ζ1−¯α0ζ)1/m,a2ζ−α21−¯α0ζ),

where , , satisfy (5), (6). By (7) and since we have , and by (5) . By (6)

 1+b4m|α1|2=b2m(1+|α1|2)+|a2|2,

that is

 |a2|2=(1−b2m)(1−b2m|α1|2).

This means that any is allowed and by (8)

 φ′(0) =(b(b2m−1)m¯α1,a2) =(b(1−b2m)rmeit,√(1−b2m)(1−b2mr2)eis),

where , . Similarly as before we have

 H=−b2(1−b2m)3rm2

and

 λ(I2)=π2∫10|H|dr=π2b2(1−b2m)32m2.

This combined with (12) finishes the proof of Theorem 1. ∎

## 3. Proof of Theorem 2

For and , where , we have by (7)

 (13) aj=⎧⎨⎩−bαj,j∈Ab,j∉A

and by (8)

 (14) φ′j(0)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2b¯α0−b(¯αj+1αj),j∈A2b(¯α0−¯αj),j∉A.

There are four possibilities for the set : , , , and . Denote the corresponding parts of by , , , and , respectively, so that

 (15) λ(IΩ(w)) =λ(I0)+λ(I1)+λ(I2)+λ(I12) =λ(I0)+2λ(I1)+λ(I12).

The case

By (5), (6) and (13)

 (16) (1b+2b)|α1||α2|+2b\rm Re(α1¯α2)=(1+|α1|2)|α2|+(1+|α2|2)|α1|.

Since the set of satisfying (16) is -invariant, let us consider only those with . If we then replace with then (16) will still be valid and will be replaced by . We thus consider

 (17) α1=reit,   α2=ρ,   r,ρ∈(0,1), t∈(0,π);

to get we will have to multiply the obtained volume by 2. The condition (16) transforms to

 (18) 1b+2b(1+cost)=r+1r+ρ+1ρ.

It will be convenient to substitute , , and consider the domain

 (19) U:={(x,y)∈(2,1b+4b−2)×(0,π):x<1b+2b(1+cosy)−2}.

We have

 α0=b(α1|α1|+α2|α2|)=b(eit+1)

and thus by (14) and (18)

 (20) φ′(0) =b(2¯α0−¯α1−1α1,2¯α0−¯α2−1α2) =(2b2(e−it+1)−b(r+1r)e−it,2b2(e−it+1)−b(ρ+1ρ)) =(2b2+b(2b−x)e−iy,bx−1−2b2isiny) =:(f(z),g(z)).

The mapping

 Δ×U∋(ζ,z)⟼ζ(f(z),g(z))

parametrizes . From Lemma 3 and (11) it follows that

 λ(I12)=π∬U|H|dλ,

where , are given by (20), by (19) (recall that again we had to multiply by 2) and we can compute that

 H=b2[1−2b2(cosy+1)][−bx2+(1+2b2(cosy+1))(x−2b)−2b(b2cos(2y)+1)].

One can check that in . The region may look as follows

aa aa

a

We set

 y0:=⎧⎪⎨⎪⎩πb≤1/4arccos(−1+4b−12b2)b>1/4,

then

 λ(I12)=π∫y00∫1/b+2b(1+cosy)−22Hdxdy.

For we will get

 (21) λ(I12)=π26(1−32b2+80b3−12b4−112b5+176b6−192b7+110b8)

and for

 (22) λ( I12)=π72(37−140b+270b2−528b3+530b4−712b5+660b6)(1−2b)√4b−1 +π6(1−32b2+80b3−12b4−112b5+176b6−192b7+110b8)arccos(−1+4b−12b2).

The case

By (13) , and by (5) . From (6) we get

 (23) 1+b2(1+2\rm Re(α1¯α2)|α1|+|α2|2)=b|α1|(1+|α1|2)+b(1+|α2|2).

We may assume that , then (23) has a solution if and only if , where

 T =1b+b(1+2\rm Reα2+|α2|2)−1−|α2|2 =1b+b−1+2bx−(1−b)(x2+y2),

and we write . This means that

 (24) ∣∣∣α2−b1−b∣∣∣<1−2b√b(1−b)

and the set will be the intersection of this disc with . By (14) and (23)

 φ′(0)=2b(b(1+¯α2)−T/2,b−(1−b)¯α2)

and therefore

 f =2b2(1+x)−bT−2b2yi, g =2b2−2b(1−b)x+2b(1−b)yi.

We can compute that

 H [−1+2b+b3−2b2(1−b)x+b(1−b)2(x2+y2)] =4(1−b)b3(b+b2−(1−b)T)(b2+2b−2+bT).

One can check that everywhere on .

If then and using the polar coordinates in and Lemma 3 we will get

 (25) λ(I1)=2π23(1−b)b2(3−9b+2b2+6b3−6b4+10b5).

For it is more convenient to use the polar coordinates in the disk (24) instead:

 x=b1−b+rcost,   y=rsint,

then

 H=4b2(1−2b)2−4b4(1−b)4r4.

For with

 1−2b1−b

the circles and intersect when , where

 (26) t(r)=arccos1−2b−(1−b)2r22br(1−b).

Therefore

 λ(I1)=π2∫(1−2b)/(1−b)0rHdr+π∫(1−2b)/(√b(1−b))(1−2b)/(1−b)r(π−t(r))Hdr.

We can compute the second integral using the following indefinite integrals:

 (27) ∫v arccos(av−v)dv=14√−a2+2av2−v4+v2 +4a+18arctan2a−2v2+12√−a2+2av2−v4+v2+v22arccos(av−v)+const, ∫v5 arccos(av−v)dv=1288(15+78a+80a2+(10+32a)v2+8v4)√−a2+2av2−v4+v2 +5+36a+72a2+32a3192arctan2a−2v