Zero Problems of the Bergman Kernel Function on the First Type of Cartan-Hartogs Domain?

Xin ZHAO An WANG

Abstract The authors give the condition that the Bergman kernel function on the first type of Cartan-Hartogs domain exists zeros.If the Bergman kernel function of this type of domain has zeros, the zero set is composed of several path-connected branches, and there exists a continuous curve to connect any two points in the non-zero set.

Keywords Cartan-Hartogs domain, Zeros of Bergman kernel function, Path connectivity

1 Introduction

The biholomorphic equivalence classification of bounded domains is one of the central problems in the study of several complex variables.The Riemann mapping theorem resolves the problem about the biholomorphic equivalence classification of the simply connected bounded domains in C∞.At the beginning of 20th century, Poincarpointed out that the unit ball and the polydisc are not biholomorphically equivalent in Cn(n>1), which indicates that Riemann mapping theorem fails to hold in higher dimensional case.In 1936,Cartan gave the whole conclusion about the biholomorphic equivalence classification of the bounded symmetric domains in Cn, i.e., any bounded symmetric domain is the topological product of irreducible bounded symmetric domains, and irreducible bounded symmetric domains consist of domains of four classical types and two exceptional ones.However,the biholomorphic equivalence classification of general bounded domains has not been resolved.It is an important research direction of several complex variables.

In 1930, Bergman introduced the concept of representative coordinates and representative domain when he studied the biholomorphic equivalence classification of bounded domains in Cn(see [5]).Let G be a bounded domain in Cn.K(z,)denotes the Bergman kernel function of G, and (Tjk(z,))denotes the Bergman metric matrix of G.For any fixed w0∈G and any z ∈G, ?w0maps z to ?w0(z):

then ?w0is called the representative mapping, and ?w0(z)is the representative coordinates of z.If ?w0(G)is a domain, ?w0(G)is called the representative domain of G.Since domains which are biholomorphically equivalent have identical representative domain, the representative domain can be used as the representative element of biholomorphically equivalence class of domains.Moreover, two domains must not be biholomorphically equivalent if their representative domains are not identical.Therefore, the representative domain could be the tool for studying the biholomorphic equivalence classification of bounded domains.In 1966, when Lu was studying that bounded domain with complete Bergman metric of negative constant holomorphic sectional curvature is biholomorphically equivalent to the unit ball, he needed to define the representative mapping globally on the bounded domain.About the above definition of representative mapping, he pointed out that ?w0is well defined on a sufficiently small neighborhood U(w0)of w0and at the point where K(z,)≠ 0 for any z ∈G.However,defining representative mapping globally on a general bounded domain depends on that the Bergman kernel function of this domain is zero-free (see [15]).The Bergman kernel functions of some specific domains do not have zeros, such as unit ball, polydisc and bounded symmetric domains, but there is no definite conclusion as to whether the Bergman kernel functions of general bounded domains have zeros or not.Therefore, Lu raised an open question of whether the Bergman kernel functions of general bounded domains are zero-free.In 1969,Skwarczy′nski formally called the question raised by Lu as Lu Qi-Keng conjecture, i.e., the Bergman kernel function of a general bounded domain is zero-free.Moreover,he proved Lu Qi-Keng conjecture does not hold with counterexample: The Bergman kernel function of an annulus has zeros in the complex plane {z ∈C : r < |z| < 1,0 < r < e?2} (see [21]).It shows that the bounded domains on which Bergman kernel functions have zeros exist.

Although Lu Qi-Keng conjecture fails, the problem of determining whether the Bergman kernel function of a bounded domain has zeros is of great significance to discuss the properties of Bergman kernel functions,the properties of bounded domains,the biholomorphic equivalence classification of bounded domains and so on.Therefore, this research has received extensive attention in the area of several complex variables, and the problem about studying whether the Bergman kernel function of a domain has zeros or not is called the Lu Qi-Keng problem.Since Skwarczy′nski, many scholars have studied Lu Qi-Keng problem and achieved fruitful results.In 1966, Rosenthal proved that the Bergman kernel function of an annulus in the complex plane has zeros, moreover there exists k (k > 2)connected domain whose Bergman kernel function has zeros in C (see [20]).In the same year, Bell showed that the Bergman kernel function of every bounded, homogeneous,complete circular domain is zero-free(see [4]).In 1976, Suita and Yamada proved that the Bergman kernel function of non-simply connected finite Riemann surface has zeros, and obtained a general conclusion, i.e., the Bergman kernel function of bounded, multiple connected, planar domain with smooth boundary has zeros (see[22]).In 1981, Greene and Krantz proved that the sufficiently small smooth perturbations of the ball has zero-free Bergman kernel function (see[14]).In 1986,Boas gave a specific example to show that the Bergman kernel function of a bounded, strongly pseudo-convex, contractible domain with C∞r(nóng)egular boundary has zeros (see [6]).In 1996, Boas proved that the bounded holomorphic domains whose Bergman kernel functions have no zeros,form a nowhere dense set in the topological space composed of all bounded holomorphic domains(see[7]).This conclusion indicates that it is a normal situation for the Bergman kernel function of a holomorphic domain to have zeros.In 1999, Boas, Fu and Straube further proved that the Bergman kernel function of the following domainhas zeros when p > 2, and gave concrete examples that the Bergman kernel functions of strongly convex domains have zeros in [9].In 2000,Boas obtained a more general conclusion that the Bergman kernel function of the domain defined as

has zeros when p2+p3+···+pn> 2 (see [8]).In 1998, Pflug and Youssfi proved that the Bergman kernel function of the minimal ball exists zeros when n ≥4 in Cn(see [19]).In 2000, Engliˇs proved that the Bergman kernel function of a kind of Hartogs domain has zeros(see [13]).In 2005, Edigarian and Zwonek got that the Bergman kernel function of symmetric double-disc G2is zero-free(see [12]), while Nikolov and Zwonek later proved that the Bergman kernel function of symmetric polydisc Gn(n ≥3)has zeros (see [18]).In 2006, Chen studied the Lu Qi-Keng problem on a pseudo-convex domain applying the asymptotic behavior of the weighted Bergman kernel (see [10]).In 2009, Zhang and Yin discussed the conditions that the Bergman kernel function of the generalized Thullen domain has zeros (see [28]).In 2012, Ahn and Park presented the algorithmic procedure to determine the condition that the Bergman kernel function is zero-free on a type of generalized Cartan-Hartogs domain (see [1]).In 2016,Beberok investigated the Lu Qi-Keng problem of the intersection of two complex ellipsoids(see[3]).We refer to [2, 11, 16, 23–24, 27] for more research results about Lu Qi-Keng problem.

In the study on Lu Qi-Keng problem, many research achievements show that the Bergman kernel functions of bounded domains have zeros.In fact, if the Bergman kernel function of a domain has zeros, these zeros are not isolated and form a lower dimensional manifolds.It is significant for discussing the geometry properties of domains and the properties of Bergman kernel functions to study the structure and topological properties of the zero-sets of the Bergman kernel functions on bounded domains.The aim of this present paper is to study the Lu Qi-Keng problem on the first type of Cartan-Hartogs domain, and discuss the topological properties of the zero set when the Bergman kernel function of this type of domain has zeros.The form of the first type of Cartan-Hartogs domain is as following (see [25]):

where RI(m,n)denotes the first type of Cartan domain, NI(z,z)denotes the generic norm of RI(m,n)in Jordan triple system.By constructing a biholomorphic invariant x on ?×?,we write the Bergman kernel function of ? into the product of a polynomial F(x)and a nonvanishing function.Then the research on the zero problem of the Bergman kernel function with multidimensional complex variable is transformed into the research on the zero problem of a polynomial function with single complex variable.From the condition that F(x)has zeros in the unit disc,we obtain that there existsμ0(m,n)>0 such that the Bergman kernel function of? has zeros when the parameterμsatisfies 0<μ<μ0(m,n)for any m and n.Furthermore,we show that the zero set of the Bergman kernel function of ? is composed of some path-connected branches, and a continuous curve is constructed to connect any two points in the non-zero set.

2 Preliminaries

In this section, we are going to outline some basic conclusions about ?, and introduce a type of special self-reciprocal polynomial.

Theorem 2.1(see [26])The transformsψ(ξ,z)= (ψ1(ξ,z),ψ2(ξ,z))constitute the holomorphic automorphism group of?, and the specific forms ofψ1andψ2are as below:

whereθ ∈R,z0∈RI(m,n),z0= φ?1(0).Aut(?)denotes the holomorphic automorphism group of?.

Now, we construct a complex value function x on ?×?, and it is a invariant under the holomorphic transformation of ?×?.

Theorem 2.2Let

for any((ξ,z),(η,w))∈?×?.Then

x((ξ,z),(η,w))=x(ψ(ξ,z),ψ(η,w))for?ψ ∈Aut(?),

and|x|<1.

ProofFor any z = (z11,z12,··· ,z1n,··· ,zm1,zm2,··· ,zmn)∈RI(m,n), the components of z can be arranged into a matrix

where z is corresponding to Z,and the first type of Cartan domain is represented as RI(m,n):={Z ∈Cmn:I ?>0}.So the point in RI(m,n)is denoted by Z in the proof of the theorem.

For any ((ξ,Z),(η,W))∈?×?, (ξ,Z)and (η,W)are mapped to (ξ?,Z?)and (η?,W?)respectively by any ψ ∈Aut(?).Applying Theorem 2.1, we have

where Z0=φ?1(0)(φ ∈Aut(RI(m,n))),

Since

and

for any Z,W ∈RI(m,n), we obtain

i.e., x is a invariant under the holomorphic transformation of ?×?.

We can prove

NI(Z,Z)NI(W,W)≤|NI(Z,W)|2

for any ((ξ,Z),(η,W))∈?×? using the conclusion in [17].Then based on the inequality, we obtain

A special kind of self-reciprocal polynomial plays an important role in investigating the Lu Qi-Keng problem on the first type of Cartan-Hartogs domain, here we give the definition of a general self-reciprocal polynomial and discuss the properties of the special kind of self-reciprocal polynomial.

Definition 2.1Letfk(x)be a real coefficient polynomial of degreek.Iffk(0)≠ 0and, thenfk(x)is called self-reciprocal polynomial of degreek.

Remark 2.1fk(x)=is a real coefficient self-reciprocal polynomial of degree k if and only if its coefficients satisfy a0≠0 and al=ak?l(l=0,1,··· ,k).

Theorem 2.3Letfk(x)be a real coefficient self-reciprocal polynomial of degreek.Ifx0is a zero offk(x), thenis also its zero.

ProofFrom Definition 2.1, this theorem can be proved straightforwardly.

Lemma 2.1Let

thenσr(α1,··· ,αl)>0 (r =1,2,··· ,l)if and only ifαj>0,j =1,··· ,l.

ProofSince the sufficiency of the lemma is obvious, we need only to prove its necessity.

We can construct an equation of degree l with roots of ?α1,?α2,··· ,?αl:

Expand the left side of (2.1), it is changed into the following form

Because αj(j = 1,··· ,l)are real numbers, and σr(α1,··· ,αl)> 0 (r = 1,2,··· ,l), the roots of (2.1)can not be 0 or positive real numbers, and they must be negative real numbers, i.e.,αj>0 (j =1,··· ,l).

Theorem 2.4Let

be a polynomial of degreekwith respect tox, whereak,ldenotes the coefficient of thel-th order term of the polynomial of degreek:

S(k+1,j)denotes the Stirling number of the second kind, i.e.,

Setak,l=0forl >kandl<0, thenfk(x)satisfies the following properties:

(1)ak,l=(k+1 ?l)ak?1,l?1+(l+1)ak?1,l(k ≥1, 0 ≤l ≤k).

(2)fk(x)(k ∈N)is a self-reciprocal polynomial, i.e.,ak,l= ak,k?l(0 ≤l ≤k), andak,0=ak,k=1.

(3)ak,l>0 (0 ≤l ≤k).

(4)ak,l?1

(5)Whenkis odd number,fk(x)= (x+1)hk?1(x)(k ≥1), andhk?1(x)is also a selfreciprocal polynomial with positive real coefficients.

(6)fk(x)= [(k ?1)2x2+(3k ?2)x+1]fk?2(x)+x(1 ?x)[(2k ?3)x+3](x)+(1 ?x)2x2(x)(k ≥2).

(7)fk(x)(k ≥2)has at least one zero in the unit disc.

Proof(1)According to (2.2)and the property of S(k+1,j), for any fixed l, we have

Therefore, we obtain the recurrent formula about the coefficients of fk(x):

About(2)–(4),these properties can be proved by applying(2.3)and mathematical induction.

(5)Because fk(x)is a self-reciprocal polynomial, i.e., ak,l=ak,k?l(0 ≤l ≤k), ?1 must be the zero of fk(x)for odd number k, and

fk(x)=(x+1)hk?1(x)(k ≥1).

We are going to prove that hk?1(x)is also a self-reciprocal polynomial with positive real coefficients.

Put hk?1(x):=, then

By comparing the above expression with,we obtain the relationship between bk?1,land ak,las following:

From (2.4)and property (2)of fk(x), it follows that

which means that hk?1(x)is a self-reciprocal polynomial.

Next, we will show that the coefficients of hk?1(x)are all positive.As hk?1(x)is a selfreciprocal polynomial, it suffices to prove bk?1,l>0 for l>.

For ak,l+1>ak,l+2>···>ak,k(l>), if l is odd, then

bk?1,l=(ak,l+1?ak,l+2)+···+(ak,k?1?ak,k)>0;

if l is even, then

bk?1,l=(ak,l+1?ak,l+2)+···+(ak,k?2?ak,k?1)+ak,k>0.

In conclusion, when k is odd number, hk?1(x)is a self-reciprocal polynomial with positive real coefficients.

(6)To prove the property (6)of fk(x), we first need to verify

According to (2.5), if k ≥2, we have

Differentiating the both sides of (2.6)with respect to x gives the following formula:

By substituting (2.6)and (2.7)back into (2.5), it follows that

(7)We first show that fk(x)has at least one zero in the unit disc for k = 2p (p ∈Z+)by negation.

If the assertion fails to hold, f2p(x)would have no zeros in the unit disc.As f2p(x)is a self-reciprocal polynomial,according to Theorem 2.3,if x0is a zero of f2p(x),thenis also the zero of f2p(x).Thereby,based on the hypothesis,f2p(x)has no zeros out of the unit disk either,which means that all of the zeros of f2p(x)are on the boundary of the unit disc.Moreover,the reciprocal of every zero is just its conjugate zero.Set the zeros of f2p(x)as following:

With the zeros of f2p(x), we can write f2p(x)as the product of 2p factors:

In the above expression of f2p(x), we put

where

However,according to the property of fk(x)and Lemma 2.1,it can be inferred that αq>0(q =1,2,··· ,p)by showing that σp?r(α1,··· ,αp)>0 (0 ≤r ≤p ?1).

In fact, because f2p(x)=xpGp(y)and y =x++2, we have

By substituting f2p(x),into (2.8), it follows that

Dividing both sides of the above equation by xp, and substituting y for x++2,we obtain

Furthermore, by calculating the r-order derivative of Gp(y), we acquire

From f2(x)=x2+4x+1,it follows that G1(y)=y+2,G1(0)=2 and G′1(0)=1.According to (2.11), we haveby recurrence method.From (2.9)and (2.10), it follows that

Finally, we obtain αq> 0 (q = 1,2,··· ,p)from Lemma 2.1.But this conclusion is contrary to αq= ?2 cosθq?2 ≤0 (q = 1,2,··· ,p).Therefore, the zeros of f2p(x)are not all on the boundary of the unit disc.Based on Theorem 2.3, f2p(x)has at least one zero in the unit disk.

For k =2p+1 (p ∈Z+), according to the property (5)of fk(x): f2p+1(x)=(x+1)h2p(x),because h2p(x)is also a self-reciprocal polynomial, we can prove that h2p(x)has at least one zero in the unit disc by the same method.

Consequently, we infer that fk(x)(k ≥2)has at least one zero in the unit disc.

3 Lu Qi-Keng Problem on ?

In this section, we discuss that the Bergman kernel function of ? has zeros in ?×? if the parameter μ of ? satisfies certain condition, and obtain the following conclusion.

Theorem 3.1Let the Bergman kernel function of?be

where

andB(α,β)denotes the Beta function(see [26, 28]).Then for anymandn, there existsμ0(m,n)>0such thatK((ξ,z),has zeros in?×?when0<μ<μ0(m,n).

ProofIn the expression of the Bergman kernel function of ?, we have

In (3.2), the part containing μ can be rewritten as following:

where

By substituting (3.3)into F(x), F(x)can be expressed as below:

F(x)=fmn(x)+μgmn(x),

where

and

By rearranging the expression of fmn(x), we get

so it is exactly a real coefficient polynomial of degree mn as the one defined in Theorem 2.4.

From Theorem 2.4(7), we have known that fmn(x)has at least one zero in the unit disc.According to the relationship between F(x)and fmn(x), we will prove that for any m and n,there exists μ0(m,n)>0 such that F(x)has zeros in the unit disc when 0<μ<μ0(m,n).

Let x0be a zero of fmn(x)in the unit disc, because the zero of a non-constant holomorphic function with a single complex variable is isolated, there exists 0 < ρ < 1 such that x0is the unique zero of fmn(x)in the closed disc

According to (3.5), for 0<μ<1, we get

It follows from (3.4)that

Therefore,

Taking μ0(m,n)=when 0<μ<μ0(m,n), we have

Thereby, F(x)= fmn(x)+μgmn(x)has one zero in D(x0,ρ)from Rouch′e theorem.Namely,F(x)has at least one zero in the unit disc when μ satisfies the above condition.

Consequently, for any fixed m,n, when 0 < μ < μ0(m,n), the Bergman kernel function of? has zeros in ?×?.

4 The Topological Properties of the Zero Set of the Bergman Kernel Function on ?

According to Theorem 3.1, there exists μ0(m,n)> 0 such that F(x)as (3.2)has zeros in the unit disc when 0<μ<μ0(m,n), and the set of these zeros is denoted by

Under the same condition, the Bergman kernel function of ? has zeros in ?×?.Put

j = 1,2,··· ,M, M < mn, then the zero set of the Bergman kernel function of ? is the union of these sets.

We are going to discuss the topological properties of the zero set of the Bergman kernel function on ?, and give the main results and proofs about the connectivity of the zero set.

Theorem 4.1If the Bergman kernel function of?has zeros, and the zero set is denoted by, where the form ofΛjis as(4.2), thenΛj(j = 1,2,··· ,M,M < mn)is apath-connected subset.

ProofWithout loss of generality, we may prove Λ1is a path-connected subset.Set Q?=because |λ1| < 1 and x(Q?)= λ1, it is easy to see that Q?is a fixed point in Λ1.Let P = ((ξ,z),(η,w))be any point in Λ1, by proving that there exists a continuous curve to connect P and Q?, we can obtain that any two points can be connected with a continuous curve in Λ1, i.e., Λ1is a path-connected subset.

According to Theorem 2.1, we select ψz0,θ∈Aut(?), where z0= z and θ = ?arg ξ, such that

Since x is an invariant under the action of holomorphic transformation Ψ of ?×?, i.e., x(P)=x(P?)=λ1, it follows P?∈Λ1, and

Furthermore, putting Q:=Ψ?1(Q?), we have Q ∈Λ1from x(Q?)=x(Q)=λ1, and

Because P and Q are respectively mapped to P?and Q?by Ψ, we will show that P and Q can be connected with a continuous curve by proving that P?and Q?can be connected with a continuous curve in Λ1.Then, a continuous curve can be constructed to connect Q and Q?.So there is a continuous curve to join P and Q?in Λ1.

Firstly, there exists

corresponding to P?in Λ1.We can construct a continuous curve as following to connect P?and:

In fact,

so ?1(s)?Λ1, and ?1(0)=,?1(1)=P?.

Next, we construct a continuous curve

Summarizing the above discussion, ?1is a continuous curve connecting P?and, and ?2is a continuous curve connecting Q?and, therefore there exists a continuous curve to connect P?and Q?, which is denoted by ?.Obviously, we obtain that

is a continuous curve connecting P and Q, which is denoted by γ1.

Furthermore, we construct a continuous curve

We have

Moreover,γ2(0)=Q?,γ2(1)=Q.Hence,γ2(s)?Λ1and γ2(s)is a continuous curve connecting Q and Q?.

In conclusion, γ1is a continuous curve connecting P and Q, and γ2is a continuous curve connecting Q and Q?, so there exists a continuous curve connecting any point P and the fixed point Q?.So Λ1is a path-connected subset.Using the same method, we can show that every Λj(j = 2,··· ,M,M < mn)is a path-connected subset of the zero set of the Bergman kernel function on ?.

Theorem 4.2If the Bergman kernel function of?has zeros, let the zero set bewhere the form ofΛjis as(4.2), then for anyP ∈ΛlandQ ∈Λk(l ≠ k), there is not a continuous curve to connectPandQinΛ.

ProofWe will prove the theorem by negation.

For any P ∈Λland Q ∈Λk(l ≠ k), suppose that there is ?(s)(s ∈[0,1])?Λ to connect P and Q, ?(0)=P, ?(1)=Q, then x(?(0))=λl, x(?(1))=λk, and λl≠λk.Because x(?(s))is uniformly continuous on [0,1], for any s0∈[0,1] and for ε =min{|λ′?λ′′|,?λ′,λ′′∈A},where A is defined as (4.1), there exists δ >0 such that

|x(?(s))?x(?(s0))|<ε, s ∈U(s0,δ)∩[0,1].

If s0takes all the values in [0,1], there is a set of open intervals {U(s0,δ),s0∈[0,1]} covering[0,1].Moreover, there are following finite open intervals to cover [0,1] in the set of open intervals:

Uα(sα,δ0), α=1,2,··· ,q.

Without loss of generality, let s1

|x(?(s))?x(?(sα))|<ε for any s ∈U(sα,δ)∩[0,1].

Because 0 ∈U(s1,δ0), x(?(0))= λland |x(?(s1))?x(?(0))| < ε, we have x(?(s1))= λl.Next, as any s ∈U(s1,δ)∩[0,1], |x(?(s))?x(?(s1))| < ε, we acquire x(?(s))= λlfor any s ∈U(s1,δ).Since any s ∈U(s1,δ)∩U(s2,δ), |x(?(s))?x(?(s2))| < ε, it is turned out that x(?(s2))= λl.Moreover, any s ∈U(s2,δ)∩[0,1], |x(?(s))?x(?(s2))| < ε, we can show that x(?(s))= λlfor any s ∈U(s2,δ).Proceeding the derivation in a similar manner, it follows that x(?(1))=λl, which is contrary to x(?(1))=λk.

Consequently, the hypothesis fails to hold, i.e., P and Q which belong to different subset Λland Λkrespectively, can not be joined with a continuous curve in Λ.

According to Theorems 4.1–4.2, if the Bergman kernel function of ? has zeros, the zero set is composed of M (M

Theorem 4.3If the Bergman kernel function of?has zeros, let the zero set bewhere the form ofΛjis as(4.2), and?×?Λdenote the complement set ofΛin?×?, then for anyP,Q ∈?×?Λ, there exists a continuous curve to connect the two points.

ProofLet O = ((0,0),(0,0)).Since x(O)= 0 and 0A, where A is defined as (4.1), we have O ∈?×?Λ.For any P = ((ξ,z),(η,w))∈?×?Λ, if we show that P and O can be joined with a continuous curve in ?×?Λ, the continuous curve can be constructed to connect P and another any point Q.

We are going to prove that there exists a continuous curve to connect P and O according to the character of P in two cases:

(i)Let P =((ξ,z),(η,w))∈?×?Λ.If x(P)= 0A, then ξ = 0 or η =0.Construct a continuous curve

γ0(s)={((sξ,sz),(sη,sw)), s ∈[0,1]}.

It is easy to verify that γ0(s)??×? for ?s ∈[0,1].Moreover, x(γ0(s))= 0, and γ0(0)= O,γ0(1)=P.Thereby, γ0(s)is a continuous curve connecting P and O in ?×?Λ.

(ii)Let P = ((ξ,z),(η,w))∈?×?Λ, if x(P)A and x(P)≠ 0, then ξ ≠ 0 and η ≠ 0.There is=((0,z),(η,w))corresponding to P in ?×?Λ.We are going to first prove that P andcan be joined with a continuous curve in ?×?Λ.

Construct a cluster of curves, which are disjoint curves with P andas endpoints:

where a,b ∈R,a≠0 and b≠0.Because

we have γk(s)??×?, and γk(0)=, γk(1)=P.Moreover,

so x(γ1(s)),x(γ2(s)),··· ,x(γM+1(s))are disjoint continuous curves connecting x(P)and origin in C.Since A is a finite set containing M elements at most, there exists k0such that x(γk0(s))∩A = ?.Namely, there exists at least one continuous curve γk0(s)to join P andin ?×?Λ.

Next, from (i), γ0(s)= {((0,sz),(sη,sw)), s ∈[0,1]} is a continuous curve to joinand O in ?×?Λ.Hence, joining γk0(s)and γ0(s), we get a continuous curve to connect P =((ξ,z),(η,w))(ξ ≠0,η≠0)and O.

In the same way, for another Q ∈?×?Λ, we also get a continuous curve to connect Q and O.So by connecting P and Q with O, we construct a continuous curve to connect any two points in ?×?Λ.