On Fridman Invariants and Generalized Squeezing Functions?

Feng RONG Shichao YANG

Abstract In this paper,the authors introduce the notion of generalized squeezing function and study the basic properties of generalized squeezing functions and Fridman invariants.They also study the comparison of these two invariants, in terms of the so-called quotient invariant.

Keywords Fridman invariant, Squeezing function, Quotient invariant

1 Introduction

Due to the lack of a Riemann mapping theorem in several complex variables, it is of fundamental importance to study the biholomorphic equivalence of various domains in Cn,n ≥2.For such a study, it is necessary to introduce different kinds of holomorphic invariants.In this paper, we study two such invariants, the Fridman invariants and the (generalized)squeezing functions.

The Fridman invariant was defined by Fridman in [5] for Kobayashi hyperbolic domainsDin Cn,n ≥1, as follows.Denote by(z,r)thekD-ball inDcentered atz ∈Dwith radiusr >0, wherekDis the Kobayashi distance onD.For two domainsD1andD2in Cn, denote byOu(D1,D2)the set of injective holomorphic maps fromD1intoD2.

Recall that a domain ??Cnis said to be homogeneous if the automorphism group of ? is transitive.For any bounded homogeneous domain ?, set

For comparison purposes, we call(z):=tanh((z))?1the Fridman invariant (see [4, 9]).

For any bounded domainD ?Cn, the squeezing function was introduced in [1] by Deng,Guan and Zhang as follows:

sD(z)=sup{r:rBn ?f(D),f ∈Ou(D,Bn),f(z)=0}.

Here Bndenotes the unit ball in Cn.Comparing with the Fridman invariant, it seems natural to consider more general squeezing functions, replacing Bnby other “model domains”.

Recall that a domain ? is said to be balanced if for anyz ∈?,λz ∈? for all|λ|≤1.Let? be a bounded, balanced and convex domain in Cn.The Minkowski functionρΩis defined as(see e.g.[6])

Note that ? = {z ∈Cn:ρΩ(z)<1}.Set ?(r)= {z ∈Cn:ρΩ(z)< r}, 0< r <1.Then for any bounded domainD ?Cn, we can define the generalized squeezing function as follows:

It is clear from the definitions that both Fridman invariants and generalized squeezing functions are invariant under biholomorphisms,and both take values in (0,1].There have been much study on the Fridman invariant and the squeezing function in recent years, and we refer the readers to two recent survey articles (see [3, 13])and the references therein for various aspects of the current research on this subject.

The main purpose of this paper is to study some basic properties of both Fridman invariants and generalized squeezing functions.Moreover,we will also discuss the comparison of these two invariants, for which we introduce the quotient invariant

whereD ?Cnis bounded and ??Cnis bounded, balanced, homogeneous and convex.

In Section 2, we study basic properties of Fridman invariants, in particular refining several results from Fridman’s original work (see [5]).In Section 3, we study basic properties of generalized squeezing functions, in particular extending various properties of the squeezing function given in [1–2] to the more general setting.In Section 4, we study the comparison of Fridman invariants and generalized squeezing functions, in particular generalizing previous results from [9–10].

2 Fridman Invariants

Throughout this section, we suppose thatDis a Kobayashi hyperbolic domain in Cnand? is a bounded homogeneous domain in Cn(unless otherwise stated).

We say thatf ∈Ou(?,D)is an extremal map atz ∈Dif

It is not known from Fridman’s original work (see [5])whether extremal maps exist.However,if we assumeDto be bounded or taut, then extremal maps do exist.

Theorem 2.1If D is bounded or taut, then an extremal map exists at each z ∈D.

For the proof of Theorem 2.1, we need two lemmas.The first is probably well-known, and we provide a short proof for completeness.

Lemma 2.1For any domain D inCn,(z,r)is a subdomain of D.

ProofSince the Kobayashi pseudodistance is continuous,(z,r)is an open subset ofD.Since the Kobayashi pseudodistance is inner, for any(z,r)there exists a piece-wiseC1-curves: [0,1]→Dsuch thats(0)=z,s(1)=wandwheredenotes the Kobayashi pseudometric.This implies thats([0,1])(z,r).Hence(z,r)is a subdomain ofD.

The next lemma is known as the generalized Hurwitz’s theorem in several complex variables(see e.g.[11]).

Lemma 2.2Let D be a domain inCn and {fi(z)} be a sequence of injective holomorphic maps from D toCn.Suppose that fi’s converge to a map f:D →Cn uniformly on compact subsets of D.Then either f is an injective holomorphic map ordet(z)≡0.

Proof of Theorem 2.1Since the proof for the taut case is similar as (and simpler than)for the bounded case, we will assume thatDis bounded.

Without loss of generality, we assume that 0∈?.By definition, there exists a sequence of holomorphic embeddingsfi: ?→Dwithfi(0)=z, and a sequence of increasing positive numbersriconvergent to arctanh((z))such that(z,ri)?fi(?).SinceDis bounded, by Montel’s theorem, there exists a subsequence {fki} of {fi} which converges to a holomorphic mapf:?uniformly on compact subsets of ?.

By Lemma 2.1,(z,ri)’s are increasing subdomains ofD.Denotegi:=(z,ri).Since ? is bounded, by Montel’s theorem, there exists a subsequence {gk’i} of {gki} which converges to a holomorphic mapg:(z,arctanh((z)))→uniformly on compact subsets of(z,arctanh((z))).

Takes >0 such that Bn(z,s)(z,r1).By Cauchy inequality for anyi,|det(z)|< cfor some positive constantc.So we have|det(0)|>for anyi.Thus,we have|det(0)|>0 and|det(z)|>0.By Lemma 2.2, bothfandgare injective.In particular,f(?)?Dandg((z,arctanh((z))))??.Sincef ?g(w)=wfor all(z,arctanh((z))), it shows thatfis the desired extremal map.

Based on Theorem 2.1, we can give another proof of [5, Theorem 1.3(2)] as follows.

Theorem 2.2If there exists z ∈D such that(z)= 1, then D is biholomorphically equivalent to?.

ProofSince ? is homogeneous,sΩ(z)≡cfor some positive numberc.Thus, by [1,Theorem 4.7], ? is Kobayashi complete, hence taut.

Without loss of generality, we assume that 0∈?.Letfi’s andgi’s be as in the proof of Theorem 2.1.Since(z)=1, we have

Since ? is taut, by [7, Theorem 5.1.5], there exists a subsequence {gki} of {gi} which converges to a holomorphic mapg:D →? uniformly on compact subsets ofD.By the decreasing property of the Kobayashi distance, forz1,z2∈Dsuch thatg(z1)=g(z2), we have forkilarge enough,

kD(z1,z2)≤kfki(Ω)(fki ?gki(z1),fki ?gki(z2))=kΩ(gki(z1),gki(z2)).

Lettingki →∞, by the continuity of the Kobayashi distance, we have

kD(z1,z2)≤kΩ(g(z1),g(z2))=0.

SinceDis Kobayashi hyperbolic, we havez1=z2.Thus,gis injective andDis biholomorphic to a bounded domain.

Now Theorem 2.1 applies and shows thatDis biholomorphically equivalent to ?.

It was shown in [5, Theorem 1.3(1)] that(z), hence(z), is continuous.For its proof,Fridman showed that forz1andz2sufficiently close,Our next result gives a “global” version of this estimate in terms of(z).

Theorem 2.3For any z1and z2in D, we have

For the proof of Theorem 2.3, we need the following basic fact, whose proof we provide for completeness.

Lemma 2.3Suppose that ti ≥0, i=1,2,3, and t3≤t1+t2.Then,

tanh(t3)≤tanh(t1)+tanh(t2).

ProofSincet3≤t1+t2, we have

Define

To show that tanh(t3)≤tanh(t1)+ tanh(t2), it suffices to show thatf(t1,t2)≤0 for allt1,t2≥0.For any fixedt1≥0, consider

Then,

Proof of Theorem 2.3Fix 0(z1), by definition there exists a holomorphic embeddingf:?→Dsuch that(z1,arctanh[(z1)?ε])?f(?).

Ifz2(z1,arctanh[(z1)?ε]), then obviously

Ifz2(z1,arctanh[(z1)?ε]), by Lemma 2.3, we have for allzwith tanh[kD(z2,z)](z1)?ε ?tanh[kD(z1,z2)] that

Thus,

This implies that(z2)(z1)?ε ?tanh[kD(z2,z1)].

Sinceεis arbitrary,we have(z2)(z1)?tanh[kD(z2,z1)].Similarly,(z1)(z2)?tanh[kD(z2,z1)].This proves the theorem.

We say that a sequence of subdomains{Dj}j≥1ofDis a sequence of exhausting subdomains if for any compact subsetK ?D, there existsN >0 such thatK ?Djfor allj >N.In this case, we also say that {Dj}j≥1exhaustsD.

Corollary 2.1Let {Dj}j≥1be a sequence of exhausting subdomains of D.(z)for all z ∈D, then the convergence is uniform on compact subsets of D.

ProofLetKbe a compact subset ofD.Then there exists 0< r <1 such thatHence there existsN1>0 such thatfor allj > N1.Fix anyε>0 and takeSince {Bn(z,δ)}z∈Kis an open covering ofK, there is a finite setsuch thatFor anyz ∈K,there is somezisuch thatz ∈Bn(zi,δ).By Theorem 2.3 and the decreasing property of the Kobayashi distance, we have

On the other hand, there existsN2>0 such that(zi)(zi)for allziandj >N2.TakeN=max{N1,N2}.Then for anyj >N, we have(z)(z)|<εfor allz ∈K.This completes the proof.

Theorem 2.4Suppose that D is bounded or taut.Let {Dj}j≥1be a sequence of exhausting subdomains of D.Then for any z ∈D,

To prove Theorem 2.4, we need the following lemma.

Lemma 2.4Let {Dj}j≥1be a sequence of exhausting subdomains of D.Then for any z ∈D and r >0(z,r)}j≥1(z,r).

ProofBy Lemma 2.1, we know that(z,r)is a subdomain ofDfor anyz ∈Dandr>0.Firstly, we show that

Consider a sequence of subdomains {Gj}j≥1such that (i)GjD, (ii)Gj ?Gj+1, (iii)By [6, Proposition 3.3.5], we have

For anyj ≥1, there existsNj >0 such thatGj ?Di, for alli > Nj.By the decreasing property of the Kobayashi distance, we get

Now we prove that for anyK(z,r), there existsN >0 such that(z,r)for allj >N.

SincekD(z,·)is continuous, there exists 00 such that.Hence,there existsN1>0 such thatfor allj >N1.

Let 0< ε < r ?r0and takeδ1=δtanh.Since {Bn(z,δ1)}z∈Kis an open covering ofK, there is a finite setsuch thatIt is clear that there existsN2>0 such that|kDj(z,zl)?kD(z,zl)for anyj > N2and 1≤l ≤m.For anyw ∈K, there is somezlsuch thatw ∈Bn(zl,δ1).SetN=max{N1,N2}.Then for allj >N,by the decreasing property of the Kobayashi distance, we have

Therefore,kDj(z,w)N.This completes the proof.

Proof of Theorem 2.4Since the proof for the taut case is similar as (and simpler than)for the bounded case, we will assume thatDis bounded.

For anyz ∈D, letbe a sequence such thatFor any 0<ε0 such thattanh(r ?ε)for allli >N1.

Without loss of generality,we assume that 0∈?.By definition,for anyli >N1,there exists an open holomorphic embeddingfli:?→Dlisuch thatfli(0)=zand(z,r ?ε)?fli(?).SinceDis bounded, by Montel’s theorem, there exists a subsequence {fki} of {fli} which converges to a holomorphic mapf:?uniformly on compact subsets of ?.

By Lemma 2.1,each(z,r?ε)is a domain.DefineBy Montel’s theorem and Lemma 2.4,we may assume that the sequencegkiconverges uniformly on compact subsets ofBkD(z,r ?ε)to a holomorphic mapg:(z,r ?ε)→

Takes >0 such that Bn(z,s)(z,r ?ε).By Lemma 2.4, there existsN > N1such that Bn(z,s)(z,r ?ε)for allli > N.Considergli|Bn(z,s),by Cauchy inequality,|det(z)|< cfor allli > N, for some positive constantc.So we have|det(0)|>for allli > N.Thus, we have|det(0))|>0 and|det(z)|>0.By Lemma 2.2, bothfandgare injective.In particular,f(?)?Dwithf(0)=zandg((z,r ?ε))?? withg(z)=0.Sincef ?g(w)=wfor all(z,r ?ε), we get(z)≥tanh(r ?ε).Sinceεis arbitrary,we have

Based on Corollary 2.1 and Theorem 2.4, we can slightly refine [5, Theorem2.1] as follows.

Theorem 2.5Suppose that D is Kobayashi complete and {Dj}j≥1exhausts D.Thenuniformly on compact subsets of D.

ProofSinceDis Kobayashi complete, thus taut, we havefor allz ∈D, by Theorem 2.4.

Forz ∈Dand 0(z),by the definition of Fridman invariant and the completeness ofD, there exists an open holomorphic embeddingf:?→Dsuch that((z)?ε)f(?).Thus, there existsδ >0 such that((z)?ε)?f((1?δ)?)D.Hence, there existsN >0 such that((z)?ε)?f((1?δ)?)?Djfor allj > N.By the decreasing property of the Kobayashi distance, we have((z)?ε)((z)?ε).So we have((z)?ε)?f((1?δ)?)for allj > N, which implies thatSinceεis arbitrary,we getand henceBy Corollary 2.1, the convergence is uniform on compact subsets ofD.

3 Generalized Squeezing Functions

Throughout this section,we suppose thatDis a bounded domain in Cnand ? is a bounded,balanced and convex domain in Cn(unless otherwise stated).

Denote bykΩandcΩthe Kobayashi and Carath′eodory distance on ?, respectively.The following Lempert’s theorem is well-known.

Theorem 3.1(see [8, Theorem 1])On a convex domain?,kΩ=cΩ.

Combining Theorem 3.1 with [6, Proposition 2.3.1(c)], we have the following key lemma.

Lemma 3.1For any z ∈?, ρΩ(z)=tanh(kΩ(0,z))=tanh(cΩ(0,z)).

We will also need the following basic fact.

Lemma 3.2ρΩis aC-norm.

ProofFor anyz1,z2∈Cn, we want to show thatρΩ(z1+z2)≤ρΩ(z1)+ρΩ(z2).

Fixε>0.Takec1=ρΩ(z1)+andc2=ρΩ(z2)+, thenandSince ? is convex, we get

Hence,ρΩ(z1+z2)≤c1+c2≤ρΩ(z1)+ρΩ(z2)+ε.Sinceεis arbitrary,we obtainρΩ(z1+z2)≤ρΩ(z1)+ρΩ(z2).

Since ? is bounded, it is obvious thatρΩ(z)>0 for allz /=0, which completes the proof.

We say thatf ∈Ou(D,?)is an extremal map atz ∈Dif ?((z))?f(D).When ?=Bn,the existence of extremal maps was given in [1, Theorem 2.1].The proof of the next theorem is very similar to that of Theorem 2.1 and [1, Theorem 2.1], based on Montel’s theorem and the generalized Hurwitz theorem, so we omit the details.

Theorem 3.2An extremal map exists at each z ∈D.

As an immediate result, we have the following corollary.

Corollary 3.1(z)= 1for some z ∈D if and only if D is biholomorphically equivalent to?.

In [1, Theorem 3.1], it was shown thatsD(z)is continuous.Moreover, it was given in [1,Theorem 3.2] without details the following inequality:

|sD(z1)?sD(z2)|≤2 tanh[kD(z1,z2)], z1,z2∈D.

Our next theorem gives the same inequality for generalized squeezing functions,and in particular shows that they are also continuous.

Theorem 3.3For any z1,z2∈D, we have

(z)is continuous.

ProofBy Theorem 3.2, there exists a holomorphic embeddingf:D →? such thatf(z1)=0 and ?((z1))?f(D).

If tanh[kD(z1,z2)](z1), then it is obvious that

Suppose now that tanh[kD(z1,z2)](z1).By the decreasing property of the Kobayashi distance and Lemma 3.1, we have

≥tanh[kΩ(f(z1),f(z2))]=tanh[kΩ(0,f(z2))]=ρΩ(f(z2)).

Define

and setg(z)=h ?f(z).Theng ∈Ou(D,?)andg(z2)=0.

For anyw ∈? with

we have

SinceρΩ(z)is a C-norm by Lemma 3.2, we get

This implies that

So we have

Hence,

Similarly,

Therefore,(z1)(z2)|≤2 tanh[kD(z1,z2)] for allz1,z2∈D.

Since the Kobayashi distance is continuous (see e.g.[6]), we get that(z)is continuous.

In case that ? is homogeneous, we have better estimates as follows.

Theorem 3.4If?is bounded, balanced, convex and homogeneous, then for any z1,z2∈D,we have

ProofBy Theorem 3.2, there exists a holomorphic embeddingf:D →? such thatf(z1)=0 and ?((z1))?f(D).

If tanh[kD(z1,z2)](z1), then it is obvious that

Suppose now that tanh[kD(z1,z2)](z1).Since ? is homogeneous, there existsψ ∈Aut(?)such thatψ ?f(z2)=0.

For anyw ∈? with

by the decreasing property of the Kobayashi distance and Lemma 2.3, we have

By Lemma 3.1, this implies thatψ?1(w)∈?((z1)).Hence,

Sinceψ ?f(z2)=0, again by Lemma 3.1, we have

Thus,

Similarly,

Therefore,(z1)(z2)|≤tanh[kD(z1,z2)] for allz1,z2∈D.

The following stability result generalizes and slightly refines [2, The orem2.1].

Theorem 3.5If {Dl}l≥1exhausts D, thenuniformly on compactsubsets of D.

ProofSince the proof of the convergence is similar to that of [2, Theorem2.1], we omit the details.

Now suppose that the convergence is not uniform on compact subsets ofD.Then, there exists a compact subsetK ?D,ε>0, a subsequence {lj} andzlj ∈K ?Dljsuch that

SinceKis compact, there exists a convergent subsequence, again denoted by {zlj}, withChooser >0 such that Bn(z,r)?D.Then, there isN1>0 such thatzlj ∈Bn(z,r)?Dljfor alllj > N1.By Theorem 3.3 and the decreasing property of the Kobayashi distance, for alllj >N1we have

It is clear that there isN2>0 such that for alllj >N2we have

SetN=max{N1,N2}.Then for alllj >Nwe have

which is a contradiction.

The notion of the squeezing function was originally introduced to study the“uniform squeezing” property.In this regard, we have the following theorem.

Theorem 3.6For two bounded, balanced and convex domains?1and?2inCn,(z)has a positive lower bound if and only if(z)has a positive lower bound.

ProofIt suffices to prove the equivalence when ?2= Bn.By Lemma 3.2,ρΩ1(z)is a C-norm.Thus,it is continuous and there existsM ≥m>0 such thatm‖z‖≤ρΩ1(z)≤M‖z‖.Then, one readily checks using the definition that

Combining Theorem 3.6 with [1, Theorems 4.5 and 4.7], we have the following theorem.

Theorem 3.7(z)has a positive lower bound, then D is complete with respect to the Carath′eodory distance, the Kobayashi distance and the Bergman distance of D.

4 Comparison of Fridman Invariants and Generalized Squeezing Functions

Since Fridman invariants and generalized squeezing functions are similar in spirit to the Kobayashi-Eisenman volume formKDand the Carath′eodory volume formCD, respectively, it is natural to study the comparison of them.For this purpose, we will always assume thatDis a bounded domain in Cnand ? is a bounded, balanced, convex and homogeneous domain in Cn.

Similar to the classical quotient invariantwe introduce the quotientwhich is also a biholomorphic invariant.When ? = Bn, we simply write

In [9], Nikolov and Verma have shown thatmD(z)is always less than or equal to one.The next result shows that the same is true for(z).

Theorem 4.1For any z ∈D, we have(z)≤1.

ProofFor anyz ∈D, by Theorem 3.2, there exists a holomorphic embeddingf:D →?such thatf(z)=0 and ?((z))?f(D).

Defineg(w):=f?1((z)w), which is an injective holomorphic mapping from ? toDwithg(0)=z.By the decreasing property of the Kobayashi distance and Lemma 3.1, we have

Thus,

This implies that(z)(z), i.e.,(z)≤1.

A classical result of Bun Wong(see[12,Theorem E])says that if there is a pointz ∈Dsuch thatMD(z)= 1, thenDis biholomorphic to the unit ball Bn.In [10, Theorem 3], we showed that an analogous result formD(z)does not hold.The next result is a generalized version of[10, Theorem 3] for(z).

Theorem 4.2If D is bounded, balanced and convex, then(0)=1.

ProofBy Theorem 2.1, there exists a holomorphic embeddingf: ?→Dsuch thatf(0)=0 and(0(0))?f(?).

Defineg(w):=f?1((0)w), which is an injective holomorphic mapping fromDto ? withg(0)=0.By the decreasing property of the Kobayashi distance and Lemma 3.1, we have

Thus,

This implies that(0)(0).By Theorem 4.1, we always have(0)(0).This completes the proof.

Corollary 4.1Let?i,i=1,2,be two bounded,balanced, convex and homogeneous domains inCn.Thenfor all z1∈?1and z2∈?2.

ProofSince both ?1and ?2are homogeneous, it suffices to show that

By Lemma 3.1,we have(0,arctanh(r))=?2(r)forr >0.Then,by definition,By Theorem 4.1, we get

We can also compare generalized squeezing functions for different model domains as follows.

Theorem 4.3Let?i, i=1,2,be two bounded, balanced, convex and homogeneous domains inCn.Then, for any z ∈D, we have

ProofFor anyz ∈D, by Theorem 3.2,there exists a holomorphic embeddingf:D →?1such thatf(z)= 0 and ?1((z))?f(D).And there exists a holomorphic embeddingg:?1→?2such thatg(0)=0 and

SetF=g ?f, thenF ∈Ou(D,?2)withF(z)= 0.Denotethen ? is a bounded, balanced and convex domain withBy the decreasing property of the Kobayashi distance and Lemma 3.1, we have

On the other hand, by Lemma 3.1, we have

We finish our study by computing explicitly some generalized squeezing functions in the next result, which generalizes [1, Corollary 7.3].

Theorem 4.4For any z ∈?{0(z)=ρΩ(z).

ProofSince ? is homogeneous, for anyz ∈?{0}, there existsψ ∈Aut(?)such thatψ(z)=0.Then, by Lemma 3.1,

ρΩ(ψ(0))=tanh[kΩ(ψ(0),0)]=tanh[kΩ(ψ(0),ψ(z))]=tanh[kΩ(0,z)]=ρΩ(z).

Next, we show that(z)≤ρΩ(z).By Theorem 3.2, there exists a holomorphic embeddingf:?{0}→? such thatf(z)=0 and ?((z))?f(?{0}).

Defineg(w):=f?1((z)w), which is an injective holomorphic mapping from ? to?{0} withg(0)=z.By the decreasing property of the Carath′eodory distance and Lemma 3.1, we have

By Riemann’s removable singularity theorem,we havecΩ{0}(z1,z2)=cΩ(z1,z2)for allz1,z2∈?{0}.Thus,

Hence,(z)≤ρΩ(z), which completes the proof.