Holomorphic Retractions of Bounded Symmetric Domains onto Totally Geodesic Complex Submanifolds

Ngaiming MOK

Abstract Given a bounded symmetric domain ? the author considers the geometry of its totally geodesic complex submanifolds S? ?.In terms of the Harish-Chandra realization ? ? Cnand taking S to pass through the origin 0∈ ?,so that S=E ∩ ? for some complex vector subspace of Cn,the author shows that the orthogonal projection ρ:?→E maps ? onto S,and deduces that S?? is a holomorphic isometry with respect to the Carath′eodory metric.His first theorem gives a new derivation of a result of Yeung’s deduced from the classification theory by Satake and Ihara in the special case of totally geodesic complex submanifolds of rank 1 and of complex dimension≥2 in the Siegel upper half plane Hg,a result which was crucial for proving the nonexistence of totally geodesic complex suborbifolds of dimension≥2 on the open Torelli locus of the Siegel modular variety Agby the same author.The proof relies on the characterization of totally geodesic submanifolds of Riemannian symmetric spaces in terms of Lie triple systems and a variant of the Hermann Convexity Theorem giving a new characterization of the Harish-Chandra realization in terms of bisectional curvatures.

Keywords Bounded symmetric domain,Harish-Chandra embedding,Holomorphic retraction,Totally geodesy

1 Introduction

Let ??Cnbe a bounded symmetric domain in its Harish-Chandra realization(cf.Theorem 2.3).Denote bythe G0-invariant K?hler metric on ? such that minimal disks on each irreducible factor of ? are of constant Gaussian curvature ?2.When ? is irreducible,is a complete K?hler-Einstein metric.In general,the choice of a G0-invariant K?hler metric on ? depends on normalizing scalar constants,one for each irreducible factor.

In this article we consider complex linear slices S of ? which are totally geodesic with respect toand prove the following theorem yielding a holomorphic retraction of ? onto S.

Theorem 1.1 Let ??Cnbe a bounded symmetric domain in its Harish-Chandra realization.Let E?Cnbe a complex vector subspace such that S:=E∩??? is a totally geodesic complex submanifold with respect to.Let ρ :Cn→ E be the orthogonal projection.Then,ρ(?)=S.

We observe that any totally geodesic complex submanifold of(?,)passing through the origin 0∈? is necessarily of the form S=E∩? for a complex vector subspace E?Cnsatisfying additional conditions(cf.Proposition 2.1),hence Theorem 1.1 yields a holomorphic retraction of ? onto any given totally geodesic complex submanifold of ? with respect to one and hence all G0-invariant K?hler metrics.

Theorem 1.1 in the special case where ? is the type-III domain,and S is biholomorphic to the complex unit ball Bmof dimension m,where M(a,b;C)stands for the complex vector space of a-by-b matrices with complex coefficients and Ztdenotes the transposed matrix of Z,was established in Yeung[12,Theorem 1]by explicitly checking according to the classification of such embeddings due to Satake[10]and Ihara[2].Theorem 1.1 in these special cases are crucial for the establishment of the following theorem in[12]concerning the open Torelli locus.For the understanding of the statement,note first of all that the type-III domainis biholomorphic via the inverse Cayley transform τ= λ(Z):= ??(Z+ ?Ig)(Z ? ?Ig)?1,where Igstands for the g-by-g identity matrix,to the Siegel upper half plane Hg:={τ:Im(τ):τt= τ,Im(τ)>0}defined by the Riemann bilinear relations,so that Ag:=Hg/PSp(g;Z)is the Siegel modular variety,the classification space of principally polarized abelian varieties.

The Torelli map tg:Mg→Ag,where Mgis the Teichmüller modular variety,i.e.,the moduli space of compact Riemann surfaces C of genus g≥2,is the holomorphic map defined for a compact Riemann surface C of genus g≥2 by tg([C])=[Jac(C)],where Jac(C)stands for the Jacobian variety of C in its natural principal polarization,and[···]is here and henceforth a notation for the class of an object in some classification space.Denote bythe Deligne-Mumford compactification,and bythe Satake-Baily-Borel compactification,then it is known that tg:Mg→ Agextends holomorphically to τg:The set T0g:=tg(Mg)is called the open Torelli locus,which is a Zariski open subset of the Zariski closed subset.Denoting by Hg? Mgthe locus of hyperelliptic curves,then Hg?Mgis Zariski closed.It is well-known that the Torelli map tg:Mg→Agis injective and that tg|Mg?Hg:Mg?Hg→ Agis immersive.The principal result of[12]is the following theorem.

Theorem 1.2 (cf.[12,Theorem 2])The set T0g?tg(Hg)?Agfor g>2 does not contain any complex hyperbolic complex ball quotient,compact or non-compact with finite volume,of complex dimension at least 2 as a totally geodesic complex suborbifold of Ag.

The above result of Yeung,in conjunction with known rigidity results in the higher rank case and existence results of Shimura curves on the open Torelli locus,yielded Yeung[12,Theorem 3],related to Oort’s Conjecture,which described all Shimura varieties(necessarily of dimension 1)contained in the open Torelli locus T0g?tg(Hg).(For the statement of[12,Theorem 3]and related background and references cf.[12,§1].)

We give in this article a proof of Theorem 1.1 in the general situation,where the target bounded symmetric domain ? may contain direct factors which are exceptional domains,and where the complex submanifold S=E∩? is of arbitrary rank as a Hermitian symmetric manifold of the semisimple and noncompact type.Our proof is free from classification theory.It exploits the Harish-Chandra realization and a variant of the Hermann Convexity Theorem defining ? in terms of inequalities involving bisectional curvatures.In Section 2 we collect basic materials on Riemannian symmetric spaces and bounded symmetric domains.In Section 3 we give in Theorem 3.1 a new description of independent interest of the Harish-Chandra realization of a bounded symmetric domain in terms of bisectional curvatures.In Section 4 we give the proof of Theorem 1.1 together with the immediate implication(Theorem 4.1)that totally geodesic complex submanifolds of ? are holomorphic deformation retracts of ?.In Section 5 we give in Theorem 5.1 an application of Theorem 1.1 to the geometry of the complex submanifold S?? in terms of the Carath′eodory metric and equivalently the Kobayashi metric,which are equal on weakly convex bounded domains according to the celebrated work of Lempert[3]and a theorem of Royden-Wong(cf.Section 5 for remarks and references).In the Appendix we give a selfcontained proof that the(infinitesimal)Carath′eodory metric and the(infinitesimal)Kobayashi metrics agree with each other on a bounded symmetric domain ? by means of Theorem 1.1.The article has been written in a somewhat expository style,supplied sometimes with more details than those are absolutely necessary,in order to make it more accessible to non-experts.

2 Background Materials

2.1 Basic materials in Lie theory and on Riemannian symmetric spaces

On a Riemannian symmetric space(M,ds2M)denote by G the identity component of the isometry group of(M,ds2M),and by e∈G its identity element.We have Te(G):=g.Here and in what follows,for real Lie groups in Roman letters we denote by the corresponding Gothic letters their associated Lie algebras,and vice versa.Let K?G be the isotropy subgroup at a reference point 0∈M,so that M=G/K as a homogeneous space,and 0=eK.Let s be the involution of(M,ds2M)as a Riemannian symmetric space at 0,s=s?1,and σ :G → G be defined by σ(g)=sgs=s?1gs,so that dσ(e):g → g,and we have the Cartan decomposition g=⊕m where(resp.m)is the eigenspace of dσ(e)associated to the eigenvalue+1(resp.?1),from which we have an identification T0(M)～=m.We have the following characterization of totally geodesic submanifolds of Riemannian symmetric spaces(cf.Helgason[1,Chapter IV,Theorem 7.2]).

Theorem 2.1 On the Riemannian symmetric space(M,ds2M)and in the notation above,let m1? m ～=T0(M)be a vector subspace.Then,denoting by Exp0:T0(M)→ M the exponential map in the sense of Riemannian geometry,S:=Exp0(m1)?M is a totally geodesic submanifold if and only if m1?g is a Lie triple system,i.e.,if and only if[m1,[m1,m1]]?m1for the Lie bracket[·,·]on g.Moreover,in the notation above,writing1=[m1,m1] ? [m,m] ?and defining g1:=m1⊕1?m⊕=g,g1?g is a Lie subalgebra,and,denoting by G1?G the Lie subgroup corresponding to the Lie subalgebra g1?g,K1?K the Lie subgroup corresponding to the Lie subalgebra1?,(S,ds2M|S)is a Riemannian symmetric space on which G1acts transitively,and S=G1/K1as a homogeneous space.

For a Cartesian product of Riemannian manifoldsthe Riemannian connection?is unchanged if the background metricof each Cartesian factor is replaced byfor some λk>0.Since a smooth submanifold Z ? N is totally geodesic if and only if its tangent bundle T(Z)is parallel along Z,which depends only on?,the latter occurs if and only if Z is totally geodesic with respect to any of the Riemannian metric h thus obtained by scaling.It is therefore not surprising that the necessary and sufficient condition in Theorem 2.1 in terms of Lie triple systems is a purely Lie-theoretic condition independent of the choice of the background metric ds2Mrendering(M,ds2M)Riemannian symmetric,noting also that the subset S=Exp0(m1),which is the union of geodesics emanating from 0,also remains unchanged by introducing the scaling constants.

2.2 Basic materials on bounded symmetric domains

Consider now the case where(M,ds2M)=(X0,g)is a Hermitian symmetric space of the noncompact type,so that X0is biholomorphic to a bounded symmetric domain.Here and henceforth by a Hermitian symmetric space of the noncompact(resp.compact)type we will mean one with negative(resp.positive)Ricci curvature,i.e.,it is implicitly assumed that the Hermitian symmetric space is of the semisimple type.

Write G0for the identity component of the isometry group of X0,which is equivalently the identity component of the group Aut(X0)of biholomorphic automorphisms of X0.Write K?G0for the isotropy subgroup at 0∈X0,0=eK.Denote by Xc:=Gc/K the Hermitian symmetric space of the compact type dual to X0.Denote by GCthe identity component of Aut(Xc)and by P?GCthe isotropy(parabolic)subgroup at 0,so that Xc=GC/P as a complex homogeneous space.Write X0=G0/K?→GC/P=Xcfor the Borel embedding identifying X0as an open subset of its compact dual Xc.Whenever appropriate,we write VC=V?RC for the complexification of a real vector space V.Write g0=+m for the Cartan decomposition of g0with respect to the involution at 0.Then,gc=stands for the corresponding Cartan decomposition of gc.

For u,v∈gcwe write ad(u)(w):=[u,w],for ad(u)∈End(gc),etc.and denote by B(u,v):=Tr ad(u)ad(v)the Killing form B(·,·)on gc.Since gcis a compact real form of a semisimple complex Lie algebra,B(·,·)is negative definite.Extend B(·,·)to the complexification gCof gcby complex bilinearity so that B(·,·)is a nondegenerate complex bilinear form on gC.Denote by(·,·)the Hermitian bilinear pairing defined by(u,v)=B(u,?τc(v)),where τcstands for the conjugation on gCwith respect to the real form gc? gC,and write.The isotropy subgroup K?Gcis reductive,and the complex structure on Xcis induced by the adjoint action of some element z belonging to the centerof.We have the Harish-Chandra decomposition gC=m+⊕C⊕m?which is eigenspace decomposition of ad(z)∈End(gC)corresponding to the eigenvalues,respectively.By considering the action of ad(z)it follows readily that[m+,m+]=[m?,m?]=0,[C,C]?C,[C,m+]? m+,[C,m?]? m?and[m+,m?]?C.In particular,the complex vector subspaces m+,m??gCare abelian subalgebras.We have m+⊕m?=mCand.Here and in what follows,for u∈gC,will be taken with respect to the conjugation τ0on the noncompact real form g0? gC.We have τ0|= τc|and τ0|mC= ?τc|mC.

The Hermitian symmetric space X0of the noncompact type can be identified as a bounded symmetric domain by means of the Harish-Chandra embedding,as follows(cf.Wolf[11]).For its formulation given the Harish-Chandra decomposition gC=m+⊕C⊕m?,we have correspondingly the abelian subgroups M+,M??GC,and the reductive subgroup KC?GC.

Theorem 2.2 (Harish-Chandra Embedding Theorem)The holomorphic map F:M+×KC×M?→GCdefined by F(m+,k,m?)=m+km?is a biholomorphism of M+×KC×M?onto a dense open subset of the complex Lie group GCcontaining G0.In particular,the map η :m+→ GC/P=Xcdefined by η(m+)=exp(m+)P is a biholomorphism onto a dense open subset of Xccontaining G0/K=X0.Furthermore,η?1(X0)=:? is a bounded domain on m+～=Cn,n=dimCX0.

The bounded symmetric domain ??Cnin its Harish-Chandra realization can be precisely described in Lie-theoretic terms as the unit ball in m+～=Cnwith respect to a Banach norm,implying in particular its convexity,by the Hermann Convexity Theorem(cf.Wolf[11]),as follows.

Theorem 2.3(Hermann Convexity Theorem)Identify Cnwith the holomorphic tangent space T0(?).Then,? ? Cnis the unit ball in Cncorresponding to the Banach norm ‖ ·‖Hon T0(?)defined by

Remark 2.1 Note that in the statement of the Hermann Convexity Theorem,the operator norm of ad(Reξ) ∈ End(gC)is unchanged when the Hermitian inner product(·,·)is rescaled,i.e.,when the Hermitian inner product on each of the simple factors of gCis replaced by a scalar multiple.

2.3 Characterization of totally geodesic complex submanifolds of bounded symmetric domains in Harish-Chandra coordinates

In this subsection,we give a characterization of totally geodesic complex submanifolds S?? of bounded symmetric domains ??Cnin terms of Harish-Chandra coordinates.From the homogeneity of ? under G0it suffices to characterize those S ? ? passing through 0.

Since the focus is now on complex manifolds,here and henceforth we adopt a convention common in complex geometry on the notation for tangent spaces.Given an n-dimensional complex manifold Z and a point x∈Z,we denote by TRx(Z)the real(2n)-dimensional tangent space at x of the real(2n)-dimensional smooth manifold underlying Z,while the notation Tx(Z)is reserved for the complex n-dimensional holomorphic tangent space at x,as opposed to the meaning of the same notation in(2.1).Writingand decomposing the(2n)-dimensional complex vector spaceas a direct sum of eigenspaces of the J-operator underlying the integrable almost complex structure of Z,the holomorphic tangent space Tx(Z)is canonically identified with the complex vector subspaceof complexified tangent vectors of type(1,0).We have the following proposition.

Proposition 2.1 Let ??Cnbe a bounded symmetric domain in its Harish-Chandra realization,?=G0/K as a homogeneous space in the notation above,and S?? be a complex submanifold passing through the origin 0 ∈ ?.Then,identifying T0(S)as an abelian complex Lie subalgebra m+1?m+?gC,S?? is totally geodesic with respect to a given invariant K?hler metric g on ? if and only if[m+1,[m+1,m?1]] ? m+1.Furthermore,S=E ∩ ? for the complex vector subspace E?Cncorresponding to m+1?m+whenever S?? is totally geodesic.

Proof Let S?? be a totally geodesic complex submanifold passing through 0∈?.Under the identification T0(?) ～=m+,we identify T0(S)with a complex vector subspace m+1? m+.The real tangent space TR0(S)is given by Re(m+1)=:m1? m=TR0(?).By Theorem 2.1,m1? g0is a Lie triple system,i.e.,(?)[m1,[m1,m1]] ? m1holds.We claim that(?)is equivalent to(??)[m+1,[m+1,m?1]]? m+1.

Starting with(?)and complexifying,we have[mC1,[mC1,mC1]] ? mC1.Since mC1=m+1⊕m?1,where m?1=m+1,and[m+1,m+1]=[m?1,m?1]=0,(?)is equivalent to[m+1,[m+,m?1]]+[m?1,[m?1,m+1]]?m+1⊕m?1.Noting that[m?1,[m?1,m+1]]=and that[m+1,[m+1,m?1]]? [m+,[m+,m?]]? [m+,C]? m+,we conclude that(?)is equivalent to(??)[m+1,[m+1,m?1]]?m+1,as claimed.

We have deduced from Theorem 2.1 that S=Exp0(m+1)?? is a totally geodesic complex submanifold if and only if(??)holds.To complete the proof of Proposition 2.1 it remains to show that S?? must be given by S=E∩? for the complex vector subspace E?Cncorresponding to m+1?m,whenever the complex submanifold S?? is totally geodesic with respect to(?.g)and it passes through the origin 0∈ ?.

For θ∈ R define μθ∈ GL(n;C)by μθ(z)=eiθz.Consider now the circle group S1={μθ:θ ∈ R} ? K,which acts on ? by scalar multiplication,so that ? is a circular domain.Write Sθ:= μθ(S).We have 0 ∈ Sθand T0(Sθ)=eiθ·S=S.Since there is exactly one totally geodesic submanifold(Z,g|Z)of(?,g)passing through 0 such that TR0(Z)=TR0(S),we have Sθ=S.Thus,for any point x ∈ S,eiθx ∈ S for any θ∈ R,hence for x/=0,the complex analytic subset Cx∩S of the open disk Cx∩? must be the whole disk as it contains the real analytic curve S1·x,so that S ? ? is a union of open disks centered at 0 on complex lines ? passing through 0.Thus,writing λ :Cn?{0}→ Pn?1for the canonical projection,there is a complex analytic subvariety A ? Pn?1such that S=(λ?1(A)∪{0})∩ ?.Finally,since S ? ? is smooth at 0,as is well-known the subvariety A ? Pn?1must necessarily be a projective linear subspace,i.e.,S must be of the form E∩? for some complex linear subspace E??(noting that E∩ ? is connected as ? is convex).Clearly E ?Cncorresponds to m+1,as desired.The proof of Proposition 2.1 is complete.

3 Characterization of Harish-Chandra Realizations of Bounded Symmetric Domains in Terms of Bisectional Curvatures

Recall that the G0-invariant K?hler metricon the bounded symmetric domain ? has been chosen so that the minimal disks of each irreducible Cartesian factor ? are of constant Gaussian curvature ?2.For an irreducible bounded symmetric domain ? =G0/K,we give a brief description of the root space decomposition of the complex simple Lie group gCrelevant to the study of bisectional curvatures,and refer the reader to[4,11]and references therein for details.Here and in what follows we use the notation of the first two paragraphs in(2.2).

In what follows,we will fix a lexicographic ordering of the roots compatible with the choice of positive Weyl chamber so that there is a unique highest root μ of g,and μ∈Φ+0 is always a long root.When gCis of type A,D or E,all roots in Φ are of equal length.

From now on,for ? irreducible,we will replace the Killing form B(·,·)on gCin the definition of the Hermitian inner product(·;·)and the Hermitian norm ‖ ·‖ by B′(·,·)=cB(·,·)for some constant c=cg>0 such that the induced Hermitian inner product(u;v):=B′(u,?τc(v))is the standard Euclidean Hermitian inner product for u,v∈ T0(?),i.e.,for the G0-invariant K?hler metric g we have gij(0)= δij,and g is precisely our choice of ds2?when ? is irreducible.The restriction B′|?his positive definite,and it defines a real linear isomorphism form:= ?h?to ?h=:hR,and we identify ? ∈ Φ in this way with an element H?∈ hR.

In the general case for gC=⊕ ···⊕we rescale the Killing form on each simple direct factor,1 ≤ i≤ s,accordingly.Observe that the operator Banach norm ‖ad(u)‖Hsuch as that appearing in the statement of the Hermann Convexity Theorem for u∈gCremains unchanged by such a replacement of B by B′(cf.Remark 2.1).

For a positive noncompact root ? we write g[?]:=g?⊕ g??⊕ [g?,g??],where[g?,g??]=CH?.Writing e?∈ g?for a unit root vector,e??=e?∈ g??,we have[H?,e?]=2e?,[H?,e??]= ?2e??and[e?,e??]=H?,so that g[?]～=sl(2,C).

The orbit of 0 ∈ ? under the Lie group G[?]? GCcorresponding to g[?]? gCis a rational curve ??:=G[?]·0 on the compact dual Xcof X0～= ?,which is totally geodesic with respect to the Gc-invariant K?hler metric gcon ? dual to, ? ? Xcbeing the Borel embedding.When ? ∈ Φ+0is a long root,??? Xcis a minimal rational curve.Defining g0[?]:=g[?]∩ g0,we have g0[?]～=su(1,1).The orbit of 0 ∈ ? under the corresponding Lie subgroup G0[?]? G0is a totally geodesic holomorphic disk D?:=G0[?]·0 on ?.When ? ∈ Φ+0is a long root,Dα=??∩? is a minimal disk on ?.

We say that two roots ?1,?2∈ Φ are strongly orthogonal if and only if neither ?1+?2nor ?1? ?2is a root.Let Ψ ={ψ1,···,ψs} ? Φ+0be a maximal strongly orthogonal subset,i.e.,a subset of maximal cardinality of mutually strongly orthogonal positive noncompact roots.Then,s=r:=rank(?).Note that ψ1,ψ2∈ Φ+0are strongly orthogonal to each other if and only if ψ1?ψ2/∈Φ,since ψ1+ψ2is never a root,observing that[m+,m+]=0.

For a strongly orthogonal set of positive noncompact roots Θ ? Φ+0we write

Then,g[Θ]is a semisimple complex Lie algebra,g[Θ]～=sl(2,C)|Θ|.Writing g0[Θ]=g[Θ]∩ g0,then g0[Θ]? g[Θ]is a semisimple Lie algebra which is a noncompact real form of g[Θ]without compact factors,g0[Θ]～=su(1,1)|Θ|,and the G0[Θ]-orbit of 0 ∈ ? is a Euclidean polydisk.

To extract a maximal strongly orthogonal subset Ψ ?,we may start with choosing ψ1= μ ∈ Φ+0being the highest root and consider the subset Σ ? Φ+0consisting of roots ? strongly orthogonal to ψ1.Then there exists a simple Lie subalgebra g′0? g0such that,putting′=∩g′0,G′0/K′0?G0/K=X0～=? is an irreducible Hermitian symmetric space of the noncompact type embedded as a totally geodesic complex submanifold of ?′? ?,and such that T0(?′)is spanned by{gσ:σ ∈ Σ}.Writing h′:=h ∩,h′C? g′Cis a Cartan subalgebra,and the restriction σ′:= σ|h′ for all σ ∈ Σ gives the set of positive noncompact roots of g′.We have a totally geodesic complex submanifold ? × ?′?→ ? such that T0(?)=Cgμand.Repeating the same procedure with ?′in place of ?,we obtain inductively a maximal strongly orthogonal subset Ψ ?of positive noncompact roots,Ψ ={ψ1,···,ψr}.

Given any ξ∈ T0(?),there exists k ∈ K such that η :=k(ξ)is tangent to the reference maximal polydisk Π0? ?.Since Aut(Π0)embeds into G0,composing with the action of(S1)rfor the circle group S1,acting according to(eiθ1,···,eiθr)·(z1,···,zr) → (eiθ1z1,···,eiθrzr)and with permutations of the r Cartesian factors,we obtain some element k′∈ K such that k′(ξ)=(a1,···,ar;0,···,0)such that all ai,1 ≤ i≤ r are real and nonnegative,and such that a1≥ ···≥ ar≥ 0.We call(a1,···,ar;0,···,0),or simply(a1,···,ar),the normal form of ξ under the action of K.

Lemma 3.1 For any irreducible bounded symmetric domain ?0? Cn0,there exists an irreducible bounded symmetric domain ? ? Cnsuch that ?0? ? as a totally geodesic complex submanifold passing through 0,and such that,writing ?=G0/K,gCis of type A,D or E.Hence,writing Φ for the set of all roots of gCwith respect to a Cartan subalgebra hC? gC,all roots ?∈Φ are of equal length.

Proof Up to biholomorphisms,the only irreducible bounded symmetric domains ? not of these types are those of types B or C.These include type II domainswhere n≥5 is odd,type III domainsof rank n,n≥3,and type IV domains of odd dimension n≥3.For,m ≥ 2,it suffices to takeFor,it suffices to takeFor,m ≥ 1,it suffices to take.All notations for bounded symmetric domains are standard ones and the embedding ?0? ? are also standard embeddings.

We will need the following lemma on the combinatorics relating the space of positive noncompact rootswith a maximal strongly orthogonal set Ψ of positive noncompact roots.

Lemma 3.2 Let ??Cnbe an irreducible bounded symmetric domain of type A,D or E.Let Ψ ={ψ1,···,ψr}be a maximal strongly orthogonal set of positive noncompact roots and pick ? ∈? Ψ.Then either

(a)there exist exactly two distinct roots ψi1,ψi2∈ Ψ such that ? ? ψj∈ Φ if and only if j=i1or j=i2;or

(b)there is exactly one root ψ ∈ Ψ such that ? ? ψ ∈ Φ.

Proof We use standard notation for irreducible bounded symmetric domains.For1≤p≤q,and for,m≥2,the lemma is obvious by using the standard representation of T0(?)as a complex vector space of matrices and the root vectors of ? ∈ Φ+0as a standard basis of such a vector space.For example,take in the case of a type I domain DIp,q,1≤p≤q the standard choice of the Cartan subalgebra hC?gCso that each of the 1-dimensional root spaces g?,? ∈ Φ+0,is spanned by an elementary matrix Eij,with(i,j)-entry being 1,and all other entries being 0,where 1≤i≤ p,1≤j≤q,and choose the maximal polydisk Π to be such that T0(Π)=SpanC{Ekk:1≤ k≤ p}.Then,case(a)occurs if and only if g?=CEij,1≤i≤p,1≤j≤p.and case(b)occurs if and only if g?=CEij,1≤i≤p,p+1≤j≤q.The case of type II domainsis very similar to the case of type I domains,except that only case(a)occurs.Since type-IV domains,m≥2 are of rank 2,the lemma is vacuous in that case.The same is true for DV,which is also of rank 2.

It remains to check the case of ?=DVI,which is of type E7and of rank 3.We will make use of the labeling of roots as in Zhong[13].In the standard notation used in[13],a maximal set of strongly orthogonal positive noncompact roots Ψ ? Φ+0is given by Ψ ={ψ1,ψ2,ψ3},in which ψ1=x1?x2,ψ2=x1+x2+x3,ψ3=d? x3,where d=x1+···+x7.For eachwe define Hν:={?∈Φ+0:ν??∈Φ}.To complete the proof of the lemma,it suffices to show that Hψ1∩Hψ2∩Hψ3= ?.Any root ν∈ Φ+0is a long root,so that[eν]∈ C0(Xc),the various of minimal rational tangents(VMRT for short)on the irreducible Hermitian symmetric space Xcof the compact type,and we have a decomposition of T0(?)into a direct sum of eigenspaces of the Hermitian bilinear form Heν(u,v)= Θeνeνuv,given by T0(?)=Ceν⊕Heν⊕Neν,where Ceν,Heνand Neνare the eigenspaces of Heνcorresponding to the eigenvalues 2,1 and 0,respectively,and Heν=Span{e?:? ∈ Hν}.Now for each unit vector α such that[α]∈ C0(Xc),Hαcan be identified with T[α](C0(Xc)).In the case of Xcdual to X0～=DVI,C0(Xc)is dual to DV,hence dimCHα=16,|Hν|=16 for each ν∈ Φ+0.Now Hψi? Φ+0?Ψ,|Φ+0?Ψ|=24 and the maximal possible cardinality of Hψ1∩Hψ2is 16+16?24=8.By direct checking we have Hψ1∩Hψ2={x1?x4,x1?x5,x1?x6,x1?x7,x1+x3+x4,x1+x3+x5,x1+x3+x6,x1+x3+x7}.Finally,ψ3=d? x3,and none of the 8 elements of the set ψ3?(Hψ1∩ Hψ2)={x2+2x4+x5+x6+x7,x2+x4+2x5+x6+x7,x2+x4+x5+2x6+x7,x2+x4+x5+x6+2x7,x2?x3+x5+x6+x7,x2?x3+x4+x6+x7,x2?x3+x4+x5+x7,x2?x3+x4+x5+x6}?is a root,hence Hψ1∩Hψ2∩Hψ3= ?,as desired.The proof of Lemma 3.2 is complete.

The given proof of Lemma 3.2 relies on some direct checking on roots.While that has the advantage of being straightforward,it is also desirable to give a more conceptual proof of the lemma.We give here such a proof which relies on some knowledge of the VMRT C0(Xc)of an irreducible Hermitian symmetric space Xcof the compact type,and on a curvature formula for C0(Xc) ? PT0(?)as a K?hler submanifold,PT0(?)being endowed with the Fubini-study metric induced by(0)on T0(?).Let gcbe the Gc-invariant K?hler metric on Xcsuch that gcagrees withat 0.(Xc,gc)is of nonnegative bisectional curvature,and,denoting by R0(resp.Rc)the curvature tensor of(?,ds2?)(resp.(Xc,gc)),we havefor α,β,γ,δ∈ T0(?).For convenience we write.We recall the following result from[4,Appendix(III.2)].Here we will write(X,g)for(Xc,gc).

Proposition 3.1 Let(X,g)be an irreducible Hermitian symmetric space of the compact type,and denote by h the Fubini-study metric on PT0(?)induced by g.Then(C0(X),h|C0(X)) ?→(PT0(?,h))is a Hermitian symmetric space of the compact type of rank ≤ 2.Moreover,denoting by S the curvature tensor of(C0(X),h|C0(X)),and identifying at each[α] ∈ C0(X),T[α](PT0(?))with the orthogonal complement of Cα with respect to ds2?(0)for a unit characteristic vector α,S is the restriction of the curvature tensor Θ of(X,g)at 0 to T[α](C0(X)),which corresponds under the aforementioned identification with Hα,the eigenspace belonging to the eigenvalue 1 of the Hermitian bilinear form Hα(u,v)= Θααuv.In particular,for bisectional curvatures we have

for all ξ,η ∈ Hα.

Using Proposition 3.1 we prove the following statement,without requiring g to be of type A,D or E,which implies Lemma 3.2.

Proposition 3.2 Let X be an irreducible Hermitian symmetric space of the compact type,h ?be a Cartan subalgebra of? g0,and Φ be the set of all hC-roots of gC.Let Ψ ={ψ1,···,ψr} ? Φ+0be a maximal set of strongly orthogonal positive noncompact roots,and ρ ∈ Φ+0be a long root. Then,there are at most two distinct elements ψ of Ψ such that ρ?ψ∈Φ.

Proof Since ρ ∈ Φ+0is a long root,the unit root vector eρis a minimal rational tangent,i.e.,α :=[eρ]∈ C0(X).Suppose there exist distinct positive integers i,j and k such that ρ?ψi,ρ?ψjand ρ?ψkare roots.ξψ:=eψmod Cα ∈ T0(X)/Cα are unit tangent vectors of type(1,0)at[α]for ψ = ψi,ψj,ψk.For brevity we write also ξ?for ξψ?,1 ≤ ?≤ r.By the definition of Ψ,we haveBy Proposition 3.1,for the curvature tensor of C0(X) ? PT0(X),we have.It follows that SpanR{Reξi,Reξj,Reξk}is a real 3-dimensional abelian subalgebra in TR0(C0(X)).Exponentiating,we get a real 3-dimensional totally geodesic flat submanifold Σ?C0(X),so that the latter must be of rank≥3 as a Riemannian symmetric space,which contradicts with the fact that rank(C0(X))≤2 as given in[4,Appendix III.2],proving the proposition.

On a complex affine line Λ ? Cnand x ∈ Λ,Tx(Λ)=Cα,where α is a unit vector,for r>0 we denote by ?α(x;r)the open disk on Λ centered at x of radius r,i.e.,?α(x;r):=Bn(x;r)∩Λ.

Theorem 3.1 Let ??Cnbe a bounded symmetric domain in its Harish-Chandra realization.Then,? is the union of open disks ?α(0;rα)? Λαon the complex lines Λα:=Cα,as α ranges over unit vectors on Cn,where

In other words, ξ∈ ? if and only iffor any unit vector ν ∈ T0(?).

Proof By Lemma 3.1 without loss of generality,we may assume that writing ?=G0/K,each irreducible factor of the semisimple complex Lie algebra gCis of type A,D or E.Let α ∈ T0(?)be a unit vector.For simplicity in the ensuing arguments we will assume also that ? is irreducible.The theorem for the general case where ? = ?1× ···× ?sand each irreducible factor ?,1≤ k≤ s is of A,D or E type follows readily from the arguments in the special case where ? is irreducible.

Writing Ψ ={ψ1,···,ψr} ? Φ+0for a maximal set of strongly orthogonal positive noncompact roots,and writing e?for a unit root vector associated to a root ? ∈ Ψ,there is a reference maximal polydisk Π0? ? passing through 0,such that T0(Π0)=Ceψ1⊕ ···⊕ Ceψr,and,for any α ∈ T0(?)there exists k ∈ K such that k(α) ∈ T0(Π0).An element c1eψ1+ ···+eψ1will be denoted as(c1,···,cr).Choosing k properly,we may take k(α)to be the normal form(a1,···,ar)such that each aiis real and nonnegative,and furthermore a1≥ ···≥ ar?1≥ ar.For the proof of Theorem 3.1 without loss of generality we consider α to be the normal form(a1,···,ar)itself.

Remark 3.1(a)The identification of each C0(X)? PT0(?)as a Hermitian symmetric space of the compact type and of rank≤ 2 can be read off from the Dynkin diagram D(g),g=gC.If the Hermitian symmetric space(X,ds2X)is of type(g,αk)in standard notation,and Σ is the set of simple positive roots adjacent to αk,then each connected component of D(g)? {αk},with a marking at σ ∈ Σ,corresponds to an ireducible factor of the Hermitian symmetric space C0(X).

(b)Theorem 3.1 can be reformulated by stating that ξ∈ ? if and only iffor any unit vector ν ∈ m+,from which the convexity of ? follows immediately.As such,Theorem 3.1 may be regarded as a variant of the Hermann Convexity Theorem.Here we prefer to formulate Theorem 3.1 as a statement concerning bisectional curvatures,with an essentially geometric and self-contained proof.

(c)One can prove Theorem 3.1 by geometric means free from classification results.As given here,Theorem 3.1 is deduced from the Polydisk Theorem and the combinatorial Lemma 3.2.The main feature of the Polydisk Theorem relevant to us is that each Cartesian factor is of radius 1,which can be derived without using any classification theory.The alternative proof of Lemma 3.2 given here relies on identifying the VMRT of X as a Hermitian symmetric space(C0(X),s)of rank≤ 2,which follows from the pinching conditionon holomorphic sectional curvatures for unit vectors η ∈ T[α](C0(X))established in[4,Appendix III.2]by elementary means,and which by Ros[7]implies the parallelism of the second fundamental form of C0(X)? PT0(?)and hence the Hermitian symmetry of(C0(X),s),as its curvature tensor S must then necessarily be parallel.By considering maximal polyspheres,the fact that(C0(X),s)must be of rank≤2 follows by observing that the Segre embedding of(P1)3into P7does not have parallel second fundamental form.

(d)The result of[7]was accompanied by a complete listing of K?hler submanifolds satisfying the aforementioned pinching condition of the projective space(Pm,ds2FS)equipped with the Fubini-study metric of constant holomorphic sectional curvature+2,according to Nakagawa-Takagi[6,Theorem 7.4],which miraculously corresponds exactly to the listing of VMRTs of(X,gc)for irreducible Hermitian symmetric spaces of the compact type(cf.[4]).

4 Proof of Theorem 1.1

We will continue to adopt notation with the meaning of symbols as defined in Section 3.For the proof of Theorem 1.1 using results of Section 4 we will need furthermore a couple of preliminary results,as follows.

Consider now the case where the normal form(a1,···,ar)of the unit vector α ∈ T0(? is arbitrary).Writing

Proof Recall that S???Cnis a totally geodesic complex submanifold passing through 0,S=E∩?,in which E ? Cnis a complex vector subspace identified with T0(S)? T0(?)～=Cn,and that ρ :? → E is the orthogonal projection with respect to the Euclidean metric ds20(0)on T0(?) ～=Cn.Let ξ∈ ?.Write ξ= ξ1+ η according to the orthogonal decomposition T0(?)=T0(S)⊕ T0(S)⊥,in which ξ1∈ T0(S)and η ∈ T0(S)⊥,where A⊥for a complex vector subspace A ? T0(?)denotes its orthogonal complement with respect to ds2?(0).By Theorem 3.1,we havefor any unit vector ν∈ T0(?).For μ ∈ T0(S),by Lemma 4.1 we have.Together with the orthogonal decomposition ξ= ξ1+ η we deduce that forμ∈T0(S)we have

where the last inequality follows from the nonnegativity of bisectional curvatures of(Xc,gc).On the other hand,by Proposition 4.1 we know that for any unit vector ν ∈ T0(?)we have

for any unit vector ν ∈ T0(?),hence ξ1∈ ? by Theorem 3.1.In other words, ρ(ξ)= ξ1∈? ∩ E=S,so that ρ(?)=S,as desired.The proof of Theorem 1.1 is complete.

Remark 4.1 With an aim towards a specific application,the special case of Theorem 1.1 where the bounded symmetric domain ? ? Cnis irreducible and S? ? is a minimal(totally geodesic)disk was proved in Mok-Ng[5].The proof there relied on the Hermann Convexity Theorem.

From Theorem 1.1,we readily have the following theorem.

Theorem 4.1 Let(X,g)be a Hermitian symmetric manifold of the noncompact type,and(Y,g|Y)?→(X,g)be a totally geodesic complex submanifold.Then,Y ?X is a holomorphic retract of X,i.e.,there exists a holomorphic mapping r:X→Y such that r|Y=idY.Moreover the identity map idXon X is homotopic through a continuous family{Ft:t∈[0,1]}of holomorphic maps Ft:X→X such that F0(x)=x and F1=r is a holomorphic retract of X on Y and such that the continuous map F:X×[0,1]→X defined by F(x,t):=Ft(x)is real analytic on X×(0,1).

Proof Since the total geodesy of Y in X does not depend on the choice of the Aut0(X)-invariant K?hler metric g on X,without loss of generality we may take g to correspond to the K?hler metricon ? ～=X.Denoting by ξ:X → ? the Harish-Chandra realization,define S:= ξ(Y).Then,writing r:X → Y to correspond to the orthogonal projection ρ:?→S,we have r:X→Y and r|Y=idY.Finally,for 0≤t≤1,define ft:?→S by ft= ρ(x)+t(x? ρ(x)),we have ft(x)∈ ?.As t ranges over[0,1],for each point x ∈ ?,ft(x)describes the closed interval joining x to ρ(x) ∈ S ? ?.Hence,writing f(x,t)=ft(x)we have defined a continuous map f:? ×[0,1]→ ?,which corresponds under the inverse of the Harish-Chandra realization η :to a continuous map F:X×[0,1]→Y yielding a deformation of the identity map to the holomorphic retract r:X→Y,as desired.

5 Holomorphic Totally Geodesic Isometric Embeddings with Respect to Carath′eodory and Koba-yashi Metrics

For a bounded domain D in a complex Euclidean space CN,we denote by ‖ ·‖CDits infinitesimal Carath′eodory metric and by ‖ ·‖KDits infinitesimal Kobayashi metric.On the unit disk ?,we denote by ‖ ·‖?its Poincar′e metric of constant Gaussian curvature ?2.By convention we have ‖ ·‖?= ‖·‖C?= ‖ ·‖K?.Concerning the Carath′edory metric and the Kobayashi metric on bounded symmetric domains and those of their totally geodesic complex submanifolds(which are themselves biholomorphic to bounded symmetric domains)we have the following theorem.

Theorem 5.1 Let(X,g)be a Hermitian symmetric manifold of the noncompact type,and(Y,g|Y) ?→ (X,g)be a totally geodesic complex submanifold.Then,the inclusion map ?:Y ?→ X is a holomorphic isometric embedding with respect to the Carath′eodory(and equivalently the Kobayashi)metric.

Proof In this proof for a complex manifold M we denote by ‖ ·‖Mthe Carath′eodory pseudonorm on M.Write n(resp.m)for the complex dimension of X(resp.Y).Let χ:? be the Harish-Chandra realization of X as a bounded domain ? ? Cn.Since(Y,g|Y) ?→(X,g)is a totally geodesic complex manifold,S:= χ(Y)=E∩? for an m-dimensional complex vector subspace E ? Cn.By a theorem of Wong-Royden based on Lempert’s theorem on extremal holomorphic Kobayashi disks on strictly convex bounded domains,we know that the infinitesimal Carath′eodory and Kobayashi metrics on ?.S=E ∩ ?,being the intersection of a bounded domain with a complex linear subspace,is itself a weakly convex domain in the complex vector space E,thus the infinitesimal Carath′eodory and Kobayashi metrics agree on S.To prove the theorem it remains to show that the inclusion ?:S ?→ ? is an isometric embedding with respect to the Carath′eodory metrics.

Given any point x ∈ S and any vector η of type(1,0)tangent to S at x,among all holomorphic maps h:S → ? of S into the unit disk ? as a consequence of Montel’s theorem and the homogeneity of? that there exists f:S → ? such that ‖?f(η)‖?realizes the supremum of all‖?h(η)‖?.Let now ρ :? → S be the holomorphic retract defined as in Theorem 1.1 as the orthogonal projection with respect to the Euclidean metric on Cn.Then,F:=f?ρ:?→?and we have ?F(η)= ?f(η)since ρ|S=idS.From the inclusion S ? ? we have

On the other hand,by the choice of f we have ‖η‖CS= ‖?f(η)‖?,while the extension F:? →?yields

Combining the two inequalities we have ‖η‖C?= ‖η‖CS,as desired.The proof of Theorem 5.1 is complete.

Remark 5.1 Regarding the Theorem of Royden-Wong referred to in the second paragraph of the proof,the original unpublished manuscript was elaborated and further developed posthumously leading to the published work of Royden-Wong-Krantz[8],and there was also a different proof by Salinas[9]using operator theory.

6 Appendix

In the proof of Theorem 5.1,in place of quoting the Theorem of Royden-Wong,for the special case of a bounded symmetric domain in its Harish-Chandra realization ??Cn,one can assert the equivalence of the infinitesimal Carath′eodory metric ‖ ·‖C?and the infinitesimal Kobayashi metric ‖·‖K?by means of Theorem 1.1 itself,thus yielding a self-contained proof of the equivalence of‖·‖C?and ‖ ·‖K?independent of Lempert’s theorem in[3],as follows.

Proposition 6.1 For any η ∈ T?we have ‖η‖C?= ‖η‖K?.

Proof Write r for the rank of ? as a bounded symmetric domain.Let Π ? ? be a maximal polydisk passing through 0 ∈ ?,. Π ? ? is totally geodesic with respect to ds2?,and we have Π=V ∩?,where V ?Cnis a complex vector subspace,dimCV=r.By the Polydisk Theorem,any tangent vector ν ∈ T?is equivalent under the action of G0to a vector η ∈ T0(Π).Then,by Theorem 1.1,the image of the orthogonal projection ρ :? → V is exactly Π.Note that on the unit disk we have ‖·‖C?= ‖ ·‖K?from the definitions.On,for any x ∈ Π and η ∈ Π written as η =(η1,···,ηr)in terms of Euclidean coordinates,we have obviously

The proof of Theorem 5.1,without quoting the theorem of Royden-Wong,shows that Π ? ? is an isometric embedding with respect to the Carath′eodory metric.Thus,‖η‖C?= ‖η‖CΠ.On the other hand,

Acknowledgement The author would like to thank Sai-Kee Yeung for communicating his work[12],in which the key issue was a special case of a question of independent interest on the Euclidean geometry of Harish-Chandra realizations of bounded symmetric domains vis-avis totally geodesic complex submanifolds,a question settled conceptually here in the general form.He would like to dedicate the current article to the memory of Professor Gu Chaohao at this juncture 10 years after he left us.