SPHERES IN THE SHILOV BOUNDARIES OF BOUNDED SYMMETRIC DOMAINS

Abstract

In this paper, we classify all nonconstant smooth CR maps from a sphere $S_{n,1}{\subset}\mathbb{C}^n$ with n > 3 to the Shilov boundary $S_{p,q}{\subset}\mathbb{C}^{p{\times}q}$ of a bounded symmetric domain of Cartan type I under the condition that p - q < 3n - 4. We show that they are either linear maps up to automorphisms of $S_{n,1}$ and $S_{p,q}$ or D'Angelo maps. This is the first classification of CR maps into the Shilov boundary of bounded symmetric domains other than sphere that includes nonlinear maps.

1. INTRODUCTION

The rigidity of holomorphic maps between open pieces of a sphere was first studied by Poincaré [13] in 2-dimensional case and later by Alexander [1] and Chern and Moser [2] for general dimensions. Then Webster [16] obtained rigidity for holomorphic maps between open pieces of spheres of different dimension, proving that any such map between spheres in and extends as a totally geodesic map between balls with respect to the Bergman metric. Later, Huang [6] generalized Webster’s result for CR maps between open pieces of spheres in and under the assumption n′ − 1 < 2(n − 1). Beyond this bound, the rigidity fails as illustrated by the Whitney map.

Unit ball is a bounded symmetric domain of Cartan type I with rank 1 and sphere is its Shilov boundary. However, comparing with rigidity of holmorphic maps between spheres mentioned above, holomorphic rigidity for maps between bounded symmetric domains D and D′ of higher rank remains much less understood. If the rank r′ of D′ does not exceed the rank r of D and both ranks r, r′ ≥ 2, the rigidity of proper holomorphic maps f : D → D′ was conjectured by Mok [12] and proved by Tsai [15], showing that f is necessarily totally geodesic (with respect to the Bergmann metric).

For the case r < r′, in [11], Zaitsev and author showed the rigidity of CR maps f : Sp,q → Sp′,q′ under the assumption that q ≥ 2 and (p′ − q′) < 2(p − q). Here, Sp,q and Sp′,q′ are the Shilov boundaries of a bounded symmetric domains of Cartan type I((See §1 for definition) and q and q′ are the ranks of Sp,q and Sp′,q′ , respectively. When (p′ − q′) = 2 (p − q), then the rigidity fails to hold, as authors introduced the generalized Whitney map as a counterexample in the same paper.

Recently, in [14], A. Seo introduced a nonlinearizable proper holomorphic maps between Sp,q and S2p−1,2q−1. Therefore, to classify all CR maps between Sp,q and Sp′,q′ when p′ − q′ ≥ 2(p − q), one should consider nonlinear maps. In [9], Huang, Ji and Xu classified all locally defined CR maps between Sn,1 and Sn′,1 under the assumption that 3 < n ≤ n′ < 3n − 3. It is proved that such map is either a linear map or a D’Angelo map.

In this paper, we generalize the result of Huang, Ji and Xu. We define D’Angelo map from a sphere into the Shilov boundary of bounded symmetric domains of type I as follows:

Definition 1.1. Let be the set of all complex p × q matrices. A map fθ : Sn,1 → Sp,q for a fixed 0 < θ ≤ π/2, is called a D’Angelo map if fθ is equivalent to the following map

up to automorphisms of Sn,1 and Sp,q, where Wθ(z) is a map from Sn,1 to S3n−3,1 defined by

and Iq−1 is the identity matrix of size (q−1).

This map is not linear after composing with any automorphisms of Sn,1 and Sp,q. For q = 1 and θ = π/2, this is the classical Whitney map between unit balls in and respectively. In this paper, we classify all locally defined CR maps from a sphere Sn,1 with n > 3 into the Shilov boundary Sp,q of a general Cartan type I bounded symmetric domain of higher rank. We showed

Theorem 1.2. Let f be a nonconstant smooth CR map from an open piece of Sn,1 into Sp,q. Assume that n > 3 and p − q < 3n − 4. Then after composing with suitable automorphisms of Sn,1 and Sp,q, f is either a linear embedding or D’Angelo map.

Note that our basic assumption p − q < 3n−4 corresponds precisely to the optimal bound n′ − 1 < 3(n − 1) in the rank 1 case (q = 1) of maps between spheres, where n −1 and n′ − 1 are the CR dimensions of the spheres.

Throughout this paper we adopt the Einstein summation convention unless mentioned otherwise.

 

2. PRELIMARIES

In this section, we review CR structure and Grassmannian frames adapted to Sp,q. For details, we refer [2] and [11] as references. In this section, we let Greek indices α, β, γ, … and Latin indices j, k, ℓ, … run over {1,…,q} and {1,…, p − q}, respectively. For q = 1, i.e., sphere case, we omit Greek indices.

A Hermitian symmetric domain Dp,q of Cartan type I has a standard realization in the space of p × q matrices, given by

where Iq is the q × q identity matrix and The Shilov boundary of Dp,q is given by

In particular, Sp,q is a CR manifold of CR dimension (p − q) × q. For q = 1, Sp,1 is the unit sphere in . We shall always assume p > q so that Sp,q has positive CR dimension.

Let Aut (Sp,q) be the Lie group of all CR automorphisms of Sp,q. By [10, Theorem 8.5], every f ∈ Aut (Sp,q) extends to a biholomorphic automorphism of the bounded symmetric domain Dp,q. Consider the standard linear inclusion

Then we may regard Sp,q as a real submanifold in the Grassmanian Gr(q, p + q) of all q-planes in and Aut (Sp,q)(= Aut (Dp,q)) becomes a subgroup of the automorphism group of Gr(q, p + q).

For column vectors u = (u1, … ,up+q)t and v = (v1, … ,vp+q)t in , define a Hermitian inner product by

A Grassmannian frame adapted to Sp,q, or simply Sp,q-frame is a frame {Z1, …, Zp+q} of with det (Z1, …, Zp+q) = 1 such that scalar product <·, ·> in basis (Z1, …, Zp+q) is given by the matrix

Now let be the set of all Sp,q-frames. Then is identified with SU(p, q) by the left action. The Maurer-Cartan form on is given by the equation

where π satisfies the trace-free condition

and the structure equation

where the capital Greek indices Λ, Γ, Ω etc. run from 1 to p + q.

From now, we will use the notation

Z := (Z1, … ,Zq), X = (X1, … ,Xp−q) := (Zq+1, … ,Zp), Y = (Y1, … , Yq) := (Zp+1 …, Zp+q)

so that the Maurer-Cartan form with respect to the basis (Z, X, Y) can be written as

with the symmetry relations

By abuse of notation, we also denote by Z the q-dimensional subspace of spanned by Z1, …, Zq. Then the defining equations of Sp,q can be written as

Sp,q = {Z ∈ Gr(q, p + q) : <·,·>|z = 0}

and hence their differentiation yields

By substituting into (1, 0) component of (2.3) we obtain, in particular,

when restricted to the (1, 0) tangent space. Comparing the dimensions, we conclude that span the space of contact forms on Sp,q, i.e.,

is the complex tangent space of Sp,q. The structure equation is given by

Moreover, since

we conclude that θαj form a basis in the space of (1, 0) forms.

There are several types of frame changes.

Definition 2.1. We call a change of frame

i) change of position if where W = (Wαβ) and V = (Vαβ) are q × q matrices satisfying V*W = Iq; ii) change of real vectors if where H = (Hαβ) is a hermitian matrix; iii) dilation if where λα > 0; iv) rotation if where (Ujk) is a unitary matrix.

Finally, we shall use the change of frame given by

such that

and

where

The new frame is an Sp,q-frame and the related 1-forms remain the same, while change to

 

3. Sp,q-FRAMES ADAPTED TO CR MAPPINGS

Let f : Sn,1 → Sp,q be a (germ of a) smooth CR mapping. We shall identify Sn,1 and its image f(Sn,1) ⊂ Sp,q. We consider the connection forms φ, θj, 𝜓, ωjk, σj, ξ with j, k = 1, …, n − 1 on Sn,1 and denote by capital letters Φαβ, Θαj, Ψαβ, ΩJK,ΣKβ, Ξαβ with α, β = 1,…,q and J, K = 1,…,p − q, their corresponding counterparts on Sp,q. We also define one forms φαβ, θαJ adapted to f as follows:

Definition 3.1. We say that f is of contact rank r if f sends any nonzero vector in TSn,1/TcSn,1 to a rank r vector in TSp,q/TcSp,q.

For a map f of contact rank r, we define φαβ, θαJ for α = 1, …, q and J = 1, …, p − q adapted to f by

φ11 = ⋯ φrr = φ,θ1j = ⋯ = θr(r−1)(n−1)+j = θj, j = 1, …, n − 1

and 0 otherwise.

In this section we show the following lemma.

Lemma 3.2. For any nonconstant local CR map f : Sn,1 → Sp,q with p − q < 3(n−1), there exist r ∈ {1, 2} and a choice of Sp,q-frames such that f is of contact rank r and the forms φαβ, θαJ adapted to f satisfy

Proof is a slight modification of the proof of Lemma 4.2 and argument in §.5 of [11]. We refer [11] for details.

Proof. Since φ and Φ = (Φαβ) are contact forms on Sn,1 and Sp,q, respectively, the pull back of Φ f is a span of φ. Choose a diagonal contact form of Sp,q and say Φ11. Then we can write

for some smooth function λ. At generic points, we may assume that either λ ≡ 0 or λ never vanishes. By differentiating (3.2) and using (2.4) we obtain

Arguing similar to [11] we conclude λ ≥ 0 and, after dilation of Φ11, we may assume that λ = 1 if λ ≢ 0.

Suppose that Φαα vanishes identically for all α. Then we obtain

Since each ΘαJ is a (1, 0) form, it follows that

ΘαJ = 0 mod φ,

i.e., f(Sn,1) is a totally real submanifold. Since Sn,1 is Levi-nondegenerate, this implies that f is a constant map, which contradicts our assumption. Hence there exists at least one diagonal term of Φ whose pullback does not vanish identically.

Choose such a diagonal term of Φ, say Φ11. Then (3.3) yields

Therefore after a suitable rotation of Sp,q, we may assume that

Write

for some smooth functions λα. Then by differentiating (3.6) and using (2.4) together with (3.4), (3.5), we obtain

Choose a suitable change of position that leaves Θ1J invariant and replaces ΘαJ with ΘαJ − λαΘ1J for α ≥ 2. This change of position leaves Φ11 invariant and transforms Φα1 into Φα1 − λαΦ11 for α ≥ 2. After performing such change of position, (3.6) becomes

Φα1 = 0, α ≥ 2,

and (3.7) becomes

Θαj ∧ θj1 = 0 mod φ, α ≥ 2.

Sincec Θαj are (1, 0) but θj are (0, 1) and linearly independent, it follows that

Next for each α ≥ 2, let

for another smooth function λα. If λα ≡ 0 for all α ≥ 2, then by differentiation, we obtain

which yields

In this case, by considering the differentiation of

Φαβ = λαβφ

and substituting (3.10), we conclude that

Φαβ = 0, (α, β) ≠ (1, 1),

which implies that df(T) modulo TcSp,q is a rank 1 vector for any T ∈ TSn,1 transversal to TcSn,1. That is to say, f is of contact rank 1 and the forms adapted to f satisfy

Φαβ − φαβ = 0,ΘαJ − θαJ = 0 mod φ.

Suppose there exists α such that λα ≢ 0. We may assume α = 2. After a dilation of Φ22, we may assume that at generic points, λ2 = 1. By differentiating (3.9) for α = 2 and substituting (3.8) we obtain

Hence after a suitable rotation

where (UKJ) is unitary matrix leaving Θαj, j = 1,…, n−1, invariant, we may assume that

Θ2n−1+j = θj mod φ, j = 1,…, n−1

and

Θ2J = 0 mod φ

otherwise. Write

for some smooth function λα. Then as before, we can choose a suitable change of position that leaves Θ1J and Θ2J invariant and replaces ΘαJ with ΘαJ − λαΘ2J for α > 2, which also leaves Φ11, Φ21 and Φ22 invariant and transforms Φα2 into Φα2 − λαΦ22 for α > 2. By (3.8), after performing such change of position, the following property

ΘαJ = 0 mod φ, α ≥ 2

still holds and (3.11) becomes

Φα2 = 0, α > 2.

By differentiating this we obtain

which yields

Write

Φαα = λαφ, α > 2

for some smooth functions λα. Suppose that λα ≡ 0 for all α. Then as before, we can obtain

ΘαJ = 0 mod φ, α > 2, ∀J, Φαβ = 0, α > 2 or β > 2,

i.e., f is of contact rank 2 and the forms adapted to f satisfy

Φαβ − φαβ = 0, ΘαJ − θαJ = 0 mod φ.

Suppose there exists α such that λα ≠ 0 We may assume α = 3. After a dilation of Φ33, we may assume that at generic points, λ3 = 1, i.e.,

Φ33 = φ.

By differentiating this, we obtain

hen by (3.8) and (3.12), we have at most p − q − 2(n −1) linearly independent (1, 0) forms on the left-hand side, while on the right-hand side we have n − 1 linearly independent (1, 0) forms. Since we assumed that p − q < 3(n − 1), this is a contradiction.

Next we will show that there exists a choice of frames such that

ΘαJ = θαJ.

Write

for some ηαJ. Consider the equations obtained by differentiating (3.13):

where

ψαα = ψ, α = 1, …,r, ψαβ = 0 otherwise

and

Let α > r. Then left-hand side of (3.14) contains at most one (1, 0) form, while the right-hand side contains (n − 1) linearly independent (1, 0) forms with n − 1 > 1 unless ηαJ = 0. Therefore we conclude that

ηαJ = 0, α > r

or equivalently

ΘαJ = 0, α > r.

Finally, define a matrix (BαJ) by

BαJ := ηαJ,

where ηαJ satisfies

ΘαJ − θαJ = ηαJφ.

Consider the change of frame of Sp,q discussed after Definition 2.1, given by

such that

CJα := −BJα

and Aαβ satisfies

Since the sum here is hermitian, one can always choose Aαβ with this property. Then Φαβ remain the same while change to

ΘαJ − ΦαβBβJ.

Therefore the new ΘαJ satisfies

ΘαJ = θαJ.

 

4. SECOND FUNDAMENTAL FORMS AND GAUSS EQUATIONS FOR CR EMBEDDINGS

In this section, we determine second fundamental forms given by ΩJK. Then we determine Ψαβ and ΣαJ. By using these forms, we construct a linear subspace of Gr(q, p+q) that contains the image of a given embedding(Lemma 4.1, Lemma 4.2). Their proofs are slight modification of the proof of Proposition 7.1 in [11].

Let f be a CR map of contact rank r with r ∈ {1, 2}. Differentiate (3.1) using the structure equations to obtain

where

σα(α−1)(n−1) + j = σj, α = 1,…,r, j = 1,…, n − 1

and 0 otherwise.

4.1. Contact rank 1 map Choose α > 1 and J = j. Then (4.1) takes the form

Ψα1 ∧ θj = 0, α > 1.

By Cartan Lemma we obtain

Ψα1 = 0 mod θj

for fixed j. Since Ψ is independent of j = 1, … ,n −1 and we assumed n −1 > 1, we obtain

We will show the following lemma.

Lemma 4.1. There exists (p−q+2)-dimensional subspace V1 and (q−1)-dimensional subspace V2 in orthogonal to each other such that Gr(1, V1) ⊕ V2 contains the image f(Sn,1).

Proof. Choose an open set M ⊂ Sn,1 where f is defined. Let Z, X, Y be constant vector fields of forming a Sp,q-frame at a fixed reference point of f(M) and let

be an adapted Sp, q-frame along f(M). Write

so that (4.3) - (4.5) take the form

Since Z, X, Y form an adapted frame at a reference point of M, we may assume that

at the reference point. Since Z, X, Y are constant vector fields, i.e., dZ = dX = dY = 0, differentiating (4.6) and using (2.1) we obtain

Next, it follows from Lemma 3.2 and (4.2) that

in particular, the span of is independent of the point in M. Hence together with (4.3) and (4.7), we conclude

Furthermore, (4.8) for α = 1 together with Lemma 3.2 and (4.2) (and with the symmetry relations analogous to (2.2)) we obtain

Now with (4.9) taken into account, (4.10) becomes

Thus each of the vector valued functions for a fixed β satisfies a complete system of linear first order differential equations. Then by the initial condition (4.7) and the uniqueness of solutions, we conclude, in particular, that

ζβ = 0 , β > 1

Hence (4.3) implies

Now setting

we still have

whereas (4.11) becomes

implying

Then we conclude that

where

V1 = span {Z1, X1,…, Xp−q, Y1}, V2 = span {Z2,…, Zq}.

4.2. Contact rank 2 map Choose α > 2 and J = j or J = n − 1 + j. Then (4.1) takes the form

Since Ψ is independent of j = 1,…, n − 1 and we assumed n − 1 > 1, by Cartan Lemma we obtain

Use (4.1) for either α = 1 and J = n − 1 + J or α = 2 and J = j or α = 1, 2 and J > 2(n −1) to obtain

By Cartan’s Lemma, we obtain

where θ is an ideal generated by θ1,…,θn−1. Since

by using (4.16), we conclude that

Moreover, since Ψ is independent of j, substituting (4.18) into (4.13) and (4.14), we obtain

Next we will determine second fundamental forms of f as in [16]. We will show that it has a trivial solution only. For details, we refer [16].

Use (4.1) for α = 1 and J = j ≤ (n − 1) to obtain

Then by Cartan Lemma, we obtain

By symmetry relation for ­Ω, we obtain

Furthermore, differentiation of

Φ11 − φ = 0

by using the structure equations yields

or equivalently

Therefore we obtain

for some pure imaginary function g.

Similar computation for (4.1) with α = 2 and J = n − 1 + j together with the relation

Φ22 − φ = 0

yields

and

for a pure imaginary function h.

Take a real vector change of Sp,q defined by

for a smooth function μ satisfying

Ψ12 = μφ

in (4.19) and fixing the rest. Then after the frame change, we obtain

By differentiating (4.20),(4.21),(4.22) and substituting (4.12) and (4.19), we obtain

Then by Cartan Lemma, we obtain

By (4.16) and symmetry relation for Σ, we obtain

Now let

Then (4.1) implies

Write

ΩjJ = hjJℓθℓ mod φ, K > 2(n − 1).

Differentiate (4.24) and substitute (4.18) to obtain

which implies

If p−q < 3(n−1), then (4.26) has trivial solution only.(See [3].) Therefore we obtain

hkJℓ = 0

or equivalently

Similar computation for using (4.25) yields

By (4.18) and (4.27), we can write

ΩkJ = ηkJφ, J > n − 1.

By differentiating this, we obtain

By (4.17) and (4.23) we can show that the left-hand side of (4.28) contains at most one (0, 1) form, while the right-hand side contains (n − 1) linearly independent (0, 1) forms unless ηkJ = 0. Hence we conclude that

ηkJ = 0

or equivalently

and therefore by substituting (4.17) and (4.23) into (4.28), we obtain

Similar computation for implies

Furthermore, by substituting (4.29) to(4.15) with J = j, we obtain

Σ2J = 0 mod φ, j ≤ n − 1.

Finally we will determine Ψ and Σ. By (4.19), we can write

Ψ21 = μφ.

By differentiating this and substituting (4.12) and (4.19), we obtain

By (4.30), this implies

μ = 0

or equivalently

Ψ21 = 0.

Let

Σ1J = μJφ, J > (n − 1).

By differentiation, we obtain

which yield

μJ = 0

or equivalently

Σ1J = 0.

Since Ξ12 is independent of j, we obtain

Similar computation for Σ2J yields

Σ2j = Σ2J = 0, j < n, J > 2(n − 1).

Summing up we obtain the following:

For any contact rank 2 local CR embedding f from Sn,1 into Sp,q, there is a choice of frames such that

We will show the following lemma.

Lemma 4.2. There exist (n+1)-dimensional subspaces V1, V2 and (q−2)-dimensional subspace V3 in orthogonal to each other such that Gr(1, V1) ⊕ Gr(1, V2) ⊕ V3 contains the image f (Sn,1).

Proof. We use the same method in Lemma 4.1. Let M ⊂ Sn,1, Z, X, Y and

be as in Lemma 4.1.

It follows from Lemma 3.2 and (4.31) that

in particular, the span of is independent of the point in M. Hence as in Lemma 4.1, we conclude

Furthermore, (4.8) implies

In particular, restricting to α = 1 and J = j ≤ n with (4.31)-(4.34) and (4.37) taken into account, we obtain

Repeating the above argument for λ and ζ instead of η, we obtain

and

Thus each of the vector valued functions for a fixed K and for a fixed β satisfies a complete system of linear first order differential equations. Then as in Lemma 4.1 we conclude, in particular, that

λ12 = 0

and

ηK = ζβ = 0, K > n, β > 1.

Hence (4.35) implies

Similar computation for implies

Now setting

we still have

whereas (4.38), (4.39) become

implying

Then together with (4.36) we conclude that

where

V1 = span {Z1, X1,…,Xn−1, Y1}, V2 = span {Z2, Xn,…,X2n−2, Y2}, V3 = span {Z3,…,Zq}

 

5. PROOF OF THEOREM 1.2

Suppose f is of contact rank 1. Then by Lemma 4.1, there exist (p − q + 2)-dimensional subspace V1 and (q − 1)-dimensional subspace V2 such that the image of f is contained in Gr(1, V1) ⊕ V2. The V2-component of f is a constant map. Therefore it is enough to show that Gr(1, V1)-component of f is either a linear map or Whitney map. But Gr(1, V1) = . Therefore by the result of [9] under the condition n > 3 and (p − q) < 3n − 4, we conclude that Gr(1,V1)-component of f is either a flat embedding or D’Angelo map.

Suppose f is of contact rank 2, then by Lemma 4.2, there exist (n+1)-dimensional subspaces V1, V2 and (q − 2)-dimensional subspace V3 such that the image of f is contained in Gr(1, V1)⊕Gr(1, V2)⊕V3. As before, it is enough to show that Gr(1, V1) and Gr(1, V2)-components of f are linear. Since V1 and V2 are of dimension (n+1), each component of f is a CR automorphism of Sn,1. Therefore, it is projective linear, which completes the proof.