Holomorphic maps from the complex unit ball to Type IV classical domains

Transcription

1 Holomorphic maps from the complex unit ball to Type IV classical domains Ming Xiao and Yuan Yuan March 9, 016 Abstract 1 Introduction The first part of this paper is devoted to establish new rigidity results for proper holomorphic maps from the complex unit ball to higher rank bounded symmetric domains. The rigidity properties have been extensively studied in the past decades for proper holomorphic maps F : Ω 1 Ω, between bounded symmetric domains Ω 1, Ω. The pioneer works are due to Poincaré [P] and later to Alexander [Al] when Ω 1, Ω are complex unit balls. In particular, any proper holomorphic self-map of the unit ball in C n is an automorphism if n [Al]. It is well-known that the rigidity properties fail dramatically for proper holomorphic maps between balls of different dimensions. For this type of results, see [HS], [L], [Fo1],[Gl], [St], [Do], [D1] and etc. However, the rigidity properties can still be expected if certain boundary regularity of the map is assumed. See [W], [Fa], [CS], [Hu1], [Hu], [HJ], [HJY], [Eb] and etc. The lists above are by no means to be complete. On the other hand, it is a widely open problem to understand proper holomorphic maps F : Ω 1 Ω between bounded symmetric domains Ω 1, Ω of higher rank. When rankω 1 rankω and Ω 1 is irreducible, it was proved by Tsai [Ts] that F must be a totally geodesic isometric embedding with respect to Bergman metrics. Tu [Tu1] proved that the proper holomorphic map between equal dimensional irreducible bounded symmetric domains must be an automorphism. When rankω rankω 1, the studies are mainly focused on the Type I classical domains and many interesting results have been established cf. [Tu], [Ng4], [KZ1], [KZ] et al. Note that the total geodesy of F fails in general although it is believed that F should take certain special forms module automorphisms cf. [Ng4] [KZ]. In this paper, we prove new rigidity results for proper holomorphic maps from the unit ball in C n to 1

2 the m-dimensional Type IV classical domain. The isometry property still survives although the total geodesy fails in our setting. In particular, we establish a classification result for CR maps between their boundaries when m = n + 1. Write B n for the unit ball in C n, and Dm IV them with the Bergman metrics ω B n, ω D IV We say a holomorphic map F : B n D IV m F ω D IV m = λω B n the boundary of D IV for the classical type IV domain in C m and equip m, respectively See section for explicit formulas. is an isometric embedding or simply an isometry if for the unit sphere in C n and for some constant λ > 0. Write B n and Dm IV m, respectively. Theorem 1.1. Assume n 4, n + 1 m n Local version Let F be a holomorphic map from a connected open set U in C n containing p B n to C m. Assume that F B n U Dm IV and F U Dm IV. Then F extends to a holomorphic isometric embedding from B n into Dm IV with F ω D IV m = m ω n+1 B n.. Global version Any algebraic proper holomorphic map from B n to Dm IV embedding with F ω D IV m = m ω n+1 B n. is an isometric Recall that a holomorphic function f over U C n is called holomorphic Nash-algebraic, or simply algebraic if there is an irreducible holomorphic polynomial P z, X in X with coefficients polynomials of z such that P z, fz 0 over U. A holomorphic map F is called algebraic if each of its components is algebraic. In the case m = n + 1, the proper holomorphic maps are classified under a weaker boundary regularity condition. Theorem 1.. Let n 4 and F be a C -smooth CR transversal map from an open piece of B n to an open smooth piece of Dn+1. IV Then F extends to a holomorphic isometry from B n to Dn+1. IV Furthermore, F is equivalent to either 1 n 1 z 1,, z n 1, z i zn + z n, 1 n 1 1 z i + zn z n ; 1 1 zn 1 zn or z 1,, z n 1, z n, 1 1 n z j. Here two proper holomorphic maps F 1, F : Ω 1 Ω are called equivalent if there exist automorphisms φ and ψ of Ω 1 and Ω respectively, such that ψ F 1 φ = F. The definition of CR transversality will be given in Section. We next list some important remarks of Theorem 1.1 and Theorem 1.. Remark It follows from Mok s result [M4] that the algebraicity assumption on F is necessary in Theorem 1.1. Moreover, a non-algebraic proper map from B n to Dn+ IV that is not an isometry will be constructed in Example.4.

3 . The statement of Theorem 1. fails if the transversality assumption is dropped. Similarly, the statement of Theorem 1.1 fails if the condition F U Dm IV is dropped. See Example The conclusion of Theorem 1.1 fails if m ncf. Example.6. The conclusion of Theorem 1. fails if n = 1 cf. Example.7. We suspect that Theorem 1.1 holds for all n, n + 1 m n 1 and Theorem 1. holds for all n. See more details in Section. 4. We state the following fact as a remark of Theorem 1.. Let G be a nonconstant local C -smooth CR map from the unit sphere in C n to Dn+1. IV Then the image of G cannot be contained in the singular set of Dn+1. IV This is due to the fact that the singular set of Dn+1 IV is a real n + 1 dimensional subvariety of {ZZ t = }. The study of proper holomorphic maps from B n to D IV m or CR maps from B n to D IV m is closely related to CR maps between hyperquadrics. Rigidity properties are explored for CR maps into hyperquadrics in [BH], [BEH1, BEH], [EHZ1, EHZ], [ES], [Ng3], etc See also [D1], [DL] for irrigidity phonemena in this setting. The crucial idea to establish Theorem 1.1 is to recognize Dm IV as an isometric submanifold of the generalized complex unit ball so that we can apply techniques from CR geometry. As an important step to approach Theorem 1.1, we establish a rigidity result first for holomorphic proper maps from the unit ball to the generalized balls. This part is motivated by the framework of Baouendi-Ebenfelt-Huang [BEH] and is proved by a very similar argument. In another direction, since the work of Bochner [B] and Calabi [C], lots of efforts have been made to understand the local holomorphic isometry F : U Ω with respect to the normalized Bergman metrics of Ω 1 and Ω respectively, i.e. F ω Ω = λω Ω1 on U, where U Ω 1 is a connected open set. This problem is largely motivated by the algebraic dynamics problem considered by Clozel-Ullmo [CU] cf. [M4] [MN] [Yu] [FHX] et al for further developments. Mok [M4] proved that F extends to an algebraic proper holomorphic isometry from Ω 1 to Ω. Assume dim C Ω 1 and Ω 1 is irreducible. Mok proved that F is totally geodesic if rankω 1 [M4]. When rankω 1 = 1, and Ω is the product of complex unit balls, F is also totally geodesic by works of Mok [M], Ng [Ng], the second author and Zhang [YZ]. However, when Ω 1 = B n and Ω is a bounded symmetric domain other than the product of unit balls, the total geodesy fails dramatically [M5]. In fact, assuming that Ω is irreducible and rankω, Mok constructed a non-totally geodesic holomorphic isometry from B n into Ω by using the theory of variety of minimal rational tangents [M5]. In the next theorem, we classify the local holomorphic isometries from B n into Dn+1. IV We refer to [XY] for the study in the general case and note that this problem is studied independently by Chan-Mok in [CM]. Theorem 1.4. Let n and F be any holomorphic isometry from U B n into D IV n+1 satisfying F ω D IV n+1 = λω B n on U 3

4 for some λ > 0. Then F is equivalent to either the map in 1 or the map in. The novelty in Theorem 1.1 is that, as long as the codimension is small, the isometry is implied by properness, which is the converse statement of Mok s theorem [M4]. This can be applied to obtain the following corollary. Corollary 1.5. Assume n 4, K 1, N 1,, N K n 3. Let F = F 1,, F K be a holomorphic map from a connected open set U B n into the product of Type IV domains Ω = DN IV 1 DN IV K satisfying the following isometric equation: λω B n = K l=1 F l ω D IV Nl = F ω Ω on U 3 for some positive constant λ. Then each F l, 1 l K, is either a constant map or extends to a holomorphic isometric embedding from B n to DN IV l with F ω D IV Nl = N l ω n+1 B n. We would like to point out that the same conclusion can be made with the slightly more general assumption m ω B n = λ l Fl ω D IV Nl on U l=1 instead of 3 for positive constants λ l. The difference in the proof is that the algebraicity in this case follows from the argument in [HY1, HY]. Acknowledgement: The authors are grateful to Professors J. D Angelo, X. Huang, N. Mok and S. Ng for helpful discussions. The second author is supported in part by National Science Foundation grant DMS and the seed grant program at Syracuse University. This work is also supported by the National Science Foundation under Grant No while the second author was in residence at the Mathematical Sciences Research Institute in Berkeley, California, during the 016 Spring semester. Preliminaries and Some Examples Hermitian symmetric manifolds of non-compact type are realized as the four types of Cartan s classical domains and two exceptional cases in complex Euclidean spacescf. [H] [M1]. In particular, the complex unit ball B n = {z = z 1,, z n C n : z < 1} in C n is a special case of the type I classical domain. The Bergman kernel is given by The type IV classical domain is defined as K B nz, z = c I 1 z n+1. D IV m = {Z = z 1,, z m C m ZZ t < and 1 ZZ t ZZt > 0}, 4

5 where Z t is the transpose of Z and the Bergman kernel function K D IV m Z, Z is explicitly given by K D IV m Z, Z = c IV 1 ZZ t ZZt m. 4 Here c I, c IV are positive constants depending on the dimensions. cf. [H] [M1]. The Bergman metric ω Ω Z := 1 log K Ω Z, Z of a bounded symmetric domain Ω is Kähler-Einstein as the Bergman kernel function is invariant under the holomorphic automorphisms. A straightforward computation shows that Dm IV = {Z C m : 1 ZZ t ZZt = 0, ZZ t }. Moreover, if m, the singular set P of Dm IV is given by which is a real m-dimensional variety. P = {Z C m 1 ZZ t ZZt = 0, ZZ t = }, In [M5], Mok constructed non-totally geodesics holomorphic isometries from the complex unit ball into the irreducible bounded symmetric domain Ω when rankω by using the theory of varieties of minimal rational tangents. For the purpose of the current paper, we formulate Mok s theorem merely in the case of Type IV domains. Theorem.1 Mok. Assume m. i If F : B n D IV m is a holomorphic isometry, then n m 1. ii There exists a non-totally geodesic holomorphic isometric embedding G : B m 1 D IV m with G ω D IV m = λω B m 1 for some λ > 0. We now make some remarks on our main theorems. The following simple fact explains why we only consider m n + 1 in our main theorems. Lemma.. Let n and F : B n D IV m be a proper holomorphic map. Then m n + 1. Proof. It suffices to show that there is no proper holomorphic map for m = n. Suppose that F : B n Dn IV is a proper holomorphic map. Note that B n and Dn IV are bounded complete circular domains. Then F is algebraic by Bell s algebraicity result [B]. It then extends holomorphically to a neighborhood U of an open piece of the boundary B n. Note that the image of B n U cannot be contained in the singular set of Dn IV. One then easily achieves a contradiction since Dn IV is Levi-degenerate at any smooth point. Remark.3. The statement of Lemma. fails if n = 1. F z = z is indeed a holomorphic isometry from the unit disc to D IV 1. 5

6 As mentioned in Section 1, the assumption that F is algebraic in Theorem 1.1 is a necessary condition by Mok s result. We now give an example of non-algebraic proper map from B n to Dn+ IV that is not an isometry. Example.4. Assume n and let H = h 1,, h n+1 : B n B n+1 be a proper holomorphic map with H0 = 0 such that H is continuous up to the boundary B n but is not twice continuously differentiable up to any open piece of boundary B n See [Do]. In particular, H is not algebraic. Define n+1 g = 1 1 h j. Then g is holomorphic in B n and satisfies It follows that g = 1 n+1 h j + g. n+1 h j + g 1 n+1 n+1 h j + g = h j = 1 on B n, 4 and thus G = H, g : B n D IV n+ is a proper holomorphic map and is not an isometry. We now recall the definition of CR transversality. Let M C n, M C N be two CR submanifolds. A CR map H : M M is called CR transversal at p M if dhct p M is not contained in V F p V F p, where V is the CR bundle of M. Example.5 shows that the statement of Theorem 1. fails if the transversality assumption is dropped, and the statement of Theorem 1.1 fails if the assumption F U Dm IV is dropped. Example.5. Let n 1, m 3. Let G be a holomorphic map from a small neighborhood U of p B n defined by 1 + ψ G =, 1 ψ, 0,..., 0, where there are m zero components and ψ is any holomorphic function in U. It is easy to see that G maps U to Dm IV and does not extend to an isometry. Example.6 show that the statement of Theorem 1.1 fails when m n. Example.7 shows that the statement of Theorem 1. fails if n = 1 and the statement of Theorem 1.1 fails for n = 1, m. Example.6. Assume n and let H = h 1,, h n 1 = z 1,, z n 1, z 1 z n,, zn : B n B n 1 be the well-known Whitney map. Define the holomorphic function n 1 g = h j

7 on B n. Note that n 1 h i + g 1 n 1 n 1 n h i + g = z 4 i + z n z i. It is easy to verify that G = H, g is a holomorphic polynomial proper map from B n to D IV n, while it is not an isometry. Example.7. Assume k 1. Let G k be the holomorphic map from the unit disc C to D IV defined by G k = z k, 1 1 z k. 5 It is easy to verify that G k is a proper holomorphic map. Moreover, G k is an isometry from to D IV if and only if k = 1 See Proposition.11. Note that all the proper map examples above are constructed from proper maps between balls. It would be reasonable to ask whether all holomorphic proper maps are obtained in this manner. More precisely, let H = h 1,..., h m : B n Dm IV be a proper holomorphic map. Does there always exist a holomorphic proper map F = f 1,..., f m : B n B m with m m, such that the following relation holds: m h i 1 m h m i 4 = f j? 6 The answer is negative in general See Example.8, while we suspect that it is true when m is small compared to n. Example.8. Assume n. Let H = h 1,..., h 4n 1 be the proper holomorphic map from B n to D4n 1 IV defined by z n, z 1 z, 1,..., z n 1, z n 1, z 1zn, z 1 zn..., z n 1zn, z n 1zn, zn 3, zn 3. By computation, we have m h i 1 m 4 h i = n z i 1 z n n z i. 7 Note that the right hand side of 7 cannot be written as sum of norm squares of holomorphic functions cf. the proof of Proposition.11. This implies there is no proper holomorphic map F : B n B m such that 6 holds. On the other hand, it would still be interesting to study how the initial boundary regularity of the map F can be relaxed in the hypotheses of Theorem 1.. We make the following conjecture along this line: 7

8 Conjecture.9. Let n. Any proper holomorphic map from B n to D IV n+1 is a holomorphic isometry and thus it is equivalent to one of the maps in Theorem 1.4. Remark.10. Example.7 also shows that the statement of Conjecture.9 fails when n = 1. We end this section with the following observation on isometric constants. Note that this result is obtained independently by Chan-Mok [CM]. Proposition.11. Let F : B n Dm IV be a local holomorphic isometric embedding with respect to the Bergman metrics of isometric constant λ > 0, i.e. Then If n, then λ = m/n + 1. If n = 1, then λ = m/ or m. F ω D IV m = λω B n. 8 Proof. Without loss of generality, we may assume F 0 = 0 by composing the automorphisms of B n and Dm IV. By the standard reduction, 8 is equivalent to 1 z λn+1/m t 1 = 1 F zf z + 4 F zf zt 9 that holds in a neighborhood U of 0. Note that the signature of the left hand side of 9 is either 1, s or, s for some integer s 0, meaning that it can be written as linear combination of 1 or sum of squares minus s sum of square of linearly independent holomorphic functions over positive real numbers. It fact, it is of signature 1, s if and only if F zf z t 0. Obviously, the left hand side is of finite rank if and only if λn+1/m N cf. [Um]. Write p = λn+1/m. We first consider the case n =. Assume p. Applying binomial formula, p n p p k u n 1 z j = 1 k z j. k k=0 Note that the monomials on the right hand side are linearly independent and thus the right hand side is of signature r, s with r 3 if n. Therefore, the left hand side and right hand side of 9 have different signatures and this is a contradicttioncf. [D] or [Um]. This implies the first part of the Lemma for n. In the case n = 1, one can similarly get a contradiction if p 3. The Lemma is then established. Remark.1. Both λ = m/ and m can be obtained when n = 1 in Proposition.11. Indeed, the map G 1 in 5 is a holomorphic isometry with isometric constant λ = 1 while the isometric map F z = z, 0 : D IV has the isometric constant λ =. Remark.13. The same argument yields that if F : B n Ω is a holomorphic isometric embedding from a unit ball into a classical symmetric domain with respect to Bergman metric of isometric constant λ, then λn + 1/p + q, λn + 1/m 1 or λn + 1/m + 1 is a positive integer when Ω = Dp,q, I Dm II or Dm III respectively. 8

9 3 On CR maps from B n to D IV m We will prove Theorem 1.1, Theorem 1. and Corollary 1.5 in this section. To this end we first introduce the generalized complex balls and their indefinite metrics. Definition 3.1. Let 0 l n, n. B n+l l = {w 1,, w l, z 1,, z n C n+l n z j is called the generalized complex ball with signature l in C n+l and = 1 l n log 1 + w j z j ω B n+l l is the indefinite Kähler metric on B n+l l. l w j < 1} Note that B n+l l is the indefinite complex hyperbolic space form and B n 0 is the complex unit ball B n and ω B n 0 is the Bergman metric on B n up to a positive constant n + 1. The holomorphic automorphism group of B n+l l can be described as follows. We first embed B n+l l into P n+l as an open set by standard coordinate embedding: Therefore B n+l l w 1,, w l, z 1,, z n [1, w 1,, w l, z 1,, z n ]. = {[s, w 1,, w l, z 1,, z n ] P n+l s 0, s + l w j n homogenous coordinates. The holomorphic automorphism group of B n+l l holomorphic automorphism group of P n+l given by z j > 0} in is a subgroup of the Un + l + 1, l + 1 = {A GLn + l + 1, C AEl + 1, n + l + 1Āt = El + 1, n + l + 1}, where El + 1, n + l + 1 is a diagonal matrix with 1 on the first l + 1 diagonal elements and then 1 on the next n diagonal elements. It is easy to check that ω B n+l is invariant under l Un + l + 1, l + 1. The connection between classical domain of Type IV and the generalized complex balls is as follows. Define L : Dm IV B m+1 1 by Lz 1,, z m = [1, 1 m z j, z 1,, z m ]. One can easily check that L is a proper holomorphic map and furthermore is a holomorphic isometric embedding up to a positive constant 1/m, i.e. L ω B m+1 = 1 ω 1 m Dm IV. The following is a key result in proving Theorem 1.1. Theorem 3.. Assume n 4, n + 1 N n and assume U is a connected open set in C n containing p B n, V is an open set in C N containing q B N 1. Let F be a holomorphic map on U such that 9

10 F U B N 1, F U B n B N 1. Then the following statement holds: If N = n + 1, then F is equivalent to 0, z 1,, z n 10 If N > n + 1, then F is equivalent to ψ, ψ, z 1,, z n, 0, 11 where 0 denotes the N n dimensional zero row vector and ψ is some holomorphic function on U. The proof follows from a very similar argument as in [BEH]. The crucial techniques in the proof include the normal form type argument, Huang s Lemma [Hu1] and its generalizations [BEH], moving point trick [Hu1], and a transverality result in [BER]. To make it easier for the readers, we will sketch a proof in Appendix. The following corollary follows from Theorem 3.. Corollary 3.3. Assume the same assumptions as in Theorem 3.. Then F is a local holomorphic isometric embeddings from B n to B N 1, i.e. F ω B N 1 = ω B n/n + 1 on U B n. Proof. It is straightforward to check that both maps in 10 and 11 are isometries from B n to B N 1, i.e. F ω B N 1 = ω B n/n + 1. Then Theorem 3.3 follows directly from Theorem 3. as ω B n and ω B N 1 are invariant under automorphisms.. We are now at the position to prove Theorem 1.1, 1. and Corollary 1.5. Theorem 1.4 will be proved in the next section. Proof of Theorem 1.1. The composition L F is a holomorphic map satisfying the assumption in Corollary 3.3. Thus L F is a local holomorphic isometry from B n to B m+1 1 by Corollary 3.3, i.e. L F ω B m+1 = ω B n on U B n. Note that 1 ω B n/n + 1 = L F ω B m+1 = F L ω 1 B m+1 = 1 1 m F ω D IV m on U B n. Therefore F extends to a global holomorphic isometry from B n to Dm IV by Mok s theorem [M4]. We conclude F ω D IV m = m ω n+1 Bn by Proposition.11. Proof of Theorem 1. assuming Theorem 1.4. An easy computation verifies that Dn+1 IV is holomorphically nondegenerate See [BER1] at any smooth point. Indeed, Dn+1 IV is -nondegenerate in the sense of Baouendi-Huang-Rothschild [BHR] at any smooth point. It 10

11 follows from a regularity result in [KLX] that F is real analytic in some open piece V of B n. F thus extends holomorphically to a neighborhood of V. We conclude that F is an isometry by Theorem 1.1. Then Theorem 1. follows from Theorem 1.4. Proof of Corollary 1.5. First we note that by the algebraic extension theorem of Mok [M4] following from the argument in [HY1, HY] in this case as well, F is an algebraic map. By holomorphic continuation, F can be extended along some path to an open set V in C n containing a generic boundary point of B n and the holomorphic isometry equation 3 is preserved along the path. Therefore, there must exist a nonconstant holomorphic map F l such that F V B n DN IV l and F V DN IV l. It follows from Theorem 1.1 that F l is a holomorphic isometric embedding. The corollary then follows from an induction argument. 4 On holomorphic isometries from B n into D IV n Examples of holomorphic isometries Write z = z 1,..., z n to be the coordinates in C n for n. Let R IV n : B n D IV n+1 be defined as R IV n = f 1,..., f n 1, f n, f n+1, 1 where f i = z i, 1 i n 1, f n = Pn, f Q n+1 = P n+1, Q P n = 1 n 1 zi zn + z n, P n+1 = 1 n 1 1 zi + zn z n, Q = 1 z n. It is easy to verify that n fi = n 1 z i 1 z n, f n + f n+1 = 1 4 n 1 i z i 1 z n + z n. It then follows that Rn IV is a holomorphic isometry from B n to Dn+1. IV In fact, we will show that Rn IV is the unique rational holomorphic isometry from B n to Dn+1 IV up to holomorphic automorphisms. For any θ [0, π/4, define h θ z = sinθz n z n cosθ I n,θ = f 1,, f n+1 be n 1 z j. Let f 1 z = z 1,, f n 1 z = z n 1 ; f n z = cos θ + 1 sin θz n cos θ h θ z ; cosθ f n+1 z = 1 sin θ + cos θz n + 1 sin θ h θ z. cosθ 13 11

12 Then I n,θ is a holomorphic isometry from B n to Dn+1. IV In particular, n I n,0 = z 1,, z n 1, 1 1 zj, z n. 14 We will also show that I n,0 is the unique irrational holomorphic isometry from B n to D IV n+1 up to holomorphic automorphisms. 4. Singularities of holomorphic isometries The rational holomorphic isometry Rn IV given in previous sections is not a totally geodesic and only produces singularities at one single point on the boundary B n. When n, one can easily avoid passing through this point by slicing B n with a complex linear hyperplane. Therefore, one obtains a holomorphic polynomial isometry from B n 1 into Dn+1. IV In particular, this answers the question raised by Mok in [M3] Question 5.. in the negative for Type IV domains and counter-examples for other types of classical domains are given in [XY]. Note that this type of examples are discovered independently by Chan-Mok [CM]. Theorem 4.1. Assume m n +. There exist non-totally geodesic holomorphic isometries from the unit ball B n into D IV m that can be extended holomorphically to C n. Proof. The map can be indeed chosen to be polynomial. The polynomial holomorphic isometries is given by n n z 1,, z n, zi, zi, 0,, 0 : B n Dm IV for m n +, n 1, 4 4 where there are m n zero components in the map. 4.3 Proof of Theorem 1.4 This subsection is devoted to establish Theorem 1.4. For that we will first describe the holomorphic automorphism group action on Dm IV in terms of the Borel embedding cf. [H1] [M1]. The hyperquadric Q m, the compact dual of Dn IV is defined by Q m := {[z 1,, z m+ ] P m+1 z1 + + zm = zm+1 + zm+}. The Borel embedding Dm IV Q m P m+1 is given by Z = z 1,, z m [z 1,, z m, ZZt, 1 1 ] ZZt. The holomorphic automorphism group of Dm IV is given by {[ ] } A B AutDm IV = Om,, R detd > 0, C D 1

13 where A Mm, m, R, B Mm,, R, C M, m, R, D M,, R. The automorphism group [ action ] is given in the following explicit way. Let Z = z 1,, z m Dm IV and T = A B AutDn C D IV. Write Z 1+ 1 = ZZt, 1 1 ZZt. Then the action of T on Dm IV is given by D IV m T Z = ZA + Z C ZB + Z D 1/, t. 1/ Rephrasing in homogenous coordinates, if the holomorphic automorphism maps Z = z 1,, z m to W = w 1,, w m Dm IV, then there exists T AutDm IV such that [w 1,, w m, 1 + 1W W t, 1 1W W t ] = In other words, there exists nonzero λ C, such that w 1,, w m, 1 + 1W W t, 1 1W W t = λ Note that the isotropy group K 0 at the origin is K 0 = [z 1,, z m, ZZt, 1 1 ] ZZt T. z 1,, z m, ZZt, 1 1 ZZt T. {[ ] } A 0 Om,, R detd = 1 = 0 D Om, R SO, R. We are now at the position to prove Theorem 1.4. We first establish a result on isotropy equivalence. Theorem 4.. Let F : B n D IV n+1 be a holomorphic isometric embedding satisfying F 0 = 0 and F ω D IV n+1 = ω B n. 15 Then F is isotropically equivalent to either the map Rn IV some θ [0, π/4. in 1 or the map I n,θ in 13 for Proof. First normalization: Write F = f 1,, f n+1. By the isometry assumption, a standard reduction yields that n+1 f i 1 n+1 4 n fi = z i. 16 We then have z 1,, z n, 1 m fj z = f 1 z,, f n+1 z U 17 for some n + 1 n + 1 unitary matrix U by a lemma of D Angelo [D]. Write U = u 1,, u n+1, where each u i for 1 i n + 1, is a column vector in C n+1. Write the first n-columns of U as U 0 = u 1,, u n. By 17, we have z 1,, z n = f 1 z,, f n z U

14 By the singular value decomposition of symmetric matrices, there exists an unitary n n matrix V such that V t U 0 t U 0 V = diag{λ 1,, λ n } for real numbers λ i satisfying λ 1 λ n 0. Apply the unitary change of coordinate in C n by letting z = z 1,, z n = z V, where z = z 1,, z n or equivalently, z = z V 1. By 18, we have z 1,, z n = f 1 zv 1,, f n+1 zv 1 Û 0, where Û 0 = U 0 V with Û t 0Û0 = diag{λ 1,, λ n }. Moreover, by 17, we have z 1,, z n, 1 n+1 fi zv 1 = f 1 zv 1,, f n+1 zv 1 Û 0, u n Let F z = F zv 1 and write F z = f 1 z,, f n+1 z. rewritten as z 1,, z n, 1 n+1 f i z Then equation 19 can be = f1 Z,, f n+1 z Û 0, u n+1. 0 Note that Û 0, u n+1 is still an unitary n + 1 n + 1 matrix. It follows from 0 that F is also a holomorphic isometry from B n to D IV n+1 and moreover, F is equivalent to F. Second normalization: In the following, we write x y = k x iy i for two k dimensional vectors x = x 1,..., x k, y = y 1,..., y k. We now consider the new map F in the new holomorphic coordinate z. But for the simplicity of notations, we still use F, z, U 0 to denote F, z, Û 0 respectively. Therefore, we have z 1,, z n, 1 m fj z = f 1 z,, f n+1 z U 0, u n+1 with Write U 0 = u 1,, u n. It follows from 1 that U t 0U 0 = diag{λ 1,, λ n }. 1 u i u i = λ i R for 1 i n, u i u j = 0 for 1 i j n. Note U t 0U 0 = I n. We have u i u i = 1 for 1 i n, u i u j = 0 for 1 i j n. 3 14

15 Write u i = a i + 1b i for 1 i n. It follows from -3 that a i b j = 0 for 1 i, j n, a i a j = 0 for 1 i j n, b i b j = 0 for 1 i j n, a i a i + b i b i = 1 for 1 i n. 4 Therefore, these n vectors {a i, b i } 1 i n in R n+1 are mutually orthogonal. This implies that at least n 1 of them are zero vectors. However, by the last equation in 4, a i and b i cannot be both zero for each 1 i n. Hence, by applying again the unitary change of coordinates in C n if necessary, we assume that for each 1 j n 1, either a j or b j is zero. Furthermore, for each fixed 1 j n 1, by applying the unitary change of coordinates z 1,, z j 1, z j, z j+1, z n z 1,, z j 1, 1z j, z j+1, z n in C n if necessary, we can always assume that b j = 0, a j 0 for 1 j n 1. Therefore, we have u j = a j R n+1 for all 1 j n 1, and moreover, u 1,, u n 1 t u 1,, u n 1 = I n 1. Extend {u 1,, u n 1 } to an orthonormal basis {u 1,, u n 1, c n, c n+1 } of R n+1 and write the n + 1 n + 1 matrix C = u 1,, u n 1, c n, c n+1. It follows that C, C t are orthogonal matrices SOn + 1 as CC t = C t C = I n+1. Define F = f 1,, f n+1 = F C t 1 or equivalently F = F C t. One can easily check that F is still a holomorphic isometry from B n into Dn+1 IV and F is equivalent to F. Furthermore, one has z 1,, z n, 1 n+1 Since C is an orthogonal matrix, then where X = [ In 1 0 n 1 f j = f 1,, f n+1 C t U 0, u n+1. C t U 0 = X, ũ n n+1 n, ] and 0 n 1 is the n 1 zero matrix. Note C t U 0 t C t U 0 = U t 0U 0 = I n, i.e. the columns of C t U 0 are orthonormal basis in C n+1. Then we conclude that ũ n = [ 0,, 0, ξ 1, ξ ] t 15

16 for ξ 1, ξ C with ξ 1 + ξ = 1. Note that C t U 0, u n+1 is an n + 1 n + 1 unitary matrix. We will again use F, U to denote F, C t U 0, u n+1 respectively for the simplicity of notations. To summarize the above, we have normalize the original holomorphic isometry to the map F satisfying z 1,, z n, 1 n+1 fj z = f 1 z,, f n+1 z U, 5 where U is n + 1 n + 1 unitary matrix and for X = [ In 1 0 n 1 U = X, u n, u n ], u n = 0 ξ 1 and u n+1 = 0 η 1 with ξ η ξ 1 + ξ = 1, η 1 + η = 1 and ξ 1 η 1 + ξ η = 0. Now Replacing F by F = e 1α F and writing F = f 1,, f n+1, we have: z 1,, z n, 1 n+1 f j z = f1 z,, f e 1α I n 1 0 n n 1 1 n+1 z 0 t n 1 1 e 1α ξ 1 e 1α η 1. 0 t n 1 1 e 1α ξ e 1α η Choose a suitable α such that the real and imaginary parts of e 1α η 1, e 1α η t are orthoganal. Applying the unitary change of coordinates ẑ = ẑ 1,, ẑ n = e 1α z 1,, z n and defining ˆF ẑ = ˆf 1 ẑ,, ˆf n+1 ẑ = F e 1αẑ, then one can easily check that ˆF satisfies: ẑ 1,, ẑ n, 1 n+1 ˆf j Ẑ = ˆf1 ẑ,, ˆf n+1 ẑ I n 1 0 n n t n 1 1 ξ 1 e 1α η t n 1 1 ξ e 1α η We will still use F, z, η i to denote ˆF, ẑ, e 1α η i for i = 1, respectively. Then 6 reads z 1,, z n, 1 n+1 I n 1 0 n n 1 1 fj z = f 1 z,, f n+1 z 0 t n 1 1 ξ 1 η 1 0 t n 1 1 ξ η 16

17 with ξ 1 + ξ = 1, η 1 + η = 1 and ξ 1 η 1 + ξ η = 0, Reη 1, η t Imη 1, η t. 7 By applying an automorphism [ ] In 1 0 n 1 0 n 1 V of Dn+1 IV with a suitable V O, we can further make the real part of vector η 1, η be of form c, 0 for some c R. Consequently, by 7 we conclude that η 1 R and η = 1η for some η R. By further applying an automorphism I n 1 0 n n n 1 ± n 1 0 ±1 of D IV n+1 if necessary, we can assume that η 1 0 and η 0. Since η 1 +η = η 1 + η = 1, write η 1 = cos θ, η = sin θ for θ [0, π/] and then η = 1 sin θ. As ξ 1 + ξ = 1, ξ 1 η 1 +ξ η = 0, write ξ 1 = 1 sin θe 1α, ξ = cos θe 1α for α [0, π. By applying an unitary transform z 1,, z n 1, z n = z 1,, z n 1, e 1α z n in C n if necessary, we may let ξ 1 = 1 sin θ, ξ = cos θ. Therefore, we have normalized the map F = f 1,, f n+1 to make it satisfy z 1,, z n, 1 n+1 I n 1 0 n n 1 1 fj z = f 1 z,, f n+1 z 0 t n sin θ cos θ 0 t n 1 1 cos θ 1 sin θ for θ [0, π/]. Denote the matrix We the proceed in two different cases. Case I: If θ = π/4. In this case, we have z 1,, z n, 1 n+1 fj z I n 1 0 n n 1 1 U = 0 t n sin θ cos θ 8 0 t n 1 1 cos θ 1 sin θ = f 1 z,, f n+1 z U, where U = I n 1 0 n n t n t n

18 By replacing F by 1I n 1 0 n n 1 1 F 0 1 n n 1 0, 1 and then apply the unitary transformation in C n : z 1,, z n = 1z 1,, 1z n 1, 1z n, we are able to make I n 1 0 n n 1 1 U = 0 t n t n 1 1 By solving this linear system, one obtains the map R IV n in 1. We then replace f 1,, f n+1 by f 1,, f n, f n+1 to obtain the map in 1. Case II: If θ [0, π/] with θ π/4. The conclusion is trivial if θ [0, π/4. In the sequel, we assume θ π/4, π/]. Write β = π/ θ [0, π/4. Then I n 1 0 n n 1 1 U = 0 t n cos β sin β. 0 t n 1 1 sin β 1 cos β By applying the automorphism I n 1 0 n n 1 1 w 1,, w n+1 = w 1,, w n+1 0 n t n of D IV n+1, we may let I n 1 0 n n 1 1 U = 0 t n 1 1 sin β 1 cos β. 1 cos β sin β 0 t n 1 1 Applying the automorphism w 1,, w n+1 = 1w 1,, w n+1 of Dn+1 IV and then applying the unitary transform z 1,, z n 1, z n = 1z 1,, 1z n 1, z n of C n, we may let I n 1 0 n n 1 1 U = 0 t n sin β cos β 0 t n 1 1 cos β. 1 sin β Finally applying the automorphism w 1,, w n 1, w n+1 = w 1,, w n 1, w n+1 of D IV I n 1 0 n n 1 1 U = 0 t n sin β cos β. 0 t n 1 1 cos β 1 sin β 18 n+1,

19 This is the matrix in 8. By solving the system z 1,, z n, 1 n+1 fj z = f 1 z,, f n+1 z U, we obtain that F is equivalent to I n,β in 13 for some β [0, π/4. This establishes Theorem 4.. Theorem 4.3. For any θ [0, π/4, I n,θ : B n D IV n+1 given in 13 is equivalent to I n,0. Proof. We first apply the Borel embedding to embed B n as an open subset of P n and Dn+1 IV as an open subset of Q n+1 P n+, where the Borel embedding is given by z = z 1,, z n+1 D IV n+1 [z 1,, z n+1, zzt, 1 1 zzt ] Q n+1 P n+. We write [z, s] = [z 1,, z n, s] to denote the homogeneous coordinates in P n. homogeneous coordinates, I n,θ is identified with Then under I n,θ z, s = [z 1,, z n 1, φ n,θ z, s, φ n+1,θ z, s, φ n+,θ z, s, φ n+3,θ z, s] 9 from P n to P n+ where φ n,θ z, s = cos θs + 1 sin θz n cos θ H θ z, s ; cosθ φ n+1,θ z, s = 1 sin θs + cos θz n + 1 sin θ H θ z, s ; cosθ φ n+,θ z, s = 1 + cosθ 1 tanθzn 1 s + Hθ z, s; cosθ cosθ 30 φ n+3,θ z, s = cosθ 1 1 tanθzn 1 s + Hθ z, s; cosθ cosθ In particular, Let H θ z, s = s + n 1 1 sinθz n s zn cosθ zj. I n,0 z, s = [ z 1,, z n 1, s H 0, z n, s ] H 0 H0,. I n cos θ B = 0 sin θ 1 cosθ cosθ Un, 1 = AutBn, sin θ 0 1 cosθ cos θ cosθ 19

20 and define În,θz, s = I n,θ z, s B. Then it follows from the straightforward calculation that [ Î n,θ = z 1,, z n 1, s cos θ H 0, z n + sin θ H 0, cos θs H 0, sin θz n ] H 0. cosθ cosθ cosθ cosθ Let T = [ ] In 1 0 n 1 4 with 0 4 n 1 V V = 1 4 sin θ/ 0 sin θ/ sin θ cosθ sin θ/ sin θ/ 0 0. sin θ 0 1 Then one can verify that T AutD IV n+1 and I n,0 T = z 1,, z n 1, s H 0, z n, s H 0 H0, T = z 1,, z n 1, s cos θ H 0, z n + sin θ H 0, cos θs H 0, sin θz n H 0 cosθ cosθ cosθ cosθ = În,θ = I n,θ z, s B. This implies that I n,θ is equivalent to I n,0. Note that a rational map from B n Dn+1 IV cannot be equivalent to an irrational map. Thus combining Theorem 4., 4.3, we obtain the classification result Theorem 1.4 for holomorphic isometries from B n into Dn+1. IV Remark 4.4. n = 1 is a special case n + 1 = n. Let F be a holomorphic isometry from C to D IV satisfying F ω D IV = ω. It follows from the same argument of Theorem 1.4 that F is either equivalent to the totally geodesic embedding z/, z/ or equivalent to the non-totally geodesic embedding I 1,0. Note that D IV is biholomorphic to bidisc C. In the Euclidean coordinate of, the first map is z z, 0 and I 1,0 is the square root embedding constructed by Mok [M4]. This classification result from to was obtained earlier by Ng cf. [Ng1] Theorem

21 Appendix Theorem 3. can be proved using the idea from [BEH] in a very similar manner. We first introduce the following notationssee more details in [BH] and [BEH]. Assume N 3 and let H N 1 C N be the hyperquadric of signature 1 defined by H N 1 = {z, w = u + 1v C N v = N where z = z 1,, z N 1. Let H n be the Heisenberg hypersurface z j z N 1 }, H n = {z, w = u + n 1 1v C n v = z j }, where z = z 1,, z n 1. To prove Theorem 3., it suffices to show the following theorem. Theorem 4.5. Assume n 4, n + 1 N n and let F be a holomorphic map from a connected open set U C n containing 0 to C N. If F U H n H N 1 and F U H N 1, then F is equivalent to z 1,, z n 1, 0, w if N = n + 1; F is equivalent to z 1,, z n, 0, ψ, ψ, w if N > n + 1, where ψ is some holomorphic function on U. To prove Theorem 4.5, we will first establish the following proposition: Proposition 4.6. Let F = f, g be a holomorphic map from an open connected set U C n containing 0 to C N with n 4, n + 1 N n. Assume that F U H n H N 1, F 0 = 0 and F is CR transversal at 0. Then g 0 > 0 and F is w equivalent to z 1,, z n 1, 0, w if N = n + 1; equivalent to z 1,, z n, 0, ψ, ψ, w if N > n + 1, where ψ is some holomorphic function on U. We merely prove the case N > n + 1 in Proposition 4.6 and the other case follows similarly. Proof of Proposition 4.6. We first show g 0 > 0. Write w F = f, g = f, ϕ, g = f 1,, f n 1, ϕ 1,, ϕ N n, g 1

22 and write the tangent vector fields of H n at 0 F U H n H N 1 implies that 1 L i = z i z i w, 1 i n 1, T = w + w. ρf, F = g ḡ N n f + ϕ i ϕ N n = 0 31 holds on H n. Here f = n 1 f i. Applying L k i to 31 and evaluating at 0, we get Applying T to 31 and evaluating at 0, we get k g 0 = 0. zi k g g 0 = w w 0. This implies g 0 = λ R. Since F is transversal at 0, we have λ 0. We now write w g = λw + O z, w, n f j = b j w + a ij z i + O z, w, where b j C, a ij C. The second equation can be rewritten as f = wb 1,, b N 1 + z 1,, z n 1 A + ˆ f, where A = a ij n 1 N 1 is an n 1 N 1 matrix and ˆ f = O z, w. We call a function h on U H n has weighted degree s, denoted by h o wt s, if htz, t z, t u lim = 0. t 0+ t s For a smooth function hz, z, u defined in U H n, we denote by h k z, z, u the sum of terms of weighted degree k in the Taylor expansion of h at 0. We also denote by h k z, z, u a weighted homogeneous polynomial of degree k. When h k z, z, u extends to a weighted holomorphic polynomial of degree k, we write it as h k z, w or h k z if it depends only on z.

23 Note that 31 implies that N n 1 g ḡ 1 = f + ϕ i ϕ N n 3 holds on w = u + 1 z near 0. We collect terms of weighted degree two on both sides of 3 to get λ z = zaen, N 1Āt z t, 33 which further implies λi n = AEN, N 1Āt. Here EN, N 1 denotes the N 1 N 1 diagonal matrix with its first N diagonal element +1 and the rest 1. It follows from linear algebra that λ > 0. We now fix some notations. For two m-tuples x = x 1,, x m, y = y 1,, y m of complex numbers, we write < x, y >= m x jy j and x =< x, x > and write < x, y > 1 = m 1 x jy j x m y m and x 1 =< x, x > 1. In this following context, to make notations easier, we will assume N = n. The proof of other cases is of no significant difference. By the same argument as in [BH] cf. Lemma., we can assume that F has the following normalization fz, w = z + 1 a1 zw + o wt 3, ϕz, w = ϕ z + o wt, gz, w = w + o wt 4, 34 with < a 1 z, z > z =< ϕ z, ϕ z > 1. By Huang s lemma cf. [Hu1] Lemma 3. or [BEH] Lemma.1, a 1 z 0 and < ϕ z, ϕ z > 1 0. Therefore, Now suppose that we have obtained f 3 0. g t 0 and f t for 3 t < s. Note that 35 is already proved for s = 5. Now by collecting terms of weighted degree s, we obtain Im{g s z, w 1 < z, f s 1 z, w >} = < ϕ s1 z, w, ϕ s 1 z, w > s 1 +s =s

24 when w = u + 1 < z, z >. We shall use the notation Lp, qz, z, u = Im{qz, w 1 < z, pz, w >} w=u+ 1<z, z>, where pz, w = p 1 z, w,, p n 1 z, w and qz, w are holomorphic polynomials. Equation 36 can be written as Lf s 1, g s z, z, u = < ϕ s1 z, w, ϕ s 1 z, w > 1 w=u+ 1<z, z>. 37 s 1 +s =s The following result is crucial in the proof of Theorem 1.. Lemma 4.7. Let F = f, ϕ, g be any normalized map as above sending an open piece M of H n near 0 into H N 1 with N n. Assume that for all 3 t s 1, f t 1 0, g t 0, < ϕ s 1, ϕ s > for any s 1, s with s 1 + s = t. Then 38 holds also for t = s and t = s for any such a map F. Once Lemma 4.7 has been proved, we conclude by induction that 38 holds for all t 3. We shall need some notation and results from [BEH] and [EHZ]. Given a real-valued power series Az, z, w, w, we use the expansion Az, z, w, w = µνγδ A µνγδ z, zw γ w δ for z, w C n 1 C where A µνγδ z, z is a bihomogeneous polynomial in z, z of bidegree µ, ν for every µ, ν, γ, δ Z 4 +. We recall from [EHZ] that Az, z, w, w belongs to the class S k for a positive integer k if A vanishes at least up to order at 0 and for every µ, ν, γ, δ Z+, 4 we have k A µνγδ z, zw γ w δ = p j z, wq j z, w, where p j, q j are homogeneous holomorphic polynomials of the appropriate degree. We recall the following result from [EHZ]. Theorem 4.8. [EHZ] Let Az, z, w, w be a real-valued weighted homogeneous polynomial of degree s 5 and assume that A S n. If pz, w = p 1 z, w,, p n 1 z, w and qz, w are weighted homogeneous holomorphic polynomials of degree s 1 and s respectively, such that Lp, qz, z, u = Az, z, w, w w=u+ 1 z. Then p 0, q 0 and A 0. 4

25 Proof of Lemma 4.7. As in [BEH], we shall first prove that 38 holds for t = s = s 1 with s 3. Recall the hypotheses in Lemma 4.7 implies that 37 holds with s = t. Note that the hypotheses also imply that < ϕ s1, ϕ s 1 > 1 0 for s 1 s 1. By a lemma of D Angelo [D], we conclude that there are constants a s 1 j such that ϕ s 1 j = a s 1 j ϕs 1 n, for 1 j n 3, s 1 s Now if s 1 + s = s 1, then mins 1, s s 1. Without of loss of generality, assume s 1 = mins 1, s. Then it follows from 39 that < ϕ s 1, ϕ s > 1 = ϕ s 1 n q s, where q s z, w = n 3 as 1 j ϕs j ϕ s n. Since 1 n, it follows that Az, z, w, w := s 1 +s =s 1 < ϕs 1 z, w, ϕ s z, w > 1 belongs to S n. It follows from 37 and Theorem 4.8 that f s 1 0, g s 0 where s = s 1 and A 0. By the definition of A, we have < ϕ s1, ϕ s > 1 0 for s 1 + s = s 1. It remains to show 38 for t = s 6. We first complexity 37 with s = s : g s z, w g s ξ, η 1 < ξ, f s 1 z, w > 1 < z, f s 1 ξ, η > = 1 k < ϕ k z, w, ϕ s k ξ, η > Write L j = + 1 ξ j z j w, 1 j n 1 be the tangent vector field of the complex hypersurface defined by w = η + 1 < z, ξ >. We will first prove f s 1 z, w 0, g s We begin by establishing the following proposition. Proposition 4.9. For 1 j n 1, we have f s 1 j z, w = a 1 j zw s 1, g s z, w = d 0 w s and a 1 z = d 0 z j. Proof of Proposition. Applying L j to 40, we have on w = η + 1 < z, ξ >, L j g s z, w 1 < ξ, L j f s 1 z, w > 1f s 1 ξ, η = 1 k < L j ϕ k z, w, ϕ s k ξ, η > 1. 5

26 Write f s 1 j z, w = a τ j j zw τ j s ; ϕ k = b µ k k zw µ k ; g s z, w = d j zw nj s, where sums run over all indices such that τ j + τs j = s 1, µ k + µ k = k, j + nj s = s with s = s. Letting w = 0, η = 1 < z, ξ > and collecting terms of degree k > in ξ and degree P in z. We obtain 1a K P j η P s = 1 k= < ϕ k z j z, 0, b µ s k s k s ξη µ s k >1 4 k =3 ξ j < ϕ k w z, 0, b µ s k s k ξη µ s k > 1, where the sum inside <, > 1 run over the indices µ s k + µ s k = K, k 1 + µ s k = P and µ s k + µ s k + 1 = K, k + µ s k = P. Note that the solution only exist when K + P + 1 = s = s. Letting K = s + p, P = s p 1, we then get µ s k = k + p, µ s k = s p k and µ s k = k + p, µ s k = s p k + 1. Since both µ s k and µ s k must be nonnegative, one has k s p, 3 k s p + 1. We now rewrite 4 as 1a p+1 j ξη s p 1 = s p 1 k= s p+1 < ϕ k z j z, 0, b k+p s k ξ > 1 η s p k 4 k =3 which can further be rewritten as 1a p+1 j ξη s p 1 = s p k= 1 < ϕ k z j z, 0, b k+p s k ξ > 1 4 ξ j < ϕ k+1 w 4 ξ j < ϕ k w z, 0, b k +p s k ξ > 1 η s +1 p k. 43 z, 0, b k+p 1 s k 1 ξ > 1 η s p k This equation is valid for all p = 0,, s 1. If p = s 1, the sum on the right hand side is void. Note that for any q s 1, we can use 39 to rewrite n 3 < ϕ q z j z, 0, b q+p s q ξ > 1 = ϕ q i,z j z, 0b q+p i,s q ξ ϕq n,z j z, 0b q+p = ϕ q n,z j z, 0Cξ n,s q ξ 6

27 with n 3 Cξ = a s jb q+p i,s q ξ bq+p n,s qξ. We can make a similar substitution in < ϕ q w z, 0, b q+p s k 1 ξ > 1 for any q s 1. Therefore, if q s 1, we then conclude by Corollary. in [BEH], as n, that When p = 1, 43 can be written as 1a 3 j ξη s = a p+1 j z 0, for j = 1,, n 1 and p =,, s s k=1 1 < ϕ k z j z, 0, b k+ s k ξ > 1 4 ξ j < ϕ k+1 w z, 0, b k+1 s k 1 ξ > < ϕ s 1 z j z, 0, b s +1 s +1 ξ > 1 4 ξ j < ϕ s w z, 0, b s s ξ > 1 We now turn to the equation 40 in which we set w = 0 and η = 1 < z, ξ >. Collecting terms of degree s in z and s in ξ, we obtain d 0 η s 1 < z, a 1 ξ > η s8 1 = 1 This can be rewritten as d 0 η s 1 < z, a 1 ξ > η s 1 s 1 1 k= < b k k s k= z, bk < ϕ k z, 0, b k s k ξ > 1 η s k. η s 1 k 45 s k ξ > ηs k = 1 < b s s z, bs s ξ > 1. Note the left hand side is divisible by η that is a summation of n 1 terms, while < b s s z, bs s ξ > 1 is a summation of n terms, by Huang s lemma cf. Lemma 3. in [Hu1] or Lemma.1 in [BEH], we have < b s s z, bs s ξ > 1 0. Therefore b js z = As j b s n s z, 1 j n 3 for some constant A s j. Here b s js is the j-th component of the vector-valued function bs s z. Using Corollary. in [BEH], we have. and < b k k z, bk s k ξ > 1 0 for k =,, s 1 46 d 0 η = 1 < z, a 1 ξ >. 47 7

28 Equation 47 implies a 1 j z = d 0 z j. If we use 46 to substitute for b s js in 45 as we did before, then it follows from Corollary. in [BEH] again that a 3 j 0 for j = 1,, n and 48 imply that f s 1 z, w = a 1 zw s 1. To show g s z, w = d 0 w s, we go back to 40 in which we again set w = 0 and η = 1 < z, ξ >. Note that we have proved that the degree of f s 1 ξ, η in ξ is s. Thus, if we collect terms of degree K = s + p in ξ with p 1, then we obtain d p ξη s p = s p 1 k= < ϕ k z, 0, b µ s k s k ξη µ s k >1, where the sum inside <, > 1 rums over µ s k + µ s k = s + p, µ s k + µ s k the equation above can be rewritten as = s k. As before, d p ξη s p = s p 1 k= < ϕ k z, 0, b k+p s k ξ > η s p k for p = 1,, s. We then substitute for ϕ k z, 0 using 39 and apply Corollary. in [BEH] to conclude d p ξ 0 for p = 1,, s. This establishes the proposition. We will need the following lemma to complete the proof of 41. Lemma Suppose that the hypotheses in Lemma 4.7 hold. Assume further that for any p M, we have fp z, w = z + a 1 p zw s 1 + O wt s gp z, w = w + d 0 p w s + O wt s + 1 where s = s. Then a 1 p z 0, d 0 p = 0 and hence fz, w = z + O wt s, gz, w = w + O wt s + 1. Proof. One can apply the identical proof of Lemma 3.5 in [BEH]. We need to employ the moving points trick that was first introduced in [Hu1] and has been extensively used in many literatures cf. [Hu1], [BH], [BEH]. 8

29 To complete the induction step in the proof of Lemma 4.7, we must show that < ϕ k z, w, ϕ s k ξ, η > 1 0 for k =,, s with s = s. This can be established by a very similar proof as [BEH] cf. pp using the same idea as above with the following difference. In our setting, one should apply Huang s lemma cf. Lemma 3. [Hu1] or Lemma.1 [BEH] while [BEH] uses Lemma.3. Here note Huang s lemma can be applied as by our assumption < ϕ, ϕ > 1 is a summation with less terms than < z, ξ >, i.e. N n < n 1. Now we are able to finish the proof of Proposition 4.6. First we may assume that F satisfies the normalization 34. By Lemma 4.7, we conclude that fz, w z, gz, w w, < ϕz, w, ϕξ, η > 1 0. By the result of D Angelo [D], there is a constant n 3 n 3 unitary matrix U such that ϕ 1,, ϕ n 3 U = ϕ n, 0,, 0 with n 4 zero components on the right hand side of the equation. Letting γ be the automorphism of H N 1 given by γz, w = z 1,, z n 1, z n,, z N U, z N 1, w, then γ F satisfies the conclusion of Proposition 4.6. This finishes the proof of Proposition 4.6. We now prove Theorem 4.5. Proof of Theorem 4.5. We set E = {p H n F is not CR transversal to H N 1 at p}. The set E is a real analytic subvariety near 0 in H n. If E contains an open neighborhood of 0, then it follows from Theorem 1.1 in [BER] cf. [BH] Lemma 4.1 that F U H N 1. Now assume F U H N 1. Then E is a proper real analytic subvariety of H n. Consequently, there exists p H n near 0 such that F is CR transversal at p. We recall from [BEH] that F is CR transversal to H N 1 at p M if and only if gp 0 0. Here F w p = f p, ϕ p, g p refer the definition of F p in [BH]. Thus replacing F by F p if necessary, we can assume p = 0. Then the result follows easily from Proposition 4.6. Recall the well-known fact that the Cayley transformation biholomorphically maps H N 1 B N 1 minus a proper subvariety. Then Theorem 3. is a consequence of Theorem 4.5. into 9

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