Proper holomorphic polynomial maps between bounded symmetric domains of classical type

0.19 MB·English

by  Aeryeogn Seo

PROPER HOLOMORPHIC POLYNOMIAL MAPS BETWEEN
BOUNDED SYMMETRIC DOMAINS OF CLASSICAL TYPE
5
1
AERYEONGSEO
0
2

n
a
J
Abstract. Weprovethattwoproperholomorphicpolynomialmapsbetween

6 bounded symmetric domains of classical type which preserve the origin are
1 equivalentifandonlyiftheyareisotropicallyequivalent. Usingthisproperty
we show that each member of a one-parameter family of maps from [19] is

] inequivalent.
V
C
1. Introduction
.
h
LetΩ , Ω be domainsinCn andCN andf, g :Ω Ω be holomorphicmaps.
t 1 2 1 2
a Wesaythatf isproper iff−1(K)iscompactforeveryc→ompactsubsetK Ω . We
m ⊂ 2
saythat f andg are equivalent if and only if f =A g B for some B Aut(Ω )
1
[ andA Aut(Ω ). ForadomainΩ,denotethegroup◦ofau◦tomorphismsfi∈xingp Ω
2
∈ ∈
1 by Isotp(Ω). Suppose that for fixed p Ω1, f(p) = g(p). Then we say that f and
∈
v g are isotropically equivalent at p if there are U Isot (Ω ) and V Isot (Ω )
p 1 g(p) 2
8 such that f =V g U. The notion of isotropic e∈quivalence coincide∈s with that of
0 ◦ ◦
unitaryequivalenceof[3]definedwhenΩ andΩ areballs. Thefollowingdomains
9 1 2
are called bounded symmetric domains of classical type:
3
0 (1) ΩI = Z MC :I ZZ∗ >0 , where s r =rank(ΩI ).
. r,s { ∈ r,s r − } ≥ r,s
1 (2) ΩII = Z MC :I ZZ∗ >0, Zt = Z , rank(ΩII)= n .
0 n { ∈ n,n n− − } n 2
5 (3) ΩInII ={Z ∈MnC,n :In−ZZ∗ >0, Zt =Z}, rank(ΩInII)=(cid:2)n.(cid:3)
1 (4) ΩIV = Z = (z ,...,z ) Cn : ZZ∗ < 1, 0 < 1 2ZZ∗ + ZZt 2 ,
v: rannk(ΩIV(cid:8))=2. 1 n ∈ − (cid:12) (cid:12) (cid:9)
i (cid:12) (cid:12)
X Here we denote by M > 0 positive definiteness of square matrix M, by MC the
r,s
r set of r s complex matrices and by I the r r identity matrix.
a × r ×
Theaimofthispaperistoprovethefollowingtheoremsasageneralizationofthe
results in [3] which are concerned with the proper holomorphic polynomial maps
between balls.
Theorem 1.1. Let Ω , Ω be bounded symmetric domains of classical type and
1 2
f, g : Ω Ω proper holomorphic polynomial maps such that f(0) = g(0) = 0.
1 2
→
Then f and g are equivalent if and only if they are isotropically equivalent at 0.
Theorem 1.2. There areuncountablymanyinequivalent proper holomorphic maps
from ΩI to ΩI for r 2, s 2.
r,s 2r−1,2s ≥ ≥
Date:January19,2015.
2010 Mathematics Subject Classification. Primary32M15,32H35.
Keywordsandphrases. Boundedsymmetricdomain,classicaltype,properholomorphicmap.
1
2 AeryeongSeo
The motivation of this paper comes from generalizingthe study onproper holo-
morphic maps between balls to bounded symmetric domains of rank greater than
or equal to two. Proper holomorphic maps between balls have been studied for a
long time since Alexander ([1]) proved that every proper holomorphic self-map of
the n-dimensional unit ball Bn with n 2 is a holomorphic automorphism. For
≥
proper holomorphic maps between balls with different dimensions, much work has
beendone,relatingthe maximumdegreeofproperholomorphicmapsto the differ-
ence of dimensions between the domain ball and the target ball. As the first work
along these lines, Webter ([22]) proved that any proper holomorphic map from Bn
toBn+1 withn 3,C3-smoothuptotheboundary,isequivalenttotheembedding
≥
f :(z ,...,z ) (z ,...,z ,0).
s 1 n 1 n
7→
Givena proper holomorphic map f from Bn to BN, we consider a proper holomor-
phicmapfromBntoBN+k definedbyz (f(z),0,...,0)withk-zerosforwhichwe
7→
willuse the same notationf if there is no confusion. When n 3 and N 2n 2,
≥ ≤ −
Faran([9]) showed that it is equivalent to f if it is extended holomorphically over
s
the boundary. Furthermore he precisely classified the equivalence classes of proper
holomorphic maps from B2 to B3 which is C3-smooth up to the boundary ([8]). In
[10], Forstneriˇc proved that any proper holomorphic map from Bn to BN which is
CN−n+1-smoothuptotheboundaryisarationalmap(p ,...,p )/q wherep and
1 N j
q are holomorphic polynomials of degree at most N2(N n+1). Since this work
−
has been done, much results fit into the framework of providing sharp bounds in
special situations. See [5, 7, 16], for more details. If N =2n 1, there is a proper
holomorphic map which is called the Whitney map f :Bn −B2n−1 defined by
w
→
f (z ,...,z )=(z , ...,z , z z , z z , ..., z2).
w 1 n 1 n−1 n 1 n 2 n
It is inequivalent to the embedding f . Moreover Huang and Ji ([13]) proved that
s
any proper rational map from Bn to B2n−1 with n 3 is equivalent to f or f
s w
≥
andanyproperholomorphicembedding whichis C2-smoothupto the boundary is
equivalent to f . If the dimension of the targetdomain is larger than 2n, there are
s
infinitelymanyinequivalentproperholomorphicmaps. Forexample,f :Bn B2n
θ
→
given by
(1.1) f (z)=(z , ..., z , cosθz , sinθz z , ..., sinθz z )
θ 1 n−1 n 1 n n n
with 0 θ π are found by D’Angelo ([3]). In [3], D’Angelo showed that any
≤ ≤ 2
two proper holomorphic polynomial map from Bn to BN preserving the origin are
equivalent if and only if they are isotropically equivalent at the origin and as a
consequence (1.1) are inequivalent for all 0 θ π. Interestingly, Hamada ([11])
≤ ≤ 2
showed that any proper rational map from Bn to B2n with n 4 is equivalent to
≥
f for some θ, 0 θ π/2 and Huang, Ji and Xu ([15]) showed that any proper
θ
holomorphic map≤from≤Bn to BN with 4 n N 3n 4 which is C3-smooth
≤ ≤ ≤ −
up to the boundary should be equivalent to f for some θ, 0 θ π/2. Recently
θ
Huang-Ji-Yin([14])provedthatanyproperrationalmapfrom≤Bn t≤oBN withn 8
and 3n+1 N 4n 7 should be equivalent to proper rational map from Bn≥to
B3n. ≤ ≤ −
As a one generalization of the unit ball, one consider bounded symmetric do-
mainswhichareHermitiansymmetricspacesofnon-compacttypewithnon-smooth
boundaries. There are several rigidity theorems on proper holomorphic maps be-
tween bounded symmetric domains of rank greater than or equal two. In contrast
properholomorphicpolynomialmapsbetweenBSDsofclassicaltype 3
with the case of the unit balls, the difference of the ranks between the domains is
more crucial than that of the dimensions. The first result on bounded symmetric
domains along these lines is the following which is due to Tsai. Let f : Ω Ω
1 2
→
be a proper holomorphicmap between irreducible bounded symmetric domains Ω
1
and Ω . If rank(Ω ) rank(Ω ), then rank(Ω ) = rank(Ω ) and f should be a
2 1 2 1 2
≥
totally geodesic isometric embedding with respect to the Bergman metrics on the
domains (see [20]). If Ω =ΩI and Ω =ΩI , then f is also a totally geodesic
1 r,r−1 2 r,r
isometric embedding (see [21]), although the rank of Ω is larger than that of Ω .
2 1
Furthermore, Ng ([18]) showed that for f : ΩI ΩI , if s 2, s r′ r
r,s → r′,s′ ≥ ≥ ≥
Z 0
and r′ 2r 1, then f is equivalent to the embedding, Z . If the
≤ − 7→ (cid:18) 0 0 (cid:19)
differenceoftheranksofthedomainsgetsbigger,thenitisexpectedthatthereare
lots of inequivalent proper holomorphic maps. In [19], one way of finding proper
holomorphicmapsbetweenboundedsymmetricdomainsoftype I issuggestedand
severalproperholomorphicmapsareconstructed. Forexample,forr′ =2r 1and
−
s′ =2s 1, there is a generalized Whitney map defined by
−
z2 z z ... z z z ... z
11 11 12 11 1s 12 1s
 z11z21 z21z12 ... z21z1s z22 ... z2s 
z11 ... z1s  ... ... ... ... ... ... ... 
(1.2)  ... ... ... 7→ z11zr1 zr1z12 ... zr1z1s zr2 ... zrs .
 zr1 ... zrs   z21 z22 ... z2s 0 ... 0 
   ... ... ... ... ... ... ... 
 
 z z ... z 0 ... 0 
 r1 r2 rs 
In this paper, as a one step to observe analogous phenomenon on proper holo-
morphic maps between bounded symmetric domains of rank greater than or equal
to two, we generalize the result of D’Angelo in [3] to the domains of classicaltype.
Acknowledgement. This research was supported by National Researcher Pro-
gram of the National Research Foundation (NRF) funded by the Ministry of Sci-
ence, ICT and Future Planning(No.2014028806).
2. Preliminaries
Inthissection,weintroduceterminologyandsomebasicbackground. Abounded
domain Ω is called symmetric if for each p Ω, there is a holomorphic automor-
phismi suchthati2 istheidentitymapofΩ∈whichhaspasanisolatedfixedpoint.
p p
All bounded symmetric domains are homogeneous domain, i.e. the automorphism
group acts transitively on the domain. In 1920’s, Cartan classified all irreducible
bounded symmetric domains. There are 4 classical types and 2 exceptional types.
The four classical types are given by (1),(2),(3) and (4) in the introduction. Note
that ΩI is the m-dimensional unit ball and ΩIII is the unit disc.
m,1 1
A B
From now on, we will use the notation M = GL(r+s,C) to split
(cid:18) C D (cid:19)∈
M into 4 block matrices with A MC , B MC , C MC and D MC . We
∈ r,r ∈ r,s ∈ s,r ∈ s,s
will denote by ASMC the set of anti-symmetric complex n n matrices and by
n,n ×
C
SM the set of symmetric complex n n matrices.
n,n ×
4 AeryeongSeo
Let U(r,s) be the subgroup of GL(r+s,C) satisfying
I 0 I 0
M − r M∗ = − r
(cid:18) 0 Is (cid:19) (cid:18) 0 Is (cid:19)
for all M U(r,s). Let SU(r,s) be the subset of U(r,s) which consists of the
∈
matrices with determinant one. Explicitly,
A B
SU(r,s)= SL(r+s,C):AA∗ BB∗ =I ,
(cid:26)(cid:18) C D (cid:19)∈ − r
(2.1)
AC∗ =BD∗, CC∗ DD∗ = I .
s
− − (cid:27)
Let O(n+2,C) be the complex orthogonalgroup of (n+2) (n+2) matrices.
×
Since every bounded symmetric domain is Hermitian symmetric space of non-
compact type, the domain can be canonically embedded into the corresponding
compact dual (Borel embedding) and every holomorphic automorphism of the do-
main can be extended to the automorphism of its compact dual. For example, let
G be the Grassmannian of r-planes in r+s dimensional complex vector space
r,s
Cr+s which is the compact dual of ΩI . For X MC of rank r, denote [X] the
r,s ∈ r,r+s
r-plane in Cr+s whichis generatedby row vectorsofX. For M GL(r+s,C), M
∈
acts on G by [X] G [XM]. Then the Borel embedding ξI of ΩI is given
r,s ∈ r,s 7→ r,s
by
ξI(Z)=[I ,Z] G
r r,s
∈
and M U(r,s) acts on ΩI by g (Z) := ξI−1(ξ(Z)M). Explicitly, for M =
∈ r,s M
A B
U(r,s), M acts on ΩI by
(cid:18) C D (cid:19)∈ r,s
(2.2) Z (A+ZC)−1(B+ZD).
7→
A B
Similarly, for M = U(n,n) acts on ΩII and ΩIII by (2.2).
(cid:18) C D (cid:19)∈ n n
IncaseofΩIV,theexplicitexpressionoftheholomorphicautomorphismislittle
n
more messy. The compact dual of ΩIV is the hyperquadric H in Pn+1 given by
n n
H := [z ,...,z ] Pn+1 : z2+ +z2 z2 z2 = 0 . Then the Borel
n { 1 n+2 ∈ 1 ··· n− n+1− n+2 }
embedding ξIV is given by
(2.3) ξIV(Z)=[ 2iZ,1+ZZt,i(1 ZZt)] H
n
− − ∈
A B
For M = O(n+2,C) U(n,2), then M acts on ΩIV by
(cid:18) C D (cid:19)∈ ∩ n
1
Z =(z ,...,z ) (2iZA Z′C)
1 n 7→ ( 2iZB+Z′D)(i,1)t −
−
where Z′ =(1+ZZt, i iZZt).
−
The automorphism groups of classical domains and their isotropy groups at the
origin are given by the following:
U 0
(1) Aut ΩI =U(r,s), Isot ΩI = :U U(r), V U(s)
r,s r,s (cid:26)(cid:18) 0 V (cid:19) ∈ ∈ (cid:27)
(cid:0) (cid:1) (cid:0) (cid:1)
properholomorphicpolynomialmapsbetweenBSDsofclassicaltype 5
0 I 0 I
(2) Aut ΩII = M U(n,n):Mt n M = n ,
n (cid:26) ∈ (cid:18) In 0 (cid:19) (cid:18) In 0 (cid:19)(cid:27)
(cid:0) (cid:1)
A 0
Isot ΩII = :A U(n)
n (cid:26)(cid:18) 0 A (cid:19) ∈ (cid:27)
(cid:0) (cid:1)
0 I 0 I
(3) Aut ΩIII = M U(n,n):Mt n M = n ,
n (cid:26) ∈ (cid:18) In 0 (cid:19) (cid:18) In 0 (cid:19)(cid:27)
(cid:0) (cid:1) − −
A 0
Isot ΩIII = :A U(n)
n (cid:26)(cid:18) 0 A (cid:19) ∈ (cid:27)
(cid:0) (cid:1)
(4) Aut(ΩIV)=O(n+2,C) U(n,2), Isot(ΩIV)=O(n) O(2)
n ∩ ×
3. Isotropically equivalent proper holomorphic polynomial maps
In this section, we will prove Theorem 1.1. Define a polynomial function SI :
r,s
ΩI R by
r,s →
SI (Z)=det(I ZZ∗)
r,s r −
C
for Z = (z ) M , as a real polynomial in Re(z ), Im(z ) where 1 i r,
ij ∈ r,s ij ij ≤ ≤
1 j s. SI is a polynomial of degree 2 in each Re(z ), Im(z ). In case of
≤ ≤ r,s ij ij
ΩII,itisknownthatdet(I ZZ∗)=sII(Z)2 forsomepolynomialsII(Z)(cf.[17]).
Dnefine SII :ΩII R andrS−III :ΩIII n R by n
n n → n n →
SII(Z)=sII(Z) for Z ASMC ,
n n ∈ n,n
SIII(Z)=det(I ZZ∗) for Z SMC
n n− ∈ n,n
and
SIV(Z)=1 2ZZ∗+ ZZt 2 for Z Cn
n − ∈
SI (Z), SII(Z), SIII(Z) and SIV(Z) are(cid:12)(cid:12)calle(cid:12)(cid:12)d the generic norm of the cor-
r,s n n n
responding domains cf.[17]. Then SII(Z) is a polynomial of degree 2 in each
n
Rez ,Imz for 1 i < j n and SIII(Z) is a polynomial of degree 4 in each
ij ij ≤ ≤ n
Rez ,Imz for 1 i<j n and of degree 2 in each Rez ,Imz for 1 i n.
ij ij ii ii
≤ ≤ ≤ ≤
Lemma 3.1.
(1) For Z =(z ) MC , the coefficient of (Rez )2 in SI (Z) is
ij ∈ r,s ij r,s
det(I Z′Z′∗)
r−1
− −
where Z′ is the (i,j) minor of Z.
(2) For Z = (z ) ASMC , the coefficient of (Rez )4,1 i < j n in
ij ∈ n,n ij ≤ ≤
SII(Z) is
n
det(I Z′′Z′′∗)
n−2
−
where Z′′ is (n 2) (n 2) matrix obtained by removing i,j-th rows and
− × −
i,j-th columns in Z.
(3) For Z = (z ) SMC , the coefficient of (Rez )4,1 i < j n and
ij ∈ n,n ij ≤ ≤
(Rez )2 for 1 i n in SIII(Z) are
ij ≤ ≤ n
det(I Z′′Z′′∗) and det(I Z′Z′∗)
n−2 n−1
− − −
respectively where Z′′ is (n 2) (n 2) matrix removing i,j-th rows and
− × −
i,j-th columns in Z and Z′ is the (i,i) minor of Z.
6 AeryeongSeo
Proof.
X
(1). For Z MC , denote Z = with X = (X′,X′′) = (x ) MC
∈ r,s (cid:18) Y (cid:19) ij ∈ r−1,s
whereX′ MC , X′′ MC andY =(y ,...,y ) MC . Weonlyconsider
∈ r−1,1 ∈ r−1,s−1 1 s ∈ 1,s
the coefficient of (Rey )2. The coefficient of (Rey )2 in SI (Z) is ∂2 SI (Z).
1 1 r,s ∂y1∂y1 r,s
Since
∂ ∂ I XX∗ XY∗
(3.1) ∂y det(Ir −ZZ∗)= ∂y det(cid:18) r−1Y−X∗ 1− YY∗ (cid:19)
1 1 − −
and YX∗ =(y x + +y x , ... ,y x + +y x ), we obtain
1 11 s 1s 1 (r−1)1 s (r−1)s
··· ···
∂ I XX∗ XY∗
∂y1 det(Ir−ZZ∗)=det(cid:18) (−x11, −rx−211−,...,−x(r−1)1) −−y1 (cid:19)
and
∂2
(3.2) det(I ZZ∗)
r
∂y ∂y −
1 1
x
11
−
 I XX∗  ..  
(3.3) = det r−1− .
  −x(r−1)1  
 ( x , x ,..., x ) 1 
 − 11 − 21 − (r−1)1 − 
I X′′X′′∗ 0
= det r−1−
(cid:18) ( x11, x21,..., x(r−1)1) 1 (cid:19)
− − − −
(3.4) = det(I X′′X′′∗).
r−1
− −
Thesecondequationcomesfromsubtractingj-throwof (3.3)byx timesther-th
j1
row of (3.3).
(2). We will only consider the coefficient of (Rez )4. By Lemma 3.1(1), for
12
W =(w ) MC , det(I WW∗)=a (Rew )2+a (Rew )+a where a are
ij ∈ n,n n− 2 12 1 12 0 i
polynomials in Rew ,Imw for i = 1, j = 2 and Imw . Since the coefficient of
ij ij 12
6 6
(Rez )4 is the coefficient of (Rew )2(Rew )2 substituted w = z for 1 i <
12 12 21 ij ij
≤
j < n, w = z for 1 i < j n and w = 0 for 1 i n, the coefficient of
ji ij ii
− ≤ ≤ ≤ ≤
(Rez )4 is det(I Z′′Z′′∗) where Z′′ =(z ) .
12 n−2 ij 3≤i≤n,3≤j≤n
(3). We can obtain−the result by similar method in (1) and (2). (cid:3)
Proposition 3.2. SI (Z), SII(Z), SIII(Z) and SIV(Z) are irreducible.
r,s n n n
Proof. In case of SI : At first, we will prove that SI (Z) is irreducible. Note
r,s r,s
that r s. We use induction with respect to k on SI . For Z MC ,
≤ k,s−r+k ∈ 1,s−r+1
1 ZZ∗ is irreducible. Suppose that SI (Z) is irreducible and SI (Z)=AB
− r−1,s−1 r,s
X
for some polynomial A and B. Denote Z = as in the proof of Lemma 3.1.
(cid:18) Y (cid:19)
The degree of Rey is 2.
1
Step 1 : Supposethatthereisanonzeromonomialof(Rey )2 inA. ThenSI (Z)
1 r,s
should be of the form
SI (Z)=(µ(Rey )2+σRey +ν)B
r,s 1 1
where µ,σ,ν and B are polynomials without Rey variable. Without loss of gen-
1
erality, we may assume that B is not a constant. Then by Lemma 3.1, µB is
det(I X′′X′′∗) which is irreducible by the induction hypothesis. Hence B
r−1
−
properholomorphicpolynomialmapsbetweenBSDsofclassicaltype 7
should be det(I X′′X′′∗) (up to constants). This implies that it consists of
r−1
−
the monomialsofvariableX′′. TheneveryvariableofX′ shouldbe inA. Consider
the coefficientof (Rex )2 whichis irreducible anddoes notcontainRex , Imx
11 1j 1j
variables. Howeverthe coefficient of (Rex )2 should containB and this induces a
11
contradiction. Therefore there is no second order term in each A and B.
Step 2 : Suppose that SI (Z)=(µRey +σ)(νRey +ρ) where µ,σ,ν,ρ has no
r,s 1 1
Rey variable. Thenµν isirreduciblebyLemma3.1andhenceµorν isaconstant.
1
If µ is a constant, ν = det(I X′′X′′∗) up to constant. Note that ν contains
r−1
−
secondordertermsofvariablesinX′′. Since thereisno Rey terminρ, νRey +ρ
1 1
shouldhavesecondorderterm. ButbyStep1,νRey +ρcannothavesecondorder
1
term. Hence SI (Z) is irreducible.
r,s
In case of SII, SIII, SIV: Since for n = 1, SIII(z) = 1 z 2 for z ∆
n n n 1 −| | ∈
which is irreducible, use the same method (induction) as in the proof of the case
SI considering factorization with respect to Rez variable. In case of SIV, we
r,s 11 n
can easily show that it is irreducible. We omit the proof.
0 a
WeonlyneedtoprovethatSII(Z)isirreducible. Forn=2,letZ = .
n (cid:18) a 0 (cid:19)
−
0 a b
Then SII(Z)=1 a2 which is irreducible. For n=3, let Z = a 0 c .
2 −| | −
b c 0
 − − 
Then SII(Z)=1 a2 b2 c2 which is also irreducible. Assume that SII(Z)
3 −| | −| | −| | 2n
and SII (Z) are irreducible. Since the even dimensional case is similar to the
2n−1
odd dimensional case, we will only consider the odd dimensional case. Since the
coefficientof(Rez )4 inSII (Z)2 isdet(I Z′′Z′′∗)asinLemma3.1(2),the
ij 2n+1 2n−1−
coefficient of (Rez )2 in SII (Z) is irreducible. Hence similar proof of the case
ij 2n+1
SI can be applied. (cid:3)
r,s
Let f : Ω Ω be a proper holomorphic polynomial map where Ω , Ω are
1 2 1 2
→
irreducible bounded symmetric domains of classical type. Let S , S be the corre-
1 2
sponding generic norms. Since f is proper, by Proposition 3.2,
S (f(Z))=0 whenever S (Z)=0.
2 1
(Note that if Z ∂ΩIV, SIV(Z)= 0 since if ZZ∗ = 1, SIV(Z)< 0. Hence we do
∈ n n n
not need to consider ZZ∗ 1 term in the definition of ΩIV.) Hence there is a real
− n
analytic map F such that
f
(3.5) S (f(Z))=S (Z)F (Z).
2 1 f
We can polarize this equation by
(3.6) S (f(Z),f(W))=S (Z,W)F (Z,W).
2 1 f
Example 3.3. (1) Let f : Ωr,s Ωr′,s′ be a proper holomorphic polynomial
→
map. Then (3.6) is expressed by
det(Ir′ f(Z)f(W)∗)=det(Ir ZW∗)Ff(Z,W).
− −
(2) Let f :ΩIV ΩIV be a proper holomorphic polynomial map. Then (3.6)
n → N
is expressed by
1 2f(Z)f(W)∗+ f(Z)f(Z)t (f(W)f(W)t)
−
(cid:0) (cid:1)
= 1 2ZW∗+ ZZt (WWt) F (Z,W).
f
(cid:16) − (cid:0) (cid:1) (cid:17)
8 AeryeongSeo
Lemma 3.4. Let g :Ω Ω and f :Ω Ω be proper holomorphic polynomial
1 2 2 3
→ →
maps. Then
F (Z,W)=F (Z,W)F (g(Z),g(W)).
f◦g g f
Proof.
S (Z,W)F (Z,W) = S (f g(Z),f g(W))
1 f◦g 3
◦ ◦
= S (g(Z),g(W))F (g(Z),g(W))
2 f
= S (Z,W)F (Z,W)F (g(Z),g(W))
1 g f
(cid:3)
A B
Lemma 3.5. Let U = be an automorphism of a domain Ω with ap-
(cid:18) C D (cid:19)
propriate block matrices A,B,C,D. Then F is given as the following:
U
(1) If Ω=ΩI or ΩIII,
r,s n
1
F (Z,W)= .
U
det(A+ZC)det(A+WC)
(2) If Ω=ΩII,
n
1
F (Z,W)= where f (Z,W)2 =det(A+ZC)det(A+WC).
U U
f (z,w)
U
(3) If Ω=ΩIV,
n
1 1
F (Z,W)=
U
−2 ( 2iZB+Z′D)(i,1)t ( 2iWB+W′D)(i,1)t
{ − }{ − }
Proof. (1 and 2) Since by (2.1),
U(Z)U(W)∗ =(A+ZC)−1(B+ZD) (A+WC)−1(B+WD) ∗
{ }
= (A+ZC)−1((A+ZC)(A+WC)∗+ZZ∗ I ) (A+WC)−1 ∗
r
−
= I (A+ZC)−1(I ZZ∗) (A+WC)−1 ∗, (cid:0) (cid:1)
r r
− −
(cid:0) (cid:1)
det(I ZW∗)
(3.7) det(I U(Z)U(W)∗)= r − .
r
− det(A+ZC)det(A+WC)
(3)NotethatforZ =(z ,...,z ),SIV(Z)= 1Q(ξ˜IV(Z))whereQ(x ,...,x )=
1 n n −2 1 n+2
x 2+... x 2 x 2 x 2 andξ˜IV(Z)=( 2iZ,1+ZZt,i(1 ZZt)). Then
1 n n+1 n+2
| | | | −| | −| | − −
SIV(g (Z)) = 1Q ξ˜IV(ξIV−1(ξIV(Z)M)
n M −2 (cid:16) (cid:17)
1 2iZA+Z′C 2iZB+Z′D
= Q − , −
−2 (cid:18)( 2iZB+Z′D)(i,1)t ( 2iZB+Z′D)(i,1)t(cid:19)
− −
1 SIV(Z)
= n
−2 ( 2iZB+Z′D)(i,1)t 2
| − |
(cid:3)
Remark 3.6. IncaseofΩII,sinceforZ ASMC ,CtA+CtZC isanti-symmetric.
n ∈ n,n
This implies that
det(CtA+CtZC) (Pf(CtA+CtZC))2
det(A+ZC)= =
det(C) det(C)
properholomorphicpolynomialmapsbetweenBSDsofclassicaltype 9
where Pf(Y) is the Pfaffian polynomial of a matrix Y and hence F is a rational
U
function.
V V
Proof of Theorem 1.1. Suppose that g U = V f for some V = 1 2
◦ ◦ (cid:18) V3 V4 (cid:19) ∈
U U
Aut(Ω ) and U = 1 2 Aut(Ω ). Then since F = F , by Lemma
2 (cid:18) U3 U4 (cid:19) ∈ 1 g◦U V◦f
3.4,
F (U(Z),U(W))F (Z,W)=F (f(Z),f(W))F (Z,W).
g U V f
By multiplying S1(Z,W) to both side, we obtain
(3.8) S2(g U(Z),g U(W))=F (f(Z),f(W))S1(f(Z),f(W)).
V
◦ ◦
For simplicity, we only consider that the case Ω and Ω are bounded symmetric
1 2
domains of type I. (3.8) is
det(Ir′ f(Z)f(W)∗)=det(Ir′ g U(Z)(g U(W))∗)
(3.9) − − ◦ ◦
det(V +f(Z)V )det(V +f(W)V ).
1 3 1 3
Put W =0 in (3.9). Then
(3.10) 1=det(Ir′ g U(Z)(g U(0))∗) det(V1+f(Z)V3)det(V1).
− ◦ ◦
IfU =0,thenby g(U−1ZU )=V f(Z),weobtain0=g(0)=V f(0)=V−1V
3 1 4 ◦ ◦ 1 2
and hence V =0. So assume U =0. Note that in this case, det(U +ZU ) is not
2 3 1 3
6
a constant. Suppose that g U(0)=0. Then det(Ir′ g U(Z)(g U(0))∗) is not
◦ 6 − ◦ ◦
a constant and hence it should be of the form p/q where p and q are non-constant
polynomial without common factors and q =det(U +ZU )l for some l. But since
1 3
productofp/qandpolynomialcannotbeaconstant,(3.10)inducesacontradiction.
Henceg U(0)shouldbezero. Thisimpliesthat0=g(U−1U )=V f(0)=V(0)=
V−1V .◦Hence V =0 (and also V =0) and hence 1=d1et(V2 +f(◦Z)V ). Put this
1 2 2 3 1 3
in (3.9).
(3.11) det(Ir′ f(Z)f(W)∗)=det(Ir′ g U(Z)(g U(W))∗).
− − ◦ ◦
Since right side of (3.11) is singular on Z Ω :det(U +ZU )=0 , U should
r,s 1 3 3
be zero. { ∈ } (cid:3)
4. Application
Inthissection,wesuggestexampleswhichare1-parameterfamilyofinequivalent
proper holomorphic maps between bounded symmetric domains of classical type.
We use Theorem 1.1 to prove that proper holomorphic maps f : Ω Ω are
t 1 2
→
inequivalent for each 0 t 1. Every example in this section is obtained in [19].
≤ ≤
As in [6], we define the following equivalence relation:
Let Ω ,Ω be domains. Consider a continuous map H : Ω [0,1] Ω . Denote
1 2 1 2
× →
H (z) = H(z,t). Suppose that H : Ω Ω is holomorphic for each t [0,1].
t t 1 2
→ ∈
Then we will say H is a continuous family of holomorphic maps from Ω to Ω .
t 1 2
Definition 4.1. Let f : Ω Ω and g : Ω Ω be proper holomorphic maps.
1 2 1 3
→ →
Then f and g are homotopic in the target domain Ω if for each t [0,1], there is a
∈
proper holomorphic maps H :Ω Ω such that
t 1
→
therearetotallygeodesicembeddingse :Ω Ωfork =2,3,withrespect
k k
• →
to their Bergman metrics,
H =e f and H =e g,
0 2 1 3
• ◦ ◦
10 AeryeongSeo
H is a continuous family of holomorphic maps from Ω to Ω.
t 1
•
4.1. 1-parameter family of proper holomorphic maps among ΩI .
r,s
Consider proper holomorphic maps f,g :ΩI ΩI which are defined by
2,2 → 3,3
z2 z z z
z z 1 1 2 2 z z
f(cid:18)(cid:18) z31 z42 (cid:19)(cid:19)= zz1z3 zz2z3 z04 , for (cid:18) z13 z24 (cid:19)∈ΩI2,2
3 4
 
z2 √2z z z2
z z 1 1 2 2
g(cid:18)(cid:18) z31 z42 (cid:19)(cid:19)= √2zz21z3 z1√z42+z zz2z3 √2zz22z4 
 3 3 4 4 
and f :ΩI ΩI be proper holomorphic maps for 0 t 1 defined by
t 2,2 → 4,4 ≤ ≤
(4.1)
z2 √2 tz z √1 tz2 √tz
1 − 1 2 − 2 2
 √2 tz z 2(1−t)z z +z z 2 1−tz z t z 
z1 z2 − 1 3 2−t 1 4 2 3 q2−t 2 4 q2−t 4 .
(cid:18) z3 z4 (cid:19)7→ √1−tz32 2q12−−ttz3z4 z42 0 
 √tz3 2−ttz4 0 0 
 q 
Then it is easily observed that f and g are homotopic in the target domain ΩI
4,4
through f .
t
Corollary 4.2. f are inequivalent for different t, 0 t 1.
t
≤ ≤
Proof. Supposethatf A=B f forsomeA U(2,2)andB U(4,4). Without
t s
◦ ◦ ∈ ∈
loss of generality, we may assume that t = 0. Then by Theorem 1.1, f (UZV) =
t
6
U U V V
Lf (Z)M for some U = 1 2 U(2), V = 1 2 U(2) and
s (cid:18) U3 U4 (cid:19) ∈ (cid:18) V3 V4 (cid:19) ∈
L = (L ), M = (M ) U(4). Denote f = f where f is a homogeneous
ij ij t t,j t,j
∈
polynomial of degree j. Then ft,j(UZV)=LfPs,j(Z)M for each j. Consider linear
part
0 0 0 √tz
2
 0 0 0 t z 
2−t 4
ft,1 = 0 0 0 q 0 .
 
 √tz t z 0 0 
 3 2−t 4 
 q 
1 0
At first, at Z = ,
1 (cid:18) 0 0 (cid:19)
0 0 0 √tU V
1 2
ft,1(UZ1V)=ft,1(cid:18) UU31VV11 UU31VV22 (cid:19)= 00 00 00 q2−t0tU3V2 
 √tU V t U V 0 0 
 3 1 2−t 3 2 
 q 
andLf (Z )M =0. ThisimpliesthatU V =U V =U V =U V =0. Suppose
s,1 1 1 2 3 2 3 1 3 2
thatU =0. ThenV =V =0. ThisisacontradictiontoV U(2). HenceU =0.
3 1 2 3
6 ∈
This implies that U =0 and U =0, hence V =V =0.
2 1 2 3
6