Proper holomorphic mappings between complex ellipsoids and generalized Hartogs triangles

0.15 MB·English

by  Pawel Zapalowski

PROPER HOLOMORPHIC MAPPINGS BETWEEN COMPLEX
ELLIPSOIDS AND GENERALIZED HARTOGS TRIANGLES
PAWEL ZAPAL OWSKI
3
1 Abstract. Theexplicitformofproperholomorphicmappingsbetweencom-
0 plexellipsoidsisgiven. Usingthisdescription,wecharacterizetheexistenceof
2 properholomorphicmappingsbetweengeneralizedHartogstrianglesandgive
n theirexplicitform. Inparticular,theautomorphismgroupofsuchdomainsis
a found.
J For any bounded domains D,G Cn let Prop(D,G) denote the set of proper
⊂
8 holomorphicmappingsF :D G,whereproper,asusual,meansF−1(K)compact
→
1 in D for every compact K G. In this context, the compactness of F−1(K) for
⊂
every compact K G is equivalent to the following condition: for any sequence
] ⊂
V (zν)ν∈N D that has no limit point in D the sequence (F(zν))ν∈N G has
⊂ ⊂
no limit point in G. Let Aut(D) denote the automorphism group of D, i.e. the
C
set of all biholomorphic self-mappings F : D D. Moreover, we shall write
h. Prop(D):=Prop(D,D). →
t
a Our aim is to characterize the sets Prop(D,G) and Aut(D) when D,G belong
m either to the class of the complex ellipsoids or the so-called generalized Hartogs
[ triangles.

Here is some notation. Let Σn denote the group of the permutations of the set
2
1,...,n . For σ Σ , z = (z ,...,z ) Cn denote z := (z ,...,z ) and
v { } ∈ n 1 n ∈ σ σ(1) σ(n)
6 Σn(z):= σ Σn :zσ =z . We shall also write σ(z):=zσ.
{ ∈ }
8 For α=(α1,...,αn)∈Rn>0 put
7
Ψ (z):=zα :=(zα1,...,zαn), z =(z ,...,z ) Cn.
0 α 1 n 1 n ∈
1. If, moreover, β = (β1,...,βn) ∈ Rn>0 we shall write αβ := (α1β1,...,αnβn) and
1 1/β :=(1/β ,...,1/β ).
1 n
2 For z C, A C let zA:= za:a A .
1 Finally∈, let U(⊂n) denote the{set of u∈nita}ry mappings U :Cn Cn.
: →
v
i 1. Complex ellipsoids
X
ar For p=(p1,...,pn)∈Rn>0, n≥2, define the complex ellipsoid
n
E := (z ,...,z ) Cn : z 2pj <1 .
p  1 n ∈ | j| 
 Xj=1 
NotethatB :=E istheunitballinCn. WeshallwriteD:=B ,T:=∂D.
n (1,...,1)  1
Moreover,if α/β Nn then Ψ Prop(E ,E ).
α/β α β
∈ ∈
The problem of characterization of Prop(E ,E ) and Aut(E ) has been investi-
p q p
gated in [9] and [6]. The question on non-emptiness of Prop(E ,E ) as well as the
p q
form of Prop(E ,E ) and Aut(E ) in the case p,q Nn was completely solved in
p q p
∈
[9]. The case p,q Rn was considered in [6], where the Authors characterized
∈ >0
non-emptiness of Prop(E ,E ) and found Aut(E ). They did not give, however,
p q p
Key words and phrases. complex ellipsoid, generalized Hartogs triangle, proper holomorphic
mapping.
1
2 PAWEL ZAPAL OWSKI
the explicit form of an F Prop(E ,E ). The previous results are formulated in
p q
∈
the following
Theorem 1 (cf. [9], [6]). Assume that n 2, p,q Rn .
≥ ∈ >0
(a) The following conditions are equivalent
(i) Prop(E ,E )=∅;
p q
6
(ii) there exists σ Σ such that p /q Nn.
n σ
∈ ∈
(b) If p,q Nn, then the following conditions are equivalent
∈
(i) F Prop(E ,E );
p q
∈
(ii) F =φ Ψ σ, where σ Σ is such that p /q Nn and φ Aut(E ).
◦ pσ/q◦ ∈ n σ ∈ ∈ q
In particular, Prop(E )=Aut(E ).
p p
(c) If 0 k n, p 1 k (R 1 )n−k, z = (z′,z ,...,z ), then the
>0 k+1 n
≤ ≤ ∈ { } × \ { }
following conditions are equivalent
(i) F =(F ,...,F ) Aut(E ),
1 n p
∈
H (z′), if j k
j
≤
(ii) Fj(z) := ζ z √1−ka′k2 1/pσ(j), if j >k, where ζj ∈ T, j > k,
 j σ(j) 1−hz′,a′i
(cid:18) (cid:19)
H =(H ,...,H ) Aut(B ), a′ :=H−1(0), and σ Σ (p).
1 k ∈ k ∈ n
In the general case thesis of Theorem 1 (b) is no longer true (take, for instance,
Ψ H Ψ Prop(E ,E ), where H Aut(B ), H(0)=0).
(2,2) (2,2) (2,2) (1/2,1/2) 2
◦ ◦ ∈ ∈ 6
Nevertheless, from the proof of Theorem 1.1 in [6] we easily derive the follow-
ing theorem which will be of great importance during the investigation of proper
holomorphic mappings between generalized Hartogs triangles.
Theorem 2. Assume that n 2, p,q Rn . Then the following conditions are
≥ ∈ >0
equivalent
(i) F Prop(E ,E );
p q
∈
(ii) F = Ψ φ Ψ σ, where σ Σ is such that p /q Nn, r Nn is
pσ/(qr) ◦ ◦ r ◦ ∈ n σ ∈ ∈
such that p /(qr) Nn, and φ Aut(E ).
σ ∈ ∈ pσ/r
In particular, Prop(E )=Aut(E ).
p p
2. Generalized Hartogs triangles
Let n,m N. For p = (p ,...,p ) Rn and q = (q ,...,q ) Rm , define
∈ 1 n ∈ >0 1 m ∈ >0
the generalized Hartogs triangle
n m
F := (z ,...,z ,w ,...,w ) Cn+m : z 2pj < w 2qj <1 .
p,q  1 n 1 m ∈ | j| | j| 
 Xj=1 Xj=1 
Note that F is nonsmooth pseudoconvex Reinhardt domain, not containing the
p,q 
origin. Moreover,if n=m=1, then F is the standard Hartogs triangle.
1,1
The problem of characterization of Prop(F ,F ) and Aut(F ) has been in-
p,q p˜,q˜ p,q
vestigated in many papers. The necessary and sufficient conditions for the non-
emptiness of Prop(F ,F ) are given in [3] for p,p˜ Nn, q,q˜ Nm, n,m 2, in
p,q p˜,q˜
∈ ∈ ≥
[4] for p,p˜ Rn , q,q˜ Rm , n,m 2, andin [10] for p,p˜ Nn, q,q˜ Nm, m=1.
∈ >0 ∈ >0 ≥ ∈ ∈
The explicit form of an F Prop(F ,F ) is presented in [10] for p,p˜ Nn,
p,q p˜,q˜
∈ ∈
q,q˜ Nm, m = 1, whereas the description of Aut(F ) may be found in [5] for
p,q
∈
p Nn, q Nm, n,m 2, and in [10] for p Nn, q Nm, m=1.
∈ ∈ ≥ ∈ ∈
In the paper we shall only consider the case n=1.
First we deal with the case m=1.
Theorem 3. Assume that n=m=1, p,q,p˜,q˜ R .
>0
∈
(a) The following conditions are equivalent
PROPER MAPPINGS BETWEEN ELLIPSOIDS AND HARTOGS TRIANGLES 3
(i) Prop(F ,F )=∅;
p,q p˜,q˜
6
(ii) there exist k,l N such that lq˜/p˜ kq/p Z.
∈ − ∈
(b) The following conditions are equivalent
(i) F Prop(F ,F );
p,q p˜,q˜
∈
ζzkwlq˜/p˜−kq/p,ξwl , if q/p / Q
(ii) F(z,w)= ∈ ,
((cid:0)ζzk′wlq˜/p˜−k′q/pB (cid:1)zp′w−q′ ,ξwl , if q/p Q
∈
where ζ,ξ (cid:16)T, k,l N, k′ (cid:16)N 0 (cid:17)are su(cid:17)ch that lq˜/p˜ kq/p Z,
∈ ∈ ∈ ∪{ } − ∈
lq˜/p˜ k′q/p Z, p′,q′ Z are relatively prime with p/q = p′/q′, and B
− ∈ ∈
is a finite Blaschke product nonvanishing at 0 (it may happen that B 1,
≡
but then k′ >0).
In particular, Prop(F ))Aut(F ).
p,q p,q
(c) The following conditions are equivalent
(i) F Aut(F );
p,q
∈
(ii) F(z,w) = wq/pφ(zw−q/p),ξw , where ξ T, and φ Aut(D) with
∈ ∈
φ(0)=0 whenever q/p / N.
(cid:0) ∈ (cid:1)
Remark 4. The counterpart of the Theorem 3 for p,q,p˜,q˜ N was proved (with
∈
minor mistakes) in [10], where it was claimed that F Prop(F ,F ) iff
p,q p˜,q˜
∈
ζzkwlq˜/p˜−kq/p,ξwl , if q/p / N, lq˜/p˜ kq/p Z
(1) F(z,w)= ∈ − ∈ ,
((cid:0)ζwlq˜/p˜B zw−q/p ,(cid:1)ξwl , if q/p∈N, lq˜/p˜∈N
where ζ,ξ T, k,(cid:0)l N, an(cid:0)d B is (cid:1)a finit(cid:1)e Blaschke product. Nevertheless, the
∈ ∈
mapping
F (z,w) z3w3B(z2w−3),w3 F ,
2,3 2,5
∋ 7→ ∈
where B is nonconstant finite Blaschke product nonvanishing at 0, is proper holo-
(cid:0) (cid:1)
morphicbutnotoftheform(1). Infact,fromtheTheorem3(b)itfollowsimmedi-
atelythatforanychoiceofp,q,p˜,q˜ NonemayfindmappingF Prop(F ,F )
p,q p˜,q˜
∈ ∈
having, as a factor of the first component, nonconstant Blaschke product nonvan-
ishing at 0.
Our result gives a negative answer to the question posed by the Authors in [8],
whether the structure of Prop(F ,F ) remains unchanged when passing from
p,q p˜,q˜
p,q,p˜,q˜ N to arbitrary p,q,p˜,q˜ R . It should be mentioned that, on the
>0
∈ ∈
otherhand,theautomorphismgroupAut(F )doesnotchangewhenpassingfrom
p,q
p,q N to p,q R .
>0
∈ ∈
In the proof of Theorem 3, however, neither the method from [4] (where the
assumption m 2 is essential) nor the method from [10] (where the assumption
≥
p,q,p˜,q˜ N is essential) can be used. Fortunately, it turns out that one may get
∈
Theorem3usingpartofthemainresultfrom[7],wherecompletecharacterizationof
nonelementaryproperholomorphicmappingsbetweenboundedReinhardtdomains
in C2 is given.
The remaining case m 2 is considered in the following result.
≥
Theorem 5. Assume that n=1, m 2, p,p˜ R , q,q˜ Rm , (z,w) C Cm.
≥ ∈ >0 ∈ >0 ∈ ×
(a) The following conditions are equivalent
(i) Prop(F ,F )=∅;
p,q p˜,q˜
6
(ii) p/p˜ N and there exists σ Σ such that q /q˜ Nm.
m σ
∈ ∈ ∈
(b) The following conditions are equivalent
(i) F Prop(F ,F );
p,q p˜,q˜
∈
(ii) F(z,w)=(ζzk,h(w)), where ζ T, k N, h Prop(E ,E ), h(0)=0.
q q˜
∈ ∈ ∈
In particular, Prop(F )=Aut(F ).
p,q p,q
(c) The following conditions are equivalent
(i) F Aut(F );
p,q
∈
4 PAWEL ZAPAL OWSKI
(ii) F(z,w)=(ζz,h(w)), where ζ T, h Aut(E ), h(0)=0.
q
∈ ∈
Theorem 5 (a) was proved in [3] (for n,m 2, p,p˜ Nn, q,q˜ Nm) and in
≥ ∈ ∈
[4] (for n,m 2, p,p˜ Rn , q,q˜ Rm ). Theorem 5 (b) was proved in [5] for
≥ ∈ >0 ∈ >0
n,m 2, p = p˜ Nn, q = q˜ Nm. Theorem 5 (c) was proved in [5] for n,m 2,
≥ ∈ ∈ ≥
p Nn, q Nm. Part(c) of Theorem5 givesan affirmativeanswerto the question
∈ ∈
posed by the Authors in [8], whether the description of the automorphism group
(c) remains true for arbitrary p Rn , q Rm (at least in the case n=1).
∈ >0 ∈ >0
Remark 6. Using Barrett’s and Bell’s results (cf. [1], [2]) one may show (cf. [3])
that any F Prop(F ,F ) extends holomorphically past any boundary point
p,q p˜,q˜
∈
(z,w) ∂F (0,0) .
p,q
∈ \{ }
Let
n m
K =K := (z,w) Cn+m :0< z 2pj = w 2qj <1 ,
p,q  ∈ | j| | j| 
 Xj=1 Xj=1 
n m
 
L=L := (z,w) Cn+m : z 2pj < w 2qj =1 .
p,q  ∈ | j| | j| 
 Xj=1 Xj=1 
Analogically we define K˜ :=K and L˜ :=L .
 p˜,q˜ p˜,q˜ 
For m>1 it is shown in [4] that
F(K) K˜, F(L) L˜.
⊂ ⊂
for any F Prop(F ,F ).
p,q p˜,q˜
∈
3. Proofs
Proof of Theorem 2. The implication (ii) (i) is obvious.
⇒
To prove (i) (ii) let F =(F ,...,F ) Prop(E ,E ).
1 n p q
⇒ ∈
Following [11] any automorphism H =(H ,...,H ) Aut(B ) is of the form
1 n n
∈
1 a 2 n
H (z)= −k k h (z a ), z =(z ,...,z ) B , j =1,...,n,
j j,k k k 1 n n
1 z,a − ∈
p−h i k=1
X
where a=(a ,...,a ) B and Q=[h ] is an n n matrix such that
1 n n j,k
∈ ×
Q¯(I a¯ta)tQ=I ,
n n
−
whereI istheunitn nmatrix,whereasA¯(resp.tA)istheconjugate(resp.trans-
n
×
pose) of an arbitrary matrix A. In particular, Q is unitary if a=0.
Itfollowsfrom[6]thatthereexistsσ Σ suchthatp /q Nn, h =0,and
n σ j,σ(j)
∈ ∈ 6
1 a 2 1/qj
(2) F (z)= −k k h zpσ(j)
j 1 zp,a j,σ(j) σ(j)
p−h i !
whenever 1/q / N.
j
∈
If 1/q N then F either is of the form (2), where p /q N, or
j j σ(j) j
∈ ∈
1 a 2 n 1/qj
F (z)= −k k h (zpk a )
j p1−hzp,ai k=1 j,k k − k !
X
where p N for any k such that h =0.
k j,k
∈ 6
Consequently, if we define r=(r ,...,r ) as
1 n
p , if a =0 or there is k =σ(j) with h =0
r := σ(j) σ(j) 6 6 j,k 6 ,
j
(pσ(j)/qj, otherwise
then it is easy to see that r Nn, p /(qr) Nn, and F is as desired. (cid:3)
σ
∈ ∈
PROPER MAPPINGS BETWEEN ELLIPSOIDS AND HARTOGS TRIANGLES 5
Remark 7. Note that in the case p,q Nn we have 1/q N iff q = 1. Hence
j j
∈ ∈
the above definition of r implies that r = p /q and, consequently, we get thesis of
σ
Theorem 1 (b).
Proof of Theorem 3. Observe,that(a)and(c)followsimmediatelyfrom(b). Thus,
it suffices to prove part (b).
The implication (ii) (i) in part (b) holds for any p,q,p˜,q˜> 0. Indeed, if F =
⇒
(G,H) is of the form given in (ii), then
z w−q/p kp˜, if q/p / Q
|G(z,w)|p˜|H(z,w)|−q˜=(cid:0)|z||w|−q/p(cid:1)k′p˜ B(zp′w−q′) p˜, if q/p∈Q.
 | || | ∈
(cid:12) (cid:12)
To prove the implication (i)⇒(cid:0)(ii) in (b)(cid:1), let(cid:12)(cid:12)F ∈Prop(F(cid:12)(cid:12)p,q,Fp˜,q˜).
Assume first that F is elementary algebraic mapping, i.e. it is of the form
F(z,w)= αzawb,βzcwd ,
where a,b,c,d Z are such that ad (cid:0) bc = 0 and α(cid:1),β C are some constants.
∈ − 6 ∈
Since F is surjective, we infer that c=0, d N, and ξ :=β T. Moreover,
∈ ∈
(3) αp˜z ap˜wbp˜−dq˜<1,
| | | | | |
whence a N, bp˜ dq˜ N, and ζ := α T. Let k := a, l := d. One may rewrite
∈ − ∈ ∈
(3) as
z p w−q kp˜/p wbp˜−lq˜+kqp˜/p <1.
| | | | | |
Since one may take sequ(cid:0)ence (zν,1(cid:1)/2)ν∈N ⊂ Fp,q with |zν|p2q → 1 as ν → ∞, we
infer that bp˜ lq˜+kqp˜/p=0, i.e.
−
lq˜ kq
b= .
p˜ − p
Consequently, F is as in the condition (ii) of the Theorem 3 (b).
Assume now that F is not elementary. Then from the Theorem 0.1 in [7] it
follows that F is of the form
F(z,w)= αzawbB˜ zp′w−q′ ,βwl ,
where a,b Z, a 0, p′,q′,l N(cid:16), p′,q′ are(cid:16)relativel(cid:17)y prim(cid:17)e,
∈ ≥ ∈
q′ q q˜ aq′+bp′
(4) = , = ,
p′ p p˜ lp′
α,β C are some constants, and B˜ is a nonconstant finite Blaschke product non-
∈
vanishing at the origin.
FromthesurjectivityofF weimmediatelyinferthatζ :=α Tandξ :=β T.
∈ ∈
If we put k′ :=a, then (4) implies
lq˜ k′q
b= ,
p˜ − p
which ends the proof. (cid:3)
Proof of Theorem 5. We shall write w = (w ,...,w ) Cm. We may assume
1 m
∈
without lossofgeneralitythatthere is 0 µ m with q˜ 1 µ (R 1 )m−µ.
>0
≤ ≤ ∈{ } × \{ }
Let
F =(G,H):F F C Cm
p,q p˜,q˜
→ ⊂ ×
be proper holomorphic mapping. It follows from [4] that F(L) L˜ and H is inde-
⊂
pendent of the variable z. Hence h:= H(0, ) Prop((E ) ,(E ) ). Consequently,
q ∗ q˜ ∗
· ∈
by Hartogs theorem h Prop(E ,E ), i.e. (Theorem 2)
q q˜
∈
h=Ψ ψ Ψ σ
qσ/(q˜r)◦ ◦ r◦
6 PAWEL ZAPAL OWSKI
forsome σ Σ with q /q˜ Nm, r Nm withq /(q˜r) Nm, andψ Aut(E )
∈ m σ ∈ ∈ σ ∈ ∈ qσ/r
with ψ(0)=0. Indeed, if a=(a ,...,a ) is a zero of h we immediately get
1 m
m
G(z,a)=0, z 2p < a 2qj,
j
| | | |
j=1
X
which is clearly a contradiction, unless a=0.
Withoutlossofgeneralitywemayassumethatthereisµ l mwith1/q˜ / N
j
≤ ≤ ∈
iff j >l. It follows from the proof of Theorem 2 that
q 1, if j =1,...,l
σ(j)
= ,
rj (q˜j, if j =l+1,...,m
whence
ψ(w)=(U(w ,...,w ),ξ w ,...,ξ z ),
1 l l+1 l+τ(1) m l+τ(m−l)
where U =(U ,...,U ) U(l) and τ Σ (q˜ ,...,q˜ ). Finally,
1 l m−l l+1 m
∈ ∈
h(w)= U1/q˜1 wqσ(1),...,wqσ(l) ,...,U1/q˜l wqσ(1),...,wqσ(l) ,
1 σ(1) σ(l) l σ(1) σ(l)
(cid:16) (cid:16) (cid:17) ξ (cid:16)wqσ(l+1)/q˜l+1,...,ξ(cid:17) wqσ(m)/q˜m .
l+1 σ(l+1) m σ(m)
(cid:17)
In particular, if we write h=(h ,...,h ),
1 m
m m
(5) h (w)2q˜j = w 2qj, w E .
j j q
| | | | ∈
j=1 j=1
X X
For w∈Cm, 0<ρw := mj=1|wj|2qj <1 let
gP(z):=G(z,w), z ρ1/(2p)D.
∈ w
g may depend, a priori, on w. Since F(K) K˜, it follows from (5) that g
⊂ ∈
Prop ρ1/(2p)D,ρ1/(2p˜)D , i.e.
w w
(cid:16) (cid:17)
(6) g(z)=ρ1/(2p˜)B zρ−1/(2p) , z ρ1/(2p)D,
w w ∈ w
where B is a finite Blaschke produc(cid:16)t. Let (cid:17)
F0 :=F C 0 σ(1)−1 C 0 m−σ(1) ,
p,q p,q∩ ×{ } × ×{ }
F0 :=F (cid:16) C2 0 m−1 . (cid:17)
p˜,qσ/r p˜,qσ/r ∩ ×{ }
Let Φ∈Aut(Fp˜,qσ/r) be defined(cid:0)by (cid:1)
Φ(z,w):= z,U−1(w ,...,w ),w ,...,w
1 l l+1 m
and let (cid:0) (cid:1)
ξ , if l =0 q˜ , if l=0
ξˆ := 1 , qˆ := 1 .
1 1
(1, if l >0 (1, if l>0
Then Φ (G,ψ Ψ σ) Prop(F0 ,F0 ) with
◦ ◦ r◦ ∈ p,q p˜,qσ/r
(7) (Φ (G,ψ Ψ σ))(z,w)= G(z,w),ξˆwqσ(1)/qˆ1,0,...,0 , (z,w) F0 .
◦ ◦ r◦ 1 σ(1) ∈ p,q
It follows from Theorem 3 that(cid:16) (cid:17)
(8) (Φ (G,ψ Ψ σ))(z,w)= Gˆ(z,w),ηws ,0,...,0 , (z,w) F0 ,
◦ ◦ r◦ σ(1) ∈ p,q
where (cid:16) (cid:17)
ζzkwsqˆ1/p˜−kqσ(1)/p, if q /p / Q
Gˆ(z,w):=ζzk′wσs(qˆ11)/p˜−k′qσ(1)/pBˆ zp′w−qσ′(1) , if qσ(1)/p∈Q,
 σ(1) σ(1) σ(1) ∈
(cid:18) (cid:19)

PROPER MAPPINGS BETWEEN ELLIPSOIDS AND HARTOGS TRIANGLES 7
ζ,η T, k,s,p′,q′ N, k′ N 0 are such that p′,q′ are relatively prime,
∈ σ(1) ∈ ∈ ∪{ } σ(1)
q /p = q′ /p′, sqˆ /p˜ kq /p Z, q /p = q′ /p′, sqˆ /p˜ kq /p Z,
σ(1) σ(1) 1 − σ(1) ∈ σ(1) σ(1) 1 − σ(1) ∈
andBˆ is a finite Blaschkeproductnonvanishingat0 (if Bˆ 1then k′ >0). Hence
≡
(9) (Φ (G,ψ Ψ σ))(z,w)=
r
◦ ◦ ◦
Gˆ(z,w)+α(z,w),wqσ(1),...,wqσ(l),ξ wqσ(l+1)/q˜l+1,...,ξ wqσ(m)/q˜m ,
σ(1) σ(l) l+1 σ(l+1) m σ(m)
(cid:16) (cid:17)
for (z,w) Fp,q, wσ(1) = 0, where α is holomorphic on Fp,q with αF0 = 0.
∈ 6 | p,q
Comparing (7) and (8) we conclude that
η =ξˆ, s=q /qˆ .
1 σ(1) 1
Since the mapping on the left side of (9) is holomorphic on F , the function
p,q
ζzkwqσ(1)(1/p˜−k/p), if q /p / Q
(10) Gˆ(z,w)=ζzk′wσq(σ1()1)(1/p˜−k′/p)Bˆ zp′w−qσ′(1) , if qσ(1)/p∈Q
 σ(1) σ(1) σ(1) ∈
(cid:18) (cid:19)
with qσ(1)(1/p˜−k/p) ∈ Z and qσ(1)(1/p˜−k′/p) ∈ Z has to be holomorphic on
F , too. Since m 2, it may happen w = 0. Consequently, q (1/p˜
p,q σ(1) σ(1)
k/p) N 0 in the≥first case of (10), whereas Bˆ(t) = tk′′ for some k′′ N wit−h
∈ ∪{ } ∈
q (1/p˜ k′/p) k′′q′ N 0 in the second case. Thus
σ(1) − − σ(1) ∈ ∪{ }
Gˆ(z,w)=ζzkwqσ(1)(1/p˜−k/p),
σ(1)
where k N, q (1/p˜ k/p) N 0 (in the second case of (10) it suffices to
σ(1)
∈ − ∈ ∪{ }
take k :=k′+p′k′′).
Observe that Gˆ+α=G. Fix w 0 σ(1)−1 C 0 m−σ(1) with 0<ρ <1.
w
∈{ } × ×{ }
Then ρw = wσ(1) 2qσ(1) and Gˆ(,w)=g on ρ1w/(2p)D, i.e.
| | ·
ζzkwqσ(1)(1/p˜−k/p) = w qσ(1)/p˜B z w −qσ(1)/p , z w qσ(1)/pD.
σ(1) | σ(1)| | σ(1)| ∈| σ(1)|
(cid:16) (cid:17)
Hence B(t)=ζtk and q (1/p˜ k/p)=0, i.e. k =p/p˜. Hence part (a) is proved.
σ(1)
−
To finish part (b), note that g(z)= ζzp/p˜. Consequently, g does not depend on w
and
G(z,w)=ζzp/p˜, (z,w) F .
p,q
∈
Part (c) follows directly from (b). (cid:3)
References
[1] D.E.Barrett,Holomorphic equivalence and proper mapping of bounded Reinhardt domains
not containing the origin,Comment.Math.Helv.59(1984), 550–564.
[2] S. R. Bell, The Bergman kernel function and proper holomorphic mappings, Trans. Amer.
Math.Soc.270(1982), 685–691.
[3] Z. H. Chen, D. K. Xu, Proper holomorphic mappings between some nonsmooth domains,
Chin.Ann.Math.2(2001), 177–182.
[4] Z. H. Chen, Proper holomorphic mappings between some generalized Hartogs triangles, in
“GeometricFunctionTheoryinSeveralComplexVariables”,C.H.Fitzgerald,S.Gong(ed.),
74–81ProceedingsofaSatelliteConferencetotheInternationalCongressonMathematicians
inBeijing2002, WorldScientificPublishingCo.Pte.Ltd.,2004.
[5] Z.H.Chen,D.K.Xu,Rigidityof proper self-mapping onsome kindsof generalized Hartogs
triangle,ActaMath.Sin.(Engl.Ser.)18(2002), 357–362.
[6] G.Dini,A.SelvaggiPrimicerio,Proper holomorphic mappingsbetweengeneralizedpseudoel-
lipsoids, Ann.Mat.PuraAppl.158(1991), 219–229.
[7] A. V. Isaev, N. G. Kruzhilin, Proper holomorphic maps between Reinhardt domains in C2,
MichiganMath.J.54(2006), 34–63.
[8] M.Jarnicki, P.Pflug,First Steps in Several Complex Variables: Reinhardt Domains, Euro-
peanMathematicalSocietyPublishingHouse,2008.
8 PAWEL ZAPAL OWSKI
[9] M. Landucci, Proper equivalence for a class of pseudoconvex domains, Trans. Amer. Math.
Soc.282(1984), 807–811.
[10] M. Landucci, Proper holomorphic mappings between some nonsmooth domains, Ann. Mat.
PuraAppl.155(1989), 193–203.
[11] E. M. Stein, Boundary behaviour of holomorphic functions of several complex variables,
Math.NotesNo.11,Princeton,1972.
Faculty of Mathematics and Computer Science, Jagiellonian University, L ojasiewi-
cza 6,30-348Krako´w, Poland
E-mail address: [email protected]