Bull. Sci. math. 126 (2002) 343–367
An algebra of differential operators and generating
functions on the set of univalent functions
Hélène Airault a , Jiagang Ren b
a INSSET, Université de Picardie Jules Verne, 48 rue Raspail, 02100 Saint-Quentin, (Aisne),
Laboratoire CNRS FRE 2270, LAMFA (Amiens), France
b Department of Mathematics, Huazhong University of Science and Technology, Wuhan,
Hubei 430074, PR China
Received August 2001
Presented by M.P. Malliavin
Abstract
With a method close to that of Kirillov [4], we define sequences of vector fields on the set
of univalent functions and we construct systems of partial differential equations which have the
sequence of the Faber polynomials (Fn ) as a solution. Through the Faber polynomials and Grunsky
coefficients, we obtain the generating functions for some of the sequences of vector fields.  2002
Éditions scientifiques et médicales Elsevier SAS. All rights reserved.
AMS classification: 17B66; 17B68; 33C80; 33E30; 35A30
Keywords: Faber polynomials; Algebra of differential operators; Univalent functions
1. Introduction
In the first part, we consider the function
g(z) = z + b1 +
∞
n=1
bn+1
1
zn
(1.1)
E-mail addresses: [email protected] (H. Airault), [email protected] (J. Ren).
0007-4497/02/$ – see front matter  2002 Éditions scientifiques et médicales Elsevier SAS. Tous droits réservés.
PII: S 0 0 0 7 - 4 4 9 7 ( 0 2 ) 0 1 1 1 5 - 6
344
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
the coefficients (b1 , b2 , . . . , bn , . . .) are in the subset M of C N such that g(z) is univalent
outside the unit disc.
∞
1
g (z) =
Fn (b1 , b2 , b3 , . . . , bn ) n
(1.2)
z
g(z)
z
n=0
where Fn (b1 , b2 , b3 , . . . , bn ) is a homogeneous polynomial of degree n, the variables b1 ,
b2 , . . . , bn have respective weight, 1 for b1 , 2 for b2 , 3 for b3 , . . . , n for bn . We have (see
for example [3])
F0 = 1,
F1 = −b1 ,
F2 = b12 − 2b2 ,
F3 = −b13 + 3b2b1 − 3b3,
F4 = b14 − 4b12b2 + 4b1b3 + 2b22 − 4b4 .
On the submanifold M, we define the partial differential operators (Zk )k1 , the variables
are b1 , b2 , . . . , bn , . . . , and ∂n = ∂b∂ n denotes the partial derivative with respect to the nth
variable bn ,
Zk = −
∞
n bn ∂n+k−1 .
(1.3)
n=1
For a function φ(z, b1 , b2 , . . . , bn , . . .), we consider the system of partial differential
∂
φ. We find that the sequence (Fn ) of the Faber polynomials is a
equations, Zk φ = ∂z
solution of an infinite system of partial differential equations involving the (Zk ). We show
how to calculate the Faber polynomials from the coefficients in the asymptotic expansion
p
of the function ( g(z)
z ) where p is an integer,
p1
g(z)p
=
1
+
Kn n
zp
z
(1.4)
n1
p
in the notation Kn , n and p are indices. We define generalized Faber polynomials (Hjk ),
j 0, k ∈ Z, associated to g by
k−j 1
g (z) g(z) k
z
=1+
Hj
(1.5)
g(z)
z
zj
j 1
and generalized Faber polynomials (Fjk ), j 0, k ∈ Z, associated to the univalent
function f (z) = z(1 + b1 z + b2 z2 + · · · + bn zn + · · ·) by
k+j
f (z) f (z) k
=1−
Fj zj = 1 + (k + 1)b1 z + · · · .
(1.6)
z
f (z)
z
j 1
When
g(z)
z
k−j
= zf ( 1z ), then Kjk = 12 (Hj
Fnk = −
k
H k−2n
k − 2n n
k+j
− Fj
) and
n
Knk .
and Fnn+k = − 1 +
k
(1.7)
With an iteration procedure, we obtain homogeneous fractions which are solutions of a
system of partial differential equations involving the (Zk )k1 .
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
345
p
j
p
The sequence of Faber polynomials and the other polynomials (Kn ), (Fn ), (Hn ) can
be obtained as the solution of many systems of partial differential equations with an infinite
number of variables. For example (see [1,4]), with the embedding map
f (z) = z 1 +
bn zn → (b1 , b2 , . . . , bn , . . .)
n1
in the submanifold M of C N , we put ∂n =
where k 1, the vector fields
Lk = ∂k +
∂
∂bn
and we associate to Lk f (z) = z1+k f (z),
∞
(n + 1)bn ∂n+k .
(1.8)
n=1
We calculate the polynomials (Fk )k0 and (Fjk ) with Lp (Fjk ) = 0 for j p, and
k−p−1
k ) = −k, L (F k ) = kF
Lp (Fp+1
p j
j −p−1 for j p + 2.
In the second part, we study the generating functions
associated to the sequence of
vector fields found by Kirillov in [4]. Let f (z) = z(1 + n1 cn zn ) be a univalent function
bn
+ · · · be a univalent function
on the unit disc, and let g(z) = b0 z + b1 + bz2 + · · · + zn−1
outside the unit disc. With the action of vector fields on the set of the diffeomorphisms of
the circle Diff(S 1 ), and a variational approach on the equation f ◦ γ = g, where γ is a
diffeomorphism of the circle, Kirillov [4] obtained the following vectors fields on the set
of univalent functions
2 2
f (z)2
dt
t f (t)
1
L−p f (z) =
= φp (z) + z1−p f (z).
(1.9)
2
p+1
2iπ
f (t) f (t) − f (z) t
∂D
The term φp (z) comes from the residue at t = 0 for the contour integral (1.9) and vanishes
if p < 0. In that case Lk f (z) = z1+k f (z) = Lk [f (z)], where k 1 and Lk is given
by (1.8). We assume p 0. We have φp (z) = Np (f (z)) and (Np (w))p0 is given by
the generating function
φf (ξ, w) =
Np (w)ξ p =
p0
w2
ξ 2 f (ξ )2
.
2
f (ξ ) f (ξ ) − w
(1.10)
In this work, we obtain (1.10) with the Faber polynomials of h(z) = 1/f ( 1z ). Moreover,
L−p f (z) =
∞
p
An zn+1 .
(1.11)
n=1
p
j
In [1], the (An ) are given in terms of the polynomials Kn and Pnk such that
f (z)j j n
=
Kn z
zj
n0
and z
2
f (z)
f (z)
2 f (z)
z
k
=1+
n1
Pnn+k zn .
(1.12)
346
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
p
In the present paper, we obtain the generating function for the (An ) in two different
manners,
θf (u, v) =
p
Ak up v k+1 =
k1 p0
u2 f (u)2
v 2 f (v)
f (v)2
+
.
2
v−u
f (u) [f (u) − f (v)]
(1.13)
p
We deduce (1.13) from (1.10) or in Part II, Section 2, we prove (1.13) writing the An in
terms of the Grunsky coeffients βj k of h(z), given by (see [3,6])
K(u, v) = log
1
f (u)
1
u
−
−
1
f (v)
1
v
=−
1
n1 k1
n
βnk un v k .
(1.14)
Section 3, Part II, is a further step towards the classification of Faber type polynomials
which was started in Part I. We express the Grunsky coefficients βkj of h(z) in terms of
j
the polynomials Kn with generating function (1.12), or in terms of the Pjk , the following
identity (1.15) is a factorization of the symmetric matrix kβj k into a product of two infinite
matrices
1
1
βkj =
K n K −n .
k
n k−n j +n
k
(1.15)
n=1
p
The expression of the (An ) found in [1] yields
1−p
Kk+p +
p−1
p
j +1−p
Pj Kk+p−j
j =1
(1.16)
= βp−1,k+1 + 2c1 βp−2,k+1 + · · · + (p − 1)cp−2 β1,k+1 .
Let the Schwarzian derivative of f , Sf (z) = ( ff " ) − 12 ( ff " )2 and the asymptotic expansion
∂2
z2 Sf (z) = 6 n0 Pn zn . Since ∂u∂v
|u=v K(u, v) = − 16 Sf (u), see [2], we obtain the
Neretin polynomials (Pn )n0 in terms of the (βj k ),
Pn = βn−1,1 + 2βn−2,2 + 3βn−3,3 + · · · + (n − 1)β1,n−1 .
(1.17)
Then, as in [4] and [1], the set of functions univalent inside the unit disc is embedded
into the submanifold M in C N , via the map
n
f (z) = z 1 +
cn z → (c1 , c2 , . . . , cn , . . .).
n1
For k 1, ∂k = ∂c∂ k . On M ∈ C N , we consider the partial differential operators (1.8) and
(1.3). We have ∂n Lj = Lj ∂n + (n + 1)∂n+j . We express ∂k in terms of the (Lp )pk with
the generating function f 1(z) = 1 + n1 Bn zn . We put L0 = n1 ncn ∂n and for p 1,
p
following [4] and [1], we associate to (1.9) the operator L−p = n1 An ∂n . We obtain
L−p in terms of the (Lj )j 1 .
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
347
bn
In the third part, with the function g(z) = b0 z + b1 + bz2 + · · · + zn−1
+ · · ·, with b0 = 0,
we consider on the subset of (b0 , b1 , . . . , bn , . . .) such that g(z) is univalent outside the unit
disc, the vector fields
R−p = b0
∂
∂
∂
− b2
− · · · − (n − 1)bn
− ···
∂bp
∂bp+2
∂bn+p
if k 0.
(1.18)
The (Rp ) are obtained in [4] from right invariant vector fields on Diff(S 1 ). We relate the
(Zk )k1 of (1.3) to the (R−k ). It permits to define Zk for negative k.
Part I
I.1. The differential operators (Zk )k1 and the Faber polynomials
Let
b1 1
1
+
h(z) = g(z) = 1 +
bn n
z
z
z
n1
and h (z) = −
n1
n bn
1
.
zn+1
(I.1.1)
We consider h(z) as a function h(z) = φ(z, b1 , b2 , . . . , bn , . . .) of the infinite number of
variables z, b1 , b2 , . . . , bn , . . . . In this way, we have
∂
1
1
∂
[h(z)] = , . . .
[h(z)] = n , . . . .
(I.1.2)
∂b1
z
∂bn
z
∂
∂
Thus zh (z) = − ∞
n=1 nbn ∂bn (h(z)) = Z1 [h(z)] with Z1 = −
n1 nbn ∂bn and the
∂
φ=
function h(z) = φ(z, b1 , b2 , . . . , bn , . . .) satisfies the partial differential equation z ∂z
bn
1
Z1 φ. In the same manner, since zk−2 h (z) = − n1 n zk+n−1 , we obtain for k 1, the
partial differential equation
1
zk−2
∂
[h(z)] = Zk [h(z)]
∂z
(I.1.3)
where (see (1.3))
Zk = −
∞
n bn ∂n+k−1 .
(I.1.4)
n=1
We have [Z1 , Z2 ] = −Z2 , . . . , [Zk , Zk+r ] = −rZ2k+r−1 . The operators (Zk )k1 defined
above are different from the (Lk )k1 of (1.8). We do not know a priori if the (Zk )k1 in
(1.3) come from an ε-perturbation on Diff(S 1 ) as it appears in [4] for the (Lk ).
Lemma. Let the function ψ(z, b1 , b2 , b3 , . . . , bn , . . .) = s0 Ps (b1 , b2 , b3 , . . . , bn , . . .) z1s ,
∂
∂
we have for any k 1, Zk ( ∂z
ψ) = ∂z
Zk (ψ). In particular, Zk [h (z)] = (Zk [h(z)]) where
= ∂ and for the derivative of order p, we denote (p) = ∂ p ,
∂z
∂zp
(p)
(p) .
(I.1.5)
Zk h (z) = Zk [h(z)]
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
The proof is straightforward.
Theorem. The Faber polynomials (Fn ) defined by (1.2) are solutions of the system of
partial differential equations
Z1 Fn = −nFn ,
Z F = 0,
2 0
Z2 F1 = 1 − F0 = 0,
Z F = −nF
for n 2,
Z2 Fn = 0 n−1
for n k − 2,
k n
Zk Fk−1 = (k − 1)(1 − F0 ) = 0,
Zk Fn = −nFn−k+1
for n > k − 1.
(I.1.6)
The equation Z1 Fn = −nFn shows that the polynomial Fn is homogeneous of degree
n and the variables b1 , b2 , b3 , . . . , bn have respectively the weight 1 for b1 , 2 for b2 , 3 for
b3 , . . . , n for bn .
The coefficient of b1n in Fn is (−1)n . With the operators Zk , k n, and with (I.1.6),
we calculate the homogeneous polynomial Fn from the Fk , k < n. For example, F5 is
homogeneous of degree 5 and Z5 F5 = −5F1 , Z4 F5 = −5F2 , Z3 F5 = −5F3 , Z2 F5 =
−5F4 . We find
F5 = −b15 + 5b13 b2 − 5b12b3 − 5b1 b22 + 5b2 b3 + 5b1 b4 − 5b5 ,
F6 = b16 + 3b32 + 6b13b3 − 12b1 b2 b3 − 6b14 b2 − 2b23
+ 9b12b22 + 6b1 b5 + 6b2 b4 − 6b12b4 − 6b6 .
Proof of (I.1.6). From h(z) = 1z g(z), we have
z
h
h
= − 1z +
g
g
and
h
F1 1
1
= −1 +
+
Fn n =
Fn n .
h
z
z
z
n0
(I.1.7)
n2
Since Zk is a derivation, with (I.1.5), and then using (I.1.3) to replace Zk [h(z)], we obtain
h (z)Zk [h(z)]
Zk [h(z)] h (z)
Zk [h (z)]
−z
=z
Zk z
= z
h(z)
h(z)
h(z)2
h(z)
1 h (z) = z k−2
.
(I.1.8)
z
h(z)
(z)
by its expansion given by (I.1.7),
We replace the expression z hh(z)
1 1
1 Zk (Fn ) n = −z k−1 + z
Fn n+k−1
z
z
z
n0
=
k−1
−
zk−1
n0
(k − 1 + n)Fn
n0
By identifying in (I.1.9) the coefficients of
1
zp ,
1
.
zn+k−1
we obtain (I.1.6).
(I.1.9)
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
349
With the same method, and from the identity Zk (φ p ) = pφ p−1 Lk (φ), we deduce
Lemma. Consider v = (z hh )p =
Zk (v) = z(k−1)(p−1)+1
1
np Hn zn
v
zp(k−1)
for any integer p, then
.
Moreover Zk (Hn ) = −(n + (p − 1)(k − 1))Hn−k+1 for n p + k − 1 and if n p + k − 2,
Zk (Hn ) = 0.
p
The Faber polynomials in (1.2), can be obtained from the following (Kn ). For any
p
p 1
p
integer p ∈ Z, let g(z)
n1 Kn zn where in the notation Kn , n and p are indices,
zp = 1 +
(see (1.4)). From (I.1.3), we have
g(z)p
1
g(z)p Zk
.
(I.1.10)
=
zp
zk−2
zp
p
Kn , n 1, p ∈ Z, is a homogeneous polynomial in the variables b1 , b2 , . . . , bn where
p
p
b1 has weight 1, b2 weight 2, . . . . Moreover Zk (Kn ) = −(n − k + 1)Kn−k+1 for n k
p
p
and Zk (Kn ) = 0 for n < k. This permits to obtain the Kn . The coefficient of b1n in
p
p!
p!
= (p−n+1)(p−n+2)···p
the polynomial Kn is (p−n)!n! b1n . Remark that the notation (p−n)!n!
n!
p
extends to any p ∈ Z. The other coefficients in Kn are calculated with the (Zp )p2 . For
p
p
p
p
b12 + αb2 and Z2 (K2 ) = −K1 with K1 = pb1 , Z2 = −b1 ∂b∂ 2 . This
example, K2 = p(p−1)
2
p
gives the value of α and K2 =
the operators Lk , see (1.8).
p
p(p−1) 2
b1
2
p
+ pb2 . In [1] (A.1.6), the Kn are obtained with
p(p − 1)(p − 2) 3
b1 ,
3!
p(p − 1) 2 p(p − 1)(p − 2) 2
= p(p − 1)b1 b3 + pb4 +
b2 +
b1 b2
2
2
p!
b4 ,
+
(p − 4)!4! 1
K3 = p(p − 1)b1 b2 + pb3 +
p
K4
p
Kn =
p!
p!
p!
b1n +
b1n−2 b2 +
bn−3 b3
(p − n)!n!
(p − n + 1)!(n − 2)!
(p − n + 2)!(n − 3)! 1
p!
p−n+3 2
b1n−4 b4 +
b2
+
(p − n + 3)!(n − 4)!
2
n−j
p!
b1n−5 b5 + (p − n + 4)b2b3 +
b 1 Vj ,
+
(p − n + 4)!(n − 5)!
j 6
where Vj with 6 j n, is a homogeneous polynomial of degree j in the variables b2 , . . . , bn .
350
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
I.2. Generalized Faber polynomials
We assume that g(z) is given by (1.1). We have
n
f (z) = z 1 +
bn z .
g(z)
z
= zf ( 1z ) where
(I.2.1)
n1
Since
b1 (b12 − b2 )
z
=1−
+
+···
g(z)
z
z2
and z
1
1 f ( z )
g (z)
=2−
g(z)
z f ( 1z )
(I.2.2)
(u)
n
= 1 + b1 u + · · · = 1 − ∞
we deduce u ff (u)
n=1 Fn (b1 , b2 , . . . , bn )u . We put
k+j
f (u) f (u) k
=1−
Fj uj = 1 + (k + 1)b1u + · · ·
u
f (u)
u
(I.2.3)
j 1
j
then Fj = Fj where Fj is the j th Faber polynomial. On the other hand, if we take k = 1
1+j
in (I.2.3), we find Fj = −(j + 1) bj . With the same polynomials Fjk , we have, for any
k ∈ Z,
f (u) k
=1−
Fjk f (u)j .
(I.2.4)
u
j 1
Definition. We call the polynomials (Fjk ), the generalized polynomials associated to the
function f (z) = z(1 + n1 bn zn ).
When k = −1 in (I.2.4), we obtain
u = f (u) −
Fj−1 f (u)j +1 .
(I.2.5)
j 1
In particular, if w = f (u) where f is univalent, we have the asymptotic expansion of the
inverse map f −1 ,
u = f −1 (w) = w −
Fj−1 wj +1 = w + b1 w2 + b2 − 2b12 w3
(I.2.6)
j 1
+ 5b13 − 5b1 b2 + b3 w4 + · · · .
We rewrite (I.2.4) with k ∈ Z as uk = f (u)k − j 1 Fj−k f (u)j +k , or
−1
k
f (w) = wk −
Fj−k wj +k = wk + kb1 wk+1 + · · · .
j 1
The homogeneous polynomials (Fjk ) can be calculated with (I.2.3) and the (Lp )p0 as
(z) f (z) k
( z ) . By
in [1] (A.3.3)–(A.6.5): We have Lp [f (z)] = zp+2 f (z). Let φ(z) = z ff (z)
commuting
∂
∂z
and Lp , we get Lp [φ(z)] = zp+2 φ (z) + (p + 1 + k)zp+1 φ(z). From
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
351
k+j
(I.2.3), Lp [φ(z)] = − j 1 Lp (Fj )zj . With the identification of the coefficients in the
asymptotic expansions (I.2.3), we obtain (compare with (I.1.6)) Lp (Fjk ) = 0 for j p,
k−p−1
k
and Lp (Fp+1
) = −k, Lp (Fjk ) = kFj −p−1 for j p + 2. This gives F1k = −kb1 ,
F2k =
F4k =
F5k =
k(3 − k) 2
b1 − kb2 ,
2
F3k =
k(k − 5)(4 − k) 3
b1 + k(4 − k)b1 b2 − kb3 ,
6
k(5 − k)(k − 6)(k − 7) 4 k(5 − k)(k − 6)
b1 +
b2 b12 + k(5 − k)b1 b3
4!
2
k(5 − k) 2
b2 − kb4 ,
+
2
k(6 − k)(k − 7)(k − 8)(k − 9) 5 k(k − 7)(6 − k)(k − 8) 3
b1 +
b1 b2
5!
3!
k(k − 7)(6 − k) 2
k(6 − k)(k − 7)
b1 b3 +
b1 b22
+
2
2
+ k(6 − k)b2 b3 + k(6 − k)b1 b4 − kb5 ,
(k − 7)!k 6
(k − 7)!k 4
(k − 7)!k 3
k(7 − k) 2
b1 −
b1 b2 −
b1 b3 +
b3
6!(k − 12)!
4!(k − 11)!
3!(k − 10)!
2
(k − 7)!k (k − 9) 2
b2 + b4 b12
+ k(7 − k) (k − 8)b2 b3 + b5 b1 −
2!(k − 9)!
2
k(7 − k)(k − 8) 3
b2 + k(7 − k)b2 b4 − kb6 ,
+
6
F6k = −
(k − 8)!k 7
(k − 8)!k 5
(k − 8)!k 4
b1 −
b1 b2 −
b b3
7!(k − 14)!
5!(k − 13)!
4!(k − 12)! 1
(k − 8)!k (k − 11) 2
(k − 8)!k b2 + b4 b13 −
−
(k − 10)b2b3 + b5 b12
3!(k − 11)!
2
2!(k − 10)!
k(8 − k)(k − 9) k − 10 3
2
b2 + b3 b1
+ k(8 − k) (k − 9)b2 b4 + b6 b1 +
2
3
k(8 − k)(k − 9) 2
b2 b3 + k(8 − k)b3 b4 + k(8 − k)b2 b5 − kb7 .
+
2
Like in (I.2.3)–(I.2.4), we find that
k−j 1
g(z) k
1
g (z) g(z) k
=1+
Hjk
and
z
=1+
Hj
(I.2.7)
j
z
g(z)
g(z)
z
zj
F7k = −
j 1
with the same polynomials
−j
Hj
j 1
Hjk ,
k ∈ Z and j 1. Because of (I.2.2), the polynomials
= Fj are the Faber polynomials.
Definition. We call (Hjk ) the generalized Faber polynomials associated g(z).
352
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
1−j
= −(j − 1) bj and Hk0 = 0 for any k 1. As in (I.2.5), we put k = −1
in the first equation of (I.2.7), this gives z = g(z) + j 1 Hj−1 g(z)1−j and if w = g(z),
then
Hj−1 w1−j .
z = g −1 (w) = w +
We have Hj
j 1
We have
k(k + 1) 2
b1 + kb2 ,
2
k(k + 1)(k + 2) 3
b1 + k(k + 2)b1 b2 + kb3 ,
H3k =
6
H1k = kb1 ,
H2k =
k(k + 1)(k + 2)(k + 3) 4 k(k + 2)(k + 3)
b1 +
b2 b12 + k(k + 3)b1b3
4!
2
k(k + 3) 2
b2 + kb3 .
+
2
k
k 1
We can obtain the Hjk as follows: Let h(z) = g(z)
n1 Kn zn . Since
z . Then h(z) = 1 +
g (z) g(z) k
n
1
k
k−1 z
= h(z) + zh(z) h (z) = 1 +
1−
Knk n ,
g(z)
z
k
z
H4k =
n1
we deduce from (I.2.7) that
n
Hnk−n = 1 −
Knk .
k
(I.2.8)
k
Proposition. (See (1.7).) We have Fnk = − k−2n
Hnk−2n and Fnn+k = −(1 + nk )Knk .
f (u)
Proof. h(z) = f (1/z)
1/z = u where u = 1/z. Thus
k
k f (z) f (z) k
1
1 ∂
1
1−k k−1
z
= z f (z)f (z)
= h
+z
. (I.2.9)
h
f (z)
z
z
k ∂z
z
Since [h(1/z)]k = 1 + n1 Knk zn , we replace in (I.2.9). If we put k = n in (1.7), we find
again Fnn = Hn−n = Fn .
Corollary.
n−1
−k
(n − p)Fn−p
Fpn = Kn−k + Fn−k .
(I.2.10)
p=1
Proof. We have
k
u
=1+
Kn−k un
f (u)
n1
and
u
f (u)
k
=1−
j 1
Fj−k
f (u)j j
u .
uj
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
353
j
f (u)j
= 1 + n1 Kn un . We identify the coefficients of equal powers of u and
uj
j
we replace Kn by
its expression (1.7) to obtain (I.2.10). If we put k = −n in (I.2.10), it
n−p n n
gives Knn + Fnn = n−1
p=1 n Fp Fn−p . In particular
We replace
1 2 2
F ,
F33 + K33 = F13 F23 ,
2 1
1 2
F44 + K44 = F24 + F14 F34 + · · · .
2
F22 + K22 =
I.3. Homogeneous fractions of the (bi )i1
(z)
(z)
Let u(z) = 1 − z gg(z)
= −z hh(z)
, then 1 − z uu = z hh − z h"
h . From (I.1.8), Zk (u) =
1
u) . Since Zk (z uu ) = z( Zku(u) ) , we obtain
z( zk−1
Theorem. Let v = 1 + z uu = n1 Gn z1n , then
1
k(k − 1)
Zk (v) = z k−1 v +
z
zk−1
and
Z1 Gn = −nGn
and for k > 1, Zk (Gn ) = 0
Zk (Gk−1 ) = k(k − 1)
Zk (Gn ) = −nGn−k+1
if n k − 2
if n k.
For n 1, the relation Z1 Gn = −nGn implies that Gn is homogeneous of degree n in
the variables b1 , b2 , . . . , bn ; Gn = b1n × Qn where Qn is a homogeneous polynomial of
1
degree 2n in the variables b1 , b2 , . . . , bn , if b1 = 0.
G1 =
F2
b2
= b1 − 2 ,
b1
b1
G2 =
2F3 F22
1
+ 2 = 2 −b14 + 2b2 b12 − 6b3b1 + 4b22 .
b1
b1
b1
Part II
In the following, f (z) is a univalent function inside the unit disc,
n
f (z) = z 1 +
cn z .
n1
Let χ(z) be a function and assume that the expansion χ(f −1 (u)) = n∈Z αn un converges,
1
χ(f −1 (u))u−n du
αn is obtained with the contour integral αn = 2iπ
u . We put u = f (z) in
the two last expressions and we obtain the expansion of χ(z) in sum of powers of f (z),
du
uf (u)
j 1
f (u)−j χ(u) .
χ(z) =
f (z)
2iπ
f (u)
u
j ∈Z
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
In this formula, instead of χ(z), let zχ(z)f (z), it gives
2 2
du
u f (u)
1−j 1
zχ(z)f (z) =
f (z)
f (u)j χ(u) = S1 + S2
2iπ
f (u)2
u
j ∈Z
with
S1 =
f (z)1−j
1
2iπ
2+j
1
2iπ
j 0
S2 =
f (z)
j 0
du
u2 f (u)2
f (u)j χ(u) ,
2
u
f (u)
du
u2 f (u)2
f (u)−1−j χ(u) = Lχ [f (z)].
f (u)2
u
If χ(z) = z−p , where p is a positive integer p 0, S1 reduces to a finite sum of powers of
f (z) and is equal to
2 2
f (z)1−p
u f (u) (f (z)p+1 − f (u)p+1 ) du
= −Np f (z) .
2
p+1
2iπ
(f (z) − f (u))
f (u)
u
From this expression, we see that the generating function for the Np (w) is given by (1.10).
In the next section, we find (1.10) through
Faber polynomials.
With the embedding f (z) = z(1 + n1 cn zn ) → (c1 , c2 , . . . , cn , . . .) in the submanifold M of C N , if χ(z) = z−p , S2 = L−p [f (z)], and expanding f (z)2+j in powers of z in
p
S2 , we find L−p = n1 An ∂n where ∂n = ∂c∂ n .
II.1. The vector fields L−p
Theorem. If p 0, there exists a function Np (w) such that
∞
p
An zn+1 .
z1−p f (z) + Np f (z) =
(II.1.1)
n=1
The Akn are homogeneous polynomials in the variables (cn )n1 , Np (w) is calculated with
(1.10). Let φp (z) = Np (f (z)), then
φp (z)ξ p =
p0
f (z)2
ξ 2 f (ξ )2
2
f (ξ ) (f (ξ ) − f (z))
and
f (z)2
2iπ
1
ξ 2 f (ξ )2
dξ
= φp (z) + z1−p f (z).
2
f (ξ ) (f (ξ ) − f (z)) ξ p+1
p
The polynomials Ak defined by (II.1.1), are given by (1.13).
(II.1.2)
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
355
We have N0 (w) = −w, N1 (w) = −1 − 2c1 w, N2 (w) = − w1 − 3c1 − (4c2 − c12 )w,
d
N3 (w) = − w12 − 4c1 w1 − (c12 + 5c2 ) − P33 w, . . . . Let Np (w) = dw
Np (w) be the derivative
of Np . From (II.1.1), we deduce that for any j 1,
dz
f "(z) dz
+
Nj f (z)
= 0 and
z
f (z) zj
(II.1.3)
Nj (f (z)) dz
f (z) dz
+
= 0.
f (z) z
f (z) zj
Proof of (II.1.1)–(1.10)–(II.1.2). Consider the function
h(z) =
1 1
2
+ 2c1 c2 − c3 − c13 2
=
z
−
c
+
c
−
c
1
2
1
1
z
z
f (z)
1
1
+ 2c1 c3 − c4 + c22 − 3c12c2 + c14 3 + · · ·
z
∞
= z+
bn+1
n=0
1
zn
(II.1.4)
∞
ξ h (ξ )
−n where F (w) are the Faber
we have (see [5] and [6]) h(ξ
n
n=0 Fn (w)ξ
)−w =
polynomials associated to the function h. In terms of f ,
ψ(z, w) =
zf (z)
=
1
+
Fn (w)zn ,
f (z) − wf (z)2
(II.1.5)
n1
F1 (w) = w + c1 , F2 (w) = w2 + 2c1 w + 2c2 − c12 , F3 (w) = w3 + 3c1 w2 + 3c2 w + c13 −
3c1 c2 + 3c3 . If we take the derivative of (II.1.5) with respect to w and then integrate with
respect to z, we obtain
f (z)
zn
=
Fn (w) ,
(1 − wf (z))
n
(II.1.6)
n1
F1 (w) = 1, 12 F2 (w) = w + c1 , 13 F3 (w) = (w + c1 )2 − (c12 − c2 )(w + c1 ),
(w + c1 )3 − 2(c12 − c2 )(w + c1 ) − (2c1 c3 − c4 + c22 − 3c12c2 + c14 ). Moreover
Fn h(z) = z +
n
∞
βnk z−k ,
1 4 F4 (w)
=
(II.1.7)
k=1
where the βnk are the Grunsky coefficients of h. Thus
Fn
∞
1
βnk zk .
= z−n +
f (z)
k=1
(II.1.8)
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
1
We rewrite (II.1.8) as z−n = Fn ( f (z)
)−
∞
k=1 βnk z
k.
On the other hand, if p > 1,
z1−p f (z) = z1−p (1 + 2c1 z + 3c2 z2 + · · · + (n + 1)cn zn + · · ·
(p − 1)cp−2
1
2c1
3c2
= p−1 + p−2 + p−3 + · · · +
+ pcp−1
z
z
z
z
+ (p + 1)cp z +
(p + k + 1)cp+k zk+1 .
(II.1.9)
k1
We replace the negative powers of z by their expressions given by (II.1.8). We obtain
1
1
1
z1−p f (z) = Fp−1
+ 2c1 Fp−2
+ 3c2 Fp−3
+ ···
f (z)
f (z)
f (z)
1
+ pcp−1 + (p + 1)cp z
+ (p − 1)cp−2 F1
f (z)
(II.1.10)
− βp−1,1 + 2c1 βp−2,1 + · · · + (p − 1)cp−2 β1,1 z
(p + k + 1)cp+k − [βp−1,k+1 + 2c1 βp−2,k+1 + · · ·
+
k1
+ (p − 1)cp−2 β1,k+1 ] zk+1 .
For p 2, we consider
Tp−1 (w) = Fp−1 (w) + 2c1 Fp−2 (w) + 3c2 Fp−3 (w) + · · ·
+ (p − 1)cp−2 F1 (w) + pcp−1 .
(II.1.11)
From the expansion of f (z) and (II.1.5), we obtain the generating function
zf (z)2
=
1
+
Tn (w)zn ,
f (z) − wf (z)2
(II.1.12)
n1
T0 (w) = 1, T1 (w) = w + 3c1 , T2 (w) = w2 + 4c1 w + (c12 + 5c2 ), T3 (w) = w3 + 5c1 w2 +
(4c12 + 6c2 )w + P34 + · · · . With (1.12), we obtain
Tp (w) =
p
p+1
Pp−j wj .
j =0
We write (II.1.10) as
z1−p f (z) − Tp−1
1
f (z)
=
p
Bk−1 zk
k1
= (p + 1)cp z − βp−1,1 + 2c1 βp−2,1 + · · ·
p
+ (p − 1)cp−2 β1,1 z +
Bk zk+1
(II.1.13)
k1
with
p
Bk = (p + k + 1)cp+k
− βp−1,k+1 + 2c1 βp−2,k+1 + · · · + (p − 1)cp−2 β1,k+1 .
(II.1.14)
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
This gives, for p 1,
z1−p f (z) − Tp−1
p
1
p
− Pp f (z) =
Ak zk+1 ,
f (z)
357
(II.1.15)
k1
where P11 = 2c1 and if p > 1,
p
Pp = − βp−1,1 + 2c1 βp−2,1
+ · · · + (p − 1)cp−2 β1,1 + (p + 1)cp
p
Ak
p
Bk
p
Pp ck .
(II.1.16)1
p
p
and for p 1, k 1, we put
=
−
With the convention c0 = 1, B0 = Pp ,
p
p
we have A0 = 0. In [1], we obtained the generating function of the (Pp ),
p
z2 f (z)2
=1+
Pp zp .
2
f (z)
(II.1.16)2
p1
1
n
In fact, since f 1(z) = 1z + ∞
n=0 bn+1 z and bn+1 = β1n = n βn1 , by taking the derivative
with respect to z, we obtain
∞
f (z)
1 =
−
βn1 zn−1
f (z)2 z2
n=1
n−1 )f (z) and (II.1.16) implies (II.1.16) . From
thus
= f (z) − ( ∞
2
1
n=1 βn1 z
(II.1.11)–(II.1.15)–(II.1.16), we obtain the generating function of
(z)2
z2 ff (z)
2
p
Mp (w) = −Tp−1 (w) − Pp
1
.
w
(II.1.17)1
We put M0 (w) = − w1 , then
Mp (w)ξ p = −
p0
1
ξ 2 f (ξ )2
2
f (ξ ) w(1 − wf (ξ ))
(II.1.17)2
and from (II.1.14), we see that
∞
1
p
1−p f (z) + Mp
An zn+1 .
z
=
f (z)
n=1
We obtain (II.1.1)–(II.1.10) with Np (w) = Mp ( w1 ).
Since φp (z) is the coefficient of ξ p in the asymptotic expansion of the function
ξ 2 f (ξ )2
f (z)2
,
f (ξ )2 (f (ξ )−f (z))
we obtain (II.1.2). We deduce (1.13) from (II.1.1) and (1.10).
This ends the proof of the theorem. If we divide (II.1.12) by (II.1.17)2, we obtain
ξ f (ξ )
p
n
n
p0 Tp (w)ξ = −
n0 wMn (w)ξ . Since ξ = 1 +
n1 cn ξ , this yields a
f (ξ )
relation between the polynomials Mp and Tp . Combining with (II.1.17)1, it gives
Tp (w) = w Tp−1 (w) + c1 Tp−2 (w) + · · · + cp−1 T0 (w)
p
p−1
+ Pp + c1 Pp−1 + · · · + cp .
In Section 2, we give a direct proof of (1.13).
(II.1.18)
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
p
II.2. The generating function for the (An )
p
The polynomials Ak defined by (II.1.1) can be calculated with (1.13).
Proof. Taking the derivative of (1.14) with respect to u yields
uf (u)
f (v)
v
1
v
=
−
ψ u,
−
f (v)
v−u
f (u) f (v) − f (u) v − u
=
βnk un v k .
(II.2.1)
n1 k1
We multiply (II.2.1) by f (u),
vf (u)
uf (u)2
f (v)
−
f (u) f (v) − f (u)
v−u
βnk + 2c1 βn−1,k + · · · + ncn−1 β1,k un v k .
=
(II.2.2)
n1 k1
p
The (Bk ) are given by (II.1.14) for p > 1 and Bk1 = (k + 2)ck+1 . We rewrite (II.2.2) as
p
Bk − (p + k + 1)cp+k up−1 v k+1
k0 p2
=−
f (v)
vf (u)
uf (u)2
+
.
f (u) f (v) − f (u)
v−u
(II.2.3)
Since Bk1 = (k + 2)ck+1 , we can write the sum in (II.2.3), starting from p = 1. On the other
hand,
vf (u) vf (v)
+
.
(p + k + 1)cp+k up−1 v k+1 = −
v−u
v−u
k0 p1
Thus
γ (u, v) =
p
Bk up v k+1 = −
k0 p1
uvf (v)
u2 f (u)2
f (v)
+
.
f (u) f (v) − f (u)
v−u
(II.2.4)
Since
p
p
p
Ak = Bk − Pp ck
(II.2.5)
p
and A0 = 0, we have
k1 p1
p
Ak up v k+1 =
uv f (v)2
u2 f (u)2
+ f (v) +
f (v).
v−u
f (u)2 f (u) − f (v)
We divide by v. Since
p
p
A k up v k =
A k up v k +
A0k v k
k1 p0
k1 p1
k1
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
359
and
A0k v k =
k1
kck v k = f (v) −
k0
f (v)
,
v
we obtain (1.13).
II.3. Identities between polynomials
From (1.12), we obtain that for any p = 0, and k = 0,
n+p+k
Pn
=
n
(j + p)(n − j + k) p k
Kj Kn−j .
kp
(II.3.1)
j =0
In the following, we prove more identities between the polynomials.
(w) = (k + 1)
Proof of (1.15). With (II.1.6), Fk+1
Fk (w) = Fk (0) +
k
k
n=1
n
k
n+1 n
n=0 Kk−n w .
Thus
n
Kk−n
wn .
(II.3.2)
We replace w by f 1(z) , with (II.1.7), we obtain (1.15). The factorization can also be obtained
from (II.2.1). We have
K21 = c2 ,
K3−1 = 2c1 c2 − c3 − c13 ,
K1−2 = 2c1 ,
K4−2 = 6c1 c3 − 2c4 + 3c22 − 12c12c2 + 5c14,
K5−3 = 12c2 c3 + 12c1 c4 − 3c5 − 30c1c22 − 30c12c3 + 60c13 c2 − 21c15
and with (1.15), it gives β3,2 = 3K21 K3−1 + 32 K1−2 K4−2 + K5−3 . In the same way,
K11 = c1 ,
K4−1 = 2c1 c3 − c4 + c22 − 3c12 c2 + c14 ,
K5−2 = 6c2 c3 + 6c1 c4 − 2c5 − 12c1c22 − 12c12c3 + 20c13 c2 − 6c15
and β2,3 = 2K11 K4−1 + K5−2 .
Proof of (1.17). We deduce (1.17) from (II.2.5)–(II.1.14) and (A.4.8) in [1]. It can also be
proved as follows.
Let Sf (z) = (f "/f ) − 12 (f "/f )2 and φ(u) = 1/f (u), then
φ (u)φ (v)
f (u)f (v)
=
2
(f (u) − f (v))
(φ(u) − φ(v))2
and Sφ (u) = Sf (u).
Thus (see [3], p. 64),
1
∂2
f (u)f (v)
log
−
=
2
2
∂u∂v
(f (u) − f (v))
(u − v)
1
f (u)
−
1
f (v)
u−v
360
and
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
∂ 2 log
∂u∂v u=v
1
f (u)
−
1
f (v)
u−v
1
= − Sf (u) = −
6
kβnk un+k−2 .
(II.3.3)
n1,k1
Theorem. Let z2 Sf (z) = 6 n0 Pn zn be the expansion of the Schwarzian derivative. We
have P0 = 0, P1 = 0, and the (Pn )n2 are given by (1.17) in terms of the coefficients
in (1.14), P2 = β11 = c12 − c2 , P3 = β21 + 2β12 = 8c1 c2 − 4c3 − 4c13 , P4 = β31 + 2β22 +
3β13 = 12c14 − 34c12c2 + 12c22 + 20c1 c3 − 10c4 , . . . .
Let bnk = kβnk , then (bnk ) is a symmetric matrix, moreover Pn = trace(An Bn ) where
An is the square matrix of order n − 1 such that an−1,1 = an−2,2 = an−j,j = 1 and all the
other coefficients are zero and Bn = (bij )1in−1,1j n−1 .
II.4. The operators (∂k )k1 and (L−k )k0 in terms of (Lk )k>0
We identify the set of univalent functions with the submanifold M in C N via the map
z(1 + n1 cn zn ) → (c1 , c2 , . . . , cn , . . .). For k 1, Lk is given by (1.8) and (Akn ) by
(II.1.1) or (1.13). We put
ncn ∂n and for k 1 L−k =
Akn ∂n .
(II.4.1)
L0 =
n1
n1
In the following, we relate the (∂k )k1 and (L−k )k0 to the (Lk )k>0 and study the action
of these operators on some of the generating functions.
Proposition. For k 1,
∂k = Lk − 2c1 Lk+1 + (4c12 − 3c2 )Lk+2 + · · · + Bn Lk+n + · · ·
where the (Bn )n1 are independent of k.
1
=
1
+
Bn zn .
f (z)
(II.4.2)
(II.4.3)
n1
Proof. We verify (II.4.3) on f (z). We have Lk [f (z)] = zk+1 f (z). Since
∂k [f (z)] = zk+1
and ∂k [f (z)] = (k + 1)zk
zk+1 f (z)
zk+2 f (z)
=
− 2c1
+ · · · + Bn
we have to prove
zk+1 and we obtain (II.4.3).
Consider the generating functions (see (II.1.12) and (II.1.5)),
zf (z)
=
1
+
Fn (w)zn
ψ(z, w) =
f (z) − wf (z)2
zk+1
n1
and
zf (z)2
=
1
+
Tn (w)zn .
f (z) − wf (z)2
n1
(II.4.4)
zk+n f (z).
We divide by
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
361
The polynomials (Tp )p0 are expressed in (II.1.11) in terms of the (Fp )p0 . Conversely,
with (II.4.3), we have
F1 = T1 − 2c1 T0 ,
F2 = T2 − 2c1 T1 + 4c12 − 3c2 T0 , . . . ,
F0 = T0 ,
Fp = Tp − 2c1 Tp−1 + 4c12 − 3c2 Tp−2 + · · · + Bk Tp−k + · · · + Bp T0 .
(II.4.5)
From the generating functions, we obtain the action of (Lk ) on the polynomials. Let
k 1, we have Lk (Tp ) = (k + p + 1)Tp−k if p k and Lk (Tp ) = 0 if p < k. From
1
1
1
k+1
Lk
− (k + 1)zk ,
=z
f (z)
f (z)
f (z)
we obtain Lk (Bp ) = (p − 2k − 1)Bp−k if k p and Lk (Bp ) = 0 if k > p. From (II.1.5),
we deduce that ψ(z, w) satisfies
∂ψ
ψ(z, w) + kzk ψ(z, w),
(II.4.6)
Lk [ψ(z, w)] = zk+1
∂z
Lk (Fn ) = nFn−k if n k and Lk (Fn ) = 0 if n < k. We obtain Lk (βn,p ) with (II.4.1) and
(II.1.5). From
1
Lk ψ z,
f (v)
1
∂
∂ψ
1
) + v p+1
+ kzk ψ z,
= zk+1 ψ z,
,
(II.4.7)
∂z
f (v)
∂v
f (v)
we deduce Lk (βn,p ) = nβn−k,p + (p − k)βn,p−k if k p − 1, k n − 1 and the formula
extends for any n 1, p 1 with the convention βmn = 0 if m 0 or n 0.
Lemma. Let (see (1.10))
φf (u, w) =
u2 f (u)2
w2
=
×
Np (w)up ,
2
f (u) − w
f (u)
(II.4.8)
p0
then for k 1,
∂φf
Lk [φf (u, w)] = 2kuk φf (u, w) + uk+1
(u, w),
∂u
k ∂
uk+1 ∂φf
u
∂k [φf (u, w)] = 2u
(u, w).
(u,
w)
+
φ
f
∂u f (u)
f (u) ∂u
(II.4.9)
(II.4.10)
Proof. The proof of (II.4.9) is straightforward, then (II.4.9) and (II.4.2) imply that
uk+1 ∂φf
1
k 1
k+1
+ 2u
(u, w).
(u,
w)
+
∂k [φf (u, w)] = 2ku φ
f
f (u)
f (u)
f (u) ∂u
Corollary. Lk (Np ) = (k + p)Np−k if p k and Lk (Np ) = 0 if p < k. Let Bn as in (II.2.3),
then if p k,
∂k Np =
p−k
j =0
(2p − j )Bp−k−j Nj (w).
(II.4.11)
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
Remark. Let (see (1.13))
k1 p0
v 2 f (v)
p
Ak up v k+1 = φf u, f (v) +
= θf (u, v),
v−u
then
f u)2
3
θf (u, v)|v=u = u2 f (u) − 2u2
+ 2uf (u),
2
f (u)
∂θf 1 2
u2 f (u)2 1
= u Sf (u) −
f (u) + u2 f (u) ,
2
∂v v=u
6
f (u)
2
(II.4.12)
(II.4.13)
∂θf 1 2
3v 2 f (v)2 = − v Sf (v) +
f (v) + v 2 f (v)
2
∂u u=v
6
f (v)
f (v)
− 1 + 4v
× vf (v) .
f (v)
In the same manner, consider (see (II.4.4))
γf (u, v) =
f (v)
uvf (v) p p k+1
u2 f (u)2
×
+
=
Bk u v
f (u)
f (u) − f (v)
v−u
k0 p1
then
f u)2
3
+ uf (u),
γf (u, v)|v=u = u2 f (u) − u2
2
f (u)
∂γf 1
u
= u2 Sf (u)f (u) + uf (u) .
∂v v=u 6
2
(II.4.14)
(II.4.15)
p
Lemma. If p 1, for any k such that k > p, ∂k Ak is independent of k and we have
1
f dz
p
∂k Ak = −
.
(II.4.16)
2iπ
f zp
If k p,
p
∂k Ak = −
1
2iπ
1
f (z) dz
+
f (z) zp 2iπ
dz
(∂k Np ) f (z) k+2 .
z
Proof. From (II.1.1), we have
∞
p
(∂k An )zn+1 .
(1 + k)z1−p+k + Np f (z) zk+1 + (∂k Np ) f (z) =
n=1
With (II.1.3), it gives (II.4.17). Since ∂k Np = 0 if k > p, we obtain (II.4.16).
(II.4.17)
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
363
With (II.4.2), we obtain L−k , k > 0, in terms of the (Lj )j 1 .
Proposition. Let L−k =
gk (z) =
k
j 1 Dj Lj ,
then gk (z) =
k j +1
j 1 Dj z
satisfies
L−k [f (z)]
φk (z)
= z1−k + ,
f (z)
f (z)
f (z)
f (z)
f (z)
,
= gk (z) + gk (z) + gk (z)
L−k
f (z)
f (z)
f (z)
L−k S(f )(z) = gk (z) + 2gk (z)S(f )(z) + gk (z) S(f )(z) .
Part III
III.1. The vector fields Rp
(z) 2 f (z) k
In [1], we related the asymptotic expansion of z2 ( ff (z)
) ( z ) in sum of powers of z
(z) f (z) k
( z ) in sum of powers of f (z). In the same way, for g(z) with expansion
to that of z ff (z)
(1.1), we have
g (z) g(z) k
1
z
=1+
Vjk
and
g(z)
z
g(z)j
j 1
(III.1.1)
2 k−j 1
g(z) k
g (z)
=1+
Vj
.
z
g(z)
z
zj
j 1
The coefficients Vjk are obtained with contour integrals. The first equation in (III.1.1) gives
z1+k g (z) = g(z)1+k +
Vj−k g(z)1+k−j +
Vj−k g(z)1+k−j .
(III.1.2)
1j k−1
j k
Let g(z) be a univalent function outside the unit disc
bn
b2
g(z) = b0 z + b1 +
+ · · · + n−1 + · · · .
z
z
(III.1.3)
We assume b0 = 0. To g, it corresponds the sequence (b0 , b1 , . . . , bn , . . .) ∈ C N .
∂
∂
1
∂
[g(z)] = z,
[g(z)] = 1, . . .
[g(z)] = n−1 , . . . .
(III.1.4)
∂b0
∂b1
∂bn
z
Thus, if k 0, we have z1−k g (z) = R−k [g(z)] where
∂
∂
∂
R−k = b0
− b2
− · · · − (n − 1)bn
.
∂bk
∂bk+2
∂bn+k
Consider the expansion
du
ug (u)
j 1
g(u)−j χ(u)
g(z)
χ(z) =
2iπ
g(u)
u
j ∈Z
(III.1.5)
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
and then replace χ(z) by zχ(z)g (z), compare with Part II,
2 2
du
u g (u)
1
zχ(z)g (z) =
g(z)1−j
g(u)j χ(u)
2
2iπ
u
g(u)
j 0
1
du
u2 g (u)2
= S1 + S2 .
+
g(z)1+j
g(u)−j χ(u)
2
2iπ
u
g(u)
(III.1.6)
j 1
The first sum,
1
S1 = Rχ g(z) =
g(z)1−j
2iπ
j 0
du
u2 g (u)2
g(u)j χ(u)
g(u)2
u
can be expanded in powers of zk , k 1. Assume that χ(z) = zp , where p ∈ Z, in
that case, there exists a unique holomorphic vector field Rp on M ⊂ C N , satisfying
Rp [g(z)] = Rχ (g(z)) where
2 2
du
up
g(z)2
u g (u)
.
S1 = Rp (g(z)) =
2iπ
g(u)2 g(z) − g(u) u
If χ(z) = z−k , with k 0, then S2 vanishes
R−k [g(z)] = z1−k g (z).
and
k
k
If χ(z) = z with k 1, S2 = Jk (g(z)) = j =1 αj g(z)1+j and Jk is a polynomial.
k
En+1
Rk g(z) = z1+k g (z) − Jk g(z) = E0k z + E1k +
.
zn
(III.1.7)
n1
Thus Rk (g(z)) = Rk [g(z)] with the partial differential operator Rk =
k ∂
n0 En ∂bn .
Proposition. The polynomials Jk are given by
1
w2
u2 g (u)2
=
Jk (w) k .
2
g(u) (g(u) − w)
u
(III.1.8)
k1
The (Enk ) are given by
Enk z−n u−k = −
k1 n0
zg (z) u2 g (u)2
g(z)2
−
.
×
2
z−u
z(g(u) − g(z))
g(u)
Compare (III.1.9) with (1.13). With (III.1.8), we obtain
w2
,
b0
w3 3b1
J2 (w) = 2 − 2 w2 ,
b0
b0
J1 (w) =
J3 (w) =
w4
b03
−
4b1
b03
w3 +
(6b12 − 5b0 b2 )
b03
w2 , . . . .
(III.1.9)
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
365
Proposition. Let (Zk )k1 the operators defined by (1.3). For k 1,
R1−k − Zk = b0
∂
∂
∂
+ b1
+ · · · + bn
.
∂bk−1
∂bk
∂bn+k−1
For any k ∈ Z, let
2
k (g(z)) = g(z)
Z
2iπ
uk
du
ug (u)
.
g(u) (g(z) − g(u)) u
(III.1.10)
(III.1.11)
k = j 0 W k ∂ on M such that
There exists a unique holomorphic vector field Z
j ∂bj
−k where
Zk [g(z)] = Zk (g(z)) is given by (II.1.11). For k 0, we have Z1+k = R−k − Z
−k
Z−k satisfies Z−k (g(z)) = z g(z).
If k 1, then
k [g(z)] = zk g(z) − νk g(z)
Z
(III.1.11)1
νk is a polynomial function,
1
ξg (ξ )
1
×
.
νk (w) k = w2
ξ
g(ξ )
g(ξ ) − w
(III.1.12)
k1
g(z)
z
Proof. Let h(z) =
∂
h(z) = 1,
∂b0
= b0 +
b1
z
+··· +
∂
1
h(z) = ,
∂b1
z
...
bn
zn
+ · · ·, we assume that b0 = 0. Then
∂
1
h(z) = n .
∂bn
z
For k 1, z2−k h (z) = Zk [h(z)] = 1z Zk [g(z)], where Zk is given by (1.3). On the other
hand, we have R−k [g(z)] = z1−k g (z) and h (z) = g z(z) − g(z)
, we deduce [R1−k −
z2
1−k
Zk ](g(z)) = z g(z), this proves (III.1.10).
For (III.1.11), let
du
1
ug (u)
g(u)−j u2+k h (u) .
z2+k h (z) =
g(z)j
2iπ
g(u)
u
j ∈Z
Since h (u) =
g (u)
we obtain z2+k h (z) = Rk [g(z)] + Jk (g(z)) − S with
1
du
ug (u)
g(u)j uk
S =
g(z)1−j
2iπ
g(u)
u
j 0
1
du
ug (u)
g(u)−j uk .
g(z)1+j
+
2iπ
g(u)
u
u
j 1
−
g(u)
,
u2
k (g(z)) = j 0 W k z1−j is given by (III.1.11).
Let S = S1 + S2 . For any k ∈ Z, S1 = Z
j
k = j 0 W k ∂ .
We put Z
j ∂bj
If k 0, S2 vanishes and the integrand (III.1.11) has a pole only at u = z, we have
1−k = R1−k − Zk and Z
1−k is given by
k (g(z)) = zk g(z). Thus, for k 1, we have Z
Z
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H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
(III.1.10). If k 1, the integrand (III.1.11) has a pole at u = z and a pole at u = 0, the
residue at u = 0 is given by
S2 =
g(z)2
2iπ
g(u)k − g(z)k du ug (u) uk
=
×
αj g(z)1+j = νk g(z)
k
g(u) g(u)
g(u) − g(z) u
k
j =1
and the polynomials νk are obtained with (III.1.12).
k satisfy
Remark. With (III.1.12) and (III.1.11)1, it is easy to verify that the (Wjk )k1 in Z
λ(ξ, z) =
Wjk z1−j ξ −k =
j 0 k1
g(z)2
zg(z) ξg (ξ )
−
×
.
ξ −z
g(ξ )
g(ξ ) − g(z)
Moreover, let Sg (ξ ) the Schwarzian derivative of g, then
∂λ 1
g (ξ )2
1
S
=
−ξg(ξ
)
(ξ
)
−
− ξg (ξ ) − g (ξ ).
g
2
∂z z=ξ
6
g(ξ )
2
III.2. The Faber polynomials (Gn )n0 of g(z). The Gn (f (z))n0
Let g(z) be given by (III.1.3), its derivative g (z) = b0 −
b2
z2
− ···−
nbn+1
zn+1
− · · · and
1
zg (z)
Gn (w) n ,
=1+
g(z) − w
z
n1
then
1
Gn g(z) = zn +
γnk k .
z
(III.2.1)
k1
Let f (z) = z(1 + n1 cn zn ), we have Gn (f (z)) = Gn (0) + k1 pkn zk . The matrix
P = (pkn ) is given by the generating function
ξg (ξ )
=1+
Gn (0)ξ −n +
pin zi ξ −n .
g(ξ ) − f (z)
n1
(III.2.2)
n1 i1
Let P the matrix complex conjugated of P and P t the transposed matrix, then the
coefficients of the matrix P P t are given by
ξg (ξ ) 2 dξ 1
=
P P t ki zi zk .
2iπ
g(ξ ) − f (z) ξ
∂D
The proof is straightforward.
i,k
H. Airault, J. Ren / Bull. Sci. math. 126 (2002) 343–367
367
Acknowledgements
This work was completed while J. Ren visited in July 2001 the INSSET at SaintQuentin (Université de Picardie). J. Ren is very grateful to the University of Picardie for
his enjoyable stay.
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