ESI The Erwin Schrodinger International Boltzmanngasse 9Institute for Mathematical Physics A-1090 Wien, Austria

Holomorphic Mappings from Balls

John P. D’Angelo

Vienna, Preprint ESI 1738 (2005) November 9, 2005

Supported by the Austrian Federal Ministry of Education, Science and CultureAvailable via http://www.esi.ac.at

Holomorphic Mappings from Balls

John P. D’AngeloDept. of MathematicsUniversity of Illinois1409 W. Green St.Urbana, IL 61801email address: jpda@math.uiuc.edu

Abstract. We merge results from two preprints into a coherent whole. First wedescribe one-dimensional versions of two problems, a monotonicity result for volumes andthe Szego Limit Theorem. We prove two versions of the monotonicity result for volumesof images of certain domains in Cn. We obtain a sharp upper bound for the volume of theimage under a proper polynomial mapping f of degree d between balls, achieved only whenf is homogeneous. We also determine the asymptotic behavior, as the order of the grouptends to infinity, of group-invariant polynomials. These polynomials arise from variousreducible representations Γ(p, q) of cyclic groups of order p in the unitary group U(n);we provide new formulas for the invariant polynomials, and we prove a result evokingthe classical Szego Limit Theorem. The results include a striking generalization of knownasymptotics for Fibonacci numbers.

Key words. Several complex variables, volumes of holomorphic images, unit ball,proper holomorphic mappings, CR Geometry, finite unitary groups, invariant polynomials,Szego Limit Theorem, roots of unity.

AMS Classification Numbers. 32H35, 30E15, 32V30, 57S25, 11C20, 42A05,51M16.

Introduction

This paper merges two of the author’s preprints [D1] and [D2]. Some material iscopied from these sources, but the exposition has been altered considerably in order toplace these papers into a coherent framework, as described in the author’s lecture at ESI.

We consider two problems which resonate nicely with classical matters from one com-plex variable; both problems arise from natural questions about holomorphic mappingsfrom balls, and the homogeneous mapping z → z⊗d plays a key role in each case.

The first problem concerns volumes of holomorphic images of balls. We prove amonotonicity result (Theorem 1) for more general domains, but the principal application(Theorem 4) is to proper mappings between balls. Consider the set of polynomial mappingsof degree d that are proper from the unit ball Bn ⊂ Cn to some BN . For p in this setthe volume of the image of Bn under p is at most dnπn

n! . Equality occurs only when p ishomogeneous. All homogeneous examples are given by p(z) = L(z⊗d), where L is linear.Such mappings are invariant under the cyclic group generated by the linear transformationz → ωz, where ω is a primitive d-th root of unity.

The second problem concerns the possible images of the sphere under Γ-invariantholomorphic polynomial mappings f from balls. Here Γ is a finite subgroup of the unitary

1

group and thus the unit sphere is mapped to itself under Γ. In this paper we emphasizerepresentations Γ(p, q) ⊂ U(n) of cyclic groups of order p depending on an n-tuple of posi-tive integers q, and we seek Γ-invariant CR mappings from spheres to hyperquadrics. Thissimple situation yields families of Hermitian symmetric polynomials with many interestingnumber-theoretic and combinatorial properties. In Section VI we study the asymptotic in-formation about these polynomials that arises by letting the order of the cyclic group tendto infinity. We discover (Theorem 5) a result which evokes the classical Szego Limit The-orem for measures on the unit circle. In Proposition 5 we obtain a striking generalizationof known facts about Fibonacci numbers.

The author acknowledges support from NSF grant DMS-0500551 and support fromthe Center for Advanced Study at UIUC. The author spoke on the work in this paper atESI in October 2005, and he thanks Friedrich Haslinger for the invitation to visit there.

I. Background in one complex dimension.

In this section we describe two intriguing elementary results in one complex variable.Let f : B1 → C be a holomorphic function and let

∑∞n=0 bnz

n denote its Taylor seriesexpansion. The area Af of the image of B1 under f , with multiplicity counted, is

Af =

∫

B1

|f ′(z)|2dV2 = ||f ′||2L2. (1)

Consider the effect on the area of the image when f is replaced by zf . The nextresult, whose proof is elementary, shows that multiplication by z increases the area. Theconclusion requires integration; the pointwise inequality |(zf(z))′| > |f ′(z)| fails in general.

Proposition 1. Let f(z) =∑∞

n=0 bnzn be a holomorphic function with f ′ in L2(B1).

Then both f and (zf)′ are in L2(B1), and we have

||f ||2L2 = π

∞∑

n=0

|bn|2n+ 1

(2)

||f ′||2L2 = π

∞∑

n=0

(n+ 1)|bn+1|2 (3)

||(zf)′||2L2 = ||f ′||2L2 + π

∞∑

n=0

|bn|2 = ||f ′||2L2 +1

2

∫

|z|=1

|f |2dθ. (4)

Thus Azf ≥ Af and equality occurs only when f vanishes identically.

We mention that the area of the image of the disk under the map z → zm, with multi-plicity counted, is mπ. Suppose f ′ ∈ L2(B1); comparison with (3) shows that

∑∞n=0 |bn|2

converges, and hence f can be extended to the circle. Hence the integral in (4) makessense and we see also that (zf)′ ∈ L2(B1). Let D denote differentiation, and let M denotemultiplication by z. Consider the operator given by (DM)∗DM − D∗D on the space ofholomorphic functions with derivative in L2. By (4) this operator is positive definite andhence of the form T ∗T for an operator T which involves only a boundary integral:

2

||DMf ||2 − ||Df ||2 = ||Tf ||2 = π

∞∑

n=0

|bn|2 =1

2

∫

|z|=1

|f(z)|2dθ. (5)

We turn to the second matter in one dimension by recalling a famous result of Szego.See [S] and [Sz] for more information. Let µ be a probability measure on the unit circlewith Fourier coefficients cn. Let Dp(µ) denote the determinant of the p+1 by p+1 matrixwith entries ck−l for 0 ≤ k, l ≤ p. It is easy to see that

ck−l = 〈zl, zk〉L2(dµ). (6)

A quantity of interest in classical complex analysis is

limp→∞(Dp(µ))1p . (7)

When µ is absolutely continuous with respect to the measure dθ2π

, that is µ = h(θ) dθ2π

for afunction h, Szego showed that

limp→∞(Dp(µ))1p = exp

(

1

2π

∫ 2π

0

log(h(θ))dθ

)

. (8)

In words, the limit equals the exponential of the average value of log(h). Manygeneralizations and different proofs of Szego’s result are known. For example, the sameconclusion holds when µ has a singular part.

For later use we make an elementary observation about limits. Let p be a positive

integer. We choose a branch of the p-th root defined by w1p = |w| 1

p eiθp for 0 ≤ θ < 2π. Let

w ∈ C and consider limp→∞(1 −wp)1p . If |w| < 1, then

limp→∞(1 −wp)1p = 1. (9)

If w = 1, then limp→∞(1 − wp)1p = 0. If |w| = 1, but w 6= 1, then limp→∞(1 − wp)

1p does

not exist. If |w| > 1, then

limp→∞|1 −wp| 1p = limp→∞

(

|w|p|1− w−p|)

1p = |w|. (10)

II. Proper Holomorphic Mappings between Balls and Tensor Products.

Recall that a continuous mapping f : X → Y between topological spaces is properif, whenever K is a compact subset of Y , then f−1(K) is a compact subset of X. WhenX and Y are domains in Cn, the condition for f being proper is that f(zν ) tends to theboundary bY of Y whenever zν tends to bX. When f has a continuous extension to bX,it follows that f is proper if and only if f : X → Y and f : bX → bY . When X and Y

are balls and f is holomorphic on the open ball X and continuous on the closed ball, theboundary condition for properness is ||f(z)||2 = 1 on ||z||2 = 1. This equation enables usto get far with simple ideas.

3

The mapping z → zm in one dimension generalizes to a mapping Hm from Bn. Asin one dimension the mapping Hm is invariant under a cyclic group of order m, and wemake that connection in Section VI. We use it now to help classify all proper polynomialmappings between balls.

Let N = N(n,m) denote the dimension of the space of homogeneous polynomials ofdegree m in n variables, and let eα form an orthonormal basis of CN . We define Hm

(essentially the result of applying a linear transformation to z⊗m) by

Hm(z) =∑

|α|=m

√

(

m

α

)

zαeα. (11)

Then Hm : Bn → BN is a proper mapping, because ||Hm(z)||2 = ||z||2m = 1 on||z||2 = 1, and ||Hm(z)||2 < 1 on the open ball. The following result begins to indicate theimportance of Hm.

Theorem. ([D4], [R2]) Let p : Cn → CK be homogeneous of degree m. Supposethe components of p are linearly independent and that ||p(z)||2 = 1 on ||z||2 = 1. ThenK = N(n,m) and there is a unitary mapping U such that p = UHm.

Next we define the tensor product and division operations with respect to a subspace.Given functions p and g with the same domain, but possibly different targets, we want tomultiply them on this subspace. Let A be a subspace of CN with orthogonal complementA⊥. Let π denote the orthogonal projection onto A. Given a function p with values in CN

we naturally writep = (1 − π)(p) ⊕ π(p).

Let g be a function with the same domain as p and with values in some Ck. We define amapping EA(p, g) by

EA(p, g) = (1 − π)(p) ⊕ (π(p) ⊗ g). (12)

In case p and g are proper mappings to balls, it follows that EA(p, g) also is. To seethis fact we observe that

||EA(p, g)||2 = ||(1− π)(p)||2 + ||π(p) ⊗ g||2 = ||(1− π)(p)||2 + ||π(p)||2 ||g||2 =

||(1 − π)(p)||2 + ||π(p)||2 + ||π(p)||2(||g||2 − 1) = ||p||2 + ||π(p)||2(||g||2 − 1). (13)

If we let ||p|| and ||g|| tend to unity, then we see from (13) that ||EA(p, g)|| does also.Hence if p and g are proper mappings to balls, then EA(p, g) also is. Notice that the targetdimension of EA(p, g) is (N − a) + ak, where a = dim(A) (hence N − a = dim(A⊥)) andk is the dimension of the target of g.

When g is the identity mapping z, the operation p → EA(p, g) is orthogonal homog-

enization. The inverse operation sending EA(p, g) → p is an analogue of tensor division.We sometimes call it undoing. When g is the identity mapping, undoing is orthogonal

dehomogenization.

4

Suppose that the subspace A is one-dimensional, say the span of the vector e1. Theoperation is analogous to multiplication by z in one dimension:

EA(f, z) = (z1f1, ..., znf1, f2, ..., fN). (14)

Undoing replaces the right-hand side of (14) with f .

The author (See [D3] and [D4]) obtained the following result on the classificationproblem for proper polynomial mappings between balls.

Theorem 0. Let p : Cn → CK be a polynomial mapping of degree m, and supposethat p : Bn → BK is proper. Then there there is an integer N ′, subspaces A0, ..., Am ofCN′

, tensor products E0, ...Em, and a linear mapping L : CN′ → CN(n,m) such that

Hm = L

m∏

j=0

Ej(p). (15)

Each tensor product Ej is of the form EAj(f, z) for the corresponding subspace Aj .

Sketch of proof. The main idea uses a Fourier series argument resulting from replacingz by eiθz. The identity ||p(eiθz)||2 = 1 holds for z on the sphere and all θ. Write p =

∑

pj

for the expansion into homogeneous parts, and plug into the identity. When p isn’t alreadyhomogeneous, this identity forces the lowest order part pν to be orthogonal to the highestorder part pm, on the sphere, and hence, by homogeneity, everywhere. We then tensoron the subspace determined by the lowest order part; doing so results in a new properpolynomial mapping of the same degree but for which the order has increased. We repeatthe procedure until we obtain something homogeneous. ♠

The choice of the subspaces is not unique, but the proof provides a canonical wayof choosing them. If we wish, by allowing more than m tensor products in (15), we maychoose all the Aj to be one-dimensional. If we wish, we can ignore the L in (15) and obtaina homogeneous proper polynomial mapping from Bn to BN′ .

We next describe a second way to build a new proper mapping from two given ones.Let f : Bn → Bk and g : Bn → Bl be proper mappings. For each t ∈ [0, 1] we consider thejuxtaposition mapping Jt(f, g) defined by

Jt(f, g) = tf ⊕√

1 − t2g. (16)

Since ||Jt(f, g)||2 = t2||f ||2 + (1− t2)||g||2, and f and g are proper mappings to balls,we see that Jt(f, g) : Bn → Bk+l also is. We may think of Jt(f, g) as providing a homotopybetween f and g by allowing the natural embedding of each target ball in a target ball inhigher dimensions. We allow composition of Jt(f, g) with a linear mapping to make thetarget dimension minimal. We give a simple example:

Put f(z, w) = (z, zw, w2) and g(z, w) = (z2,√

2zw, w2). Let ut = LJt(f, g), where

ut(z, w) = (tz,√

1 − t2z2,√

2 − t2zw, w2). (17)

5

We obtain a one-parameter family of proper mappings from B2 to B4.

III. Monotonicity for volumes of holomorphic images

We next investigate the volumes of holomorphic images. Our main interest is in theball, but we prove a more general result in this section.

Let f : Ω ⊂ Cn → CN be a holomorphic mapping. In (18) and (19) we provide threeformulas for the 2n-dimensional volume, with multiplicity counted, of the holomorphicimage f(Ω). Let ω denote the usual (1, 1) form on CN given by

ω =i

2

N∑

j=1

dzj ∧ dzj .

The factor i2 arises because dz∧dz = −2idx∧dy in C. We let ωn denote the n-fold wedge

product of ω with itself.

Definition 1. (2n-dimensional volume) Let Ω be an open subset in Cn, and supposethat f : Ω → CN is holomorphic. We define V2n(f,Ω) by

V2n(f,Ω) =

∫

Ω

(f∗ω)n

n!=

∫

Ω

u dV. (18)

We call V2n(f,Ω) the 2n-dimensional parameterized volume. In (18) the upper star nota-tion denotes pullback. We next give equivalent formulas for the integrand u dV in (18).The expression involving L2 norms of Jacobians evokes the one-dimensional case, whereasthe expression involving the complex Hessian H(||f ||2) is also useful.

u dV =∑

|J(fI)|2dV = det(H(||f ||2))dV. (19)

The sum in (19) is taken over all choices of n component functions, and J(fI) denotes theJacobian determinant of the mapping fI with components fi1 , ..., fin

. This sum equals thedeterminant of the complex Hessian H(||f ||2) of the function ||f ||2. By (19) the volume ofthe image is finite if and only if each Jacobian J(fI) is in L2(Ω), which holds if and onlyif det(H(||f ||2)) is in L1(Ω).

Let P : Cn → CM be a holomorphic mapping, and let r(z, z) = ||P (z)||2. We assume

that the level set r = 1 is compact. Note that r is plurisubharmonic and hence ∂∂r−2i

is anonnegative (1, 1)-form. The domain Ω = z : r(z, z) < 1 is bounded and pseudoconvex.

Let f : Ω → CN be a holomorphic mapping and assume that f extends to be contin-uously differentiable on bΩ. We recall the tensor product operation from Section II:

EPf = g = (p1f1, ..., pMf1, f2, ..., fN) (20)

Let Vf and VEP f denote the 2n-dimensional volumes of the images of Ω under f andEPf . We have the following monotonicity result.

Theorem 1. Let Ω be the bounded pseudoconvex domain defined by ||P ||2 < 1, withP as above. Let f : Ω → CN be a holomorphic mapping and assume that f is continuouslydifferentiable on bΩ. Then

VEP f ≥ Vf .

6

Proof. We write

VEP f =∑

∫

Ω

|J(gI)|2dV, (21)

where the multi-indices I are of three types. The terms with multi-indices of type I arethose whose Jacobians involve none of the pjf1 terms. Terms of type I occur also in theformula for Vf and can be ignored. Terms of type III are those for which at least twocomponents occur involving the pjf1. These terms arise only in the formula for VEP f and,because they are positive, can also be ignored. Terms of type II are all of the form

∫

Ω

|J(pvf1, fi2 , ..., fin)|2dV.

To prove the theorem it therefore suffices to show that

∑

ν

∫

Ω

|J(pvf1, fi2 , ..., fin)|2dV ≥

∫

Ω

|J(f1, fi2 , ..., fin)|2dV (22)

for all multi-indices I′ = (i2, ..., in). For notational convenience we write I′ = (2, ..., n) andin proving (22) we now assume f = (f1, ..., fn). Let η denote the nonnegative (n−1, n−1)form given by

1

(−2i)n−1df2 ∧ df2 ∧ ... ∧ dfn ∧ dfn. (23)

Since f is continuously differentiable on bΩ we may apply Stokes’s theorem and obtain

Vf =

∫

Ω

|J(f1, ..., fn)|2dV =

∫

bΩ

f1

(−2i)∧ df1 ∧ η. (24)

We replace f1 by Pνf1 in (24) and use the product rule to obtain

∫

Ω

|J(Pνf1, f2, ..., fn)|2dV =

∫

bΩ

|Pν|2f1

(−2i)df1 ∧ η +

∫

bΩ

|f1|2Pν

(−2i)dPν ∧ η. (25)

We sum (25) on ν to find VEP f . Note that r =∑ |Pν|2 = 1 on bΩ, and

∂r =∑

ν

PνdPν. (26)

Putting these together and using (24) we obtain

VEP f = Vf +

∫

bΩ

|f1|2−2i

∂r ∧ η. (27)

Since r is plurisubharmonic and the (n − 1, n− 1)-form η is nonnegative, ∂r(−2i)

∧ η is

a nonnegative multiple of the surface area form. The surface integral in (27) is thereforenonnegative. Theorem 1 follows. ♠.

7

IV. Explicit Volume Computations.

This section combines calculus with some subtle multi-index notation. See [D1] fordetails of the computations. We first recall that the dimension of the vector space ofhomogeneous polynomials of degree d in n (commuting) variables is

(

n+d−1n−1

)

.Let K+ denote the part of the unit ball in Rn lying in the first orthant; that is

K+ = x :∑

x2j ≤ 1 and xj ≥ 0 for all j. Let α be an n-tuple of nonnegative real

numbers. We denote by |α| the sum of the αj, (the length of the multi-index α). For t > 0we let Γ(t) denote the usual Gamma function; for example Γ(n + 1) = n! when n is aninteger, and n ≥ 0. When each αj > 0, we define an n-dimensional analogue of the EulerBeta function by

B(α) =

∏

Γ(αj)

Γ(|α|) . (28)

There is a beautiful integral formula for this B function:

B(α) = 2n|α|∫

K+

r2α−1dV (r). (29)

We note that one can find the volumeΓ(1

2 )n

Γ(n2 +1)

of the unit ball in Rn instantly from

(29) by setting each αj equal to 12 . We evaluate additional integrals using (29). First we

note two uses of multi-index notation:

||z||2d = (

n∑

j=1

|zj|2)d =∑

|α|=d

(

d

α

)

|z|2α (30)

|z|2α =

n∏

j=1

|zj|2αj (31)

Lemma 1. Let d be a nonnegative integer, and let α be a multi-index of nonnegativereal numbers. Let Bn denote the unit ball in Cn. Then

∫

Bn

||z||2ddV =πn

(n − 1)!(n+ d). (32)

∫

Bn

|z|2αdV =πn

(n+ |α|)B(α + 1). (33)

Let p be an n-tuple of positive integers, and put Ω(p) = z :∑ |zj|2pj < 1. Then

∫

Ω(p)

|z|2αdV =πn

∏

pj

B(α+1p

)

[α+1p

]. (34)

For use in checking Example 1 we write (33) when n = 2 and a, b are integers:

8

∫

B2

|z|2a|w|2bdV4 =π2a!b!

(a + b+ 2)!. (35)

The techniques in this section allow us compute the volume of the image of Bn undera homogeneous polynomial proper mapping. We leave the proof to the reader.

Theorem 2. Let f : Bn → BK be a proper holomorphic homogeneous polynomialmapping of degree m. The 2n-dimensional parametrized volume Vf is given by

Vf = mnπn 1

n!. (36)

We do not assume in Theorem 2 that f = Hm. By Theorem 0, however, f = L(Hm)for some linear L. Even though L is not necessarily an equi-dimensional mapping (andhence not unitary) the volumes of the images under f and Hm are the same. The reasonis that ||f ||2 = ||Hm||2 and hence the volume forms in (19) are equal.

Proper holomorphic mappings f, g from Bn to BN are spherically equivalent if thereare automorphisms φ of Bn and ψ of BN such that f = ψ g φ. Faran [Fa] proved thereare precisely four distinct spherical equivalence classes of proper rational mappings fromB2 to B3; each equivalence class contains a monomial mapping. See (37) through (40)below. The mapping in (42) is spherically equivalent to the one in (38). For fun we findthe volumes of the images for these maps and also for one-parameter families of maps toB4 and to B5 .

Example 1. We give the 4-dimensional volumes of the image of B2 under variousproper mappings to B3, B4, and B5.

(z, w) → (z, w, 0). V =π2

2. (37)

(z, w) → (z, zw, w2). V =7π2

6. (38)

(z, w) → (z2,√

2zw, w2). V = 2π2. (39)

(z, w) → (z3,√

3zw, w3). V =63π2

20(40)

(z, w) → (tz,√

1 − t2z2,√

2 − t2zw, w2). V =(t4 − 6t2 + 12)π2

6. (41)

(z, w) → 1√2(z − w, z2 + zw, zw +w2). V =

7π2

6. (42)

(z, w) → 1√2(z,−w, z2,

√2zw, w2). V =

9π2

8. (43)

9

(z, w) → (z3, w3,√

3√

1 − t2z2w,√

3√

1 − t2zw2, t√

3zw). V = π2 90 − 12t2 − 15t4

20. (44)

V. An Alternative Proof of the Monotonicity Result for Balls and Eggs.

We continue by outlining a direct computational approach to the monotonicity resultthat avoids Stokes’s Theorem, and does not assume that the mappings are continuouslydifferentiable at the boundary. Both approaches are consistent with Proposition 1 in onedimension. The Stokes’s Theorem approach generalizes the integral that appears in (4),whereas the approach here generalizes the infinite sum appearing in (4).

We state several lemmas from [D1] but omit the (essentially computational) proofs.The first result uses the orthogonality of the distinct monomials in L2(Bn). From therewe expand in power series and use formulas for the Jacobian of a monomial mapping.

Lemma 2. Suppose that h : Bn → C is holomorphic and square integrable, withpower series expansion h(z) =

∑

cαzα. Then

∫

Bn

|h(z)|2dV =

∫

Bn

∑

α

|cα|2|z|2αdV = πn∑

α

|cα|2B(α + 1)

|α|+ n. (45)

Lemma 3. Let f : Cn → Cn be the monomial mapping f = (m1, ..., mn), and letJ(f) denote the Jacobian determinant of f . Then

J(f) =

∏

j mj∏

j zj

det(akl), (46)

where akl is the exponent of zl in mk.

Lemma 4. Let f : Bn → Cn be a monomial mapping f = (m1, ..., mn). Using thenotation from Lemma 3, the 2n-dimensional parametrized volume Vf is given by

Vf = πn|det(akl)|2∏

k(∑

j ajk − 1)!

(∑

j

∑

k ajk)!. (47)

Let f : Bn → Cn be a holomorphic mapping. Then

f(z) = (∑

cα1zα1 , ...,

∑

cαnzαn),

where each αj is itself a multi-index. By multilinearity, the Jacobian J(f) satisfies

J(f) =∑

(

n∏

j=1

cαj)J(zα1 , ..., zαn).

10

By combining the previous lemmas we can compute ||J(f)||2L2. First we have

||J(f)||2L2 =

∫

Bn

|J(f)|2dV =

∫

Bn

|∑

(

n∏

j=1

cαj)J(zα1 , ..., zαn)|2dV.

Define new multi-indices m by mj =∑

k αjk − 1. We obtain

∫

Bn

|J(f)|2dV =

∫

Bn

|∑

m

Amzm|2dV = πn

∑

m

|Am|2B(m + 1)

|m| + n, (48)

where Am =∑

(∏n

j=1 cαj)det(αkl), and the sum is restricted by the definition of m. For

later convenience we rewrite the m-th term in (48):

πn|Am|2B(m+ 1)

|m| + n= πn|Am|2

∏

j mj !

(|m| + n)!

= |∑

(n∏

j=1

cαj)det(αkl)|2

πn

(∑∑

αjk)!

n∏

j=1

((∑

k

αkj − 1)!). (49)

We need to see the effect on (49) of replacing f by a tensor product. The followinglemma about determinants arises because of Lemma 3.

Lemma 5. Let L = (ajk) be an n-by-n matrix of real numbers. Suppose that eachcolumn sum vj of L is nonnegative. Let Dj = (δ1j) denote the matrix whose entires arezero except that the entry in the j-th column and first row is unity. Let Cj be the cofactordeterminant (with sign included) obtained by deleting the j-th column and first row andtaking the signed determinant. The following hold:

det(L) =

n∑

j=1

Cjvj. (50)

n∑

j=1

det(L+Dj)2vj = det(L)2(2 +

∑

j

vj) +∑

j

C2j vj. (51)

det(L)2(2 +∑

j

vj) ≤n∑

j=1

det(L+Dj)2vj . (52)

We can now show by computation, without assuming any boundary regularity, thatparametrized volume increases under tensor product. As before there is generally nopointwise inequality on the volume forms; the inequality requires integration. We note onenew issue in higher dimensions; Example 2 below shows that the inequality fails if we allowmultiplication by only a single coordinate function. Example 3 and Theorem 1 indicatethat we must tensor with something closely related to the geometry of the boundary.

11

Theorem 3. Let f : Bn → CN be holomorphic, and suppose that the 2n-dimensionalparametrized volume Vf of the image f(Bn) is finite. Let g = Ef be defined by

g = (z1f1, ..., znf1, f2, ..., fN).

Then Vg ≥ Vf . Equality occurs if and only if f1 = 0.More generally, let p = (p1, ..., pn) be an n-tuple of positive integers, let Ω(p) be the

egg domain z :∑n

j=1 |zj|2pj < 1, and let f : Ω(p) → CN be a holomorphic mapping.Assume that the volume of the image of Ω(p) is finite. Define g = Ef by

g = (zp1

1 f1, ..., zpn

n f1, f2, ..., fN).

Write V pg and V

pf for the corresponding volumes of images of Ω(p). Then V p

g ≥ Vp

f , andequality occurs if and only if f1 = 0.

Sketch of Proof. We first describe the proof in case of the ball, and at the end of thesketch we indicate how the proof generalizes to the eggs.

The volume Vg is the sum of the integrals of all |J(gI)|2. As in the proof of Theorem1 we separate the n-tuples gI into three types. Type I consists of those formed from n ofthe functions f2, ..., fN. The corresponding terms occur also in the expression for Vf , andhence can be ignored when proving the inequality between Vf and Vg . Type II consists ofthose n-tuples for which exactly n−1 of the components are chosen from among f2, ..., fN.Type III consists of those n-tuples for which at most n−2 are chosen from among f2, ..., fN.Those of type III contribute positively to Vg, but do not contribute to Vf and hence canbe ignored. We will therefore prove that

∫

Bn

∑

|J(fI)|2dV ≤∫

Bn

∑

|J(gI)|2dV (53)

where the sums on both sides are taken over those n-tuples of type II.To verify (53) it suffices to prove, for any choice of n components f = (f1, ..., fn) (after

relabeling), the following inequality:

∫

Bn

|J(f)|2dV ≤∫

Bn

n∑

ν=1

|J(zνf1, f2, ..., fn)|2dV. (54)

The left-hand side has been computed in (48). We compute the right-hand side inthe same fashion, obtaining a sum of n terms. As before we define multi-indices m bymj =

∑

k αjk − 1. The sums in Am and Km(ν) below are taken over all n-tuples ofmulti-indices such that this definition holds. The m-th term of the left hand side of (54) is

πn|Am|2∏

j mj !

(|m| + n)!.

Computing ||J(zνf1, f2, ..., fn)||2L2 in the same way we obtain

πn|Km(v)|2(mν + 1)

∏

j mj !

(|m|+ n+ 1)!,

12

for the m-th term, where

Km(ν) =∑

(

n∏

j=1

cαj)det(αjk +Dv). (55)

Using the ideas of Lemma 5 we write

det(αjk +Dv) = det(αjk) + C(α, ν),

where C(α, ν) is an appropriate cofactor determinant. Hence

Km(ν) = Am +∑

(

n∏

j=1

cαj)C(α, ν). (56)

We will show for each multi-index m that

πn|Am|2∏

j mj !

(|m| + n)!≤∑

ν

πn|Km(v)|2(mν + 1)

∏

j mj !

(|m| + n+ 1)!. (57)

Summing (57) on m implies (54). After cancelling common factors (57) is equivalent to

|Am|2(|m| + n+ 1) ≤∑

ν

(mν + 1)|Km(v)|2. (58)

By (56) the right-hand side of (58) becomes

∑

ν

(mν + 1)|Km(v)|2 = |Am|2(∑

ν

(mν + 1)) +∑

ν

(mν + 1)|∑

(

n∏

j=1

cαjC(α, v))|2

+2Re(Am

∑

(

n∏

j=1

cαj

∑

ν

C(α, v))(mν + 1)). (59)

Recall that |m| + n =∑

ν(mν + 1) by multi-index notation. Since mν + 1 =∑

ανj,the definition of Am and (50) yield

Am =∑

(n∏

j=1

cαj)∑

ν

C(α, v))(mν + 1)).

Thus the last term in (59) is 2|Am|2. We may drop the nonnegative middle term in (59).The right-hand side of (59) is thus at least |Am|2(|m| + n+ 2), and (58) follows.

In other words the determinantal inequality from Lemma 5 holds for eachm. Summingon m we obtain (57) and hence (54). Recalling that it suffices to prove the inequality fortype II terms, we have proved that Vf ≤ Vg .

If f1 = 0, then Vf = Vg . If f1 6= 0, (and n ≥ 2) then there is a term of type III of theform |J(z1f1, ..., znf1)|2; such a Jacobian is not identically zero, and hence Vf < Vg .

13

Suppose next that Ω(p) is an egg domain. The monomials remain a complete or-thogonal system for L2(Ω(p)), and a holomorphic mapping f has a power series expansionvalid on all Ω(p). Thus the structure of the proof is identical. The calculations are quitesimilar, except that the various factorials used in case of the ball become values of Gammafunctions at non-integral values. Recalling (34) we have

||zα||2L2(Ω(p)) =πn

∏

pj

B(α+1p

)

[α+1p

]. (60)

In forming the tensor product we multiply zα by each zpνν . The effect is to alter the

multi-index α+1p

appearing in (60) by adding 1 in the ν-th slot. We then use the formula

Γ(x + 1) = xΓ(x) in both the numerator and denominator of the resulting terms, ratherthan using j! = j(j − 1)! as in the proof for the ball. After these modifications the proofgoes through as before. ♠.

We leave it to the reader to formulate a Hilbert space interpretation of Theorem 3.We continue with one of its main consequences, an extremal property of the homogeneousproper mappings between balls,

Theorem 4. Let p : Bk → BK be a proper polynomial mapping of degree d. Thenthe parametrized volume Vp is at most dkπk 1

k! . Equality occurs only if p is homogeneous.Proof. We saw in Theorem 2 that Vp = VHd

= dkπk 1k! , so equality holds if p is

homogeneous. In case p is not homogeneous, we can repeatedly tensor p on one-dimensionalsubspaces as in the proof of Theorem 0 from Section II until it is homogeneous. By Theorem3, each tensor operation increases the volume, so we obtain Vp < VHd

. ♠Example 2. The conclusion of Theorem 3 fails in general if we multiply a component

of f by only a single coordinate function. The simplest example is given by f(z, w) =(zw, w). In this case ||J(f)||L2 > 0 whereas ||J(g)||L2 = 0 if g(z, w) = (zw, zw).

Example 3. The conclusion of Theorem 3 for the ball fails in general if we tensor witha mapping other than z = (z1, ..., zn), even when the mapping defines a zero-dimensionalvariety. Let n = 2, and put f(z, w) = (z4, w). Suppose we tensor by multiplying the secondcomponent by both z3 and w3, obtaining g(z, w) = (z4, z3w,w4). Then the parametrizedvolume decreases:

Vf =4π2

5>

24π2

35= Vg.

On the egg Ω(p) one must tensor by (zp1

1 , ..., zpnn ) in order for the conclusion to be valid.

V. Asymptotics for Invariant Polynomials

Let Γ be a finite subgroup of the unitary group U(n). We first ask whether there isan N and a nonconstant Γ-invariant proper holomorphic mapping f : Bn → BN . Theanswer, perhaps surprisingly, depends on the regularity of f at the boundary sphere. Theanswer is yes (for all Γ) if we demand only that f be continuous on the closed ball. If,instead, we demand that f be smooth on the closed ball, then the answer is no for mostΓ. When n ≥ 2, f is a proper mapping between balls, and f is smooth on the sphere, then

14

f must be rational [F1]. The survey [F2] describes some restrictions on Γ in case thereis a nonconstant Γ-invariant rational proper holomorphic mapping between balls. It isproved in [L] that, if there is a nonconstant Γ-invariant rational proper mapping betweenballs, then Γ must be cyclic. See [DL] or [D4] for precisely which representations of cyclicgroups arise in this context. Given a general Γ we find nonconstant Γ-invariant mappingsby replacing the target sphere with a hyperquadric.

Let G be a finite group of order p, and let η : G → U(n) be a faithful representation.The representation Γ = η(G) is allowed to be reducible and to have fixed points. Associatedwith this representation is an invariant polynomial ΦΓ:

ΦΓ(z, z) = 1 −∏

γ∈Γ

(1 − 〈γz, z〉). (61)

Because each γ is unitary it follows that ΦΓ is Hermitian symmetric. Furthermore itis uniquely determined by the following result. [D5]

Proposition 1. The function ΦΓ given by (61) is the unique polynomial satisfyingthe following four properties:

1) ΦΓ(0, 0) = 0.2) ΦΓ(z, z) = 1 on the unit sphere, i.e., for ||z||2 = 1.3) ΦΓ(γz, z) = ΦΓ(z, z) for all γ ∈ Γ and for all z.4) ΦΓ is a polynomial in (the first slot) z of degree at most the order of G.

The polynomial 1 − ΦΓ is positive at the origin. By (61) it vanishes at (z, z) only if〈γz, z〉 = 1 for some γ in Γ. Since γ is unitary, |〈γz, z〉| < 1 on the open unit ball, andhence 1 − ΦΓ(z, z) > 0 if ||z||2 < 1.

Using the Hermitian symmetry and the invariance property 3) from Proposition 1,one can write

ΦΓ(z, z) = ||A(z)||2 − ||B(z)||2

for holomorphic Γ-invariant polynomial mappings A and B. The mapping A ⊕ B thendefines a Γ-invariant mapping sending the unit sphere to a hyperquadric. The numbers Pand Q of positive and negative eigenvalues in the defining equation for the hyperquadric aredetermined by the number of independent components of A and B. Property 2) guaranteesthat the sphere is mapped into a hyperquadric defined by

||w1||2 − ||w2||2 = 1,

where w1 ∈ CP and w2 ∈ CQ.In most of the rest of this paper we assume that Γ is cyclic; this case already reveals

some interesting information.Let Gp be cyclic of order p. We can represent Gp in many ways, and the invariant

polynomials depend on the representation; it is natural to investigate this dependence.First consider a “universal” special case. Fix a primitive p-th root of unity ω. Considerthe cyclic subgroup Γ(p) of U(p) generated by a diagonal matrix whose eigenvalues are ωj

for 1 ≤ j ≤ p. More generally let q = (q1, ..., qn) be an n-tuple of positive integers with

15

1 ≤ q1 < q2 < ... < qn ≤ p. Thus p ≥ n. Let Γ(p, q) denote the cyclic subgroup of U(n)generated by a diagonal matrix whose eigenvalues are ωqj . There is no loss in generalityin assuming that the exponents qk are distinct; the case with multiple exponents followsin a routine way from the case with distinct exponents.

Definition 1. Let q = (q1, ..., qn) be an n-tuple of positive integers. The critical

polynomial Qq(ζ, z) for Γ(p, q) is defined by

Qq(ζ, z) = 1 −n∑

k=1

|zk|2ζqk . (62)

For each fixed z the function Qq(ζ, z) is a polynomial in the single complex variable ζ.It does not depend on p. We writeQ(ζ, z) for the critical polynomial in case q = (1, 2, ..., n).

Remark. The critical polynomial Qq(ζ, z) for z ∈ Cn can be recovered from thespecific critical polynomial Q(ζ, Z) for Z ∈ Cp by substitution. We have

Q(ζ, Z) = 1 −p∑

l=1

|Zl|2ζl. (63)

For each integer l with 1 ≤ l ≤ p we set Zl = 0 in (63) if l 6= qk for some k. We set Zl = zk

if l = qk. We obtain the critical polynomial Qq.

Next we put the invariant polynomials ΦΓ(p,q) into one framework. Let Tp be adiagonal linear transformation on Cp whose eigenvalues are the p-th roots of unity ωj :1 ≤ j ≤ p. We note that Tp is conjugate to a permutation matrix Lp whose entries arezeroes and ones. The characteristic polynomial of Tp is

det(Tp − λI) = (−1)p(λp − 1). (64)

Corresponding to the group Γ(p, q), we let Tp,q denote the diagonal n by n matrix whoseeigenvalues are ωqk .

Proposition 2. Let p be a positive integer. Let Γ(p, q) be the cyclic subgroup ofU(n) of order p generated by Tp,q and let Qq(ζ, z) be its critical polynomial. Let Tp be asabove. Then the invariant polynomial ΦΓ(p,q) satisfies (65):

ΦΓ(p,q)(z, z) = 1 − det(Qq(Tp, z)). (65)

Proof. The eigenvalues of Tp are ωj. Let Q be a polynomial in one variable. By thespectral mapping theorem, the linear transformation Q(Tp) has eigenvalues Q(ωj). Hence

det(Q(Tp)) =

p∏

j=1

Q(ωj). (66)

In (66) replace Q by the critical polynomial Qq(ζ, z). We obtain

det(Qq(Tp, z)) =

p∏

j=1

(1 −n∑

k=1

|zk|2ωjqk ). (67)

16

On the other hand, by hypothesis the group Γ is generated by the diagonal matrix Tp,q.For each power j we have

〈T jp,qz, z〉 =

n∑

k=1

ωjqk |zk|2.

Therefore the right-hand side of (67) is

p∏

j=1

(1 − 〈T jp,qz, z〉), (68)

and we obtain the desired conclusion (65) from (61) and (68). ♠.

Corollary 1. The coefficients of ΦΓ(p,q)(z, z) are integers.

Proof. The matrix Tp is conjugate to a permutation matrix Lp of zeroes and ones.By (9),

1 − ΦΓ(p,q)(z, z) = det(Qq(Tp, z)) = det(Qq(Lp, z)). (69)

Each entry of Qq(Lp, z) is an integer multiple of |zj|2 for some j, and hence its determinantis a polynomial in the variables |zj|2 with integer coefficients. ♠

Thus 1 − ΦΓ(p,q) is nicely expressed as a determinant. The special case where Γ isgenerated by Tp is universal in the sense that we can recover the formulas for Γ(p, q)by substitution as in the Remark. We will let p tend to infinity while keeping the n-tuple q = (q1, ..., qn) fixed. As in the Szego limit theorem, the size of the matrix whosedeterminant we compute grows with p. We need also a formula for 1−ΦΓ(p,q) as a productof a fixed number of terms.

Proposition 3 below expresses 1−ΦΓ(p,q) in terms of the roots of the critical polynomialQq, which does not depend on p. For each z we note that Qq(0, z) = 1 6= 0 and that itscoefficients are real. Its non-real roots therefore occur in conjugate pairs, and we can write

Qq(ζ, z) =

d∏

l=1

(1 − cl(z)ζ) =

d1∏

j=1

(1 − aj(z)ζ)(1 − aj(z)ζ)

d2∏

k=1

(1 − bk(z)ζ). (70)

If necessary we repeat terms in (70) to account for multiplicity. The roots are the recipro-cals of the cl(z). The aj(z) and their conjugates are the reciprocals of the non-real rootsand the bk(z) are the reciprocals of the real roots.

Proposition 3. Let Γ(p, q) be the cyclic subgroup generated by Tp,q . The invariantpolynomial is determined explicitly by the (reciprocals cl of the) roots of Qq:

ΦΓ(p,q)(z, z) = 1 −d∏

l=1

(1 − cl(z)p). (71)

17

Proof. We write cj for cj(z) in the proof. By Proposition 2, (70), the definition ofcharacteristic polynomial, and (64), we compute

1 − ΦΓ(p,q)(z, z) = det(Qq(Tp, z)) =

d∏

j=1

det(I − cjTp) =

d∏

j=1

(−cj)pdet(Tp − c−1j I)

=

d∏

j=1

(−cj)p(−1)p(c−pj − 1) =

d∏

j=1

(1 − cpj ). ♠ (72)

To motivate our study of asymptotics, consider the special case where Γ(p, q) is thesubgroup of U(2) generated by (73) below. In this special case a combinatorial interpreta-tion of the integer coefficients of 1 − ΦΓ and asymptotic information about the maximum

coefficient appear in [LWW]. Here we study the asymptotics of the polynomials 1−ΦΓ formore general Γ. In Proposition 5 we obtain information about the sum of the coefficientsof ΦΓ. We begin with the special case.

Let n = 2, let ω be a primitive p-th root of unity, and let q = (1, r) for some r with1 ≤ r ≤ p. Consider the diagonal representation Γ(p, q) of Gp whose generator is given by

(

ω 00 ωr

)

. (73)

For a fixed p the invariant polynomials for Γ(p, q) depend on r. The interest is forr ≥ 2, but we pause to discuss r = 1. When r = 1 the polynomial is given by

Φ(z, z) = (|z1|2 + |z2|2)p.

We noted earlier that it is unnecessary to consider repeated exponents; if we had set n = 1,then the invariant polynomial would have been |z1|2p. We obtain the result for n = 2 bysubstituting |z1|2 + |z2|2 for |z1|2.

When r = 2 we have the following explicit formula from [D4]:

Φ(z, z) = (−1)p+1 |z2|2p+

(

|z1|2 +√

|z1|4 + 4|z2|22

)p

+

(

|z1|2 −√

|z1|4 + 4|z2|22

)p

. (74)

Despite its appearance, (74) defines a polynomial in |z1|2 and |z2|2 with integer coefficients.The explicit formula enables us to express the coefficient sum Sp in terms of Fibonaccinumbers and to determine its asymptotic behavior: (74) yields

Sp = Φ((1, 1), (1, 1)) = (−1)p+1 + (1 +

√5

2)p + (

1 −√

5

2)p,

and hence

limp→∞(Sp)1p =

1 +√

5

2. (75)

18

We will determine the asymptotic behavior of the coefficient sum in general without useof an explicit formula. It is probably impossible to find an explicit formula such as (74)for the polynomial ΦΓ(p,q) for general r and q = (1, r). The asymptotics for the maximumcoefficient and for the coefficient sum differ in general, because cancellation occurs. Whenr = 2, however, the coefficients (with one exception when p is even) have the same sign,and the polynomials defined by (74) are “unimodal”. In this special case the asymptoticsfor the maximum and for the coefficient sum are the same.

First we study Bp(z) = (1−ΦΓ(p,q)(z, z))1p . The expression (1−ΦΓ(p,q)(z, z)) is real.

For z ∈ C with argument in [0, 2π) we define the p-th root by z1p = |z| 1

p eiθp .

We study Bp(z) as p tends to infinity by considering the roots of the critical polyno-mial. The n-tuple q is fixed. In particular the maximum exponent qn does not depend onp, and therefore the degree of Qq does not grow with p.

Our next result gives a rather complete analysis of the asymptotics of Bp(z) as p tendsto infinity. Statement 4) of Theorem 5 provides the link with the Szego limit theorem.

Theorem 5. Let 1 ≤ q1 < q2 < ... < qn be an n-tuple of integers. For each p ≥ n, let

ωp = e2πi

p . Let Γ(p, q) denote the cyclic subgroup of U(n) generated by a diagonal matrixwhose eigenvalues are ω

qj

p , and let Qq(ζ, z) denote its critical polynomial:

Qq(ζ, z) = 1 −n∑

k=1

|zk|2ζqk =

d∏

l=1

(1 − cl(z))ζqk . (76)

1) Assume no cl(z) lies on the unit circle. The limit in (77) then exists and has theindicated value:

limp→∞(

1 − ΦΓ(p,q)(z, z))

1p =

′∏

|cl(z)|. (77)

In (77) the product is taken over those l for which |cl(z)| > 1.2) Suppose ||z||2 = 1. Then 1 − ΦΓ(p,q)(z, z) = 0, and the limit is 0.3) Suppose ||z||2 < 1. Then 1 − ΦΓ(p,q)(z, z) > 0 and

limp→∞(

1 − ΦΓ(p,q)(z, z)) 1

p = 1. (78)

4) Assume no cl(z) lies on the unit circle. Then

limp→∞(

1 − ΦΓ(p,q)(z, z))

1p = exp

(

1

2π

∫ 2π

0

log(|Qq(eiθ, z)|)dθ

)

. (79)

Proof. We first note that 2) holds for all finite subgroups of U(n); it is immediatefrom the definition (61) of the invariant polynomial.

We provide two proofs of 3). We first note, if ||z||< 1 and |ζ| ≤ 1, then

∣

∣

∣

∣

∣

n∑

k=1

|zk|2ζqk

∣

∣

∣

∣

∣

≤n∑

k=1

|zk|2 < 1;

therefore Qq(ζ, z) 6= 0 on |ζ| ≤ 1.

19

First proof of 3). Assume that 1) has been proved. Since Qq(ζ, z) 6= 0 on the closeddisk, |ck(z)| < 1 for each k. The product in (77) therefore equals unity, and 3) holds.

Second proof of 3). Since Qq(ζ, z) 6= 0 for |ζ| ≤ 1, there is a branch of ζ →log(Qq(ζ, z)) holomorphic for |ζ| < 1 and continuous for |ζ| ≤ 1. Hence

1

plog(

1 − ΦΓ(p,q)(z, z))

=

∑pj=1 log(Qq(ω

jp, z))

p. (80)

Since w → log(Qq(w, z)) is continuous, the right hand side of (80) tends to its averagevalue I(z) on |w| = 1. Thus we obtain

limp→∞(

1 − ΦΓ(p,q)(z, z)) 1

p = exp(I(z)). (81)

The average value I(z) is given by the integral

I(z) =1

2π

∫ 2π

0

log(Qq(eiθ, z))dθ. (82)

Since log(Qq) is holomorphic on the closed unit disk, we can evaluate (82) by theCauchy integral formula:

I(z) =1

2π

∫ 2π

0

log(Qq(eiθ, z))dθ =

1

2πi

∫

|w|=1

log(Qq(w, z))

wdw = log(Qq(0, z)) = log(1).

(83)Exponentiating (83) gives 3).

The proof of 4) is similar to the second proof of 3). We begin with (66) and use ourbranch of the p-th root to obtain

(

1 − ΦΓ(p,q)(z, z))

1p =

p∏

j=1

|Qq(ωj)| 1

p eiarg(Qq(ωj ))

p .

Noting that the argument term tends to unity, and then taking logarithms we obtain

limp→∞log(

1 − ΦΓ(p,q)(z, z))

1p = limp→∞

∑pj=1 log(|Qq(ω

j)|)p

. (84)

Since Qq(ζ, z) 6= 0 on the unit circle |ζ| = 1, the logarithm ζ → log(|Qq(ζ, z)|) iscontinuous there. Therefore the limit on the right-hand side of (84) exists and equals theaverage value J(z) of the function ζ → log|Qq(ζ, z)|. Exponentiating yields 4).

Finally we turn to the proof of 1). By Proposition 2 we have

1 − ΦΓ(p,q)(z, z) =

d∏

l=1

(1 − cl(z)p) (85)

and hence

20

(

1 − ΦΓ(p,q)(z, z)) 1

p =

d∏

l=1

(1 − cl(z)p)

1p =

d1∏

l=1

|1 − al(z)p|

2p

d2∏

k=1

(1 − bk(z)p)1p . (86)

By hypothesis no cl(z) lies on the unit circle, and hence the limit exists. Furthermoreno factor is zero, and the number of factors does not grow with p. By (9) each cl(z) insidethe unit circle contributes unity to the limit. By (10) each cl(z) outside the unit circlecontributes its absolute value. Notice that |al(z)| arises twice, corresponding to the twoconjugate roots. We conclude that the limit in (87) exists and has the indicated value:

limp→∞(

1 − ΦΓ(p,q)(z, z))

1p =

′∏

|cl(z)|. (87)

The product includes only those l for which |cl(z)| > 1. Thus 1) holds. ♠

Consider next the limit of ΦΓ(p,q)(z, z)1p . We first note, using the simplest possible

example, that the limits of (ΦΓ)1p and (1 − ΦΓ)

1p are not equal in general.

Example 4. Let Γ be the cyclic subgroup of U(1) generated by a p-th root of unity;

then ΦΓ = |z1|2p, and (ΦΓ(z, z))1p = |z1|2. On the other hand,

(1 − (ΦΓ(z, z))1p → L(z),

where L(z) = |z1|2 if |z1|2 > 1 but L(z) = 1 for |z1|2 < 1. If |z1|2 = 1, then L(z) = 0.

Proposition 4 computes the limit of ΦΓ(p,q)(z, z)1p . From it we determine the asymp-

totics of the coefficient sum in Proposition 5.

Proposition 4. Assume the hypotheses of Theorem 1, and assume also that |cl(z)| >1 for at least one l. Then

limp→∞(

ΦΓ(p,q)(z, z))

1p =

′∏

|cl(z)|, (88)

where again the product is restricted to those l for which |cl(z)| > 1.Proof. We suppress the dependence on z. By Proposition 2,

ΦΓ(p,q) = 1 −d∏

l=1

(1 − cpl ) =

d∑

j=1

cpj −

∑

j<k

(cjck)p + ...±d∏

k=1

cpk

. (89)

The number of terms in the complete sum in (89) is independent of p. Under the hypothesisthat |cl(z)| > 1 for at least one l, there must be at least one term inside the parentheseson the far right-hand side of (89) whose modulus exceeds unity. Consider the productC =

∏′cl(z)

p where each |cl(z)| > 1. Factor out |C| from the big sum in (89) to obtain

ΦΓ(p,q)(z, z) = |C|p(ξ + ...), (90)

21

where |ξ| = 1, and the other terms are of the form wp for |w| < 1. We obtain

limp→∞(

ΦΓ(p,q)(z, z))

1p = |C|. ♠

It follows from Theorem 5 and Proposition 4 that the two limits are equal wheneverQq has at least one root in the unit disk. In Example 4 Qq(ζ, z) = 1− |z1|2ζ, and the onlyroot is 1

|z1|2 . The conclusion is thus consistent with the result of Example 4.

Recall that ΦΓ(p,q)(z, z) is a polynomial with integer coefficients. The sum Sp,q ofthese integers is obtained from ΦΓ(p,q)(z, z) by setting each zj equal to unity. We write 1

for z in this case. Proposition 4 determines the asymptotic behavior of Sp,q whenever noneof the roots of Qq(ζ, 1) lie on the circle and at least one lies inside. This condition musthold when n ≥ 2. When n = 1 the coefficient sum is 1, by Example 4. Assume n ≥ 2; then

Qq(ζ, 1) = 1 −n∑

k=1

ζqk .

Note that Qq(0, 1) > 0 and Qq(1, 1) < 0. By continuity there is a root in the interval(0, 1), and Proposition 4 applies. We therefore obtain

Proposition 5. Let Sp,q denote the sum of the coefficients of ΦΓ(p,q). Then

limp→∞S1pp,q = |C|,

where C is the reciprocal of the product of roots of Qq(ζ, 1) lying inside the unit circle.

Example 5. For each p let Γ be generated by (73) with q = (1, 2). The limit of S1pp,q

can be obtained easily from Proposition 5. In this case Qq(ζ, z) = 1− |z1|2ζ − |z2|2ζ2. Forgeneral z the roots are real:

|z1|2 ±√

|z1|4 + 4|z2|2−2|z2|2

.

The root with the minus sign lies inside the unit circle, and its reciprocal is

|z1|2 +√

|z1|4 + 4|z2|22

.

Evaluating at z = (1, 1) yields the golden mean 1+√

52

.

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