Proper holomorphic mappings of the spectral unit ball
arXiv:0704.0614v2  [math.CV]  14 Jun 2007
PROPER HOLOMORPHIC MAPPINGS
OF THE SPECTRAL UNIT BALL
W lodzimierz Zwonek
Abstract. We prove an Alexander type theorem for the spectral unit ball Ωn show-
ing that there are no non-trivial proper holomorphic mappings in Ωn, n ≥ 2.
Let Mn denote the space of n× n complex matrices.
In order to avoid some trivialities and ambiguities we assume in the whole paper
that n ≥ 2.
Let ρ(A) := max{|λ| : λ ∈ Spec(A)} be the spectral radius of A ∈ Mn. Denote
also by Spec(A) := {λ ∈ C : det(A − λIn) = 0} the spectrum of A ∈ Mn, where
the eigenvalues are counted with multiplicities (In denotes the identity matrix).
We also denote the spectral unit ball by Ωn := {A ∈ Mn : ρ(A) < 1}. Note that
Ωn is an unbounded pseudoconvex balanced domain in C
with the continuous
Minkowski functional equal to ρ. For A ∈ Mn denote PA(λ) := det(λIn − A) =
j=1(−1)
jσj(A)λ
n−j , A ∈ Mn. Denote also σ := (σ1, . . . , σn). We put
Gn := σ(Ωn). The domain Gn is called the symmetrized polydisc. Note that
σ ∈ O(Mn,Gn). Denote also Jn := πn({(ζ1, . . . , ζn) : ζj = ζk for some j 6= k}),
where πn,j(ζ1, . . . , ζn) :=
1≤k1<...<kj≤n
ζk1 · . . . · ζkj , ζl ∈ D, l = 1, . . . , n (D
denotes the unit disc in C). Note that Gn \Jn is a domain and Gn \Jn is dense in
Note that Ωn =
Tz, where Tz := {A ∈ Ωn : σ(A) = z}, z ∈ C
n. The sets
Tz, z ∈ C
n are pairwise disjoint analytic sets. Note that if the matrix A ∈ Tz is non-
degoratory then A is a regular point of Tz – recall that in such a case rankσ
′(A) = n
– it is the largest possible number. For definition and basic properties of non-
derogatory matrices see [Nik-Tho-Zwo 2007] and references there. One of possible
definitions of a non-derogatory matrix is that different blocks in the Jordan normal
form correspond to different eigenvalues (or equivalently all eigenspaces are one-
dimensional). We shall deliver some properties of the sets Tz (see Lemma 5, Lemma
6 and Corollary 7). It is also simple to see that T0 is a cone which contains at least
n2 − n + 1 linearly independent vectors: for instance the ones consisting of one
1 lying not on the diagonal (and with other entries equal to 0) and the matrix
A = (aj,k)j,k=1,... ,n such that a1,1 = 1, a1,2 = −1, a2,1 = 1, a2,2 = −1 (and with all
other entries equal to 0). Consequently, we shall see that 0 is not a regular point
of T0. On the other hand the sets Tπn(ζ1,... ,ζn), where the points ζ1, . . . , ζn ∈ D are
The research was partially supported by the Research Grant No. 1 PO3A 005 28 of the Polish
Ministry of Science and Higher Education.
2000 Mathematics Subject Classification. Primary: 32H35. Secondary: 15A18, 32C25, 47N99
keywords: spectral unit ball, proper holomorphic mappings, symmetrized polydisc
Typeset by AMS-TEX
http://arxiv.org/abs/0704.0614v2
2 W LODZIMIERZ ZWONEK
pairwise different, are submanifolds – it follows from the fact that in this case all
elements of Tπn(ζ1,... ,ζn) are non-derogatory.
It is well-known that for a given mapping F ∈ O(Ωn,Ωn) there exists a mapping
F̃ ∈ O(Gn,Gn) such that σ(F (A)) = F̃ (σ(A)) (see e.g. [Edi-Zwo 2005]).
If f ∈ O(D,D) then one may well-define the following holomorphic mapping
Ff : Ωn ∋ A 7→ f(A) :=
(j)(0)
Aj ∈ Ωn. Note that F̃f (σ(A)) = σ(Ff (A)) =
πn(f(λ1), . . . , f(λn)), where σ(A) = πn(λ1, . . . , λn). In particular, Fa ∈ Aut(Ωn)
for any a ∈ AutD. On the other hand the function FB , where B(λ) = λ
2, λ ∈ D is
a mapping of the form Ω2 ∋ A 7→ A
2 ∈ Ω2, which is not a proper holomorphic one
– it maps T0 into 0.
The structure of the group of automorphisms of Ωn has been been studied in
several papers (see e.g. [Ran-Whi 1991] and [Ros 2003]). However, it is still not
understood completely. Let us mention only that Aut(Ωn) is not transitive. Moti-
vated by the results of the mentioned papers we are going to examine the structure
of the class of proper holomorphic self-mappings of the spectral unit ball. It turns
out that we get an analogue of the theorem of Alexander on proper holomorphic self
mappings of the Euclidean ball in Cn stating that there are no non-trivial proper
holomorphic self maps in the unit ball Bn, n ≥ 2 (see [Ale 1977]).
In the paper we need some properties of proper holomorphic mappings between
complex analytic sets that could be found in [Chi 1989] and [ Loj 1991]. The book
[Rud 1980] may serve as another reference on proper holomorphic mappings (mostly
between open sets in Cn).
Theorem 1. Let F : Ωn 7→ Ωn be a proper holomorphic mapping, n ≥ 2. Then F
is an automorphism.
The following necessary form of proper holomorphic mappings of the spectral
ball, which is a simple consequence of the description of the set of proper holomor-
phic self-mappings of the symmetrized polydisc, will be crucial in our considerations
and justifies the introducing of the condition (1) below.
Proposition 2 (see Theorem 17 in [Edi-Zwo 2005]). Let F : Ωn 7→ Ωn be a proper
holomorphic mapping. Then there is a non-constant finite Blaschke product B such
that σ(F (A)) = πn(B(ζ1), . . . , B(ζn)), where A ∈ Ωn and σ(A) = πn(ζ1, . . . , ζn),
ζ1, . . . , ζn ∈ D.
In view of Proposition 2 it is natural that we study below the holomorphic
mappings F : Ωn 7→ Ωn such that there is a function f ∈ O(D,D) with the property
(1) σ(F (A)) = πn(f(ζ1), . . . , f(ζn)), A ∈ Ωn is such that σ(A) = πn(ζ1, . . . , ζn).
We start with the following lemma.
Lemma 3. Let F ∈ O(Ωn,Ωn) be such that F (0) = 0 and (1) is satisfied for
f ∈ O(D,D) (then necessarily f(0) = 0) with f ′(0) 6= 0. Then F ′(0) is a linear
isomorphism (of Mn).
Proof. Put α := f ′(0). Fix V ∈ Mn. Let πn(µ) = σ(V ) for some µ = (µ1, . . . , µn) ∈
n. We first prove that
F ′(0)(V ) ∈ σ(V ).
PROPER HOLOMORPHIC MAPPINGS OF THE SPECTRAL UNIT BALL 3
Actually, σ(ζV ) = πn(ζµ), ζ ∈ C. Consequently,
σ(F (ζV )) = πn(f(ζµ1), . . . , f(ζµn))
for sufficiently small ζ ∈ D and then
F (ζV )
(f(ζµ1), . . . , f(ζµn))
Passing with ζ to 0 we get that F ′(0)(V ) ∈ Tπn(αµ). Therefore, Φ :=
F ′(0) :
Mn 7→ Mn is a linear mapping such that
(3) Φ(Tz) ⊂ Tz , z ∈ C
To finish the proof of the lemma it is sufficient to show that Φ is a monomorphism.
Suppose that it does not hold. Then there is an N ∈ Mn, N 6= 0 such that
Φ(N) = 0. Because of (3) we get that N ∈ T0. But then there is an M ∈ T0 such
that N + M 6∈ T0. In particular,
T0 6∋ Φ(N + M) = Φ(N) + Φ(M) = Φ(M) ∈ T0
– contradiction.
Lemma 4. Let F ∈ O(Ωn,Ωn) be such that (1) is satisfied with f(0) = 0 and
f ′(0) 6= 0. Then F−1(0) ∩ T0 ⊂ {0}.
Proof. Suppose that there is an A ∈ T0, A 6= 0 such that F (A) = 0. It follows
from the Jordan decomposition theorem that there are linearly independent vectors
v1, v2 ∈ Mn such that A(v2) = v1, A(v1) = 0 (at the moment it is essential that
n ≥ 2). Let (v1, . . . , vn) be a vector base of C
n. Define the linear mapping V : Cn 7→
n (equivalently an element from Mn) as follows V (v2) := v1, V (v1) := v2 and
V (vj) := 0, j = 3, . . . , n. Then (A+ ζV )
2(vj) = ζ(1 + ζ)vj , j = 1, 2. Consequently,
the properties of the spectral radius imply that
|ζ||1 + ζ| ≤ ρ((A + ζV )2) ≤ ρ2(A + ζV ).
For any ζ ∈ C there are µj(ζ) ∈ C, j = 1, . . . , n such that πn(µ1(ζ), . . . , µn(ζ)) =
σ(A + ζV ). Then σ(F (A + ζV )) = πn(f(µ1(ζ)), . . . , f(µn(ζ))) for ζ ∈ D small.
We also know that max{|µj(ζ)| : j = 1, . . . , n} = ρ(A + ζV ) ≥
|1 + ζ| and
ρ(A + ζV ) → 0 as ζ → 0.
Note that ρ(
F (A+ζV )
) → ρ(F ′(A)(V )) as ζ → 0. But on the other hand
F (A + ζV )
= max
|f(µj(ζ))|
: j = 1, . . . , n
which tends to infinity as ζ → 0 because f ′(0) 6= 0 – a contradiction. �
Note that the results proven so far referred to a larger class of mappings than
only proper ones. It is possible that they may have application to the study of more
general mappings than only the proper holomorphic ones.
First we show simple results on the geometry of the sets Tz .
4 W LODZIMIERZ ZWONEK
Lemma 5. The set of non-derogatory matrices is dense in Tz for any z ∈ Gn.
Proof. Fix z ∈ Gn. Let A ∈ Tz. Without loss of generality assume that A is not
non-derogatory. Choose a vector base B in which A has Jordan normal form. Let
us study two different blocks corresponding to the same λ (and the corresponding
vectors from B: v1, . . . , vk, w1, . . . , wl, k, l ≥ 1). Let Av1 = λv1, Avj = λvj + vj−1,
j = 2, . . . , k, Aw1 = λw1, Awj = λwj + wj−1, j = 1, . . . , l. For ǫ > 0 define
Bv1 := λv1 + ǫwl and for all other elements of the base B define Bv := Av, v ∈ B,
v 6= v1. This easily gives an approximation of A with matrices still in Tz having one
block corresponding to the eigenvalue λ less than in the original matrix. Repeating
this procedure for all Jordan blocks having the same eigenvalues we easily construct
a sequence of non-derogatory matrices in Tz tending to A. �
Lemma 6. The set of non-derogatory matrices in Tz is connected and open in Tz
for any z ∈ Gn.
Proof. Fix z ∈ Gn. The non-trivial part of the lemma is the connectedness. Let
us fix a system of numbers (ζ1, . . . , ζn) and the sequence of indices 1 = k1 < k2 <
. . . < kl+1 = n + 1 where ζkj = ζkj+1 = . . . = ζkj+1−1, j = 1, . . . , l (and such that
no other equalities between different ζj ’s hold) and πn(ζ1, . . . , ζn) = z. And now
for any vector base (v1, . . . , vn) of C
n we define the matrix A (more precisely, an
element in Tz ⊂ Mn) as follows Avt = ζjvt + vt−1, j = 1, . . . , l, kj + 1 ≤ t < kj+1,
Avkj = ζjvkj , j = 1, . . . , l. Note that the above mapping is continuous and its
image equals the set of non-derogatory matrices in Tπn(ζ1,... ,ζn). This together with
the fact that the set of all vector basis is connected in Mn completes the proof. �
As a simple corollary of the results on the set of non-derogatory matrices in the
sets Tz we get the following.
Corollary 7. For any z ∈ Gn the set Tz is an analytic irreducible set of codimen-
sion n.
At the moment we are ready to move to the proof of our main result.
Proof of Theorem 1. First recall that when F : Ωn 7→ Ωn is a proper holomor-
phic mapping then there is a finite non-constant Blaschke product B such that
σ(F (A)) = πn(B(ζ1), . . . , B(ζn)), where σ(A) = πn(ζ1, . . . , ζn). In particular,
F (Tπn(ζ1,... ,ζn)) ⊂ Tπn(B(ζ1),... ,B(ζn)), ζj ∈ D, j = 1, . . . , n. But the proper-
ness of F implies even that the equality F (Tπn(ζ1,... ,ζn)) = Tπn(B(ζ1),... ,B(ζn)),
ζj ∈ D, holds – it is sufficient to note that Tz is always connected. Even more,
F|Tπn(ζ1,... ,ζn)
: Tπn(ζ1,... ,ζn) 7→ Tπn(B(ζ1),... ,B(ζn)) is open and proper for any ζj ∈ D,
j = 1, . . . , n.
We claim that for any λ0 ∈ D such that B
′(λ0) 6= 0 (note that such points exist)
the function
(4) F|Tπn(λ0,... ,λ0)
is injective.
Actually, making use of the automorphisms of Ωn and the properties of Blaschke
products we may assume that λ0 = 0, B(0) = 0 and B
′(0) 6= 0. It follows from
Lemma 4 that F−1(0) ∩ T0 = {0}. In particular, F (0) = 0. Now Lemma 3 applies
and we get that F ′(0) is an isomorphism. Consequently, F is locally invertible near
0. Note that there is a neighborhood V of 0 such that #F−1(A) ∩ T0 = 1 for any
A ∈ V ∩ T0. Otherwise there would exist T0 ∋ A
ν , Ãν such that Aν 6= Ãν and
PROPER HOLOMORPHIC MAPPINGS OF THE SPECTRAL UNIT BALL 5
F (Aν) = F (Ãν) → 0. But the properness of F implies that (taking if necessary a
subsequence) either both sequences (Aν), (Ãν) converge to 0 or at least one of the
sequences converges to a non-zero element Ã from T0 such that F (Ã) = 0. In the
first case we contradict the local invertibility of F near 0 and in the second case we
get two points in F−1(0) ∩ T0 – a contradiction, too.
Now the analyticity of the set {A ∈ T0 : #F
−1(A)∩T0 = 1} (see e.g. [ Loj 1991],
Section V.7.1) (the mapping F|T0 : T0 7→ T0 is proper and open) and the fact that
T0 is a cone shows that the mapping F|T0 : T0 7→ T0 is a one-to-one mapping.
Now we prove the following property.
(5) Let zν → z0 ∈ Gn, where z
ν ∈ Gn be such that F|Tzν is not injective for any ν
then F|T
is not injective.
Actually, to prove (5) note that because of the properties of proper holomorphic
mappings we may assume that there are two sequences of non-derogatory matrices
(Aν), (Ãν) with Aν 6= Ãν lying in Tzν , F (A
ν) = F (Ãν) and tending to matrices
A, Ã ∈ Tz0 such that A is non-derogatory and F|T
is locally invertible in A. In
the case A 6= Ã we are done, so assume that A = Ã. The local invertibility of F|T
near A implies that there is a neighborhood U of A such that F|Tzν∩U is invertible
for ν large enough, which contradicts the equality F (Aν) = F (Ãν).
We claim that
(6) for any z ∈ Gn the mapping F|Tz is injective.
Put U := {z ∈ Gn : F|Tz is injective}. The fact that F|T0 is injective shows
that U is not empty. The property (5) shows that U is open. To see that U is
closed in Gn take a sequence U ∋ z
ν → z ∈ Gn. Suppose that z 6∈ U . Then
there are different non-derogatory matrices A1, . . . , Ak ∈ Tz , C ∈ Tw with k ≥ 2
such that F−1(C) ∩ Tz = {A1, . . . , Ak}. We may choose arbitrarily small open
connected neighborhoods U1, . . . ,Uk,V of A1, . . . , Ak, C such that Ul ∩ Up = ∅ for
l 6= p, Uj ∩ Tz̃ is connected, V ∩ Tw̃ is connected for any z̃, w̃ ∈ Gn, j = 1, . . . , k
and F−1(V) ⊂
j=1 Uj . Consequently, for any ν there are pairwise disjoint sets
F (Uj ∩ Tzν )∩V , j = 1, . . . , k that are open in V ∩ Twν and that are non-empty for
ν large enough. Now the properness of F shows that for V sufficiently small the
sets F (Uj ∩ Tzν ), j = 1, . . . , k cover the whole set V ∩ Twν for ν large enough; thus
contradicting the connectedness of V ∩ Twν .
Since Gn is connected we get that U = Gn, so (6) is satisfied.
Let m denote the degree of B. We claim that m = 1. Suppose that m ≥ 2. Note
that taking instead of F the composition of many F ’s we may assume that m ≥ n.
There is a point ζ0 ∈ D such that B
−1(ζ0) = {ζ1, . . . , ζm}. Composing, if necessary,
with automorphisms of Ωn we may assume that ζ0 = 0. Recall that Tπn(ζ1,... ,ζn) is
an n2 − n dimensional submanifold. Choose A ∈ Tπn(ζ1,... ,ζn) such that F (A) = 0.
Let f := F|Tπn(ζ1,... ,ζn)
: Tπn(ζ1,... ,ζn) 7→ T0. Then f is a holomorphic bijective
mapping. Let us fix a regular point C in T0. Then the function
ϕ : C ∋ λ 7→ f−1(λC) ∈ Tπn(ζ1,... ,ζn)
is holomorphic on C \ {0} (the points λC, λ ∈ C \ {0}, are regular in T0) and
continuous at 0 with ϕ(0) = A (use the properness and injectivity of f). Con-
sequently, ϕ is holomorphic on C. Note that (F ◦ ϕ)(λ) = λC, λ ∈ C, so the
tangent space to Tπn(ζ1,... ,ζn) at A i.e. TA(Tπn(ζ1,... ,ζn)) is mapped onto H :=
F ′(A)(TA(Tπn(ζ1,... ,ζn))), which contains the vector F
′(A)(ϕ′(0)) = (F ◦ϕ)′(0) = C.
6 W LODZIMIERZ ZWONEK
Consequently, H contains all regular points of T0, so it contains the whole T0, which
contains n2−n+ 1 linearly independent vectors contradicting the fact that H is at
most n2 − n dimensional vector space.
Consequently, we have proven that #F−1(C) = 1 for C ∈ Ωn showing that F is
an automorphism. �
Acknowledgment. The author wishes to express his gratitude to Witold Jarnicki
for fruitful conversations on the properties of proper holomorphic mappings between
analytic sets.
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Instytut Matematyki, Uniwersytet Jagielloński, Reymonta 4, 30-059 Kraków,
Poland
E-mail address: Wlodzimierz.Zwonek@im.uj.edu.pl
