Proper holomorphic self-maps of symmetric powers of balls

TL;DR

This paper proves that proper holomorphic self-maps of symmetric powers of higher-dimensional balls are actually automorphisms derived from automorphisms of the original ball, revealing a strong structural rigidity.

Contribution

It establishes a rigidity result for proper holomorphic self-maps of symmetric powers of balls, showing they are induced by automorphisms of the ball itself.

Findings

Proper holomorphic self-maps are automorphisms

Induced by automorphisms of the original ball

Results hold for balls of dimension at least two

Abstract

We show that each proper holomorphic self map of a symmetric power of the unit ball is an automorphism naturally induced by an automorphism of the unit ball, provided the ball is of dimension at least two.

Equations

$${\Sigma }^m\mathbb{D}=\left\{({\sigma }_1(z),{\sigma }_2(z),\dots ,{\sigma }_m(z))\in {\mathbb{C}}^m|z\in {\mathbb{D}}^m\right\}$$

$$f(\sigma (z^1,\dots ,z^m)=\sigma (B(z^1),\dots ,B(z^m)).$$

$$g_{\mathrm{Sym}}^m(⟨z^1,z^2,\dots ,z^m⟩)=⟨g(z^1),g(z^2),\dots ,f(z^m)⟩,$$

$$f=g_{\mathrm{Sym}}^m,$$

$${\phi }_a(z)=\frac{a-P_{a}z-s_a{Q}_{a}z}{1-⟨z,a⟩},$$

$$\{(x^1,\dots ,x^m),x^j\in X,j=1,\dots ,m\}.$$

$$\sigma \cdot x=(x^{\sigma (1)},\dots ,x^{\sigma (m)}).$$

$$\pi :X^m\to X_{\mathrm{Sym}}^m$$

$$v^{\sigma (1)}\otimes v^{\sigma (2)}\otimes \cdots \otimes v^{\sigma (m)}\sim v^1\otimes v^2\otimes \cdots \otimes v^m$$

$${\mathbf{e}}_{\mu }=e_{i_1}\odot e_{i_2}\odot \cdots \odot e_{i_m}$$

$$\dim V^{\odot m}=\left(\genfrac{}{}{0ex}{}{m+s}{m}\right).$$

$$⟨v^1,v^2,\dots ,v^m⟩\mapsto v^1\odot v^2\odot \cdots \odot v^m,$$

$$\psi :(\mathbb{P}(V){)}_{\mathrm{Sym}}^m\to \mathbb{P}(V^{\odot m})$$

$$\psi (⟨[v^1],[v^2],\dots ,[v^m]⟩)=\left[v^1\odot v^2\odot \cdots \odot v^m\right]\in \mathbb{P}(V^{\odot m}),$$

$$N(m,s)=\left(\genfrac{}{}{0ex}{}{m+s}{m}\right)-1={\dim }_{\mathbb{C}}\left(({\mathbb{C}}^{s+1}{)}^{\odot m}\right)-1.$$

$$i_{\mathrm{Sym}}^m:E_{\mathrm{Sym}}^m\to (\mathbb{C}{\mathbb{P}}^s{)}_{\mathrm{Sym}}^m,$$

$$\psi \circ i_{\mathrm{Sym}}^m:E_{\mathrm{Sym}}^m\to \mathbb{C}{\mathbb{P}}^{N(m,s)}$$

$${\Sigma }^{m}E=\psi \circ i_{\mathrm{Sym}}^m(E_{\mathrm{Sym}}^m)\subset \mathbb{C}{\mathbb{P}}^{N(m,s)}.$$

$$z\mapsto ({\sigma }_1(z),\dots ,{\sigma }_m(z)),$$

$$i_{\ast }:{\pi }_1(M\setminus A,x)\to {\pi }_1(M,x)$$

$$i_{\ast }:{\pi }_1(X\setminus Y,x)\to {\pi }_1(X,x)$$

$$A=\bigcup _{i=1}^n{A}_i,$$

$$M_k=M\setminus B_k$$

$$M_k=M_{k-1}\setminus A_k,$$

$$⟨x^1:m_1,x^2:m_2,\dots ,x^k:m_k⟩$$

$$\alpha =⟨x^1:m_1,\dots ,x^k:m_k⟩,$$

$$\tilde{V}(m_1,\dots ,m_k)={\pi }^{-1}(V(m_1,\dots ,m_k))\subset {{\mathbb{B}}_s}^m.$$

$$\pi :\tilde{V}(m_1,\dots ,m_k)\to V(m_1,\dots ,m_k)$$

$$\frac{m!}{m_1!m_2!\cdots m_k!}.$$

$$U_{\alpha }=\{⟨{\alpha }^1:m_1,\dots ,{\alpha }^k:m_k⟩\in V(m_1,\dots ,m_k)\mid {\alpha }^i\in U_i\}.$$

Full text

Debraj Chakrabarti was partially supported by a grant from the NSF (#1600371), a collaboration grant from the Simons Foundation (# 316632), and an Early Career internal grant from Central Michigan University. Christopher Grow was partially supported by a Research Assistantship from the Central Michigan University mathematics department.

1. Introduction

Let $\mathbb{D}^{m}$ denote the $m$-dimensional polydisc in $\mathbb{C}^{m}$, and for $1\leq k\leq m$, denote by $\sigma_{k}$ the $k$-th elementary symmetric polynomial in $m$ variables. The subset of ${\mathbb{C}}^{m}$ given by

[TABLE]

is known as the symmetrized polydisc of $m$-dimensions. It turns out that $\Sigma^{m}\mathbb{D}$ is a pseudoconvex domain ${\mathbb{C}}^{n}$ with remarkable function theoretic properties, and applications to engineering (see [1] and the work inspired by it.) Of particular interest are the symmetries and mapping properties of these domains. In [11], Jarnicki and Pflug determined the biholomorphic automorphisms of $\Sigma^{m}\mathbb{D}$. More generally, we have the following result of Edigarian and Zwonek on proper self-maps of $\Sigma^{m}\mathbb{D}$:

Theorem 1** (See [7, 8]).**

Let $f:\Sigma^{m}\mathbb{D}\to\Sigma^{m}\mathbb{D}$ be a proper holomorphic map, and let $\sigma=(\sigma_{1},\dots,\sigma_{m}):\mathbb{C}^{m}\to\mathbb{C}^{m}$ be as in (1.1). Then, there exists a proper holomorphic map $B:\mathbb{D}\to\mathbb{D}$ such that

[TABLE]

Recall that a proper holomorphic map $B:\mathbb{D}\to\mathbb{D}$ is represented by a finite Blaschke product, so the above result gives a complete characterization of proper holomorphic self-maps of $\Sigma^{m}\mathbb{D}$, so that each such map is induced by a proper holomorphic self-map of the disc.

In this note we prove an analogous result for proper self-maps of complex analytic spaces analogous to the symmetrized polydisc, where the disc is replaced by a higher dimensional ball. To define these spaces, let $\mathbb{B}_{s}\subset{\mathbb{C}}^{s}$ denote the unit ball in ${\mathbb{C}}^{s}$, and for some positive integer $m$, let $(\mathbb{B}_{s})^{m}_{\rm Sym}$ denote the $m$-fold symmetric power of $\mathbb{B}_{s}$, i.e., the collection of all unordered $m$-tuples $\langle z^{1},z^{2},\dots,z^{m}\rangle$, where each $z^{j}\in\mathbb{B}_{s}$. Note that the construction of the symmetric power is functorial: given any map $g:\mathbb{B}_{s}\to\mathbb{B}_{s}$, there is a naturally defined map $g^{m}_{\rm Sym}:(\mathbb{B}_{s})^{m}_{\rm Sym}\to(\mathbb{B}_{s})^{m}_{\rm Sym}$ given by

[TABLE]

which is easily seen to be well-defined. See below in Section 2 for a discussion of this notion, and further details. Then, $(\mathbb{B}_{s})^{m}_{\rm Sym}$ is a complex analytic space, and $\Sigma^{m}\mathbb{D}$ is biholomorphic to $\mathbb{D}^{m}_{\rm Sym}$. The main result of this paper is:

Theorem 2**.**

Let $s\geq 2$, $m\geq 2$, and let $f:(\mathbb{B}_{s})^{m}_{\rm{Sym}}\to(\mathbb{B}_{s})^{m}_{\rm{Sym}}$ be a proper holomorphic map. Then, there exists a holomorphic automorphism $g:\mathbb{B}_{s}\to\mathbb{B}_{s}$ such that

[TABLE]

that is $f$ is the $m$-th symmetric power of $g$.

It follows in particular that each proper self-map of $(\mathbb{B}_{s})^{m}_{\rm{Sym}}$ is in fact an automorphism. Recall also (see [13, p. 25] that an automorphism of $\mathbb{B}_{s}$ is of the form

[TABLE]

where $a\in\mathbb{B}_{n}$, $P_{a}$ is the orthogonal projection from $\mathbb{C}^{n}$ onto the one dimensional complex linear subspace spanned by $a$, $Q_{a}=\operatorname{id}-P_{a}$ is the orthogonal projection from $\mathbb{C}^{n}$ onto the orthogonal complement of the one dimensional complex linear subspace spanned by $a$, and $s_{a}=(1-|a|^{2})^{\frac{1}{2}}$.

Note also that the domain $\mathbb{B}_{s}$ may be replaced in Theorem 2 by any strongly pseudoconvex domain, without any change in the proof. We prefer to state it in this special case for simplicity.

2. Symmetric Powers

2.1. Complex Symmetric powers

Recall that, informally, a complex analytic space is made of local analytic subsets of ${\mathbb{C}}^{n}$ glued together analytically, just as a complex manifold of dimension $s$ is made of open sets of ${\mathbb{C}}^{s}$ analytically glued together (see [6, 14, 10] for more information). Recall also that an analytic subset of ${\mathbb{C}}^{n}$ is given near each of point of ${\mathbb{C}}^{n}$ by the vanishing of a family of analytic functions, and a local analytic subset of ${\mathbb{C}}^{n}$ is an open subset of an analytic subset of ${\mathbb{C}}^{n}$.

Let $X$ be a complex manifold, or more generally, a complex analytic space. let $X^{m}$ denote the $m$-th Cartesian power of $X$, which is by definition the collection of ordered tuples

[TABLE]

$X^{m}$ is then a complex manifold in an obvious way. The symmetric group $S_{m}$ of bijective mappings of the set $\{1,\dots,m\}$ acts on $X^{m}$ in as biholomorphic automorphisms: for $\sigma\in S_{m}$, and $x=(x^{1},\dots,x^{m})\in X^{m}$ we set

[TABLE]

The $m$-th symmetric Power of $X$, denoted by $X^{m}_{\rm Sym}$ is the quotient of $X^{m}$ under the action of $S_{m}$ defined above, i.e, points of $X^{m}_{\rm Sym}$ are orbits of the action of $S_{m}$ on $X^{m}$. We denote by

[TABLE]

the natural quotient map. It follows from the general theory of complex analytic spaces that $X^{m}_{\rm Sym}$ has a canonical structure of an analytic space, i.e., the quotient analytic space of $X^{m}$ under the action of $S_{m}$ as biholomorphic automorphisms (see [3, 4] and [10]). When $X^{m}_{\rm Sym}$ is given this complex structure, the map $\pi$ becomes a proper holomorphic map.

In our application, we are interested in the case when $X=\mathbb{B}_{s}$, the unit ball in ${\mathbb{C}}^{s}$. The symmetric power $(\mathbb{B}_{s})^{m}_{\rm Sym}$ is then biholomorphic to a local analytic set, in fact to an open subset of a certain affine algebraic variety in ${\mathbb{C}}^{N}$ for some large $N$ depending on $m$ and $s$. Though logically not needed for the proof of Theorem 2, we give a short account of this construction in order to explain the relation of Theorem 2 with Theorem 1, as well as to emphasize the elementary nature of the constructions.

2.2. Embedding of symmetric powers of projective spaces

Recall that the $m$-fold symmetric tensor product of a (complex) vector space $V$, denoted $V^{\odot m}$ is defined as $V^{\odot m}=V^{\otimes m}/\sim$, where $V^{\otimes m}=V\otimes V\otimes\dots\otimes V$ is the $m$-fold tensor product of $V$ with itself, and the equivalence relation $\sim$ is defined by:

[TABLE]

for all $v^{1},v^{2},\dots,v^{m}\in V$ and all $\sigma\in S_{m}$. We denote the equivalence class of $v^{1}\otimes v^{2}\otimes\dots\otimes v^{m}$ under this equivalence relation (which is an element of $V^{\odot m}$) by $v^{1}\odot v^{2}\odot\dots\odot v^{m}.$ Just as the tensor product of vector spaces is itself a vector space, the symmetric tensor product $V^{\odot m}$ of a vector space is again a vector space, with a natural linear structure as a quotient vector space of $V^{\otimes m}$. Suppose $V$ is finite-dimensional of dimension $s+1$, and let $\{e_{0},\dots,e_{s}\}$ be a basis of $V$. Let $\mu=(\mu_{0},\mu_{2},\dots,\mu_{s})\in\mathbb{N}^{s+1}$ be a multi-index with $|\mu|=\sum_{j=0}^{s}\mu_{j}=m$, and let $\mathbf{e}_{\mu}$ denote the element of $V^{\odot m}$ given by

[TABLE]

where exactly $\mu_{j}$ of the $e_{i_{k}}$ are equal to $e_{j}$. It is not difficult to see that $\{\mathbf{e}_{\mu}\}_{|\mu|=m}$ gives a basis for $V^{\odot m}$. Therefore the dimension of $V^{\odot m}$ is the number of solutions of $\sum_{j=0}^{s}\mu_{j}=m$, i.e.

[TABLE]

Given a vector space $V$, there is a natural mapping from the symmetric power, $V^{m}_{\rm Sym}$ (which is just a set), to the symmetric tensor product $V^{\odot m}$ (which is a vector space), given by

[TABLE]

which is easily seen to be well-defined.

Denote by $\mathbb{P}(V)$ the projectivization of the vector space $V$, and for $v\in V$, denote by $[v]$ its equivalence class in the projectivization $\mathbb{P}(V)$. There is a natural map

[TABLE]

induced by the map (2.4) given by

[TABLE]

with $v^{j}\in V$. We call this the Segre-Whitney map (see Appendix V of [14]). It is easily seen to be well-defined and injective, and is a symmetric version of the well-known classical Segre embedding of the product of two or more projective spaces as a projective variety in a higher dimensional projective space.

Thanks to classical results in complex algebraic geometry, the image $\psi\left(\mathbb{P}(V))^{m}_{\rm Sym}\right)$ is a projective algebraic variety in the projective space $\mathbb{P}(V^{\odot m})$, which, if $V={\mathbb{C}}^{s+1}$ is of dimension

[TABLE]

2.3. Embedding of $(\mathbb{B}_{s})^{m}_{\rm Sym}$

Let $E$ be a subset of $\mathbb{C}\mathbb{P}^{s}=\mathbb{P}({\mathbb{C}}^{s+1})$, and let $i:E\to\mathbb{C}\mathbb{P}^{s}$ be the inclusion map. Then we have a natural inclusion

[TABLE]

so that we can think of $E^{m}_{\rm Sym}$ as a subset of $(\mathbb{C}\mathbb{P}^{s})^{m}_{\rm Sym}$. Composing with the Segre-Whitney embedding $\psi$ of $(\mathbb{C}\mathbb{P}^{s})^{m}_{\rm Sym}$ in $\mathbb{C}\mathbb{P}^{N(m,s)}$, where $N(m,s)$ is defined as in (2.6), we obtain an embedding

[TABLE]

which we will again call the Segre-Whitney embedding. In this way we can think of symmetric powers as sitting in some projective space. We will denote this embedded version of the $m$-th symmetric power of $E$ by $\Sigma^{m}E$, i.e.,

[TABLE]

Two special cases of this construction are relevant here. The first is when $E$ is an affine piece of $\mathbb{C}\mathbb{P}^{s}$, so that $E$ can be identified with ${\mathbb{C}}^{s}$. Then, explicitly working through the computations, one can verify that $\Sigma^{m}E$ is an affine algebraic variety in an affine piece of $\mathbb{C}\mathbb{P}^{N(m,s)}$. So we can think of $\Sigma^{m}\mathbb{C}^{s}$ as an affine algebraic variety in $\mathbb{C}^{N(m,s)}$. For details of this construction, see [14, Appendix V].

Now, if $E=\mathbb{B}_{s}$ is the unit ball in an affine piece of $\mathbb{C}\mathbb{P}^{s}$, then it is easy to see that $\Sigma^{m}E=\Sigma^{m}\mathbb{B}_{s}$ is an open subset of the affine algebraic variety $\Sigma^{m}\mathbb{C}^{s}$. In this way, $(\mathbb{B}_{s})^{m}_{\rm Sym}$ is realized as a local analytic set, i.e., an open set of an analytic subset of $\mathbb{C}^{N(m,a)}$ (actually an open set of an affine algebraic variety).

2.4. The case $s=1$

When $s=1$, we have $N(m,s)=m$, and it is not difficult to see that the Segre-Whitney map ${\mathbb{C}}^{m}_{\rm Sym}\to{\mathbb{C}}^{m}$ is actually a biholomorphism, given by

[TABLE]

where $\sigma_{k}(\langle z_{1},\dots,z_{m}\rangle)$ is the $k$-th elementary symmetric polynomial in the variables $(z_{1},\dots,z_{m})$. Then the image of $\mathbb{D}^{m}_{\rm Sym}$ is a pseudoconvex domain $\Sigma^{m}\mathbb{D}$ in ${\mathbb{C}}^{m}$, called the symmetrized polydisc. See [5] for more details. Consequently, the symmetric power $\mathbb{D}^{m}_{\rm Sym}$ is biholomorphically identified with the domain $\Sigma^{m}\mathbb{D}$ in ${\mathbb{C}}^{m}$, which shows that Theorem 2 is indeed an extension of Theorem 1.

3. Proper mappings of Cartesian to Symmetric powers

The first step in the proof of Theorem 2 is the following result, which is interesting in its own right:

Theorem 3**.**

Let $s\geq 2$, $m\geq 2$, and let $f:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}_{\rm{Sym}}$ be a proper holomorphic map. Then, there exists a proper holomorphic map $\widetilde{f}:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}$ such that $f=\pi\circ\widetilde{f}$.

In other words, the map $f$ can be lifted to a proper holomorphic map $\widetilde{f}$ such that the following diagram commutes:

(\mathbb{B}_{s})^{m}$$(\mathbb{B}_{s})^{m}_{\rm{Sym}}$$(\mathbb{B}_{s})^{m}$$f$$\pi$$\widetilde{f}

Note however, that the map $\pi$ is not a covering map, so that the classical theory of lifting maps into a covering space is not directly applicable. However, as we will see, we can reduce this problem to a problem involving covering maps by removing the ramification locus and the branching locus of the map $\pi$ from $(\mathbb{B}_{s})^{m}$ and $(\mathbb{B}_{s})^{m}_{\rm Sym}$ respectively.

3.1. Fundamental group of complements of analytic sets

The proof of Theorem 3 will involve the computation of some fundamental groups, for which we need the following fact. We include a proof for completeness.

Proposition 3.1**.**

Let $M$ be a connected complex manifold without boundary, $A\subset M$ be an analytic subset, $x$ a point in $M\setminus A$, and $i:M\setminus A\to M$ the inclusion map. Then if the complex codimension of $A$ is at least 2, then the homomorphism of the fundamental groups

[TABLE]

is an isomorphism.

Proof.

We begin by recalling the following fact from Differential Topology: Let $X$ be a connected $\mathcal{C}^{\infty}$ differentiable manifold without boundary, $Y\subset M$ be closed submanifold, $x$ a point in $X\setminus Y$, and $i:X\setminus Y\to X$ the inclusion map. Then if the codimension of $Y$ is at least 3, then the group homomorphism

[TABLE]

*is an isomorphism. *

For a proof, see [9, Théorème 2.3, page 146]. Essentially, this is a reflection of the fact that thanks to the low codimension of $Y$, by a standard transversality argument, there is no problem in homotopically deforming a loop based at $x$ to a loop based at $x$ and not intersecting $Y$, and further, given two loops based at $x$ homotopic in $X$, there is no problem in homotopically deforming them to each other in $X\setminus Y$.

Now, since $A$ is an analytic subset of a complex manifold $M$, there is a stratification of $A$ by local analytic subsets (see [6] for details). More precisely, there exist pairwise disjoint local analytic subsets $A_{j}$ of $A$ such that

[TABLE]

where the set $B_{k}=\bigcup_{i=1}^{k}A_{i}$ is an analytic subset of $M$ and $A_{k}$ is a closed submanifold of $M\setminus B_{k-1}$ for each $1\leq k\leq n$. Note that, since $A$ has codimension at least $2$ in $M$, then $A_{k}$ also has codimension at least $2$ in $M\setminus B_{k-1}$ for each $1\leq k\leq n$. Thus, assuming $A\neq M$,

[TABLE]

is a open submanifold of $M$ for each $1\leq k\leq n$. Hence, since

[TABLE]

each inclusion in the following chain

M\setminus A=M_{n}$$M_{n-1}$$\dots$$M_{1}$$M_{0}=M$$i$$i$$i$$i

is an isomorphism of groups by the fact from Differential Topology quoted in the first paragraph, since we may take $X=M_{k}$ and $Y=A_{k}$ and the conditions are satisfied. The conclusion follows. ∎

3.2. Branching behavior of $\pi$

Since $\pi$ is proper holomorphic map of equidimensional complex analytic sets, it follows that $\pi$ must be a covering map when the analytic sets over which it is branched are removed from the source and the target. However, given the elementary nature of the considerations here, one can be much more explicit in this special case. Let $m_{1},\dots,m_{k}$ be a partition of $m$, i.e., $m_{j}$ be positive integers such that $\sum_{j=1}^{k}m_{j}=m$. We denote by

[TABLE]

the element of $(\mathbb{B}_{s})^{m}_{\rm Sym}$ in which $x^{j}$ is repeated $m_{j}$ times. Let $V(m_{1},\dots,m_{k})$ be the set of points $\alpha$ in ${\mathbb{B}_{s}}^{m}_{\rm Sym}$ such that there are distinct $x^{1},\dots,x^{k}\in{\mathbb{B}_{s}}$ such that

[TABLE]

where we use the notation (3.1), that is $x^{j}$ is repeated $m_{j}$ times. Also let

[TABLE]

Proposition 3.2**.**

The restricted map

[TABLE]

is a holomorphic covering map of degree

[TABLE]

Proof.

Let $\alpha\in V(m_{1},\dots,m_{k})$. We will show that there is an open set $U_{\alpha}\subset V(m_{1},\dots,m_{k})$, containing $\alpha$, such that $\pi^{-1}(U_{\alpha})$ is a disjoint union of open subsets $\widetilde{U}_{\alpha}^{i}\subset\widetilde{V}(m_{1},\dots,m_{k})$, with the restriction of $\pi$ to each $\widetilde{U}_{\alpha}^{i}$ a homeomorphism onto its image $\pi(\widetilde{U}_{\alpha}^{i})$.

We can write $\alpha=\langle x^{1}:m_{1},\dots,x^{k}:m_{k}\rangle$ for some distinct $x^{1},\dots,x^{k}\in{\mathbb{B}_{s}}$. Since ${\mathbb{B}_{s}}$ is Hausdorff, there exist disjoint open subsets $U_{i}\subset{\mathbb{B}_{s}}$ with $x^{i}\in U_{j}$ if and only if $i=j$. Now, let

[TABLE]

Note that $U_{\alpha}$ is open in in the subspace topology defined on $V(m_{1},\dots,m_{k})$, since

[TABLE]

where $U_{i}^{m_{i}}$ denotes the $m_{i}$-fold Cartesian power of $U_{i}$. Let $\sigma(y^{1},\dots,y^{m})$ denote $(y^{\sigma(1)},\dots,y^{\sigma(m)})$ for $\sigma\in S_{m}$ and $(y^{1},\dots,y^{m})\in{\mathbb{B}_{s}}^{m}$. Then, $\pi^{-1}(\langle y^{1},\dots,y^{m}\rangle)=\{\sigma(y^{1},\dots,y^{m})\mid\sigma\in S_{m}\}$, and we have

[TABLE]

Since the sets $\sigma(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})$ are open in ${\mathbb{B}_{s}}^{m}$, it follows that $\pi^{-1}(U_{\alpha})$ is open in $\widetilde{V}(m_{a},\dots,m_{k})$ with the subspace topology. Also, since $U_{i}\cap U_{j}=\emptyset$ for $i\neq j$, the sets $\sigma(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})$ and $\tau(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})$ are either identical or disjoint for each $\sigma,\tau\in S_{m}$, and the number of distinct sets $\sigma(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})$ is equal to the number of distinct preimages $\pi^{-1}(\langle\alpha^{1}:m_{1},\dots,\alpha^{k}:m_{k}\rangle)$, which is exactly

[TABLE]

Now, the restricted map $\pi:\sigma(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})\to\pi(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})$ is one-to-one, and hence a homeomorphism, as $\pi$ is a quotient map. Thus, by restricting $\pi$ to the subspace $\sigma(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})\cap\widetilde{V}(m_{1},\dots,m_{k})$, we have $\pi:\sigma(U_{1}^{m_{1}}\times\cdots\times U_{k}^{m_{k}})\cap\widetilde{V}(m_{1},\dots,m_{k})\to U_{\alpha}$ is a homeomorphism as well. ∎

Proposition 3.3**.**

Let $X$ and $Y$ be analytic subsets of $\Omega_{1}\subset\mathbb{C}^{n}$ and $\Omega_{2}\subset\mathbb{C}^{m}$, respectively, and let $f:X\to\Omega_{2}$ be a proper holomorphic map such that $f(X)\subset Y$. Then,

1. (1)

If $f(X)=Y$, then $\dim X=\dim Y$. 2. (2)

If $Y$ is irreducible and $\dim X=\dim Y$, then $f(X)=Y$.

Proof.

By Remmert’s Theorem, $f(X)$ is an analytic subset of $\Omega_{2}$, with $\dim f(X)=\dim f$. Since $f$ is proper, we cannot have $\dim f<\dim X$. Otherwise, the Rank Theorem would imply that for some $y\in f(X)$, $f^{-1}(y)$ is a compact analytic subset of $X$ with positive dimension, which is impossible. Hence, we conclude that $\dim f(X)=\dim X$.

Suppose first $f(X)=Y$. Then, evidently, $\dim X=\dim f(X)=\dim Y$, establishing (1).

Now suppose $\dim X=\dim Y$ and suppose that $f(X)\neq Y$. Then, by well-known properties of analytic sets, $\overline{Y\setminus f(X)}$ is an analytic set, and since $Y$ is closed in $\Omega_{2}$, $\overline{Y\setminus f(X)}$ is contained in $Y$. Additionally, since $\dim f(X)=\dim Y$, $\overline{Y\setminus f(X)}$ cannot be all of $Y$. Now, since $Y=f(X)\cup\overline{Y\setminus f(X)}$, $Y$ is reducible. Hence, if $Y$ is irreducible and $\dim X=\dim Y$, then $f(X)=Y$, completing the proof of (2).

∎

We now prove the following lemma:

Lemma 3.4**.**

Let A=\{(z^{1},\dots,z^{m})\in(\mathbb{B}_{s})^{m}:z^{i}=z^{j}\text{ for some i\neq j}\}. Then $A$ is an analytic subset of $(\mathbb{B}_{s})^{m}$ of codimension $s$ and $\pi(A)$ is an analytic subset of $(\mathbb{B}_{s})^{m}_{\rm Sym}$ of codimension $s$. The restricted map

[TABLE]

is a holomorphic covering map of complex manifolds.

Proof.

Let $A_{ij}$ be the linear subspace of $(\mathbb{C}^{s})^{m}\cong\mathbb{C}^{sm}$ given by

[TABLE]

Then $A_{ij}$ is defined by the vanishing of $s$ linearly independent linear functionals $z\mapsto z^{i}_{k}-z^{j}_{k}$ where $1\leq k\leq s$, and consequently $A_{ij}$ is of codimension $s$ in $(\mathbb{C}^{s})^{m}$. Since

[TABLE]

it now follows that $A$ is an analytic subset of codimension $s$ in $(\mathbb{B}^{s})^{m}$. Since the finite-to-one quotient map $\pi:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}_{\rm Sym}$ is proper and holomorphic, by Remmert’s Theorem, $\pi(A)$ is an analytic subset of $(\mathbb{B}_{s})^{m}_{\rm Sym}$. Since $\pi^{-1}(\pi(A))=A$, it follows that $\pi|_{A}$ is a proper map $A\to\pi(A)$, and since $\pi$ is also surjective, it is easy to see that we must have $\dim A=\dim\pi(A)$. Since $\dim((\mathbb{B}_{s})^{m})=\dim((\mathbb{B}_{s})^{m}_{\rm Sym})$, it follows that $\pi(A)$ has the same codimension in $(\mathbb{B}_{s})^{m}_{\rm Sym}$ as $A$ has in $(\mathbb{B}_{s})^{m}$, which is $s$. ∎

Recall that, for a partition $m_{1},\dots,m_{k}$ of $m$, $V(m_{1},\dots,m_{k})$ is the set of points $\alpha$ in $(\mathbb{B}_{s})^{m}_{\rm Sym}$ such that there are distinct $z^{1},\dots,z^{k}\in\mathbb{B}_{s}$ such that

[TABLE]

where this notation is as in (3.1), i.e., $z^{j}$ is repeated $m_{j}$ times. Let us also set

[TABLE]

Then, $(\mathbb{B}^{s})^{m}\setminus A$ is precisely the set $\widetilde{V}(m_{1},\dots,m_{k})$ and $(\mathbb{B}^{s})^{m}_{\rm Sym}\setminus\pi(A)$ is precisely the set $V(m_{1},\dots,m_{k})$.

We are now ready to prove Theorem 3.

Proof of Theorem 3.

Let $f:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}_{\rm{Sym}}$ be a proper holomorphic map and let $A$ be as defined in Lemma 3.4. Since $\pi(A)$ is an analytic subset in $(\mathbb{B}_{s})^{m}_{\rm{Sym}}$, $(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)$ is an open, connected set.

Since $(\mathbb{B}_{s})^{m}\setminus A=\widetilde{V}(1,1,\dots,1)$ and $(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)=V(1,1,\dots,1)$, by Proposition 3.2, $\pi|_{(\mathbb{B}_{s})^{m}\setminus A}$ is a holomorphic covering $(\mathbb{B}_{s})^{m}\setminus A\to(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)$. Since a holomorphic covering is a local biholomorphism, $(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)$ is an $sm$-dimensional complex manifold. Moreover, since $\pi(A)$ is an analytic subset of $(\mathbb{B}_{s})^{m}_{\rm{Sym}}$, we have that $(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)$ is connected and dense in $(\mathbb{B}_{s})^{m}_{\rm Sym}$. Since $\operatorname{reg}((\mathbb{B}_{s})^{m}_{\rm Sym})$ is an open subset of $(\mathbb{B}_{s})^{m}_{\rm Sym}$ containing $(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)$, $\operatorname{reg}((\mathbb{B}_{s})^{m}_{\rm Sym})$ must be connected, and hence $(\mathbb{B}_{s})^{m}_{\rm Sym}$ is an irreducible analytic set, with $\dim((\mathbb{B}_{s})^{m}_{\rm Sym})=\dim((\mathbb{B}_{s})^{m}_{\rm Sym}\setminus\pi(A))$. Since $f$ is a proper holomorphic map from $(\mathbb{B}_{s})^{m}$, which is a manifold of dimension $sm$, to $(\mathbb{B}_{s})^{m}_{\rm Sym}$, which is an irreducible analytic set of dimension $sm$, by Theorem 3.3, $f$ is surjective.

Clearly $f|_{f^{-1}(\pi(A))}$ is a proper holomorphic map from $f^{-1}(\pi(A))$, which is an analytic subset of $(\mathbb{B}_{s})^{m}$, onto $\pi(A)$, and so by Proposition 3.3, $\dim(f^{-1}(\pi(A))=\dim(\pi(A))$. Since we also have $\dim((\mathbb{B}_{s})^{m})=\dim((\mathbb{B}_{s})^{m}_{\rm{Sym}})=sm$, and we know from Lemma 3.4 that $\pi(A)$ has codimension at least $s$, $f^{-1}(\pi(A))$ must have codimension at least $s$ in $(\mathbb{B}_{s})^{m}$.

Let $U=(\mathbb{B}_{s})^{m}\setminus f^{-1}(\pi(A))$. Since $A$ and $f^{-1}(\pi(A))$ have complex codimension at least $s\geq 2$, by Proposition 3.1, both $(\mathbb{B}_{s})^{m}\setminus A$ and $U$ are simply-connected. Hence, $f|_{U}$ has a holomorphic lift $\widetilde{f}|_{U}:U\to(\mathbb{B}_{s})^{m}\setminus A$, where $f|_{U}=\pi\circ\widetilde{f}|_{U}$. Since $\widetilde{f}|_{U}$ is bounded on $U$ and $f^{-1}(\pi(A))$ is an analytic set, by Riemann’s Continuation Theorem $\widetilde{f}|_{U}$ extends to a holomorphic function $\widetilde{f}:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}$ with $f=\pi\circ\widetilde{f}$.

U$$(\mathbb{B}_{s})^{m}_{\rm{Sym}}\setminus\pi(A)$$(\mathbb{B}_{s})^{m}\setminus A$$f|_{U}$$\pi$$\widetilde{f}|_{U}$$(\mathbb{B}_{s})^{m}$$(\mathbb{B}_{s})^{m}_{\rm{Sym}}$$(\mathbb{B}_{s})^{m}$$f$$\pi$$\widetilde{f}

It remains to show that $\widetilde{f}:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}$ is a proper map. Note that $f=\pi\circ\widetilde{f}$ is a proper map, and $f$ is proper. If $\widetilde{f}$ were not proper, one could find a sequence $\{z_{n}\}_{n=1}^{\infty}$ with no limit points in $(\mathbb{B}_{s})^{m}$ such that $\widetilde{f}(z_{n})\to w\in(\mathbb{B}_{s})^{m}$. But this composing with $\pi$, we see that $f$ is not proper, which is a contradiction. Therefore $\widetilde{f}$ is proper. ∎

4. Proof of Theorem 2

Let $f:(\mathbb{B}_{s})^{m}_{\rm{Sym}}\to(\mathbb{B}_{s})^{m}_{\rm{Sym}}$ be a proper holomorphic map. Then, since $\pi$ is proper and holomorphic, $h=f\circ\pi$ is a proper holomorphic map $(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}_{\rm{Sym}}$. By Theorem 3, $h$ lifts to a proper holomorphic map $\widetilde{h}:(\mathbb{B}_{s})^{m}\to(\mathbb{B}_{s})^{m}$ with $h=\pi\circ\widetilde{h}$. By a classical application of the methods of Remmert and Stein (see [12, page 76]), we conclude that there exist proper holomorphic self-maps of the ball $\mathbb{B}_{s}$, $\widetilde{h}_{i}$ for $i=1,\dots,m$ and a permutation $\sigma$ of $\{1,2,\dots,m\}$ such that $\widetilde{h}$ has the structure

[TABLE]

We get the following commutative diagram:

(\mathbb{B}_{s})^{m}_{\rm{Sym}}$$(\mathbb{B}_{s})^{m}_{\rm{Sym}}$$(\mathbb{B}_{s})^{m}$$(\mathbb{B}_{s})^{m}$$f$$\pi$$\pi$$h$$\widetilde{h}

Since $f\circ\pi=\pi\circ\widetilde{h}$, and the left-hand side is invariant under the action of $S_{m}$ on $(\tau^{1},\tau^{2},\dots,\tau^{m})$, we must have

[TABLE]

for every $(\tau^{1},\tau^{2},\dots,\tau^{m})\in(\mathbb{B}_{s})^{m}$ and every $\sigma\in S_{m}$. Such a relation cannot hold unless there is a self map $g$ of $\mathbb{B}_{s}$ such that for each $j=1,\dots,m$, we have

[TABLE]

since otherwise we could choose a $\sigma$ for which the two sides would be distinct. Now, since $f\circ\pi=h$, we have

[TABLE]

Since $f$ is proper and $f=g^{m}_{\rm{Sym}}$, it follows that $g$ is a proper holomorphic self-map of the ball $\mathbb{B}_{s}$. Thanks to a classical result of Alexander (see [2], and also [13, page 316]), for $s\geq 2$, the only proper holomorphic self-mappings of $\mathbb{B}_{s}$ are the automorphisms of $\mathbb{B}_{s}$, and it is known that the automorphisms are given by certain multi-dimensional fractional linear maps (see [13]). Hence, $g$ is an automorphism of $\mathbb{B}_{s}$ (of the form (1.2) and the proof is complete.