False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08

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The claim that the ball and the polydisc are biholomorphic

Statement

False claim. For every $m \geq 2$ the unit ball $B^{m}$ and the unit polydisc $D^{m}$ in $C^{m}$ are biholomorphic; more generally, any two bounded simply connected domains in $C^{m}$ , $m \geq 2$ , are biholomorphic.

Facts & Assumptions

[A1]

The only choice assumption is ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), inherited through the Bergman-geometric suppliers and used for the topological conventions below; no full Axiom of Choice is asserted beyond the published statements.

[F1]

$B^{m} = {z : \sum_{j < m} ∣ z_{j} ∣^{2} < 1}$ and $D^{m} = {z : ∣ z_{j} ∣ < 1}$ are nonempty open subsets of $C^{m} ≅ R^{2 m}$ ; $B^{m}$ is bounded and $D^{m} \subseteq B^{m} (0, \sqrt{m})$ is bounded (Balls, polydiscs and the distinguished boundary in $C^{m}$ , Complex $m$ -space and its real coordinate dictionary).

[F2]

For $z, w \in C^{m}$ the Euclidean norm satisfies $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ and $∥ λ u ∥ = ∣ λ ∣ ∥ u ∥$ , and for $a, b \in C$ the modulus satisfies $∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣$ (The inner-product norm is definite, homogeneous, and satisfies the triangle inequality, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, Complex $m$ -space and its real coordinate dictionary).

[F3]

A subset $C$ of a normed space is convex when $(1 - t) x + t y \in C$ for all $x, y \in C$ and $t \in [0, 1]$ ; the straight segment and the straight-line homotopy $F (s, t) = (1 - t) γ (s) + t x_{0}$ are continuous whenever $γ$ is, by continuity of the vector operations (Paths, path-connected spaces and path components, Vector addition and scalar multiplication are continuous in a normed space).

[F4]

A space is simply connected when it is nonempty, path-connected, and its fundamental group at every basepoint is trivial; loops are tested up to homotopy relative to the endpoints, and the identity element of $π_{1} (X, x_{0})$ is the class of the constant loop (Simply connected topological spaces, Based loops and the fundamental group, Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints, Loop classes form the group $π_{1} (X, x_{0})$ under concatenation).

[F5]

For every $m \geq 2$ there is no biholomorphism $B^{m} \to D^{m}$ , and a biholomorphism is a bijective holomorphic map with holomorphic inverse (Poincaré's theorem: the ball and the polydisc are not biholomorphic for $m \geq 2$ , Biholomorphic maps between open sets in $C^{m}$ ).

[F6]

The invariant quotients $\det g_{B^{m}} / K_{B^{m}} = (m + 1)^{m} π^{m} / m!$ and $\det g_{D^{m}} / K_{D^{m}} = 2^{m} π^{m}$ agree under biholomorphisms and are distinct for $m \geq 2$ (The determinant quotient $\det g_{Ω} / K_{Ω}$ is a biholomorphic invariant, Determinants and kernel quotients of the model Bergman metrics).

Refutation

technique · direct; a counterexample pair of bounded simply connected domains that are not biholomorphic

Given: ${A C}_{ω}$ and an integer $m \geq 2$ .

1.1A1F1F2given

The ball $B^{m}$ is convex: for $z, w \in B^{m}$ and $t \in [0, 1]$ , [F2] gives $∥ (1 - t) z + t w ∥ \leq (1 - t) ∥ z ∥ + t ∥ w ∥ < 1$ . The polydisc $D^{m}$ is convex coordinatewise: $∣ (1 - t) z_{j} + t w_{j} ∣ \leq (1 - t) ∣ z_{j} ∣ + t ∣ w_{j} ∣ < 1$ for every $j$ . Both are nonempty and bounded by [F1], so both are bounded convex nonempty subsets of $R^{2 m}$ .

1.2F5given

By [F5] there is no biholomorphism $B^{m} \to D^{m}$ for $m \geq 2$ . The first clause of the claim is therefore false.

2.1F3F4step 1.1

A nonempty convex set $C$ is path-connected, since $t \mapsto (1 - t) x + t y$ is a continuous path in $C$ between any two of its points by [F3]. It is simply connected: for $x_{0} \in C$ and a loop $γ$ at $x_{0}$ , the straight-line homotopy $F (s, t) : = (1 - t) γ (s) + t x_{0}$ of [F3] lies in $C$ , is continuous, satisfies $F (s, 0) = γ (s)$ , $F (s, 1) = x_{0}$ and $F (0, t) = F (1, t) = x_{0}$ , hence is a homotopy relative to the endpoints from $γ$ to the constant loop at $x_{0}$ . By [F4] its class is the identity of $π_{1} (C, x_{0})$ , so that group is trivial; therefore $C$ is simply connected. Applying this to the convex sets of step 1.1, $B^{m}$ and $D^{m}$ are bounded simply connected domains in $C^{m}$ .

3.1F5step 2.1step 1.2

The second clause is false as well: by step 2.1 the pair $(B^{m}, D^{m})$ consists of bounded simply connected domains in $C^{m}$ , and by step 1.2 they are not biholomorphic. This is a counterexample witness with the failed conclusion " $B^{m}$ and $D^{m}$ are biholomorphic", so the universal assertion "any two bounded simply connected domains in $C^{m}$ , $m \geq 2$ , are biholomorphic" fails.

4.1F6step 3.1∎

The obstruction is explicit: by [F6] the quotient $\det g_{Ω} / K_{Ω}$ is a biholomorphic invariant, and on the two model domains it takes the distinct constants $(m + 1)^{m} π^{m} / m!$ and $2^{m} π^{m}$ ; in particular no biholomorphism can identify them. The claim is the naive several-variable analogue of the Riemann mapping theorem, and the ball-polydisc pair above shows that this analogue fails in $C^{m}$ for $m \geq 2$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Jiří Lebl, Tasty Bits of Several Complex Variables (book) (standard reference, not scraped)