TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08

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Poincaré's theorem: the ball and the polydisc are not biholomorphic for $m \geq 2$

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For every $m \geq 2$ the unit ball $B^{m}$ and the unit polydisc $D^{m}$ in $C^{m}$ are not biholomorphic. (For $m = 1$ both are the unit disc.)

Facts & Assumptions

[A1]

The only choice assumption is ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), inherited through the Bergman metric, kernel and determinant suppliers; no full Axiom of Choice is used.

[F1]

The unit ball and unit polydisc are $B^{m} = {z : \sum_{j < m} ∣ z_{j} ∣^{2} < 1}$ and $D^{m} = {z : ∣ z_{j} ∣ < 1}$ ; for $m = 1$ both equal the unit disc ${∣ z ∣ < 1}$ (Balls, polydiscs and the distinguished boundary in $C^{m}$ ).

[F2]

A biholomorphism $F : Ω \to Ω^{'}$ is a bijective holomorphic map whose inverse is holomorphic, and the determinant quotient satisfies $\frac{\det g_{Ω} (z)}{K_{Ω} (z, z)} = \frac{\det g_{Ω^{'}} (F (z))}{K_{Ω^{'}} (F (z), F (z))}$ for every $z \in Ω$ ; if each quotient is constant on its domain, the two constants are equal (Biholomorphic maps between open sets in $C^{m}$ , The determinant quotient $\det g_{Ω} / K_{Ω}$ is a biholomorphic invariant).

[F3]

The model quotients are the constants $\frac{\det g_{B^{m}}}{K_{B^{m}}} = \frac{(m + 1)^{m} π^{m}}{m!}$ and $\frac{\det g_{D^{m}}}{K_{D^{m}}} = 2^{m} π^{m}$ , and these constants are distinct for every $m \geq 2$ (Determinants and kernel quotients of the model Bergman metrics).

Proof

technique · direct, comparing the biholomorphically invariant quotient of the two model domains

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

1.1A1F2F3given

Suppose, for contradiction, that $F : B^{m} \to D^{m}$ is a biholomorphism. Both domains are bounded, so [F2] applies and gives $\frac{\det g_{B^{m}} (z)}{K_{B^{m}} (z, z)} = \frac{\det g_{D^{m}} (F (z))}{K_{D^{m}} (F (z), F (z))}$ for every $z \in B^{m}$ . By [F3] the left-hand side is the constant $\frac{(m + 1)^{m} π^{m}}{m!}$ and the right-hand side the constant $2^{m} π^{m}$ ; hence, by the constant-quotient clause of [F2], the two constants are equal.

2.1F2F3step 1.1

However, [F3] states that $\frac{(m + 1)^{m} π^{m}}{m!} \neq 2^{m} π^{m}$ for every $m \geq 2$ . This contradicts step 1.1, so no biholomorphism $B^{m} \to D^{m}$ exists; the same argument applies to a biholomorphism in either direction, by symmetry of the biholomorphism relation.

3.1F1given∎

For $m = 1$ , [F1] gives $B^{1} = D^{1} = {∣ z ∣ < 1}$ , so the two domains coincide; this is why the theorem is stated for $m \geq 2$ only, and no inequivalence is asserted in dimension one.

Depends on

Used by

The claim that the ball and the polydisc are biholomorphic False statement

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

Poincaré's theorem: the ball and the polydisc are not biholomorphic for m≥2m≥2