TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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The holomorphic implicit function theorem

Statement

Let $m, n \geq 1$ , let $U \subseteq C^{m} \times C^{n}$ be open, let $f : U \to C^{n}$ be holomorphic, and let $(a, b) \in U$ satisfy $f (a, b) = 0$ . Assume the complex Jacobian with respect to the second block is invertible:

$\det {(\frac{\partial f_{j}}{\partial w_{k}} (a, b))}_{1 \leq j, k \leq n} \neq 0.$

Then there are neighbourhoods $A$ of $a$ and $B$ of $b$ , and a unique holomorphic map $φ : A \to B$ , such that

$f (z, φ (z)) = 0 (z \in A),$

and, after shrinking $A \times B$ if needed,

$f (z, w) = 0 ⟺ w = φ (z) .$

Facts & Assumptions

Given: The holomorphic map $f : U \to C^{n}$ , the point $(a, b) \in U$ with $f (a, b) = 0$ , and the invertible $w$ -Jacobian at $(a, b)$ .

[L1]

A holomorphic map with invertible complex Jacobian at a point is biholomorphic between neighbourhoods of that point and its image (The holomorphic inverse function theorem in several complex variables).

[L2]

Holomorphic maps into $C^{m + n}$ are read componentwise, so the first $m$ output coordinates of a holomorphic inverse are holomorphic too (A map into $C^{n}$ is holomorphic exactly when each of its components is).

Proof

technique · direct

1.1

Define $H (z, w) : = (z, f (z, w)) \in C^{m} \times C^{n} .$ Its complex differential at $(a, b)$ is $(u, v) ⟼ (u, \partial_{z} f (a, b) u + \partial_{w} f (a, b) v) .$ Because $\partial_{w} f (a, b)$ is invertible, this linear map has inverse $(ξ, η) ⟼ (ξ, \partial_{w} f (a, b)^{- 1} (η - \partial_{z} f (a, b) ξ)),$ so $J_{C} H (a, b)$ is invertible.

givenalgebra

2.1

By [L1], after shrinking to neighbourhoods of $(a, b)$ and $(a, 0)$ the map $H$ is biholomorphic. Write its inverse as $H^{- 1} (z, η) = (z, ψ (z, η));$ this form is forced because the first $m$ coordinates of $H$ are exactly $z$ , and [L2] makes $ψ$ holomorphic.

step 1.1L1L2construct

3.1

Define $φ (z) : = ψ (z, 0)$ . Then $H (z, φ (z)) = H (H^{- 1} (z, 0)) = (z, 0),$ so $f (z, φ (z)) = 0$ . Conversely, if $(z, w)$ is in the shrunken source neighbourhood and $f (z, w) = 0$ , then $H (z, w) = (z, 0) = H (z, φ (z))$ ; injectivity of the biholomorphism from step 2.1 forces $w = φ (z)$ . This also proves the uniqueness of $φ$ .

step 2.1algebra∎

Depends on

Used by

Dependency tree · two levels

33 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

Jaap Korevaar and Jan Wiegerinck, Several Complex Variables, Section 4.2 (standard reference, not scraped)
Jiří Lebl, Tasty Bits of Several Complex Variables, Chapter 5 (standard reference, not scraped)