TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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A plane domain is homologically simply connected exactly when every harmonic function has a global conjugate

Statement

Let $Ω \subseteq C$ be a complex domain. Then the following are equivalent.

$Ω$ is homologically simply connected.
Every harmonic function $u : Ω \to R$ has a harmonic conjugate on $Ω$ .

Facts & Assumptions

Given: A complex domain $Ω$ .

[L1]

On a homologically simply connected complex domain, every harmonic function has a harmonic conjugate (Harmonic conjugates exist on homologically simply connected plane domains).

[L2]

For a complex domain, homological simple connectivity is equivalent to the statement that for every point $p \in C ∖ Ω$ the function $z \mapsto 1 / (z - p)$ has a primitive on $Ω$ (Equivalent characterisations of a homologically simply connected domain).

[L3]

A harmonic conjugate of a harmonic function $u$ is a real-valued function $v$ such that $u + i v$ is holomorphic (Harmonic conjugates, Plane harmonic functions).

[L4]

If $L$ and $h$ are holomorphic with $\exp \circ L = h$ , then $L^{'} = h^{'} / h$ (A holomorphic logarithm is a primitive of the logarithmic derivative).

[L5]

The complex exponential is entire with derivative itself, satisfies $\exp (z + w) = \exp z \exp w$ , and compositions and nonvanishing quotients of holomorphic functions are holomorphic (The complex exponential is entire and its complex derivative is itself, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, The chain rule for complex derivatives, Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L6]

A nonconstant holomorphic function on a complex domain is open (Open mapping theorem for holomorphic functions).

Proof

technique · direct

1.1L1given

Assume condition 1. Then [L1] gives condition 2 immediately.

1.2L3givenchooseconstruct

Assume condition 2. Fix $p \in C ∖ Ω$ and define $u_{p} (z) = \log ∣ z - p ∣ .$ Direct differentiation gives $\frac{\partial u_{p}}{\partial x} = \frac{x - Re p}{∣ z - p ∣^{2}}, \frac{\partial u_{p}}{\partial y} = \frac{y - Im p}{∣ z - p ∣^{2}},$ and then $\frac{\partial^{2} u_{p}}{\partial x^{2}} + \frac{\partial^{2} u_{p}}{\partial y^{2}} = 0$ on $Ω$ , because $p \notin Ω$ . So $u_{p}$ is harmonic on $Ω$ . By condition 2 and [L3], choose a harmonic conjugate $v_{p}$ on $Ω$ and put $F_{p} = u_{p} + i v_{p},$ which is holomorphic on $Ω$ .

2.1step 1.2L5L6algebra

The function $G_{p} (z) = \frac{\exp (F_{p} (z))}{z - p}$ is holomorphic on $Ω$ by [L5]. Its modulus is $∣ G_{p} (z) ∣ = \frac{∣ \exp (F_{p} (z)) ∣}{∣ z - p ∣} = \frac{e^{Re F_{p} (z)}}{∣ z - p ∣} = \frac{e^{u_{p} (z)}}{∣ z - p ∣} = 1,$ so $G_{p} (Ω)$ lies on the unit circle. By [L6], $G_{p}$ cannot be nonconstant, hence it is constant: $G_{p} (z) \equiv c_{p}, ∣ c_{p} ∣ = 1.$ Therefore $L_{p} : = F_{p} - \log c_{p}$ satisfies $\exp (L_{p} (z)) = z - p$ on $Ω$ .

3.1L4L2step 2.1discharge-construct∎

By [L4], each $L_{p}$ from step 2.1 is a primitive of $1 / (z - p)$ on $Ω$ . Since $p \in C ∖ Ω$ was arbitrary, condition 4 of [L2] holds for $Ω$ , so [L2] gives condition 1. Together with step 1.1, this proves the equivalence.

Depends on

Used by

For a plane domain, the complement, homology, primitive, logarithm, conjugate, conformal, homotopy, and contractibility conditions are equivalent Theorem

Dependency tree · two levels

44 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

J. Lebl, Guide to Cultivating Complex Analysis, Ch. 4, §4.3 (standard reference, not scraped)
Jeremy Orloff, MIT 18.04 Topic 5 (standard reference, not scraped)