Conformal Indistinguishability — the Riemann Mapping Theorem
Lesson 8 · Phase A — every proper, nonempty, simply connected Ω ⊊ ℂ is biholomorphically the unit disk 𝔹; genuine univalent maps ft:𝔹→Ωt, plus the Liouville reason ℂ itself fails
preset
lobes n
𝔹 (source)
probe not applicable — there is no Riemann map here to probe; that is the entire point of this preset.
—
Ωt(image)EXACT MAP
t
—
ζ₀
—
|ft′(ζ₀)|
—
arg ft′(ζ₀)
—
output scale (auto-fit)
—
boundary radius range
—
—measured angle between the images of two perpendicular tangent arms at ζ₀, computed as ft′(ζ₀)·v — the readout that proves this is a real conformal map.
RMT gives no formula, in generalThe general Riemann Mapping Theorem proof (normal families + Montel's theorem) proves f exists and is unique — it hands you no closed-form expression for it. The two families here are the exception, not the rule: ft and kt are known, closed-form univalent maps, chosen precisely so this visualization can show genuine Riemann maps instead of a fake interpolation. Don't generalize from “I can see the formula here” — almost never true for an arbitrary Ω.RMT itself gives no formula — these are known closed-form exceptions, so this piece shows genuine maps, not a fake interpolation.
Why Ω ⊊ ℂ is a real hypothesis (Liouville)If f:ℂ→𝔹 were a biholomorphism, f would be an entire function with |f| < 1 — bounded. By Liouville's theorem (which Max proved himself — see 05-liouville-fta-chain.html), a bounded entire function is constant, contradicting bijectivity. So ℂ ≢ 𝔹, and “Ω ⊊ ℂ” is a genuine hypothesis of the RMT, not a technicality. The animation below is deliberately illustrative: gs(z) = s·z is just linear scaling, not an attempted biholomorphism — it exists only to make the boundedness obstruction visible. No matter how small s gets, the threshold circle |z| = 1/s always maps exactly onto ∂𝔹, and everything further out in ℂ (which always exists) keeps escaping.f:ℂ→𝔹 would be entire & bounded ⇒ constant (Liouville) ⇒ not bijective. So ℂ ≢ 𝔹. The scaling below is illustrative only.