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The punctured disc has infinite conformal parameter, unlike every finite annulus

Sources

Mikhail Lyubich, Conformal Geometry and Dynamics of Quadratic Polynomials, vol. I, Ch. 1 §§6.3.1 and 6.3.6, printed pp. 121–124. Proposition 6.6 and Corollaries 6.11–6.12 compute the annular family values; Corollary 6.20 records the degeneration of nested annuli.
Lars Ahlfors and Arne Beurling, Conformal Invariants and Function-Theoretic Null-Sets, §§4–5, printed pp. 115 and 119–121, for the extremal-distance computations and conformal invariance conventions.

Example

Assume Countable Choice and use the conformal parameter and path-family conventions of The conformal parameter of a round annulus is a complete invariant.

(a) For $0 < r < R < \infty$ , the round annulus $A (r, R)$ has finite conformal parameter $M (A (r, R)) = \frac{1}{2 π} \log \frac{R}{r},$ and its connecting-family modulus is $μ (Γ_{r, R}) = \frac{2 π}{\log (R / r)} > 0.$

(b) For the punctured disc $D^{*} = {0 < ∣ z ∣ < 1}$ and its puncture-to-outer-circle family $Γ_{D^{*}}$ , the conformal parameter is infinite: $M (D^{*}) = λ (Γ_{D^{*}}) = + \infty, μ (Γ_{D^{*}}) = 0.$ For each integer $n \geq 2$ , every path in $Γ_{D^{*}}$ contains a subpath joining the circles $∣ z ∣ = 1 / n$ and $∣ z ∣ = 1$ . Consequently the nested finite-annulus parameters force this divergence.

(c) There is no conformal equivalence between $D^{*}$ and any finite round annulus $A (r, R)$ , nor between $D^{*}$ and $D$ or $C$ ; likewise $C^{*} = C ∖ {0}$ is not conformally equivalent to a finite round annulus, by The conformal parameter of a round annulus is a complete invariant(iii).

(d) Quantitatively, on $A (1 / n, 1)$ the density $ρ (z) = 1 / (2 π ∣ z ∣)$ gives every path joining the two boundary circles length at least $\log n / (2 π)$ and has area $\log n / (2 π)$ . Its extremal-length quotient is therefore at least $\log n / (2 π)$ .

Facts & Assumptions

Given: Countable Choice, the punctured disc, the finite round annuli, and the density/length conventions.

[F1]

Extremal length is monotone under overflow: if each path in $Γ_{0}$ contains a subpath in $Γ_{1}$ , then $λ (Γ_{0}) \geq λ (Γ_{1})$ . Extending densities by zero makes the family comparison independent of the ambient domain (Conformal invariance, monotonicity, and the series and parallel laws for extremal length, The rho-length and the extremal length are well defined).

[F2]

For every $0 < r < R < \infty$ , the connecting family of $A (r, R)$ has $λ (Γ_{r, R}) = \frac{1}{2 π} \log \frac{R}{r}, μ (Γ_{r, R}) = \frac{2 π}{\log (R / r)} .$ The density $1 / (∣ z ∣ \log (R / r))$ gives the lower extremal-length bound, while weighted Cauchy–Schwarz on radial segments gives the upper bound (Extremal length of the rectangle and of the round annulus).

[F3]

The conformal parameter is the extremal length of the connecting family; the punctured-disc path family has endpoints at $0$ and the unit circle and has interior in $D^{*}$ ; finite annuli have the values in [F2]; and the non-equivalence assertions in (c) are proved by winding families and the Liouville/logarithm obstructions (The conformal parameter of a round annulus is a complete invariant).

[F4]

For a rectifiable path crossing the boundary circles of $A (r, R)$ , $\int_{γ} \frac{∣ d z ∣}{∣ z ∣} \geq \log (R / r),$ and polar change of variables gives $\int_{A (r, R)} g d A = \int_{0}^{2 π} \int_{r}^{R} g (s e^{i θ}) s d s d θ$ for every nonnegative Borel $g$ (Extremal length of the rectangle and of the round annulus).

Verification

technique · nested-annulus overflow, with the explicit radial extremal metric as a quantitative check

1.1F2F3

For $0 < r < R < \infty$ , clause (i) of [F3] and [F2] give $M (A (r, R)) = (2 π)^{- 1} \log (R / r)$ and $μ (Γ_{r, R}) = 2 π / \log (R / r)$ . Since $R / r > 1$ , the logarithm is finite and positive, so the displayed parameter and reciprocal are finite and positive.

1.2F1F2F3given

Fix $n \geq 2$ and a path $γ : [0, 1] \to C$ in $Γ_{D^{*}}$ , so $γ (0) = 0$ , $∣ γ (1) ∣ = 1$ , and $0 < ∣ γ (t) ∣ < 1$ for $0 < t < 1$ . By continuity the set $T_{n} = {t \in [0, 1] : ∣ γ (t) ∣ = 1 / n}$ is nonempty and compact; let $t_{n} = \max T_{n}$ . For every $t \in (t_{n}, 1)$ one has $∣ γ (t) ∣ > 1 / n$ : it cannot be smaller without a later intermediate hit of $1 / n$ , and it is less than $1$ by the path hypothesis. Thus $γ ∣_{[t_{n}, 1]}$ is a subpath in the connecting family of $A (1 / n, 1)$ . Regard both families in the ambient plane; [F1] and [F2] give $λ (Γ_{D^{*}}) \geq λ (Γ_{1 / n, 1}) = \frac{\log n}{2 π} .$ As $n \to \infty$ the right side tends to $+ \infty$ , hence $M (D^{*}) = λ (Γ_{D^{*}}) = + \infty$ and $μ (Γ_{D^{*}}) = 0$ .

1.3F3

The finite-annulus, disc and plane non-equivalence claims, and the punctured-plane claim, are exactly clause (iii) of [F3], proved there using compact-trace winding families and the Liouville/logarithm obstructions.

2.1F2F4givenalgebra∎

On $A (1 / n, 1)$ , [F4] gives $A (ρ) = \int_{0}^{2 π} \int_{1 / n}^{1} \frac{1}{4 π^{2} s^{2}} s d s d θ = \frac{\log n}{2 π} .$ The crossing estimate in [F4] gives $ℓ_{ρ} (Γ_{1 / n, 1}) \geq \log n / (2 π)$ , so the quotient is at least $(\log n / (2 π))^{2} / (\log n / (2 π)) = \log n / (2 π)$ . This is the quantitative lower bound used in step 1.2.

Depends on

Used by

A modulus obstruction to quasiconformal equivalence of round annuli Example

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Sources

Mikhail Lyubich, Conformal Geometry and Dynamics of Quadratic Polynomials, vol. I (book draft, Stony Brook) (standard reference, not scraped)
Lars Ahlfors and Arne Beurling, Conformal Invariants and Function-Theoretic Null-Sets, Acta Mathematica 83 (1950), 101–129 (standard reference, not scraped)