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Extremal Length and Planar Quasiconformality: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples and counterexamples exercise the extremal-length and quasiconformality machinery of extremal-length-and-planar-quasiconformality on explicit maps and domains. Extremal length of a rectangle and of a round annulus by hand computes the rectangle and round-annulus constants and their similarity invariance, while The punctured disc has infinite conformal parameter, unlike every finite annulus contrasts the vanishing reciprocal modulus of the punctured disc with the positive finite modulus of a round annulus.

The affine ellipse field, its inverse coefficient and the radial stretch supply The affine ellipse map and its Beltrami coefficient, The Beltrami coefficient of the inverse of an affine quasiconformal map and The radial stretch is quasiconformal with K equal to max of alpha and one over alpha, including the exact constant-coefficient formulae, the semiaxes $1 \pm ∣ μ ∣$ and the annular equality case $\max (α, 1 / α)$ . Composition is treated by Composition of two affine quasiconformal maps and the multiplicative dilatation bound, whose equality condition $ν \overset{ˉ}{μ} \in [0, \infty)$ is checked together with the affine-after-radial case, and A modulus obstruction to quasiconformal equivalence of round annuli converts the annular modulus bound into a quantitative obstruction.

Finally An orientation-reversing homeomorphism need not be quasiconformal shows that complex conjugation preserves every extremal length and satisfies the quadrilateral modulus bounds with $K = 1$ , yet is excluded by the orientation clause of the geometric definition and fails the analytic Beltrami inequality for every finite $K$ . All examples use the reciprocal-modulus convention of the A page and inherit its Countable Choice, respectively Axiom of Choice, assumptions.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08Open item page →

Extremal length of a rectangle and of a round annulus by hand

Sources

Christopher J. Bishop, Quasiconformal Mappings, Ch. 1 §1, printed pp. 4–5 (PDF pp. 9–10). Lemmas 1.6 and 1.7 give the rectangle and round-annulus constants by explicit test metrics and Cauchy–Schwarz on the respective foliations.
Mikhail Lyubich, Conformal Geometry and Dynamics of Quadratic Polynomials, vol. I, Ch. 1 §6.3.1, printed pp. 121–122. Proposition 6.6 computes the vertical family of an annulus, and Exercise 6.8 gives its dual circular family.

Example

Assume Countable Choice and use the conventions of Extremal length and the curve-family modulus of a path family.

(a) For $Π = (0, 2) \times (0, 1)$ and the family $Γ$ of paths joining the two vertical sides, $λ (Γ) = 2, μ (Γ) = \frac{1}{2} .$ The constant density $ρ = \frac{1}{2}$ gives $ℓ_{ρ} (Γ) \geq 1$ , has area $\frac{1}{2}$ , and has quotient $2$ . For every finite-positive-area Borel density, the horizontal slices and Cauchy–Schwarz give the matching upper bound.

(b) For $A = A (1, e^{2 π})$ , the connecting family $Γ_{A}$ has $λ (Γ_{A}) = 1, μ (Γ_{A}) = 1.$ The density $ρ (z) = 1 / (2 π ∣ z ∣)$ gives every connecting path length at least $1$ and has area $1$ . The radial Cauchy–Schwarz estimate gives the matching upper bound.

(c) Let $Θ_{A}$ be the family of closed paths in $A$ with winding number $1$ about $0$ . For every $0 < r < R$ , $λ (Γ_{A (r, R)}) λ (Θ_{A (r, R)}) = 1.$ For $A (1, e^{2 π})$ , both values are $1$ ; for $A (1, 2)$ , they are $\log 2 / (2 π)$ and $2 π / \log 2$ , respectively.

(d) Similarities $z \mapsto c z$ with $c \neq 0$ preserve the connecting-family extremal length of round annuli. In particular, $z \mapsto z / r$ sends $A (r, R)$ to $A (1, R / r)$ , and the conformal parameter $M$ defined in The conformal parameter of a round annulus is a complete invariant is $(2 π)^{- 1} \log (R / r)$ .

Facts & Assumptions

Given: Countable Choice; the rectangle and annulus curve families, and their Borel densities and area conventions.

[F1]

Borel densities are extended by zero outside their domain; nonrectifiable paths have infinite length. The path length equals the integral against arc length, is additive on subpath intervals, and agrees with the absolute line integral for continuous densities (Extremal length and the curve-family modulus of a path family, The rho-length and the extremal length are well defined).

[F2]

A Borel function on an open subspace extends by zero as a Borel function; Borel sets in a Euclidean product are product-measurable, and the product of the one-dimensional Lebesgue measures agrees with planar Lebesgue measure on Borel sets (The Borel sigma-algebra of a subspace is the trace of the ambient Borel sigma-algebra, The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}, On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, A continuous map has Borel preimages of Borel sets, Arithmetic and lattice operations preserve measurability whenever they are defined).

[F3]

Tonelli interchanges nonnegative product integrals, and Cauchy–Schwarz applies to square-integrable slices (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

The polar map $P (s, θ) = s e^{i θ}$ is $C^{1}$ with determinant $s > 0$ . The Euclidean inverse-function theorem gives local $C^{1}$ inverses; uniqueness of the polar angle, shifted to the branch $0 < θ < 2 π$ , makes $P$ a global diffeomorphism onto the annulus with its positive radial cut removed (The Euclidean inverse function theorem, Every nonzero complex number has a unique polar form $r (\cos θ + i \sin θ)$ with $r > 0$ and $- π < θ \leq π$ ). The cut is a planar null set under Countable Choice (A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in $R^{n}$ ). The Borel change-of-variables theorem therefore gives, for every nonnegative Borel $g$ on $A (1, e^{2 π})$ , $\int_{A (1, e^{2 π})} g d A = \int_{0}^{2 π} \int_{1}^{e^{2 π}} g (s e^{i θ}) s d s d θ .$ (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, A continuous map has Borel preimages of Borel sets, Arithmetic and lattice operations preserve measurability whenever they are defined)

[F5]

A rectifiable path crossing the two circles of a round annulus has $\int_{γ} ∣ d z ∣ / ∣ z ∣ \geq \log (R / r)$ . To prove this, cover its compact trace by discs avoiding zero, subdivide so each subpath lies in one disc, take a holomorphic logarithm of $z$ on each disc, and add the primitive integrals of $1 / z$ ; the real endpoint increment is $\log R - \log r$ . The modulus of a complex line integral is bounded by the absolute line integral (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset, Every open cover of a compact metric space has a Lebesgue number: a $δ > 0$ such that every nonempty subset of diameter less than $δ$ lies inside a single member of the cover, A nonvanishing holomorphic function on a disc has a holomorphic logarithm, A holomorphic logarithm is a primitive of the logarithmic derivative, The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path, The fundamental inequality: the modulus of the integral is at most the absolute line integral for rectifiable contours, Complex line integrals change sign under reversal and add under concatenation, The complex line integral over a rectifiable path as a componentwise Riemann–Stieltjes integral, The absolute line integral over a rectifiable path using its arc-length function, Continuous integrands have complex and absolute line integrals along every rectifiable path, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , The natural logarithm as the inverse of the exponential function).

[F6]

For the Borel test density $ρ (z) = 1 / (2 π ∣ z ∣)$ on $A (1, e^{2 π})$ , the Cauchy–Schwarz upper bound on radial slices is computed by $\int_{1}^{e^{2 π}} d s / s = 2 π$ ; also $\log (e^{2 π}) = 2 π$ and $π > 0$ (The natural logarithm as the inverse of the exponential function, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Pi as twice the smallest positive zero of cosine).

[F7]

The round-annulus connecting-family extremal length is the conformal parameter $M (A (r, R)) = (2 π)^{- 1} \log (R / r)$ , and the winding-one closed-family value is its reciprocal (Extremal length of the rectangle and of the round annulus, The conformal parameter of a round annulus is a complete invariant). Similarities preserve the connecting-family value by the Borel change-of-variables formula and arc-length scaling (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, A $C$ -Lipschitz map multiplies path length by at most $C$ ; isometries preserve length and scalar dilation multiplies it by the absolute scale).

Verification

technique · explicit test densities, slice estimates and the annulus formula

1.1F2F3F4given

For a Borel density on $Π$ , extend $ρ^{2} 1_{Π}$ by zero to $R^{2}$ . [F2] makes it product-measurable, and [F3] gives $A (ρ) = \int_{0}^{1} \int_{0}^{2} ρ (x, y)^{2} d x d y .$ For the annulus, [F4] gives the displayed polar area formula for arbitrary nonnegative Borel functions, not just continuous densities.

1.2F1F2F3givenalgebra

The constant rectangle density $ρ = \frac{1}{2}$ gives every crossing path Euclidean length at least $2$ , hence $ℓ_{ρ} (Γ) \geq 1$ . The box area formula gives $A (ρ) = 2 \cdot 1 \cdot (1 / 2)^{2} = 1 / 2$ , so the quotient is $2$ . For an arbitrary Borel density with $0 < A (ρ) < \infty$ , put $L = ℓ_{ρ} (Γ)$ . Each horizontal segment belongs to $Γ$ , so for every $y \in (0, 1)$ , $L \leq \int_{0}^{2} ρ (x, y) d x$ . Finite area and Tonelli give a full-measure set of $y$ with finite square integral; choosing one such $y$ first shows $L < \infty$ . On almost every such slice, [F3] gives $L^{2} \leq 2 \int_{0}^{2} ρ (x, y)^{2} d x$ . Integrating in $y$ gives $L^{2} \leq 2 A (ρ)$ . Thus every quotient is at most $2$ , and $λ (Γ) = 2$ , $μ (Γ) = 1 / 2$ .

1.3F1F4F5F6givenalgebra

Let $γ$ be any rectifiable path joining the boundary circles of $A (1, e^{2 π})$ . By [F5], $ℓ_{ρ} (γ) = \frac{1}{2 π} \int_{γ} ∣ d z ∣ / ∣ z ∣ \geq 1$ for $ρ (z) = 1 / (2 π ∣ z ∣)$ ; nonrectifiable paths have infinite length by [F1]. The polar area formula [F4] gives $A (ρ) = \int_{0}^{2 π} \int_{1}^{e^{2 π}} \frac{1}{4 π^{2} s^{2}} s d s d θ = 1.$ Hence this explicit metric gives quotient $1$ .

1.4F6F7algebra

For every $0 < r < R$ , [F7] gives $λ (Γ_{A (r, R)}) = (2 π)^{- 1} \log (R / r)$ and $λ (Θ_{A (r, R)}) = 2 π / \log (R / r)$ . Their product is $1$ . Substituting $R / r = e^{2 π}$ gives both values $1$ ; substituting $R / r = 2$ gives $\log 2 / (2 π)$ and $2 π / \log 2$ .

1.5F1F7givenalgebra

If $c \in C^{\times}$ , the similarity $f (z) = c z$ maps $A (r, R)$ bijectively onto $A (∣ c ∣ r, ∣ c ∣ R)$ . For a Borel density $τ$ on the target, $ρ (z) = ∣ c ∣ τ (c z)$ has the same $ρ$ -lengths on source paths as $τ$ has on their images, because arc length scales by $∣ c ∣$ ; its area is unchanged by the Jacobian $∣ c ∣^{2}$ and the Borel change-of-variables formula. The inverse similarity gives a bijection of the finite-positive-area metrics, so the extremal lengths agree. Taking $c = 1 / r$ and applying [F7] gives $M (A (r, R)) = (2 π)^{- 1} \log (R / r)$ .

2.1F1F3F4F6step 1.3given

For any Borel density $σ$ on the annulus with $0 < A (σ) < \infty$ , put $L = ℓ_{σ} (Γ_{A})$ . Every radial segment is in $Γ_{A}$ , so $L \leq \int_{1}^{e^{2 π}} σ (s e^{i θ}) d s$ for every $θ$ . By [F3, F4], a full-measure set of angles has finite weighted square integral; choosing one first shows $L < \infty$ . For almost every $θ$ , weighted Cauchy–Schwarz gives $L^{2} \leq (\int_{1}^{e^{2 π}} \frac{d s}{s}) \int_{1}^{e^{2 π}} σ (s e^{i θ})^{2} s d s = 2 π \int_{1}^{e^{2 π}} σ (s e^{i θ})^{2} s d s .$ Integrating in $θ$ and using [F3, F4] gives $2 π L^{2} \leq 2 π A (σ)$ , so every quotient is at most $1$ . With step 1.3, this proves $λ (Γ_{A}) = 1$ and $μ (Γ_{A}) = 1$ .

3.1F1F4F6given∎

The hypotheses always have $w = 2$ , $h = 1$ , and $1 < e^{2 π} < \infty$ ; the horizontal rectangle segments, radial annulus segments, and once-traversed circle show the assigned families are nonempty. All testing densities have positive finite area, all stated endpoints are boundary endpoints with zero arc-length mass, and no empty, zero-area, or degenerate-radius case is included. Countable Choice is used only through the explicitly declared measure and length interfaces; no full AC is used.

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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.

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The affine ellipse map and its Beltrami coefficient

Statement

Assume the Axiom of Choice. Fix $μ \in C$ with $∣ μ ∣ < 1$ and define $f : C \to C$ by $f (z) = z + μ \overline{z}$ . Verify:

(a) $f$ is a homeomorphism with inverse $f^{- 1} (w) = \frac{w - μ \overline{w}}{1 - ∣ μ ∣^{2}},$ is orientation-preserving, and has $f_{z} = 1$ and $f_{\overline{z}} = μ$ . Consequently its Beltrami coefficient is $μ_{f} \equiv μ$ and its analytic maximal dilatation is $K_{f} = \frac{1 + ∣ μ ∣}{1 - ∣ μ ∣}$ (The Beltrami coefficient and the maximal dilatation, The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions).

(b) The unit circle maps to an ellipse with semiaxes $1 + ∣ μ ∣$ and $1 - ∣ μ ∣$ . Their ratio is $K_{f}$ ; the map is conformal exactly when $μ = 0$ .

(c) For every round annulus $A (r, R) = {r < ∣ z ∣ < R}$ with $0 \leq r < R \leq \infty$ (Annuli in the complex plane), its image is the ring between homothetic ellipses when $0 < r < R < \infty$ . When $r = 0$ its inner complementary component is the puncture ${0}$ ; when $R = \infty$ its outer complementary component in the sphere is ${\infty}$ . Let $Γ_{r, R}$ join the two annular ends, using the end-path convention of An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K (equivalently the boundary-joining family for finite positive radii). Then $\frac{1}{K_{f}} λ (Γ_{r, R}) \leq λ (f Γ_{r, R}) \leq K_{f} λ (Γ_{r, R}), \frac{1}{K_{f}} μ (Γ_{r, R}) \leq μ (f Γ_{r, R}) \leq K_{f} μ (Γ_{r, R})$ by An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K, where $μ (Γ) = 1 / λ (Γ)$ denotes the library's curve-family modulus. In particular, for the Beltrami parameter $μ = 1 / 3$ and $A (1, e^{2 π})$ , $K_{f} = 2$ and $λ (Γ_{1, e^{2 π}}) = μ (Γ_{1, e^{2 π}}) = 1$ , so the guaranteed distortion interval is $[1 / 2, 2]$ .

(d) The map and its restriction $f ∣_{Ω} : Ω \to f (Ω)$ for every complex domain $Ω$ are $K_{f}$ -quasiconformal in both the analytic and geometric definitions (The ACL and Sobolev analytic definition of quasiconformality, Orientation-preserving homeomorphisms and the geometric definition of quasiconformality).

Facts & Assumptions

Given: Choice, $μ \in C$ with $κ : = ∣ μ ∣ < 1$ , the displayed real-linear map, and the path-family conventions of Extremal length and the curve-family modulus of a path family.

[F1]

Solving $w = z + μ \overline{z}$ together with $\overline{w} = \overline{z} + \overline{μ} z$ gives the stated inverse because $1 - κ^{2} > 0$ . The real determinant is $1 - κ^{2}$ , so the map is invertible and orientation-preserving (A real linear isomorphism preserves or reverses orientation according to the sign of its determinant, Smooth orientation sign is the local integral homology multiplier).

[F2]

Direct Wirtinger differentiation gives $f_{z} = 1$ and $f_{\overset{ˉ}{z}} = μ$ . Thus $μ_{f} = μ$ and the analytic maximal dilatation is $(1 + κ) / (1 - κ)$ (The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions, The Beltrami coefficient and the maximal dilatation).

[F3]

Writing $μ = κ e^{i θ}$ and rotating the output by $e^{- i θ / 2}$ gives $e^{- i θ / 2} f (e^{i t}) = (1 + κ) \cos (t - θ / 2) + i (1 - κ) \sin (t - θ / 2) .$ These are the semiaxes of the image ellipse; their ratio equals the value in [F2].

[F4]

Analytic $K_{f}$ -quasiconformality gives both quadrilateral and annular inequalities for extremal length and its reciprocal modulus with constant $K_{f}$ (An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K). For $0 < r < R < \infty$ , $λ (Γ_{r, R}) = (2 π)^{- 1} \log (R / r)$ and $μ (Γ_{r, R}) = 1 / λ (Γ_{r, R})$ (Extremal length of the rectangle and of the round annulus).

[F5]

The image of an open connected set under this invertible linear homeomorphism is open and connected, hence a complex domain; the analytic inequality and the geometric quadrilateral bounds restrict to that image (A complex domain is a nonempty connected open subset of $C$ , The ACL and Sobolev analytic definition of quasiconformality, Orientation-preserving homeomorphisms and the geometric definition of quasiconformality, [F1], [F2], [F4]).

Proof

technique · solve the real-linear inverse system, compute the Wirtinger data and ellipse axes, then apply the exact annular modulus theorem

1.1F1givenalgebra

The equations $w = z + μ \overline{z}$ and $\overline{w} = \overline{z} + \overline{μ} z$ imply $w - μ \overline{w} = (1 - ∣ μ ∣^{2}) z$ , so [F1] gives the displayed inverse. Since $1 - ∣ μ ∣^{2} > 0$ , the real determinant is positive; thus $f$ is an orientation-preserving invertible real-linear map and therefore a homeomorphism of $C$ onto itself.

1.2F2F3given

Write $μ = κ e^{i θ}$ . The parametrization in [F3] identifies the image of the unit circle with an ellipse of semiaxes $1 + κ$ and $1 - κ$ , so their ratio is $(1 + κ) / (1 - κ) = K_{f}$ . Also $f_{\overset{ˉ}{z}} = μ$ ; therefore $f$ is holomorphic exactly when $μ = 0$ , in which case it is the identity and conformal.

2.1F2step 1.1algebra

Differentiating $f (z) = z + μ \overset{ˉ}{z}$ gives $f_{z} = 1$ and $f_{\overset{ˉ}{z}} = μ$ . Since this smooth map belongs to $W_{l o c}^{1, 2}$ , the inequality $∣ f_{\overset{ˉ}{z}} ∣ / ∣ f_{z} ∣ = κ < 1$ makes it analytically $K_{f}$ -quasiconformal, and [F2] yields $μ_{f} \equiv μ$ and $K_{f} = (1 + κ) / (1 - κ)$ .

3.1F1F3F4step 2.1givenalgebra

For $0 < r < R < \infty$ , linearity and [F3] send the two boundary circles to homothetic ellipses; homeomorphism sends the region between them onto the region between those ellipses. If $r = 0$ , the omitted origin remains the origin. The bound $∣ f (z) ∣ \geq (1 - κ) ∣ z ∣$ shows that $f$ extends to infinity with $f (\infty) = \infty$ , so $R = \infty$ gives an ellipse exterior, or the punctured plane when also $r = 0$ . The map transports the two annular ends and their path families, and [F4] gives both distortion bounds in every case with its end-path convention. For $μ = 1 / 3$ , $K_{f} = 2$ ; the finite radii $r = 1$ , $R = e^{2 π}$ give $λ = μ (Γ) = 1$ , so each target quantity lies in $[1 / 2, 2]$ .

4.1F1F2F4F5given∎

By [F5], the restriction to any complex domain remains a homeomorphism onto a complex domain, retains the same constant Wirtinger derivatives and analytic inequality, and satisfies the geometric quadrilateral bounds for every quadrilateral compactly contained in that domain. Hence both definitions hold with constant $K_{f}$ on the plane and on every such restriction.

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The radial stretch is quasiconformal with K equal to max of alpha and one over alpha

Statement

Assume the Axiom of Choice. For $α > 0$ define $f_{α} (0) = 0$ and $f_{α} (z) = ∣ z ∣^{α - 1} z$ for $z \neq 0$ . Verify:

(a) $f_{α}$ is a homeomorphism of $C$ onto itself, orientation-preserving, with inverse $f_{1 / α}$ , and belongs to $W_{l o c}^{1, 2}$ . For $z \neq 0$ , $(f_{α})_{z} = \frac{α + 1}{2} ∣ z ∣^{α - 1}, (f_{α})_{\overset{ˉ}{z}} = \frac{α - 1}{2} ∣ z ∣^{α - 1} \frac{z}{\overset{ˉ}{z}} .$ Consequently $μ_{f_{α}} (z) = \frac{α - 1}{α + 1} \frac{z}{\overset{ˉ}{z}} (z \neq 0), ∣ μ_{f_{α}} ∣ = \frac{∣ α - 1 ∣}{α + 1}, K_{f_{α}} = \max (α, \frac{1}{α}),$ and $f_{α}$ is analytically and geometrically $K_{f_{α}}$ -quasiconformal.

(b) $f_{α}$ maps $∣ z ∣ = r$ onto $∣ w ∣ = r^{α}$ and each ray onto itself. It is conformal exactly when $α = 1$ .

(c) The extremal-length distortion bound is sharp on every annular connecting family. For $A (r, R)$ (Annuli in the complex plane) with $0 < r < R < \infty$ , let $Γ_{r, R}$ be the paths joining its boundary circles. Then $λ (f_{α} Γ_{r, R}) = \frac{1}{2 π} \log \frac{R^{α}}{r^{α}} = α λ (Γ_{r, R}) .$ If $α \geq 1$ , then $K_{f_{α}} = α$ and this attains the upper extremal-length bound; if $0 < α < 1$ , then $K_{f_{α}} = 1 / α$ and the ratio $α = 1 / K_{f_{α}}$ attains the lower bound. The reciprocal modulus bounds are attained at the corresponding opposite endpoints (An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K).

Facts & Assumptions

Given: Choice, $α > 0$ , the ACL/Sobolev convention, and the annulus path-family conventions.

[F1]

The map has polar form $f_{α} (r e^{i t}) = r^{α} e^{i t}$ . The function $r \mapsto r^{α}$ is a strictly increasing homeomorphism of $[0, \infty)$ with inverse $s \mapsto s^{1 / α}$ , so $f_{α}$ is a homeomorphism with inverse $f_{1 / α}$ .

[F2]

On $C ∖ {0}$ , direct Wirtinger differentiation gives the derivatives in the Statement. In particular the real Jacobian is $J_{f_{α}} = ∣ (f_{α})_{z} ∣^{2} - ∣ (f_{α})_{\overset{ˉ}{z}} ∣^{2} = α ∣ z ∣^{2 α - 2} > 0.$

[F3]

On each horizontal or vertical line not passing through $0$ , $f_{α}$ is smooth. On the two coordinate lines through $0$ , its components are constant multiples of $g (t) = sgn (t) ∣ t ∣^{α}$ , which is absolutely continuous on compact intervals because $g^{'} (t) = α ∣ t ∣^{α - 1} \in L_{l o c}^{1}$ for $α > 0$ .

[F4]

The function is locally bounded, and its classical first partial derivatives off $0$ are bounded by $C_{α} ∣ z ∣^{α - 1}$ . Since $\int_{∣ z ∣ < R} ∣ z ∣^{2 α - 2} d A = 2 π \int_{0}^{R} r^{2 α - 1} d r < \infty$ for $α > 0$ , they are locally square-integrable (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma). The excluded point $0$ is null because it lies in boxes of arbitrarily small area (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included). The ACL characterization therefore gives $f_{α} \in W_{l o c}^{1, 2}$ and identifies these almost-everywhere classical derivatives with its weak derivatives (Absolute continuity on almost every coordinate line, The ACL characterisation of $W^{1, p}$ ).

[F5]

The ratio $∣ (f_{α})_{\overset{ˉ}{z}} ∣ / ∣ (f_{α})_{z} ∣ = ∣ α - 1 ∣ / (α + 1) < 1$ off $0$ , and $\frac{1 + ∣ α - 1 ∣ / (α + 1)}{1 - ∣ α - 1 ∣ / (α + 1)} = \max (α, \frac{1}{α}) .$ The modulus-distortion lemma gives the quadrilateral inequalities for analytic maps; together with orientation preservation this is the geometric definition (An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K, Orientation-preserving homeomorphisms and the geometric definition of quasiconformality).

[F6]

For every finite round annulus, $λ (Γ_{r, R}) = (2 π)^{- 1} \log (R / r)$ and $μ (Γ_{r, R}) = 1 / λ (Γ_{r, R})$ (Extremal length of the rectangle and of the round annulus).

[F7]

At a point where the real derivative is invertible, the inverse-function theorem makes the map a local diffeomorphism; for a smooth local diffeomorphism its local-homology orientation multiplier is the sign of its determinant (The Euclidean inverse function theorem, Smooth orientation sign is the local integral homology multiplier, R-orientation of a topological manifold).

Proof

technique · use the polar form for the homeomorphism and annulus images, compute the Wirtinger derivatives off the origin, establish local Sobolev regularity by ACL, then compare the resulting constants

1.1F1F2F7given

By [F1], $f_{α}$ is a homeomorphism with inverse $f_{1 / α}$ . At every $z \neq 0$ , [F2] gives a positive Jacobian, so the Euclidean inverse function theorem makes $f_{α}$ a local diffeomorphism there; the smooth-to-local-homology orientation lemma identifies its local orientation multiplier with this positive determinant sign. The local orientation sign of the homeomorphism is locally constant on connected $C$ by Orientation-preserving homeomorphisms and the geometric definition of quasiconformality, so the sign at $0$ is positive as well.

1.2F2F4givenalgebra

Write $f_{α} (z) = z ∣ z ∣^{α - 1}$ for $z \neq 0$ . Using $\partial_{z} ∣ z ∣ = \overset{ˉ}{z} / (2 ∣ z ∣)$ and $\partial_{\overset{ˉ}{z}} ∣ z ∣ = z / (2 ∣ z ∣)$ gives $(f_{α})_{z} = ∣ z ∣^{α - 1} + \frac{α - 1}{2} ∣ z ∣^{α - 3} z \overset{ˉ}{z} = \frac{α + 1}{2} ∣ z ∣^{α - 1},$ $(f_{α})_{\overset{ˉ}{z}} = \frac{α - 1}{2} ∣ z ∣^{α - 3} z^{2} = \frac{α - 1}{2} ∣ z ∣^{α - 1} \frac{z}{\overset{ˉ}{z}} .$ Thus [F2] and [F4] provide the stated almost-everywhere derivatives and $W_{l o c}^{1, 2}$ regularity; the point $0$ is a null set.

2.1F4F5step 1.1givenalgebra

Since $∣ (f_{α})_{z} ∣ = (α + 1) ∣ z ∣^{α - 1} / 2$ and $∣ (f_{α})_{\overset{ˉ}{z}} ∣ = ∣ α - 1 ∣ ∣ z ∣^{α - 1} / 2$ , the Beltrami coefficient has constant modulus $∣ α - 1 ∣ / (α + 1)$ . If $α \geq 1$ , the quotient $∣ μ ∣ = (α - 1) / (α + 1)$ gives $K_{f_{α}} = α$ ; if $0 < α < 1$ , it gives $K_{f_{α}} = 1 / α$ . The analytic inequality holds almost everywhere; [F5] and step 1.1 then give geometric $K_{f_{α}}$ -quasiconformality. If $α \neq 1$ , its $\overset{ˉ}{\partial}$ derivative is nonzero on $C ∖ {0}$ , so it is not holomorphic; if $α = 1$ , it is the identity. This proves (a) and the conformality claim in (b).

3.1F1F6step 2.1givenalgebra∎

The polar formula in [F1] gives $∣ f_{α} (z) ∣ = ∣ z ∣^{α}$ and preserves the argument, so $A (r, R)$ maps to $A (r^{α}, R^{α})$ and its connecting family maps onto the target connecting family. By [F6], $λ (f_{α} Γ_{r, R}) = \frac{1}{2 π} \log \frac{R^{α}}{r^{α}} = α \frac{1}{2 π} \log \frac{R}{r} = α λ (Γ_{r, R}) .$ The two cases in step 2.1 show this is the upper endpoint for $α \geq 1$ and the lower endpoint for $0 < α < 1$ . Taking reciprocals shows the corresponding modulus endpoint is also attained.

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Composition of two affine quasiconformal maps and the multiplicative dilatation bound

Statement

Assume the Axiom of Choice. Let $f (w) = w + μ \overset{ˉ}{w}$ and $g (z) = z + ν \overset{ˉ}{z}$ with $∣ μ ∣, ∣ ν ∣ < 1$ , so $g$ is $K_{1}$ -quasiconformal and $f$ is $K_{2}$ -quasiconformal, where $K_{1} = (1 + ∣ ν ∣) / (1 - ∣ ν ∣)$ and $K_{2} = (1 + ∣ μ ∣) / (1 - ∣ μ ∣)$ (The affine ellipse map and its Beltrami coefficient). Verify:

(a) The composite is affine, with $μ_{f \circ g} = \frac{ν + μ}{1 + μ \overset{ˉ}{ν}}, K_{f \circ g} = \frac{1 + ∣ \frac{ν + μ}{1 + μ \overset{ˉ}{ν}} ∣}{1 - ∣ \frac{ν + μ}{1 + μ \overset{ˉ}{ν}} ∣} \leq K_{1} K_{2},$ in agreement with Composition and inversion of quasiconformal maps and their Beltrami coefficients(i).

(b) Equality holds exactly when $ν \overset{ˉ}{μ}$ is a nonnegative real number, including the cases $μ = 0$ or $ν = 0$ . When equality holds, $∣ \frac{ν + μ}{1 + μ \overset{ˉ}{ν}} ∣ = \frac{∣ ν ∣ + ∣ μ ∣}{1 + ∣ ν ∣ ∣ μ ∣},$ and $\frac{1 + t}{1 - t} = K_{1} K_{2}$ for $t = (∣ ν ∣ + ∣ μ ∣) / (1 + ∣ ν ∣ ∣ μ ∣)$ .

(c) For the radial stretch $g = f_{α}$ from The radial stretch is quasiconformal with K equal to max of alpha and one over alpha followed by $f (z) = z + μ \overset{ˉ}{z}$ , the product bound is sharp for every $α > 0$ and $∣ μ ∣ < 1$ : $K_{f \circ f_{α}} = K_{f} \max (α, 1 / α)$ . At points where the ellipse fields are not aligned, the pointwise coefficient estimate is strict; the radial field nevertheless takes aligned values along rays, so the essential supremum reaches the product bound.

Facts & Assumptions

Given: Choice, $f (z) = z + μ \overset{ˉ}{z}$ , $g (z) = z + ν \overset{ˉ}{z}$ , and the affine/radial examples on this page pair.

[F1]

Expanding $f (g (z))$ gives $(1 + μ \overset{ˉ}{ν}) z + (ν + μ) \overset{ˉ}{z}$ . Since $∣ μ \overset{ˉ}{ν} ∣ < 1$ , the coefficient $1 + μ \overset{ˉ}{ν}$ is nonzero, so the Beltrami coefficient is $(ν + μ) / (1 + μ \overset{ˉ}{ν})$ (The Beltrami coefficient and the maximal dilatation, The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions).

[F2]

Put $r = ∣ ν ∣$ and $s = ∣ μ ∣$ . If $x = Re (ν \overset{ˉ}{μ})$ , then ${∣ \frac{ν + μ}{1 + μ \overset{ˉ}{ν}} ∣}^{2} = \frac{r^{2} + s^{2} + 2 x}{1 + r^{2} s^{2} + 2 x}, - r s \leq x \leq r s .$ The right side is increasing in $x$ when $r, s > 0$ , since its derivative is $2 (1 - r^{2}) (1 - s^{2}) / (1 + r^{2} s^{2} + 2 x)^{2} > 0$ ; when $r s = 0$ it is constant. Therefore its maximum occurs at $x = r s$ , equivalently $ν \overset{ˉ}{μ} \in [0, \infty)$ , and its maximum modulus is $(r + s) / (1 + r s)$ .

[F3]

For $0 \leq r, s < 1$ and $t = (r + s) / (1 + r s) < 1$ , $\frac{1 + t}{1 - t} = \frac{(1 + r) (1 + s)}{(1 - r) (1 - s)} .$

[F4]

For $f_{α}$ , the coefficient is $ν (z) = q z / \overset{ˉ}{z}$ with $q = (α - 1) / (α + 1)$ , and $g_{z} = (α + 1) ∣ z ∣^{α - 1} / 2$ is positive real for $z \neq 0$ . Thus the composition formula of Composition and inversion of quasiconformal maps and their Beltrami coefficients(i) reduces to $μ_{f \circ f_{α}} (z) = (ν (z) + μ) / (1 + μ \overline{ν (z)})$ almost everywhere.

Proof

technique · expand the affine composite, maximize its coefficient over the relative phase, and check the radial field reaches the maximizing phase on rays

1.1F1givenalgebra

Since $g (z) = z + ν \overset{ˉ}{z}$ , one has $\overline{g (z)} = \overset{ˉ}{z} + \overset{ˉ}{ν} z$ . Therefore $f (g (z)) = g (z) + μ \overline{g (z)} = (1 + μ \overset{ˉ}{ν}) z + (ν + μ) \overset{ˉ}{z}$ . The coefficient denominator is nonzero because $∣ μ \overset{ˉ}{ν} ∣ < 1$ , and [F1] gives the displayed Beltrami coefficient.

1.2F2given

Let $r = ∣ ν ∣$ , $s = ∣ μ ∣$ , and $x = Re (ν \overset{ˉ}{μ})$ . By [F2], the squared coefficient modulus is increasing in $x$ when $r s > 0$ , so it is largest at $x = r s$ , exactly when $ν \overset{ˉ}{μ}$ is nonnegative real. If $r = 0$ or $s = 0$ , then $x = r s = 0$ and equality holds automatically. Thus the equality condition and the maximum modulus in (b) hold, including the degenerate cases.

2.1F2F3step 1.1step 1.2givenalgebra

The function $u \mapsto (1 + u) / (1 - u)$ is increasing on $[0, 1)$ , so [F2] bounds the composite dilatation by its value at $t = (r + s) / (1 + r s)$ . The identity in [F3] turns that value into $((1 + r) / (1 - r)) ((1 + s) / (1 - s)) = K_{1} K_{2}$ , proving (a) and (b).

3.1F2F3F4givenalgebra∎

For the radial stretch, [F4] makes $ν (z) = q z / \overset{ˉ}{z}$ range through the full circle of radius $∣ q ∣$ as $\arg z$ varies. If $q μ \neq 0$ , there are rays on which $ν (z) \overset{ˉ}{μ} = ∣ q ∣ ∣ μ ∣ > 0$ , so [F2] gives pointwise equality there. By continuity in the angle, every neighborhood of each such ray contains a positive-area sector where the coefficient modulus is arbitrarily close to $(∣ q ∣ + ∣ μ ∣) / (1 + ∣ q ∣ ∣ μ ∣)$ . Hence its essential supremum equals this maximum. If $q μ = 0$ , the coefficient modulus is constant and equals the same formula. Applying [F3] gives $K_{f \circ f_{α}} = ((1 + ∣ q ∣) / (1 - ∣ q ∣)) ((1 + ∣ μ ∣) / (1 - ∣ μ ∣)) = \max (α, 1 / α) K_{f}$ . At nonaligned points the strict increase in [F2] gives strict pointwise inequality, while the essential supremum remains sharp.

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A modulus obstruction to quasiconformal equivalence of round annuli

Statement

Assume the Axiom of Choice. Write $λ$ for extremal length and $μ$ for its reciprocal curve-family modulus (Extremal length and the curve-family modulus of a path family). Call two domains $K$ -quasiconformally equivalent if a $K$ -quasiconformal homeomorphism carries one onto the other, using the equivalent geometric and analytic definitions on this page (Orientation-preserving homeomorphisms and the geometric definition of quasiconformality, The ACL and Sobolev analytic definition of quasiconformality, The geometric and analytic definitions of quasiconformality agree). Then:

(a) If the round annuli $A (1, 2)$ and $A (1, R)$ (Annuli in the complex plane) are $K$ -quasiconformally equivalent for finite $R > 1$ , then $\frac{1}{2 π} \log R \leq K \frac{1}{2 π} \log 2,$ so $\log R \leq K \log 2$ . In particular, for every finite $K \geq 1$ , the annuli $A (1, 2)$ and $A (1, 2^{K + 1})$ are not $K$ -quasiconformally equivalent. For example, $A (1, 2)$ and $A (1, 8)$ are not $2$ -quasiconformally equivalent, though this estimate does not exclude equivalence for $K \geq 3$ .

(b) The inverse direction gives the lower bound as well: any such equivalence satisfies $\frac{1}{K} \log 2 \leq \log R \leq K \log 2,$ equivalently the ratio $M (A (1, R)) / M (A (1, 2))$ of their conformal parameters from The conformal parameter of a round annulus is a complete invariant lies in $[1 / K, K]$ .

(c) No finite round annulus $A (r, R)$ with $0 < r < R < \infty$ is $K$ -quasiconformally equivalent to the punctured disc $D^{*} = {0 < ∣ z ∣ < 1}$ for any finite $K \geq 1$ (The punctured disc has infinite conformal parameter, unlike every finite annulus).

Facts & Assumptions

Given: Choice, $K \geq 1$ , round annuli, their connecting curve families, and the punctured-disc path-family conventions.

[F1]

The two definitions of quasiconformality agree with the same constant, and inverses of analytic quasiconformal maps are quasiconformal with the same maximal dilatation (The geometric and analytic definitions of quasiconformality agree, Composition and inversion of quasiconformal maps and their Beltrami coefficients).

[F2]

For $0 < r < R < \infty$ , the connecting-family modulus is $μ (Γ_{r, R}) = 2 π / \log (R / r) > 0$ (Extremal length of the rectangle and of the round annulus). For $D^{*}$ the corresponding family has $μ (Γ_{D^{*}}) = 0$ (The punctured disc has infinite conformal parameter, unlike every finite annulus).

[F3]

An analytically $K$ -quasiconformal homeomorphism and its inverse distort the connecting-family modulus of any doubly connected domain by at most the factor $K$ (An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K, part (ii)).

Proof

technique · apply the two annular modulus inequalities in both directions, then compare the positive finite-annulus modulus with the zero punctured-disc modulus

1.1F1F2F3givenalgebra

Suppose $h : A (1, 2) \to A (1, R)$ is a $K$ -quasiconformal equivalence. By [F1], its inverse is analytically $K$ -quasiconformal. Apply [F3] to $h^{- 1} : A (1, R) \to A (1, 2)$ to obtain $μ (Γ_{1, 2}) \leq K μ (Γ_{1, R})$ . Using [F2], this is $2 π / \log 2 \leq K (2 π / \log R)$ , so $\log R \leq K \log 2$ .

2.1F1F2F3step 1.1givenalgebra

Apply [F3] to $h$ itself to get $μ (Γ_{1, R}) \leq K μ (Γ_{1, 2})$ . By [F2], $2 π / \log R \leq K (2 π / \log 2)$ , hence $\log R \geq (\log 2) / K$ . If $R = 2^{K + 1}$ , then $\log R = (K + 1) \log 2 > K \log 2$ , contradicting step 1.1; for $R = 8$ and $K = 2$ this is the stated example. The two inequalities together prove (b).

3.1F1F2F3givenalgebra∎

Suppose $h : A (r, R) \to D^{*}$ were $K$ -quasiconformal. By [F1], $h^{- 1} : D^{*} \to A (r, R)$ is analytically $K$ -quasiconformal. Applying [F3] to the doubly connected source $D^{*}$ gives $μ (Γ_{r, R}) \leq K μ (Γ_{D^{*}}) = 0$ . But [F2] makes the left side $2 π / \log (R / r) > 0$ , a contradiction.

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The Beltrami coefficient of the inverse of an affine quasiconformal map

Statement

Assume the Axiom of Choice. Let $f (z) = α z + β \overline{z}$ with $∣ β ∣ < ∣ α ∣$ , so $f$ is the affine map of The affine ellipse map and its Beltrami coefficient and $μ_{f} = β / α$ . Put $Δ = ∣ α ∣^{2} - ∣ β ∣^{2} > 0$ . Then $g (w) = f^{- 1} (w) = \frac{\overline{α} w - β \overline{w}}{Δ},$ and $g_{w} = \frac{\overline{α}}{Δ}, g_{\overline{w}} = - \frac{β}{Δ}, μ_{g} = - \frac{β}{\overline{α}} = - μ_{f} \frac{f_{z}}{\overline{f_{z}}}, ∣ μ_{g} ∣ = ∣ μ_{f} ∣, K_{g} = K_{f},$ in agreement with Composition and inversion of quasiconformal maps and their Beltrami coefficients(ii). For $α = 2$ and $β = i / 2$ , this gives $μ_{f} = i / 4$ , $K_{f} = 5 / 3$ , $g (w) = (2 w - \frac{i}{2} \overline{w}) / (15 / 4)$ , and $μ_{g} = - i / 4$ (The Beltrami coefficient and the maximal dilatation, The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions).

Facts & Assumptions

Given: Choice, $α, β \in C$ with $∣ β ∣ < ∣ α ∣$ , and the coefficient conventions of The Beltrami coefficient and the maximal dilatation.

[F1]

The system $w = α z + β \overset{ˉ}{z}$ , $\overset{ˉ}{w} = \overset{ˉ}{α} \overset{ˉ}{z} + \overset{ˉ}{β} z$ has determinant $Δ = ∣ α ∣^{2} - ∣ β ∣^{2} > 0$ and solving it gives $z = (\overset{ˉ}{α} w - β \overset{ˉ}{w}) / Δ$ .

[F2]

Wirtinger differentiation gives $f_{z} = α$ , $f_{\overset{ˉ}{z}} = β$ , $g_{w} = \overset{ˉ}{α} / Δ$ , and $g_{\overset{ˉ}{w}} = - β / Δ$ (The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions).

[F3]

For an analytic quasiconformal affine map, $μ_{f} = f_{\overset{ˉ}{z}} / f_{z}$ and $K_{f} = (1 + ∣ μ_{f} ∣) / (1 - ∣ μ_{f} ∣)$ ; the inverse theorem states $μ_{f^{- 1}} (f (z)) = - μ_{f} (z) f_{z} / \overline{f_{z}}$ and $K_{f^{- 1}} = K_{f}$ (The Beltrami coefficient and the maximal dilatation, Composition and inversion of quasiconformal maps and their Beltrami coefficients, The affine ellipse map and its Beltrami coefficient).

Proof

technique · solve the two-coordinate real-linear system, differentiate the inverse, and compare its coefficient with the general inverse formula

1.1F1givenalgebra

Since $∣ β ∣ < ∣ α ∣$ , one has $Δ > 0$ . Multiplying the first equation by $\overset{ˉ}{α}$ and subtracting $β$ times the second gives $\overset{ˉ}{α} w - β \overset{ˉ}{w} = Δ z$ , so the displayed formula for $g$ is the inverse; the same invertible linear system gives both inverse identities.

1.2F2given

Differentiating $g (w) = (\overset{ˉ}{α} w - β \overset{ˉ}{w}) / Δ$ and $f (z) = α z + β \overset{ˉ}{z}$ by [F2] yields $g_{w} = \overset{ˉ}{α} / Δ$ , $g_{\overset{ˉ}{w}} = - β / Δ$ , $f_{z} = α$ , and $f_{\overset{ˉ}{z}} = β$ . As $α \neq 0$ , division gives $μ_{g} = g_{\overset{ˉ}{w}} / g_{w} = - β / \overset{ˉ}{α}$ and $μ_{f} = β / α$ .

2.1F3step 1.2algebra

Substituting $μ_{f} = β / α$ and $f_{z} = α$ into the right side of the inverse identity in [F3] gives $- (β / α) (α / \overset{ˉ}{α}) = - β / \overset{ˉ}{α} = μ_{g}$ , so this concrete calculation agrees with Composition and inversion of quasiconformal maps and their Beltrami coefficients(ii). Also $∣ μ_{g} ∣ = ∣ β ∣ / ∣ α ∣ = ∣ μ_{f} ∣$ ; applying the formula in [F3] gives $K_{g} = K_{f}$ .

3.1step 1.1step 1.2step 2.1algebra∎

For $α = 2$ , $β = i / 2$ , one has $Δ = 4 - 1 / 4 = 15 / 4$ , $μ_{f} = (i / 2) / 2 = i / 4$ , and $K_{f} = (1 + 1 / 4) / (1 - 1 / 4) = 5 / 3$ . The inverse formula becomes $g (w) = (2 w - \frac{i}{2} \overset{ˉ}{w}) / (15 / 4)$ and its coefficient is $- i / 4$ , as asserted.

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An orientation-reversing homeomorphism need not be quasiconformal

Statement

Assume the Axiom of Choice. Statement refuted. Every homeomorphism $f : Ω \to Ω^{'}$ that preserves the moduli of all quadrilateral families up to factor $K$ is $K$ -quasiconformal; no orientation hypothesis is needed.

Counterexample. Take $Ω = Ω^{'} = C$ and $f (z) = \overline{z}$ . This is an orientation-reversing Euclidean isometry. It preserves the extremal length and reciprocal curve-family modulus of every path family exactly, so it satisfies all quadrilateral modulus inequalities with $K = 1$ . But it violates the orientation clause of Orientation-preserving homeomorphisms and the geometric definition of quasiconformality and is not analytically $K$ -quasiconformal for any finite $K$ : $f_{z} = 0$ and $f_{\overset{ˉ}{z}} = 1$ , contradicting $∣ f_{\overset{ˉ}{z}} ∣ \leq k ∣ f_{z} ∣$ for every $k = (K - 1) / (K + 1) < 1$ (The ACL and Sobolev analytic definition of quasiconformality, The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions).

Facts & Assumptions

Given: Choice, the plane with its Borel area measure, and the extremal-length convention of Extremal length and the curve-family modulus of a path family.

[F1]

The reflection $T (x, y) = (x, - y)$ is continuous and hence measurable, and it is an orthogonal linear map. Since $T^{- 1} = T$ , orthogonal invariance of Lebesgue measure makes it a measure-preserving transformation of the plane (A measurable function between measurable spaces, Lebesgue measure on $R^{n}$ is invariant under every orthogonal linear map, Measure-preserving transformations and systems).

[F2]

For every path $γ$ , distances along $T γ$ equal those along $γ$ , so their arc-length functions agree. Rectifiability is preserved, and nonrectifiable paths have infinite density length under both conventions (The arc-length function $s_{γ} (t) = L (γ ∣_{[a, t]})$ of a rectifiable path, Extremal length and the curve-family modulus of a path family).

[F3]

If $σ$ is a nonnegative Borel density on $T (Ω)$ and $ρ = σ \circ T$ , then $ρ$ is Borel and the measure-preserving integral theorem gives $A_{Ω} (ρ) = A_{T (Ω)} (σ)$ , allowing infinite values (Integral invariance under measure-preserving maps, [F1]).

[F4]

The real derivative of conjugation is $(\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ , whose determinant is $- 1$ , so the local orientation sign is negative (A real linear isomorphism preserves or reverses orientation according to the sign of its determinant, Smooth orientation sign is the local integral homology multiplier). Its Wirtinger derivatives are $f_{z} = 0$ and $f_{\overset{ˉ}{z}} = 1$ (The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions). The analytic definition requires the ACL/Sobolev and Wirtinger-inequality conditions (The ACL and Sobolev analytic definition of quasiconformality).

[F5]

The library's geometric definition requires orientation preservation in addition to the two-sided quadrilateral modulus inequalities (Orientation-preserving homeomorphisms and the geometric definition of quasiconformality).

Proof

technique · transport admissible densities by the reflection, then check orientation and the analytic inequality directly

1.1F1F2F3given

Let $T (z) = \overset{ˉ}{z}$ , $Ω^{'} = T (Ω)$ , and $Γ^{'} = T Γ$ . Given any Borel density $σ$ on $Ω^{'}$ , set $ρ = σ \circ T$ on $Ω$ . By [F2], each rectifiable path has the same density length before and after reflection, while nonrectifiable paths have infinite length on both sides. Thus $ℓ_{ρ} (Γ) = ℓ_{σ} (Γ^{'})$ . By [F3], the two densities also have equal area, with $0 < A < \infty$ on one side exactly when it holds on the other. Since $T$ is an involution, this is a bijection of the admissible density classes; taking the defining supremum gives $λ (Γ^{'}) = λ (Γ)$ and hence $μ (Γ^{'}) = μ (Γ)$ , including empty, zero, and infinite cases.

2.1F4step 1.1F5given

Every quadrilateral family and its reflected image therefore satisfy the two-sided modulus inequalities with constant $K = 1$ . But [F4] shows that $T$ reverses the local orientation, so it fails the orientation-preserving clause in the library's geometric definition.

3.1F4given∎

The map is smooth and belongs to $W_{l o c}^{1, 2}$ , but [F4] gives $∣ f_{\overset{ˉ}{z}} ∣ = 1$ and $∣ f_{z} ∣ = 0$ . For every finite $K \geq 1$ , the analytic definition has $k = (K - 1) / (K + 1) < 1$ and would require $1 \leq k \cdot 0 = 0$ , which is impossible. Thus the modulus bounds alone do not imply either orientation-preserving geometric or analytic quasiconformality.