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Composition and inversion of quasiconformal maps and their Beltrami coefficients

Sources

Mikhail Lyubich, Conformal Geometry and Dynamics of Quadratic Polynomials, vol. I, Ch. 2 §12.5, printed p. 188, Proposition 12.15; §11.1 for the composition and inverse coefficient identities.
Christopher J. Bishop, Quasiconformal Mappings, Ch. 2 §1, printed pp. 49–51, for the linear dilatation product bound and the general Beltrami chain identity.

Statement

Assume the Axiom of Choice. Let $Ω \overset{g}{\to} Ω^{''} \overset{f}{\to} Ω^{'}$ be analytically quasiconformal homeomorphisms, with $g$ $K_{1}$ -quasiconformal and $f$ $K_{2}$ -quasiconformal (The ACL and Sobolev analytic definition of quasiconformality).

(i) Composition. The composite $f \circ g$ is analytically $K_{1} K_{2}$ -quasiconformal, with $K_{f \circ g} \leq K_{1} K_{2}$ , and almost everywhere $μ_{f \circ g} (z) = \frac{g_{\overset{ˉ}{z}} (z) + μ_{f} (g (z)) \overline{g_{z} (z)}}{g_{z} (z) + μ_{f} (g (z)) \overline{g_{\overset{ˉ}{z}} (z)}} .$

(ii) Inverse. The inverse $f^{- 1}$ is $K_{2}$ -quasiconformal, $K_{f^{- 1}} = K_{f}$ , and $μ_{f^{- 1}} (f (z)) = - μ_{f} (z) \frac{f_{z} (z)}{\overline{f_{z} (z)}} for almost every z .$

Consequently quasiconformal homeomorphisms are closed under inverses and composition, and the $1$ -quasiconformal self-maps of a domain form a group.

Facts & Assumptions

Given: The Axiom of Choice, two analytic quasiconformal homeomorphisms as in the Statement, and their Beltrami representatives.

[F1]

Analytic and geometric quasiconformality agree with the same least constant; geometric quasiconformality is closed under composition because the two modulus inequalities multiply, and orientation signs multiply (The geometric and analytic definitions of quasiconformality agree, Orientation-preserving homeomorphisms and the geometric definition of quasiconformality).

[F2]

The inverse of an analytic $K$ -quasiconformal map is analytic $K$ -quasiconformal with the same maximal dilatation; the proof gives the a.e. inverse Beltrami formula (The inverse of a quasiconformal map is quasiconformal with the same dilatation).

[F3]

The real chain rule holds at common differentiability points. A real-linear map $z \mapsto a z + b \overset{ˉ}{z}$ has Wirtinger coefficients $a, b$ ; composition and inversion are calculated by the two Wirtinger equations (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The Wirtinger chain rule for compositions of real-differentiable complex-valued maps, The Wirtinger derivatives $\partial_{z} f$ and $\partial_{\overset{ˉ}{z}} f$ , and antiholomorphic functions). The Beltrami coefficient and least-dilatation conventions are those of The Beltrami coefficient and the maximal dilatation.

[F4]

Analytic quasiconformal homeomorphisms and their inverses map area-null Borel sets to null sets by the area clause of An analytically quasiconformal homeomorphism distorts quadrilateral moduli by at most K. This allows the exceptional differentiability sets of the factors to be pulled back when applying the a.e. chain rule. The supplier proves the area formula through independent inverse regularity and signed local winding, so this is an earlier proved interface.

[F5]

For $0 \leq a, b < 1$ and $∣ ω ∣ = 1$ , $∣ \frac{a_{0} + ω b_{0}}{1 + \overline{a_{0}} ω b_{0}} ∣ \leq \frac{∣ a_{0} ∣ + ∣ b_{0} ∣}{1 + ∣ a_{0} ∣ ∣ b_{0} ∣} (a_{0}, b_{0} \in C, ∣ a_{0} ∣ = a, ∣ b_{0} ∣ = b) .$ Indeed, the squared ratio is $(a^{2} + b^{2} + 2 a b \cos θ) / (1 + a^{2} b^{2} + 2 a b \cos θ)$ , increasing in $\cos θ$ ; its maximum is at $\cos θ = 1$ . Moreover, if $t = (a + b) / (1 + a b)$ , then $(1 + t) / (1 - t) = ((1 + a) / (1 - a)) ((1 + b) / (1 - b))$ .

Proof

technique · compose the geometric modulus inequalities, then compute the almost-everywhere Beltrami chain rule

1.1F1givenalgebra

By [F1], $f$ and $g$ are geometrically quasiconformal with constants $K_{2}$ and $K_{1}$ . Applying their two-sided modulus bounds successively to any quadrilateral gives the two-sided bound with constant $K_{1} K_{2}$ for $f \circ g$ ; the orientation signs multiply, so the composite is geometrically $K_{1} K_{2}$ -quasiconformal. The equivalence in [F1] makes it analytically $K_{1} K_{2}$ -quasiconformal.

2.1F2F3F4step 1.1given

By [F2], $g = f^{- 1}$ in the inverse case has the same analytic maximal dilatation and the stated inverse coefficient identity. For the composition formula, take the full-measure set where $g$ is differentiable, $f$ is differentiable at $g (z)$ , and the weak derivatives agree with the classical derivatives. The exceptional set for $f$ pulls back to a null set by [F4]. The composite is analytic by step 1.1, so its weak and classical derivatives also agree almost everywhere.

3.1F2F3F4F5step 1.1step 2.1given∎

At each point of the common set, the real chain rule gives $\begin{aligned} (f \circ g)_{z} & = f_{w} (g) g_{z} + f_{\overset{ˉ}{w}} (g) \overline{g_{\overset{ˉ}{z}}}, \\ (f \circ g)_{\overset{ˉ}{z}} & = f_{w} (g) g_{\overset{ˉ}{z}} + f_{\overset{ˉ}{w}} (g) \overline{g_{z}} . \end{aligned}$ Writing $μ_{f} (g) = f_{\overset{ˉ}{w}} (g) / f_{w} (g)$ and dividing the second equation by the first yields the displayed formula in (i); the denominator is nonzero almost everywhere because each analytic quasiconformal homeomorphism has positive Jacobian almost everywhere. Put $ν = μ_{g} (z)$ , $μ = μ_{f} (g (z))$ , and $ω = \overline{g_{z}} / g_{z}$ , so $∣ ω ∣ = 1$ . The composition formula becomes $μ_{f \circ g} = \frac{ν + ω μ}{1 + ω \overline{ν} μ} .$ By [F5], $∣ μ_{f \circ g} ∣ \leq (k_{1} + k_{2}) / (1 + k_{1} k_{2})$ , where $k_{j} = (K_{j} - 1) / (K_{j} + 1)$ . The identity in [F5] converts this to $K_{f \circ g} \leq K_{1} K_{2}$ , consistent with step 1.1. Part (ii) and the group assertion follow from [F2] and the identity map's coefficient $0$ .

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Sources

Mikhail Lyubich, Conformal Geometry and Dynamics of Quadratic Polynomials, vol. I (book draft, Stony Brook) (standard reference, not scraped)
Christopher J. Bishop, Quasiconformal Mappings (Stony Brook Math 627 lecture notes) (standard reference, not scraped)