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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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Sources

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