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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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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)