Past exam of the complex analysis: conformal maps and geometry course of the University of Oxford / 2023 / 2

By conformal invariance of extremal length, work first on $1<|w|<R$. For a conformal metric $\rho$, let $L_{\rho }$ be the infimum of lengths of curves joining the two circles, and $A_{\rho }$ its area. Each radial segment is admissible, so the Cauchy-Schwarz inequality gives, for every angle,Integrating in angle gives $2\pi L_{\rho }^2\le A_{\rho }\log R$. Conversely, $\rho (w)=1/|w|$ has shortest joining length $\log R$ and area $2\pi \log R$, attaining the bound. Thus the conformal modulus of an annulus givesThe second formula is the pullback of $c/|w|$ by $\psi$. Pullback preserves curve lengths and metric area, so every displayed positive multiple is an extremal metric in $\Omega$.

The symmetry rule for extremal length says that if a domain and its curve family are invariant under reflection, the supremum defining extremal length may be taken over reflection-invariant conformal metrics. More generally, for a conformal or anticonformal involution $T$, use metrics satisfying $T^{\ast }\rho =\rho$.

To see this, replace $\rho$ by ${\rho }_s=(\rho +T^{\ast }\rho )/2$. Every curve in the invariant family and its image have length at least $L_{\rho }$, so $L_{{\rho }_s}\ge L_{\rho }$. Convexity of the square and change of variables give $A_{{\rho }_s}\le A_{\rho }$. Therefore symmetrizing cannot reduce the length-squared to area ratio.

Regard the upper half-disc, with cyclic boundary vertices $-1,0,x,1$, as a conformal quadrilateral. Map it to a rectangle of width $w$ and height $h$, so the boundary arcs $[-1,0]$ and $[x,1]$ become its vertical sides. The rectangle formulas for the conformal modulus of a quadrilateral giveThe remaining two boundary arcs, $(0,x)$ and the semicircle, become the horizontal sides. Thus these are precisely the dual joining families in the same quadrilateral.

Use the simply connected domain shown in the diagram. A Riemann mapping theorem map followed by an automorphism of the unit disk sends its marked points to $0$ and $x\in (0,1)$. Split this disc along its real diameter. Map the upper conformal quadrilateral from c to a rectangle $[0,w]\times [0,h]$, with $[-1,0]$ and $[x,1]$ as vertical sides and $(0,x)$ as its bottom. The lower half is a second identical rectangle. The full disc is obtained by gluing the two rectangles along their two vertical sides and their bottom sides; the two top sides form the original outer boundary.

In this doubled rectangle extremal-length duality, loops enclosing both marked bottom corners have shortest flat length $2w$. They must meet both vertical seams, and a closed curve's total horizontal travel is at least $2w$; the doubled horizontal segments attain it. Crosscuts separating the two corners must meet the bottom seam, and run from outer boundary back to outer boundary. Their total vertical travel is at least $2h$, attained by doubled vertical segments. The total flat area is $2wh$.

These flat metrics are extremal, not merely lower bounds. For an arbitrary metric, average the lengths of the doubled horizontal segments over their height and apply the Cauchy-Schwarz inequality to get $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Averaging the doubled vertical segments gives $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. ConsequentlyThe conformal metric on the doubled rectangles pulls back to an admissible metric on the original punctured domain; the two marked corner points have zero area. Here ${\Gamma }^{\ast }$ consists of crosscuts with endpoints on the outer boundary, as drawn. Boundedness alone, without the illustrated domain topology and the specified curve families, would not justify this construction.