Conformal quadrilateral ($(\Omega ;a,b,c,d)$)

A Jordan domain together with four distinct boundary points in cyclic order, regarded up to a conformal map preserving those marked points. Such a domain maps to a rectangle with its marked points at the four corners. The opposite boundary-arc pairs specify two dual curve-joining problems whose extremal lengths are reciprocal.

Map a conformal quadrilateral to a rectangle of width $w$ and height $h$, with the chosen opposite boundary arcs mapped to its vertical sides. Their extremal length is $M=w/h$; the other pair has extremal length $1/M$. The constant conformal metric is extremal. For an arbitrary density, average horizontal joining lengths over height and apply the Cauchy-Schwarz inequality to obtain $L_{\rho }^{2}h\le w{A}_{\rho }$, proving the upper bound. Vertical joining segments prove the dual formula. Conformal invariance of extremal length transports both equalities back to the quadrilateral.

Glue two copies of $[0,w]\times [0,h]$ along their left, bottom and right sides. Their unglued top sides form the outer boundary of a disk; the two bottom corners are marked interior points. Let $\Gamma$ be the loops surrounding both points, and ${\Gamma }^{\ast }$ the boundary-to-boundary crosscuts separating them. Then $\lambda (\Gamma )\lambda ({\Gamma }^{\ast })=4$.

Each vertical seam joins a marked point to the outer boundary, so the Jordan curve theorem forces a loop surrounding both points to meet both seams. In the flat conformal metric, its total horizontal travel is therefore at least $2w$. Doubled horizontal segments attain this length. Each separating crosscut meets the bottom seam and starts and ends at height $h$, so its total vertical travel is at least $2h$; doubled vertical segments attain it. The total area is $2wh$.

For an arbitrary density, integrate the lengths of the doubled horizontal segments over their height. The Cauchy-Schwarz inequality yields $L_{\rho }(\Gamma {)}^{2}h\le 2w{A}_{\rho }$. Doubled vertical segments similarly give $L_{\rho }({\Gamma }^{\ast }{)}^{2}w\le 2h{A}_{\rho }$. These upper bounds equal the flat-metric ratios, proving both displayed formulas. Mapping the upper and lower halves of a disc to these rectangles gives the corresponding duality for a simply connected domain with two marked interior points.