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For a conformal map $f$ and a conformal metric $\rho$ on its image, the pullback density is $\rho (f(z))|f^{\prime }(z)|$. The change of variables formula preserves its metric area, and the chain rule preserves the length of each corresponding curve. Applying this to $f$ and its inverse proves the title equality. This allows extremal length computations on rectangles, discs and annuli to be transported to other domains.

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.

For a path family $\Gamma$ on a Riemann surface, take the supremum of $L_{\rho }(\Gamma {)}^2/A_{\rho }$ over measurable conformal metrics of finite positive area. This is unchanged under conformal equivalence, and a quasiconformal map of maximal dilatation $K$ changes it by a factor between $K^{-1}$ and $K$.

Let $\alpha$ be a real closed differential form on a Riemann surface, of finite positive conformal energy $E=\int |\alpha {|}^{2}dA$. If its period on a loop homotopy class $\gamma$ is $p$, then $\lambda (\gamma ,X)\ge p^2/E$. Indeed the conformal metric $\rho =|\alpha |$ has area $E$ and each loop in the class has length at least $|\int \alpha |=|p|$. Energy is independent of the auxiliary smooth conformal metric used to compute the pointwise norm: the inverse scaling of the squared norm cancels the area scaling.

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.

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.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

Use the Gaussian curvature $-1$ normalizationof the Poincaré half-plane model. This is a complete conformal metric, invariant under real Möbius transformations. Its quotient by the free effective principal congruence subgroup is again complete: a geodesic lifts to the complete covering plane and extends there for all time. Transporting the quotient metric through the biholomorphism supplies the required complete conformal metric.

A nonconstant closed geodesic corresponds to a hyperbolic Möbius transformation $A$ of $\Gamma (2)$, and its hyperbolic translation length is $2\mathrm{arcosh}(|\mathrm{tr}A|/2)$. Indeed its eigenvalues have magnitudes $\lambda ,{\lambda }^{-1}$ with $\lambda >1$, and its axis quotient has length $2\log \lambda$. For $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ in $\Gamma (2)$, $bc\equiv 0(\mathrm{mod}4)$ implies $ad\equiv 1(\mathrm{mod}4)$. The odd numbers $a,d$ therefore have the same residue modulo four, andA hyperbolic Möbius transformation must have $|\mathrm{tr}A|>2$, so its smallest possible absolute trace is six. It is attained byConsequentlyThis element is primitive: a proper power would have a shorter root represented by a hyperbolic Möbius transformation, contradicting the trace bound. No essential simple closed geodesic exists on the three-punctured sphere, since every essential simple loop is peripheral and corresponds to a parabolic Möbius transformation. The minimizing closed geodesic is therefore nonsimple. Finally, the PDF leaves the numerical Gaussian curvature unspecified: the displayed length uses Gaussian curvature $-1$; for Gaussian curvature $-{\kappa }^2$, it is divided by $\kappa$.

A generalized conformal metric is locally $d{s}_{\rho }=\rho (z)|dz|$, where $\rho$ is nonnegative and measurable and transforms as a length density under a holomorphic change of coordinate. Its area is $A_{\rho }={\int }_X{\rho }^{2}dxdy$. Use the usual convention of using locally rectifiable paths for a path family $\Gamma$, and put $L_{\rho }(\Gamma )={\inf }_{\gamma \in \Gamma }{\int }_{\gamma }\rho |dz|$. ThenThis is extremal length. Zeros and isolated singularities of an admissible density are allowed; requiring a smooth strictly positive Riemannian metric would unnecessarily restrict the definition. Line integrals have their extended nonnegative values; if an arbitrary family is supplied, use its members that are locally rectifiable paths. An empty path family has infinite infimal length, whereas a family containing a constant path has extremal length zero.

Both numerator and denominator scale quadratically when $\rho$ is multiplied by a positive constant. The coordinate transformation of the area element makes the quotient unchanged under conformal equivalence.

The normalization relevant later is worth deriving. On the conformal cylinder $(\mathbb{R}/\mathbb{Z})\times (0,M)$, let $\Gamma$ contain the loops going once around it. For the horizontal loop at height $t$, Cauchy-Schwarz inequality givesIntegrating in $t$ shows $L_{\rho }(\Gamma {)}^2\le A_{\rho }/M$. The constant density achieves equality, since every winding-one loop has Euclidean length at least one. HenceHere $M$ is the height divided by circumference, the conformal modulus of an annulus; the reciprocal is used for the family joining its boundary components.

In local holomorphic coordinates, write the two positive singular values of $Df$ as $s_1\ge s_2$. Preservation of orientation gives $J_f=s_1{s}_2$, and quasiconformality gives $s_1/s_2\le K$.

For an admissible conformal metric $\rho |dw|$ on $Y$, define a density on $X$ by $\sigma (z)=\rho (f(z))s_1(z)$. Along each path,so $L_{\rho }(f(\Gamma ))\le L_{\sigma }(\Gamma )$. The change of variables formula and $s_1^2\le K{J}_f$ giveAlso $s_1^2\ge J_f$, so $A_{\sigma }\ge A_{\rho }>0$. Thus $\sigma$ has positive finite area and is admissible, andTaking the supremum over $\rho$ provesApplying the same argument to $f^{-1}$, which has the same bound on its maximal dilatation, also gives $K^{-1}\lambda (\Gamma ,X)\le \lambda (f(\Gamma ),Y)$. The inequalities remain valid for extended extremal lengths; no extremizing density need exist.

The given $\gamma$ is a simple closed curve. Assume it is essential; for a contractible class both infimal lengths are already zero. Write ${\ell }_n={\ell }_{X_n}(\gamma )$. The collar lemma supplies an embedded annulus with coordinates $r\in (-w_n,w_n)$ and $t\in \mathbb{R}/\mathbb{Z}$, and metricSet $y(r)={\ell }_n^{-1}{\int }_0^r\mathrm{sech}sds$. This changes the metric to a positive scalar multiple of $d{t}^2+d{y}^2$, so the conformal modulus of an annulus isThe extremal length of its core curves is $1/M_n$. Allowing all curves homotopic to $\gamma$ in $X_n$ can only decrease infimal lengths, while the area of a metric on all of $X_n$ is at least its area on the collar. Thus $\lambda (\gamma ,X_n)\le 1/M_n$. Use the particular conformal metric $|q_n{|}^{1/2}$, whose area of a quadratic differential is one:This proves the required implication uniformly over all the area-one holomorphic quadratic differentials on these surfaces.

The converse is false. Here is an explicit slit connected sum of translation tori. Start with square flat copies of a torus $T_1=\mathbb{C}/(\mathbb{Z}+i\mathbb{Z})$ and $T_{\delta }=\mathbb{C}/(\delta \mathbb{Z}+i\delta \mathbb{Z})$, where $\delta \to 0$. Cut a horizontal slit of physical length ${\delta }^2$ in each, centred in an interior coordinate disk, and cross-glue the banks by translation. The resulting surface has genus two. Its two slit endpoints have cone angle $4\pi$, so the locally defined $dz$ extends to a holomorphic one-form ${\omega }_{\delta }$ with two simple zeros. Its flat area is $1+{\delta }^2$. Setwhich is a holomorphic quadratic differential of area one. Let $\gamma$ be a horizontal generator in the small torus, taken away from the slit and fixed by the marking of this small handle. Then

To verify that its hyperbolic length does not tend to zero, construct a uniform lower bound on extremal length. On the unit square torus choose a disk $D$ about the eventual slit centre and a smooth cutoff function $\chi$ equal to one on a smaller disk and supported in $D$. Let $x$ denote a local real coordinate on $D$; in the first term below, $dx$ is the globally defined torus one-form, while $\chi x$ is extended by zero outside $D$. The real closed differential formis globally defined, vanishes on the smaller disk, and has period one on the horizontal generator. Pull it to $T_{\delta }$ by the rescaling map $z\mapsto z/\delta$, and extend it by zero across the slit and over the other torus. For sufficiently small $\delta$, the slit is inside the region where the form vanishes. Hence this extension ${\alpha }_{\delta }$ is a smooth closed differential form on the connected sum, with ${\int }_{\gamma }{\alpha }_{\delta }=1$.

Define a nonnegative conformal metric density by the pointwise norm of ${\alpha }_{\delta }$ relative to the flat metric. Two-dimensional scale invariance givesindependently of $\delta$. For every representative $\tilde{\gamma }$ homotopic to $\gamma$,because the period of a closed differential form is unchanged by homotopy. This is the extremal length lower bound from a closed one-form; therefore $\lambda (\gamma ,X_{\delta })\ge 1/C>0$. If ${\ell }_{X_{\delta }}(\gamma )$ tended to zero along any subsequence, the collar estimate would force $\lambda (\gamma ,X_{\delta })\to 0$, a contradiction. In fact its hyperbolic lengths are uniformly bounded away from zero. ThusFor every fixed $g>2$, replace $T_1$ by the area-one translation surface $X_{2g-2}$ constructed in the polygon argument; its genus is $g-1$. Cut its slit inside a nonsingular flat coordinate disk. The new connected sum has genus $g$, and the area normalization, small-handle length bound and closed-one-form energy argument are unchanged. Thus the converse fails at every fixed genus $g\ge 2$.

If a conformal map or anticonformal map $T$ is an involution preserving a curve family, its extremal length can be computed using invariant conformal metrics. Replace any density by ${\rho }_s=(\rho +T^{\ast }\rho )/2$. Every curve and its image have length at least $L_{\rho }$, hence $L_{{\rho }_s}\ge L_{\rho }$. Convexity of the square and change of variables give $A_{{\rho }_s}\le A_{\rho }$, proving the assertion. For reflection in the real axis, invariant means $\rho (z)=\rho (\bar{z})$.^[ref]