Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 132 / 3

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.

Use the closed genus $g\ge 2$ version of Teichmüller's uniqueness theorem. Let $f_0:X\to Y$ be a Teichmüller map: for a nonzero holomorphic quadratic differential $q$, normalized by ${\int }_X|q|=1$, and $0<k_0<1$,The value at a zero of $q$ is irrelevant to the Beltrami coefficient, which is defined almost everywhere. For every quasiconformal map $f:X\to Y$ preserving orientation and homotopic to $f_0$, the conclusion isThe normalized $q$ is also uniquely determined when $k_0>0$.

We use the following analytic input, stated with its hypotheses. The Reich–Strebel inequality says that if $f_0$ has the displayed Beltrami coefficient and a competitor $f$ has the same target and homotopy class, then, for $\mu ={\mu }_f$,This is the standard fundamental inequality for an integrable holomorphic quadratic differential; it also holds on a finite-type punctured surface with the homotopy fixing the punctures. Its formulation is given in Gardiner and Hu, §5. We quote this analytic inequality as the lecture result used in the proof.

Put $k=\Vert \mu {\Vert }_{\infty }$. Pointwise, away from the isolated zeros of $q$,Integrating against the probability density function $|q|$ proves $K_0\le K(f)$. If $K(f)=K_0$, all these inequalities are equalities almost everywhere. The strictly increasing last function forces $|\mu |=k_0$ almost everywhere; equality in the triangle inequality then forces $\mu q/|q|=k_0$. ThereforeTwo quasiconformal maps with the same Beltrami coefficient differ by postcomposition with a biholomorphism, by the local chain rule for the Beltrami equation. Consequently $h=f\circ f_0^{-1}$ is a biholomorphism from $Y$ to itself homotopic to the identity.

For completeness, such a biholomorphism is the identity when $g\ge 2$. Apply the uniformization theorem and choose the lift of the homotopy to $\mathbb{H}$ starting at the identity. Its endpoint lift $\tilde{h}$ commutes with every deck transformation. It is a real Möbius transformation. The compact quotient has no parabolic Möbius transformations in its deck group, and freeness excludes elliptic Möbius transformations. Two distinct hyperbolic axes exist: a discrete free group preserving just one axis would be cyclic, contradicting the fundamental group of a closed surface of genus at least two. Commutation with two deck transformations represented by hyperbolic Möbius transformations having distinct axes makes it fix their boundary endpoints individually; there are at least three such endpoints. A Möbius transformation fixing three points is the identity. Hence $h=\mathrm{id}$ and $f=f_0$.

If $q_1$ is another unit-area differential for the same map, then $|q_1|/q_1=|q|/q$, so $q_1/q$ is positive real wherever defined. This meromorphic function is constant by the open mapping theorem, and normalization makes the constant one. Without normalization, positive multiples of $q$ describe the same map. At $k_0=0$, uniqueness of the map still holds in genus at least two, but there is no distinguished $q$. In genus one, translations supply nontrivial biholomorphisms homotopic to the identity, so equality determines the map only up to those biholomorphisms; the genus hypothesis cannot be omitted.