7
$\begingroup$

The Wikipedia article "Quadratic Differential" opens with the following text:

In mathematics, a quadratic differential on a Riemann surface is a section of the symmetric square of the holomorphic cotangent bundle. If the section is holomorphic, then the quadratic differential is said to be holomorphic. The vector space of holomorphic quadratic differentials on a Riemann surface has a natural interpretation as the cotangent space to the Riemann moduli space or Teichmüller space.

Could somebody explain how exactly can one view the space of quadratic differentials as a cotangent space either of the Riemann moduli space or of the Teichmüller space?

The references list three books (by Strebel, Imayoshi-Taniguchi and Gardiner), I tried to find an explanation there but failed.

$\endgroup$
5
  • 1
    $\begingroup$ I think it should be the tangent space to the moduli space? See section 2.4 of Tromba's Teichmüller Theory in Riemannian Geometry. $\endgroup$ May 9, 2018 at 14:19
  • 2
    $\begingroup$ Actually the OP is correct, quadratic differentials are identified with the cotangent space. The tangent space is Beltrami differentials on the Riemann surface modulo an equivalence relation; a Beltrami differential is the infinitesimal form of a deformation of the conformal metric, which is what a tangent vector should be tangent to. I don't know the literature or the textbooks well enough, though, to find where this is explicitly stated. $\endgroup$
    – Lee Mosher
    May 9, 2018 at 20:57
  • $\begingroup$ You may also try "compact Riemann surface" by Jost. $\endgroup$
    – user99914
    May 10, 2018 at 1:41
  • 1
    $\begingroup$ One way of understanding why it should be the cotangent space instead of the tangent space is to understand the Bers embedding. My favorite reference for that is the book Teichmüller Theory Volume 1 by John Hubbard. $\endgroup$ Jun 14, 2018 at 13:36
  • $\begingroup$ @MaximeScott Many thanks for the suggestion, it is a wonderful book indeed! $\endgroup$ Jun 14, 2018 at 21:51

1 Answer 1

11
$\begingroup$

This question is rather old, but I will give it a shot anyways (for further reference). My reference is the lesser known textbook "An Introduction to Teichmüller Theory" by Imayoshi and Taniguchi. The identification is not immediate but requires a series of intermediate steps. Note that there probably are quicker ways or at least differentily formulated ways.

Let $H$ denote the upper half plane and $H^*$ the lower half plane. Firstly, take a Fuchsian model $\Gamma$ for the surface. I will denote by $QC(\Gamma)$ the set of quasi-conformal maps from $H / \Gamma$ to some other Riemann surface. Then the Teichmüller space of the surface takes the form $$Teich(\Gamma) = \{ [f] ~|~ f \in QC(\Gamma) \},$$ where $f,g \in QC(\Gamma)$ are equivalent if the composition $f \circ g^{-1}$ is isotopic to a conformal map. By means of the measurable Riemann mapping theorem, you can show that this Teichmüller space is naturally bijective to the set $$\{ [w_{\mu}] ~|~ \mu \in BC(\Gamma)_1 \},$$ where these objects are defined as follows: $BC(\Gamma)$ is the set of Beltrami coefficients on $\Gamma$, i.e. bounded measurable maps $\mu \colon H \rightarrow \mathbb{C}$ that satisfy $$(\mu \circ \gamma) \frac{\overline{\gamma'}}{\gamma'} = \mu$$ for every $\gamma \in \Gamma$. This "equivariance" condition corresponds to the change of coordinate behaviour. $BC(\Gamma)_1$ denotes the unit ball with respect to the supremums norm. Given $\mu \in BC(\Gamma)_1$, define a new coefficient by $$\mu^*(z) = \begin{cases} \mu(z), & \text{if } z \in H \\ 0, & \text{if } z \in \overline{H^*}. \end{cases}$$ The measurable Riemann mapping theorem provides you with a solution to $\mu^*$ that is uniquely determined by fixing 0, 1 and $\infty$, which we denote by $w_{\mu}$. Lastly, we say $w_{\mu}$ and $w_{\nu}$ are equivalent if they are equal on $H^*$.

Next, given a conformal map $f$ defined on a domain in $\mathbb{C}$, we define the "Schwarzian derivative of $f$ at $z$" as $$s(f)(z) = \frac{f'''(z)}{f'(z)} - \frac{3}{2} \left(\frac{f''(z)}{f'(z)}\right)^2.$$ This definition is motivated by the fact that $s(f)$ vanishes if and only if $f$ is a Möbius transformation. Let us define a map that takes an element $\mu \in BC(\Gamma)_1$ and sends it to $$S_{\mu} \colon H^* \rightarrow \mathbb{C}, ~z \mapsto s(w_{\mu})(z).$$ You can check that for every $\gamma \in \Gamma$ and every $z \in H^*$ we have $$\left( S_{\mu} \circ \gamma \right)(z) \cdot (\gamma'(z))^2 = S_{\mu}(z)$$ and that $[w_{\mu}] = [w_{\nu}]$ in $Teich(\Gamma)$ if and only if $S_{\mu} = S_{\nu}$. Thus, we get a well-defined map $$Teich(\Gamma) \rightarrow QD^*(\Gamma), ~[w_{\mu}] \mapsto S_{\mu},$$ where $QD^*(\Gamma)$ is the set $$\{ \phi \colon H^* \rightarrow \mathbb{C} \text{ holomorphic} ~|~ (\phi \circ \gamma) \cdot (\gamma')^2 = \phi \text{ for all } \gamma \in \Gamma \}$$ of holomorphic quadratic differentials (on $H^*$). You can show that this map is continuous so that by the Invariance of Domain theorem $Teich(\Gamma)$ is homeomorphic to an open subset of $QD^*(\Gamma)$.

Next, given an element $\phi \in QD^*(\Gamma)$, we get an element in $BC(\Gamma)$ defined by $$H[\phi](z) = \mathcal{I}(z)^2 \phi(\overline{z}),$$ where $\mathcal{I}(z)$ is the imaginary part. We call the image of $QD^*(\Gamma)$ under this map $H$ the vector space of harmonic Beltrami coefficients and denote it by $HBC(\Gamma)$. Since $QD^*(\Gamma)$ is a vector space, we get an isomorphism $$T_0(Teich(\Gamma)) \simeq QD^*(\Gamma) \simeq HBC(\Gamma).$$

We are almost done. Consider the pairing $$\Lambda \colon BC(\Gamma) \rightarrow QD^*(\Gamma)^*, ~\mu \mapsto \Lambda_{\mu},$$ where $QD^*(\Gamma)^*$ is the dual space and $$\Lambda_{\mu}(\phi) = \int_F \mu(z) \overline{\phi(\overline{z})} ~dA(z),$$ where $F \subset H$ is any fundamental domain of $\Gamma$. If $N(\Gamma)$ denotes the kernel of $\Gamma$, then $BC(\Gamma)$ splits into $HBC(\Gamma) \oplus N(\Gamma)$. Thus, $$image(\Lambda) \simeq BC(\Gamma) / N(\Gamma) \simeq HBC(\Gamma) \simeq QD^*(\Gamma)$$ and, by a dimension argument, $\Lambda$ must be surjective. Hence, $$T_0^*(Teich(\Gamma)) \simeq HBC(\Gamma)^* \overbrace{\simeq}^{via \Lambda} (QD^*(\Gamma)^*)^* \simeq QD^*(\Gamma).$$ This is what we wanted (at least for the cotangent space at the base point). For a general point $p = [w_{\nu}] \in Teich(\Gamma)$, we have $$T_p^*(Teich(\Gamma)) \simeq QD^*(\Gamma_{\nu}),$$ where $\Gamma_{\nu} = w_{\nu} \Gamma w_{\nu}^{-1}$.

$\endgroup$
4
  • $\begingroup$ Thanks a lot for this most thorough answer! Sorry but I still cannot see one of the main points: where does the symmetric square of the holomorphic cotangent bundle of a Riemann surface come from? The Riemann surface itself must correspond to the point in which we take the cotangent to the Teichmüller space; but where is the cotangent space of that Riemann surface itself, represented by that point? A priori $QD^*(\Gamma)$ is a space of functions on $H^*$, how do these functions correspond to sections of a bundle on the corresponding Riemann surface? $\endgroup$ Jun 9, 2018 at 14:06
  • 1
    $\begingroup$ I sneakily called $QD^*(\Gamma)$ the set of hol. quadratic differentials. Usually, these are caled automorphic forms of $\Gamma$. What is a holom. section of the symmetric square of the cotangent bundle? It is an object $q$ such that in each chart $(U,z)$ it takes the form $q(z) = \phi(z) dz^2$, $\phi$ holom. Moreover, on the intersection of two charts we have two representations $q(z) = \phi(z) dz^2 = \psi(w) dw^2$ so that $\phi(z) \frac{dz^2}{dw^2} = \psi(w)$. But this exactly corresponds to $(\phi \circ \gamma)(w) \gamma'(w)^2 = \phi(w)$ when we convert to the language of Fuchsian models. $\endgroup$
    – Florian R
    Jun 9, 2018 at 14:25
  • $\begingroup$ Thanks for the explanation, and sorry, my confusion is not entirely gone. An authomorphic form (in particular modular form) is a section of a line bundle on a modular curve, or on some similar moduli space of some kind of Riemann surfaces, rather than on a Riemann surface itself, no? There also are Jacobi forms, which can be viewed as compatible families, parametrized by the moduli space, of sections of bundles on all Riemann surfaces at once, but this is still something different I believe. $\endgroup$ Jun 9, 2018 at 15:33
  • $\begingroup$ To be honest, more general moduli space theory goes beyond my expertise. The space $QD^*(\Gamma)$ is called space of automorphic forms in the book I referenced; I just wanted to point out that I was a little sly calling it the space of holomorphic quadratic differentials. Hopefully someone else can give you more details on this. $\endgroup$
    – Florian R
    Jun 9, 2018 at 15:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .