How to interpret the space of quadratic differentials as a cotangent space? 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.
 A: 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 $\Gamma_{\nu} = w_{\nu} \Gamma w_{\nu}^{-1}$.
