Question on Weil-Petersson metric on Teichmuller space I'm reading Ahlfors' original articles about Weil-Petersson metric:  "Some remarks on Teichmüller's space of Riemann surfaces" and "Curvature properties of Teichmüller's space". 
The tangent space at a point $C$ (here identified with a Riemann surface of genus $g$) of the Teichmuller space can be identified with the space of Beltrami differentials on $C$ and the author defines the Weil-Petersson hermitian form as $\langle \nu,\mu\rangle=\int_C\nu\varphi(\mu)dxdy$, for every Beltrami differentials $\nu,\mu$ on $C$, where $\varphi(\mu)$ is the quadratic differential corresponding to $\mu$.
On the Teichmuller space $\mathcal{T}_g$ there are complex coordinates $t_1,\dots,t_n$, because every Beltrami differential $\mu$ can be written $\mu=t_1\nu_1+\dots+t_n\nu_n$, where $\nu_1,\dots,\nu_n$ is a base. Then author then wants to write the Weil-Petersson form in this coordinates as $\sum_{\alpha,\beta}g_{\alpha\overline{\beta}}(\mu)dt_\alpha d\overline{t}_\beta$.
What I don't get is why he writes $g_{\alpha\overline{\beta}}(\mu)=\langle L^\mu\nu_\alpha,L^\mu\nu_\beta\rangle$
The definition of $L^\mu\nu$ is the following. Consider $f^\mu:\hat{\mathbb{C}}\rightarrow\hat{\mathbb{C}}$ such that $\partial_{\overline{z}}f^\mu=\mu\partial_zf^\mu$, then, given three Beltrami differentials $\mu,\rho$ and $\lambda$, we write $\rho=\mu|\lambda$ iff $f^\mu=f^\rho\circ f^\lambda$ which is true iff $\rho\circ f^\lambda=\frac{\mu-\lambda}{1-\overline{\lambda}\mu}(\partial_zf^\lambda/|\partial_zf^\lambda|)^2$.
Then, by abuse of notation, we can define the function on Beltrami differentials $\rho:\mu\mapsto \mu|\lambda$ and $\frac{\partial}{\partial \nu}\rho(\lambda):=lim_{t\rightarrow 0}\frac{\rho(\lambda+t\nu)-\rho(\lambda)}{t}=L^\lambda\nu=(\nu\frac{(\partial_zf^\lambda)^2}{|\partial_zf^\lambda|^2-|\partial_\overline{z}f^\lambda|^2})\circ(f^\lambda)^{-1}$ 
Why this definition of $L^\mu\nu$ should imply that, imposing $g_{\alpha\overline{\beta}}(\mu)=\langle L^\mu\nu_\alpha,L^\mu\nu_\beta\rangle$, we get the metric $\int_C\nu\varphi(\mu)dxdy$?
 A: One way to define Teichmueller space is to fix a Riemann surface
$X$, with a fixed complex structure, and define the space of all
Beltrami differentials $\mathcal{M}(X)$ on $X$. Notice that in
this way we have fixed a base point $X$ of Teichmueller space with
a fixed "background" complex structure. By a Beltrami differential
I mean a $(-1,1)$ form $\mu(z)\frac{d\bar{z}}{d z}$ such that
$\|\mu\|_{\infty} < 1$. By an infinitesimal Beltrami differential
I simply mean a $(-1,1)$ form without the latter inequality, all
of which form the set $\mathcal{IB}(X)$. Then one way to define
Teichmueller space is $\mathcal{T}(X) = \mathcal{M}(X) / \sim$,
where two Beltramies $\mu_1$ and $\mu_2 \in \mathcal{M}(X)$ are
equivalent, i.e. $\mu_1 \sim \mu_2$, exactly when there is a
quasi-conformal homeomorphism $f : X \to X$ homotopic to identity
such that
$$\mu_1 = f^*\mu_2 = \frac{\bar{\partial} f + (\mu_2\circ f)
\overline{\partial f}}{{\partial} f + (\mu_2 \circ f)
\partial\bar{f}}.
$$ Furthermore, the tangent space $T_X\mathcal{T}(X)$ of
$\mathcal{T}(X)$ at the base point $X$ can be thought of as
$\mathcal{IB}(X)/\sim$.
Now, if by $X^{\mu}$ I denote the Riemann surface with complex
structure induced by the Beltrami $\mu$ on $X$, then the tangent
space of the Teichmueller sapce at that point is
$T_{X^{\mu}}\mathcal{T}(X) = T_{[\mu]}\mathcal{T}(X) =
\mathcal{IB}(X^{\mu})/\sim$. The Wiel-Petersson hermitian form,
calculated at the point $[\mu]$, is defined as $$\langle\nu_1,
\nu_2\rangle_{\mu} = \int_{X^{\mu}} \nu_1 \varphi(\nu_2),$$ where
$\nu_1$ and $\nu_2 \in \mathcal{IB}(X^{\mu})$ are infinitesimal
Beltramies on $X^{\nu}$ (not on $X$!). This is the reason for
which one needs to use the formulas you are asking about. One
needs to look at all the infinitesimal Beltramies $\nu =
\sum_{\alpha} t_{\alpha}\nu_{\alpha}$ on the base point Riemann
surface $X$ from the perspective of the Riemann surface determined
by $[\mu] \in \mathcal{T}(X)$, i.e. to look at the infinitesimal
Beltriemies $\nu|_{\mu} = \sum_{\alpha}
t_{\alpha}\nu_{\alpha}|_{\mu}$ on the surface $X^{\mu},$ i.e.
$\nu|_{\mu} \in \mathcal{IB}(X^{\mu})$. In order to do that, first
one needs a way to move all Beltramies from $\mathcal{M}(X)$ to
$\mathcal{M}(X^{\mu})$ by establishing an appropriate isomorphism.
Then, by differentiating that isomorphism at the point $[\mu]$ one
would find out how infinitesimal Beltramies are moved from $X$ to
$X^{\mu}$ isomorphically, i.e. one would obtain the tangent
isomorphism. Consequently, one would have an identification
between $\mathcal{IB}(X^{\mu})/\sim$ and $\mathcal{IB}(X)/\sim$.
Given a Beltrami $\lambda$ denote by $f^{\lambda} : X \to
X^{\lambda}$ the quasi-conformal map (unique up to post-composition
with a holomorphic isomorphism) that determines the Beltrami
$\lambda = \frac{\bar{\partial} f^{\lambda}}{\partial
f^{\lambda}}$ on $X$. Also denote by $f^{\lambda|_{\mu}} : X^{\mu}
\to X^{\lambda}$ the quasi-conformal map that determines the
(intermidiate) Beltrami $\lambda|_{\mu} = \frac{\bar{\partial}
f^{\lambda|_{\mu}}}{\partial f^{\lambda|_{\mu}}}$ on $X^{\mu}$
(not on $X$!). Then, (technically up to holomorphic isomorphism)
$f^{\lambda} = f^{\lambda|_{\mu}} \circ f^{\mu} : X \to
X^{\lambda},$ where $ f^{\lambda|_{\mu}} : X^{\mu} \to
X^{\lambda}.$ After differentiating the latter identity between
the three quasi-conformal maps, $df^{\lambda} =
df^{\lambda|_{\mu}} \circ df^{\mu}$, and then combining it with
the expressions of the type $df(z) = \partial f(z) dz +
\bar{\partial} f(z) d\bar{z}$, one obtains again the formula,
already featured above,
$$\lambda = f^{\mu*}(\lambda|_{\mu}) =
\frac{\bar{\partial} f^{\mu} + (\lambda|_{\mu} \circ f^{\mu})
\overline{\partial f^{\mu}}}{{\partial} f^{\mu} + (\lambda|_{\mu}
\circ f^{\mu})
\partial\overline{f^{\mu}}}.
$$
By solving for $\lambda|_{\mu} \circ f^{\mu}$ and using the fact
that $\bar{\partial} f^{\mu} = \mu \partial f^{\mu}$, one obtains
exactly the expression
$$\lambda|_{\mu} \circ f^{\mu} = \frac{\lambda - \mu}
{1 - \bar{\mu}\lambda}\left(\frac{\partial f^{\mu}}{|\partial
f^{\mu}|}\right)^2.$$ This is basically the map from
$\mathcal{M}(X)$ to $\mathcal{M}(X^{\mu})$ sending Beltramies on
$X$ to Beltramies on $X^{\mu}$. To obtain a tangent vector at
$[\mu] \in \mathcal{T}(X)$, take an infinitesimal Beltrami $\nu$
on $X$ and form the one parameter family $\mu_t = \mu + t \nu$. On
$X^{\mu}$ this family becomes
$$\mu_t|_{\mu} \circ f^{\mu} = \frac{\mu + t \nu - \mu} {1 -
\bar{\mu}(\mu + t \nu)}\left(\frac{\partial f^{\mu}}{|\partial
f^{\mu}|}\right)^2 =  t \frac{\nu } {1 - \bar{\mu}(\mu + t
\nu)}\left(\frac{\partial f^{\mu}}{|\partial f^{\mu}|}\right)^2.$$
To obtain the tangent vector $\nu|_{\mu}$ at the point $\mu$
simply differentiate the last identity with respect to $t$ and
then set $t=0$. The result is $$\nu|_{\mu} \circ f^{\mu} = \nu
\frac{(\partial f^{\mu})^2} {|\partial f^{\mu}|^2 -
|\bar{\partial} f^{\mu}|^2}.$$ After pre-composing with $(f^{\mu})^{-1}$,
one obtains exactly the formula you mention
$$L^{\mu}\nu = \nu|_{\mu} = \left(\nu
\frac{(\partial f^{\mu})^2} {|\partial f^{\mu}|^2 -
|\bar{\partial} f^{\mu}|^2}\right)\circ (f^{\mu})^{-1}.$$ That is
why the Weil-Petersson form should look like
$$\langle\nu_{\alpha}|_{\mu},
\nu_{\beta}|_{\mu}\rangle_{\mu} = \langle L^{\mu}\nu_{\alpha},
L^{\mu}\nu_{\beta}\rangle_{\mu} = \int_{X^{\mu}}
\nu_{\alpha}|_{\mu} \varphi(\nu_{\beta}|_{\mu}) = \int_{X^{\mu}}
L^{\mu} \nu_{\alpha} \varphi(L^{\mu} \nu_{\beta}),$$ for any
$\nu_{\alpha}$ and $\nu_{\beta}$ from $\mathcal{IB}(X)$ (notice
that the base point is $X$!).
