According to [Riemann surfaces, dynamics and
geometry](http://people.math.harvard.edu/%7Ectm/home/text/class/harvard/275/09/html/base/rs/rs.pdf) by C. McMullen (Course notes), the definition for a quadratic differential $\phi$ on a Riemann surface $X$ is given by
$$
\|\phi\|\_p = \left(\int\_X \rho^{2-2p} |\phi|^p\right)^{1/p}, \|\phi\|\_\infty = \sup\_X \frac{|\phi|}{\rho^2},
$$


where $\rho |dz|$ is the hyperbolic metric.


For a Beltrami coefficient $\mu,$
$$
\|\mu\|\_p = \left(\int\_X \rho^{2} |\mu|^p\right)^{1/p}, \|\mu\|\_\infty =\sup\_X |\mu|.
$$


I am confused about the mysterious exponents like $2-2p$ and I am not sure how the definition of $L^p$ can pass "smoothly" to $L^\infty.$ I also do not know why we are dividing $\rho^2$ in the first expression but not in the second one.


**Why must we put that certain power of $\rho$ into the expression?**



Motivation for the definition of $L^p$ norm for quadratic and Beltrami differentials
The powers of $\rho$ are necessary to make the integrals well-defined. In probably excessive detail:


1. **Because $X$ is a Riemann surface, the integrands here need to be $(1,1)$-forms.** This is just the familiar fact (change of variables theorem) that integrals over (oriented) manifolds are only well-defined on objects $\omega$ which are given in local coordinates by functions $\omega(x)$ satisfying the transformation rule
$$
\omega(x) = \det(D\psi\_x) \omega(y)
$$
whenever $y = \psi(x)$ a smooth change of coordinates. If $M = X$ is a Riemann surface, the coordinate changes $w = \psi(z)$ are holomorphic, so that $\det(D\psi) = |\psi'|^2$ (Cauchy-Riemann), and thus the above rule can be rewritten as
$$
\phi(z) = |\psi'(z)|^2 \phi(w) = \psi'(z) \overline{\psi'(z)} \ \phi(w). 
$$
This condition can be summarized in the language of Riemann surface $(n, m)$-tensors (see Lyubich's [book](http://www.math.stonybrook.edu/%7Emlyubich/book.pdf), p. 99) by saying that $\phi$ can be integrated over $X$ iff it is a $(1,1)$-form, i.e. an object with local form
$$
\phi = \phi(z) \ dz \ d\bar{z} = \phi(z) \ |dz|^2. 
$$
2. **Quadratic differentials are $(2, 0)$-forms. Beltrami forms are $(-1, 1)$-forms.** Take $\phi$ a quadratic differential. Then $|\phi|^p$ is a $(p, p)$-form. In local coordinates:
$$
\phi = \phi(z) \ dz^2 \implies
|\phi|^p = |\phi(z) dz^2|^p = |\phi(z)|^p |dz|^{2p} = |\phi(z)|^p dz^p d\bar{z}^p. 
$$
In particular, $|\phi|$ is a $(1, 1)$-form, so can be integrated over $X$. This shows that the $L^1$-norm on quadratic differentials is canonically defined, even when $X$ is not hyperbolic (and thus has no preferred conformal metric), in the way you'd expect:
$$
||\phi||\_1 = \int\_X |\phi|. 
$$
When $p \neq 1$, however, $|\phi|^p$ is *not* a $(1,1)$-form, so its integral over $X$ is not defined, and thus the naive definition of the $L^p$-norm for quadratic differentials will not work! Fortunately, this issue can be remedied in the presence of a conformal metric $\rho = \rho(z) |dz|$, because we can adjust the tensor type of our candidate integrand using powers of $\rho$. In particular, multiplying by $\rho^{2-2p}$ gives
$$
\rho^{2-2p} |\phi|^p = \rho(z)^{2-2p} |dz|^{2-2p} |\phi(z)|^p |dz|^{2p} = \rho(z)^{2-2p} |\phi(z)|^p |dz|^{2}, 
$$
which *is* a $(1,1)$-form, and so can be integrated over $X$. Therefore we can define an $L^p$-norm on quadratic differentials by the expression you gave above:
$$
||\phi||\_p = \left( \int\_X \rho^{2-2p} |\phi|^p \right)^{1/p}.
$$
Moreover, though this definition appears to depend on an arbitrary choice of $\rho$, when $X$ is a *hyperbolic* Riemann surface there is a canonical choice, namely the unique conformal metric of constant curvature -1. So, in the hyperbolic case with $\rho$ taken to be this distinguished metric, it makes sense to refer to the above definition as *the* $L^p$-norm on quadratic differentials. The Beltrami case works similarly.


The construction here is exactly analogous to the problem of defining $L^p$-norms for spaces of *functions* on a smooth manifold $M$. The $L^\infty$-norm of a function is, of course, defined as usual by
$$
||f||\_\infty = \displaystyle\text{ess sup}\_M |f|. 
$$
But without further structure, the $L^p$-norms for $p < \infty$ are not well-defined, precisely because functions cannot be integrated over $M$. After fixing a Riemannian metric on $M$, however, the $L^p$-norm of a function can be defined in the obvious way by integrating against the associated volume form:
$$
||f||\_p = \left( \int\_M |f|^p \ \text{dVol} \right)^{1/p}
$$
Taking this approach in our context leads to the same result as above. That is, we could have alternatively begun by modifying $\phi$ by $\rho$ to obtain a *function* on $X$, then applied the Riemannian definition given above to our special case. In detail, the local computation
$$
\dfrac{|\phi|}{\rho^2} = \dfrac{|\phi(z)||dz|^2}{\rho(z)^2 |dz|^2} = \dfrac{|\phi(z)|}{\rho(z)^2}, 
$$
shows that $\dfrac{|\phi|}{\rho^2}$ is a function on $X$. Then we can define
$$
||\phi||\_\infty = \displaystyle\text{ess sup}\_X \dfrac{|\phi|}{\rho^2}$$
and, using the volume form $\rho^2 = \rho(z)^2 |dz|^2$ associated to the conformal metric $\rho$, define
$$||\phi||\_p = \left( \int\_X \left( \dfrac{|\phi|}{\rho^2} \right)^p \ \rho^2 \right)^{1/p} = \left( \int\_X \rho^{2-2p} |\phi|^p\right)^{1/p}
$$
for $p < \infty$. Voila!


