According to Riemann surfaces, dynamics and geometry 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?


1 Answer 1


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, 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 $p < \infty$.


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