Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Scharwz Lemma in its general form says that any holomorphic map between hyperbolic surfaces is contracting.

Noting that Riemann surfaces admit a unique metric of constant curvature -1, I wonder if we can rewrite this result completely in terms of conformal or holomorphic structure, without reference to a hyperbolic metric. Indeed, the original result is in terms of the Euclidean metric on the plane.


More generally, I am wondering whether it is possible - at least in the case of Riemann surfaces - to rephrase any result concerning metrics and curvature, in terms of conformal or holomorphic structure.

share|improve this question
add comment

2 Answers 2

I think the short answer to your question is no, since the notion of a holomorphic map is purely local, whereas the existence of a hyperbolic metric (uniformization) depends on the global structure of the Riemann surface. Except at branched points (or for constant maps), there is no local invariant of a holomorphic map. I think another way to say this is that on a disk, there may be many different incomplete Riemannian metrics of curvature -1, which are incomparable.

On the other hand, if you take the lift of the holomorphic map to the universal covers of the two Riemann surfaces, and identify them holomorphically with the unit disk, then you can reduce the general Schwarz lemma to the classic one, which may be stated without reference to the hyperbolic metric on the unit disk. However, you are implicitly using the global structure of the Riemann surface (and its universal cover) and uniformization. The map is either an isometry (a conformal diffeomorphism of the universal covers), or it has branched points (is not locally 1-1), or it is not proper. This last option shows again the non-locality of the question.

It's possible I haven't understood the sort of answer that you are looking for though?

share|improve this answer
add comment

Pretty much what Agol said, but I wonder if you've seen the generalization of the Schwarz-Ahlfors-Pick theorem (that shows how Schwarz lemma may be understood in terms of the curvature of a suitable conformal metric as you said) to other manifolds of higher dimension. Systems of pseudometrics on complex spaces that satisfy the Schwarz–Pick lemma have been studied extensively. A Schwarz-Pick system is a functor $X\to d_X$ that assigns to each complex Banach manifold $X$ a pseudometric $d_X$ so that the following conditions hold:

(a) The pseudometric assigned to the unit Disk $\mathbb D$ is the Poincare metric $$d_{\mathbb D}(x_1,x_2)=\tan^{-1}\left(\frac{x_1-x_2}{1-x_1\overline{x_2}}\right)$$

(b) If $X,Y$ are two complex Banach manifolds then $$d_Y(f(x_1),f(x_2))\le d_X(x_1,x_2)$$ for all $x_1,x_2\in X$ and $f\in \mathcal O(X,Y)$ (holomorphic maps $X\to Y$).

Now $\mathcal O(\mathbb D,X)$ and $\mathcal O(X,\mathbb D)$ provide upper and lower bounds for $d_X$ which correspond to the Kobayashi and Caratheodory pseudometrics. To have a meaningful notion of the Schwarz lemma you would need the Kobayashi pseudometric to be non-degenerate

There is certain notions of hyperbolicity around. Call a complex manifold $X$ Brody hyperbolic if there are no non-constant holomorphic maps $\mathbb C\to X$. Call $X$ Kobayashi hyperbolic if its Kobayashi pseudometric is non-degenerate. At least for compact complex manifolds these two notions are equivalent (sketch of proof here).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.