Let D be the open unit disk in the complex plane. The Schwarz lemma says that if f is an analytic function from D to D with f(0) = 0, then
for all z in D. (The lemma also says more, but this post will focus on just this portion of the theorem.
The Schwarz-Pick theorem generalizes the Schwarz lemma by not requiring the origin to be fixed. That is, it says that if f is an analytic function from D to D then
The Schwarz-Pick theorem also concludes more, but again we’re focusing on part of the theorem here. Note that if f(0) = 0 then the Schwarz-Pick theorem reduces to the Schwarz lemma.
The Schwarz lemma is a sort of contraction theorem. Assuming f(0) = 0, the lemma says
This says applying f to a point cannot move the point further from 0. That’s interesting, but it would be more interesting if we could say f is a contraction in general, not just with respect to 0. That is indeed what the Schwarz-Pick theorem does, though with respect to a new metric.
For any two points z and w in the open unit disk D, define the Poincaré distance between z and w by
It’s not obvious that this is a metric, but it really is. As is often the case, most of the properties of a metric are simple to confirm, but the proving the triangle inequality is the hard part.
If we apply the monotone function tanh-1 to both sides of the Schwarz-Pick theorem, then we have that any analytic function f from D to D is a contraction on D with respect to the Poincaré metric.
Here we’re using “contraction” in the lose sense. It would be more explicit to say that f is a non-expansive map. Applying f to a pair of points may not bring the points closer together, but it cannot move them any further apart (with respect to the Poincaré metric).
By using the Poincaré metric, we turn the unit disk into a hyperbolic space. That is D with the metric d is a model of the hyperbolic plane.
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