the Schwarz-Pick lemma

The plain Schwarz lemma has a blemish: it only speaks about maps that fix the center of the disk. Pick's insight was to remove that crutch. If you measure distance inside the disk not with the ordinary ruler but with a special, distance-warping ruler — the hyperbolic ruler — then every holomorphic self-map of the disk shrinks or preserves distances, no matter where it sends the center. The disk, viewed through the right metric, is a place where holomorphic maps can never spread points apart.

Here is the precise statement, in two equivalent forms. Let f be holomorphic from the unit disk to itself. The differential (infinitesimal) form says: for every z, |f'(z)| / (1 - |f(z)|^2) <= 1 / (1 - |z|^2). The integrated (finite) form says: for any two points z_1, z_2, the hyperbolic distance from f(z_1) to f(z_2) is at most the hyperbolic distance from z_1 to z_2. Equality at one pair (or in the differential form at one point) forces f to be a conformal automorphism of the disk — a Mobius map that moves the whole disk rigidly. You recover the original Schwarz lemma by specializing to z_1 = 0, f(0) = 0. The proof is exactly the original Schwarz lemma sandwiched between two disk automorphisms: pre-compose to send z_1 to 0, post-compose to send f(z_1) to 0, apply Schwarz in the middle, then translate back.

This is the doorway from rigidity into geometry. The quantity 1/(1 - |z|^2) that appears is the density of the Poincare (hyperbolic) metric, and the lemma is the statement that holomorphic self-maps are distance-non-increasing for that metric — they are 1-Lipschitz contractions of the hyperbolic plane. A subtle point worth stating honestly: 'distance-non-increasing' does not mean f is a contraction in the strict sense everywhere; an automorphism preserves hyperbolic distance exactly, and only genuine non-automorphisms strictly shrink. So the lemma is sharp, and the automorphisms are precisely the isometries it allows.