Originally published at 狗和留美者不得入内. You can comment here or there.
I’ve been asked to prove the Big Picard theorem, assuming the fundamental normality test. Assuming the latter, it is a very short proof, and I could half-ass with that. I don’t like writing up stuff that I don’t actually understand for the sake of doing so. There’s little point, and if I’m going to actually write up a proof of it, I’ll do so for real, which means that I go over the fundamental normality test in its entirety.
Theorem 2.28 (Riemann mapping theorem). Let be simply-connected and
. Then there exists a conformal homeomorphism
onto the unit disk
.
Proof: Linked here.
Theorem 2.30. Suppose n is bounded, simply-connected, and regular. Then any conformal homeomorphism as in Theorem 2.28 extends to a homeomorphism .
Schwarz reflection principle. Suppose that is an analytic function which is defined in the upper half-disk
. Further suppose that
extends to a continuous function on the real axis, and takes on real values on the real axis. Then
can be extended to an analytic function on the whole disk by the formula
and the values for reflected across the real axis are the reflections of
across the real axis.
We begin by presenting the standard “geometric” procedure by which the covering map may be obtained. Here
are distinct points. This then leads naturally to the “little” and “big” Picard theorems, which are fundamental results of classical function theory.
The construction takes place in the Poincaré disk. In the above figure, we have a circle reflected about
. The configuration is such that
intersects
perpendicularly. We are reflecting
across
. The intersection points must be fixed, and the reflection must preserve the orthogonality. Moreover, reflection preserves the geodesic nature, and under the hyperbolic metric, geodesics are generalized circles. From this, we can deduce that
goes to itself, with its two arcs relative to
interchanged.
To construct the map, we start with a triangle inside the unit circle consisting of circular arcs that intersect the unit circle at right angles. Reflect
across each of its sides and one gets three more triangles with circular arcs intersecting the unit circle at right angles.
The above figure shows how the unit disk is partitioned by triangles as a result of iterating these reflections indefinitely. To obtain the sought after covering map, we start from the Riemann mapping theorem which gives us a conformal isomorphism , the upper half-plane. This map extends as a homeomorphism to the boundary by Theorem 2.30. Thus, the three circular arcs of
get mapped to the intervals
, respectively. By the Schwarz reflection principle, the map
extends analytically to the region obtained by reflecting
across each of its sides as just explained above. There is that
are on the boundary of the unit disk and thus omitted. There is also that complex conjugation as specified in the Schwarz reflection principle reflects the upper half plane to the lower half plane. This way, we obtain a conformal map onto
defined on the entire unit disk that is a local isomorphism and a covering map.
Theorem 4.18. Every entire function which omits two values is constant.
Proof. Indeed, if is such a function, we may assume that it takes its values in
. But then we can lift
to the universal cover of
to obtain an entire function
into
. By Liouville’s theorem,
is constant. ▢
Theorem 4.19 (Fundamental normality test). Any family of functions in
which omits the same two distinct values in
is a normal family.
Theorem 4.20. If has an isolated essential singularity at
, then in any small neighborhood of
the function
attains every complex value infinitely often, with one possible exception.
Proof. Suppose without loss of generality that and define
for an integer
. We take
so large that
is analytic on
by making the neighborhood about
sufficiently small. Suppose by contradiction every neighborhood of
omits the some two points. Then every function in this family, defined via
, omits the same two points. Thus, by the fundamental normality test, some subsequence of the family
uniformly on
where either
is analytic or
by Weierstrass’s theorem (see here). By the maximum principle, in the former case,
is bounded near
, which means it’s removable. In the latter case, convergence to
implies that
is a pole, contradicting that
has an essential singularity there. ▢
References
- Schlag, W., A Course in Complex Analysis and Riemann Surfaces, American Mathematical Society, 2014, pp. 70-72,81,160-164.