In this post, we will deal with the proof of Picard’s Great and Little theorem via the method of differential geometry.

## Ahlfors-Schwarz Lemma

Assume and are 2 domains in , and is an analytic function. If is a metric on and , then the pull-back of , defined on , is . Recall that the definition of Curvature with respect to metric is , so

**Lemma 0: **If notations as above, then

Proof: This follows from a directly computation.

**Theorem 1(Ahlfors-Schwartz Lemma)** is a holomorphic function on , f maps to . If there is a metric on , that is, , such that its curvature is not bigger than , then , where . In another word,

Proof: fix any and define a metric on as . Obviously, for any point , its curvature is . Define a function on . It is easy to notice that is bounded on and it is on the boundry. So must attain its maximum at an interior point . So all we are left to do is prove

1), then , there’s nothing to say.

2) , then is defined, by the definition . Since is the maximal point, then

In another word, this is , or .

In the above theorem, if we choose and , it agrees with the Schwartz-Pick Lemma.

Further, if we define on , here , then this metric has curvature at any point in .

**Theorem 2(General form of Ahlfors-Schwartz Lemma)** is a holomorphic function on , maps to . If there is a metric on , that is, , such that its curvature is not bigger than , then for each and

## Generalized Liouville Theorem

In classical Complex Analysis, we know that a bounded entire function must be constant. Now by the Ahlfors-Schwartz Lemma, there is a corresponding theorem with respect to curvature.

**Theorem 3(Generalized Liouville Theorem)** Assume to be entire, and be a metric on , such that for any point , for some positive constant . Then must be a constant.

Proof: for any , maps to . By our assumption, and general form of Ahlfors-Schwartz Lemma, we have

Since , then we may conclude that . Thus , this force to be a constant.

**Deduction of Liouville thoerem**: If is bounded by , then $g=\frac{1}{M}f$ maps . If we choose a metric on such that its curvature is . So choose in the above theorem, then is a constant, so is .

## Picard’s Little Theorem

With the theorem above, we may prove Picard’s little theorem.

**Lemma 4:** If is an open set, with at least 2 point in , then we can introduce a metric such that for each , where is some positive constant.

Proof: Choose a linear transform to map 2 fixed point in to , denote , and introduce a metric by

By directly computation we have

In this case, we know

So is bounded

**Theorem 5(Picard’s Little Theorem):** If entire function maps to , and contains at least 2 points, then is a constant.

Proof: Directly by theorem 1 and lemma 2.

## Picard’s Great Theorem

**Def 6.1 **If is a sequence of functions defined on ,it is called normally convergence if for any and a compact subset, there must exists a natural number ,(depends on and ), such that for all we have . That is if inner-closed uniformly convergence, then it is normally convergence.

And we call it compactly divergence, if for any compact subset and , there exists a natural number ,(depends on and ), such that for all we have . That is if uniformly diverge to on any compact subset, then it is compactly divergent.

**Def 6.2** If is a familly complex function defined on such that any sequence in has a subsequence normally convergent or compactly divergent, then we call a normal family.

**Def 6.3** If is a familly meromorphic function defined on normally convergent with respect to the sphere metric, is called a normal family.

**Theorem 7(Marty):** If is a family of meromorphic function defined on , then is normal iff the set are uniformly bounded on any compact set, where is the sphere metric. In another word, there exists a constant for each compact set such that .

Proof: If uniformly bounded, then

where is a curve connecting and contained in , , and is the strait line connecting . In this case, the sphere metric is bounded and under this metric is uniformly continuous, so by A-A lemma, is normal.

On the other hand, If is normal, and assume the boundedness is not true, then there is a compact set and a sequence such that is unbounded on . Then by assumption, there is a subsequence uniformly converging to on .

At each point of , there can be a closed circle , with either or is holomorphic on . For the first case, is bounded on . Since converges under sphere metric, so when large enough, has no poles on . So converges uniformly to on a slightly smaller circle that . Since is continuous, then are bounded on the slightly smaller circle. The same argument can show that are bounded, and by the fact . Cover by finitely many such circles, we thus deduced a contradiction that are bounded on .

**Theorem 8(Montel): **If is a family of meromorphic function on , are three different point. If each function in missed the three points, then the family is normal.

Proof: Use linear transformation to transform to . So we just need to show All holomorphic function families with values in are normal. This suffices to prove is normal for any , and W.L.O.G. we may assume .

Let , multiply it by a constant such that the upper bound of its curvature is , and still denote it by . By General Ahlfors-Schwartz Lemma, for any , we have , that is for each

compare the sphere metric with in . It is easy to conclude that( directly computation)

So there is a positive constant such that , so for each we know

This means that is bounded on any compact subset of uniformly(not dependent of the choice of ), by Marty’s theorem, we proved this theorem.

**Theorem 8′(Montel)** If is a holomorphic familly on , such that each function in miss 2 fixed point in , then is normal familly.

Proof: In fact, this is what we proved in theorem 8.

**Theorem 9(Picard’s Great Thoerem):** If is holomorphic on , and is its essential sigular point, then contains at most one point for any neighborhood of .

Proof: Otherwise, assume is holomorphic on and do not contain . Claim $z=0$ is its removable sigular point or polar point.

Define . Construct holomorphic family , with image contained in . By theorem 8′ we know it is normal family, so there is a subsequence , which is normally convergent or compactly divergent.

- If normally convergent, then are inner-closed uniformly convergent, thus is bounded on any compact subset of . In partucularly, are bounded on by , that is is bounded by on . By the maximal mudulus theorem, is bounded by on . So is removable, a contradiction.
- If compactly divergent, consider , we can prove that , then is a polar point, a contradiction.