Picard's Little Theorem and Twice-Punctured Plane

Introduction

Let be a holomorphic function. By Liouville’s theorem, if is bounded, then has to be a constant function. However, there is a much stronger result. In fact, if differs from exactly points, then is a constant. In other words, suppose is non-constant, then the equation for all except at most one . To think about this, if is a non-constant polynomial, then always has a solution (the fundamental theorem of algebra). If, for example, , then has no solution only if .

The proof will not be easy. It will not be proved within few lines of obvious observations, either in elementary approaches or advanced approaches. In this post we will follow the later by studying the twice-punctured plane . To be specific, without loss of generality, we can assume that and are not in the range of . Then . Next we use advanced tools to study in order to reduce the question to Liouville’s theorem by constructing a bounded holomorphic function related to .

We will find a holomorphic covering map and then replace with the unit disc using the Cayley transform . Then the aforementioned will be lifted to a holomorphic function , which has to be constant due to Liouville’s theorem, and as a result is constant.

With these being said, we need analytic continuation theory to establish the desired , and on top of that, (algebraic) topology will be needed to justify the function .

Analytic Continuation

For a concrete example of analytic continuation, I recommend this post on the Riemann function. In this post however, we only focus on the basic language of it in order that we can explain later content using analytic continuation.

Our continuation is always established “piece by piece”, which is the reason we formulate continuation in the following sense.

Definition 1. A function element is an ordered pair where is an open disc and . Two function elements and are direct continuation of each other if and on . In this case we write

The notion of ordered pair may ring a bell of sheaf and stalk. Indeed some authors do formulate analytic continuation in this language, see for example Principles of Complex Analysis by Serge Lvovski.

The relation is by definition reflective and symmetric, but not transitive. To see this, let be the primitive -th root of unity. Let be open discs with radius and centres . Since the are simply connected, we can always pick such that , and and but on one has . Indeed there is nothing mysterious: we are actually rephrasing the fact that square root function cannot be defined at a region containing .

Definition 2. A chain is a finite sequence of discs such that for . If is given and if there exists function elements such that for , then is said to be the analytic continuation of along .

A chain is said to cover a curve with parameter interval if there are numbers such that is the centre of , is the centre of , and

If can be continued along this to , we call an analytic continuation of along ; is then said to admit an analytic continuation along .

Either way, it is not necessary that . However, unicity of is always guaranteed. We will sketch out the proof on unicity.

Lemma 1. Suppose that , and , then .

Proof. By assumption, in , and in . It follows that in , which is open and non-empty. Since and are holomorphic in and is connected, we have in . This is because on a open connected set , the zero of is not discrete. Therefore has to be everywhere on .

Theorem 1. If is a function element and is a curve which starts at the centre of , then admits at most one analytic continuation along .

Sketch of the proof. Let and be two chains that cover . If can be analytically continued along to a function element and along to , then in .

We are also given partitions and such that

and function elements and for and with . The poof is established by showing that the continuation is compatible with intersecting intervals, where lemma 1 will be used naturally. To be specific, if and , and if , then .

The Monodromy Theorem

The monodromy theorem asserts that on a simply connected region , for a function element with , we can extend it to all if can be continued along all curves. To prove this we need homotopy properties of analytic continuation and simply connected spaces.

Definition 1. A simply connected space is a path connected topological space with trivial fundamental group for all .

The following fact is intuitive and will be used in the monodromy theorem.

Lemma 2. Let be a simply connected space and let and be two closed curves with and . Then and are homotopic.

Proof. Let be the curve defined by for . Then

where is the identity of .

Next we prove the two-point version of the monodromy theorem.

Monodromy theorem (two-point version). Let be two points on and be a function element where is centred at . Let be a homotopy class indexed by a map with the same origin and terminal . If admits analytic continuation along each , to an element , then .

In brief, analytic continuation is faithful along homotopy classes. By being indexed by we mean that . We need the uniform continuity of .

Proof. Fix . By definition, there is a chain which covers , with , such that is obtained by continuation of along . There are numbers such that

For each , define

The makes sense and is always positive, because is always compact and is an open set. Then pick any . Since is uniformly continuous, there exists a such that

We claim that also covers . To do this, pick any . Then because

Therefore by theorem 1, we have . Notice that for any , there is a segment such that for all . Since is compact, there are finitely many that cover . Since is connected, we see, after a finite number of steps, we can reach .

Momodromy theorem. Suppose is a simply connected open subset of the plane, is a function element with , and can be analytically continued along every curve in that starts at the centre of . Then there exists such that for all .

Proof. Let and be two curves in from the centre of to some point . Then the two-point monodromy theorem and lemma 2 ensures us that these two curves lead to the same element , where is a circle with centre at . If intersects , then can be obtained by continuing to , then along the segment connecting and . By definition of analytic continuation, in . Therefore the definition

is a consistent definition and gives the desired holomorphic extension of .

Modular Function

Let be the open upper half plane. We will find a function whose image is and is in fact the (holomorphic) covering space of . The function is called a modular function.

As usual, consider the action of on given by

Definition 2. A Modular function is a holomorphic (or meromorphic) function on which is invariant under a non-trivial subgroup of . That is, for any , one has .

In this section, we consider this subgroup:

It has a fundamental domain

Basically, is bounded by two vertical lines and vertically, and two semicircles with centre at and with diameter , but only the left part contains boundary points. The term fundamental domain will be justified by the following theorem.

Theorem 4. Let and be as above.

(a) Let be two distinct elements of , then .

(c) is generated by two elements

Sketch of the proof. Let be the subgroup of generated by and , and show (b’):

Then (a) and (b’) would imply that and (b) is proved. To prove (a), one will replace with the identity element and discuss the relationship between and for . To prove (b’), one need to notice that

For , by picking that maximises , only to show that and therefore .

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We are now allowed to introduce the modular function.

Theorem 5. Notation being above, there exists a function such that

(a) for every .

(b) is one-to-one on .

(d) has the real axis as its natural boundary. That is, has no holomorphic extension to any region that properly contains .

Proof. Consider

This is a simply connected region with simple boundary. There is a continuous function which is one-to-one on and is holomorphic in such that , , and . This is a consequence of conformal mapping theory.

The Schwartz reflection principle extends to a continuous function on which is a conformal mapping of (the interior of ) onto the plane minus the non-negative real axis, by the formula

Note the extended is one-to-one on , and is defined in (c).

On the boundary of , the function is real. In particular,

and that

We now define

for and . This definition makes sense because for each , there is one and only one such that . Properties (a) (b) and (c) follows immediately.

Notice is continuous on

and therefore on an open set containing . Cauchy’s theorem shows that is holomorphic in . Since is covered by the union of the sets for , and since , we conclude that .

Finally, the set of all numbers is dense on the real axis. If could be analytically continued to a region which properly contains , the zeros of would have a limit point in this region, which is impossible since is not constant.

We are now ready for the pièce de résistance of this post.

Picard’s Little Theorem

Theorem (Picard). If is an entire function and if there are two distinct complex numbers and such that are not in the range of , then is constant.

The proof is established by considering an analytic continuation of a function associated with . The continuation will be originated at the origin and validated by monodromy theorem. Then by Cayley’s transformation, we find out the range of is bounded and hence is constant, so is .

Proof. First of all notice that without loss of generality, we assume that and , because otherwise we can replace with . That said, the range of is in theorem 5. There is a disc with centre at so that lies in a disc .

For every disc , there is an associated region such that in theorem 5 is one-to-one on and ; each such intersects at most two of the domains . Corresponding to each choice of , there is a function such that for all .

Now let be the function such that as above. Define for . We claim that can be analytically continued to an entire function.

If is another disc in with , we can choose a corresponding so that . Then and are direct analytic continuations of each other. We can proceed this procedure all along to find a direct analytic continuation of with . Note for all .

Let be a curve in the plane which starts at . The range of is a compact subset of and therefore can be covered by a chain of discs, say , so that each is in a disc . By considering function elements , composing with on each (this is safe because is entire), we get an analytic continuation of along the chain . Note again.

Since is arbitrary, we have shown that can be analytically continued along every curve in the plane. The monodromy theorem implies that extends to an entire function. Thus proving our claim given before.

Note the range of the extended on every possible has range lying inside . Therefore . It follows that

has range in the unit disc. By Liouville’s theorem, is a constant function. Thus is constant too.

Now we move back to by looking at . Since is one-to-one on and is not empty and open, has to be a singleton. Thus is constant on . If we represent as a power series on a disc lying inside , we see has to be a constant.

Note we have also seen that the range of a non-constant function cannot be half of a plane. But this result is useless because we can find two points on a large chunk of a plane after all.

Reference

• Walter Rudin, Real and Complex Analysis.
• Tammo tom Dieck, Algebraic Topology.