Kenneth Kuttler

1850 CHAPTER 58. ELLIPTIC FUNCTIONS

Proof: For z = z0 the equation holds. Let

A ={

z ∈ I : gm (z) = φ

(λ−1T ◦ f (z)

)}.

Thus z0 ∈ I. If z ∈ I and if w is close enough to z, then w ∈ I also and so both sides of58.1.40 with w in place of z are in λ

−1m ( f (I)) . But by construction, λ is one to one on this

set and since λ is invariant with respect to φ ,

λ (gm (w)) = λ

(λ−1T ◦ f (w)

)= λ

(φ ◦λ

−1T ◦ f (w)

)and consequently, w ∈ A. This shows A is open. But A is also closed in I because thefunctions are continuous. Therefore, A = I and so 58.1.40 is obtained.

Letting f (z) ∈ f (B(γ (T )) ,r) ,

(φ

(λ−1T ( f (z))

))= λ

(λ−1T ( f (z))

)= f (z)

and so φ ◦λ−1T is a local inverse for λ on f (B(γ (T )) ,r) . Let the new function element be

gm+1︷︸︸︷φ ◦λ

−1T ◦ f ,B(γ (T ) ,r)

 . This has shown the initial function element can be continued

along every curve through a.By the monodromy theorem, there exists g analytic on V such that g has values in P+

and g = g0 on B(a,r0) . By the construction, it also follows λ ◦ g = f . This last claim iseasy to see because λ ◦g = f on B(a,r0) , a set with a limit point so the equation holds forall z ∈V . This proves the lemma.

58.2 The Picard Theorem AgainHaving done all this work on the modular function which is important for its own sake,there is an easy proof of the Picard theorem. In fact, this is the way Picard did it in 1879. Iwill state it slightly differently since it is no trouble to do so, [65].

Theorem 58.2.1 Let f be meromorphic on C and suppose f misses three distinct points,a,b,c. Then f is a constant function.

Proof: Let φ (z) ≡ z−az−c

b−cb−a . Then φ (c) = ∞,φ (a) = 0, and φ (b) = 1. Now consider

the function, h = φ ◦ f . Then h misses the three points ∞,0, and 1. Since h is meromorphicand does not have ∞ in its values, it must actually be analytic. Thus h is an entire functionwhich misses the two values 0 and 1. If h is not constant, then by Lemma 58.1.29 thereexists a function, g analytic on C which has values in the upper half plane, P+ such thatλ ◦ g = h. However, g must be a constant because there exists ψ an analytic map on theupper half plane which maps the upper half plane to B(0,1) . You can use the Riemannmapping theorem or more simply, ψ (z) = z−i

z+i . Thus ψ ◦g equals a constant by Liouville’stheorem. Hence g is a constant and so h must also be a constant because λ (g(z)) = h(z) .This proves f is a constant also. This proves the theorem.

1850 CHAPTER 58. ELLIPTIC FUNCTIONSProof: For z = zo the equation holds. LetA= {2€1:8m(2)=6 (Az! of())}.Thus zp € J. If z € J and if w is close enough to z, then w € J also and so both sides of58.1.40 with w in place of z are in A;,' (f (I)). But by construction, J is one to one on thisset and since A is invariant with respect to @,A (8m(w)) =A (Az! of (w)) =A (Gods! of(w))and consequently, w € A. This shows A is open. But A is also closed in J because thefunctions are continuous. Therefore, A = J and so 58.1.40 is obtained.Letting f(z) € f(B(Y(T)).7),A(0(ar'(f(2))) =4 (Ar (F)) =F)and so @oA;' isa local inverse for A on f (B(y(T)) ,r). Let the new function element be&m+1@oAz!of,B(y(T),r) | . This has shown the initial function element can be continuedalong every curve through a.By the monodromy theorem, there exists g analytic on V such that g has values in Pyand g = go on B(a,ro). By the construction, it also follows A og = f. This last claim iseasy to see because A og = f on B(a,ro), a set with a limit point so the equation holds forall z € V. This proves the lemma.58.2. The Picard Theorem AgainHaving done all this work on the modular function which is important for its own sake,there is an easy proof of the Picard theorem. In fact, this is the way Picard did it in 1879. Iwill state it slightly differently since it is no trouble to do so, [65].Theorem 58.2.1 Let f be meromorphic on C and suppose f misses three distinct points,a,b,c. Then f is a constant function.Proof: Let ¢(z) = a Then @ (c) = , @ (a) = 0, and @ (b) = 1. Now considerthe function, h = @ 0 f. Then h misses the three points °,0, and |. Since / is meromorphicand does not have © in its values, it must actually be analytic. Thus / is an entire functionwhich misses the two values 0 and 1. If / is not constant, then by Lemma 58.1.29 thereexists a function, g analytic on C which has values in the upper half plane, P, such thatA og =h. However, g must be a constant because there exists y an analytic map on theupper half plane which maps the upper half plane to B(0,1). You can use the Riemannmapping theorem or more simply, y(z) = a. Thus yo g equals a constant by Liouville’stheorem. Hence g is a constant and so h must also be a constant because A (g(z)) = A(z).This proves f is a constant also. This proves the theorem.