Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Ahlfors' complex analysis text, page 301 he defines an algebraic function the following way:

Definition 3. A global analytic function $\mathbf{f}$ is called an algebraic function if all its function elements $(f,\Omega)$ satisfy a relation $P(f(z),z) = 0$ in $\Omega$, where $P(w,z)$ is a polynomial which does not vanish identically.

In order to make this well defined, we obviously must have that all such function elements "generate" the same global analytic function. The proof of this fact is postponed to the end of the chapter:

It has not yet been proved that all elements $(f_i,\Omega)$ belong to the same global analytic function. For this part of the proof it is necessary to study the behavior at the critical points $c_k$ in greater detail.

Later, on page 306, comes the proof:

It is now easy to settle the point which was left open in Sec. 2.2. Suppose that the function element $(f,\Omega)$ satisfies the equation $P(f(z),z) = 0$ where $P$ is irreducible and of degree $n$ in $w$. Then the corresponding global analytic function $\mathbf{f}$ has only algebraic singularities and a finite number of branches. According to what we have just shown $\mathbf{f}$ will satisfy a polynomial equation whose degree is equal to the number of branches. It will hence satisfy an irreducible equation whose degree is not higher. But the only irreducible equation it can satisfy is $P(w,z) = 0$, and its degree is $n$. Therefore the number of branches is exactly $n$, and we have shown that all solutions of $P(w,z) = 0$ are branches of the same analytic function.

I don't see how this last part proves the previous statement. Could anyone please explain this proof further? I also have some specific questions regarding the proof:

  1. Why does $\mathbf{f}$ has only algebraic singularities? Couldn't the function element $(f,\Omega)$ induce singular points which are situated in a more involved fashion? for example where $\Omega$ is bounded and $\partial \Omega$ is the natural boundary.

  2. Overall, he shows that there are $n$ branches. However, I was expecting him to show that all branches lead to the same global analytic function by means of analytic continuations between all of them. How could it be that determining the number of branches is sufficient by itself?

If you have the book at hand feel to include references in your answers. Thanks!

share|improve this question
    
I think you should make clear in the body of the question that you're trying to understand why $P$ defines a unique algebraic function. Also, Ahlfors gives a proof of the claim that $\mathbf{f}$ has only algebraic singularities on pp. 304-305. Have you attempted reading it? (I haven't, so I can't explain it yet.) –  epimorphic Jan 21 at 2:01
    
@epimorphic I have read the proof on pp. 304-305. However, I fail to see how it all comes together. He appears to talk about the individual germs and less about how these glue together. I will try to address your answer shortly. –  user1337 Jan 22 at 18:58

1 Answer 1

up vote 1 down vote accepted

I'll attempt to explain the proof and answer your second question. Let $\mathbf{f}_i$ be the global analytic function determined by $(f_i,\Omega)$. Then

  • $\mathbf{f}_i$ has only algebraic singularities (pp. 304-305), the number of which are finite (since they correspond to $\infty$ or the roots of one of two polynomials: $a_0(z)$ or the discriminant);
  • The number of branches of $\mathbf{f}_i$ at each non-singular point is at most $n$ (the set of branches has to be a subset of the function elements from Lemma 1);
  • $\mathbf{f}_i$ has a well-defined continuation along any arc which does not pass through the singularities (second-to-last paragraph of p. 303);
  • Every function element of $\mathbf{f}_i$ satisfies $P(w,z)=0$ (to prove, continue the analytic function $P\big(f_i(z),z\big)$ along the same arc).

Because of the first three bullet points, the "converse" that begins in the middle of p. 305 is applicable to $\mathbf{f}_i$; the result is that the number of branches is constant (call it $n_i\leq n$), and that $\mathbf{f}_i$ satisfies $Q_i(w,z)=0$ where $Q_i$ is a polynomial of degree $n_i$ in $w$. It follows that $\mathbf{f}_i$ also satisfies $Q_{i,k_i}(w,z)=0$ for some irreducible factor $Q_{i,k_i}$ of $Q_i$ (beginning of p. 302). Since $P(w,z)=0$ is another irreducible equation satisfied by $\mathbf{f}_i$, it must be that $Q_{i,k_i} = aP$ for some $a\in\mathbb{C}$ (the following paragraph on p. 302). Therefore, $$ n = \deg_w P = \deg_w Q_{i,k_i} \leq \deg_w Q_i = n_i \leq n \implies n_i = n. $$ All that remains to be resolved now is your second question: Why does the fact that $\mathbf{f}_i$ has $n$ branches imply that $P$ defines a unique algebraic function? Well, recall that global analytic functions are equivalence classes of function elements under analytic continuation. Given any non-singular point $z_0$, Lemma 1 establishes that the algebraic functions that satisfy $P(w,z)=0$ share a pool of $n$ possible branches at $z_0$ to choose from. What $n_i=n$ implies is that all $n$ of those possible branches belong to the same equivalence class, namely $\mathbf{f}_i$. Moreover, given another non-singular point $z_1$, each branch at $z_0$ can be analytically continued to at least one of the $n$ potential branches at $z_1$. Therefore by transitivity, all function elements satisfying $P(w,z)=0$ belong to the same algebraic function.

share|improve this answer
    
@yiorgosssmyrlis I think $\mathbf{f}_i$ is preferable over $\boldsymbol{f}_i$ as both Ahlfors and the asker use the bold upright letter to denote global analytic functions. I'll roll back for now (partly to get the badge!). Nonetheless, I welcome any input that you wish to give. Do you feel that $\boldsymbol{f}_i$ is more readable or pleasing to the eye? Is this a modern convention that has developed after Ahlfors wrote the book? –  epimorphic Jan 21 at 23:40
    
Could you please tell me what does "$\mathbf{f_i}$ has $n$ branches" means? According to Ahlfors pp. 285 any function element $(f, \Omega) \in \mathbf{f_i}$ is a branch. Thus there are infinitely many. –  user1337 Jan 22 at 19:39
    
Sorry, was busy with other things. I wasn't aware that Ahlfors actually defined "branch"! That does make Ahlfors guilty of abuse of terminology when he talks about "branches at a point". As I understand it, you can think of the "branches of $\mathbf{f}$ at a point $z_0$" as the equivalence classes determined by a certain relation on the set $\{(f,\Omega)\in\mathbf{f} : z_0 \in \Omega\}$. The relation is defined as follows: $(f_1,\Omega_1) \sim (f_2,\Omega_2)$ iff there exists a neighborhood $U \subset \Omega_1 \cap \Omega_2$ of $z_0$ such that $\left.f_1\right|_{U} = \left.f_2\right|_{U}$. –  epimorphic Jan 26 at 1:14
    
Note that any two elements of the same equivalence class are direct analytic continuations of each other. So it follows by Lemma 1 that algebraic functions have $n$ such equivalence classes at each non-singular point. –  epimorphic Jan 26 at 1:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.