1
$\begingroup$

I had to show what a one-to-one analytic function from the plane to itself could possibly be.

So, I studied the behavior of such a function at infinity:

Case 1: Such a function cannot have no singularity (or a removable singularity) at infinity, since then the function would be entire and bounded, and by Liouville's Theorem, is constant, which contradicts the one-to-one assumption of f(z).

Case 2: Such a function cannot have an essential singularity at infinity, since then the function would attain every complex value (except for maybe one value) infinitely often in a (every) neighborhood of infinity, which again contradicts the one-to-one assumption of f(z). This comes from Picard's theorem.

Cases 1 & 2 imply that the function must have a pole at infinity.

We conclude that f(z) must be a polynomial.

Is this ok?

My concern is that I feel that I've made a jump from saying that a 1-1 entire function with a pole at infinity is a polynomial -- as if it were a definition. Can I treat this property of a polynomial as a definition? (Which would then make my proof complete.)

Thanks,

$\endgroup$
2
  • 1
    $\begingroup$ Related: math.stackexchange.com/q/29758 (there are also 21 questions linked to or from that one). $\endgroup$ Commented Jul 13, 2015 at 23:16
  • $\begingroup$ Ok, thanks for the link, @JonasMeyer! $\endgroup$
    – User001
    Commented Jul 13, 2015 at 23:20

1 Answer 1

1
$\begingroup$

A meromorphic function over the Riemann sphere is necessarily a rational function (subtract its Laurent pieces and apply Liouville to the resulting function), so indeed, if your entire function is meromorphic at infinity, it should have a pole at infinity, and is hence a polynomial.

You may want to see it as a meromorphic function that is holomorphic at infinity: then it has a finite number of poles, which lie on the complex plane. This you can do by simple inversion ($z\mapsto 1/(z-\alpha)$).

Then you might want to see it as a differentiable automorphism of the Riemann sphere, and its degree is exactly the degree of the polynomial. C'est fini!

I forgot to say that, given a rational function $f(z)=P(z)/Q(z)$ with irreducible fraction, one can see that the degree of $f$ as an analytic map of the Riemann sphere onto itself is precisely $\mbox{max}(\mbox{deg }P,\mbox{deg }Q).$ In fact, Lüroth's theorem (which is an exercise in Elementary Abstract Algebra) shows precisely that in terms of field extensions: if $T$ is a transcendental variable over a field $k,$ then every subextension $$k(T)|L|k$$ is either trivial or $L=k(\frac{P(T)}{Q(T)})$ with $P$ or $Q$ nonconstant, where the degree $$[k(T):L]=\mbox{max}(\mbox{deg }P,\mbox{deg }Q).$$

$\endgroup$
2
  • $\begingroup$ Thus, Lüroth's theorem shows that an automorphism of the Riemann sphere is necessarily given by a Möbius transformation. This you can obtain also by the complex-analytic arguments provided. $\endgroup$ Commented Jul 13, 2015 at 23:33
  • $\begingroup$ Ok, got it. Thanks so much @TheonAlexander!! :-) $\endgroup$
    – User001
    Commented Jul 13, 2015 at 23:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .