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From Harvard qualification exam, 1990.

Let $f,g$ be two entire holomorphic functions satisfy the property $$f(z)^{2}=g(z)^{6}-1,\forall z\in \mathbb{C}$$ Prove that $f,g$ are constant functions. Would this be the same if $f,g$ are allowed to be meromorphic functions?

The problem comes with a hint that I should think about the algebraic curve $$y^{2}=x^{6}-1$$but I do not see how they are related. I know this curve is hyperellipitic (from Riemann-Hurwitz or simply the wiki article). But how this help?(this curve should be of genus 2). Taking a short look at the entire function article also seems to be no help. Is the author implying $f,g$ is not best to be treated by classical Riemann Surface applications (as opposed to algebraic geometry ones)?

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If $f, g$ are non-constant holomorphic functions such that $f(z)^2 = g(z)^6 - 1$, then $(f(z), g(z))$ defines a non-constant holomorphic map from... something to... something. Can you fill in the rest? –  Qiaochu Yuan Aug 5 '12 at 22:55
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Then $(f(z),g(z))$ defines a non-constant holomorphic map from $\mathbb{C}$ to $X=\{z^{2}=w^{6}-1\}$. This is impossible since by Riemann-Hurwitz formula we can only map higher genus curves to lower ones. I do not know if this is an appropriate proof. –  Bombyx mori Aug 5 '12 at 23:17
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Could any one please answer? I prefer a detailed proof to obscure hints. –  Makoto Kato Aug 6 '12 at 0:05
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This hint might be too useful, so I encoded it in ROT13: Jung vf gur havirefny pbire bs gung ulcreryyvcgvp pheir? –  David Speyer Aug 6 '12 at 11:39
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@Makoto: no. If the OP finds the solution, the OP can post an answer. –  Qiaochu Yuan Aug 6 '12 at 15:37
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1 Answer 1

Enlightened by various hints, here is a 'proof' which is most likely to be wrong somewhere. I have not touched entire functions and covering maps for a long time. So suggestions on improvement is welcome.

If $f,g$ are meromorphic, then we can write them as quotients of rational functions. Thus we can extend $f,g$ to the Riemann Sphere by allowing $\infty$. Riemann-Hurwitz would imply we cannot map from low genus surface to a higher genus one: therefore there is no map from $S^{2}$ to $X$. And thus $f,g$ must be constants.

Assume $f,g$ are holomorphic over $\mathbb{C}$ with possible non-removable singularity at $\infty$. Since $f,g$ are both open maps if they are non-constant, together $(f,g)$ should map the open set $\overline{\mathbb{C}}-\{\infty\}$ to a connected open component to $X=\{(z,w),z^{2}=w^{6}-1\}$. $X$ is a Riemann surface that can be compactified to be homeomorphic to a two hole torus. Since $X$ is connected the map must be surjective. Further, by inverse mapping theorem since $f,g$ are assumed to non-constant, $(f',g')$ are not zero except in a discrete set of points. Ignore this for now (should be tractable by using local biholomorphic transformation to a locally ramified function) we may view the map $$F=(f,g):\mathbb{C}\rightarrow X$$ as a covering map since we have a discrete inverse image at every neighborhood of $X$ according to $F$'s degree at that point.

The universal cover of $X$ is a closed disk $D^{2}$ by the fundamental diagram. Therefore by the covering property we have a unique lift $p$ from $\mathbb{C}$ to $D^{2}$ that preserves $F$ such that $p:D^{2}\rightarrow \mathbb{C}$ satisfies $$p\circ F=q$$ where $q$ is the covering map from $D^{2}$ to $X$. Further $p$ is holomorphic. But this is contradictory since by mean value theorem (or maximal modulo principle) a non-constant holomorphic function attains its maximal absolute at the boundary. Thus $p$ must be bounded by some constant and cannot reach the whole complex plane. This showed at least one of $f,g$ must be a constant. And by definition this showed both $f,g$ are constants.

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Choosing a branch doesn't define a meromorphic function on $\mathbb{C}$, it defines a meromorphic function on the complement of the branch cut. –  Qiaochu Yuan Aug 7 '12 at 2:21
    
Hopefully fixed now. –  Bombyx mori Aug 7 '12 at 5:10
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