# An entire function with finite covering group is a polynomial.

Let $f$ be an entire function. Think of it as a covering space of $\mathbb{C}$ (perhaps with isolated punctures) to $\mathbb{C}$ (perhaps with isolated punctures). Suppose we know there is only a finite number of covering transformations, $\{\varphi\}_{i=1}^n$: $$f(\varphi_i(z)) = f(z).$$ How to show $f$ is a polynomial ?

Partial answer: The function $\frac{f(z)-f(w)}{z-w}$ is entire in both variables. It is only zero when $z=\varphi_i(w)$ for some $i$, so we may write $$\frac{f(z)-f(w)}{z-w} = e^{g(z,w)} \prod_{i=1}^n (z-\varphi_i(w)).$$ Now setting $w=0$, and denoting $w_i = \varphi_i(0)$, we obtain: $$\frac{f(z)-f(0)}{z} = e^{g(z,0)} \prod_{i=1}^n (z-w_i)=$$ $$e^{g(z,0)} (z^n - (\sum w_i) z^{n-1} + \ldots + (-1)^n \prod w_i).$$ To show this is a polynomial amounts to showing $g(z,0)$ is a constant.

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I haven't thought this through, but this is what I would try: if $f$ is not a polynomial, then it has an essential singularity at $\infty$, and therefore the fibers are $f^{-1}(\{w\})$ are infinite. If we knew that the group of covering transformation is transitive on fibers, we would be done. Sadly, I never learned if an entire function must be a regular cover... – user53153 Feb 12 '13 at 0:46
Entire functions are in general not regular covers. – Lukas Geyer Feb 28 '13 at 15:08
Are you assuming anything special about your entire function? Just curious, because there are examples (constructed by Gross, of infinite order) where the set of asymptotic values is the whole plane, i.e., where $f$ is not a covering over any point in the plane. – Lukas Geyer Feb 28 '13 at 15:13