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I'm trying to show the following:

Let $f:X\to Y$ be a proper holomorphic map between connected, non-empty Riemann Surfaces. Show that a map $g:Y\to\mathbb{C}$ is holomorphic if and only if its composition with f is holomorphic.

So far I know that $f$ is surjective, since it's proper and holomorphic. My next step was to look at the charts, but it's there, I get stuck.

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    $\begingroup$ But if $X$ is compact, a constant $f$ would be proper, and then we can't conclude anything about $g$. We need non-compact surfaces or an explicit condition that $f$ is non-constant. $\endgroup$ Commented Aug 4, 2015 at 17:17

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Assume $g\circ f$ is not constant and $f$ is surjective and proper.As holomorphicity is a local condition we can assume $X$ and $Y$ are open subsets of $\Bbb C$. Since the zeroes of a holomorphic function are isolated, the points $x$ where $(g\circ f) '(x)=0$ are isolated in $X$. Call this set of isolated points $S$.

Away from these points $g\circ f$ is a local biholomorphism and in particular $g$ is holomorphic on $Y-f(S)$ with $g'(z)=\frac1{f'(g(z))}$ in some local holomorphic coordinates. I claim $g|_{Y-f(S)}$ is bounded locally around each point of $f(S)$ and can hence be extended over each point of $f(S)$ holomorphically in a unique way.

If $g$ is not bounded around a point in $y \in f(S)$, then it either has a pole or an essential singularity at $y$ by the classification of isolated singularities.

If $g$ has a pole at $y$, locally $g(z)=\frac1{(z-y)^k}g_1(z)$ where $g_1(z)$ is holomorphic and injective on a neighborhood of $y$ and $g_1(y)\ne 0$. Now locally around $x$ with $f(x)=y$ we have $g(f(z))= \frac1{(f(z)-y)^k}g_1(f(z))$ which has a pole of order $qk$ at $x$ where $q$ is the order of vanishing of $f(z)-y$. In particular $q\geq 1$ and $g \circ f$ has a pole at $x$, which is a contradiction.

If $g$ has an essential singularity at $y$, then the image of any punctured neighborhood of $y$ under $g$ is dense in $\Bbb C$ by the Casorati–Weierstrass theorem. As $f$ is locally surjective at point $x$ with $f(x)=y$, the image of any neighborhood of $x$ under $g\circ f$ must be dense in $\Bbb C$, but then $g\circ f$ is not continuous, so we have a contradiction and the points $f(S)$ are isolated removable singularities of $g|_{Y-f(S)}$.

To see that $g$ must coincide with this extension, it is enough to show $g$ is continuous. Let $y_n$ be a cauchy sequence with $y_n$ converging to $y=f(x)$ but $g(y_n)$ not converging to $g(y)$. As $f$ is holomorphic, it is locally surjective (near $y$) and for sufficiently large $y_n$, $y_n=f(x_n)$ for some sequence $x_n$. By properness $x_n$ has a convergent subsequence (converging to $x_0$) and by continuity of $g\circ f$, we have $g(y_n)=g(f(x_n)) \to g(f(x_0))=g(y)$. Hence $g$ is continuous.

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By hyphotesis $f$ is holomorphic, but what does holomorphic mean? In general a continuous function between Riemann Surfaces is holomorphic if $f$ is holomorphic in local charts.

Let $p\in X$, $(U,z)$ a local chart around $p$ and $q=f(p)\in Y$ and $(V,w)$ a local chart around $q$. Then $w\circ f\circ z^{-1}$ is holomorphic.

Let now $g:Y\longrightarrow \mathbb{C}$ a continuous map. Clearly the complex plane is a Riemann Surface, and its atlas is given by the single chart $(\mathbb{C},id)$. So $g$ is holomorphic if and only if for every chart $(V,w)$ around an arbitrary point, the function $g \circ w^{-1}$ is holomorphic. (Note that this a particular case of the definition above).

Clearly if $g$ is holomorphic than $g\circ f$ is holomorphic because is holomorphic in local charts: $g\circ w^{-1}\circ w\circ f\circ z^{-1}=g\circ f\circ z^{-1}$.

For the converse ($g\circ f$ holomorphic $\Longrightarrow$ $g$ holorphic) suppose that $g$ is not holomorphic, then $g\circ f$ cannot be holomorphic, in local chart

$$\overline{\partial}(g\circ f)=\frac{\partial g}{\partial z}\frac{\partial f}{\partial \overline{z}}+\frac{\partial g}{\partial \overline{z}}\frac{\partial \overline{f}}{\partial \overline{z}},$$

so if $g$ is not holomorphic then $\frac{\partial g}{\partial \overline{z}}$ is not zero, moreovere also $\frac{\partial \overline{f}}{\partial \overline{z}}\neq0$ since $f$ is holomorphic. This imply that the composition is not holomorphic.

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  • $\begingroup$ Why can $g\circ f$ not be holomorphic if $g$ is not holomorphic? That's the whole point of the exercise. $\endgroup$ Commented Aug 4, 2015 at 18:09
  • $\begingroup$ I edit my post. $\endgroup$
    – InsideOut
    Commented Aug 4, 2015 at 18:57
  • $\begingroup$ Thanks for the answer first of all! But I have a question, why is it $\frac{\partial g}{\partial\overline{z}}$ in the last step, instead of $\frac{\partial\overline{g}}{\partial\overline{z}}$? $\endgroup$
    – s.fire
    Commented Aug 4, 2015 at 22:19
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Suppose that $f$ is non constant, you already know it is surjective. So for a $y\in Y$. The preimage is finite for the points whose preimage is just one point that imply the local neighbourhoods around the points say $x\in X$ such that $f(x)=y$ is isomorphic hence $g\circ f$ being holomorphic imply g is holomorphic around the point $y$. For branch points if $g$ has a singularity it will give arise to a singularity of $g\circ f$, which will be a contradiction.

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    $\begingroup$ What about points where $f$ looks like $z \mapsto z^k$ for $k > 1$? $\endgroup$ Commented Aug 4, 2015 at 17:31
  • $\begingroup$ Yes I have mentioned for those points will have a finite covering and restriction to each cover is isomorphic hence $g\circ f$ being holomorphic imply g is holomorphic. $\endgroup$
    – GGT
    Commented Aug 4, 2015 at 18:23
  • $\begingroup$ No, $f$ need not be a local isomorphism. $f$ is a branched covering, and if $y\in Y$ is such that every point of $f^{-1}(y)$ is a ramification point, how do you argue then? $\endgroup$ Commented Aug 4, 2015 at 18:27

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