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In Schlag's book on Riemann surfaces, we have Problem $4.15$, which is:

Let $M, N$ be compact Riemann surfaces and suppose $f: M\setminus\mathcal{S} \to N\setminus\mathcal{S}'$ is an isomorphism, where $\mathcal{S}$ and $\mathcal{S}'$ are finite sets. Show that $f$ extends to an isomorphism from $M\to N$.

This is also Exercise 8.2 in Forster's text. I think we can prove this as follows (but I am not totally sure about this either).

Let $a\in \mathcal{S}$ and pick a sequence $a_n \subset M \setminus > \mathcal{S}$ such that $a_n \to a$. Then, by compactness of $N$, there is a subsequence $f(a_{n_k})$ converging to some $p \in N$.

Suppose $p \in f(M \setminus \mathcal{S})$, and let $K:= \{f(a_{n_k}), > p\}$. Since $f$ is continuous, we have $f^{-1} (K)$ is closed, hence $f^{-1} (K)$ is compact (by compactness of $M$). And, since $f$ is injective, we have $f^{-1} (K) = \{a_{n_k}, f^{-1} (p)\}$.

Let $U_k$ be a neighborhood of $a_{n_k}$ containing none of the other elements of sequence $a_n$. Then for any neighborhood $U$ of $f^{-1}(p)$ , the cover $\{U_k, U\}$ has a finite subcover, so all but finitely many elements of $a_{n_k}$ are in $U$. Thus, $a_{n_k} \to f^{-1} (p)$ , which implies $f^{-1} (p) = a$, contradiction.

So, $p \in \mathcal{S}'$. Now, by the lemma in this answer and discreteness of $\mathcal{S}'$, it follows that $f(a_n) \to p$. Further, if $b_n \to a$ is a different sequence, then discreteness again implies $f(b_n) \to p$. Thus, we can continuously extend $f$ to $M$, hence we can holomorphically extend.

Now, we have an injective holomorphic map $M \to N$ of compact Riemann surfaces. By the open mapping theorem, it has to be an isomorphism.

However: in an old problem set of Curt McMullen, a similar problem is given:

Let $X$ be a compact Riemann surface and $A \subset X$ a finite set. Show that any injective holomorphic map $f: X\setminus A \to Y$, where $Y$ is another compact Riemann surface, may be extended to an isomorphism $X\to Y$.

The difference here is that we have to show ourself that $Y \setminus f(X\setminus A)$ must be discrete. How can we do this?

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  • $\begingroup$ Did not you ask this question few days ago and got a very good hint from Daniel Fischer? $\endgroup$ – Moishe Kohan Dec 6 '17 at 16:47
  • $\begingroup$ Yes, but I have had trouble seeing how to implement that hint and am trying a different approach. $\endgroup$ – Sameer Kailasa Dec 6 '17 at 16:48
  • $\begingroup$ I see. Then you should start by looking closely at your solution of 4.15 and finding a mistake (hint: your solution is correct if $S'$ is a singleton). As for McMullen's problem, it all depends on how much of the Riemann surface theory you already know. For instance, do you already know that every compact Riemann surface embeds biholomorphically in some $CP^n$? $\endgroup$ – Moishe Kohan Dec 6 '17 at 17:14
  • $\begingroup$ Or, maybe you already know the uniformization theorem: every simply-connected noncompact Riemann surface is boholomorphic to the unit disk or to the complex plane. McMullen might be assuming that you know either one of these two facts (neither one is easy, by any means). $\endgroup$ – Moishe Kohan Dec 6 '17 at 17:33
  • $\begingroup$ I am guessing the mistake in my solution of 4.15 is that we cannot immediately conclude every convergent subsequence of $f(a_n)$ converges to $p$. Can this be salvaged? As for McMullen's version, I think he probably does assume you know at least uniformization. $\endgroup$ – Sameer Kailasa Dec 6 '17 at 17:40
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Various forms of this question were asked several times at MSE. Any solution, as far as I know, will use some nontrivial results in the theory of Riemann surfaces.

Theorem. Let $M, N$ be compact connected Riemann surfaces, $X\subset M$ a finite subset. Then every biholomorphic embedding $f: M-X\to N$ will extend to a biholomorphic map $M\to N$.

(Remark: In fact, one can drop the assumption that $N$ is compact, but this would make a proof harder.)

Proof. I assume that you are already familiar with the Riemann-Roch theorem (Forster's book covers it); actually, I will need only a weak for of the RRT. From that, you learn that given a point $z_0$ on $N$ there exists a rational function $g: N\to {\mathbb C}\cup \{\infty\}$ which has a (high order) pole at $z_0$ and is holomorphic away from $z_0$. That's all what we will need. Now, pick a point $z_0\in f(M)$ and consider the holomorphic function $$ h=g\circ f: M - f^{-1}(z_0) \to {\mathbb C}. $$ It is clear that for each $x\in X$, for any sequence $x_k\to x$, $x_k\in M-X$, the image sequence $(f(x_k))$ does not accumulate at $z_0$ (since $f: M-X\to N$ is injective), hence, $g\circ f(x_k)$ is a bounded sequence in ${\mathbb C}$. Therefore, taking $D$ to be a small open disk in $M$ containing any $x\in X$ and not containing $f^{-1}(z_0)$, we conclude that the holomorphic function $h|D -\{x\}$ is bounded. Therefore, $h|D -\{x\}$ extends to a holomorphic function to the entire disk $D$. It follows that $f|D-\{x\}$ also extends to a (necessarily injective) holomorphic function $D\to N$. Injectivity of the extension is clear. Thus, $f: M-X\to N$ extends to a holomorphic function $F: M\to N$. Injectivity of $F$ follows from that of $f$. Surjectivity of $F$ follows from the fact that the image of $F$ is open (since $F$ is nonconstant and holomorphic) and closed (since $M$ is compact). qed

Addendum. Here is an alternative solution, which does not assume compactness of $N$, but requires more tools. As in the above proof, take a small open disk $D\subset M$ containing $x\in X$ and no other points of $X$. I will show that the restriction of $f$ to the punctured disk $D^*= D-\{x\}$ extends holomorphically to the disk $D$. Note that $\pi_1(D^*)$ is infinite cyclic, let $p: L\to N- Y$ denote the covering map corresponding to the subgroup $$ \Gamma:= f_*(\pi_1(D^*))< \pi_1(N-Y). $$ (At this stage, I expect you to know the basic algebraic topology: Fundamental groups and covering spaces.) I equip $L$ with the complex structure obtained by pull-back of the complex structure on $N-Y$ via $p$. Then $f|D^*$ lifts to a holomorphic map $\tilde{f}: D^*\to L$. Furthermore, the fundamental group of $L$ is cyclic, since $\Gamma$ is the image of the cyclic group $\pi_1(D^*)$. Hence, either $L$ is simply connected or $\pi_1(L)\cong {\mathbb Z}$.

I now apply the Uniformization Theorem (which is harder than RRT theorem as far as I am concerned) to the surface $L$. The conclusion is that $L$ is biholomorphic to a domain $A$ in the complex plane. This domain can be assumed to be bounded and equals either the unit disk $\Delta$ or the punctured unit disk $\Delta^*=\{z: 0<|z|<1\}$ or an annulus $$ \{z: r^{-1}< |z|< r\} $$
for some $r>1$. I will therefore identify $L$ and $A$. The holomorphic map $\tilde{f}: D^*\to A\subset {\mathbb C}$ is again bounded and, hence, $\tilde{f}$ extends holomorphically to the disk $D$. Let $h: D\to \tilde{D}\subset L$ be the biholomorphic extension, where $\tilde{D}$ is an open disk in $L$. One then checks that the restriction $p|\tilde{D}$ is 1-1 and, hence, by composing with $p$ we obtain a biholomorphic embedding $F: D\to Y$ extending $f$. The rest of the argument is as above.

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