7

I am currently studying Riemann surfaces and I find a theorem stating "Every compact Riemann surface is an algebraic curve." but I don't find a proof. Thanks in advance for helping by providing a proof...

  • 2
    Google finds a reference in wikipedia ... https://en.wikipedia.org/wiki/Compact_Riemann_surface – GEdgar Jun 22 '17 at 16:24
  • 2
    There is no one line proof, of course depending on what you know and assume. A proof can be found in Forster's book (and many others) on Riemann surfaces. Most of them will depend on proving finite dimensionality of cohomologies and thus a version of Riemann-Roch. – Mohan Jun 22 '17 at 16:25
  • 5
    @GEdgar: the linked wikipedia article is false and completely misreperesents GAGA, which assumes that you start with an algebraic variety over $\mathbb C$ and deduces all sorts of results about its analytification. But GAGA nowhere gives a criterion for a compact complex analytic manifold to be algebraic. – Georges Elencwajg Jun 22 '17 at 16:48
  • 1
    The short answer: Use Riemann-Roch to show that a divisor of large degree is very ample, then use the corresponding sections to construct an embedding. The construction isn't hard, but (like everything else in algebraic geometry) it requires a fair bit of algebraic geometry machinery. – anomaly Jun 22 '17 at 17:30
  • 5
    @anomaly: Riemann-Roch for a compact Riemann surface is not a theorem in algebraic geometry but a difficult theorem in analysis. It requires the introduction of $L^2$-norms on spaces of cochains of the structural holomorphic sheaf. The proof moreover makes use of a theorem of Laurent Schwartz on the perturbation of a surjective continuous linear map between Fréchet spaces by a compact operator. To say as you do that "the construction isn't hard" displays a considerable dose of optimism, to say the least. – Georges Elencwajg Jun 22 '17 at 21:46
  • 1
    @GeorgesElencwajg: Ah, right: I remembered the construction via Fredholm operators and Sobolev spaces (which are in fact not algebraic geometry at all), but not many of the details. Oh well. – anomaly Jun 22 '17 at 22:03
  • Moreover, all the above comments, and the answer below, are concerned with the problem of embedding a compact Riemann surface into a projective space. But then it remains to show that the image is an algebraic curve. The reference cited here do not prove that either, e.g. Foster). I don't consider this question as answered. – Visitor Nov 30 '21 at 16:38

1 Answers1

7

There is a proof in Griffiths and Harris (page 215) that uses nothing other than the Kodaira vanishing theorem. But the proof of the Kodaira vanishing theorem (page 154) uses a version of the Hodge decomposition theorem for line bundles (page 152), and the proof of Hodge decomposition uses the usual tools for elliptic PDEs (Sobolev inequalities, Rellich compactness, spectral theorem, regularity lemma, pages 85-96).

The basic idea is to show that there exists a line bundle $\mathcal L$ on the Riemann surface $C$ whose global sections separate points and separate tangent vectors. The global sections of such a line bundle $\mathcal L$ will then define an embedding $C \hookrightarrow \mathbb P^{H^0(\mathcal L) - 1}$.

The statement that the global sections of $\mathcal L$ separates points $p, q \in C$ (with $p \neq q$) is equivalent to the statement that the restriction map $$r_{p,q} : \mathcal L \to \mathcal L_p \oplus \mathcal L_q$$ induces a surjection on global sections (where $\mathcal L_p$ and $\mathcal L_q$ are the stalks of $\mathcal L$ at $p$ and $q$, which can be though of as skyscraper sheaves supported at these points).

In view of the short exact sequence, $$ 0 \to \mathcal L(-p-q) \to \mathcal L \overset{r_{p,q}}{\to} \mathcal L_q \oplus \mathcal L_q \to 0, $$ the induced map $r_{p,q} : H^0(\mathcal L) \to H^0(\mathcal L_q \oplus \mathcal L_q)$ is surjective if $$ H^1(\mathcal L(-p-q)) = 0,$$ and this is true if ${\rm deg \ } \mathcal L - 2 > {\rm deg \ } \mathcal K_C$, by Kodaira vanishing. So $\mathcal L$ separates points if $${\rm deg \ } L> {\rm deg \ } \mathcal K_C + 2.$$ Under the same inequality, it can also be proved that $\mathcal L$ separates tangent vectors. (The proof is almost a special case of the previous one, but with $p = q$.)

The conclusion is that the global sections any line bundle with sufficiently high degree define an embedding of $C$ into projective space, hence $C$ must be an algebraic curve.

Kenny Wong
  • 33,403
  • 1
    Thanks for a very nice and lucid argument. –  Jun 23 '17 at 09:19
  • But there this something missing in the proof. In the last sentence: there is an "embedding of into projective space, hence must be an algebraic curve." Not any subset of a projective space is an algebraic curve, obviously, so how to complete the argument? – Visitor Nov 30 '21 at 16:33
  • Then you need to quote Chow's theorem, see p184 in "Coherent analytic sheaves" by Grauert and Remmert. – Li Yutong May 08 '22 at 02:39