Via a very ample divisor one can embed a Riemann surface holomorphically into some $\mathbb{P}^n$. Now, we can then project the Riemann surface to $\mathbb{P}^3$, and we can even go until $\mathbb{P}^2$ if we allow nodes.

If we have a Riemann surface $X$ and an embedding $f:X\to \mathbb{P}^2$ (where the image is smooth), then we can find the equation for $f(X)$ (you pull back the meromorphic functions $x_0/x_1$ and $x_1/x_2$ via $f$, where $x_0,x_1,x_2$ are the homogeneous coordinates of $\mathbb{P}^2$; these are algebraically dependent over $\mathbb{C}$, and thus they must satisfy some polynomial. The projective closure of the zeroes of this polynomial is exactly $f(X)$).

I've been reading Shafarevich's book Algebraic Geometry I, and on page 86 he describes how to find the equations for a mapping to $\mathbb{P}^n$. He simply states that if you have a map from $X$ to $\mathbb{P}^n$, then since we can project, we can reduce to the case that the map goes to $\mathbb{P}^2$, and we're basically done.

If the Riemann surface is in $\mathbb{P}^3$, for example, I don't see why the above method guarantees that it be defined by equations... It's image in $\mathbb{P}^2$ would possibly have nodes... Any ideas?

I'm looking to show that a compact Riemann surface in $\mathbb{P}^3$ is algebraic, that is, can be defined by polynomial equations.

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    $\begingroup$ Dear Robert, beware that the canonical map is never an embedding for hyperelliptic Riemann surfaces of genus $g\geq 2$ $\endgroup$ Nov 14, 2011 at 22:43
  • $\begingroup$ Dear Georges, thank you; when I wrote the post I was thinking very ample but instead wrote canonical. $\endgroup$
    – rfauffar
    Nov 14, 2011 at 23:19

2 Answers 2


A compact Riemann surface $X$ can always be embedded holomorphically into $\mathbb P^3( \mathbb C)$ . This is a highly non-trivial theorem.

Claim: It cannot in general be embedded into $\mathbb P^2( \mathbb C)$.
The simplest argument to support this claim is to recall that a smooth complex curve of degree $d$ in $\mathbb P^2( \mathbb C)$ has genus $g=\frac{(d-1)(d-2)}{2}$.
Since there exist Riemann surfaces of any genus $g\geq 0$ on one hand, and since on the other hand most integers are not of the form $\frac{(d-1)(d-2)}{2}$ , the claim is proved.

However any Riemann surface can be immersed, but not necessarily injectively, into $\mathbb P^2( \mathbb C)$ by skilfully (or skillfully if you prefer American English) composing an embedding into $\mathbb P^3( \mathbb C)$ with a projection onto a plane, so that the immersed curve will have nodal singularities at worst.

An excellent reference for these questions is Miranda's Algebraic Curves and Riemann Surfaces.
If you want to see the complete proofs of the embedding theorem of compact Riemann surfaces into $\mathbb P^3( \mathbb C)$ (which implies algebraicity of said Riemann surfaces), look at Forster's Lectures on Riemann Surfaces (Springer) or at Narasimhan's Compact Riemann Surfaces (Birkhäuser).
(Caution: the analysis used in these two books is not for the faint-hearted!)

  • $\begingroup$ Or, once we know that $X$ can be embedded into $\mathbb P^3(\mathbb C)$, there is non-constant meromorphic function $f$ on $X$. Use $f$ to define a finite holomorphic map to $\mathbb P^1(\mathbb C)$ and apply GAGA on $\mathbb P^1(\mathbb C)$, correct ? $\endgroup$
    – user18119
    Nov 14, 2011 at 22:42
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    $\begingroup$ Dear @QiL, yes, this is correct. However, as you write, this supposes that you have already embedded the Riemann surface into projective space.If you start from scratch it requires some really tough analysis to show that there exist non-constant meromorphic function on a Riemann surface.It seems to me that this issue is not clearly addressed in Miranda's book, which alternates between Riemann surfaces and algebraic curves without stating that every Riemann surface is algebraizable. Forster gives a masterful tratment of existence of meromorphic functions in his Lectures on Riemann Surfaces. $\endgroup$ Nov 14, 2011 at 23:18
  • $\begingroup$ Thanks ! I had hard time to read Siegl's book for this result. $\endgroup$
    – user18119
    Nov 14, 2011 at 23:21
  • $\begingroup$ Dear Georges, thank for your answer. However, I don't think I made my question very clear (my fault!). What I'm really looking for is how to show that compact Riemann surfaces are in fact algebraic (ie, show that there are equations that define the surface) $\endgroup$
    – rfauffar
    Nov 14, 2011 at 23:26
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    $\begingroup$ Dear @Robert,I'm sure you are right: I was concentrating on the difficulty of embedding a Riemann surface into projective space. Chow's theorem then ensures that any holomorphic submanifold of a projective space is indeed Zariski-closed. You can find a reasonbly self-contained proof of Chow's theorem in Fritzsche-Grauert's From Holomorphic Functions to Complex Manifolds, Theorem 5.13, page 217. $\endgroup$ Nov 16, 2011 at 16:55

Compose the map with projections centered at points. To understand that, you can read the book Algebraic Curves and Riemann Surfaces by Rick Miranda, pages 98-102. That is a beautiful book. You will understand this perfectly.

  • $\begingroup$ Dear Srivatsan, I'm sorry, I didn't exactly clarify what the question really was. I am aware of how the projection is. My real question has to do with finding equations for a compact riemann surface embedded in projective space. I basically want to show that compact Riemann surfaces are in fact algebraic. $\endgroup$
    – rfauffar
    Nov 14, 2011 at 23:22

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