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I found the following claim in a paper by Griffiths and Harris :

Start with a complex torus $\mathbb{C}/\Lambda$. The vector space of meromorphic functions having period lattice $\Lambda$ and a pole of order at most $n$ at $u=0$ has dimension $n$ (a consequence of Abel's theorem). Given a basis $f_1, f_2, \ldots , f_n$ of this vector space, the mapping

$F(u) = [f_1(u), \ldots, f_n(u)]$

induces a projective embedding $\mathbb{C}/\Lambda \rightarrow \mathbb{P}^{n-1}$ whose image is easily proved to be a smooth algebraic curve of degree $n$. Thus, for $n=3$ we have a plane cubic, for $n=4$ the intersection of two quadrics in $\mathbb{P}^3$, etc.

I am trying to figure out why this statement is true. For the case $n=3$, you can choose the Weierstrass functions as a basis $(1,\wp,\wp')$ and they satisfy a polynomial equation of degree three, but that is nowhere near as general as the statement they make and I wouldn't say that it is "easily" proved.

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Look at… especially at page 157. – Ehsan M. Kermani Jan 18 '12 at 6:34
up vote 3 down vote accepted

Denote the given map by $\phi: \mathbb C/\Lambda \to \mathbb P^{n-1}$.

Let $H$ be a hyperplane in $\mathbb P^{n-1}$, cut out by the equation $a_1 X_1 + \cdots + a_n X_n = 0$. The preimage $\phi^{-1}(H)$ is then equal to the subset of points of $\mathbb C/\Lambda$ satisfying the equation $a_1 f_1 + \cdots a_n f_n = 0$.

First of all, since $n \geq 3$, Riemann--Roch says that given any two points $u$ and $v$ of $\mathbb C/\Lambda$, we may choose the $a_i$ so that $u$ lies in $\phi^{-1}(H)$ while $v$ does not. This implies that $\phi$ is injective. (Note that here and below I am being a little careless about analyzing what happens at the point $u = 0$ here, where we have to modify the formula for $\phi$; I'll leave these details to you.)

In fact, Riemmann--Roch also implies that we can choose $H$ so that $a_1 f_1 + \cdots a_n f_n$ vanishes precisely to first order at any given point $u$. This show that the image of $\phi$ is smooth at $\phi(u)$; since $u$ was arbitrary, we see that the image of $\phi$ is smooth, and hence that $\phi$ is an embedding onto its image. (I have just sketched the verification that $\phi$ "separates points and tangent vectors", which is the standard condition to check that a map defined in the manner of $\phi$ is a projective embedding; this is discussed e.g. in Hartshorne Chapter 2, Section 7, and also --- in the specific case of curves --- in the part of Chapter 4 that discusses projective embeddings of curves.)

Since $\mathbb C/\Lambda$ is compact, its image is a closed analytic submanifold of $\mathbb P^{n-1}$, which is then necessarily algebraic (by Chow's theorem if you like, although you could also prove this directly, by showing ----- by Riemann--Roch or by direct elementary arguments --- that the $f_i$ necessarily satisfy lots of algebraic relations among themselves, indeed so many that the image of $\phi$ is cut out by algebraic equations. A more systematic way to describe the situation is to note that $\mathbb C/\Lambda$ has a unique structure of a projective algebraic curve --- e.g. the one coming from the Weierstrass equation arising from the case $n = 3$ --- and with respect to this structure, the $f_i$ are rational funtions; the map $\phi$ is then a morphism of algebraic varities from $\mathbb C/\Lambda$ to $\mathbb P^{n-1}$; since its source is projective, its image is necessarily a closed algebraic subvariety.)

Now all that remains is to compute the degree of the image of $\phi$. For this, we have to compute the number of intersection points of this image with a generic hyperplane $H$. That is, we need to determine the size of $\phi^{-1}(H)$. But this is the number of zeroes of $a_1 f_1 + \cdots + a_n f_n$. Since this function has a pole of order $n$ (at least if $a_n \neq 0$, which is the generic situation), it has $n$ zeroes, which will be distinct genericaly, and so $\phi^{-1}(H)$ is generically of order $n$.

Thus the image of $\phi$ has degree $n$.

If you're not used to these sort of computations, you may want to look at some of the examples and discussion of Chapter 4 of Hartshorne. What I just did above is the standard way of analyzing projective embeddings of curves, and in that sense it is easy (once you know how it goes).

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Thank you very much! I am interested in the "elementary argument". For example, for $n=3$ I can see that $f_1, f_2, f_3$ must respectively have a pole of order $0$, $2$ and $3$ at $o$ and so there must be an equation of the type $c_1 f_2^3 - c_2 f_3^2 - c_3 f_2 f_3 - c_4 f_2^2 = c_5 f_1 + c_6 f_2 + c_7 f_3$, because you can choose the coefficients on the left hand side to make the pole at $o$ of the left hand side of order $3$ or less and therefore the function is a linear combination of $f_1,f_2,f_3$. Doing similar things for higher $n$, when do you know when you have enough equations? – Vhailor Jan 24 '12 at 18:47

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