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As is very well known, the algebraic variety $S^2$ is isomorphic to projective variety $\mathbb{CP}^1$ as a complex manifold. As is also well known, the coordinate ring of $S^2$ is given by $< x,y,z > / < x^2 + y^2 +z^2 - 1 >$ and the function field of $\mathbb{CP}^1$ is $\mathbb{C}(\mathbb{CP}^1)$ (its coordinate ring being $\mathbb{C}$). My question is how the coordinate ring of $S^2$ and the function field of $\mathbb{CP}^1$ are related? Presumably this relation is a special case of a general variety-complex manifold relationship.

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Any projective variety is also a real affine variety, by using the real and imaginary parts of the coordinates $x_{jk} = z_j\overline{z_k}$. You should first normalize the projective coordinates to have Hermitian-Euclidean length 1. Ordinarily the projective coordinates are sections of a line bundle that is only defined up to a scalar. But with the length normalization, the only ambiguity is the phase. The phase cancels out in the definition of $x_{jk}$, so it is a well-defined function rather than just a section.

This realification of $\mathbb{CP}^n$ is a map to $(n+1) \times (n+1)$ Hermitian matrices. It is important in quantum probability: The image of the map is the set of pure states of a quantum system; its convex hull is the set of all states. This convex region is the quantum analogue of the simplex of states (= measures = distributions) for classical probability on a finite set. The matrices have unit trace, so the image lies in an $(n^2+2n)$-dimensional real subspace. You can check that it is a sphere when $n=1$.

Another viewpoint that is important is that of toric varieties. The phase part of the toric action on $\mathbb{CP}^1$ is rotation about the $z$ axis, and the moment map is the projection onto the $z$ coordinate. This too generalizes to any projective toric variety.

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Thanks for your answer, it seems to be what I'm looking for, but could you please explain the first part of your comment a little more (I'm very new to AG). For me a complex algebraic variety is the set of points in $\mathbb{C}^n$ that vanish for all elements of an ideal $I$, whereas a projective variety is the set of points in $\mathbb{CP}^n$ that vanish for all elements of an ideal $J$. I don't see why multiplying together coordinates (which as far as I understand are not even globally defined - hence I think your line bundle comment) changes a projective variety into an algebraic one. –  Aston Smythe Nov 21 '09 at 11:00
    
That's exactly the point: The projective coordinate $z_j$ is not globally defined, but the product $x_{jk}$ is globally defined, provided that the vector of projective coordinates $\vec{z}$ is rescaled to have length 1. If a variety is described by global coordinates, then it is an affine variety. –  Greg Kuperberg Nov 21 '09 at 15:47
    
It's beginning to make sense, just two last concrete questions: (1) What exactly do you mean by the normalisation of $z_i$? (2) How, given global coordinates, can a projective variety be expressed as an affine variety? –  Aston Smythe Nov 21 '09 at 19:00
    
... or perhaps just a reference, I seem to be missing more background than is fair for you to explain. –  Aston Smythe Nov 21 '09 at 20:09
    
1. $\vec{z}$ is a vector in $\mathbb{C}^{n+1}$, and complex vectors have lengths (en.wikipedia.org/wiki/Inner_product). Normalization means rescaling to unit length. 2. A list of $d$ global, real functions on a set $X$ can be viewed as a function from $X$ to $\mathbb{R}^d$. That is how I am using real and imaginary parts of $x_{jk}$. In this case the image of $X$ happens to be an affine variety. 3. Google finds some references, but they are too advanced to help here, or they don't fit. –  Greg Kuperberg Nov 21 '09 at 20:50

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