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.


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.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Aston Smythe Nov 21 '09 at 11:00
  • $\begingroup$ 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. $\endgroup$ – Greg Kuperberg Nov 21 '09 at 15:47
  • $\begingroup$ 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? $\endgroup$ – Aston Smythe Nov 21 '09 at 19:00
  • $\begingroup$ ... or perhaps just a reference, I seem to be missing more background than is fair for you to explain. $\endgroup$ – Aston Smythe Nov 21 '09 at 20:09
  • $\begingroup$ 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. $\endgroup$ – Greg Kuperberg Nov 21 '09 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.