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This works over the reals but not over the complex field. Consider the set of all $n\times n$ matrices $A$ such that 1. $A^2=A$ 2.$A^T=A$ 3. $\mathrm{Trace}(A)=1$

The first condition makes $A$ a projection to a subspace of $\mathbb{R}^n$. The second ensures that $A$ is diagonalizable so its eigenvalues are all 0 or 1.The third guarantees that the image of the projection is one-dimensional. Such matrices are in a 1-1 correspondence with one-dimensional subspaces and so constitute $\mathbb{R}P^{n-1}$.

This seems to represent real projective spaces as affine varieties, with plenty of induced nonzero regular functions. How do we reconcile this with the fact that projective spaces have only constant regular functions?

This does not work over the complexes since $A$ would have to equal its conjugate transpose (to be guaranteed diagonalizable) and conjugation is not algebraic.

This is puzzling me to no end...

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Dear Justin, Here is a more basic example of the same phenomenon to consider: it is a general fact that the only invertible regular functions on $\mathbb A^1$, over any field $k$, are the non-zero constant functions. Algebraically, this is the statement that the only invertible elements of $k[x]$ are non-zero constants. Geometrically, it is the statement that a non-constant polynomial in $x$ has at least one zero. But it may be that zero is not defined over $k$, even though the polynomial is. E.g. the functions $x^2+1$ is invertible as a function on $\mathbb R$, although it is not ... – Emerton Jul 10 '12 at 17:41
... invertible as a regular function on $\mathbb A^1$; it just happens that its zeroes (the points $\pm i$) are not individually defined over $\mathbb R$. In general, there is a (big!) difference between a variety over $k$, and the $k$-valued points of that variety. (The latter can even be empty; e.g. consider the variety $x^2 +y^2 +1 = 0$, which is defined over $\mathbb R$ but has no $\mathbb R$-valued points.) Regards, – Emerton Jul 10 '12 at 17:42
up vote 10 down vote accepted

The theorem that projective spaces are not affine varieties is a theorem over the complex numbers. As you note, your construction fails over the complex numbers, so there is no contradiction.

To give an even simpler example over the reals, $\mathbb {RP}^1=S^1$ is the vanishing set of $x^2+y^2-1=0$.

I think you're slightly confused about what the functions of your conditions are. Every matrix satisfying condition $1$ is diagonalizable. Indeed, all matrices satisfying a polynomial equation without repeated roots are diagonalizable, and $A^2-A=0$ certainly has no repeated roots. The second condition ensures that the kernel is the orthogonal complement of the image, which is the only way to ensure that there is a unique projection with a given image. Remove it, and you have an affine bundle on $\mathbb P^n$, which is of course perfectly alright.

If you take $AA^T=0$ to not be the conjugate-transpose in the complex case, then you get an affine subvariety of $\mathbb P^n$ - the complement of the hypersurface of points corresponding to lines where a certain bilinear form on $\mathbb C^{n+1}$ is nonzero, that is, the vanisihing set of a degree two polynomial equation.

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Oh my god, what just happend? I've made a silly edit to your (accepted) answer (replaced "complex" by "real"), and meanwhile you seem to have made another edit, and this produced just a new answer. Sorry for that confusion. – Martin Brandenburg Jul 9 '12 at 20:31
The proofs that I've seen were sloppy and left out crucial details. The point is that only nonzero functions on an open affine extend to other open affines (since coordinates get inverted), and the Nullstellensatz implies that these are constant. Does this construction imply that all real projective varieties are affine? – Justin Smith Jul 9 '12 at 20:33
Shouldn't the second sentence read "this fails over the real numbers", not "this fails over the complex numbers"? – MTS Jul 9 '12 at 21:16
@Martin: I'm glad it's not what I previously suspected, that I failed so hard at pressing the "edit" button that I managed to produce a new answer instead. @Justin: Yes. The only thing to check is that the vanishing set of a homogeneous polynomial is actually the vanishing set of some number of polynomials in the matrix coefficients. But this is true, as you can evaluate the polynomial on all the columns. The original polynomial on $\mathbb P^n$ vanishes iff all $n+1$ polynomials on $\mathbb A^{n+1}$ vanish. @MTS: Debatable. Clarified. – Will Sawin Jul 9 '12 at 21:36

The fundamental problem seems to be the confusion between the notions of "real algebraic variety" and "algebraic variety over the reals". Given the similarity in words, it is easy to make this mistake.

You can find a discussion of the distinction in chapter 2 of Coste's lecture notes on real algebraic sets. Both objects are topological spaces with sheaves of commutative rings, but the former is made by gluing algebraic subsets of $\mathbb{R}^n$, while the latter is made by gluing prime spectra of finite type commutative rings over $\mathbb{R}$.

There is a functor that sends an algebraic variety $X/\mathbb{R}$ to the real algebraic set $X^{ras}$ whose points are the real points of $X$. Your example demonstrates that this functor does not reflect the property of being affine (i.e., if $X^{ras}$ is affine, $X$ is not necessarily affine). Neither sheaves of regular functions nor the topological space are preserved by this functor, so you can't reasonably expect global functions to be preserved.

As it happens, there is a more general statement, also mentioned in Coste's lecture notes:

All quasi-projective real algebraic varieties are affine.

In particular, for any quasi-projective variety $Y/\mathbb{R}$, there exists an affine variety $X/\mathbb{R}$, such that $X^{ras} \cong Y^{ras}$.

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A number of "proofs" that regular functions on projective space must be constants use a hand-waving argument that appears to to be valid for projective space over arbitrary fields or even rings ("polynomials on open affines induce homogeneous polynomials on $\mathbb{A}^{n+1}$, QED"). The main point is that the canonical map $\pi:\mathbb{A}^{n+1}\setminus\{0\}\to k\mathbb{P}^n$ is an open mapping so that regular functions on $k\mathbb{P}^n$ must factor through it. This proof uses the projective Nullstellensatz in an essential way --- so it requires the field to be algebraically closed.

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Dear Justin, I think the main point is that there is a difference between what you are calling $k\mathbb P^1$, which is the set of $k$-valued points of the variety $\mathbb P^1$ over $k$, and the actual variety $\mathbb P^1$. This latter variety has many more points than just its $k$-valued points. A non-constant rational function on $\mathbb P^1$, defined over $k$ or not, will necessarily have poles, but it may happen that none of these poles are defined over $k$ (even if the function is), so that it induces a regular function when restricted to $k\mathbb P^1$; there is no mystery here. – Emerton Jul 10 '12 at 17:38

More thoughts: On $\mathbb{R}P^1$ functions that are locally polynomial must be constant (the standard proof still applies since it doesn't use the field being algebraically closed). Here is a map that identifies $\mathbb{R}P^1$ with $S^1$: $f:[x:y]\mapsto \left(\frac{x^2-y^2}{x^2+y^2},\frac{2xy}{x^2+y^2}\right)$. Regular functions on $S^1$ pull back to rational functions (homogeneous of degree 0) on $\mathbb{R}P^1$ whose denominators never vanish. If we regard such functions as regular, we get plenty of nonconstant regular functions on $\mathbb{R}P^1$. So the key is how we define regular functions: if they must be local polynomials, they are constants. If they are nonsingular rational functions, it is easy to find regular functions that are not constant. When the field is algebraically closed, the Nullstellensatz implies that nowhere-singular rational functions are polynomials. (I may be opening myself up to flames for asking stupid questions, but I'm writing a book on algebraic geometry and want these little issues beaten to death). Thanks for your answers and comments!

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Thanks! I edited my post. – Justin Smith Jul 11 '12 at 12:48

The example ${\mathbb R}{\mathbb P}^1 = V(x^2+y^2 -1)$ seems compelling as are the other arguments, that real projective space can have nonconstant regular functions.

But what about Hartshorne, Algebraic Geometry, (II, Theorem 5.19) which says, that for $k$ a field, $A$ a finitely generated $k$-Algebra, $X$ a projective scheme over $A$ and $\mathcal F$ a coherent ${\mathcal O}_X$-module on $X$, the $A$-module $\Gamma(X,{\mathcal F})$ is finitely generated.

There seems to be no restriction to $k$ algebraically closed there, and also such an restriction is not made in (I, Theorem 3.9A) on which the proof of (II, 5.19) depends.

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Dear Jurgen, What you are calling $\mathbb{RP}^1$ is the real points of $\mathbb P^1$, which is not at all the same thing as the variety or scheme $\mathbb P^1$ over $\mathbb R$. Thus there is not contradiction. (If we consider your stated isomorphism, the non-constant functions $x$ and $y$ are defined over $\mathbb R$, and they will have poles on $\mathbb P^1$, but not at any real point; rather the poles will be at complex conjugate pairs of complex points.) Regards, – Emerton Jul 10 '12 at 17:30

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