Projective Varieties

Yesterday, there was talk of affine varieties, and I pointed out the fact that intersections don’t work out nicely in this case. Today we’ll talk about an elegant solution to this problem: the notion of projective space.

We begin by looking at $\mathbb{C}^{n+1}$ . Then we define $\mathbb{P}^n$ to be the collection of lines through the origin in $\mathbb{C}^{n+1}$ , and call it $n$ -dimensional projective space. This seems a bit odd at first to most people, so let’s examine it directly before trying to do anything complicated.

First thing’s first, each nonzero vector in $\mathbb{C}^{n+1}$ defines a point in $\mathbb{P}^n$ . But more than one vector can define the same point, so we should try to figure out the relationship. If $v,v'$ are two vectors that define the same projective point, then they must lie on a common line through the origin. In this case, we must have $v=\lambda v'$ where $\lambda$ is a nonzero complex number. So as long as we’re willing to work up to nonzero constant multiples, we can use the coordinates on $\mathbb{C}^{n+1}$ on our projective space. These coordinates are written as $(x_0:\ldots:x_n)$ and are called homogeneous coordinates.

One of the nice things about projective space is that is contains our old space $\mathbb{C}^n$ and just adds a bunch of points which we often say are “at infinity.” To see this look at the set of vectors in $\mathbb{C}^{n+1}$ such that $x_0\neq 0$ . Then we can divide all the coordinates by $x_0$ and we get actual coordinates rather than homogeneous coordinates, and there are $n$ of them, so here is a copy of $\mathbb{C}^n$ inside $\mathbb{P}^n$ .

So we have projective space, what’s the big deal? The first big deal is that now EVERY pair of lines intersects in $\mathbb{P}^2$ . Sticking with geometry and keeping the algebra out of it (well, mostly), a line in projective space would be a plane through the origin in $\mathbb{C}^{3}$ . Then the intersection of two of these planes must be a line, just because there isn’t any room to have four coordinates that are independent (this is made rigorous by the Rank-Nullity Theorem of linear algebra).

So this hints that projective space has nice intersection properties, and indicates that we might want to find a way to extend algebraic geometry to this object. This turns out not to be too hard. We must merely identify what sorts of polynomials we can make sense of on lines through the origin. Here’s a problem, though: if we restrict a polynomial to a line, then we get a polynomial in one variable, so it won’t be well defined unless it is a constant polynomial. However, if we stick to homogeneous polynomials, that is, polynomials with all the terms of the same total degree, then we can do some work.

A homogeneous polynomial has a nice property, that, if the degree of the polynomial is $d$ , then we have $F(\lambda x)=\lambda^dF(x)$ . So though we still can’t make sense of the value of an arbitrary homogeneous polynomial on a line, we can at least say whether or not one is zero! So our original notion of algebraic sets can be extended. We define a projective algebraic set to be the common zeroes of finitely many homogeneous polynomials, and we define a projective variety to be an irreducible projective algebraic set. From here on out, we will refer to the algebraic sets from last time as affine algebraic sets and affine varieties in affine space to distinguish them from projective algebraic sets and projective varieties in projective space.

Using the projective algebraic sets, we again get a topology and we will again call it the Zariski topology, and the basis theorem still holds, and even the Nullstellensatz still works, if you modify it to be ideals generated by homogeneous polynomials and if you throw out the ideal $(x_0,\ldots,x_n)$ .

We should now step back and mention one of the big problems in mathematics: how do you define the notion of “dimension”? This is a tricky problem and many areas have many different solutions. In algebraic geometry, we defined the dimension of a variety (affine or projective) to be the length of the longest chain of varieties each properly contained in the last. So a point is zero dimensional, because it contains no varieties properly, a line has one dimension, because it contains points, a plane is two, etc. The dimension of an algebraic set is the largest of the dimensions of the varieties that make it up. For simplicity in the future, a one dimensional variety will be called a “curve,” two dimensional will be called a “surface,” and higher dimensional varieties will sometimes be called $n$ -varieties (reserving the standard $n$ -fold for a later definition when we get to singular and nonsingular varieties).

So for today, just one more thing: any affine variety fits into a projective variety in a nice way. We call the projective variety that does this the projective closure, and geometrically, we can define it to be the smallest projective variety such that if we look at the part with $x_0\neq 0$ , we get the affine variety back. Algebraically, we have a slightly greater amount of trouble.

First, we note that we can turn ANY polynomial into a homogeneous polynomial in one more variable by the map $f(x_1,\ldots,x_n)\mapsto x_0^df(x_1/x_0,\ldots,x_n/x_0)$ where $d$ is the degree of the polynomial. So to get the projective closure, first we take the ideal of the affine variety, which will be generated (probably) by nonhomogeneous polynomials. Then we homogenize them with the above trick. Now, this won’t be an ideal any more (though if we evaluate the polynomials so that $x_0=1$ , we get the old ideal back) but it does generate another ideal. The projective closure is given by the common zeroes of all of the polynomials in this ideal.

Tomorrow: Either more projective geometry or defining a morphism of varieties (I haven’t decided yet).

5 Responses to Projective Varieties

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Farid says:

June 30, 2015 at 11:39 am

Excellent and clear introduction Charles. I am learning these concepts for the first time and you are very clear. I am part of the REU Math Program at Rutgers working with Dr. Buch.

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