The Projective Plane: An Algebraic Exploration II

This is the third post in a series discussing the projective plane. For the first, see here.

The previous post explained how certain types of polynomials, namely the homogeneous polynomials, define curves in the projective plane called projective varieties. This post will explicate the relation between projective varieties and affine varieties (typical curves in the plane) and indicate how projective varieties are in a way the extensions of affine varieties to include their points at infinity.

First, we consider how projective varieties naturally give rise to normal affine plane curves. Consider the projective variety defined by the equation F(x:y:z) = 0, where F is a homogeneous polynomial. In the first post of this series, we saw that the plane z = 1 in three-dimensional space can represent the subset of the projective plane that corresponds to the normal affine plane (i.e., without the points at infinity). We repeat the image from the first post for convenience, where each line through the origin (a point of projective space) is represented by the point at which it intersects the plane.

Since we obtain the affine plane by setting z = 1, it seems reasonable that we should be able to "collapse" projective varieties algebraically by setting z = 1 in the equation F(x:y:z) = 0. This is indeed the case, since substituting z = 1 yields a polynomial in only two variables: f(x,y) = F(x,y,1). For example, if F(x:y:z) = x²y + 2yz² - 5z³ (note that F is homogeneous and therefore defines a projective variety), then f(x,y) = F(x,y,1) = x²y + 2y*1² - 5*1³ = x²y + 2y - 5. The projective variety F(x,y,z) = 0 therefore corresponds to an affine variety f(x,y), as desired.

There is also an algebraic process that does the reverse by taking a polynomial f(x,y) and producing a corresponding homogeneous polynomial in three variables, F(x:y:z). The process works as follows:

1. Add up the powers of x and y in each term of f and let n be the greatest degree that appears
2. Multiply each term by z^n-k, where k is the degree of the term (this ensures that the resulting polynomial is homogeneous)

Let us take the polynomial f(x,y) = x²y + 2y - 5 from above and apply the new process. The degrees of the three terms are 3, 1, and 0, respectively (-5 = -5x⁰y⁰). The maximum of these is 3, so we multiply the first term by z^3-3 = z⁰ = 1, the second by z^3-1 = z², and the third by z^3-0 = z³. The resulting polynomial is F(x:y:z) = x²y + 2yz² - 5z³, exactly the same as the original polynomial! It is easy to verify that these two processes are mutually inverse in general, except in certain special cases such as when F is only a function of z.

Now we may apply these algebraic tools to solve the problems introduced in the last post that cannot be solved visually. First, regarding the hyperbola, algebra confirms our intuition. To see this, take the equation xy - 1 = 0 and transform it into the corresponding projective variety. The result is easily calculated as xy - z² = 0. The asymptotes x = 0 and y = 0 to this hyperbola (see image in previous post) are unchanged by the process since they only have one term, clearly of maximal degree.

Next, recall that the points at infinity in the projective plane are those for which z = 0 in the homogeneous coordinates (x:y:z). This can be seen in the above visualization, where only points of the form (a:b:1) belong to the affine subset of the projective plane. Now any (x:y:z) can be scaled to this form by multiplying each component by 1/z (remember, only the ratio of the coordinates matters), but only when z is nonzero. Therefore, we substitute z = 0 and solve the equations to see which points at infinity each curve intersects. For the hyperbola, this gives xy = 0, so x = 0 or y = 0. Therefore, the two points at infinity the hyperbola intersects are (0:1:0) and (1:0:0). Other coordinate triples satisfying xy = 0 such as (3:0:0) differ only by a scale factor from one of the two solutions above and therefore define the same point in the projective plane. It follows that (0:1:0) and (1:0:0) are the only solutions. But the asymptotes x = 0 and y = 0 hit exactly the same points, (0:1:0) and (1:0:0), respectively! This confirms our intuition: a hyperbola and its asymptotes really do intersect at infinity.

The cubic y - x³ = 0 has no asymptote, but clearly goes off to infinity in some manner. We may use our algebraic tools to investigate the function's behavior in the projective plane. The highest degree term is x³, of degree 3, so we must multiply the other term (namely the expression y, of degree 1) by z^3-1 = z². The projective variety corresponding to the cubic is therefore defined by the equation yz² - x³ = 0. Substituting z = 0 yields x³ = 0, which has the single point (0:1:0) in the projective plane as a solution (since z is already set to 0). Note that even though the cubic goes to infinity in both the positive and negative directions, it meets only one point at infinity because opposite directions are identified (see the representation of the projective plane in a sphere in the first post). This indicates that the projective variety induced by the cubic meets that induced by the y-axis with equation x = 0 at infinity. Indeed, this makes some intuitive sense: as x becomes very large, it becomes insignificant relative to y = x³ and therefore the point (x,y) is "close" to the y-axis x = 0 (this can also be seen by zooming out a graph of the cubic - the graph eventually becomes nearly indistinguishable from the y-axis). We can also visualize the projective variety yz² - x³ = 0 that extends the cubic on the sphere (see below).

This image shows the cubic curve in the affine plane as well as its projection (via lines through the origin, the center of the sphere) onto the surface of the sphere. It differs slightly from our earlier sphere representation since the plane is below and not above the sphere, but this makes little difference. At the bottom of the sphere, the origin of the plane touches the sphere (which is a point on the curve). At first, the path veers away from the y-axis (the grid line from top to bottom through the origin), but notice how when the curve approaches the equator of the sphere (infinity), it comes back to hover above the y-axis. Images like these help to interpret the results of our algebraic manipulations.

The projective plane has very elegant geometric properties (every two lines in the plane intersect in exactly one point, for example) and gives us a sturdy mathematical grounding for the slippery concept of behavior "at infinity." Generalizations of this concept are crucial in the study of polynomial curves and their corresponding equations.

Sources: http://voltage.typepad.com/.a/6a00e55375ef1c8833014e610f8df7970c-pi

The Projective Plane: An Algebraic Exploration I

This is the second post in a series on projective space. For the first, see here.

The idea of "adding points at infinity" to the plane introduces new behavior to the study of the intersections of lines and curves. Since there are different points of infinity for each direction in the affine plane (as discussed in the last post) and parallel lines intersect at infinity, it is reasonable to suppose that certain lines and curves also intersect at infinity (see below).

For example, the hyperbola above is given by the equation xy = 1. Away from the origin, the two branches of this curve approach the x- and y-axes, defined by the equations y = 0 and x = 0, respectively. Since the distance between the curve and these lines (called asymptotes of the curve) approaches 0 far from the origin, it makes sense to suppose that the hyperbola intersects with these asymptotes at infinity. On the other hand, for other curves that clearly "go off to infinity" like the cubic curve y = x³ shown below, there is no asymptote. What point at infinity, if any, does the cubic intersect?

Answering this question is difficult in our as yet fuzzy picture of the structure of the projective plane. However, the algebraic definition of the projective plane provides the tools necessary for solving this and many other related problems. Introducing this machinery is the purpose of this post.

Lines, parabolas, cubics, hyperbolas and many other curves in the plane may be expressed in the following algebraic form: f(x,y) = 0, where f is a polynomial function of x and y. This means that it is a sum of terms of the form a*xⁱy^j where i and j are nonnegative integers and a is a constant coefficient. For example, the equation for the hyperbola written above may be written xy - 1 = 0 and the equation for the cubic y - x³ = 0. Any curve defined by an equation of this form is known as an affine variety.

The previous post introduced the projective plane as the set of lines through the origin in three-dimensional space. It then illustrated two different ways in which certain representatives may be chosen from the lines to get a "picture" of the projective plane in three-dimensional space. We show that for equations of certain forms, it does not matter which representative we choose from a given line through the origin. First, let P = (x,y,z) be a point distinct from the origin in three-dimensional space (that is, at least one of x, y, and z is nonzero). Then since any two distinct points determine a line, P determines a unique line L through the origin. The point (ax,ay,az) is then on L for any constant a and every point on L is of this form. In other words, only the ratio of the coordinates to one another is required to determine on which line through the origin a given point lies. This fact can more easily be seen in two dimensions, as in the figure below.

The line above has equation y = 2/3*x. It passes through the origin and has slope 2/3, so any point (x,y) for which y/x = 2/3 is on the line (as demonstrated by the construction of a suitable triangle, as above). With this fact in mind, we introduce the concept of homogeneous coordinates. Homogeneous coordinates (x:y:z), where at least one is nonzero, define a point of the projective plane, with the understanding that only the ratio of x, y, and z matters. Thus (1:2:3) = (3:6:9), for example. With these identifications in mind, every point in the projective plane may be assigned homogeneous coordinates (though in many equivalent ways).

Next we consider projective varieties, i.e. certain types of curves in the projective plane. As before, they are defined as the set of points satisfying a certain polynomial equation, but in three variables instead of two: F(x,y,z) = 0. However, in light of the equivalence between points with different x-y-z coordinates, we must consider only polynomial equations that have the same points of the projective plane as solutions for any coordinate representation of the given points. These are called homogeneous polynomials. A polynomial is homogeneous if each one of its terms, or monomials, is of the same degree, meaning that the sum of the exponents in each term are the same. For example, F(x,y,z) = x²yz³ is trivially homogeneous of degree 6 because it has only one term and the sum of its powers are 2 + 1 + 3 = 6. F(x,y,z) = xy² + z³ is homogeneous of degree 3 because the sum of the exponents of the xy² term is 1 + 2 = 3 and is obviously also 3 for the second term, z³. The crucial property of homogeneous polynomials is that if F(x,y,z) = 0, then F(ax,ay,az) = 0 for any constant a:

The crucial fact used in the proof (click to enlarge) is that the exponents of each term (the p's, q's, and r's) must always add up to the same degree n. The aⁿ term can then be factored out, confirming that F(x,y,z) = 0 always implies F(ax,ay,az) = 0. This means that for any point in the projective plane, a homogeneous polynomial that is zero on one representative is zero on all. Conversely, if F(ax,ay,az) = 0, then the same proof (using 1/a) shows that F(x,y,z) = 0 so long as a is not zero. All this manipulation distills down to the following crucial statement: it is meaningful to say that a homogeneous polynomial is zero at a point in the projective plane since any representative gives the same result. We can thus denote projective varieties by the equation F(x:y:z) = 0 in homogeneous coordinates.

It follows that a homogeneous polynomial in three coordinates has a solution set of points (a curve) in the projective plane. These solution sets are the projective varieties. The next post (coming soon) continues to fill in the algebraic picture of the projective plane and relates affine varieties to projective ones, ultimately answering the questions posed at the beginning of this post.

Sources: http://intmstat.com/plane-analytic-geometry/xyis1.gif, http://www.s-cool.co.uk/assets/learn_its/gcse/maths/graphs/algebraic-graphs/g-mat-graph-dia04.gif