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Projective Geometry

Coordinate Projective Geometry



If one starts with an ordinary Euclidean plane in which points are addressed with Cartesian coordinates, (x,y), this plane can be converted to a projective plane by adding a "line at infinity." This is accomplished by means of homogeneous coordinates, (x1,x2,x3) where x = x1/x3 and y = x 2/x3. One can go back and forth between Cartesian coordinates and homogeneous coordinates quite easily. The point (7,3,5) becomes (1.4,.6) and the point (4,1) becomes (4,1,1) or any multiple, such as (12,3,3) of (4,1,1).



One creates a point at infinity by making the third coordinate zero, for instance (4,1,0). One cannot convert this to Cartesian coordinates because (4/0,1/0) is meaningless. Nevertheless it is a perfectly good projective point. It just happens to be "at infinity." One can do the same thing with equations. In the Euclidean plane 3x - y + 4 = 0 is a line. Written with homogeneous coordinates Figure 1. Illustration by Hans & Cassidy. Courtesy of Gale Group. 3x1/x3 - x2/x3 + 4 = 0 it is still a line. If one multiplies through by x3, the equation becomes 3x 1 - x2 + 4x3 = 0. The point (1,7) satisfied the original equation; the point (1,7,1) satisfies the homogeneous equation. So do (0,4) and (0,4,1) and so on.

In the Euclidean plane the lines 3x - y + 4 = 0 and 3x - y + 10 = 0 are parallel and have no point in common. In homogeneous coordinates they do. In homogeneous coordinates the system 3x1 - x2 + 4x3 = 0 3x1 - x2 + 10x3 = 0 does have a solution. It is (1,3,0) or any multiple of (1,3,0). Since the third coordinate is zero, however, this is a point at infinity. In the Euclidean plane the lines are parallel and do not intersect. In the projective plane they intersect "at infinity." The equation for the x-axis is y = 0; for the y-axis it is x = 0. The equation for the line at infinity is correspondingly x3 = 0. One can use this equation to find where a curve crosses the line at infinity. Solving the system 3x1 - x2 + 4x3 = 0 x3 = 0 yields (1,3,0) or any multiple as a solution. Therefore 3x1 - x 2 + 4x3 = 0, or any line parallel to it, crosses at that point, as we saw earlier.

Conic sections can be thought of as central projections of a circle. The vertex of the cone is the center of the projection and the generatrices of the cone are the rays along which the circle's points are projected. One can ask where, if at all, the projection of a circle crosses the line at infinity.

A typical ellipse is x2 + 4y 2 = 1. In homogeneous coordinates it is x 2 + 4x 2 1 1 = 0 yields x 2 + 4x 2 2 2 - x3 = 0. Solving this with x3 2 = 0, which has no solution other than (0,0,0), which is not a point in the projective plane.

Figure 2. Illustration by Hans & Cassidy. Courtesy of Gale Group.

A typical parabola is x2 - y = 0. In homogeneous coordinates this becomes x 2 1 - x2x3 = 0. Solving this with x3 = 0 yields x1 = 0 and x2 = any number. The parabola intersects the line at infinity at the single point (0,1,0). In other words it is tangent to the line at infinity.

In a similar fashion it can be shown that a hyperbola such as x 2 - y2 = 1 crosses the line at infinity at two points, in this case (1,1,0) and (1,-1,0). These points, incidentally, are where the hyperbola's asymptotes cross the line at infinity.


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