# 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, (x_{1},x_{2},x_{3}) where x = x_{1}/x_{3} and y = x 2/x_{3}. 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
3x
_{1}/x_{3} - x_{2}/x_{3} + 4 = 0 it is still a line. If one multiplies through by x_{3}, the equation becomes 3x 1 - x_{2} + 4x_{3} = 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 3x_{1 }- x_{2} + 4x_{3} = 0 3x_{1} - x_{2} + 10x_{3} = 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 x_{3} = 0. One can use this equation to find where a **curve** crosses the line at infinity. Solving the system 3x_{1} - x_{2} + 4x_{3} = 0 x_{3} = 0 yields (1,3,0) or any multiple as a solution. Therefore 3x_{1} - x 2 + 4x_{3} = 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 x^{2} + 4y^{ 2} = 1. In homogeneous coordinates it is x 2 + 4x^{ 2} 1 1 = 0 yields x 2 + 4x^{ 2 } 2 2 - x_{3} = 0. Solving this with x_{3} 2 = 0, which has no solution other than (0,0,0), which is *not* a point in the projective plane.

A typical **parabola** is x^{2} - y = 0. In homogeneous coordinates this becomes x 2 1 - x_{2}x_{3} = 0. Solving this with x_{3} = 0 yields x_{1} = 0 and x_{2} = 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} - y^{2} = 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.

## Additional topics

Science EncyclopediaScience & Philosophy: *Positive Number* to *Propaganda - World War Ii*Projective Geometry - Desargues' Theorem, Coordinate Projective Geometry, Cross Ratio