Introductory Remarks Projective geometry was developed through artists' attempts to paint three-dimensional scenes realistically onto a two-dimensional canvas. In analyzing the relations between objects in the scenes and their images on the canvas under central projection to the eye, it simplifies the matter if we choose to analyze scenes from one scenery plane (for instance, the ground plane) at a time. So our task is to study the relations between figures in the scenery plane and their images on the canvas plane under central projection to the eye, also called perspectivity centered at the eye. It is not hard to form the following observations.

• Straightlines in the scenery planes will appear as straightlines on the canvas.
• Parallel straightlines in the scenery planes may appear to be converging on the canvas.
• Parabolas and hyperbolas in the the scenery planes may appear to be ellipses in the canvas plane---just imagine your eyes are placed at the vertex of the cone defining the conic sections, and the canvas and scenery planes are two planes used to make the conic section cut, then one cut may be a hyperbola, the other one may be an ellipse, but through the vertex, they look the same.

To obtain a satisfactory mathematical theory, we make a mathematical generalization at this point: we assume the canvas is not a finite piece of flat panel, but an entire plane, and the artist's eye can see in 360 degrees, i.e. scenes in front of as well as behind the artist's eye are all projected onto the canvas plane. Then we observe one phenomenon: the straightline in the scenery plane right underneath the artist's eye can not be projected onto the canvas plane (as light rays from those points to the eye are parallel to the canvas plane and will not project to any finite portion of the canvas plane).

One can also view this phenomenon by flipping the roles of the canvas and scenery planes: Draw some figures on the canvas plane which is now made of transparent material, and place a flash light (of point light source) at the artist's eye position, then these figures should have their images on the ground plane(the scenery plane). However, the horizon line does not shed any image on the ground plane, and if two straightline on the canvas plane meet at a point on the horizon, then their images in the ground plane will appear to be parallel.

Mathematically we view this phenomenon in the following way: We think of both the canvas plane and the scenery plane as two different representations (two snap shots from different angles) of an underlying entity, called the projective plane. Since each of the above two planes can not represent all points in this projective plane (exactly one entire line is missing in each representation), we can think of the projective plane as the usual plane adding an additional ideal line at infinity. We use the word infinity, because when we can not represent this line in the current plane, we can place a second plane and find that it is represented on the second plane as an ordinary straightline, and as points move closer to this line in the second plane, their images in the first plane appear to diverge to infinity. With this abstraction (and counting the ideal line at infinity as an additional line and understanding that two ordinary straightlines meet at a point at infinity iff they are parallel in the finite portion of the plane), we observe that

• Any two straightlines in this projective plane intersect at exactly one point (thus there is no parallel concept in the projective plane). We reason by considering the following three possibilities:
  1. both given lines are ordinary lines and meet at one point.
  2. both given lines are ordinary lines and are parallel (in the finite portion of the projective plane), and
  3. one of the given lines is the ideal line at infinity,
  In the first case, we are only left to argue that they do not meet at any other point, including the newly introduced ideal points at infinity. If they did, they would have to be parallel (in the finite plane portion), which is not the case. In case 2 one can choose a different plane to view the images of these two lines and find that they are as in case 1, thus they do meet at one point and that point appears as an ideal point in the first plane. In this case, we say the two parallel lines (in the ordinary plane) meet at an ideal point at infinity. From this argument, we can see that parallel lines (in the ordinary plane) meet exactly at one point at infinity. From another point of view, we say there is an ideal point to each direction (if an opposite direction is counted as the same, as it produces the same family of parallel lines). In case 3, again if we choose a different plane to view the images of these two lines then they meet at exactly one point.
• Any two points determine a unique line in the projective plane. This is easier, as the only case to be argued is when one or both of the points are ideal points at infinity.

The above approach has its advantages: one can view most of the geometric happenings in this projective plane through an ordinary plane, but to understand what happens at a point (or line) at infinity, one has to project the portion of the plane near infinity to a different plane to study it. This need to deal with the exceptional situations can not be avoided. As we will see, this flexibility of viewing the geometry from a different plane also has its advantages. Such planes will be called Embedding planes. For some purposes, in particular for algebraic representations, it is advantageous to develop an approach that does not need to deal with exceptional cases. We will do that next.

Projective points and projective plane From the perspective of a fixed point O (as the center of the lens), all points on an Euclidean line (E-line) p through O are considered the same---this is based on the optical fact is that an observer can not distinguish between points on the light rays directly towards the observer's eye, points on such a light ray will be appear the same to the artist's eye and will be treated as the same point and represented by the same point (on the canvas) . So we could use any point on p other than O, or the line p itself to represent this projective point P (also called Point in the text) in the projective plane, which consists of all such points.

Embedding plane Often we use an Euclidean plane (E-plane) π not through O to take a "snap shot" of geometric figures in the projective plane.

• With the exception of those Points represented by E-lines parallel to π, any other Point P is represented by an Euclidean point on π, the intersection of P with π.
• An E-line l in π generates a projective line L, represented by the E-plane containing l and O. In general, we make the definition: Any E-plane through O represents a projective line, also called Line. Thus only one projective line, that represented by the E-plane through O and parallel to π is not represented on π.
• We can represent the projective plane by points in π, together with the set of Points not represented on π; the latter forms a projective line. Points on this projective line are called ideal Points(or points at infinity) for π, and this projective line is called the ideal Line (or line at infinity) for π. Any such E-plane, together with its ideal Points, forms an embedding plane. In the situation of the above item, the projective line L consists of the E-line l, together with an ideal Point.
• Keep in mind the perspectivity relation between two different embedding planes π and π'. Recall that under this perspectivity, an E-line l in π is sent to an E-line in π' (except when l is parallel to π'). Notice that given a Line L, a point in its representation l in π may correspond to the ideal Point for its representation l' in π'. A triangle in π is mapped to a triangle, or a pair of parallel half-lines joined by a finite straight line segment in π'. One advantage of our introduction of the abstract concept of projective plane is that we no longer need to discuss the exceptional cases separately. For instance, it is meaningful to discuss the notion of triangles and other polygons in the projective plane.
• Note that in the projective plane Any two distinct Lines meet at a unique Point---this is Theorem 3 on p.146. The text gives an algebraic argument. Here is a geometric argument: if both Lines are represented by E-lines in π, then either the E-lines meet at a unique point in π, in this case, the two projective lines also meet uniquely at this point; or the E-lines are parallel, but using an different, well chosen embedding plane, the images of these E-lines do meet at a finite point of the new embedded plane, and this means these two Lines do meet, it's just that their representation in π happens to meet at an ideal Point for π. If one of the give Lines is the Line at infinity for π, then the other is represented by an E-line in π. In this case, again these two lines meet at an ideal Point for π.

Duality between Points and Lines Recall that any Point can be represented by a non-zero vector (a, b, c), or any non-zero multiple of this vector. We refer to this point as [a, b, c]. Any Line can be represented an E-plane, which can be represented by an equation of the form ax+by+cz=0. The only pertinent information is contained in the vector (a, b, c); also recall that ax+by+cz=0 and sax+sby+scz=0 for s non-zero, represent the same Line, so (sa, sb, sc) and (a, b, c) represent the same Line. In other words, the same homogeneous coordinates [a, b, c] can be used to represent either a Point or a Line. More such duality occurs.

• If [a, b, c] and [d, e, f] are two distinct Points, then they determine a Line with equation gx+hy+iz=0. (g, h, i) is a normal vector to the E-plane gx+hy+iz=0. From ga+hb+ic=0 and gd+he+if=0 we see that (g, h, i) is perpendicular to both (a, b, c) and (d, e, f), so we may as well take (g, h, i) to be (a, b, c) x (d, e, f).
• By duality, [a, b, c] can represent the Line ax+by+cz=0 , and [d, e, f] can represent the Line dx+ey+fz=0. The two E-planes ax+by+cz=0 and dx+ey+fz=0 meet along an E-line whose direction vector is perpendicular to both (a, b, c) and (d, e, f), so we may as well take it to be (a, b, c) x (d, e, f). In other words, the same algebra (a, b, c) x (d, e, f) can be used to find the Line determined by the Points [a, b, c] and [d, e, f], and can also be used to find the intersection Point of the Lines ax+by+cz=0 and dx+ey+fz=0.
• If [a, b, c], [d, e, f], and [g, h, i] represent three collinear Points, then the three Lines represented by [a, b, c], [d, e, f], and [g, h, i] are concurrent, and vice versa. This is because [a, b, c], [d, e, f], and [g, h, i] are collinear if and only if the three vectors (a, b, c), (d, e, f) and (g, h, i) lie in the same E-plane through O; algebraically this criterion translates into the determinant of the matrix whose three rows consist of these three vectors is equal to zero. While the Lines represented by [a, b, c], [d, e, f], and [g, h, i] are concurrent, if and only if the normal vectors to the corresponding E-planes, (a, b, c), (d, e, f) and (g, h, i) are perpendicular to a common vector, the direction vector of the line of intersection of three planes, which again means that (a, b, c), (d, e, f) and (g, h, i) are collinear.

Theorem 3 of 3.3 on p.162, the fundamental theorem of projective geometry, allows us the flexibility of transforming a given problem into a situation where one of the Lines is now appearing as the Line at infinity in our embedded plane. This often allows us to understand the geometry more easily. We will do some examples of this kind.

A projective transformation also preserves a numeric quantity, the cross-ratio of four Points along a Line, as explained by Theorems 3 and 4 on pp.184. Here again we will emphasize the more geometric approach as in 3.5.2-3.

Here are some more notes on perspectivity, perspective transformations, and projective transformations. Recall that, there is a perspectivity between any two planes π and π' that do not pass through the observer O. Imagine π has its own grids (coordinate system) and a figure on it, and π' is simply another copy of π placed in its current location by a linear transformation P --> A P. The figure on π' also has its image onto π under the perspectivity between π and π', We now want to compare the two images on π and find their relations (we could think we had a figure on π, made a copy and moved it to its position on π', and then flashed that image onto π through lightrays from O. We now want to put the two figures on the same sheet π to compare them). In fact, there is a procedure for setting up a map from π to itself: any point P on π has an identical copy P' on π', which then has its image point P'' on π under the perspectivity. The map from P to this final image point P'' can be used to describe the relations between any original figure on π (and on π' ) with its image on π under the perspectivity. Such a transformation is called a perspective transformation.

To represent the perspective transformation above algebraically, the map from P to P' can be represented by an invertible linear transformation: P'= A P, the action of which between π and π' is actually a rigid motion in the 3-dimensional Euclidean space; the perspectivity p between π and π' simply says that p(P') :=P'' is on π and [p(P') ] = [P']. Thus we can represent this map from P to P'' algebraically as

[P] --> [P'']:=[ t(P) ] := [p(A P)]= [ A P ].

Once we start to consider the composition of such transformations, we will find that we can allow any invertible matrix A in the above formula. Such transformations are called projective transformations.

The text deals with projective transformations only using the homogeneous representation [x] --> [ Ax]. What we described earlier was to understand the action of this projective transformation through the embedding plane π.

Suppose that A =

a_1	b_1	c_1
a_2	b_2	c_2
a_3	b_3	c_3

and P=[X,Y,Z] (as a column vector), P'=t(P)=[X',Y',Z'] (as a column vector). Then

X'=a_1X+b_1Y+c_1Z

Y'=a_2X+b_2Y+c_2Z

Z'=a_3X+b_3Y+c_3Z

Suppose that the embedding plane π is given by {(x,y,1): (x,y) in R^2}, and π'=t(π). Then the representative (x',y',1) on π of (X', Y', Z')=t([x,y,1]) is given by

x'=X'/Z' = (a_1x+b_1y+c_1)/(a_3x+b_3y+c_3)

y'=Y'/Z'=(a_2x+b_2y+c_2)/(a_3x+b_3y+c_3)

This is a fractional linear transformation, whose definition at points where a_3x+b_3y+c_3=0 needs clarification. It turns out that we can interpret that points on the line a_3x+b_3y+c_3=0 are mapped to the line at infinity of π! The algebra of working with fractional linear transformations is not as easy as that of direct matrix multiplications, which is why math textbooks tend to favor working with direct matrix multiplications of the homogeneous coordinates as the definition of projective transformations, as is done in the text.

Example: The map t: [x, y, z] --> [x-2z, -x+2y-3z, x-y+5z] is a projective transformation with matrix A

1	0	-2
-1	2	-3
1	-1	5

The inverse to t can be represented through A^-1, or adj (A), which is

7	2	4
2	7	5
-1	1	2

Thus t^-1: [x', y', z'] --> [7x'+2y'+4z', 2x'+7y'+5z', -x'+y'+2z']. A projective line x+2y+3z=0 becomes (7x'+2y'+4z')+2( 2x'+7y'+5z')+3 (-x'+y'+2z')=0, which is 8x'+19y'+20z'=0.

Another way to find the image of x+2y+3z=0 under t is the following: Find two specific points on x+2y+3z=0, for example, [2, -1, 0] and [3, 0, -1], and find their images under t. In our case t([2, -1, 0])=[2, -2-2, 2+1]=[2, -4, 3], t([3, 0, -1])=[3+2, -3+3, 3-5]=[5, 0, -2]. So the image of x+2y+3z=0 under t must be the projective line passing through [2, -4, 3] and [5, 0, -2], which turns out to be the same: 8x+19y+20z=0.

Let's examine the action of t on the imbedding plane z=1, i.e., for any point (x, y, 1) on z=1, compute t([x, y, 1])=[X', Y',Z' ], then find its representative [x', y', 1] on the plane z=1. So x'= X'/Z', and y' = Y'/Z' . For example, t([0, 0, 1])=[-2, -3, 5] =[-2/5, -3/5, 1]. In general, if we set t([x, y, 1])=[x', y', 1], then

x' = (x-2)/(x-y+5), y'=(-x+2y-3)/(x-y+5). ---(NH)

If we use our computation earlier, the projective line x+2y+3z=0 is represented by x+2y+3=0 on the imbedding plane z=1, which is sent to the projective line 8x+19y+20z=0, represented on z=1 by 8x+19y+20=0. One could also derive this directly by solving (x, y) in terms of (X, Y), and substitute these relations into x+2y+3=0. It turns out that

x=(7X+2Y+4)/(-X+Y+2), y=(2X+7Y+5)/(-X+Y+2).

So the equation x+2y+3=0 becomes (7X+2Y+4)/(-X+Y+2)+2(2X+7Y+5)/(-X+Y+2)+3=0, which reduces to 8X+19Y+20=0. However it is not easy to solve (x, y) in terms of (X, Y) directly from (NH), this is one reason people often favor working with the homegeneous representations as in the text than the more geometric representations such as (NH).

The viewpoints in the above paragraph are often used to transform a given geometric situation into one where things become simpler. For instance, to prove three points P, Q, R are collinear, one chooses a perspectivity t so that t(P), t(Q) appear as Points at infinity; if t(R) can be proven also to be a Point at infinity, then the original points must be collinear. To prove three lines l₁, l₂, l₃ are concurrent, we choose a perspectivity t so that t(l₁), t(l₂) appear as parallel E-lines, and try to prove t(l₃) is parallel to them. This is often good strategy, because it is easier to extract information from the set up that t(l₁), t(l₂) are parallel. We will illustrate this idea with Desargues' Theoreme. The text discusses some examples in 3.4.1 along similar ideas, except that text transforms a give situation into one which is easier to handle by the coordinate method, instead of tackling the transformed case by direct geometric argument. Note that a necessary ingredient in either our approach or the text's approach is the ability to tranform a given situation into a disired situation. That is guaranteed by Theorem 3 on p. 162. Here is a link that contains applets to demonstrate Desargues' Theorem.