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Relating Affine Geometry to Projective Geometry

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Relating Affine Geometry to Projective Geometry Chapter 1 Relating Affine Geometry to Projective Geometry 1.1 Embedding Affine Planes in Projective Planes The most familiar example of an affine geometry (a geometry with the parallel 2 postuatle) is R2: We can embed the Euclidean plane R2 in RP so that the 2 image of R2 is all of RP except the famous "line at infinity." [Draw a picture.] R2 is certainly in bijection with the plane z = 1 in R3 via (x; y) $ (x; y; 1): For (x; y:1) in this plane, we set `(x;y;1) to be the line on the 2 origin and (x; y; 1). We define α : R2 ! RP by α(x; y) = `(x;y;1): In the upper hemisphere model 2 = S2 =v ∼ −v, we have α(x; y) = p 1 (x; y; 1): RP + x2+y2+1 2 α is clearly an injection (so we say that α embeds R2 in RP ). The image of 2 α is all of RP except lines in the xy-plane. Since these lines sweep out a plane ~ 2 2 on 0, these lines (= points in RP ) lie on a line in RP . We call this line `1, the line at infinity, and elements of `1 are points at infinity. Thus 2 2 RP = α(R ) [ `1: Since the analytic approach is so useful in R2, it's natural to look for good 2 ~ coordinates on RP : We give `(x;y;z), the line on 0 and (x; y; z), homogenous coordinates [x; y; z]: Of course, we don't allow (x; y; z) = (0; 0; 0): Since `(x;y;z) = ∗ ell(λx,λy,λz) for λ 2 R , we must identify the homogeneous coordinates: ∗ [x; y; z] = [λx, λy; λz] for λ 2 R : 2 It may seem upsetting that a point in RP has an infinite number of coordi- nates. However, we've learned to live comfortably with the origin having polar coordinates (0; θ) for all θ, so this isn't really a new phenomenon. 1 2CHAPTER 1. RELATING AFFINE GEOMETRY TO PROJECTIVE GEOMETRY For z 6= 0, hx y i [x; y; z] = ; ; 1 : z z (Confusing remark: It may also be upsetting to think of a continuous family of lines `i passing through (x0; y0; zi) with zi > 0 and zi ! 0, since the third component of the homogeneous coordinates seems to jump discontinuously from 1 to 0 as i ! 1: However, if we don't "divide by zi, there is no problem. The point is that there are better and worse representatives of homogeneous coordinates depending on the situation.) In particular, α is particularly simple in homogenous coordinates: α(x; y) = [x; y; 1]: Moreover, the points at infinity have homogeneous coordinates [x; y; 0]. For example, the y-axis line passes through the points (0; ±1; 0), so its homogeneous coordinate is [0; 1; 0] = [0; −1; 0]: which is one point on ell1: Homogeneous coordinates are actually easy to work with. Example. 1. What happens when we go out a vertical line in R2? We want to compute limb!1(7; b): Of course, the limit does not exist. 2 2. Let's redo the computation in RP . Now we compute 7 1 lim α(7; b) = lim [7; b; 1] = lim ; 1: = [0; 1; 0]: b!1 b!1 b!1 b b So the limit exists, and is a point on `1: Note that limb→−∞ α(7; b) = [0; −1; 0] = [0; 1; 0], the same point on `1: If you draw the line ` : x = 7; z = 1 in the z = 1 plane, you should be able to see that α(`) is heading 2 towards (0; ±1; 0) on S+ as b ! ±∞: 3. If we redo this computation for α(10; b), we see that α of the line x = 10; z = 1 also has the limit point [0; 1; 0] 2 `1: This leads to the common sloppy language: parallel lines meet at infinity. The precise statement is: α of parallel lines meet at a point on `1: 4. Let's check this on two other parallel lines, y = ax + b0; y = ax + b1. Then ax + b0 1 1 lim α(x; ax+b0) = lim [x; ax+b0; 1] = lim 1; ; = [1; a; 0] = ; 1; 0 : x→±∞ x→±∞ x→±∞ x x a This is independent of b0; b1, so these parallel lines meet at a point on `1: 1 Note that as the line goes vertical, i.e. a ! ±1, the point a ; 1; 0 2 `1 approaches [0; 1; 0], as it should. However, we can get fooled and incorrectly argue that 1 lim ; 1; 0 = lim [1; a; 0] a→±∞ a a→±∞ doesn't exist. So we have to be careful. 1.1. EMBEDDING AFFINE PLANES IN PROJECTIVE PLANES 3 5. The equation for a plane on ~0 in R3 is f(x; y; z) 2 R3 : ax + by + cz = 0g for some a; b; c 2 R not all zero. The lines in this plane are exactly f[x; y; z]: ax + by + cz = 0g, where we note that this equation is well- defined: ax+by +cz = 0 iff a(λx)+b(λy)+c(λz) = 0 for λ 2 R∗: Thus the equation of a line in RP 2, i.e. a plane on ~0 in R3, is f[x; y; z]: ax+by+cz = 0g. In particular, `1 has equation z = 0: `1 = f[x; y; 0] : (x; y) 6= (0; 0)g: 2 2 Can we repeat this process for C or Zp or for any affine plane? If we take the synthetic approach, we want to add one new point "at infinity" for every (maximal) family of parallel lines, and join all the points at infinity on a new line, the line at infinity. This works: Theorem. Any affine plane embeds into a projective plane. The proof is an easy homework problem. You should check that if the affine 2 plane is Z2, the associated projective plane is our 7 point geometry! If we want to reverse this procedure, we want to start with a projective plane, declare one line to be the line at infinity, and delete it and all its points. But which line should we delete? Experimenting with the 7 point geometry, you'll see that no matter which line you delete, the result is a copy of Z2. This motivates the following converse result: Theorem. A projective plane minus the points on any fixed line `1 is an affine plane. This is also not hard. For example, to show the parallel postulate, take 0 0 P 62 ` as usual. Let Q = ` \ `1 and set ` = PQ. Then in A = P n `1, ` \ ` = ; 00 0 00 0 00 (why?). If P 2 ` 6= ` in A and ` ` = ;, then Q = ` \ `1 6= Q. But then 0 ` \ `1 ⊃ fQ; Q g, a contradiction. Moral: Any line in a projective plane can be the line at infinity for an embedded affine plane. To approach the same problem analytically, we have to know when an affine plane is of the form k2 for a field k. (OK, this is out of order, as fields are defined later. We know what we need: a set k with two operations +; ·, where (k; +) is an abelian group and (k∗; ·) is a group, where k∗ is k minus the +-identity element. So e.g. Zp is a field iff p is prime.) The beautiful solution to this question is in Artin's Geometric Algebra. p 2 2 2 2 For specific examples like k = C ; Zp; Q ; Q[ −5] , we can introduce homo- geneous coordinates as before. In each case, we set kP2 to be the lines on the 3 origin (0; 0; 0) (where 0 is the +-identity element) in k : These lines `(x;y;z) are 3 def given by nonzero vectors (x; y; z) 2 k : `(x;y;z) = fλ(x; y; z) = (λx, λy; λz): λ 2 k∗g: We give these lines homogeneous coordinates [x; y; z], where as before [x; y; z] = [λx, λy; λz]: We define α : k2 ! kP2 by "projecting from the z = 1 plane:" α(x; y) = [x; y; 1]: Then kP2 = α(k2) [ f[x; y; 0] : (x; y) 6= (0; 0)g: 4CHAPTER 1. RELATING AFFINE GEOMETRY TO PROJECTIVE GEOMETRY We define lines in kP2 to be planes on the origin in k3, i.e.f(x; y; z): ax + by + cz = 0g for fixed a; b; c 2 k not all zero. Then e.g. the set of all points at infinity lies on the line at infinity z = 0: All set of parallel lines in k2 such as y = a0x + b, for fixed a0 2 k and all b 2 k, meet at [1=a0:1; 0] for a 6= 0: 2 We can draw a picture of Z3P . There are 13 lines and 13 points: the points at infinity are [1; 0; 0] = [2; 0; 0]; [0; 1; 0] = [; 2; 0]; [1; 1; 0] = [2; 2; 0]; [1; 2; 0] = [2; 1; 0]: Every line has four points, three of which are finite points and one of which is on `1. A final useful remark: we can dumb this down to kP1 by taking α from the y = 1 line to the set of lines in k2: α(x; 1) = [x; 1]: Every line in kP2 has the shape of a kP1: Check this: a line in k2 is in bijection with k, so a line in kP2 is given by adding a point at infinity, and this is precisely how we go from k to kP1: Let's use the Moral above to prove concordance theorems in Euclidean ge- ometry by "moving a line to infinity." Consider the y-axis ` in the z = 1 plane in R3: Under the α map, ` maps to 2 the great circle C = fx = 0g \ S2 in S2.
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