Relating Affine Geometry to Projective Geometry

Home , Affine geometry, Configuration (geometry), Line at infinity, Point at infinity

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) = √ 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 `∞, the line at inﬁnity, and elements of `∞ are points at inﬁnity. Thus

2 2 RP = α(R ) ∪ `∞.

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 ∈ R , we must identify the homogeneous coordinates:

∗ [x, y, z] = [λx, λy, λz] for λ ∈ R .

2 It may seem upsetting that a point in RP has an inﬁnite number of coordinates. 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 → ∞. 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 ell∞. 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→∞(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→∞ b→∞ b→∞ b b

So the limit exists, and is a point on `∞. Note that limb→−∞ α(7, b) = [0, −1, 0] = [0, 1, 0], the same point on `∞. 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] ∈ `∞. 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 `∞.

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 Note that as the line goes vertical, i.e. a → ±∞, the point a , 1, 0 ∈ `∞ 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 {(x, y, z) ∈ R3 : ax + by + cz = 0} for some a, b, c ∈ R not all zero. The lines in this plane are exactly {[x, y, z]: ax + by + cz = 0}, where we note that this equation is well- deﬁned: ax+by +cz = 0 iﬀ a(λx)+b(λy)+c(λz) = 0 for λ ∈ R∗. Thus the equation of a line in RP 2, i.e. a plane on ~0 in R3, is {[x, y, z]: ax+by+cz = 0}. In particular, `∞ has equation z = 0: `∞ = {[x, y, 0] : (x, y) 6= (0, 0)}.

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 `∞ is an affine plane. This is also not hard. For example, to show the parallel postulate, take 0 0 P 6∈ ` as usual. Let Q = ` ∩ `∞ and set ` = PQ. Then in A = P \ `∞, ` ∩ ` = ∅ 00 0 00 0 00 (why?). If P ∈ ` 6= ` in A and ` ` = ∅, then Q = ` ∩ `∞ 6= Q. But then 0 ` ∩ `∞ ⊃ {Q, Q }, 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. √ 2 2 2 2 For specific examples like k = C , Zp, Q , Q[ −5] , we can introduce homogeneous 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) ∈ k : `(x,y,z) = {λ(x, y, z) = (λx, λy, λz): λ ∈ k∗}. 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) ∪ {[x, y, 0] : (x, y) 6= (0, 0)}. 4CHAPTER 1. RELATING AFFINE GEOMETRY TO PROJECTIVE GEOMETRY

We define lines in kP2 to be planes on the origin in k3, i.e.{(x, y, z): ax + by + cz = 0} for fixed a, b, c ∈ 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 ∈ k and all b ∈ 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 `∞. 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 geometry 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 = {x = 0} ∩ S2 in S2. If β : {x = 1} → RP is the map from 2 the tangent plane {x = 1} analogous to α, then RP = β({x = 1}) ∪ C. Thus C is the line at inﬁnity from the {x = 1} plane’s point of view. The map

−1 2 2 β ◦ α : R \ ` → R certainly takes lines to lines, except that the map is not defined on `. We say that β−1 ◦ α has moved ` to the line at infinity. In particular, lines in {z = 1} that meet on ` map under β−1 ◦ α to parallel lines in {x = 1}. Of course, we can start with any line ` in the {z = 1} and replace the {x = 1} plane with the tangent plane to S2 parallel to α(`). Thus any line can be moved to infinity. The proofs of many concordance theorems in Euclidean geometry can be simplified by moving the right line to infinity. Strictly speaking, we draw the configuration of lines and points in the {z = 1} plane, and not that the theorem 2 holds iff it holds for α of the configuration in RP . In turn, the theorem holds 2 in RP if it holds in any tangent plane under the appropriate β−1 map. We can choose this second tangent plane, which we call the β-plane, as we wish. As an example, we prove Pappus’ Theorem from around 200 B.C. by moving a line to infinity. A direct proof is tricky.

Theorem. (Pappus’ Theorem) Let P1,P2,P3 ∈ `1 and Q1,Q2,A3 ∈ `2 Then

T = P1Q2 ∩ P2Q1,S = P1Q3 ∩ P3Q1,R = P2Q3 ∩ P3Q2 are collinear. 1.1. EMBEDDING AFFINE PLANES IN PROJECTIVE PLANES 5

Draw the picture. Set ` = RS. We have to show T ∈ `. Move ` to infinity, and call the points and lines in the new configuration (like −1 −1 β ◦ α(P ) and β ◦ α(`1)) by their old labels P1, `1, etc. In the β-plane, we have P1Q3kP3Q1,P2Q3kP3Q2. We want to show that T is now also on the line at infinity, i.e. P1Q2kP2Q1. Let U = `1 ∩ `2. (You do the case `1k`2.) We have ∆UP2Q3 ∼ ∆UP3Q2 and ∆UP1Q2 ∼ ∆UP2Q1, which implies |UP | |UP | |UP | |UP | 2 = 3 , 1 = 2 , |UQ3| |UQ2| |UQ2| |UQ1| From this we can easily conclude |UP | |UP | 2 = 1 , |UQ1| |UQ2| which implies P2Q1kP1Q2. Warning: If we try to be too clever and move U to infinity, we will no longer have P1Q3kP3Q1,P2Q3kP3Q2. Recall Desargues’ Theorem (ca. 1625): Theorem. (Desargues’ Theorem) Let ∆ABC and ∆A0B0C0 be in perspective from O (i.e. AA0,BB0,CC0 meet at O). Then R = AB ∩ A0B0,S = BC ∩ B0C0,T = AC ∩ A0C0 are collinear. [Draw a picture.] We can prove this by moving O to (the line at) infinity, as then we are in the easy case when AA0,BB0,CC” are parallel. Alternatively, we can move RS to the line at infinity (i.e. assume that ABkA0B0,BCkB0C0) and prove that 0 0 T ∈ `∞, (i.e. ACkA C ). In fact, the converse of Desargues’ Theorem holds: if R, S, T are collinear, then ∆ABC and ∆A0B0C” are in perspective. Again, we can reduce the proof to the easy case when R, S, T ∈ `∞. However, we’ll prove this using the classic perspective proof, which must have been the original motivation.

Proof. Think of ∆ABC in a plane P in R3. With our (one) eye placed at O, we want to paint ∆ABC as ∆A0B0C0 on our canvas, which is a conveniently placed plane P 0 of plexiglass. We ”paint” by setting A0 = OA ∩ P 0, etc. We claim that R, S, T ∈ ` = P ∩ P 0. (There is a special case when P kP 0.) O, A, A0,B,B0 determine a plane Q, so R = AB ∩ A0B0 ∈ Q. (There is a special case ABkA0B0, but at least these lines are not skew.) Since AB ⊂ P, A0B0 ⊂ P 0, we know R ∈ P ∩ P 0 = `. Similarly S, T ∈ `. Now we have seemed to have proven a harder theorem about triangles in perspective in R3. But appearances are deceiving, as our picture is actually in the plane of the page. In other words, since Desargue’s Theorem is a concordance theorem, our proof in R3 projects to a proof in the plane of the page. 6CHAPTER 1. RELATING AFFINE GEOMETRY TO PROJECTIVE GEOMETRY

The 3D proof also make the converse of Desargue plausible: if we swivel P keeping ` ﬁxed, or move our vantage point O but keep ∆ABC ﬁxed, we produce many examples of the converse.

1.2 Duality in Projective Geometry

Recall from the homework that the axioms of projective geometry are as follows:

Axiom 1: Two distinct points determine a unique line. Axiom 2: Two distinct lines determine a unique point. Axiom 3: (A nontriviality axiom)

Note that Axioms 1 and 2 switch if we switch ”point” and ”line.” This means that the statement of any theorem has a dual theorem, given by switching ”point” and ”line.” Since in practice we state the axioms as ”Two points join to a line, and two lines meet in a point,” our dual statement will involve switching ”join” and ”meet” as needed. Even better, if a theorem is derivable from the axioms (maybe with some additional assumptions to restrict the class of examples, such as assuming that the number of points is ﬁnite), then the dual theorem is also derivable, just by switching the use of ”point” and ”line” in each step of the proof. So we get theorems for free in projective geometry, which is a huge advantage over aﬃne geometries.

Example. 1. As a first example, we see that if we prove that a finite affine plane has n2 + n lines, then the corresponding projective plane has n2 + n + 1 lines. By duality, every finite projective plane then has n2 + n + 1 points. 2. Less trivially, let’s find the theorem dual to Desargues’ Theorem. A triangle is a configuration of three lines that meet in pairs. The dual configuration is three points which join in pairs, i.e. another triangle. So ∆ABC, ∆A0B0C0 determine dual triangles ∆LMN, ∆L0M 0N 0. More precisely, if the dual to A is the 0 0 line À, etc., then M = `B ∩ `C , etc. Similarly AA , the join of A and A , dual- 0 0 0 izes to the meet (i.e. intersection) of À and À0 . The statement ”AA ,BB ,CC all meet at the same point O” dualizes to ”the meet of À and À0 , the meet of `B and `B0 , and the meet of `C and `C0 all join on the same line Ò.” In other words, these three intersections are collinear. In other words, we have two dual triangles, such that the corresponding sides meet in three collinear points, which exactly play the role of R, S, T in the original Desargues setup. The conclusion of Desargues’ Theorem, that R = AB ∩ A0B0, S, T join on a single line, dualizes to (À ∩ `B) ∪ (À0 ∩ `B0 ) (where ∪ means join) and the other two lines meet on a single point. Untangling the notation, we get that LL0,MM 0,NN 0 meet at a point O0. Thus the dual of Desargues’ Theorem is the converse of Desargues’ Theorem, which we have now proved just by duality. 3. Sadly, you can check that the dual to Pappus’ Theorem is just Pappus’ Theorem. 1.3. CONICS AND DUALITY 7

1.3 Conics and Duality

After linear equations, the simplest equations are quadratic, those of the form

ax2 + by2 + cx + dy + exy = f = 0, (1) with at least one of a, b, e nonzero. It is well known that all conics have graphs which are either ellipses (including circles), parabolas, hyperbolas, and the ”degenerate” cases of two lines (e.g. x2 − y2 = 0), one line (x2 = 0), a single point (x2 + y2 = 0), or the empty set (x2 + y2 + 1 = 0). It is not hard to show that all quadratics occur as the intersection of a plane in R3 with a cone x2 + y2 = α2z2. Here α2 ∈ R+ determines the cone angle θ formed by the z-axis and the ”generating line” x = αz which lies on the cone, as clearly tan θ = α. In other words, for each conic (1), there exists an α, a plane P and coordinates (r, s) on P such that the intersection has the equation ar2 + bs2 + cr + ds + ers = f = 0 in P . Equivalently, the intersection of the plane and the cone projects to a quadratic equation in the xy-plane. (However, if P is vertical, we project into the xz or yz plane.) Of course, this is the reason graphs of quadratic are called conic sections. We leave the proof of this second claim to the reader. For the proof that all conics appear in P , we first intersect the cone x2 +y2 = z2 with a vertical plane. By the symmetry of the cone around the z-axis, we may assume that the plane P is y = h for some h ∈ R. The intersection has the equation x2 − z2 = h2, which in P with its natural (x, z) coordinates is either two lines (h = 0) or a hyperbola (h 6= 0). (So in this case we are identifying (r, s) with (x, z).) Clearly, by shifting the origin in P , we can produce hyperbolas 2 2 2 with linear terms, such as (x − x0) − (z − z + 0) = h , so we can pick up linear terms this way. We also pick up the cross term ers by rotating the coordinates. [The proof will be supplied later.] For nonvertical planes, [proof to be supplied later]. We claim that all nondegenerate conics embed via α to topological circles in 2 RP , at least once the points at infinity are added in. For any nondegenerate conic section hits each generating line exactly once, except that a parabola misses one generating line. [Prove this for hyperbolas given by nonvertical planes.] Thus α of a conic section equals α of a cross section circle, maybe with the addition of a point at infinity. In particular, a configuration of lines and a (from now on always nondegenerate) conic in a plane can be mapped into a configuration of lines and a circle, 2 either in the cross section plane or in RP . Thus any concordance theorem about lines and conics can be reduced to the case of lines and a circle. With this in mind, we might wonder if Pappus’ Theorem about six points on two lines (a degenerate conic) can be extended to nondegenerate conics. This is the content of Pascal’s Theorem (ca. 1600; note the 2000 years between the proofs of these two theorems!).

Theorem. (Pascal’s Theorem) Let P1,P2,P3,Q1,Q2,Q3 be six points on a 8CHAPTER 1. RELATING AFFINE GEOMETRY TO PROJECTIVE GEOMETRY conic. Set

R = P2Q3 ∩ P3Q2,S = P1Q3 ∩ P3Q1,T = P1Q2 ∩ P2Q1.

Then R, S, T are collinear.

Warning: We might think that the more difficult Pascal’s Theorem should imply Pappus’ Theorem. For if Pascal’s Theorem holds on the hyperbola x2 − y2 = , it is tempting to think it holds in the limit as → 0, i.e. on two intersecting lines. Moving the intersection to infinity gives Pappus for all lines. Thus in any affine plane with coordinates (i.e. the plane looks like k × k for some field k, Pascal’s Theorem would imply Pappus’ Theorem. Our proof of Pascal’s Theorem will depend on Desargues’ Theorem and a Lemma which involves angles. If there is a proof of the Lemma in any coordinate plane, then we would conclude that Desargues’ Theorem implies Pappus’ Theorem. But there exist coordinate affine planes (k∗, ·) nonabelian, and by Artin’s Geometric Algebra, in such affine planes Desargues’ Theorem holds and Pappus’ Theorem fails. Conclusion: At least in R2, we get a terrible proof involving conics that Desargues’ Theorem implies Pappus’ Theorem.