Notes on Advanced Geometry Dr. John Sarli Approximate Schedule of Topics:
Week 1 Affi ne transformations and conics
Week 2 Introduction to projective space
Week 3 Projective transformations; Fundamental Theorem of Projective Geometry
Week 4 Theorems of Desargues, Pappus, Brianchon; Cross-ratio
Week 5 Projective conics; tangents and secants; midterm exam
Week 6 Affi ne conics revisited
Week 7 Poles and polars; LaHire’sTheorem
Week 8 Standard forms; determination of conics
Week 9 Theorems on tangents and secants
Week 10 Pascal’sTheorem, its dual and converse
4 Conics Defined by Collineations
Let T be a collineation of the Euclidean plane. A basic theorem tells us that T is an affi ne transformation, which means it can be represented as a transformation of R2 in the form a b x p T (x, y) = + , ad bc = 0 c d y q − 6 The condition that δ = ad bc = 0 ensures that T is invertible, as every trans- formation must be. In fact,− 6
1 1 a b − x p T − (x, y) = c d y − q 1 1 = (dx by + bq dp) , (ay cx aq + cp) δ − − δ − − Let P be any point and consider the pencil of lines L concurrent at P . We define the conic at P afforded by T to be
(T,P ) = L T (L): P L E { ∩ ∈ } To understand this definition we should make the connection with the familiar description of conics by Cartesian curves. The fundamental principle that applies is as follows:
Suppose is a figure in the plane described by a Cartesian equa- tion f(x, y) =F 0. For any transformation T the image figure T ( ) is 1 F described by the Cartesian equation f (T − (x, y)) = 0.
Now suppose, for example,that the point P in the definition is O = (0, 0), the origin of the Cartesian plane. Any line L through O is a figure we want to transform by T . Since L is either the line x = 0 or is described by a Cartesian equation of the form y mx = 0 −
5 the image line T (L) is either 1 (dx by + bq dp) = 0 δ − − dx by = dp bq − − or is described by
(ay cx aq + cp) = m (dx by + bq dp) − − ay cx− aq + cp− m = − − dx by + bq dp − − For any given m we can find the intersection of L with T (L) and the collection of all such intersection points must satisfy ay cx aq + cp y = mx = − − x dx by + bq dp − − and so = (T,O) is the Cartesian curve E E cx2 + (d a) xy by2 = (cp aq)x + (dp bq)y − − − − We say is an ellipse if ∆ > 0, a parabola if ∆ = 0, a hyperbola if ∆ < 0, whereE
∆ = 4δ τ 2 − τ = a + d
This conic might be degenerate. For example, if
1 1 x 1 T (x, y) = + 0− 1 y 0 then is the curve y2 y = 0, which consists of the the two parallel lines y = 0 and yE= 1. Since ∆ = 0−, this is an example of a degenerate parabola.
First Graded Assignment
6 We say that T preserves orientation if δ > 0 and reverses orientation if δ < 0. If δ = 1 then T is an isometry (preserves distance between points). For any isometry± there is an angle θ such that
a b cos θ sin θ = , if δ = 1 c d sin θ −cos θ cos θ sin θ = , if δ = 1 sin θ cos θ − − a) If T reverses orientation show that is a hyperbola. E b) If T preserves orientation then can be any affi ne type. Find a transformation T for each affi ne type and sketch theE resulting conic. c) If T is an isometry that reverses orientation show that is a rectangular hyperbola (asymptotes are perpendicular, so eccentricity is √2E). Then provide a non-degenerate example and sketch the hyperbola with its asymptotes. d) If T is an isometry that preserves orientation show that is a circle unless θ = 0 (T is a translation) or θ = π (T is a half-turn). Find theE center and radius of the circle in the general case. What happens if T is a translation or half-turn? e) Find a collineation T such that the conic E 4x2 + 4xy + y2 + x + y = 0 is (T,O). Sketch the conic. Let L be the line y = x. Show where L and T (L) intersectE on . E
7 Degenerate Affi ne Conics
Degenerate affi ne conics are described qualitatively by examining the extreme cases of a plane intersecting a cone. For example, if the plane intersects the cone only at its vertex we obtain a single point. Since such an intersection is the limiting case of a plane that intersects the cone in an ellipse, a single point could be considered a degenerate ellipse. By similar reasoning, a single line could be a degenerate parabola since a plane tangent to the cone is the limiting case of a plane parallel to a generator of the cone; and a pair of intersecting lines could be a degenerate hyperbola, the limiting case of a plane that intersects both branches of the cone. What about a pair of parallel lines? To understand how this and the other cases occur we need to describe affi ne conics algebraically.
We know from algebraic geometry that any affi ne conic satisfies a Cartesian rela- tion of the form Ax2 + Bxy + Cy2 + Dx + Ey + K = 0 The set of points (x, y) that satisfy this relation may be empty, for example, x2 + y2 + 1 = 0. We will assume that the conic consists of at least one point. Then, without loss of generality we can assume K = 0, by applying a translation if necessary. Working with the equation
F (x, y) = Ax2 + Bxy + Cy2 + Dx + Ey = 0 where we assume not every coeffi cient is zero, we have at least the point (0, 0) on the conic.
When will (0, 0) be the only point on the conic?
Writing (Ax + D) x + (Cy + E) y = Bxy we see that if Ax + D = 0 has a non- − zero solution x0 then (x0, 0) is on the conic; similarly, (0, y0) is on the conic if Cy +D = 0 has a non-zero solution y0. If Ax+D = 0 has only x = 0 as a solution then D = 0, whereby A = 0. Finally, if Ax + D = 0 has no solution then A = 0 and D = 0; but then we6 can solve for x as a rational function of y and obtain 6
8 infinitely many points on the conic. Applying a similar analysis to the solutions of Cy + E = 0 we conclude that (0, 0) must be the only solution to
Ax2 + Bxy + Cy2 = 0 in other words, Ax2 + Bxy + Cy2 is a definite quadratic form.
Theorem. The conic F (x, y) = 0 consists of a single point provided D = E = 0 and 4AC > B2 Corollary. A single point is a degenerate ellipse.
What other degenerate geometric figures are possible? First, if A = B = C = 0 then Dx + Ey = 0 is a single line, which we would interpret to be a degenerate parabola since 4AC = B2. Otherwise, if the equation represents a degenerate form that is not a single point then F (x, y) must factor in one of the following ways: i) the square of a linear polynomial; ii) the product of distinct linear polynomials. Case i) again gives us a single line. Case ii) gives us a pair of lines but there are two possibilities: The two lines could be parallel or they could intersect. So there are three possibilities which we need to analyze to determine which type of conic each is a degenerate form of.
Case i): F (x, y) = (αx + βy)2, for some choice of α, β not both 0. Then A = α2, B = 2αβ±, C = β2 (and D = E = 0) so 4AC = B2, which is consistent± with F±(x, y) = 0 being± a degenerate parabola.
Case iia): F (x, y) = (αx + βy)(αx + βy + ν), for some choice of α, β not both 0 and ν = 0. This± is a pair of parallel lines, and expanding the product we see that the6 triple (A, B, C) is a non-zero multiple of (D2, 2DE,E2) with at least one of D,E non-zero. Again we have 4AC = B2, which is consistent with F (x, y) = 0 being a degenerate parabola, though we cannot obtain this figure as the intersection of a cone and a plane if the vertex of the cone is in R3. Note also that AE2 BDE + CD2 = 0 − 9 Exercise. If the cross product of Ai + Bj + Ck and D2i + 2DEj + E2k is 0 show that AE2 BDE + CD2 = 0. −
Case iib): F (x, y) = (αx + βy)(λx + µy + ν), which will be a pair of intersecting lines provided βλ = αµ. Expanding the product this time we find 6 4AC B2 = (αµ βλ)2 < 0 − − − AE2 BDE + CD2 = 0 − In particular, a pair of intersecting lines is a degenerate hyperbola. Conversely, these two conditions imply that the conic is a pair of intersecting lines since they allow us to factor F (x, y).
Apparently, the condition AE2 BDE + CD2 = 0 is equivalent to degeneracy for the conic F (x, y) = 0. This suggests− that this cubic form in the coeffi cients has an invariant representation. Let
2ABD Q = B 2CE DE 0 Then F (x, y) = 0 can be written
x x y 1 Q y = 0 1 We say that the symmetric matrix Q represents the affi ne conic. Further, since det Q = 2 (AE2 BDE + CD2) we have: − −
Theorem. The conic F (x, y) = 0 is degenerate if and only if det Q = 0.
Corollary. The conic (T,O) is degenerate if and only if f (p, q) = 0, where f (p, q) = cp2 + (dE a) pq bq2. Proof. We have − −
10 2c d a aq cp Q = d a −2b bq −dp aq − cp bq− dp −0 − − whose determinant is 2 (ad bc) f(p, q). Since ad bc = 0 the corol- − − − 6 lary follows.
In particular, (T,O) is degenerate if (p, q) = (0, 0), but (p, q) = T (0, 0) so in this case T (O)E = O. We will return to this special case, but first we ask, for given a, b, c, d if there are (p, q) = (0, 0) such that (T,O) is degenerate? Since f (p, q) is a quadratic form in p, q 6this can only happenE if ∆ 0 (otherwise f is a definite form). ≤
Consider first the case ∆ = 0. Then any (p, q) such that f (p, q) = 0 makes (T,O) a degenerate parabola. If the matrix part of T is not a dilation (a = d, b =Ec = 0), then the set of (p, q) such that f (p, q) = 0 is itself the degenerate parabola in R2, specifically, the line
(a d) p + 2bq = 0, if b = 0 − 6 2cp + (a d) q = 0, if c = 0 − 6 and so (T,O) is degenerate when T (O) is on this line. (If the matrix part is a dilationE then, as we will see, (T,O) is degenerate for every choice of (p, q).) For example, (T,O) is degenerateE for E 3 2 x p T (x, y) = + 2 1 y q − − when p + q = 0. For various points (p, q) on this line we obtain a family of degenerate parabolas (x + y)(p + 2x + 2y) = 0 each consisting of a pair of parallel lines, one of which is the line through (0, 0) and (p, q). When (p, q) = (0, 0) the two lines collapse to this single line.
11 1.0 y 0.8
0.6
0.4
0.2
•1.0 •0.8 •0.6 •0.4 •0.2 0.2 0.4 0.6 0.8 1.0 •0.2 x
•0.4
•0.6
•0.8
•1.0 (x + y) (1 + 2x + 2y) = 0
Now suppose ∆ < 0. Then f (p, q) = 0 is the degenerate hyperbola
a d √ ∆ p + 2bq = 0, if b = 0 − ± − 6 p (cp + (d a) q) = 0, if b = 0 − which is a pair of distinct intersecting lines in each case because b = 0 implies a = d. For example, (T,O) is degenerate for 6 E 1 2 x p T (x, y) = + 2 1 y q − 1 √ when q = 2 1 5 p. For various points (p, q) on these two lines through the origin we obtain− ± a family of degenerate hyperbolas 1 x2 xy y2 = p 5 √5 x 2√5y = 0 − − 4 ± ± One of these degenerate hyperbolas consists of the two lines that comprise the zeros of the quadratic form f :
12 5 y 4
3
2
1
•5 •4 •3 •2 •1 1 2 3 4 5 •1 x
•2
•3
•4
•5 x2 xy y2 = 1 √5 + 1 x + y 1 √5 1 x y = 0 − − 2 2 − − Finally, we look at the situation when T (O) = O, i.e., (p, q) = (0, 0), in more detail. We can understand this case in terms of eigenvectors of the matrix M = a b x because T (x, y) = M . The eigenvalues of M are c d y 1 τ √ ∆ 2 ± − If ∆ > 0 we have a degenerate ellipse consisting of only the point O. Every line L through O is taken to another line T (L) through O such that T (L) = L (otherwise we would have L T (L) = L and L would belong to (T,O) ). For6 example, M ∩ E could represent a rotation about O. This is consistent with the fact that M does not have real eigenvalues if ∆ > 0. If ∆ = 0 we saw that the degenerate parabola consisted of a single line. This means there is exactly one line L through O such that T (L) = L, corresponding 1 to the eigenvector for the single eigenvalue 2 τ. If ∆ < 0 we have a degenerate hyperbola consisting of two lines through O. This means there are two lines through O invariant under T , corresponding to the distinct eigenvalues 1 τ √ ∆ . 2 ± − 13 Empty Conics
Empty conics in R2 result from the fact that the real numbers are not algebraically closed. Our definition of the conic (T,P ) implicitly eliminates the empty conics. This is because (T,P ) containsE the points P and T (P ): If T (P ) = P this is clear; otherwise,E if L is the line through P and T (P ) then L T (L) contains 1 ∩ T (P ), and similarly, L0 T (L0) contains P if L0 = T − (L). This implies that we cannot have a collineation∩ T that takes every line through P to a parallel line without at least one line being invariant under T :
If a dilation about P is followed by a non-zero translation then the vector that produces the translation will be a direction vector for some line through P , whereby that line will be invariant.
For example, if
a 0 x p T (x, y) = + 0 a y q then (T,O) is the conic E qx = py a degenerate parabola that consists of the invariant line.
Eccentricity of Affi ne Conics
The eccentricity of a conic is a parameter that was originally derived to charac- terize the intersection of a plane with a cone obtained as a surface of revolution. Given an axis line, any line intersecting this axis is a generator of the cone obtained by revolving the generator about the axis. Let R be any plane perpendicular to the axis. For example, in R3 we might take the axis to be the z-axis and R to be the xy-plane. Suppose the line L is a generator and let S be any plane that we
14 intersect with the cone to obtain a conic. If α is the dihedral angle between R and S, and β is the angle between R and L then sin α ε = sin β π is defined to be the eccentricity of the conic. Note that it suffi ces to take 0 α 2 π ≤ ≤ and 0 < β < 2 . The affi ne type of the conic is determined by ε : 0 ε < 1 : ellipse ≤ ε = 1 : parabola ε > 1 : hyperbola Two affi ne conics are related by a similarity transformation if and only if they have the same eccentricity. In particular, all parabolas are similar. Hyperbolas of arbitrarily large eccentricity are possible, but for any given cone the eccentricity is bounded by csc β because sin α 1. For example, a right circular cone has π ≤ β = 4 so the largest eccentricity that can be obtained from slicing this cone is √2; the asymptotes of such a hyperbola are perpendicular. Eccentricity can be computed algebraically from the symmetric matrix Q that represents the quadratic part of the conic equation. We can write 2AB Q = B 2C Since Q is symmetric its two eigenvalues are real:
λ = A + C (A C)2 + B2 ± − q Note that the eigenvalues λ1 and λ2 are distinct unless A = C and B = 0, in which case the conic is a circle. The product of these two eigenvalues is det Q, the discriminant ∆. If the conic is not a hyperbola then ∆ 0 so the eigenvalues have ≥ the same sign (or one of them is 0). In this case we order them so that λ1 λ2 . Then it can be shown that | | ≤ | | λ ε = 1 1 λ r − 2 In particular, if ∆ = 0 then ε = 1, and we define, in agreement with our previous analysis, ε to be 1 if A = B = C = 0 and at least one of D,E is non-zero. If ∆ < 0 the situation is slightly more complicated because hyperbolas occur in conjugate pairs that share common asymptotes.
15 5 y 4
3
2
1
•5 •4 •3 •2 •1 1 2 3 4 5 •1 x
•2
•3
•4
•5 x2 3y2 = 1 and x2 3y2 = 1 − − − The eigenvalues of each Q are the same, but there are two eccentricities
λ1 1 λ ε = − 2 q λ2 1 λ − 1 q both > 1. They can be sorted out from the angle between the asymptotes where the hyperbola is situated. For example, in the above example we have λ1 = 1, λ2 = 3 so the two eccentricities are − 2 ε = √3 ε0 = 2
π The angle between the asymptotes of the red hyperbola is φ = 3 and the angle 2π 2 between the asymptotes of the blue hyperbola is π φ = 3 . Note that 1+cos π = − 3 2 ε and 2π = ε0. q 1+cos 3 q
16 Exercise. Look up the formula for the eccentricity of a hyperbola represented in standard form and use it to show that
2 ε = 1 + cos φ r where φ is the angle between the asymptotes.
17 The Real Projective Plane
2 RP is a planar geometry constructed from the subspaces of R3. A planar geometry consists of points and lines , and a rule that determines when a given point P 2 L is on a given line. In RP , the points are the 1-spaces (lines through the origin) 3 3 of R and the lines are the 2-spaces (planes through the origin) of R . If P1 and P2 are distinct members of then there is a unique L that contains both P ∈ L P1 and P2. Also, if L1 and L2 are distinct members of then there is a unique L P that is on both L1 and L2. Thus, the following two axioms apply to the real∈ projective P plane:
1) There is a unique line through any two distinct points. 2) Any two distinct lines intersect in a unique point.
A third axiom is usually assumed in order for the projective plane to have enough structure to contain the affi ne plane as a sub-geometry. This axiom is automati- 2 cally satisfied in RP :
3) There exist four points no three of which are collinear.
2 Definition. A quadrilateral is a set of four points in RP , no three of which are 1 0 0 1 collinear. The set X = 0 ,Y = 1 ,Z = 0 ,U = 1 is called the 0 0 1 1 quadrilateral of reference(qor).
2 The construction of RP from the subspaces of R3 is called the linear projective model. It has the advantage of using the notation of basic linear algebra. We will adopt the following notational conventions:
Points. Any member of will be written as a 3 1 projective x P × matrix, P = y , (x, y, z) = (0, 0, 0). The square brackets indicate z 6 that P is the 1-space consisting if all multiples of the vector with components xi + yj + zk.
18 Lines. Any member of will be written as a 1 3 projective matrix, L = a b c , (a, b,L c) = (0, 0, 0). The brackets× remind us to think of L as the plane with Cartesian6 equation ax + by + cz = 0. Thus, any non-zero multiple of ai + bj + ck is a normal vector for this plane.
It follows that the point P is on the line L if and only if LP = 0
The Cross-Product Map
Given p R3, consider the linear map x p x. If p = ai + bj + ck then the matrix representing∈ this map in terms of the7→ standard× basis is 0 c b c −0 a b a −0 − 2 It is occasionally useful to identify a point P of RP with a projective matrix that a takes any point Q = P to the line through P and Q. Thus, if P = b and 6 c x Q = y then the line LQ through P that contains Q is z
t LQ = Q P × 0 c b − where P × = c 0 a . Similarly, given the line L = a b c we can b a −0 − 0 c b − associate it with the matrix L× = c 0 a . Then the point L M, b a −0 ∩ − where M is any line distinct from L, is t PM = L×M
19 Projective Transformations
2 Any collineation T of RP has a natural representation as an invertible linear map from R3 to R3. Therefore it can be represented as a 3 3 non-singular matrix A with respect to the standard basis. However, any non-zero× multiple of this matrix also represents T , so we identify T with the projective matrix A, whereby
T (P ) = AP for any P . For any non-singular matrix M ∈ P 1 1 adj M − = M det M where M adj, the adjugate of M, is the transpose of the cofactor matrix for M. 1 Since A is a projective matrix it follows that T − is represented by the projective 2 matrix Aadj. The group of collineations of RP will be denoted P (2). Given L , how do we compute T (L) ? As in the affi ne plane, we use the fundamental∈ L principle of transformation: If point P is on the line T (L) 1 then T − (P ) is on the line L, so
1 LT − (P ) = 0 LAadj P = 0 adj T (L) = LA For example, suppose T is represented by
1 0 1 A = 2 3− 1 −2 3 1 and L = 5 4 2 . Then LAadj = − 0 3 3 5 4 2 4− 3 1 = 40 3 13 − 12 3 3 − − and so T (L) is the line represented by the 2-space 40x + 3y + 13z = 0.
20 Fundamental Theorem of Projective Geometry (FTPG)
Theorem. Given quadrilaterals ABCD and A0B0C0D0, there is a unique T ∈ P (2) such that T : ABCD A0B0C0D0. →
Proof. It suffi ces to show there is a unique collineation T1 : XYZU ABCD. → Then, there will also be a unique collineation T2 : XYZU A0B0C0D0, from → a1 b1 1 which T = T2T − : ABCD A0B0C0D0. So, let A = a2 ,B = b2 ,C = 1 → a3 b3 c1 d1 c2 ,D = d2 . Then c3 d3 λa1 µb1 νc1 M = λa2 µb2 νc2 λa3 µb3 νc3 λa1 µb1 νc1 a1 b1 c1 represents a collineation since λa2 µb2 νc2 = λµν a2 b2 c2 = 0. In fact, 6 λa3 µb3 νc3 a3 b3 c3
M represents the most general collineation that sends XYZ ABC . We want 7→ this collineation to send U D 7→ λ and so we can determine µ from the equation ν
adj λ a1 b1 c1 d1 µ = a2 b2 c2 d2 ν a3 b3 c3 d3 which is obtained from the vector equation λa + µb + νc = d Note that λµν = 0 since no three of the vectors a, b, c, d form a dependent set. 6 Thus, M is determined as a projective matrix and represents the collineation T1.
21 Exercise. Find all T P (2) that leave the qor invariant. ∈
22 Collinearity and Concurrence
a1 b1 Given two distinct points A = a2 and B = b2 , the line they determine a3 b3 is L = a2b3 a3b2 a3b1 a1b3 a1b2 a2b1 − − − c1 For any third point C = c2 note that c3 a1 b1 c1 LC = a b c 2 2 2 a3 b3 c3
It follows that C is on L precisely when this determinant is 0, which is just the triple-product test for the linear dependence of three vectors in R3.
Dually, given two distinct lines L1 = a1 b1 c1 and L2 = a2 b2 c2 , their intersection is the point b1c2 b2c1 − P = a2c1 a1c2 − a1b2 a2b1 − since the vector that spans this point is orthogonal to the normal vector that defines each line. For any third line L3 = a3 b3 c3 , a1 b1 c1 L P = a b c 3 2 2 2 a3 b3 c3
and so L3 is concurrent with L1 and L 2 precisely when this determinant is 0, that is, when the three normal vectors that define the lines are linearly dependent.
23 Perspective Transformations
Let T P (2) be represented by the matrix M : R3 R3. The image of any plane ∈ → Π is another plane Π0. If M :Π Π0 is an isometry for some choice of Π and Π0 7→ 2 then T is a perspective transformation of RP . Examples. 1) Let T be represented by an orthogonal matrix, for example
cos θ sin θ 0 M = sin θ −cos θ 0 0 0 1 3 For any plane Π in R , Π0 is the rotation of Π about the z-axis through the angle θ. Thus T is a perspective transformation since Π and Π0 are always isometrically related by M.
2) Let T be represented by
1 0 0 M = 0 1 0 0 0 c Not every plane is isometrically related to its image in this case, but if Π is parallel to the xy-plane then every point on Π has coordinates (x, y, a) for some given a. The image of this point by M is (x, y, ca), so Π0 is also parallel to the xy-plane. The distance between (x1, y1, a) and (x2, y2, a) is the same as the distance between their images in Π0. Thus T is a perspective transformation.
Not every T is a perspective transformation. However, it was discovered empiri- cally that every collineation can be factored into perspective transformations.
Theorem. Every T P (2) is the composite of three or fewer perspective trans- formations. ∈
Proof. For any non-zero vector x R3 let ∈ x x = u x | | 24 and for any two independent vectors x and y let x, y be the 2-space that they span. Let T be represented by the matrix M andh supposei
M : i, j, k a, b, c 7→
First, let M1 be the matrix such that
M1 : i, j, k a, b , c csc θ bu cot θ 7→ u u − where θ is the angle between b and c. Note that the second and third columns of M1 are orthogonal unit vectors.
Now M1 maps the plane x = 1 isometrically onto its image:
(i+p1j+q1k) (i+p2j+q2k) = (p1 p2) j+ (q1 q2) k | − | | − − | whereas
M1 (i+p1j+q1k) M1 (i+p2j+q2k) = | − | (p1 p2) M1j+ (q1 q2) M2k = | − − | (p1 p2) bu + (q1 q2)(cu csc θ bu cot θ) = (p1 p2) j+ (q1 q2) k | − − − | | − − | because bu and (cu csc θ bu cot θ) are orthogonal unit vectors as are j and k. −
Next, let M2 be the matrix such that
M2 : a, b , c csc θ bu cot θ a, b , c u u − 7→ u
Then M2 maps the plane
Π = a, b + cu csc θ bu cot θ h i − isometrically onto the plane Π0 = a, b + c h i because M2 fixes vectors in the 2-space a, b and so translates Π to Π0. h i
Finally, let M3 be the matrix such that
M3 : a, b , c a, b, c u 7→
25 Then M3 maps the plane a, c + bu h i isometrically onto the plane a, c + b h i Since we have M = M3M2M1 it follows that T is the composite T3T2T1 of the perspective transformations represented by these three matrices.
To find the actual matrices M1,M2,M3 it is important to represent them with respect to the same basis i, j, k. Writing vectors as columns we have
M = a b c a b q M1 = u where q = cu csc θ bu cot θ. Since T2T1 : i, j, k a, b , c we have − 7→ u
M2M1 = a bu c 1 M2 = a bu c a bu q − and since T3T2T1 : i, j, k a, b, c we have 7→
M = M3M2M1 = a b c 1 1 M3 = MM1− M2− 1 = M a bu c − The decomposition of a collineation into perspective transformations is not unique. The above procedure is one way to achieve it. For example, to find a decomposition 1 1 1 of T = 0 1 1 into three or fewer perspective transformations we would start 0 0 1 with a = i, b = i + j, c = i + j + k
26 Then b = 1 (i + j) and q = k, so u √2
1 1 0 √2 M = 0 1 0 1 √2 0 0 1 1 1 0 1 − M1− = 0 √2 0 0 0 1 1 1 1 1 1 0 1 0 1 √2 − M = 0 1 1 0 √2 0 = 0 1 1 2 √2 0 0 1 0 0 1 0 0 1 1 1 1 1 1 0 1 √2 1 √2 + 1 − − − M = 0 1 1 0 √2 √2 = 0 √2 √2 + 1 3 0 0 1 0 0− 1 0 0− 1 1 1 1 and 0 1 1 = M3M2M1. 0 0 1 Exercise. For each matrix factor, what pair of planes are related isometrically? Find Cartesian equations for these planes.
27 Desargues’sTheorem
2 Let ABC and A0B0C0 be triangles in RP such that the lines AA0,BB0,CC0 are concurrent, and let AB A0B0 = R,BC B0C0 = P,CA C0A0 = Q. Then the points P, Q, R are collinear.∩ ∩ ∩
Proof. First, note that if A = A0,B = B0 or C = C0 then the theorem is trivial, so assume A0 = A, B0 = B,C0 = C. Note that ABC and the point of concurrence O form a quadrilateral6 6 since6 the intersections of corresponding sides of the two triangles are single points. Thus, by the FTPG we can let ABCO = XYZU. Since p the line through X and U is 0 1 1 we have A0 = 1 for some choice of − 1 p because A0 = A. Similarly, the line through Y and U is 1 0 1 so we have 6 − 1 B0 = q for some choice of q, and the line through Z and U is 1 1 0 so 1 − 1 we have C0 = 1 for some choice of r. It is now straightforward to find P, Q, R r since the lines through X and Y , Y and Z, and Z and X are, respectively, 0 0 1 1 0 0 0 1 0 and the lines through A0 and B0, B0 and C0, and C0 and A0 are, respectively,
1 q 1 p pq 1 − − − qr 1 1 r 1 q − − − 1 r pr 1 1 p − − − 0 1 p p 1 Thus, P = 1 q , Q = −0 , R = 1 − q which are seen to be r − 1 r 1 −0 − − collinear by the determinant test.
28 Exercise. Prove the converse of Desargues’s Theorem: Let ABC and A0B0C0 2 be triangles in RP such that the points AB A0B0,BC B0C0,CA C0A0 are ∩ ∩ ∩ collinear. Then the lines AA0,BB0,CC0 are concurrent.
Note that the converse of Desargues’s Theorem is also its dual. Consequently, Desargues’s Theorem can be stated more generally, which is closer to the way Desargues discovered it:
Two triangles are in perspective from a point if and only if they are in perspective from a line.
2 The model RP for the real projective plane can be generalized by using coor- 2 dinates from any field. For example, CP is a model for the complex projective plane, the natural domain for the study of algebraic curves. Desargues’sTheorem is true for all of these projective planes. It also remains true when the coordi- nates come from a skew field such as H, the quaternions, for which multiplication is associative but not commutative. In fact, Desargues’s Theorem is true for a projective plane if and only if the plane can be coordinatized by a field or skew field.
Corollary to Desargues’sTheorem. Let ABC be a triangle in the affi ne plane and let LA,LB,LC be cevians through the respective vertices with LA BC = ∩ A0,LB CA = B0,LC AB = C0. Then AB A0B0,BC B0C0,CA C0A0 are ∩ ∩ ∩ ∩ ∩ collinear if and only if LA,LB,LC are concurrent.
What is the interpretation of this corollary if, for example, BC B0C0 ? What k does the corollary say if LA,LB,LC are the medians of ABC ?
29 Pappus’sTheorem
Let A, B, C be distinct points on line L and let A0,B0,C0 be distinct points on line L0 = L. Let P = BC0 B0C,Q = CA0 C0A, R = AB0 A0B. Then P, Q, R are collinear.6 ∩ ∩ ∩ Proof. We assume L L0 / A, B, C, A0,B0,C0 so that P, Q, R are distinct. Then: ∩ ∈ { }
A, A0,Q cannot be collinear, otherwise C0 = A0 and C = A. A, A0,R cannot be collinear, otherwise B0 = A0 and B = A. A, Q, R cannot be collinear: AQ L0 = C0 and AR L0 = B0, so if ∩ ∩ AQ = AR then C0 = B0. Similarly, A0, Q, R cannot be collinear.
It follows that AA0QR is a quadrilateral, so by the FTPG, can assume AA0QR = XYZU. r The line AR can now be represented by 0 1 1 , and so B0 = 1 for some − 1 1 choice of r. Similarly, A0R is represented by 1 0 1 , and so B = s − 1 for some choice of s. To find P , we first note that C = AB A0Q which is ∩ 0 represented by the intersection of 0 1 s and 1 0 0 . Thus C = s . − 1 Similarly, C0 = A0B0 AQ, the intersection of 1 0 r and 0 1 0 . Thus ∩ − r C0 = 0 . Finally, P = BC0 B0C, the intersection of s r 1 rs and 1 ∩ − − rs 1 s r rs . Therefore, P = rs . Since − − r + s 1 − rs 0 1 rs 0 1 = 0
r + s 1 1 1 −
30 we conclude that P, Q, R are collinear.
Corollary. In the affi ne plane, if AB0 A0B then PQ belongs to their parallel pencil. k
Pappus’sTheorem holds in a projective plane if and only if the plane can be coor- 2 2 dinatized by a field. Thus, it holds in CP but not in HP , because multiplication is not commutative in H, a skew field. The dual of Pappus’sTheorem is known as Brianchon’sTheorem.
Let L, M, N be distinct lines concurrent at P and let L0,M 0,N 0 be distinct lines concurrent at P 0 = P . Let 6 (M N 0)(M 0 N) = A, (N L0)(N 0 L) = B, (L M 0)(L0 M) = C. Then the∩ lines A,∩ B, C are concurrent.∩ ∩ ∩ ∩
Brianchon’s Theorem is not the converse of Pappus’s Theorem, which would state that if the points P, Q, R produced from the bipartite intersection of two triples of points are collinear then each triple is collinear. This converse statement is not true in general. The most general arrangement of the six points is one of the classical theorems and will be proved after the study of projective conics.
31 Brianchon’sTheorem
In the figure above, the black lines are concurrent at some point P 00. Is it possible for P 0,P 0,P 00 to be collinear? Dually, is it possible for the lines ABC, A0B0C0, and P QR to be concurrent? Yes, and this special case is related to the next topic.
32 Cross-Ratio and Embedding Planes
In the affi ne plane it makes sense to define ratios along a given line but this definition may not agree with ratios determined on another line. For example, in perspective drawing one might transfer locations on a given line to locations on another line using a point of perspective. In the figure below, the x-axis is in perspective with the line x + y = 10 from the point (3, 4). Note, however, that if O = (0, 0) ,P = (2, 0) ,Q = (4, 0) then
OP 1 = OQ 2 but O P 3 0 0 = O0Q0 7
10 y 9
8
7
6
5
4
3
2
1
0 0 1 2 3 4 5 6 7 8 9 10 x
Perspective from (3, 4) in R2
Since affi ne collineations preserve ratio even if the matrix part is not a dilation this means that a perspective relation betwen two lines generally does not extend to a collineation of the plane. This raised a problem in perspective drawing that
33 was remedied by the concept of cross-ratio, an idea that is easily described by first defining it in the projective plane.
If A, B, C, D are collinear points in RP 2 (we will assume they are distinct) then the vectors that span the 1-spaces representing them are linearly dependent. In fact, any two of these vectors span the 2-space that represents the line. Let A = [a] and B = [b]. Then
C = [αa + βb] D = [γa + δb] for some choice of coeffi cients α, β, γ, δ. The ratio βγ (ABCD) = αδ is independent of the vectors we choose to span the 1-spaces because any non-zero multiples would cancel out. We call (ABCD) the cross-ratio of the ordered set of points A, B, C, D. Since there are 4! = 24 permutations of the four points the cross-ratio could have 24 different values for four given collinear points. However, since the group of permutations is generated by transpositions it is easy to show that if (ABCD) = χ then
1 (BACD) = (ABDC) = χ (ACBD) = (DBCA) = 1 χ − It follows that each value is stable under at least four of the permutations, so there are at most six different values. Therefore, for most purposes we consider the cross-ratio to be an equivalence class of values
1 1 1 χ [χ] = χ, , 1 χ, , 1 , χ − 1 χ − χ χ 1 − − 1 1 2 3 1 For example, [3] = 3, 3 , 2, 2 , 3 , 2 . The cross-ratio [ 1] = 1, 2, 2 is unique for having only three{ members− − in the} equivalence class.− Points{− that produce} this cross-ratio are said to form a harmonic set.
Cross-ratio is invariant if we apply any collineation T P (2) since T is represented by a projective linear transformation: ∈
34 (ABCD) = (A0B0C0D0), where P 0 = T (P ) for any point P .
The invariance is easily proved using elementary linear algebra.
Theorem. If two sets of four points, collinear on distinct lines, are in perspective from a point then they have the same cross-ratio.
Proof. Let A, B, C, D be distinct collinear points and let A0,B0,C0,D0 be distinct and collinear on a different line, with AA0,BB0,CC0,DD0 concurrent at E. We will show that this configuration determines a collineation T P (2). Since cross-ratio ∈ is a projective invariant it will follow that (ABCD) = (A0B0C0D0). First, note that BCB0C0 and B0C0BC are quadrilaterals and so, by the FTPG, there is a unique collineation
T : BCB0C0 B0C0BC 7→ 1 Note also that T is an involution (T = T − ) because the composition of T with itself leaves the quadrilateral BCB0C0 fixed and so, again by the FTPG, this composition is the identity transformation. In particular, T interchanges the lines BC and B0C0 and so their point of intersection is fixed. Call this point F . Also, the lines BB0 and CC0 are invariant under T so the point E is also fixed. (Note that F is fixed because it is the intersection of lines that are interchanged, whereas E is fixed because it is the intersection of lines that are invariant.) Let P = T (A). Then P is on B0C0 since T is a collineation. Let G = AP BB0 ∩ and H = AP CC0. Then G and H are fixed by T , each being the intersection of a pair of invariant∩ lines. (T is an involution so AP is an invariant line.) We now know that E, F, G, H are all fixed points, so if they form a quadrilateral T would be the identity transformation, which it is not. But EF GH is a quadrilateral unless P = A0:
If H is on EG then H = EG CC0 = E in which case P = A0. ∩ If H is on EF then H = EF CC0 = E in which case P = A0. ∩ Similarly, G on EF implies P = A0. If F is on GH then F = A = A0 which contradicts the fact that the points are distinct.
Thus T (A) = A0 and, similarly, T (D) = D0. It follows that (ABCD) = (A0B0C0D0).
35 The assignment A A0,B B0,C C0,D D0 is called a perspectivity. We say that the two sets7→ of points7→ are in7→ perspective7→ from the point E.
Corollaries to the proof. 1) Any perspectivity extends uniquely to a collineation, in fact, an involution. 2) Let T be an involution that takes two vertices V1,V2 of a quadrilateral to the other two vertices V3,V4, respectively, and let Q = L M where L = V1V3 and ∩ M = V2V4. If P is on either V1V2 or V3V4 then the line through P and T (P ) passes through Q.
Theorem. The involution determined by a perspectivity is a perspective transfor- mation.
Proof. First, note that the lines XY and ZU are in perspective from the point 1 E = 0 . This perspectivity extends to the involution 1 0 1 1 T = 0 1− 0 1 1 0 − 0 1 1 The matrix M = 0 1− 0 acts isometrically on the embedding plane 1 1 0 − x y + z = 2 because this plane is − v, w + p h i where p = i + k, a vector fixed by the matrix, and v = i + j and w = i k − are eigenvectors each with eigenvalue 1. In general, suppose the involution T 0 is determined by the perspectivity P QRS− RSPQ. By the FTPG there is → T1 P (2) such that T1 : P QRS XYZU. Then ∈ → 1 T 0 = T1− TT1
1 3 1 1 and the matrix M 0 = M1− MM1 representing T 0 on R fixes M1− p and has M1− v 1 and M1− w as eigenvectors each with eigenvalue 1. Thus M 0 acts isometrically on the embedding plane −
1 1 1 M1− v,M1− w + M1− p
36 Cross-ratio can be used to parameterize points on a line in relation to three given points on the line, because (ABCX) = (ABCY ) if and only if X = Y . As a con- sequence, if A, B, C, D and A, E, F, G are two sets of collinear points on different lines and (ABCD) = (AEF G) then the lines BE,CF,DG are concurrent. This observation provides:
Another Proof of Pappus’sTheorem. Let V be the intersection of the two given lines. Let D = BA0 AC0 and ∩ E = BC0 CA0. Then V,A0,B0,C0 are in perspective with B,A0,R,D from A, ∩ and in perspective with B,E,P,C0 from C. Thus
(VA0B0C0) = (BEPC0) = (BA0RD) and so A0E,RP,DC0 are concurrent lines, at Q because A0E DC0 = Q. Thus ∩ RP contains Q since A0E = A0C and DC0 = AC0.
Exercise. Show that the lines ABC, A0B0C0 and P QR are concurrent if and only if the lines AA0, BB0 and CC0 are concurrent.
Affi ne Interpretation of Cross-Ratio
The idea of cross-ratio was originally used to assign affi ne coordinates to points on a line. Affi ne transformations preserve ratios on lines even though they do not in general preserve distance. This is related to the usual parameterization of a line in terms of vectors. Suppose vectors a, b, c, d represent the collinear points A, B, C, D in R2. If we write c = λa + (1 λ) b − d = µa + (1 µ) b − PQ and write QR for the signed ratio of distances between any three collinear points 1 λ AC 1 µ AD P, Q, R, then −λ is the ratio CB and −µ is the ratio DB , and these ratios were
37 interpreted as the affi ne coordinates of C and D relative to the points A and B. Further, µ (1 λ) (ABCD) = − λ (1 µ) − AC DB = CB AD Thus, the cross-ratio in the affi ne plane, along with its permuted values, can be calculated simply from a consistent assignment of a unit of distance on the given line. Note that the ratios are oriented and so must be interpreted consistently as signed values. The affi ne coordinate 0 indicates that the point to be located is one of the given points; for example, λ = 1 implies C = A. The connection with projective geometry arises when we note that, for example, λ = 0 implies C = B. The affi ne coordinate is undefined in this case, whereby it is convenient to assign it the value . ∞ More generally, assume that A, B, C, D are points on an affi ne line obtained as 2 images in an embedding plane of points on a line of RP . If the projective line were the 2-space parallel in R3 to the embedding plane then there would be no affi ne line containing these points. If the projective line is not parallel to the embedding plane, exactly one of the points on the projective line is the ideal point for the corresponding affi ne line. Without leaving the affi ne line we can still compute the cross-ratio when one of A, B, C, D is this ideal point. The computation is consistent with our idea of limits for rational functions; in particular, we must be consistent with the signed ratios:
AC DB (ABCD) = CB AD DB = , if A ideal CB AC = , if B ideal AD BD = , if C ideal AD AC = , if D ideal BC Returning to the simple perspective relation between two affi ne lines, we can use cross-ratio to answer a basic question that simple ratios cannot address. In our
38 figure, suppose point R0 is on the segment between O0 and Q0 and the distance from O0 to R0 is √2. How far is R from O? A simple proportion will not solve this problem but we know that R0 is the unique point such that 49 (O0P 0R0Q0) = 22 so R must be such that (OR)(QP ) 49 (OPRQ) = = (RP )(OQ) 22
Apparently R is between P and Q (why?) so RP = OP OR. Thus − 49 (OR)( 2) = − 22 (2 OR) (4) 49− OR = 19
10 y 9
8
7
6
5
4
3
2
1
0 0 1 2 3 4 5 6 7 8 9 10 x
49 R is the point 19 , 0 Exercise. Let be an affi ne hyperbola or an ellipse that is not a circle. On the line throughE the foci let C be the center of the conic, A be one of the foci,
39 B be the vertex closer to A, and let D be the ideal point. Show that (ABCD) is the eccentricity of the conic. How could cross-ratio be used to determine the eccentricity of a circle or parabola?
Important Skills for Midterm Exam
This exam will cover topics in I, II, III below.
2 I. Points and Lines of RP Representation of projective points and lines; determine incidence Find line determined by collinear points Find point of intersection of concurrent lines
II. Projective Collineations Find image of a point, T (P ), and of a line, T (L), for T P (2) Find the inverse of a projective transformation ∈
III. Basic Theorems and Their Affi ne Interpretations Fundamental Theorem of Projective Geometry: Determine a projective transformation from a quadrilateral and its image Desargues’s Theorem: Determine whether two triangles are in perspective; find the perspective point or line IV. Pappus’s Theorem: Find Pappus line for two collinear triples Duality and Brianchon’sTheorem Perspectivity: Find the collineation that extends a given perspectivity
V. Cross-Ratio 2 Calculation in RP Invariance under projective transformation Effect of permutation of points Calculation in affi ne plane
40 Projective Conics
2 Let P be a point of RP and let T P (2). ∈
The conic at P afforded by T is
(T,P ) = L T (L): P L E { ∩ ∈ }
As with the definition of affi ne conics, this projective conic is never empty because it always contains P and T (P ). If T is the identity collineation then (T,P ) would contain every line through P and hence be the entire projective plane;E note that this was the case in the affi ne plane when T was a dilation about P with no translation component. We will say that (T,P ) is undetermined if it consists of 2 E the entire plane. In RP it is also possible for (T,P ) to be the entire plane if T is not the identity but P is a fixed point of T . WeE give an example below, but first 1 1 1 1 we carry out the construction for the example P = 0 and T = 0 1 1 , 1 0 0 1 2 whereby T (P ) = 1 . Since T (L) = LT adj it follows that L T (L) is found by 1 ∩ taking the cross-product of L and LT adj. Since any line through P can be written as L = p q p for some choice of p, q we have − 1 1 0 T (L) = p q p 0− 1 1 − 0 0− 1 = p p + q p q − − − 0 p q p p2 + q2 t Then L T (L) = L×T (L) = p 0 p p + q = pq and so ∩ −q p −0 −p q −p2 − − − p2 + q2 (T,P ) = pq : p2 + q2 = 0 E − 2 6 p 41 In particular, P can be obtained by setting q = 0, and T (P ) by setting q = p. By setting p = 0 we see that X is on this conic. We can describe all other points− on the conic in terms of a single parameter t by first writing 2 2 p q p + q q + p pq = 1 − 2 −p p q p Now let t = q and we have 1 2 t + t t + 1 1 = t −t −t2 As t varies we obtain every point on the conic, except for P because that would require t = 0 which gives us the point X. This is typical of so-called rational curves in the projective plane, of which conics are an example: They can be parameterized by a single parameter t but no single parameterization will catch every point on the curve; for conics, any parameterization by polynomials will miss one point. Now that we have a parametric description of (T,P ) we can describe it as a sur- E face in R3. But this surface consists of 1-spaces, lines through the origin, which is why we needed only one parameter. We suspect this is a quadric surface since the coordinates are quadratic functions of t. A quadric surface consisting of lines in R3 through a single point in R3 is called a cone, which is why we call (T,P ) a projective conic. We often describe quadric surfaces with a single CartesianE equa- tion in x, y, z. To find such an equation we look at the most general quadratic polynomial that could describe the cone, necessarily homogeneous quadratic be- cause if (x, y, z) is on the cone then so is (λx, λy, λz) for any λ. This homogeneious polynomial is Ax2 + Bxy + Cy2 + F xz + Gyz + Hz2 = 0 Substituting x = 1 + t2, y = t, z = t2 we obtain − t4 (A + F + H) t3 (B + G) + t2 (2A + C + F ) Bt + A = 0 − − This must be an identity in t so the coeffi cients must all be zero. Solving this linear system yields A = B = G = 0 H = C = F = 0 − 6 42 Thus, the Cartesian equation for the cone is
: y2 + z2 xz = 0 E − By choosing embedding planes we obtain various affi ne representations of the projective conic . E
4
2
•4 •2 0 •4 0 0 •2 2 2 4 4
•2
•4
z = 1 : Affi ne Parabola; z 2x = 4 : Affi ne Ellipse E ∩ { } E ∩ { − }
Finding the Cartesian equation from a parametric equation is straightforward but ineffi cient for most purposes. We will interpret the construction of (T,P ) so as to obtain a direct matrix representation of the projective conic. First,E consider 1 1 1 what can happen if T (P ) = P . Let P = X and let T = 0 1 0 . Then 0 0 1 1 1 1 T (P ) = P and T adj = 0− 1− 0 . Every line through P is L = 0 p q 0 0 1 for some p, q. However, LT adj = L in this case so L T (L) = L for every line through P , that is, (T,P ) is undetermined. This does∩ not always happen if P is fixed by T . E
43 Exercise. Let P = X and let T P (2). ∈ a) If T leaves three distinct lines through P invariant show that it leaves every line through P invariant. 1 0 0 2 b) If T = 0 1 0 show that (T,P ) consists of a single line in RP . 0 1 1 E 1 0 0 2 c) If T = 0 1 1 show that (T,P ) consists of a single point in RP . 0 1− 1 E 1 1 0 2 d) If T = 0 1 1 show that (T,P ) consists of two distinct lines in RP . 0 0 2 E 1 0 0 2 e) If T = 0 1 0 show that (T,P ) consists of two distinct lines in RP . 1 0 1 E Note that in this case T (P ) = P . 6
Since P and T (P ) are both on (T,P ), the conic cannot consist of a single point unless P is fixed by T . If T (P )E= P , let L be the line through these two points. Suppose that T (L) = L, so that6 L is contained in (T,P ). Let Q be any point not on L and let M be the line through P and Q.E Then T (M) = L because L is invariant and T is invertible. Thus, the point M T (M) is not6 on L and so (T,P ) does not consist of the single line L, in fact,∩ it consists of two distinct Elines. (See Graded Assignment 2, below.)
Now suppose T (P ) = P and T (L) = L. Without loss of generality we can assume P = X, T (P ) = Y6 , and T (Y ) = 6 Z. Then the most general projective matrix representing T is
0 0 ν λ 0 b 0 µ c with λµν = 0. Any line through P is of the form 0 p q and the image of 6
44 this line by T is qµ cp 0 pν . These two lines intersect at the point − p2ν E = q (qµ cp) p (qµ− cp) − − When p = 0 we have E = T (P ). For all other points we can divide by p2 and set q p = t to obtain the parameterization ν E(t) = (µt c) t c −µt − If (T,P ) contained a line then there would be three distinct collinear points E E (t1) ,E (t2) ,E (t3). But the determinant obtained from three such points is
2 µ ν (t1 t2)(t2 t3)(t3 t3) − − − which is not zero unless two of the points are identical. A similar determinant calculation shows that T (P ) is not collinear with any two other points on the conic. Putting this all together we have proved the following theorem.
Theorem. If T (P ) = P then (T,P ) is either a single point, a single line, two distinct lines, or is undetermined.E If T (P ) = P but the line L through P and T (P ) is invariant, then (T,P ) consists of two6 distinct lines. If T (P ) = P and T (L) = L then (T,P ) doesE not contain any line. 6 6 E
The theorem allows us to focus primarily on the case where T (P ) = P and T (L) = L, though our development of the representation of projective conics6 will remain6 general. The other cases are examples of degenerate conics, for which we will develop a recognition theorem.
Extending Affi ne Conics
Recall the construction of the affi ne conic at P afforded by T . If P = O = (0, 0) a b x p and T (x, y) = + then the Cartesian equation for (T,O) c d y q E is
45 cx2 + (d a) xy by2 + (aq cp)x + (bq dp)y = 0 − − − − If we view the Cartesian plane as the embedding plane z = 1, the point O cor- 0 2 responds to 0 in RP and the corresponding projective collineation is repre- 1 sented by a b p T = c d q 0 0 1 The pencil through O consists of the lines L = s t 0 and T (L) = d b bq dp s t 0 c− a cp − aq −0 0 ad− bc − = ct ds bs at (aq cp) t + (dp bq) s − − − − t (aqt bqs cpt + dps) Then L T (L) = s (aqt− bqs− cpt + dps) . If s = 0 then t = 0 and we are ∩ − bs2 −ct2 −ast + dst 6 − − cp aq at the point −0 . Otherwise, the points on the conic are parameterized c by (cp aq) t2 + (bq dp) t (aq− cp) t + (dp− bq) ct−2 + (a d) t −b − − Exercise. Show that these points satisfy the homogeneous equation
cx2 + (d a) xy by2 + (aq cp)xz + (bq dp)yz = 0 − − − − The ideal points of this conic relative to the embedding plane z = 1 are found by setting z = 0 to obtain cx2 + (d a) xy by2 = 0 − −
46 Exercise. Let ∆ be the discriminant of this quadratic form. Show that there are no ideal points if ∆ > 0, there is one ideal point if ∆ = 0, and there are two ideal points if ∆ < 0.
Transformation of Conics
2 Given a conic in RP and a collineation S P (2) we would expect that S( ) is also a conic. ThisE follows from the general transformation∈ principle for a figureE in R3 described by a Cartesian equation. For example, let be the projective conic E y2 + z2 xz yz = 0 − − and let 0 1 1 S = 1 0 1 −1 3 2 − − x 3 5 1 1 adj − If the point P = y is on S( ) the S− (P ) is on . Since S = 1 1 1 z E E 3 1− 1 − the homogeneous triple 3x 5y + z x +− y z 3x y−+ z − must satisfy the Cartesian equation for . After substituting and simplifying we find that S( ) is the conic E E x2 6xy + y2 z2 = 0 − − 1 Note that S( ) cannot be degenerate unless is, for otherwise S− would trans- form a degenerateE conic into a non-degenerateE conic, and this is not possible since 1 a degenerate conic contains a line and S− is a collineation.
Symmetric Matrix Representation of Conics
47 There is a more direct way to obtain the transformation of a conic. Since is a E homogeneous quadric in R3 it is the zeros of a Cartesian equation Ax2 + Bxy + Cy2 + F xz + Gyz + Hz2 = 0 which can be written in matrix form 2ABF x x y z B 2CG y = 0 FG 2H z Thus can be identified with the projective symmetric matrix E 2ABF Q = B 2CG FG 2H and we can abbreviate its Cartesian equation as P tQP = 0 x where P = y . Now if S P (2) and P is a point on then S(P ) is on S( ). z ∈ E E We can summarize this relation as follows: t S(P ) Q0S(P ) = 0 where Q0 is the symmetric matrix representing S( ). To find Q0 note that S(P ) is the matrix product SP and so S(P )t = P tSt whichE yields
t t P S Q0S P = 0
t and this equation holds for all P if and only if S Q0S = Q . It follows that ∈ E
t 1 1 adj Q0 = S − QS− = ScoQS
adj t where Sco = S , the cofactor matrix of S. For example, if is the conic E 0 0 1 0 1 1 y2 + z2 xz yz = 0 then Q = 0 2 −1 . If S = 1 0 1 then − − 1 1− 2 −1 3 2 adj − − − − Q0 = ScoQS =
48 3 1 3 0 0 1 3 5 1 1 3 0 5 1 1 0 2 −1 1− 1 1 = 3− 1 0 −1 1− 1 1 1− 2 3 1− 1 −0 0 1 − − − − − corresponding to the Cartesian form x2 6xy + y2 z2 = 0, in agreement with the previous calculation. − − Now we can put this all together with our original definition of as (T,P ). E E First, note that if we choose specific matrix representations of T and P × then
adj P ×T TcoP × − t is a symmetric matrix. If A is a point other than P then A P × is the line LA through A and P , so t adj adj A P ×T = LAT = T (LA)
If A is also on T (LA), so that A is a point on (T,P ), we can write E t adj A P ×T A = 0
t adj t adj t and so A P ×T A + A P ×T A = 0. Since P × is anti-symmetric we now have
t adj A P ×T TcoP × A = 0 − We have proved the following theorem:
adj Theorem. Let Q = P ×T TcoP ×. Then Q is a symmetric matrix that represents (T,P ). − E
1 1 1 1 Consider again the example with P = 0 and T = 0 1 1 . Then 1 0 0 1 adj P ×T TcoP × = − 0 1 0 1 1 0 1 0 0 0 1 0 1− 0 1 0− 1 1 1 1 0 1− 0 1 0 1− 0 0 0− 1 − −0 1 1 0 1− 0 − 0 0 1 = 0 2 0 = Q 1− 0 2 − 49 0 0 1 The corresponding projective matrix Q = 0 2 0 is identified with the 1− 0 2 − conic (T,P ), and so we have obtained the Cartesian equation y2 + z2 xz = 0 withoutE having to parameterize the cone. − 1 Exercise. If = (T,P ) and S P (2) show that S( ) = (STS− ,S(P )). E E ∈ E E
50 MATH 529-01: Second Graded Assignment
Select one of the following and present a clear, complete solution.
1. Let T P (2) be represented by the projective matrix ∈ 1 a b 0 1 c 0 0 1 Express T as the composition of three or fewer perspective transformations. Ex- plain why each factor is a perspective transformation by exhibiting the planes in R3 that are related isometrically. Suggestion: To simplify computation, set ν = a2 (1 + c2) 2abc + b2 + 1 and note that ν 1. − ≥ 2. If T (P ) = P and the line L through P and T (P ) is invariant, show that (T,P ) consists6 of two distinct lines. Suggestion: Without loss of generality youE can assume P = X and T (P ) = Y , so that L = 0 0 1 . Find the most general projective matrix representing T and carry out the locus construction to find (T,P ). Show that the resulting parameterization yields L and another line. E 3. Let ABC be an affi ne triangle and L a line not through any vertex. If P = L BC,Q = L CA, R = L AB then a result often attributed to Menelaus of Alexandria,∩ who lived∩ about 400∩ years after Euclid, says that AR BP CQ = 1 RB PC QA − Use this result along with Desargues’sTheorem to prove Ceva’sTheorem:
Let LA,LB,LC be cevians of ABC, with A0 = LA BC,B0 = ∩ LB CA, C0 = LC AB. If LA,LB,LC are concurrent then ∩ ∩ AC BA CB 0 0 0 = 1 C0B A0C B0A
51 In your proof you should set up perspectivities, from the points that determine the Desargues line, that relate the lines through the vertices of ABC. The equality of cross-ratios will show that the four points on each side form a harmonic set. Then use the result of Menelaus.
2 4. Let A, B, C be distinct points on the line L in RP . Let V be a point not on L and let P be a point on line VC other than V or C. Let Q = AP VB, ∩ R = BP VA, and let C0 = L QR . Show that C0 does not depend on the choice of V∩ and P . Suggestion:∩Without loss of generality, let A = X,B = Y and compute the cross-ratio (ABCC0).
5. In Pappus’s Theorem, let V = L L0. Show that the Pappus line (the line ∩ through P, Q, R) contains V if and only if the lines AA0,BB0,CC0 are concurrent. Suggestion: Set up triangles in perspective from V and use Desargues’sTheorem.
6. Prove synthetically (without using coordinates) that if T leaves three distinct lines invariant that are concurrent at P then T leaves every line through P invari- ant. Suggestion: Choose a point Q1 = P on one of the invariant lines L1 and let 6 M be any line through Q1 different from L1. Let Q2 = M L2 and Q3 = M L3, ∩ ∩ where L2 and L3 are the other two invariant lines. Let L4 be any line through P . First show that you can assume T fixes at most one of Q1,Q2,Q3. Now show that T (Q4) is on L4 by proving that T is the extension of a perspectivity.
52 Congruence, Degenerate Conics, Empty Conics
2 The conic x2 + y2 + z2 = 0 is empty in RP and so cannot be congruent to the conic x2 + y2 z2 = 0. If they were congruent there would be S P (2) such that − ∈ 1 0 0 1 0 0 adj adj adj Sco 0 1 0 S = ScoS = Sco 0 1 0 S 0 0 1 0 0 1 − adj However, Sco and S are transposes of each other so this would imply that the dot product of a column in the matrix with itself is negative, and that cannot happen over the real numbers. These two conics would be congruent in the complex 2 projective plane CP , the usual domain for the theory of algebraic curves, so we do not want to call this empty conic degenerate. Is there a test for when a conic is degenerate?
The answer to this question leads to the important result that any two non- degenerate conics are projectively congruent. We use a key theorem from linear algebra.
Theorem (Diagonalization of Symmetric Matrices). If Q is a real sym- metric matrix then all of its eigenvalues are real and R3 has an orthonormal basis consisting of eigenvectors for Q. If M is a matrix whose columns are these eigenvectors then M tQM is a diagonal matrix.
0 1 1 As an example, let Q = 1 0 1 . The characteristic polynomial for Q is 1 1 0 χ3 3χ 2 = (χ 2) (χ + 1)2 − − − so the eigenvalues are λ = 2, µ = 1. In this case the eigenvalue µ has multiplicity − 2 and the corresponding eigenvectors span a 2-space in R3 with unit normal vector 1 (i + j + k), itself an eigenvector for λ. Thus, we get an orthonormal basis by √3
53 choosing any two unit vectors in the 2-space for µ that are orthogonal to each other. For example, if 1 1 1 √3 √6 − √2 M = 1 2 0 √3 √6 1 −1 1 √3 √6 √2 1 the matrix M − QM rewrites Q relative to the basis consisting of the columns of 1 t M. But M − = M (M is an orthogonal matrix) and so 2 0 0 t M QM = 0 1 0 = Q0 0− 0 1 − represents Q relative to a basis of its eigenvectors. If we define a collineation S by setting M = Sadj then we have
adj ScoQS = Q0 that is, S maps the conic xy + yz + xy = 0 to the conic 2x2 y2 z2 = 0. − − Now let Q be any symmetric matrix and apply the theorem to obtain a diagonal form λ 0 0 Q0 = 0 µ 0 0 0 ν where λ, µ, ν are the eigenvalues of Q.
We want to focus on the case where λµν = 0, but first consider what happens when at least one of these eigenvalues is zero,6 say ν = 0. Then the conic represented by Q has been mapped to the conic λx2 + µy2 = 0 If λ = µ = 0 as well then Q was the zero matrix to start with and the conic is undetermined.
Remark. In algebraic geometry one interprets a symmetric matrix Q as a vector in R6. If Q is not the zero vector its span is a 1-space and so conics are studied as points in projective 5-space. For this reason we exclude the interpretation of the zero matrix as a conic.
54 If also µ = 0 but λ = 0 the conic Q0 is 6 x = 0 which is the single line 1 0 0 . If λµ = 0 we have two cases: λµ > 0 and 6 λµ < 0. In the first case the conic Q0 is the single point Z. In the second case λx2 + µy2 = 0 is equivalent to
λ x + µ y λ x µ y = 0 | | | | | | − | | p p p p and so the conic consists of two distinct lines. This covers all possibilities assuming a zero eigenvalue. Since det Q = det Q0 = λµν we have the answer to our question:
Theorem. Let be a projective conic with symmetric matrix Q. Then is degenerateE if and only if det Q = 0. E
Corollary. Let 1 and 2 be non-degenerate, non-empty projective conics. Then E E there exists T P (2) such that T ( 1) = T ( 2). ∈ E E Proof. A non-degenerate conic is empty if and only if the eigenvalues λ, µ, ν of its matrix Q all have the same sign, for then it can be mapped to the conic with Cartesian equation λx2 + µy2 + νz2 = 0 2 which is satisfied by no point of RP . Without loss of generality we assume, then, that λ > 0, µ > 0 and ν < 0 for a given conic represented by Q. We can find M an orthogonal matrix of eigenvectors and setEM = Sadj so that
λ 0 0 adj ScoQS = 0 µ 0 = Q0 0 0 ν Setting λ = a2, µ = b2, ν = c2 consider the collineation − a bc 0 0 D = b p0 ac 0 0q 0 c ab p 55 1 0 0 adj Then DcoQ0D = 0 1 0 and so the collineation D S maps to the 0 0 1 ◦ E − conic x2 + y2 z2 = 0 − Choosing corresponding collineations S1 and D1 for 1, S2 and D2 for 2 we have E E adj adj S D D1 S1 : 1 2 2 ◦ 2 ◦ ◦ E → E
Determination of a Conic from Given Points
A basic result from the theory of non-degenerate algebraic curves states that the number of points in common with any given line is at most the degree of the curve. If the curve is a conic we can see this directly. We have seen that if (T,P ) is non-degenerateE then it acquires a parameterization that describes the Elocus of intersections. If the parameter is t then, with the possible exception of a single point, the coordinates are given by quadratic polynomials in t. When these coordinates are substituted into the linear equation for a given line the result is a quadratic equation in t, which of course has at most two real solutions. For example, since any two non-degenerate conics are projectively congruent, assume t2 + t that is xy + yz + zx = 0, which is parameterized by t + 1 along with the E t − x point X. If L = a b c then L is the set of points y such that ∩ E z ax + by + cz = 0 xy + yz + zx = 0 and so
a t2 + t + b (t + 1) ct − = at2 + t (a + b c) + b = 0 −
56 If a = 0 then X is on L and L contains one additional point if b = c, in which case we say that L is a secant line;∩ E if b = c then only X belongs to the6 intersection and we say that L is a tangent line. If a = 0 the discriminant of the quadratic determines whether L contains 0, 1 or 26 points. If the intersection is empty we say that L is an exterior∩Eline. It follows that any four points on a non-degenerate conic form a quadrilateral, since no three of them can be collinear.
Theorem. A non-degenerate conic is determined by any five distinct points.
Proof. By the FTPG, and the fact that any two non-degenerate conics are projectively congruent, we can let four of the five given points be the XYZU. Any conic that contains XYZ is of the form
Bxy + F xz + Gyz = 0 and if it contains U then B + F + G = 0 Note that BFG = 0 because the conic is non-degenerate. The fifth point cannot be collinear with6 any two points of the quadrilateral XYZU and so it must have 1 distinct coordinates λ , λµ = 0, λ = 1, µ = 1, λ = µ. Then µ 6 6 6 6 λB + µF = λµ (B + F ) and so the conic is represented by 0 µ (λ 1) λ (1 µ) Q = µ (λ 1) 0− µ −λ λ (1 −µ) µ λ −0 − − Note that det Q = 2µλ (λ 1) (µ 1) (µ λ) = 0, as expected. − − − − 6
Another theorem should be established at this juncture. We know that any two non-degenerate conics are projectively congruent but that the collineation trans- forming one to another is not unique. Given the non-degenerate conics 1 and 2, E E how specific can we be in prescribing a mapping from 1 to 2? As we have seen, E E 1 and 2 may both contain XYZU yet be distinct conics, so the collineation E E determined by specifying the images of four points may not map 1 to 2. E E 57 What is the maximum number of points on 1 whose images we can prescribe on E 2 ? E
Theorem. Given P1Q1R1 on 1 and P2Q2R2 on 2 there exists T P (2) with E E ∈ T ( 1) = 2 such that T : P1Q1R1 P2Q2R2. E E →
Proof. Without loss of generality we can assume 2 is the conic xy+yz+xz = E 0 and that P2Q2R2 = XYZ. If T1 : P1Q1R1 XYZ then T1( 1) is of the form → E Bxy + F xz + Gyz = 0, with BFG = 0 since 1 is non-degenerate. Let 6 E 1 G 0 0 1 T2 = 0 F 0 1 0 0 B and T = T2 T1. Then T : P1Q1R1 XYZ and T ( 1) = 2. ◦ → E E Exercise. Is the collineation T unique?
58 Polarity
Let be a non-degenerate conic represented as the level surface s(x, y, z) = 0 of the functionE
s(x, y, z) = Ax2 + Bxy + Cy2 + F xz + Gyz + Hz2
Then 2ABF s (x, y, z) = x y z B 2CG ∇ FG 2H Since the gradient of a scalar function is perpendicular to its level sets at the point of evaluation it follows that the tangent line to at P is E
L = P tQ
because the coordinates of L are the components of the normal vector that defines L as a 2-space in R3. 2 For example, if s(x, y, z) = xy + xz + yz then the tangent line to at 2 is E 1 − 0 1 1 2 2 1 1 0 1 = 1 1 4 − 1 1 0 In calculus we would call the 2-space x + y + 4z = 0 the tangent plane to the surface xy + xz + y = 0 at the point (2, 2, 1), which is the same tangent plane for any point (2λ, 2λ, λ). − − The process of associating a point on with a line determined by can be 2 E E generalized to any point in RP , resulting in a map
P LP 7→ 59 and this induces a dual map L PL 7→ The conic determines this map that interchanges points and lines (called the polarity inducedE by ) according to the following property: E
If P L then PL LP . We say that the polarity reverses inclu- sion. It also∈ follows that∈ a polarity induced by is an involution. We E call LP the polar of P and PL the pole of L.
Theorem. Let be a non-degenerate conic represented by the symmetric matrix E 2 t adj t Q. For any point P of RP let LP = P Q and for any line L let PL = Q L . Then the correspondence P LP , L PL is a polarity. 7→ 7→
Proof. We must show that if L is a line through P then PL is on LP , that is, LP PL = 0 if LP = 0. Now
t adj t t adj t LP PL = P Q Q L = P QQ L
adj t t t t But QQ is non-zero because det Q = 0, so LP P L = P L . However, P L = t 6 (LP ) = 0 because LP = 0. This is the polarity induced by the non-degenerate conic . What is it’sgeometric E interpretation? First, if P then LP is the tangent line to at P , and if L is a ∈ E E tangent to then PL is its point of contact. The conic segregates the remaining E 2 E points of RP into those outside and those inside. Definition. We say P is outside if there exists a line L through P that is E exterior to , that is, such that L = ∅. We say P is inside if every line through P Eintersects in two distinct∩ E points, that is, every L throughE P is a secant. E
Exercise. Let P be a point not on . If P is outside choose a representation of Q such that P tQP > 0. Then EtEQE > 0 for any pointE E that is outside , and EtQE < 0 for any point E that is inside . For example, the point P = X Eis outside the conic : x2 + y2 z2 = 0 becauseEL = 0 0 1 does not intersect E − 1 0 0 . Let Q = 0 1 0 . Then E 0 0 1 − 60 λ λ 0 0 Q 0 = λ2 > 0 0 a For E = b we have c EtQE = a2 + b2 c2 − so E is outside if a2 + b2 c2 > 0 and inside if a2 + b2 c2 < 0. E − E − Tangents through a Given Point
Let P be outside . Suppose we want to find a point E on such that the line through P is tangentE to at E, that is, we want to find a pointE E such that E t t E P × = E Q
t Note that this equation only makes sense if E is on because E P ×E = 0 and so E we must have EtQE = 0. In terms of vectors in R3 the equation becomes t E P × + λQ = 0 for some λ = 0. Thus, we can find E by determining the values of λ such that 6 det P × + λQ = 0 and then finding the kernel of P × + λQ. This determinant is a cubic polynomial in λ which has 0 as a root because det P × = 0. The other two roots are real and distinct, corresponding to the two points E1 and E2 on such that the lines E through P and Ej are tangents to . The line through E1 and E2 will be LP , that E is, if P is outside its polar will be a secant. Conversely, if L is a secant then PL is outside , beingE the intersection of the two tangents at L . E ∩ E 1 For example, if is the conic x2 + y2 z2 = 0 and P = 1 then E − 1 λ 1 1 − P × + λQ = 1 λ 1 1 1 −λ − − 61 whose determinant is λ 1 λ2 . For λ = 1 the eigenvalue 0 corresponds to the 1-space − − 1 E1 = 0 1 and for λ = 1 the eigenvalue 0 corresponds to the 1-space
0 E2 = 1 1 Then
t t E1Q = E1P × = 1 0 1 t t − E Q = E P × = 0 1 1 2 2 − are the tangents to through P , and LP is the secant 1 1 1 . E −
62 Poles of Exterior Lines
If L is a line exterior to we expect its pole PL to be inside . Dually, if P is E E inside then LP should be an exterior line. For such a point suppose a given E line through P intersects at E1 and E2. Then the polar of E1is the tangent line t E t E1Q and the polar of E2 is the tangent line E2Q. By taking the cross-product we obtain the intersection of these tangent lines. The following result from linear algebra is straightforward to verify:
If V is an ordered triple and Q is a symmetric matrix then (QV )× = adj adj Q V ×Q . t t As usual, the square brackets indicate projective matrices. Since EjQ = QEj we can express the intersection of the two tangents as adj adj (QE1)× QE2 = Q E1×Q QE2 adj = Q E1×E2
t Now LP = P Q and
t adj t P Q Q E1×E2 = P E1×E2 = 0
t because E1×E2 is the line through E1 and E2 and this line was assumed to pass through P . It follows that the polar of P is the line exterior to consisting of the intersections of all pairs of tangents whose points of contact are theE intersections of with lines through P , and that the pole of any secant line to is the intersection ofE the tangents at its two points in common with . Further, ifEL is a line exterior to then the polars of all of its points are concurrentE at a point inside . E E This geometric interpretation of polarity induced by a non-degenerate conic is usually referred to as La Hire’s Theorem. Polarity often suggests proofs to theorems about conics that do not require reduction to coordinates:
Theorem. Let A, B, C, D be points on a non-degenerate conic , and suppose that the tangents to at A and C meet at P on the line BD.E Show that the tangents to at B andE D meet on the line AC. E 63 Proof. Since P = LA LC = PAC is on BD it follows that LB LD = PBD ∩ ∩ is on AC. Theorem. Let A, B, C be points on a non-degenerate conic , and let E = E LA BC, F = LB CA, G = LC AB. Then E,F,G are collinear. ∩ ∩ ∩
Proof. We obtain the dual of this theorem by polarity:
Let L, M, N be tangents to E with PL = A, PM = B,PN = C. Then the lines PBC A, PCAB,PABC are concurrent.
This dual theorem is easily proved by transforming to a conic whose affi ne E representation in an embedding plane is a circle. Then PBC A, PCAB,PABC are cevians of the triangle PBC PCAPAB and the following segments are equal:
PABA = BPAB
PBC B = CPBC
PCAC = APCA
Then P A P C P B AB CA BC = 1 APCA · CPBC · BPAB and so PBC A, PCAB,PABC are concurrent by Ceva’s Theorem. The original theorem now follows by duality.
Diagonal Triangle of a Quadrilateral
Definition. The diagonal triangle of a quadrilateral ABCD is EFG, where E = AB CD,F = AC BD,G = AD BC. ∩ ∩ ∩ Theorem. Let the non-degenerate conic contain the vertices of a quadrilateral ABCD. Then the tangents to at any pairE of points of the quadrilateral intersect on a line of its diagonal triangle.E
64 Proof. The line LE contains the pole of AB and the pole of CD. However, the pole of AB is LA LB, the intersection of the tangents at A and B, and the ∩ pole of CD is LC LD, the intersection of the tangents at C and D. Similarly, LG ∩ is the line through LB LC and LA LD, and LF is the line through LA LC and ∩ ∩ ∩ LB LD. We can transform to the conicxy + yz + zx = 0 with ABC XYZ. ∩ t2 +Et 7→ Then D is mapped to t + 1 for some t / 0, 1 and it is readily seen that t ∈ { − } − LE = F G, LF = GE, LG = EF .
Dual Construction of a Conic
The concept of polarity allows us to define a conic as the collection of its tangents. This collection is usually called a line conic. We obtain line conics by dualizing our original construction of a conic at the point P afforded by the collineation T .
2 Definition. Let L be a line in RP and let T P (2). The collection of lines (T,L) = PT (P ): P L is the line conic at L∈afforded by T . E { ∈ }
This line conic should be represented by a symmetric matrix Q0 such that if M t is a line of (T,L) then MQ0M = 0. If Q is the symmetric matrix representing E t adj the conic consisting of the poles PM of the lines M then , since PM = MQ it follows by the usual matrix argument that
adj t Q0 = Q = L×T TL× −
1 1 1 Exercise. Let L = 1 0 0 and let T = 0 1 1 . Show that (T,L) 0 0 1 E consists of the lines q2 q (p + q) p (p + q) . Find the Cartesian equation of the conic consisting of the− poles of these lines. Harmonic Sets of Points
65 Numerous examples of harmonic sets of points occur among the intersection prop- erties of conics. Here is one of the most basic theorems. Theorem. Let P be any point outside the non-degenerate conic and let M E be any secant of through P . Let C = LP M and R,S = M. Then P,C,R,S are a harmonicE set. ∩ { } E ∩ Proof. Let A and B be the points of tangency from P and choose a collineation that maps to the conic xy + yz + xz = 0 and that takes A X and B E 7→ 1 7→ Y . Then LP is mapped to 0 0 1 and P is mapped to P 0 = 1 , 1 − the pole of 0 0 1 . Any point on other than X,Y is of the form E = E λ2 + λ λ + 1 for some λ = 0. If λ = 1 then E = Z. Otherwise we can λ 6 − − now find a collineation T such that leaves the conic invariant and such that λ 0 λ (λ + 1) T (X) = X, T (Y ) = Y and T (E) = Z. Specifically, T = 0 λµ µ (λ + 1) 0 0 λν 0 1 1 adj 2 with TcoQT = Q = 1 0 1 , which requires µ = λ and ν = λ. Then 1 1 0 − 1 0 λ + 1 2 T = 0 λ λ (λ + 1) . Since T leaves 0 0 1 invariant we have T (P 0) = 0 0 λ − P 0 but the secant M has been mapped to the line 1 1 0 through P 0 and − 1 Z. The intersection of these two lines is 1 , and the second intersection of this 0 2 secant with is 2 . Since collineations preserve cross-ratio we have shown E 1 − 1 1 2 0 that we can let P,C,R,S be the points 1 , 1 , 2 , 0 . The 1 0 1 1 − − cross-ratio of this set is either 1, 2, 1 depending on the order of computation. − 2
66 Pascal’sTheorem
2 Let A, B, C, A0,B0,C0 be six distinct points on a non-degenerate conic in RP E and let AB0 BA0 = P,BC0 CB0 = Q0CA0 AC0 = R. Then P, Q, R are ∩ ∩ collinear. ∩
Proof. Since no three of the points are collinear we can assume is the conic E t2 + t xy + yz + zx = 0 and that A = X,B = Y,C = Z. Then A0,B0,C0 are t + 1 t − with t = p, q, r, respectively, with p, q, r distinct and not equal to 0 or 1. Then − q (p + 1) q (r + 1) p (r + 1) P = q + 1 ,Q = r + 1 ,R = r + 1 q q r − − − which are readily seen to be collinear. Pascal’sTheorem is a generalization of Pappus’sTheorem since two distinct lines 2 in RP constitute a degenerate conic. It is the strongest possible generalization in the following sense.
2 Theorem. Let A, B, C, A0,B0,C0 be six distinct points in RP , no three of which are collinear, and let AB0 BA0 = P,BC0 CB0 = Q0CA0 AC0 = R. If P, Q, R are collinear then the six points∩ are on a non-degenerate∩ conic.∩
p Proof. By the FTPG, we can assume ABCA0 = XYZU. Let B0 = q ,C0 = r s t where necessarily p, q, r are distinct and non-zero, as are u r ps t s, t, u. Then P = q ,Q = qs ,R = t . Since X,Y,Z,U are on any r pu u non-degenerate conic axy + bxz+ cyz= 0 with a +b + c = 0, we must show that
67 B0 and C0 are on a conic of this form. Let a = u(t s); b = t(s u); c = s(u t). − − − Then axy + bxz + cyz = 0 is non-degenerate and contains C0. Since P, Q, R are collinear we have r ps t 0 = q qs t
r pu u
= prt (s u) + pqu (t s) + qrs (u t) − − − = apq + bpr + cqr
and so B0 is also on this conic. Corollary to the proof. Since any five of the six points determine the conic it follows that r(q p)xy + q(p r)xz + p(r q)yz = 0 is the same conic. − − − The dual of Pascal’sTheorem is:
Brianchon’s Theorem. Let L, M, N, L0,M 0,N 0 be six distinct tangents to a 2 non-degenerate conic in RP and let
L1 = (L M 0)(L0 M) ∩ ∩ L2 = (M N 0)(M 0 N) ∩ ∩ L3 = (N L0)(N 0 L) ∩ ∩
Then L1,L2,L3 are concurrent.
Charles J. Brianchon (1785-1864) originally stated this result as an affi ne theorem, where the duality with Pascal’sTheorem expresses a basic fact about hexagons:
The three diagonals of a hexagon circumscribed about a conic are concurrent, and the three intersections of the opposite sides of a hexagon inscribed in a conic are collinear.
68 Important Skills for Final Exam
Find the matrix Q for conic (T,P ) Determine whether a projectiveE conic is degenerate Find the image of a projective conic after applying a collineation E Given 1 and 2, find P (2) such that T ( 1) = 2 with images of three points specifiedE E ∈ E E Find the conic through five given points Polarity: Given , find LP for P and PL for L Given PE outside , find the tangents through P and their points of contact Verification of Pascal’sTheoremE and its application to affi ne situations
Study the exercises on pages 252-255 of Brannan/Esplen/Gray. Some of the final items may be chosen similar to these. Print out as much of the web notes as you need and bring these to the final. Try to create and work examples related to the items in the above outline, with particular attention to correct notation.
69