Prologue: Prologue: Theorem of Cayley Bacharach Chasles What is a - PowerPoint PPT Presentation

Prologue:

Prologue: Theorem of Cayley Bacharach Chasles

What is a point? Jürgen Richter-Gebert, TU Munich Article at ADG 2010 ADG=Automated Deduction in Geometry What‘s the Point? Ramans suggestion

From Micheluccis paper From Complex matroids : Below, Krummeck, JRG

From Micheluccis paper

A first example Consider all circles through a point 0 Call them „lines“ two Points ≠ 0 uniquely determine a line two lines have (usually) one intersection ≠ 0 Pappos‘s Theorem holds (kind of)

One way to think about it after rotation Line at ∞ is Affine version of contracted to a point Pappos‘s Theorem

Second way to think about it a · ( x 2 + y 2 ) + b · xz + c · xz + d · z 2 → ( a, b, c, d ) 4 ad − b 2 − c 2 < 0 0 1 0 1 x 2 0 + y 2 a 0 * + b x 0 z 0 B C B C   A , = 0 x 0 B C B C c y 0 z 0 y 0 @ @ A   z 2 d z 0 0 4 ad − b 2 − c 2 < 0 two-dimensional projective subspace in RP 3

Second way to think about it a · ( x 2 + y 2 ) + b · xz + c · xz + d · z 2 → ( a, b, c, d ) 4 ad − b 2 − c 2 < 0 0 1 0 1 x 2 0 + y 2 a 0 * + b x 0 z 0 B C B C   A , = 0 x 0 B C B C c y 0 z 0 y 0 @ @ A   z 2 d z 0 0 4 ad − b 2 − c 2 < 0 two-dimensional projective subspace in RP 3

Connecting the spaces mapping the points representing „lines“ by Φ : RP 2 → RP 3 with ( a, b, c, d ) x 2 + y 2   0 1 0 x 2 0 + y 2 1 a   0 x * + b x 0 z 0 xz   B C B C  7! A , = 0 y B C B C    c y 0 z 0 yz @ @ A   z 2 z d z 2 0 Point 0 is singular Line at infinity is dehomogenized picture contracted to a single point Blow up situation in 0 by distinguishing line   1   directions x 0    7! y    0   0 This makes the space a 0 projective plane

Circles as conics ax 2 + by 2 + cz 2 + dxy + exz + fyz = 0 in RP 5 ( a, b, c, d, e, f ) Circles are conics through     1 1 Φ ( p ) := p 2 Φ : RP 2 → RP 5 i − i I = J =     0 0     1 0     1 1 − 1 0 − 1     − 1         0 0     I 2 + J 2 ∼     0 I 2 − J 2 ∼ 0 I 2 = J 2 =         0 1     i     − i         0 0     0     0     0 0 0 0 d = 0 a = b

linear I conditions 0 J 2 directions to go: general linear conditions higher degree curves both

Degree 3 curves Cubic through 7 points Two more points fix the cubic (in general)

Degree 3 curves Cubic through 7 points Two more points fix the cubic (in general) Problem: too many intersections

Conics with three general linear constraints ax 2 + by 2 + cz 2 + dxy + exz + fyz = 0 allow arbitrary linear constraints on ( a, b, c, d, e, f ) Simple example d = 0 , e = 0 , f = 0 ax 2 + by 2 + cz 2 = 0 Four intersections!!

Conics of the form ax 2 + by 2 + cz 2 = 0

Conics of the form ax 2 + by 2 + cz 2 = 0 Symmetric „shadow points“ All quadrants are identical Idea: form equivalence classes of points

Linear pencils of curves Intersections of A C C = λ A + µB B and are also on B C = λ A + µB p A q „planes“ in RP 5 A, B, C p and q are equivalent if through 3 points (constraints) c 1 ∨ c 2 ∨ c 3 ∨ Φ ( p ) ∨ Φ ( q ) = 0 this implies C = λ A + µB

Interesting contstraints       1 0 1 − 1 0 0             0 0 − 1       c 1 = c 2 = c 3 =       0 1 0             0 0 0       0 0 0 | {z } circles perpendicular to unit circle x 2 + y 2 + z 2 + exz + fyz = 0 equivalent points related by inversion one „half“ is the hyperbolic plane

RP 5 Circle inversion interpreted in     x x  7! y y    x 2 + y 2 1 constraints and mapped points   x 2 x 2 1 0 1 y 2 y 2 − 1 0 0   x 4 + y 4 + 2 x 2 y 2   0 0 − 1 1     0 1 0 xy xy   x 3 + xy 2   0 0 0 x   x 2 y + y 3 0 0 0 y all 5x5 minors vanish, check it :-)

Incidence theorems hold (for many reasons)

Interesting contstraints       1 0 1 − 1 0 0             0 0 − 1       c 1 = c 2 = c 3 =       0 1 0             0 0 0       0 0 0 | {z } circles perpendicular to unit circle x 2 + y 2 + z 2 + exz + fyz = 0 equivalent points related by inversion one orbit: hyperbolic plane

Interesting contstraints       1 1 0 0 − 1 0             1 0 0       c 3 = c 1 = c 2 =       0 0 1             0 0 0       0 0 0 | {z } circles through antipodals of unit circle x 2 + y 2 − z 2 + exz + fyz = 0 equivalent points related by projective double cover one „half“ is the projective plane

Incidence theorems hold (for many reasons)

degree 3 curves λ A+ μ B Cubic through 7 points A B Two more points fix the cubic (in general) Problem: Too many intersections Again: these are equivalent points c 1 ∨ . . . ∨ c 7 ∨ Φ ( p ) ∨ Φ ( q ) = 0

degree 3 curves λ A+ μ B A B Interesting problem: How to calculeate the equivalent point? c 1 ∨ . . . ∨ c 7 ∨ Φ ( p ) ∨ Φ ( q ) = 0

Interesting problem: //these are the points pts=[A,B,C,D,E,F,G,H]; //Defining a function that gives the cubic parameters pars(p):=( p.x^3, How to calculeate the p.x^2*p.y, p.x*p.y^2, p.y^3, p.x^2, p.x*p.y, p.y^2, equivalent point? p.x, p.y, 1 ); //Setup a matrix consisting of all rows being the cubic parameters of the eight points mat=transpose(apply(pts,pars(#))); //We now need a point whose cubic parameters are linearly dependent //with the rows of this matrix //In Other words the exterior product of the cubic parameters of the point we search for c 1 ∨ . . . ∨ c 7 ∨ Φ ( p ) ∨ Φ ( q ) = 0 //With all the other cubic parameters must vanish //This gives all together 10 cubic equations //Fortunately they are highly dependent, so already two of them span define the set of solutions //Here are the parameters of two of these equtions (one of them without a x^3 coefficient the other without y^3) a1=0; b1=-det(mat_(3,4,5,6,7,8,9,10)); c1=det(mat_(2,4,5,6,7,8,9,10)); d1=-det(mat_(2,3,5,6,7,8,9,10)); e1=det(mat_(2,3,4,6,7,8,9,10)); f1=-det(mat_(2,3,4,5,7,8,9,10)); g1=det(mat_(2,3,4,5,6,8,9,10)); h1=-det(mat_(2,3,4,5,6,7,9,10)); i1=det(mat_(2,3,4,5,6,7,8,10)); j1=-det(mat_(2,3,4,5,6,7,8,9)); a2=det(mat_(2,3,5,6,7,8,9,10)); b2=-det(mat_(1,3,5,6,7,8,9,10)); c2=det(mat_(1,2,5,6,7,8,9,10)); d2=0; e2=-det(mat_(1,2,3,6,7,8,9,10)); f2=det(mat_(1,2,3,5,7,8,9,10)); g2=-det(mat_(1,2,3,5,6,8,9,10)); h2=det(mat_(1,2,3,5,6,7,9,10)); i2=-det(mat_(1,2,3,5,6,7,8,10)); j2=det(mat_(1,2,3,5,6,7,8,9)); //Now comes the tricky part we must solve this system of two cubic equations. //We already know eight solutions (the eight old points) So in principle it must be possibly //by factoring out the known solutions. // //Here comes the recipe: // --> Use Mathematica to calculate the resultant of these two general cubic equations w.r.t. x and y // --> Since we already know a lot about the solution, // we only need the highest and the lowest coefficient of these resultants // --> as a matter of fact the highest coefficients of the two resultants are identical for the above parameters // Here are these coefficients (Calculated by Mathematica and already applying a1=0, and d2=0; // The highest coefficient zz= a2*b1*b2*c1*c2*d1 - a2^2*c1^2*c2*d1 - a2*b1^2*c2^2*d1 - a2*b1*b2^2*d1^2 + a2^2*b2*c1*d1^2 + 2*a2^2*b1*c2*d1^2 - a2^3*d1^3; // The lowest of one resultant yy= a2*e1*e2*h1*h2*j1 - a2^2*h1^2*h2*j1 - a2*e1^2*h2^2*j1 - a2*e1*e2^2*j1^2 + a2^2*e2*h1*j1^2 + 2*a2^2*e1*h2*j1^2 - a2^3*j1^3 - a2*e1*e2*h1^2*j2 + a2^2*h1^3*j2 + a2*e1^2*h1*h2*j2 + 2*a2*e1^2*e2*j1*j2 - 3*a2^2*e1*h1*j1*j2 - a2*e1^3*j2^2; // The lowest of the other resultant xx= -(d1*g2^2*i1*i2*j1) + d1*g1*g2*i2^2*j1 - d1^2*i2^3*j1 + d1*g2^3*j1^2 + d1*g2^2*i1^2*j2 - d1*g1*g2*i1*i2*j2 + d1^2*i1*i2^2*j2 - 2*d1*g1*g2^2*j1*j2 + 3*d1^2*g2*i2*j1*j2 + d1*g1^2*g2*j2^2 - 2*d1^2*g2*i1*j2^2 - d1^2*g1*i2*j2^2 + d1^3*j2^3; // Assuming the normalizing the highest terms to 1 one gets as lowest terms xx/zz and yy/zz // These two terms are the products of the x, (resp. y) coordinates of all zeros of the system. // So we only have to divide by the known solutions. Et viola: x=xx/zz/(product(pts,#.x));