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

Prologue:

Theorem of Cayley Bacharach Chasles Prologue:

What is a point?

Jürgen Richter-Gebert, TU Munich

Ramans suggestion

What‘s the Point?

Article at ADG 2010

ADG=Automated Deduction in Geometry

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)

• ne intersection ≠0

Pappos‘s Theorem holds (kind of)

One way to think about it

Line at ∞ is contracted to a point

after rotation

Affine version of Pappos‘s Theorem

Second way to think about it

a·(x2 + y2) + b·xz + c·xz + d·z2

→ (a, b, c, d)

  x0 y0 z0  

* B B @ a b c d 1 C C A , B B @ x2

0 + y2

x0z0 y0z0 z2 1 C C A + = 0 4ad − b2 − c2 < 0 4ad − b2 − c2 < 0

two-dimensional projective subspace in RP3

Second way to think about it

a·(x2 + y2) + b·xz + c·xz + d·z2

→ (a, b, c, d)

* B B @ a b c d 1 C C A , B B @ x2

0 + y2

x0z0 y0z0 z2 1 C C A + = 0 4ad − b2 − c2 < 0 4ad − b2 − c2 < 0

two-dimensional projective subspace in RP3

  x0 y0 z0  

Connecting the spaces

  x y z   7!     x2 + y2 xz yz z2    

Line at infinity is contracted to a single point

  x y   7!     1    

* B B @ a b c d 1 C C A , B B @ x2

0 + y2

x0z0 y0z0 z2 1 C C A + = 0

(a, b, c, d)

representing „lines“ mapping the points

Φ : RP2 → RP3

with

Point 0 is singular

by

dehomogenized picture Blow up situation in 0 by distinguishing line directions This makes the space a projective plane

Circles as conics

ax2 + by2 + cz2 + dxy + exz + fyz = 0

J =   1 −i   I =   1 i  

Circles are conics through

I2 =         1 −1 i         J2 =         1 −1 −i         I2 + J2 ∼         1 −1         I2 − J2 ∼         1        

RP5

in

(a, b, c, d, e, f)

Φ : RP2 → RP5

Φ(p) := p2

a = b d = 0

I J

linear conditions

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

ax2 + by2 + cz2 + dxy + exz + fyz = 0

allow arbitrary linear constraints on

(a, b, c, d, e, f)

ax2 + by2 + cz2 = 0

d = 0, e = 0, f = 0

Simple example Four intersections!!

Conics of the form ax2 + by2 + cz2 = 0

Symmetric „shadow points“ All quadrants are identical Idea: form equivalence classes of points

Conics of the form ax2 + by2 + cz2 = 0

A B

Linear pencils of curves

C A, B, C

RP5

„planes“ in through 3 points (constraints)

C = λA + µB

Intersections of

B

and

A

are also on

C = λA + µB C = λA + µB

this implies

c1 ∨ c2 ∨ c3 ∨ Φ(p) ∨ Φ(q) = 0

p and q are equivalent if p q

Interesting contstraints

equivalent points related by inversion

c1 =         1 −1         c2 =         1         c3 =         1 −1        

x2 + y2 + z2 + exz + fyz = 0

| {z }

circles perpendicular to unit circle

• ne „half“ is the hyperbolic plane

Circle inversion interpreted in

  x y 1   7!   x y x2 + y2  

RP5

all 5x5 minors vanish, check it :-) constraints and mapped points

        1 1 x2 x2 −1 y2 y2 −1 1 x4 + y4 + 2x2y2 1 xy xy x x3 + xy2 y x2y + y3        

Incidence theorems hold (for many reasons)

Interesting contstraints

equivalent points related by inversion

c1 =         1 −1         c2 =         1         c3 =         1 −1        

x2 + y2 + z2 + exz + fyz = 0

| {z }

circles perpendicular to unit circle

• ne orbit: hyperbolic plane

Interesting contstraints

equivalent points related by projective double cover

c1 =         1 −1         c2 =         1        

| {z }

circles through antipodals of unit circle

• ne „half“ is the projective plane

c3 =         1 1        

x2 + y2 − z2 + exz + fyz = 0

Incidence theorems hold (for many reasons)

degree 3 curves

B A λA+μB Again: these are equivalent points

c1 ∨ . . . ∨ c7 ∨ Φ(p) ∨ Φ(q) = 0

Cubic through 7 points Two more points fix the cubic (in general) Problem: Too many intersections

degree 3 curves

B A λA+μB Interesting problem: How to calculeate the equivalent point?

c1 ∨ . . . ∨ c7 ∨ Φ(p) ∨ Φ(q) = 0

c1 ∨ . . . ∨ c7 ∨ Φ(p) ∨ Φ(q) = 0

Interesting problem: How to calculeate the equivalent point?

//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, p.x^2*p.y, p.x*p.y^2, p.y^3, p.x^2, p.x*p.y, p.y^2, 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 //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));

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));

Arthur Cayley

(123456)·{....}·{....} · 6 + (123457)·{....}·{....} · 7 + (123458)·{....}·{....} · 8 = 9

Grading: 8,8,8,8,8,9,9,9

{...}·1 + {...}·2 + {...}·3 + {...}·4 + {...}·5 + {...}·6 + {...}·7 + {...}·8

Question: is 8,8,8,8,8,8,8,8 possible

– [123][647][857][473][428][178][123][573][526][176]

+ [123][647][857][478][128][173][423][573][526][176] + [123][647][857][473][428][178][576][126][173][523] + [123][657][847][573][528][178][123][473][426][176] – [123][657][847][578][128][173][523][473][426][176] – [123][657][847][573][528][178][476][126][173][423]

{....} =

Cayley decifered

(123457) = [123][453][426][156] – [456][126][153][423]

{...}·1 + {...}·2 + {...}·3 + {...}·4 + {...}·5 + {...}·6 + {...}·7 + {...}·8

X

π∈S8

sign(π) [. . .][. . .][. . .][. . .][. . .][. . .] . . . [. . .] | {z }

27 brackets, grading 88888887

·π(8)

Brutally symmetric Ansatz

X

π∈S8

sign(π) [. . .][. . .][. . .][. . .][. . .][. . .] . . . [. . .] | {z }

27 brackets, grading 88888887

·π(8)

[8, 1, 2] · [ , , ] · [ , , ] [8, 2, 3] · [ , , ] · [ , , ] [8, 3, 4] · [ , , ] · [ , , ] [8, 4, 5] · [ , , ] · [ , , ] [8, 5, 6] · [ , , ] · [ , , ] [8, 6, 7] · [ , , ] · [ , , ] [8, 7, 1] · [ , , ] · [ , , ] · 8

Brutally symmetric Ansatz

X

π∈S8

sign(π) [. . .][. . .][. . .][. . .][. . .][. . .] . . . [. . .] | {z }

27 brackets, grading 88888887

·π(8)

[8, 1, 2] · [1, 2, 4] · [ , , ] [8, 2, 3] · [2, 3, 5] · [ , , ] [8, 3, 4] · [3, 4, 6] · [ , , ] [8, 4, 5] · [4, 5, 7] · [ , , ] [8, 5, 6] · [5, 6, 1] · [ , , ] [8, 6, 7] · [6, 7, 2] · [ , , ] [8, 7, 1] · [7, 1, 3] · [ , , ] · 8

Brutally symmetric Ansatz

X

π∈S8

sign(π) [. . .][. . .][. . .][. . .][. . .][. . .] . . . [. . .] | {z }

27 brackets, grading 88888887

·π(8)

[8, 1, 2] · [1, 2, 4] · [1, 2, 6] [8, 2, 3] · [2, 3, 5] · [2, 3, 7] [8, 3, 4] · [3, 4, 6] · [3, 4, 1] [8, 4, 5] · [4, 5, 7] · [4, 5, 2] [8, 5, 6] · [5, 6, 1] · [5, 6, 3] [8, 6, 7] · [6, 7, 2] · [6, 7, 4] [8, 7, 1] · [7, 1, 3] · [7, 1, 5] · 8

Brutally symmetric Ansatz

Back to conics

RP5 RP2

Veronese (dim 2) 5–2 constraints span (dim 5–2)

RP2 RP5

Veronese (dim 2) 5–2 constraints span (dim 5–2)

RP2 RP5

Veronese (dim 2) 5–2 constraints span (dim 5–2)

Φ(   x y z  ) =         x2 y2 z2 xy xz yz         ∂Φ ∂x =         2x y x         ∂Φ ∂y =         2y x z         ∂Φ ∂z =         2z x y        

Calculating the mirror lines

det ⇣ c1, c2, c3, ∂Φ

∂x , ∂Φ ∂y , ∂Φ ∂z

⌘ = 0

show some pictures

Questions

Interplay real complex Topology of underlying spaces Fast visualization Iterate the process

Thanks for your attention