Cyclic Structures and Incidence Theorems - math.tu … · Pappus’s Theorem (123) (159) (186)...

Cyclic Structures andIncidence Theorems

Jürgen Richter-GebertTechnical University Munich

Jürgen Richter-GebertTechnical University Munich

Stupid Proofs forInteresting (?) Theorems

Incidence Theorems

Oriented Matroids (Pseudoline arrangements)

Zonotopal Tilings (Penrose like tylings)

Stresses/Liftings/Reciprocal Figures

Bracket Polynomials (Algebra of projective Geometry)

Circle Patterns

DDG

Topics we will meet

Running Example1

32

654

9 8 7

The collinearity of (123), (456), (159), (168), (249), (267), (348), (357)implies the collineartity of (7,8,9).

Pappus‘s Theorem

a

x

b

y

c

z

A Warmup

Area Method

Claim:

area(a,b,c)+ area(x,z,y)+ area(a,x,y,b)+ area(b,y,z,c)+ area(a,x,z,c) = 0

A Warmup

Area Method

Claim:

area(a,b,c)+ area(x,z,y)+ area(a,x,y,b)+ area(b,y,z,c)+ area(a,x,z,c) = 0

a

x

b

y

c

z

A Warmup

Area Method

Claim:

area(a,z,b,w)+ area(a,y,d,z)+ area(b,z,d,x)+ area(a,w,c,y)+ area(b,x,c,w)+ area(c,x,d,y) = 0

Works for any Manifold

w

z

y

x

d

c

b

a

A Warmup

Area Method

Claim:

Decomposing the cube

area(a,z,b,w)+ area(a,y,d,z)+ area(b,z,d,x)+ area(a,w,c,y)+ area(b,x,c,w)+ area(c,x,d,y) = 0

A Warmup

Area MethodDecomposing the cube

==> Discrete Königs NetsBobenko, Suris

A Warmup

Area MethodReciprocal Diagrams

A Warmup

Area MethodReciprocal Diagrams

You never have a problem with the last edge

Realizability / StrechabilityArrangements of Pseudolines

- Topological lines in RP- Two cross exactly once

Is there anequivalent

line arrangement?

2

Realizability / Strechability


line arrangement?

Arrangements of Pseudolines


2

Excursion: How to draw pseudoline arrangements?

1st Method:- fix boundary points- make a Tutte embedding- draw smooth splines

ConjectureEvery such pseudolineis „a function graph“

==> no curls allowed!

http://blog.cinderella.de/archives/173-PapSpr.cdy.html








line arrangement?



2


1st Method:- fix boundary points- make a Tutte embedding- draw smooth spines

ConjectureEvery such pseudolineis „a function graph“

==> no curls allowed!



line arrangement?



2


2nd Method:- Draw two copies on a sphere- Circle pack- connect the dots



line arrangement?



2


2nd Method:- Draw two copies on a sphere- Circle pack- connect the dots

FactUnique (!) representationof the arrangement

QuestionWhat are the special properties?



line arrangement?



2



line arrangement?

No !!

Pappus’s Theorem1 O Oa b cd e fO 1 Og h ij k lO O 1m n op q r

123456789

Another proof by algebraic cancellation

132

654

9 8 7


123456789

(123)(159)(168)(249)(267)(348)(357)(456)(789)


132

654

9 8 7

==> ce=bf==> iq=hr==> ko=ln==> ar=cp==> bj=ak==> fm=do==> dh=eg==> gl=ij==> mq=np


123456789

(123)(159)(168)(249)(267)(348)(357)(456)(789)


132

654

9 8 7

==> ce=bf==> iq=hr==> ko=ln==> ar=cp==> bj=ak==> fm=do==> dh=eg==> gl=ij<== mq=np


123456789

(123)(159)(168)(249)(267)(348)(357)(456)(789)

132

654

9 8 7


Pappus’s Theorem

(123)

(159)

(186)

(429) (726)

(483)

(753) (456)

(789)

Structure of the proof

13

2

654

9 8 7

Pappus’s Theorem

(123)

(159)

(186)

(429) (726)

(483)

(753) (456)

(789)

c ab

e

d

f

r

q

p

j

l

k

h ig

m no

e r jb

g

o

Structure of the proof

13

2

654

9 8 7

Pappus’s Theorem

(123)

(159)

(186)

(429) (726)

(483)

(753) (456)

(789)

c ab

e

d

f

r

q

p

j

l

k

h ig

m no

e r jb

g

o

Structure of the proof ==> a torus

13

2

654

9 8 7

From: Self-Dual Configurations and Regular Graphs Coxeter, 1950

For Desargues’s Theorem there is no such choice of a basis

Generalizing this proof

special choice of basis

2x2 determinants

cancellation pattern

==> ce=bf==> iq=hr==> ko=ln==> ar=cp==> bj=ak==> fm=do==> dh=eg==> gl=ij<== mq=np

(123)(159)(168)(249)(267)(348)(357)(456)(789)

Problems:

Grassmann-Plücker Relations

1

2

3

x

y

[123][1xy]-[12x][13y]+[12y][13x] = OIn every configuration of five points 1,2,3,x,y

is satisfied ( with [abc]=det(a,b,c) ).

(123) collinear ==>

[12x][13y]=[12y][13x]

[12x][13y]=[12y][13x] ==> (123) collinear or (1xy) collinear

Pappos’s Theorem

(123) ==> [124][137]=[127][134](159) ==> [154][197]=[157][194](168) ==> [184][167]=[187][164](249) ==> [427][491]=[421][497](456) ==> [457][461]=[451][467](348) ==> [487][431]=[481][437](267) ==> [721][764]=[724][761](357) ==> [751][734]=[754][731](789) <== [781][794]=[784][791]

132

654

9 8 7

Same cancellation pattern as before

An automatic method

Generate all “binomial equations” that come from the hypotheses

Generate all “binomial equations” that come from the conclusion

Find a suitable combination of the conclusion by the hypotheses

A linear problem:Is the conclusion in the span of the hypotheses ?

(binomial-proofs)

Desargues’s by cancellation

==> (768) or (745)

(479) ==> [471][496] = [476][491](916) ==> [914][962] = [912][964](259) ==> [256][291] = [251][296](240) ==> [248][203] = [243][208](083) ==> [082][035] = [085][032](570) ==> [573][508] = [578][503](213) ==> [215][234] = [214][235](418) ==> [412][487] = [417][482](536) ==> [532][567] = [537][562]

54

821

09

876

[764][785] = [765][784]

A non-realizable configuration

jh

d

g

e f

c

bai

(abi) ==> [abh][agi]=+[abg][ahi](acf) ==> [adf][acj]=-[acd][afj](adh) ==> [abd][afh]=-[abh][adf](bce) ==> [bcd][bej]=-[bde][bcj](bdg) ==> [abg][bde]=-[abd][beg](cdj) ==> [acd][bcj]=+[acj][bcd](efj) ==> [afj][egj]=+[aej][fgj](egi) ==> [aeg][ghi]=-[agi][egh](fhi) ==> [ahi][fgh]=-[afh][ghi](ghj) ==> [egh][fgj]=+[egj][fgh]

[aeg][bej]=+[aej][beg]

==> (abe) or (egj)

Six points on a conic1

2

3

4

5

6

[123][156][426][453] = [456][126][254][423]

Six points 1,2,3,4,5,6 are on a conic <==>

Pascal’s Theorem

[159][257] = -[125][579] [126][368] = +[136][268] [245][279] = -[249][257] [249][268] = -[246][289] [346][358] = +[345][368] [135][589] = -[159][358][125][136][246][345] = +[126][135][245][346]

12

3

4

5

6

98 7

[289][579] = +[279][589]

Problems of this method

Usually large search space

What is the “structure” of the proof

How to cut down the search space in advance

(binomial-proofs)

Problems of this method

Usually large search space

What is the “structure” of the proof

How to cut down the search space in advance

(binomial-proofs)

A ambitious dream:Look at a theorem.... “see” its structure........ and produce a proof immediately!!

Coxeter’s proof of Pappos’s Theorem

The Theoremsof Ceva and Menelaos

Ceva and Menelaosa1a2b1b2c1c2ABCDEFG

a1a2b1b2c1c2ABCDEH

A B

Y

C

Z

X

A B

Y

C

Z

X

|AZ|·|BX|·|CY||ZB|·|XC|·|YA| = 1 |AZ|·|BX|·|CY|

|ZB|·|XC|·|YA| = -1

Ceva and Menelaosa1a2b1b2c1c2ABCDEFG

a1a2b1b2c1c2ABCDEH

A B

Y

C

Z

X

A B

Y

C

Z

X

|AZ|·|BX|·|CY||ZB|·|XC|·|YA| = 1 |AZ|·|BX|·|CY|

|ZB|·|XC|·|YA| = -1

· ·[CDA] [ADB] [BDC][CDB] [ADC] [BDA]= -1

[XYA] [XYB] [XYC][XYB] [XYC] [XYA]· · = 1

D

Glueing

|AX|·|BY|·|CZ||XB|·|YC|·|ZA| = 1

ABCDEFGHKLM

A

B

C

DX

Y

Z

S

R

|AZ|·|CR|·|DS||ZC|·|RD|·|SA| = 1

Glueing

|AX|·|BY|·|CZ||XB|·|YC|·|ZA| = 1

ABCDEFGHKLM

A

B

C

DX

Y

Z

S

R

|AZ|·|CR|·|DS||ZC|·|RD|·|SA| = 1

|AX|·|BY|·|CR|·|DS||XB|·|YC|·|RD|·|SA| = 1

Glueing

ABCDEFGHKLM

a1

b2

b3

b4

b1

a2a3

a4

= 1a1

b2 b3 b4b1

a2 a3 a4

Glueinga1a2b1b2c1c2d1d2f2e1e2f1

a1

b2

b3

b4

b5

b6

b1

a2

a3 a4

a5

a6 = 1a1

b2 b3 b4 b5b6b1

a2 a3 a4 a5a6

Glueing

ABCDEFHKMP

ABCDEFHKM

ABCDEFKMP+ =

A “factory” for geometric theorems

A theorem on a tetrahedron

ABCDEFKMP

Front: two triangles with Ceva


ABCDEFHKM

Back: two triangles with Ceva


ABCDEFHKM

ABCDEFKMP

After glueing: an incidence theorem

4 Times Ceva

spatial interpretations

interesting degenerate situations

A special case of Pappus’s Theorem as special case

Ceva on four sides of a tetrahedron

A census of incidence theorems

MM M M+

CC C C+

CC

MM

+

CM M C+

MM C C+

MM C C+==> harmonic points

==> many interesing degenerate cases

==> Desargues’s Thm.

Harmonic QuadruplesCC

MM

+

Front: two triangles with Ceva


MM

+

Back: two triangles with Menelaos


MM

+

Glued: Uniques of harmonic point construction

Desargues by GlueingMM M M+

Front: two triangles with Menelaos


Back: two triangles with Menelaos


Glued: Desargues’s Theorem

Six triangles

1

1

12

1

34

5

1

1

1 2

2

2

Double pyramid over triangle

Degenerate torus

...and other degenerate spheres

Six triangles

Six times Ceva

Six triangles

Six times Ceva

Folding the triangles

Six triangles

Six times Ceva -->after identification

Six triangles

Pappus’s Theorem

Moving points to infinity

Pappus affine Pappus

Grid theorems

+1

-1

-1

-1

+1

+1

“row sums” = “colums sums” = “diagonal sums” = 0

Another Theorem+1

-1

+1

+1

+1

-1

-1

-1

“row sums” = “colums sums” = “diagonal sums” = 0

Larger Grid Theorems

+ =

Composition of grid theorems

“space of such theorems” is a vector space

“little” Pappus configurations are a basis

Larger Grid Theorems

Composition can be interpreted topologically

„Geometrie der Waben“compare Blaschke 1937

http://blog.cinderella.de/archives/172-Waben.cdy.html


















































































































Arrangements of pseudolines

Rombic Tilings with three Directions

Conversion of proofs

12

3

4

4

4

5

6

7

7

8

9

A

AB

B

C

C

7

[124][137]=[127][134][154][197]=[157][194][184][167]=[187][164][427][491]=[421][497][457][461]=[451][467][487][431]=[481][437][721][764]=[724][761][751][734]=[754][731][781][794]=[784][791]

Conversion of proofs

12

3

4

4

4

5

6

7

7

8

9

A

AB

B

C

C

7

[124][137]=[127][134][154][197]=[157][194][184][167]=[187][164][427][491]=[421][497][457][461]=[451][467][487][431]=[481][437][721][764]=[724][761][751][734]=[754][731][781][794]=[784][791]

427479

478

347

457

467

146 137154 157

178

148

134167

127

124149 197

- Works in general- Use Tutte-Groups- and homotopy

A different Story

From: Self-Dual Configurations and Regular Graphs Coxeter, 1950

Graph: vertices -> brackets, edges -> bracktes differing by one letter

Glue versus matter

[12x][13y]-[12y][13x]

Grassmann Menelaus Ceva

Glue versus matter

Glue versus matter

BFP <--> CM

And Conics ?a1

b3

b1

a3

b5

a5b2 b4

a2

a4

b6

a6

= 1a1

b2 b3 b4 b5b6b1

a2 a3 a4 a5a6

Carnot’s Theorem

A Conic Theorem

Works for any orientable triangulated manifold

Pascal’s Theorem

B

A

B

CA

C

MM

M

M

CarB

A

C

+

Loose Ends on Circles

K

L

M


P7

P8


P7

P8

http://blog.cinderella.de/archives/170-Bottles.cdy.html




Cyclic Structures and Incidence Theorems - math.tu … · Pappus’s Theorem (123) (159) (186)...

Documents

Transcript of Cyclic Structures and Incidence Theorems - math.tu … · Pappus’s Theorem (123) (159) (186)...