Permutahedra and the Saneblidze-Umble Diagonal By Stephen Weaver Directed by Dr. Ron Umble

Computational Geometry the study of algorithms to solve problems in geometry

Permutahedra Geometric shapes based on permutations of the partitions of a set {1,2}  {3} Each permutation corresponds to a face of some permutahedron We use bar notation for convenience {3}  {1,2} = 3|12 The single partition {1,2,3,…,n} corresponds to the top dimensional face

Permutahedra The boundary of a face consists of adding a single bar in every possible way 1|2 2|1 12 Two faces are adjacent if their boundaries intersect 2|1|3 1|3|2 12|3 1|23 1|2|3

Permutahedra - Examples 1|2 2|1 12 1 P3 3|12 3|1|2 3|2|1 P4 13|2 23|1 123 1|3|2 2|3|1 1|23 2|13 1|2|3 12|3 2|1|3

Permutahedra - Examples 1|2 2|1 12 1 P3 3|12 3|1|2 3|2|1 P4 13|2 23|1 123 1|3|2 2|3|1 1|23 2|13 1|2|3 12|3 2|1|3

Permutahedra - Examples 1|2 2|1 12 1 P3 3|12 3|1|2 3|2|1 P4 13|2 23|1 123 1|3|2 2|3|1 1|23 2|13 1|2|3 12|3 2|1|3

Permutahedra - Examples 1|2 2|1 12 1 P3 3|12 3|1|2 3|2|1 P4 13|2 23|1 123 123|4 1|3|2 12|3|4 2|3|1 12|34 1|23 2|13 1|2|3 12|3 2|1|3

Diagonal Given a set S, Diagonal of S  S { (x,x) | x є S } S S

Diagonal on P2  P2 12  2|1 1|2  12 1|2  2|1 2|1  2|2 1|2  1|2 2|1  1|2

Step Matrix Example Reading a Step Matrix 5 6 4 7 8 1 9 2 3 1 2 3 4 14|2|3  4|123

Step Matrix Example Reading a Step Matrix 5 6 4 7 8 1 9 2 3 1 2 3 4 14|2|3

Step Matrix Example Reading a Step Matrix 5 6 4 7 8 1 9 2 3 1 2 3 4 14|2|3  4|123

Transforming a Step Matrix 1 3 1 3 2 5 2 5 4 4 1 3 1 2 2 3 4 5 4 5

S-U Diagonal on P3 1 1 2 1 3 1 2 2 1 2 3 1 3 2 2 3 3 3 1|2|3123 1|23  13|2 12|32|1 3 2|1323| 1 13|23|1 2 1233|2| 1 1 1 2 2 3 3 12|323|1 1|233|12

Calculating the S-U Diagonal Skip step matrix stage – Use permutations and strings One – to – one correspondence between permutations and step matrices

Calculating the S-U Diagonal Skip step matrix stage – Use permutations and strings One – to – one correspondence between permutations and step matrices 2 4132 1 3 4 A=4 B=4

Calculating the S-U Diagonal Skip step matrix stage – Use permutations and strings One – to – one correspondence between permutations and step matrices 2 4132 1 3 4 A=41 B=4|1

Calculating the S-U Diagonal Skip step matrix stage – Use permutations and strings One – to – one correspondence between permutations and step matrices 2 4132 1 3 4 A=41|3 B=4|13

Calculating the S-U Diagonal Skip step matrix stage – Use permutations and strings One – to – one correspondence between permutations and step matrices 2 4132 1 3 4 A=41|32 B=4|13|2 14|23  4|13|2

S-U Diagonal Acts multiplicatively w.r.t. the bar operation (12|34) = (12) | (34) = (1|212 + 122|1) | (3|434 + 344|3) = (1|2|3|412|34) + (1|2|3412|4|3) + (12|3|42|1|34) + (12|342|1|4|3)

Generic Diagonal abc Three Element 2|134 Diagonal a|b|c  abc a|bc  ac|b ab|c  b|ac b|ac  bc|a ac|b  c|ab abc  c|b|a ab|c  bc|a a|bc  c|ab (2|134) = (2)| (134) = 2|1|3|4  2|134 2|1|34  2|14|3 2|13|4  2|3|14 2|3|14  2|34|1 2|14|3  2|4|13 2|134  2|4|3|1 2|13|4  2|34|1 2|1|34  2|4|13

Associativity (ab)c = a(bc) m( m(a,b) , c ) = m( a , m(b,c) ) m(m x id)(a,b,c) = m(id x m)(a,b,c) S-U Diagonal takes one input and produces two outputs – “comultiplication” (  id) (X) = (id  ) (X) ? Is  “coassociative?”

Not Coassociative - Example (x1) (123) + (1x) (123) = 2|1|32|1323|1 + 1|2|32|1323|1 +1|3|213|23|12 + 1|2|313|23|12 +12|32|133|2|1 + 12|32|132|3|1 +12|32|1|323|1 + 12|32|3|123|1 +1|2313|23|2|1 + 1|2313|23|2|1 +1|231|3|23|12 + 1|233|1|23|12 (mod 2) This measures the error from being coassociative.

Homotopy Coassociativity Let Vi be the Z2-vector space whose basis is the i dimensional faces of Pn Let  : Vi  Vi-1 be the boundary operation Let H : Vi  (V*  V*  V*) i+1 such that H+H=(id) + (id) H acts multiplicatively with respect to bar H(13|24) = H(13) | H(24)

Homotopy Function H(1) = 0 H(12) = 0 H(123) = 12|3x2|13x23|1 + 1|23x13|2x3|12 H(1234) = ?

Calculating H(1234) H = (id) + (id) + H X = H(1234) є (V*  V*  V*)4 (X) = [(id) + (id) + H](1234)

Calculating H(1234) H = (id) + (id) + H X = H(1234) є (V*  V*  V*)4 (X) = [(id) + (id) + H](1234) (V*  V*  V*)4 (V*  V*  V*)3

Calculating H(1234) H = (id) + (id) + H X = H(1234) є (V*  V*  V*)4 (X) = [(id) + (id) + H](1234) (V*  V*  V*)4 120,960 x 73,729 (V*  V*  V*)3

One Solution for H(1234) H(1234) = 12|34x24|13x4|2|3|1 + 12|34x24|13x4|3|2|1 + 123|4x3|24|1x34|2|1 + 123|4x3|2|14x34|2|1 + 123|4x3|2|14x3|24|1 + 124|3x4|2|13x4|23|1 + 12|34x24|1|3x4|2|13 + 12|34x24|3|1x4|23|1 + 12|34x2|14|3x24|3|1 + 12|34x2|14|3x2|4|13 + 12|34x2|14|3x4|23|1 + 12|34x2|4|13x4|23|1 + 13|24x34|1|2x4|3|12 + 13|24x3|14|2x34|2|1 + 13|24x3|14|2x3|4|12 + 14|23x4|13|2x4|3|12 + 1|234x14|3|2x4|13|2 + 1|234x14|3|2x4|3|12 + 1|234x4|13|2x4|3|12 + 23|14x3|24|1x34|2|1 + 2|134x24|3|1x4|23|1 + 12|34x2|1|4|3x24|13 + 12|34x2|4|1|3x24|13 + 13|24x3|1|4|2x34|12 + 13|24x3|4|1|2x34|12 + 12|3|4x23|14x34|2|1 + 12|3|4x23|14x3|24|1 + 12|3|4x2|134x24|3|1 + 12|3|4x2|134x4|23|1 + 12|4|3x24|13x4|23|1 + 13|2|4x3|124x34|2|1 + 1|23|4x134|2x4|3|12 + 1|23|4x13|24x34|1|2 + 1|23|4x13|24x3|14|2 + 1|23|4x13|24x4|3|12 + 1|23|4x3|124x34|2|1 + 1|24|3x14|23x4|13|2 + 1|24|3x14|23x4|3|12 + 1|2|34x124|3x4|23|1 + 1|2|34x124|3x4|2|13 + 1|2|34x14|23x4|13|2 + 1|2|34x14|23x4|3|12 + 1|2|34x24|13x4|23|1 + 1|3|24x134|2x4|3|12 + 2|13|4x23|14x34|2|1 + 2|13|4x23|14x3|24|1 + 2|14|3x24|13x4|23|1 + 12|3|4x2|13|4x234|1 + 12|3|4x2|13|4x23|14 + 12|3|4x2|1|34x24|13 + 12|3|4x2|3|14x234|1 + 12|4|3x2|14|3x24|13 + 13|2|4x3|1|24x34|12 + 13|4|2x3|14|2x34|12 + 1|23|4x13|2|4x34|12 + 1|23|4x13|2|4x3|124 + 1|23|4x1|3|24x134|2 + 1|23|4x3|14|2x34|12 + 1|23|4x3|1|24x34|12 + 1|24|3x14|2|3x4|123 + 1|2|34x14|2|3x4|123 + 1|2|34x1|24|3x14|23 + 1|2|34x1|24|3x4|123 + 1|2|34x2|14|3x24|13 + 1|3|24x3|14|2x34|12 + 2|13|4x2|3|14x234|1

Future Work Finding an H with minimal number of terms Modifying / Parallelizing the row reduction algorithm to calculate H for n > 4 Picture on second page found at: http://www.lightstorm3d.de/portfolio/back_to_gaya/stills/programming/collisionDeformer.jpg Cross  Partial  Union  Delta 