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

2. Computational Geometry
the study of algorithms to solve problems in geometry

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

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

10. 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

11. 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

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

13. 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

14. Transforming a Step Matrix1 3 1 3 2 5 2 5 4 4 1 3 1 2 2 3 4 5 4 5

15. 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

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

17. 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

18. 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

19. 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

20. 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

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

22. 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

23. 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?”

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

25. 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)

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

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

28. 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

29. 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

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

31. 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:
Cross 
Partial 
Union 
Delta 