Embeddings with all triangles faces

Dan Archdeacon

The University of Vermont

Common goal: Embed a simple graph such that

every face is a triangle

Why?• Minimizes the genus of the embedding• Examples include n = 0,3,4,7 (mod 12) in the Map

Color Theorem, giving triangular embeddings of Kn on orientable surfaces

• Similar nonorientable embeddings exist for n = 0,1 (mod 3), except n = 3,4,7.

Change the goal: Instead of “every face is a triangle”,

try “every triangle is a face”Trouble:

In an embedding every edge is on at most 2 faces, but now every edge should be on n-2 triangles

Solution: Replace Kn with a multigraph Kn

(n-2)/2

Note: n must be even

The Main Result

Theorem:For all even n, there exists an embedding of Kn

(n-2)/2 such that every triangle is a face. Moreover, if n ≥ 6 the surface can be chosen orientable or nonorientable.

Note: The orientable genus is (n + 3)(n - 2)(n - 4)/24The nonorientable genus is (n + 3)(n - 2)(n - 4)/12(a handle is worth two crosscaps)

Proof: use induction on n in steps of size 2

Start of the induction: For n = 4 use the tetrahedron

The inductive step:We use the existing embedding of Kn

(n-2)/2 and an extender Tn to construct the embedding of Kn+2

n/2.

The key idea:Create 2 embeddings with vertices labeled 1,…,n+2 such that every triangle appears as a labeled face. Equivalently, if we identify all vertices with the same label, construct an embedding of Kn+2

n/2 on a pinched surface with every triangle as a face.

Yet to do:Describe how to identify vertices with the same label maintaining a surface embedding. Equivalently, describe how to modify our pinched embedding to remove the number of sheets at a pinch point.

The extender Tn triangulating a sphereAdd 2 new vertices with the following properties:

• Call them N,S for the north and south pole of a sphere. • Connect N,S with n/2 meridians separating the sphere into n/2 sectors• The remaining vertices all lie on the equator and are labeled 1,…,n such that every

pair appears as an equatorial edge• Every triangle with both N and S appears incident with a meridian• Every triangle with only one of N or S appears using an equitorial edge• This is possible because the last edge in the long path and the independent edges

form a matching M, and the remaining edges in the long path form an Eulerian cycle in Kn – M

Equivalently: Tn is a triangulation of a sphere with n pinch points whose faces are all triangles involving at least one of N,S

Example: T6

Identifying vertices (removing sheets)To glue u1 to u2 and v1 to v2 first create digons along edges u1v1, u2v2:

Then add a handle between these digons:

Contract the cylinder laterally to identify the boundaries of the digons

Star Trek: The Game (106th episode of “The Next Generation”)

Effect on pinch points when adding a handle

The construction described topologically as adding a handle merges two sheets into one, similar to the representation of the complex function f(z)=z2 around the origin

Adding a handle as a permutation of sides of edges• Consider the sides of the edges incident with vertices a & b. Two kinds of relations: 1) corners at

a face, 2) left-right side of a given edge. The orbits correspond to the sheets at these vertices• 10,10b and 11,11b are the two sides of two edges• Modify them so that 10,11b and 10b,11 are the two sides of two edges• Net result is that orbits merge at both a and b, reducing the number of sheets

AT THE START: a = (10, 30b 30, 40 40b, 10b) (11, 41 41b, 31b 31, 11b) (12, 42 42b, 32b 32, 12b) (13, 43 43b, 33b 33, 13b) (14, 44 44b, 34b 34, 14b) b = (10, 20 20b, 50 50b, 10b) (11b, 51 51b, 21b 21, 11) (12b, 52 52b, 22b 22, 12) (13b, 53 53b, 23b 23, 13) (14b, 54 54b, 24b 24, 14) AFTER ADDING A HANDLE WITH IDENTIFICATION: 10 10b & 11 11b 10 11b & 10b 11 a = (10, 30b 30, 40 40b, 10b 11, 41 41b, 31b 31, 11b) (12, 42 42b, 32b 32, 12b) (13, 43 43b, 33b 33, 13b) (14, 44 44b, 34b 34, 14b) b = (10, 20 20b, 50 50b, 10b 11, 21 21b, 51b 51, 11b) (12b, 52 52b, 22b 22, 12) (13b, 53 53b, 23b 23, 13) (14b, 54 54b, 24b 24, 14)

Combinatorial view as GEMs (Graph Encoded Manifolds)

• This model encodes maps as operations on properly 3-edge-colored cubic graphs

• Vertices of the GEM are flags of the embedding• Permuted by three fixed-point-involutions: rotations, lateral

along edges, and transversal across edges• Embedding is orientable iff the GEM is bipartite• Changes are only on the transversally colored edges

• Different viewpoints can yield different insights as well as double-checks

Summarizing the result of adding a handle• Number of vertices decreases by 2• Number of edges remains the same• Number of faces remains the same

The Euler characteristic decreases by 2 (so Euler genus increases by 2)

If we deal with surfaces some of these operations connect different components. The labeled faces remain the same

Equivalently, the number of sheets decreases by 2, one each at u,v

Repeat n2/4 times using the bold equatorial edges of Tn and equivalently labeled edges in Kn

(n-2)/2

A variation: adding an antihandle

As before create digons along edges u1v1, u2v2, then add a handle between these digons using opposite orientations. The resulting surface is always nonorientable

Similar to adding a handle or antihandle to the sphere forming the torus or Klein bottle respectively

Save this step until last to yield a nonorientable surface

Conclusion: can identify all vertices with the same label (equivalently eliminate all pinch points), creating both orientable and nonorientable

embeddings with every triangle appearing as a face

The End

The End ?

Embeddings with all quadrilaterals faces

Dan Archdeacon

The University of Vermont

The Main Result

Theorem:For all n, m = (n – 2)(n – 3)/2, there exists an embedding of Kn

m on a nonorientable surface such that every quadrilateral is a face

Conjecture: The surface can be orientable, provided thatn = 0,2,5,7 (mod 8)

Proof: use induction on n (steps of size 1)

Start of the induction: For n = 4 use the tetrahedron in the projective plane, called the hemicube

The inductive step:We use the existing embedding of Kn

(n-2)(n-3)/2 and n(n-1)(n-2)/6 copies of K4 in the projective plane to construct the embedding of Kn+1

(n-1)(n-2)/2. Each of the projective K4’s use the vertex n+1

The key idea:Create 1+n(n-1)(n-2)/6 embeddings with vertices labeled 1,…,n+1 such that every quadrilateral appears as a labeled face. Equivalently, if we identify all vertices with the same label, construct an embedding of Kn+1

(n-1)(n-2)/2 on a pinched surface with every quadrilateral as a face.

Yet to do:Describe how to identify vertices with the same label maintaining the desired property. Equivalently, describe how to modify our embedding to remove the number of sheets at a pinch point.

Identifying vertices: adding a crosscap

• Let v be adjacent to u1 and u2. • As before form digons as shown on the

upper right, this time both incident with v

• Identify e1’ with e2’ and e1’’ with e2’’. The result is shown on the lower right, where the shaded disk is removed and its boundary points identified

• Reduces v by 1, e and f remain the same, decreasing the Euler characteristic (increasing Euler genus)by 1 with the new surface nonorientable. The faces are still all quadrilaterals

• Equivalent to a P3 where the two incident edges lie on different pinch sheets. Construction reduces the number of sheets at u by 1.

Helaman Ferguson

Born August 11, 1940

Professor Brigham Young University

“I find that sculpture is a very powerful way to convey mathematics, and mathematics is a very powerful design language for sculpture.”Torus with Cross-

cap

Iterate the vertex identifications

• We next do a sequence of the operations “adding a handle” and “adding a crosscap” to complete the vertex identifications

• Equivalent to slowly reducing the total number of sheets at all vertices

• Must be done with care to ensure the appropriate constructions can be applied. I’ll spare you the details, but they’re not bad

Conclusion: can identify all vertices with the same label (equivalently eliminate all pinch points),

creating nonorientable embeddings with every quadrilateral appearing as a face

The End

The End ?

Embeddings with all pentagons faces

Dan Archdeacon

The University of Vermont

Change the goal again: try “every pentagon is a face”

Easy calculations:• The graph is now Kn

m with m = (n - 2)(n - 3)(n - 4)/2. There is no restriction on the parity of n

• The surface is nonorientable with genus2 – n + 3n(n – 1)(n – 2)(n – 3) (n – 4)/20

The Main Result

Theorem:For all n, m = (n – 2)(n – 3) (n – 4)/2, there exists an embedding of Kn

m on a nonorientable surface such that every pentagon is a face

Conjecture: The surface can be orientable for even n

Proof: Again by induction on n

Start of the inductionStart of the induction:

For n = 5 use the following (very hard to find) net

The inductive step

• Again uses the embedding Knm to construct the

embedding for order n + 1• This time we need to use n(n – 1)(n – 2) (n – 3)/24

copies of the embedding for K5

• Use a sequence of “adding handles” and “adding crosscaps” to identify vertices/remove pinch points, with caution about where they apply

Conclusion: can identify all vertices with the same label (equivalently eliminate all pinch points),

creating nonorientable embeddings with every pentagon appearing as a face

The End

The End ?

Embeddings with all k-gons faces

Dan Archdeacon

The University of Vermont

Change the goal yet again: try “every k-gon is a face”

Conjecture:For all (n,k) ≠ (4,3) there is a nonorientable embedding of a complete multigraph Kn

m such that every k-gon is a face. The same holds for orientable embeddings provided the parity condition on the Euler characteristic is satisfied

The difficulty: starting the induction

The case n = k: Perhaps the most interesting case is embedding complete multigraphs such that every Hamiltonian cycle is a face. These are unknown for k ≥ 6. They correspond to the start of an induction for all k-gons faces.

Theorem: If there exists an embedding of a complete multigraph of order k with all Hamiltonian cycles faces, then there is an embedding a complete multigraph of order n with all k-cycles faces for all n ≥ k.

Further variations:

• Require every triangle appears r ≥ 2 times (n even)• Every quadrilateral, or pentagon r ≥ 2 times• Every triangle and quadrilateral a face (n even)• Every triangle, pentagon, and quadrilateral a face

(n even)

Further variations:

✔Require every triangle appears r ≥ 2 times (n even)• Every quadrilateral, or pentagon r ≥ 2 times• Every triangle and quadrilateral a face (n even)• Every triangle, pentagon, and quadrilateral a face

(n even)

Further variations:

✔Require every triangle appears r ≥ 2 times (n even)✔Every quadrilateral, or pentagon r ≥ 2 times• Every triangle and quadrilateral a face (n even)• Every triangle, pentagon, and quadrilateral a face

(n even)

Further variations:

✔Require every triangle appears r ≥ 2 times (n even)✔Every quadrilateral, or pentagon r ≥ 2 times✔Every triangle and quadrilateral a face (n even)• Every triangle, pentagon, and quadrilateral a face

(n even)

Further variations:

✔Require every triangle appears r ≥ 2 times (n even)✔Every quadrilateral, or pentagon r ≥ 2 times✔Every triangle and quadrilateral a face (n even)✔Every triangle, pentagon, and quadrilateral a face (n even)

The End ?

Embeddings with directed k-gons faces

Dan Archdeacon

The University of Vermont

Change the goal again: try “every directed k-cycle is a face”

Surface must be orientable to talk about the faces as directed cycles

Easy calculations:

The order n must be even

The graph is now Knm with m = (n - 2)! / (n – k)!

The Main Result

Theorem:For all even n, k ε {3,4,5}, m = (n – 2)!/(n – k)!, there exists an embedding of Kn

m on an orientable surface such that every directed k-gon is a face

Proof: Use a double cover of the nonorientable embedding with undirected k-gons faces

All directed k-gons for larger k?

• Most probably exist (n even)

Theorem: They exist for n,k even

• Good area for future research when k is odd

The End?

The End?

This time for real

Any questions?