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)/24
The 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+2n/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+2n/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 sphere
Add 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-edgecolored 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 Knm on a nonorientable surface such
that every quadrilateral is a face
Conjecture:
The surface can be orientable, provided that
n = 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
Torus with Cross-cap
“I find that sculpture is a very powerful
way to convey mathematics, and
mathematics is a very powerful design
language for sculpture.”
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 Knm with m = (n - 2)(n - 3)(n - 4)/2.
There is no restriction on the parity of n
• The surface is nonorientable with genus
2 – 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 Knm 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 induction
Start 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 Knm 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 kcycles 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 Knm 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?