Download Betti Numbers and Parallel Deformations

How is a graph like a manifold?
Ethan Bolker
Mathematics - UMass Boston
[email protected]
www.cs.umb.edu/~eb
University of Florida, Gainesville
March 19, 2002
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Acknowledgements
• Joint work with (so far)
Victor Guillemin
Tara Holm
• Conversations with
Walter Whiteley
Catalin Zara
and others
• Preprint and slides available at
www.cs.umb.edu/~eb/betti
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Plan
• Combinatorics  topology  combinatorics
• f vectors, the McMullen conjectures
• Topological ideas for embedded graphs
– Geodesics and connections
– Morse theory and Betti numbers
• McMullen revisited
• Parallel redrawings
• Examples, pretty pictures, open questions
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Counting faces of a polytope
• Euler: fk = number of faces of dimension k
• Define i by
k
fn-k =

i=0
( n-i
k-i ) i
• McMullen conjectures: For simple polytopes,
i are palindromic and unimodal
• Stanley:
Poincare duality  palindromic
hard Lefshetz theorem  unimodal
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Dodecahedron
f = (20, 30, 12, 1)
 = (1, 9, 9, 1)
What do the i count?
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Subject matter
• Connected d-regular graph  embedded in real
Euclidean n-space
• Every pair of edges at a vertex determines a planar
cycle of edges
• These are the geodesics
• 1-skeleton of any simple polytope (since any pair
of edges at a vertex determines a 2-face)
– simplex, cube in any dimension
– dodecahedron, not icosahedron
• More examples from topology …
• More examples not from topology …
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Johnson graphs J(n,k)
• Vertices are the k element subsets of an n-set
• v,w are adjacent when #(vw) = n-1
{1,2} =
• Represent vertices as bit vectors to
(1,1,0,0)
embed on a hyperplane in n-space
• J(n,1) = Kn (complete graph)
{1,3}
{1,4}
• J(4,2) is the octahedron
• J(n,2) is not the cross polytope
{2,4}
{2,3}
• Topology: Grassmannian
manifold of k-planes in n-space
{3,4}
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Johnson graph geodesics
• Each pair of edges at a
vertex determines
a geodesic
• Geodesics are
triangles
squares
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Permutahedra
•
•
•
•
Cayley graphs of the symmetric groups Sn
Vertices are the permutations of an n-set
v,w are adjacent when v w-1 is a transposition
Represent vertices as permutations of (1,…n)
to embed on a hyperplane
in n-space
(1,2,3)
(1,3,2)
• “Internal” edges matter
• S3 is the complete (2,1,3)
bipartite graph
(2,3,1)
K(3,3) in the plane
• Topology: flag manifolds
(3,1,2)
(3,2,1)
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Geodesics for S3
(1,2,3)
(2,3)
(1,3,2)
(2,1,3)
(1,3)
(1,3)
(2,3)
(1,3)
(2,3,1)
(2,3)
(3,1,2)
(3,2,1)
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••
• • •
Cayley graph of S4 •
•
•
•
• •
•
••
•
Simplicial geometry and transportation polytopes,
Trans. Amer. Math. Soc. 217 (1976) 138.
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Geodesics for Sn
(1,2,3,4)
(2,1,3,4)
• Hexagons on S3
slices
• Rectangles
on Klein
4-group
slices
(1,2,4,3)
(2,1,4,3)
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Betti numbers
i() = number of vertices with down degree i
= ith Betti number

down degree 2
 = (1, m2, 1)
for convex m-gon
1
1
down degree 1
0
When is  = (0, 1, …) independent of  ?
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… convex not required
 = (2,1,2)
 = (2,2,2)
 = (k, m2k, k) for (convex) m-gon
winding k times (k < m/2, gcd(k,m)=1)
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… nor need vertices be distinct
 = (2,4,2)
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… polygon not required
 = (1, 4, 4, 1)
 = (1, 2, 2, 1)
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… some hypothesis is necessary
 = (1, 2, 1)
 = (2, 0, 2)
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Inflection free geodesics
• A geodesic is inflection free if it winds
consistently in the same direction in its plane
• All our examples have inflexion free geodesics
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Betti number invariance
Theorem: Inflection free geodesics
 Betti numbers independent of 
down degrees
v:3, w:2
v
v:2, w:3
w
 Poincare duality (replace  by - )
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Projections help a lot
• Generic projection to R3 preserves our axioms
• Once you know the geodesics are coplanar in
R3 you can make all Betti number calculations
with a generic plane projection!
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McMullen reprise
• Theorem: Our Betti numbers are McMullen’s
• Proof: Every k-face has a unique lowest point,
number of down edges at a point determines
the number of k-faces rooted there
2C2 = 1 of these at
each of the 1 = 9
vertices with 2 up
edges
3C2 = 3 of these at
the 0 = 1 vertex
with 3 up edges
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McMullen reprise
• Betti number invariance implies the first
McMullen conjecture (palindromic)
• With our interpretation of the Betti numbers
how hard can it be to prove they are unimodal?
• Think of our plane pictures as a rotation
invariant Hasse diagram for a poset?
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Parallel redrawing
• Attach velocity vector to each vertex so that
when the vertices move the new edges are
parallel to the originals
• There are always at least n+1 linearly
independent parallel redrawings:
n translations and the dilation
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Theorem: A 3-independent embedded graph in Rn
with convex (hence inflection free) geodesics has
n0 + 1 = n + 1
independent parallel redrawings. n+1 of these are
trivial, 11 are interesting.
Proof:
Adapted from Guillemin and Zara
argument in equivariant cohomology of GKM
manifolds
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Simple convex polytopes
• n 0 +  1 = fn-1 = number of faces
• One parallel redrawing for each face
(includes translations and dilation)
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More examples
• J(n,2)
 = (1,1,2,2,3,3,4,…,4,3,3,2,2,1,1)
11 = 0, so no nontrivial parallel redrawings
• Symmetric groups
S3
 = (1,2,2,1)
S4
 = (1,3,5,6,5,3,1)
(Mahonian numbers count permutations by number
of inversions)
11 = n 2 nontrivial parallel redrawings
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Parallel redrawing in the plane
• Parallel redrawings correspond to infinitesimal
motions (rotate velocities 90°)
• Plane m-gon is braced by m3 diagonals, so
has m3+3 = m infinitesimal motions when we
count the rotation and two translations
•  = (k, m2k, k) so we expect 2k+m 2k = m
parallel redrawings when we count the dilation
and the two translations
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One parallel redrawing for each edge,
whether or not convex or inflection free
dilation and translations are
combinations of these
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When 3-independence fails
• Need extra awkward hypothesis:
geodesics must be exact
• Suggests parallel redrawing …
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Desargues’ configuration K2  K3
 = (1, 2, 2, 1), 11 = 1
motion
parallel deformation
(we need the exactness hypothesis)
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K(3,3)
 = (1, 2, 2, 1)
11 = 1
exactness
 inscribed in conic (converse of Pascal)
 has a motion (Bolker-Roth)
(infinitesimal) motion , parallel deformation
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The Petersen graph
An exact embedding
with two inflection
free geodesics.
= (1, 4, 4, 1)
6 redrawings
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Cuboctahedron
Ink on paper. Approximately 8" by 11".
Image copyright (c) 1994 by Andrew Glassner.
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http://mathworld.wolfram.com/
GreatStellatedDodecahedron.html
Inflection free geodesics
are pentagrams
 = (5, 5, 5, 5)
30 + 1 = 20 = f2,
so behaves as if simple and convex
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Small Stellated Dodecahedron
http://amath.colorado.edu/appm/staff/fast/
Polyhedra/ssd.html
Inflection free geodesics
are pentagrams and
triangles
= (3,1,2,2,1,3)
Unimodularity fails
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Great Dodecahedron
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Great Icosahedron
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Great Truncated Cuboctahedron
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Open questions
• Prove the Betti numbers unimodular
• Find the natural boundaries
– Understand the non-3-independent cases
– Understand 0 > 1 (stellations)
• Interpret strange examples topologically
• Make the projective invariance visible
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