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1 How is a graph like a manifold? Ethan Bolker Mathematics - UMass Boston eb@cs.umb.edu www.cs.umb.edu/~eb University of Florida, Gainesville March 19, 2002

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2 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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3 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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4 Counting faces of a polytope Euler: f k = number of faces of dimension k Define i by f n-k = ( ) i McMullen conjectures: For simple polytopes, i are palindromic and unimodal Stanley: Poincare duality palindromic hard Lefshetz theorem unimodal n-i k-i k i=0

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5 Dodecahedron f = (20, 30, 12, 1) = (1, 9, 9, 1) What do the i count?

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6 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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7 Johnson graphs J(n,k) Vertices are the k element subsets of an n-set v,w are adjacent when #(v w) = n-1 Represent vertices as bit vectors to embed on a hyperplane in n-space J(n,1) = K n (complete graph) J(4,2) is the octahedron J(n,2) is not the cross polytope Topology: Grassmannian manifold of k-planes in n-space {1,2} = (1,1,0,0) {1,3} {3,4} {2,4} {1,4} {2,3}

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8 Johnson graph geodesics Each pair of edges at a vertex determines a geodesic Geodesics are triangles squares

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9 Permutahedra Cayley graphs of the symmetric groups S n 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 “Internal” edges matter S 3 is the complete bipartite graph K(3,3) in the plane Topology: flag manifolds (1,2,3) (1,3,2) (2,3,1) (3,2,1) (3,1,2) (2,1,3)

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10 Geodesics for S 3 (1,2,3) (2,1,3) (2,3,1) (3,2,1) (3,1,2) (1,3,2) (1,3) (2,3) (1,3) (2,3) (1,3)

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11 Simplicial geometry and transportation polytopes, Trans. Amer. Math. Soc. 217 (1976) 138. Cayley graph of S 4

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12 Geodesics for S n Hexagons on S 3 slices Rectangles on Klein 4-group slices (1,2,3,4) (1,2,4,3) (2,1,4,3) (2,1,3,4)

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13 Betti numbers i ( ) = number of vertices with down degree i = i th Betti number 1 0 1 down degree 1 down degree 2 = (1, m 2, 1) for convex m-gon When is = ( 0, 1, …) independent of ?

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14 = (k, m 2k, k) for (convex) m-gon winding k times (k < m/2, gcd(k,m)=1) = (2,1,2) … convex not required = (2,2,2)

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15 … nor need vertices be distinct = (2,4,2)

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16 … polygon not required = (1, 2, 2, 1) = (1, 4, 4, 1)

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17 … some hypothesis is necessary = (1, 2, 1) = (2, 0, 2)

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18 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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19 Theorem: Inflection free geodesics Betti numbers independent of down degrees v:3, w:2 Betti number invariance v:2, w:3 v w Poincare duality (replace by - )

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20 Projections help a lot Generic projection to R 3 preserves our axioms Once you know the geodesics are coplanar in R 3 you can make all Betti number calculations with a generic plane projection!

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21 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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22 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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23 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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24 Theorem: A 3-independent embedded graph in R n 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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25 Simple convex polytopes n 0 + 1 = f n-1 = number of faces One parallel redrawing for each face (includes translations and dilation)

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26 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 S 3 = (1,2,2,1) S 4 = (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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27 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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28 One parallel redrawing for each edge, whether or not convex or inflection free dilation and translations are combinations of these

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29 When 3-independence fails Need extra awkward hypothesis: geodesics must be exact Suggests parallel redrawing …

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30 motion parallel deformation (we need the exactness hypothesis) Desargues’ configuration K 2 K 3 = (1, 2, 2, 1), 1 1 = 1

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31 (infinitesimal) motion, parallel deformation K(3,3) = (1, 2, 2, 1) 1 1 = 1 exactness inscribed in conic (converse of Pascal) has a motion (Bolker-Roth)

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32 The Petersen graph An exact embedding with two inflection free geodesics. = (1, 4, 4, 1) 6 redrawings

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33 Cuboctahedron Ink on paper. Approximately 8" by 11". Image copyright (c) 1994 by Andrew Glassner.

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34 http://mathworld.wolfram.com/ GreatStellatedDodecahedron.html Inflection free geodesics are pentagrams = (5, 5, 5, 5) 3 0 + 1 = 20 = f 2, so behaves as if simple and convex

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35 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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36 Great Dodecahedron

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37 Great Icosahedron

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38 Great Truncated Cuboctahedron

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39 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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