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Motivation

Motivation. 5. Example 1: Fano configuration (projective plane). 6. 7. 3. 1. 4. 2. Exercises. N1: Determina all lines passing through the point 3 of Fano plane. N2: How many lines pass through 4 and 6? N3: How many lines pass through 1, 2, 3?

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Motivation

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  1. Motivation 5 Example 1: Fano configuration (projective plane). 6 7 3 1 4 2

  2. Exercises • N1: Determina all lines passing through the point 3 of Fano plane. • N2: How many lines pass through 4 and 6? • N3: How many lines pass through 1, 2, 3? • N4*: Is it possible to draw Fano plane with straight lines? • N5*: Describe various properties of Fano plane. For example: Each number from 1 to 7 appears exactly three times in the table.

  3. Incidence structure • An incidence structureCis a triple • C = (P,L,I) where • P is the set ofpoints, • Lis the set of blocks or lines • I  P Lis anincidence relation. • Elements from I are calledflags. • The bipartite incidence graph G(C) with black vertices P, white verticesLand edges I is known as theLevi graph of the structure C.

  4. Pappus Configuration 1 2 3 2’’ 3’’ 1’’ 2’ 3’ 1’

  5. Naloge • N1: Koliko točk in koliko premic ima Papusova konfiguracija • N2: Koliko premic poteka skozi posamezno točko Papusove konfiguracije? • N3: Koliko je največ in koliko je najmanj premic, ki potekajo skozi par različnih točk Papusove konfiguracije? • N4: Zapiši tabelo (kombinatorični opis) za Papusovo konfiguracijo. • N5*: V sadovnjak zasadi 9 dreves v 10 vrst, tako da bodo v vsaki od desetih vrst po 3 drevesa!

  6. Še en zgled • Slika na levi vsebuje 4 točke 1,2,3,4 in štiri krožnice a,b,c,d. Kombinatorični opis daje konfiguracijska tabela: 1 2 4 3

  7. Še en zgled - nadaljevanje • Konfiguracijsko tabelo lahko prikažemo “grafično”.

  8. Še en zgled - konec B • Naloge: • N1: Napiši konfiguracijsko tabelo oglišč in lic četverca (tetraedra). • N2: Napiši konfiguracijsko tabelo za konfiguracijo oglišč in robov četverca. • N3: Pokaži, da prva in zadnja konfiguracija ne razločujeta. 4 3 1 A D 2 C

  9. Miquelova konfiguracija • Konfiguracija 4 točk in 6 krožnic na levi se imenuje Miquelova konfiguracija. [Tu pojmujemo tudi premice za krožnice!] • Konfiguracija izhaja iz Miquelovega izreka.

  10. Miquelova konfiguracija - nadaljevanje Naloge: • N1: Ali je tudi konfiguracija na levi Miquelova? Preveri s primerjavo kombinatoričnih opisov. • N2: Koliko točk in koliko krivulj imata konfiguraciji?

  11. Examples • 1. Each graph G = (V,E) is an incidence structure: P = V, L = E, (x,e) 2 I if and only if x is an endvertex of e. • 2. Any family of sets FµP(X) is an incidence structure. P = X, L = F, I = 2. • 3. A line arrangement L = {l1, l2, ..., ln} consisting of a finite number of n distinct lines in Euclidean plane E2 defines an incidence structure. Let V denote the set of points from E2 that are contained in at least two lines from L. Then: P = V, L = L and I is the point-line incidence in E2.

  12. Exercises • N1. Draw the Levi graph of the incidence structure defined by the complete bipartite graph K3,3. • N2. Draw the Levi graph of the incidence structure defined by the powerset P({a,b,c}). • N3. Determine the Levi graph of the incidence structure, defined by an arrangemnet of three lines forming a triangle in E2.

  13. Incidence geometry • Incidence geometry (G,c)of rank k is a graph G with a proper vertrex coloring c, where k colors are used. • Sometimes we denote the geometry by (G,c,I,~). Here c:VG ! I is the coloring and |I| = k is the number of colors, also known as the rank of G. Also ~ is the incidence. • I is the set of types. Note that only object of different types may be incident.

  14. Examples • 1. Each incidence structure is a rank 2 geometry. (Actually, look at its Levi graph.) • 2. Each 3 dimensional polyhedron is a rank 3 geometry. There are three types of objects: vertices, edges and faces with obvious geometric incidence. • 3. Each (abstract) simplicial complex is an incidence geometry. • 4. Any complete multipartite graph is a geometry. Take for instance K2,2,2, K2,2,2,2, K2,2, ..., 2.

  15. Incidence geometries of rank 2 • Incidence geometries of rank 2 are simply bipartite graphs with a given black and white vertex coloring. • Rank 2 geometries that are in addition connecte: d and the valence of each verteix is at least 2d(G) >1 Pasini geometries.

  16. Example of Rank 2 Geometry • Graph H on the left is known as the Heawood graph. • H is connected • H is trivalent: d(H) = D(H) = 3. • H je bipartite. • H is a Pasini geometry.

  17. Another View • Geometry of the Heawood graph H has another interpretation. • Rank = 2. There are two types of objects in Euclidean plane, say, points and curves. • There are 7 points, 7 curves, 3 points on a curve, 3 curves through a points. • The corresponding Levi graph is H!

  18. In other words ... • The Heawood graph (with a given black and white coloring) is the same thing as the Fano plane (73), the smallest finite projective plane. • Any incidence geometry can be interpeted in terms of abstract points, lines. • If we want to distinguish geometry (geometric interpretation) from the associated graph (combinatorial description) we refer to the latter the Levi graph of the corresponding geometry.

  19. Simlest Rank 2 Geometries Cycle (Levi Graph) • “Simplest” geometries of rank 2 in the sense of Pasini are even cycles. For instance the Levi graph C6 corresponds to the triangle. Triangle (Geometry)

  20. Rank 3 • Incidence geometries of rank 3 are exactly 3-colored graphs. • Pasini geometries of rank 3 are much more restricted. Currently we are interested in those geometries whose residua are even cycels. • Such geometries correspond to Eulerian surface triangulations with a given 3-vertex coloring.

  21. Reye Configuration • Reye Configuration of points, lines and planes in the 3-dimensional projective space consists of • 8 + 1 + 3 = 12 points (3 at infinity) • 12 + 4 = 16 lines • 6 + 6 = 12 planes.

  22. Theodor Reye • Theodor Reye (1838 - 1919), German Geometer. • Known for his book :Geometrie der Lage(1866 in 1868). • Published this configuration in 1878. • Posed “the problem of configurations.”

  23. Centers of Similitude • We are interested in tangents common to two circles in the plane. • The two intersections are called the centers of similitudes of the two circles. The blue center is called the internal (?), the red one is the external.(?) • If the radii are the same, the external center is at infinity.

  24. Residual geometry • Each incidence geometry • G =(G, ~, c, I) • (G,~) a simple graph • c, proper vertex coloring, • I collection of colors. • c: VG! I • Each element x 2 VG determines a residual geometry Gx. defined by an induced graph defined on the neighborgood of x in G. G Gx x

  25. Reye Configuration -Revisited • Reye configuration can be obtained from centers of similitudes of .... (check Hilbet ...)

  26. Flags and Residuals • In an incidence geometry G a clique on m vertices (complete subgraph) is called a flag of rank m. • Residuum can be definied for each flag F ½ V(G). G(F) = Å{G(x) = Gx |x 2 F}. • A maximal flag (flag of rank |I|} is called a chamber. A flag of rank |I|-1 is called a wall. • To each geometry G we can associate the chamber graph: • Vertices: chambers • Two chamber are adjacent if and only if they share a common wall. • (See Egon Shulte, ..., Titts systems)

  27. The 4-Dimensional Cube Q4. 0010 0001 0000 0100 1000

  28. The Geometry of Q4. • Vertices (Q0) of Q4: 16 • Edges (Q1)of Q4: 32 • Squares (Q2) of Q4: 24 • Cubes (Q3) of Q4: 8 • Total: 80 • The Levi graph of Q4 has 80 vertices and is colored with 4 colors.

  29. Residual geometries of Q4.

  30. Exercises • N1: Determine all residual geometries of Reyeve configuration • N2: Determine all residual geometries of Q4. • N3: Determine all residual geometries of Platonic solids. • N4: Determine the Levi graph of the geometry for the grup Z2£Z2£Z2, with three cyclic subgroups, generated by 100, 010, 001, respectively. • (Add Exercises for truncations!!!)

  31. COMBINATORIAL CONFIGURATIONS

  32. (Combinatorial) Configuration • A (vr,bk) configuration is an incidence structure C = (P,L,I) of points and lines,such that • v = |P| • b = |L| • Each point lies on r lines. • Each line contains k points. • Two lines intersect in at most one point.. • Pozor: Levijev graf je semiregularen z ožino 6

  33. Symmetric configurations • A (vr,bk) configuration is symmetric, if • v = b (this is equivalent to r = k). • A (vk,bk) configuration is usually denoted by (vk) configuration.

  34. Small Configurations • Triangle is the only (32)configuration. • Pasch configuration (62,43) and its dualcomplete quadrangle (43,62) share the same Levi graph S(K4).

  35. Desargues Configurationand itsLevi Graph G(10,3)

  36. Problem: Find a Model! • We know that G(10,3) embeds in double torus. • Is there a nice model out there? • Here are the faces: • 7, 18, 19, 10, 11, 12, 13, 8 • 6, 15, 14, 13, 12, 17, 18, 7 • 5, 6, 7, 8, 9, 10, 19, 20 • 2, 3, 14, 15, 16, 17, 12, 11 • 1, 2, 11, 10, 9, 4, 5, 20 • 1, 16, 15, 6, 5, 4, 3, 2 • 1, 20, 19, 18, 17, 16 • 3, 4, 9, 8, 13, 14

  37. Exercises

  38. CYCLIC CONFIGURATIONS AND HAAR GRAPHS

  39. Haarov graf naravnega števila Naravnemu številu n priredimo graf na naslednji način: Naj k označuje število števk v dvojiškem zapisu števila n in naj bodo {bk-1, bk-2, ..., b1, b0} dvojiške števke števila n. Graf H(n) = H(k; n), ki mu rečemo Haarov graf števila n, ima vozlišča ui, vi, i=0,1,...,k-1. Vozlišče ui povežemo z vozliščem vi+j, če in samo če je bj = 1. Opomba: Če za k vzamemo število, ki je večje od števila števk, dobimo v seveda drugačen graf!

  40. Zgled Poiščimo H(37). Dvojiški seznam: • {1,0,0,1,0,1} Število k = 6. H(37) = H(6,37) je graf na levi!

  41. Dipol qn • Dipol qn ima dve vozlišči, ki ju povezuje n povezav. Če želimo vozlišči posebej označiti, imenujemo prvo vozlišče črno, drugo pa belo. Na levi vidimo q5. • Vsak dipol je dvodelen graf, zato so tudi krovi nad dipoli dvodelni. Dipol q3 je kubičen graf, ki mu nekateri pravijo kar theta grafq.

  42. Ciklični krov nad dipolom • Vsak Haarov graf je tudi ciklični krov nad dipolom. Na prejšnjem zgledu se lahko naučimo recept: • H(37) je določen s številom 37, oziroma z dvojiškim nizom • (1 0 0 1 0 1) • Dolžina niza je k=6, od tod grupa Z6. Pod niz zapišemo indekse: • (0 1 2 3 4 5) • Enice se pojavljajo na mestih 0, 3 in 5. To pa določa napetosti na levi. Pripadajoči krov je H(37). 0 3 5 Z6

  43. Naloge • Graf na levi strani se imenuje Heawoodov graf. Dokaži: • Je dvodelen • Je Haarov. (Poišči ustrezno število n!) • Zapiši ga v obliki cikličnega krova nad dipolom. • Ne vsebuje nobenega cikla dolžine < 6. • Je najmanjši kubični graf brez ciklov dolžine < 6.

  44. Povezani Haarovi grafi • Graf G je povezan, če lahko najdemo pot med poljubnima vozliščema. • Med Haarovimi grafi obstajajo tudi nepovezani. Tak je na primer H(10). • Pravimo, da je število povezano, če je pripadajoči Haarov graf povezan. • Primeri majhnih nepovezanih števil: 2,4,8,10,16,32,34,36,40,42,64.

  45. Naloge • Dokaži, da so vse pozitivne potence števila 2 nepovezana števila. • Pokaži, da je Möbius-Kantorjev graf G(8,3) Haarov graf nekega naravnega števila. Katero število je to? • Poišči vse posplošene Petersenove grafe, ki so Haarovi grafi kakšnega števila. • Pokaži, da obstajajo Haarovi grafi, ki so tudi cirkulanti. • Pokaži, da obstajajo Haarovi grafi, ki niso cirkulanti.

  46. Naloge, nadaljevanje • Dokaži, da so vsi Haarovi graf vozliščno tranzitivni. • Dokaži, da je vsak Haarov graf Cayleyev graf za diedrsko grupo. • Dokaži, da obstajajo dvodelni Cayleyevi grafi diedrskih grup, ki niso Haarovi (npr. graf na levi je tak).

  47. Naloge, konec • Števili n in m sta ciklično ekvivalentni, če in samo če lahko dvojiški niz, ki pripada prvemu prevedemo na dvojiški niz drugega števila s “cikličnimi” avtomorfizimi. To pomeni, da lahko nize ciklično permutiramo, zrcalimo ali pa indekse množimo s številom tujim proti dolžini niza. • Števili n in m sta Haarovo ekvivalentni, če sta njuna Haarova grafa izomorfna: H(n) = H(m). • Dokaži, da ciklična ekvivalenca implicira Haarovo ekvivalenco. • Z računalnikom preveri, da sta števili 137331 in 143559 Haarovo ekvivalentni, nista pa ciklično ekvivalentni.

  48. Graf Marka Watkinsa • Kubični Haarov graf H(536870930) ima zanimivo lastnost. Je najmanjši povezan Haarov graf, ki med ciklično ekvivalentnimi nima nobenega lihega števila. • Dokaži, da je vsak Haarov graf lihega števila H(2n+1) hamiltonski in zato povezan.

  49. Ožina povezanih Haarovih grafov • K2 je edini povezani enovalentni Haarov graf. • Sodi cikli C2n so povezani dvovalentni Haarovi grafi. • Izrek: Naj bo H povezan Haarov graf valence d > 2. Tedaj je njegova ožina bodisi 4 bodisi 6.

  50. Ciklične konfiguracije • Simetrično (vr) konfiguracijo, ki jo določa prvi stolpec s tabele in dobimo preostale stolpce z zaporednim prištevanjem enice (po izbranem modulu m) imenujemo ciklična konfiguracija: Cyc(m;s). • Na levi sliki je ciklična Fanova konfiguracija Cyc(7;1,2,4) = Cyc(7;0,1,3).

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