k -belts on fullerenes

Nikolay ErokhovetsMoscow State University

erochovetsn@hotmail.com(Joint work with Victor M.Buchstaber)

3rd Workshop onAnalysis, Geometry and Probability

September 28 - October, 2, 2015Ulm University, Germany

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Fullerenes

A (mathematical) fullerene is a simple 3-polytope withall 2-facets pentagons and hexagons.

Fullerene C60 Truncated icosahedron

For any fullerene p5 = 12,

f0 = 2(10 + p6), f1 = 3(10 + p6), f2 = (10 + p6) + 2

There exist fullerenes with any p6 6= 1.

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k -belts

Let P be a simple convex 3-polytope.

A 3-belt is a set of three facets (Fi ,Fj ,Fk ) such thatFi ∩ Fj ,Fj ∩ Fk ,Fk ∩ Fi 6= ∅, and Fi ∩ Fj ∩ Fk = ∅.A k-belt, k > 4, is a cyclic sequence (F1, . . . ,Fk ) of facets,such that Fi ∩Fj 6= ∅ if and only if Fi and Fj are successive.

W1

W2

Fi

Fj

Fk

Fl

4-belt of a simple 3-polytope.

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Flag polytopes

A simple n-polytope is called flag if any set of pairwiseintersecting facets Fi1 , . . . ,Fik : Fis ∩ Fit 6= ∅, s, t = 1, . . . , k, hasa nonempty intersection Fi1 ∩ · · · ∩ Fik 6= ∅.

Flag polytope Non-flag polytope

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Non-flag 3-polytopes

Simple 3-polytope P is not flag if and only ifeither P = ∆3,or P contains a 3-belt.

W1

W2

Fi

Fj

Fk

If we remove the 3-belt from the surface of a polytope, weobtain two parts W1 and W2, homeomorphic to disks.

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Non-flag 3-polytopes as connected sums

The existence of a 3-belt is equivalent to the fact that P iscombinatorially equivalent to a connected sum P = Q1#v1,v2Q2of two simple 3-polytopes Q1 and Q2 along vertices v1 and v2.

PQ1 Q2

v1v2

The part Wi appears if we remove from the surface of thepolytope Qi the facets containing the vertex vi , i = 1,2.

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Fullerenes as flag polytopes

Theorem (Erokhovets,15)Any fullerene has no 3-belts, that is it is a flag polytope.

The proof is based on the following result about fullerenes.Let the 3-belt (Fi ,Fj ,Fk ) divide the surface of a fullerene P intotwo parts W1 and W2, and W1 does not contain 3-belts. Then Pcontains one of the following fragments

Fi

Fj

Fk

Fs Fi

Fj

Fk

Fp Fq Fi

Fj

Fk

FpFq Fi

Fj

Fk

FpFqFr Fr

(1,1,1) (1,2,2) (2,2,2) (1,2,3)

This is impossible since each fragment has a triangle or aquadrangle.

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Polytopes with 4-, 5-, and 6-gonal facets

Theorem (Erokhovets,15)Let P be a polytope with all facets quadrangles, pentagons andhexagons. If P it is not flag, then it is combinatorially equivalentto the connected sum of k cubes in opposite vertices.

Q1 Q2 Q3 Q4

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Non-flag polytopes with 4-, 5-, and 6-gonal facets

a) b) c)

Non-flag polytopes with all facets quadrangles, pentagons andhexagons can be also described in the following way:

take the fragment a);add arbitrary number k > 0 of hexagonal 3-belts;glue up the opposite end by the fragment a).

Proposition

If P contains a), then either P ' I3, or it is non-flag.

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4-belts of fullerenes

Theorem (Buchstaber-Erokhovets,15)Any fullerene has no 4-belts.

W1

W2

Fi

Fj

Fk

Fl

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5-belts of fullerenes

Theorem (Buchstaber-Erokhovets,15)Any pentagonal face is surrounded by a 5-belt.There are 12 belts of this type.

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5-belts of fullerenes

Theorem (Buchstaber-Erokhovets,15)If there is a 5-belt not surrounding a pentagon, then

1 it consists of hexagons;2 the fullerene belongs to the infinite family F1:

k consecutive hexagonal 5-belts with hexagons incidentwith neighbors by opposite edges glued up by caps a).

3 the number of 5-belts is 12 + k.If a fullerene contains the cap a), then it belongs to F1.

a) b) c) 12 / 22

6-belts of fullerenes

Theorem (Erokhovets,15)The fragments a), b), c), d), f) are surrounded by a 6-belt.The fragment e) is surrounded by a 6-belt iff F is a 6-gon.

a) b) c)

F

d) e) f)13 / 22

6-belts of fullerenes

TheoremIf there is a 6-belt not surrounding one of the fragments a)-f),then (1) it consist of hexagons and (2) the fullerene belongs toone of the 5 infinite families: F1, F2, F3, F4, F5.

A fullerene from F2 consists of k consecutive hexagonal 6-beltswith hexagons incident with neighbors by opposite edgesglued up on both sides by one of the 5 caps:

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Family F3

a) b) c) d)

1 Start with a).2 At each step add one hexagon incident to the facet with a

single edge on the boundary.3 the boundary facets have still (1,3,2,2,2,2) edges on the

boundary.4 In the end glue up the fragment a).

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Family F4

a) b) c) d)

1 Start with a).2 At each step add two hexagons incident to the faces with

single edges on the boundary.3 the boundary facets have still (1,2,3,1,2,3) edges on the

boundary.4 In the end glue up the fragment a).

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Family F5

a) b) c) d)

1 Start with a).2 At each step add three hexagons incident to the faces with

single edges on the boundary.3 the boundary facets have still (1,3,1,3,1,3) edges on the

boundary.4 In the end glue up the fragment a).

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IPR-fullerenes

DefinitionAn IPR-fullerene (Isolated Pentagon Rule) is a fullerene withoutpairs of adjacent pentagons.

CorollaryAny 5-belt or 6-belt of an IPR-fullerene P surrounds a facet.In particular, P has 12 five-belts and p6 six-belts.

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k -belts

Conjecture-announcementFor any k there exist a finite families S1 and S2 of simplepartitions of a disk into 5- and 6-gons, and a list of infinitefamilies of fullerenes F1, . . . ,FNk , such that

1 Any disk from S1 can be surrounded by a k-belt on afullerene.

2 If there is a k-belt not surrounding a disk from S1, thenit consist of hexagons;the fullerene belongs to one of the families F1, . . . ,FNk .

3 Fullerenes in each family consist of consecutive k-belts ofhexagons glued up at the ends by two disks from S2.

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Toric topology

Canonical correspondence

Simple polytope P moment-angle manifold ZPdim P = n −→ dimZP = m + n

number of facets = m canonical T m-action

P1 × P2 −→ ZP1 ×ZP2

The cohomology ring H∗(ZP) is bigraded and even multigradedand has strong connection to combinatorics of P.

1 β−1,6 is equal to the number of 3-belts.2 If β−1,6 = 0, then β−2,8 is equal to the number of 4-belts.3 If β−1,6 = β−2,8 = 0, Then β−3,10 is the number of 5-belts.

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Theorem (Buchstaber-Erokhovets,15)For a fullerene P

β−1,6 = 0 – the number of 3-belts.β−2,8 = 0 – the number of 4-belts.β−3,10 = 12 + k, k > 0 – the number of 5-belts.If k > 0, then p6 = 5k.

β−1,4 = (8+p6)(9+p6)2 ;

β−2,6 = (6+p6)(8+p6)(10+p6)3 ;

β−3,8 = (4+p6)(7+p6)(9+p6)(10+p6)8 .

Corollary

The product map H3(ZP)⊗ H3(ZP)→ H6(ZP) is trivial.

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References I

Victor BuchstaberToric Topology of Stasheff PolytopesMIMS EPrint: 2007.232

V.M.Buchstaber, T.E.Panov,Toric TopologyAMS Math. Surveys and monogrpaphs. vol. 204, 2015. 518 pp.

V.M.Buchstaber, N. Erokhovets,Graph-truncations of simple polytopesProc. of Steklov Math Inst, MAIK, Moscow, vol. 289, 2015.

M.Deza, M.Dutour Sikiric, M.I.Shtogrin,Fullerenes and disk-fullerenesRussian Math. Surveys, 68:4(2013), 665-720.

V.D.Volodin,Combinatorics of flag simplicial 3-polytopesRussian Math. Surveys, 70:1(2015); arXiv: 1212.4696.

B. Grünbaum,Convex polytopesGraduate texts in Mathematics 221, Springer-Verlag, New York, 2003.

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