k -belts on fullerenes
Nikolay Erokhovets
Moscow State University
erochovetsn@hotmail.com
(Joint work with Victor M.Buchstaber)
3rd Workshop on
Analysis, Geometry and Probability
September 28 - October, 2, 2015
Ulm University, Germany
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Fullerenes
A (mathematical) fullerene is a simple 3-polytope with
all 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 that
Fi ∩ 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.
W2
Fi
Fj
W1
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 pairwise
intersecting facets Fi1 , . . . , Fik : Fis ∩ Fit 6= ∅, s, t = 1, . . . , k , has
a 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 if
either P = ∆3 ,
or P contains a 3-belt.
W2
Fi
Fj
W1
Fk
If we remove the 3-belt from the surface of a polytope, we
obtain 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 is
combinatorially equivalent to a connected sum P = Q1 #v1 ,v2 Q2
of two simple 3-polytopes Q1 and Q2 along vertices v1 and v2 .
v1
v2
Q1
Q2
P
The part Wi appears if we remove from the surface of the
polytope 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 into
two parts W1 and W2 , and W1 does not contain 3-belts. Then P
contains one of the following fragments
Fi
Fj
Fs
Fk
(1,1,1)
Fj
Fi
Fp F
q
Fk
(1,2,2)
Fp
Fi F
r
Fj
Fq
Fp
Fi
Fk
(2,2,2)
Fj
Fr Fq
Fk
(1,2,3)
This is impossible since each fragment has a triangle or a
quadrangle.
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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 and
hexagons. If P it is not flag, then it is combinatorially equivalent
to 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 and
hexagons 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 ' I 3 , or it is non-flag.
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4-belts of fullerenes
Theorem (Buchstaber-Erokhovets,15)
Any fullerene has no 4-belts.
W2
Fi
Fj
W1
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 incident
with 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)
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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)
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6-belts of fullerenes
Theorem
If there is a 6-belt not surrounding one of the fragments a)-f),
then (1) it consist of hexagons and (2) the fullerene belongs to
one of the 5 infinite families: F1 , F2 , F3 , F4 , F5 .
A fullerene from F2 consists of k consecutive hexagonal 6-belts
with hexagons incident with neighbors by opposite edges
glued up on both sides by one of the 5 caps:
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Family F3
a)
1
2
3
4
b)
c)
d)
Start with a).
At each step add one hexagon incident to the facet with a
single edge on the boundary.
the boundary facets have still (1, 3, 2, 2, 2, 2) edges on the
boundary.
In the end glue up the fragment a).
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Family F4
a)
1
2
3
4
b)
c)
d)
Start with a).
At each step add two hexagons incident to the faces with
single edges on the boundary.
the boundary facets have still (1, 2, 3, 1, 2, 3) edges on the
boundary.
In the end glue up the fragment a).
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Family F5
a)
1
2
3
4
b)
c)
d)
Start with a).
At each step add three hexagons incident to the faces with
single edges on the boundary.
the boundary facets have still (1, 3, 1, 3, 1, 3) edges on the
boundary.
In the end glue up the fragment a).
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IPR-fullerenes
Definition
An IPR-fullerene (Isolated Pentagon Rule) is a fullerene without
pairs of adjacent pentagons.
Corollary
Any 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-announcement
For any k there exist a finite families S1 and S2 of simple
partitions of a disk into 5- and 6-gons, and a list of infinite
families of fullerenes F1 , . . . , FNk , such that
1
2
Any disk from S1 can be surrounded by a k -belt on a
fullerene.
If there is a k-belt not surrounding a disk from S1 , then
it consist of hexagons;
the fullerene belongs to one of the families F1 , . . . , FNk .
3
Fullerenes in each family consist of consecutive k -belts of
hexagons glued up at the ends by two disks from S2 .
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Toric topology
Canonical correspondence
Simple polytope P
dim P = n
number of facets = m
−→
moment-angle manifold ZP
dim ZP = m + n
canonical T m -action
P1 × P2
−→
ZP1 × ZP2
The cohomology ring H ∗ (ZP ) is bigraded and even multigraded
and 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 =
β −2,6 =
β −3,8 =
(8+p6 )(9+p6 )
;
2
(6+p6 )(8+p6 )(10+p6 )
;
3
(4+p6 )(7+p6 )(9+p6 )(10+p6 )
.
8
Corollary
The product map H 3 (ZP ) ⊗ H 3 (ZP ) → H 6 (ZP ) is trivial.
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References I
Victor Buchstaber
Toric Topology of Stasheff Polytopes
MIMS EPrint: 2007.232
V.M.Buchstaber, T.E.Panov,
Toric Topology
AMS Math. Surveys and monogrpaphs. vol. 204, 2015. 518 pp.
V.M.Buchstaber, N. Erokhovets,
Graph-truncations of simple polytopes
Proc. of Steklov Math Inst, MAIK, Moscow, vol. 289, 2015.
M.Deza, M.Dutour Sikiric, M.I.Shtogrin,
Fullerenes and disk-fullerenes
Russian Math. Surveys, 68:4(2013), 665-720.
V.D.Volodin,
Combinatorics of flag simplicial 3-polytopes
Russian Math. Surveys, 70:1(2015); arXiv: 1212.4696.
B. Grünbaum,
Convex polytopes
Graduate texts in Mathematics 221, Springer-Verlag, New York, 2003.
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