From the Icosahedron to Natural Triangulations of ℂP 2 and S 2×S 2
Discrete Comput Geom
From the Icosahedron to Natural Triangulations of CP 2 and S 2 × S 2
Bhaskar Bagchi 0 1
Basudeb Datta 0 1
0 B. Datta ( ) Department of Mathematics, Indian Institute of Science , Bangalore 560 012 , India
1 B. Bagchi Theoretical Statistics and Mathematics Unit, Indian Statistical Institute , Bangalore 560 059 , India
We present two constructions in this paper: (a) a 10vertex triangulation CP120 of the complex projective plane CP 2 as a subcomplex of the join of the standard sphere (S42) and the standard real projective plane (RP 2, the decahedron), its 6 automorphism group is A4; (b) a 12vertex triangulation (S2 × S2)12 of S2 × S2 with automorphism group 2S5, the Schur double cover of the symmetric group S5. It is obtained by generalized bistellar moves from a simplicial subdivision of the standard cell structure of S2 × S2. Both constructions have surprising and intimate relationships with the icosahedron. It is well known that CP 2 has S2 × S2 as a twofold branched cover; we construct the triangulation CP120 of CP 2 by presenting a simplimciaalpreisalaizsaitmiopnliocfiatlhissucbodviveirsiniognmoafpthSe2 s×taSnd2a→rd CcePll2s.tTruhcetudroemoafinS2of×thSis2,sidmifpfleirceinatl from the triangulation alluded to in (b). This gives a new proof that Kühnel's CP 2 9 triangulates CP 2. It is also shown that CP120 and (S2 × S2)12 induce the standard piecewise linear structure on CP 2 and S2 × S2 respectively. The research of B. Datta was supported by UGC grant UGCSAP/DSAIV.
Triangulated manifolds; Complex projective plane; Product of 2spheres; Icosahedron

1 Introduction and Results
It is well known that the minimal triangulation RP62 of the real projective plane arises
naturally from the icosahedron. Indeed, it is the quotient of the boundary complex of
the icosahedron by the antipodal map. In this note, we report the surprising result that
there is a small triangulation (using only 10 vertices) of the complex projective plane
which is also intimately related to the icosahedron. Indeed, this simplicial complex
CP120 occurs as a subcomplex of the simplicial join S42 ∗ RP62. Our starting point is the
beautiful fact that CP 2 is homeomorphic to the symmetrized square (S2 × S2)/Z2
of the 2sphere, where Z2 acts by coordinate flip. So, letting S42 denote the 4vertex
triangulation of S2 (i.e., the boundary complex of the tetrahedron), we look for a
Z2stable simplicial subdivision of the product cell complex S42 × S42, without introducing
extra vertices. In order to ensure that the quotient complex (after quotienting by Z2)
does triangulate the quotient space (S2 × S2)/Z2 = CP 2, the Z2action on this
simplicial subdivision must be “pure” (cf. Definition 2 and Lemma 7 in Sect. 5). It turns
out that the following (S2 × S2)16 is the unique 16vertex triangulation satisfying
these requirements.
Description of (S2 × S2)16 The vertices are xij , 1 ≤ i, j ≤ 4. The full automorphism
group is A4 × Z2, where A4 acts on the indices and Z2 acts by xij ↔ xji . Modulo
this group the facets (maximal simplices) are the following:
x11x22x33x12x13, x11x22x12x14x34, x11x22x14x24x34, x11x22x21x24x31,
x11x22x24x31x34.
The full list of facets of (S2 × S2)16 may be obtained from these five basic facets by
applying the group A4 × Z2. Under this group, the first three basic facets form orbits
of length 24 each, while each of the last two forms an orbit of length 12, yielding
a total of 3 × 24 + 2 × 12 = 96 facets. It may be verified that the face vector of
(S2 × S2)16 is (16, 84, 216, 240, 96).
Description of CP120 Quotienting the above (S2 × S2)16 by the group Z2 generated
by the automorphism xij ↔ xji , we get the CP120 mentioned above. Its vertices are
xij , 1 ≤ i ≤ j ≤ 4. Let α, β be the generators of the alternating group A4 given by
α = (123), β = (12)(34). Then α, β act on the vertices of CP120 by:
α ≡ (x11x22x33)(x23x13x12)(x24x34x14),
β ≡ (x11x22)(x33x44)(x24x13)(x14x23).
The following are the basic facets of CP120 modulo A4 = α, β :
x11x22x33x12x13, x11x22x12x14x34, x11x22x14x24x34, x11x22x12x13x24,
x11x22x13x24x34.
The full list of facets of CP120 may be obtained from these five basic facets by
applying the group A4. Under this group, the first three basic facets form orbits of length
12 each, while each of the last two forms an orbit of length 6, yielding a total of
3 × 12 + 2 × 6 = 48 facets. The complex CP120 is 2neighborly and its face vector is
(10, 45, 110, 120, 48).
Here we prove the following:
Theorem 1 There are exactly two 16vertex simplicial complexes which (i) are
simplicial subdivisions of the cell complex S42 × S42, (ii) retain the selfhomeomorphism
α : (x, y) → (y, x) of S4  × S42 as a simplicial automorphism, and (iii) the action
2
of Z2 = α is pure (cf. Definition 2). These two complexes are isomorphic and one
of them is (S2 × S2)16.
Corollary 2 The complex CP120 := (S2 × S2)16/Z2 is a 10vertex triangulation of
CP 2. Its full automorphism group is A4.
Let T and I denote the solid tetrahedron and the icosahedron in R3 respectively.
Thus, the cell complex S42 × S42 alluded to above is a subcomplex of the boundary
complex of the product polytope T × T in R6. Although we do not present the details
in this paper, Theorem 1 can be strengthened (following the same line of arguments)
to show that there is a unique simplicial subdivision S156 of the cell complex ∂(T × T )
which is Z2stable with a pure Z2action. To our utter surprise, it turns out that as an
abstract simplicial complex, S156 is isomorphic to the combinatorial join S4 ∗ S122 of
2
the boundary complexes of T and I respectively.
Remark 1 This last fact has the following geometric interpretation. Let T I denote
the convex hull of T ∪ I, where T and I sit in two (threedimensional) affine
subspaces of R6 meeting at a point which is in the interior of both polyhedra. Then T I
is a simplicial 6polytope and the boundary complex of this polytope is
combinatorially isomorphic to a simplicial subdivision of the boundary complex of T × T . This
geometric result cries out for a geometric explanation; but we have none.
By the construction, (S2 × S2)16 is a subcomplex of S42 ∗ S122. Since the decahedron
RP62 is the quotient of S122 = ∂I by Z2, and Z2 acts trivially on S42 (the latter being
the combinatorial child of the “diagonal” S2 in S2 × S2, i.e., the S42 in Fig. 1), on
passing to the quotient, we find the surprising inclusion
CP120 ⊆ S42 ∗ RP62.
Indeed, S42 and RP62 occur as induced subcomplexes of CP120 on a complementary
pair of vertex sets. Since both S42 and RP62 are classical objects, and the combinatorial
join is such a well known operation on simplicial complexes, this inclusion says that
CP120 was all along sitting there right before our eyes!
The number 10 obtained here is not optimal. It is well known (cf. [
2–4, 10
]) that
any triangulation of CP 2 requires at least nine vertices, and there is a unique 9vertex
triangulation CP92 of this manifold, obtained by Kühnel [
11, 12
]. But, our
construction is natural in that it is obtained by a combinatorial mimicry of a topological
construction of CP 2. It shares this naturalness with another 10vertex triangulation, say
K140, of CP 2 available in the literature, namely the “equilibrium” triangulation of
Banchoff and Kühnel [
7
]. Here we prove the following:
K140.
Theorem 3 The simplicial complex CP120 is bistellar equivalent to both CP92 and
Corollary 4 (a) Kühnel’s 9vertex simplicial complex CP 2 triangulates CP 2.
9
(b) Both CP120 and CP92 induce the standard plstructure on CP 2.
Of course, in principle these ideas generalize to arbitrary dimensions. In general,
the d dimensional complex projective space CP d is the symmetric d th power of S2,
i.e., the quotient of (S2)d by the symmetric group Sd acting by coordinate
permutations. Unfortunately, even in the next case d = 3, it is not possible to subdivide the
cell complex S42 × S42 × S42 into a simplicial complex, with a pure S3action,
without adding more vertices. Indeed, we found that we need to add 60 more vertices
to obtain an (S2 × S2 × S2)124. On quotienting, we obtain a CP330—again with full
automorphism group A4. The details are so complicated that we decided to postpone
publication. We are presently trying to see if one can apply bistellar moves to this
CP330 to reduce the number of vertices. It is known that any triangulation of CP 3
requires at least 17 vertices (cf. [
2
]).
After we submitted a preliminary version of this paper to arXiv:1004.3157v1,
2010, Ulrich Brehm [
9
] communicated to us that he had the idea of obtaining CP120
as a quotient of a 16vertex S2 × S2 in the 1980’s; however, he never published the
details.
We obtain a second simplicial subdivision (S2 × S2)16 of S42 × S42.
Description of (S2 × S2)16 This is a second simplicial subdivision of the cell
complex S42 × S42. It has the same vertex set and automorphism group A4. Modulo the
group A4, its basic facets are:
x11x12x13x21x31, x11x12x14x21x31, x11x13x14x21x31, x12x13x23x31x32,
x12x14x21x24x31, x12x14x24x31x34, x12x21x24x31x32, x12x24x31x32x34.
Each facets is in an orbit of length 12, yielding a total of 8 × 12 = 96 facets.
The complex (S2 × S2)16 has the same face vector as (S2 × S2)16, namely,
(16, 84, 216, 240, 96).
We perform a finite sequence of generalized bistellar moves on (S2 × S2)16 and
obtain the following 12vertex triangulation (S2 × S2)12 of S2 × S2.
Description of (S2 × S2)12 The vertices are xij , 1 ≤ i = j ≤ 4. Its
automorphism group 2S5 is generated by the two automorphisms h = (x12x14x21x24x31) ×
(x13x42x43x32x34) and g = (x12x21x24x42x14x41x43x34x13x31x32x23). Modulo this
group, (S2 × S2)12 is generated by the following two basic facets:
x12x14x21x24x31, x12x13x14x21x31.
The first basic facet is in an orbit of size 12, while the second is in an orbit of size 60,
yielding a total of 72 facets. Its face vector is (12, 60, 160, 180, 72).
Theorem 5 The simplicial complex (S2 × S2)12 is a triangulation of S2 × S2. Its full
automorphism group is 2S5, the nonsplit extension of Z2 by S5.
The complex (S2 × S2)12 has many remarkable properties. Its automorphism
group is transitive on its vertices and edges. All its vertices have degree 10 and all
its edges have degree 8. Indeed, the link of each edge is isomorphic to the 2sphere
S82 obtained from the boundary complex of the octahedron by starring two vertices in
a pair of opposite faces. Also, all triangles of (S2 × S2)12 are of degree 3 or 5. The
automorphism group is transitive on its triangles of each degree. The degree 3
triangles constitute a weak pseudomanifold whose strong components are two icosahedra.
Thus, we find a pair I1, I2 of icosahedra sitting canonically inside the 2skeleton of
(S2 × S2)12. These two icosahedra are “antimorphic” in the sense that the identity
map is an antimorphism between them (cf. Definition 1 below). The structure of
(S2 × S2)12 is completely described in terms of this antimorphic pair of icosahedra.
The full automorphism group 2S5 of (S2 × S2)12 is a double cover of the common
automorphism group of these two icosahedra.
Again, the number 12 here is not optimal. In [
12
], Kühnel and Laßmann have
shown that any triangulation of S2 × S2 needs at least 11 vertices, and in [
14
], Lutz
finds (via computer search) several 11vertex triangulations of S2 × S2, all with trivial
automorphism groups. Surprisingly, even though (S2 × S2)12 is not minimal, it does
not admit any proper bistellar moves. Thus, there is no straightforward way to obtain
a minimal triangulation of S2 × S2 starting from (S2 × S2)12.
In [
17
], Sparla proved two remarkable inequalities on the Euler characteristic
χ of a combinatorial 4manifold M satisfying certain conditions. His first result
is that if there is a centrally symmetric simplicial polytope P of dimension d ≥ 6
such that M ⊆ ∂ P and skel2(M ) = skel2(P ), then 10(χ − 2) ≥ 43 (d−31)/2 .
Equality holds here if and only if P is a cross polytope (i.e., dual of a hypercube). His
second result is: if M has 2d vertices and admits a fixed point free involution then
10(χ − 2) ≤ 43 (d−31)/2 . Equality holds if and only if M also satisfies the
hypothesis of the first result for a cross polytope P . Notice that, in view of the Dehn–
Sommerville equations, equality in either inequality determines the face vector of
M in terms of d alone. To obtain an example of equality (in both results) with d = 6,
Sparla searched for (and found) a 4manifold with the predicted face vector under the
assumption of an automorphism group A5 × Z2. To determine the topological type
of the resulting 12vertex 4manifold, he had to compute its intersection form and
then appeal to Freedman’s classification of simply connected smooth 4manifolds.
We believe that our approach to Sparla’s complex not only elucidates its true genesis,
but also reveals its rich combinatorial structure and contributes to an elementary
determination of its topological type. Note, however, that Sparla’s approach reveals yet
another remarkable property of (S2 × S2)12. It provides a tight rectilinear embedding
of S2 × S2 in R6.
Remark 2 If X is a triangulated 4manifold on at most 12 vertices, then its vertex
links are homology 3spheres on at most 11 vertices, and hence (cf. [
5
]) are
combinatorial spheres. Thus all triangulated 4manifolds on at most 12 vertices are
combinatorial manifolds. (More generally, this argument yields: All triangulated d manifolds
on at most d + 8 vertices are combinatorial manifolds.) In particular, both CP120 and
(S2 × S2)12 are combinatorial manifolds. Actually, an old result of Bing [
8
] says that
all the vertex links of any triangulated 4manifold are simply connected triangulated
3manifolds. Therefore, in view of Perelman’s theorem (Poincaré conjecture) [
15
],
all triangulated 4manifolds are combinatorial manifolds, irrespective of the number
of vertices.
Remark 3 In [
1
], Akhmedov and Park have shown that S2 × S2 has countably
infinite number of distinct smooth structures. Since there is an one to one correspondence
between the smooth structures and plstructures on a 4manifold (cf. [16, p. 167]), it
follows that S2 × S2 has infinitely many distinct plstructures. Since (S2 × S2)16 and
(S2 × S2)16 are simplicial subdivisions of S42 × S42, it follows that the plstructures
given by (S2 × S2)16 and (S2 × S2)16 are standard. Again, (S2 × S2)12 is
combinatorially equivalent to (S2 × S2)16 (cf. Remark 5) and hence gives the same plstructure
as (S2 × S2)16. So, all the triangulations of S2 × S2 discussed here give the standard
plstructure on S2 × S2.
2 Preliminaries
All simplicial complexes considered here are finite and the empty set is a simplex (of
dimension −1) of every simplicial complex. We now recall some definitions here.
For a finite set V with d + 2 (d ≥ 0) elements, the set ∂ V (respectively, V¯ ) of
all the proper (resp. all the) subsets of V is a simplicial complex and triangulates
the d sphere Sd (resp. the (d + 1)ball). The complex ∂ V is called the standard d
sphere and is also denoted by Sdd+2(V ) (or simply by Sd
the standard (d + 1)ball and is also denoted by Dddd+++212()V. T)h(eorcosmimpplelyx Vb¯yisDcdda++ll21e)d.
(Generally, we write X = Xnd to indicate that X has n vertices and dimension d .)
For simplicial complexes X, Y with disjoint vertex sets, their join X ∗ Y is the
simplicial complex whose simplices are all the disjoint unions A ∪ B with A ∈ X,
B ∈ Y .
If σ is a simplex of a simplicial complex X then the link of σ in X, denoted by
lkX(σ ), is the simplicial complex whose simplices are the simplices τ of X such that
τ ∩ σ = ∅ and σ ∪ τ is a simplex of X. The number of vertices in the link of σ is
called the degree of σ . Also, the star of σ , denoted by starX(σ ) or star(σ ), is the
subcomplex σ¯ ∗ lkX(σ ) of X.
For a simplicial complex X, X denotes the geometric carrier. It may be described
as the subspace of [
0, 1
]V (X) (where V (X) is the vertex set of X) consisting of all
functions f : V (X) → [
0, 1
] satisfying (i) Support(f ) ∈ X and (ii) x∈V (X) f (x) =
1. If a space Y is homeomorphic to X then we say that X triangulates Y . If X is a
topological manifold (respectively, d sphere) then X is called a triangulated manifold
(resp. triangulated d sphere). If X is a pl manifold (with the plstructure induced
by X) then X is called a combinatorial manifold. For 1 ≤ d ≤ 4, X is a combinatorial
d manifold if and only if the vertex links are triangulated (d − 1)spheres.
The face vector of a d dimensional simplicial complex is the vector (f0, f1, . . . ,
fd ), where fi is the number of idimensional simplices in the complex.
If X is a d dimensional pure simplicial complex (i.e., every maximal simplex is
d dimensional) and D, D are triangulations of the d ball such that (i) ∂D = ∂D =
D ∩ X, and (ii) D ⊆ X, then the simplicial complex X := (X \ D) ∪ D˜ is said to
be obtained from X by a generalized bistellar move (GBM) with respect to the pair
(D, D). Clearly, in this case, X and X triangulate the same topological space and if
u is a vertex in ∂D then lkX(u) is obtained from lkX(u) by a GBM (cf. [
6
]).
In particular, let A be a simplex of X whose link in X is a standard sphere ∂B.
Suppose also that B ∈ X. Then, we may perform the GBM with respect to the pair
of balls (A ∗ ∂B, B ∗ ∂A). Such an operation is called a bistellar move, and will be
denoted by A → B. Also, if C is any simplex of X and x is a new symbol, then we
may perform the GBM on X with respect to the pair (C¯ ∗ lkX(C), (x¯ ∗ ∂C) ∗ lkX(C)).
The resulting simplicial complex X is said to be obtained from X by starring the
vertex x in the simplex C. In case C is a facet, this is a bistellar move—the only sort
of bistellar move which increases the number of vertices. All other kinds of bistellar
moves are said to be proper.
Two pure simplicial complexes are called bistellar equivalent if one is obtained
from the other by a finite sequence of bistellar moves. If X is obtained from Y by
the bistellar move A → B then the complex Z obtained from Y by starring a new
vertex u in B is a subdivision of both X and Y . This implies that bistellar equivalent
complexes induce same plstructure on their common geometric carrier.
The group Z2 acts on S2 × S2 by coordinate flip. The following proposition is
well known to algebraic geometers (cf. [
13
]):
Proposition 6 The quotient space (S2 × S2)/Z2 is homeomorphic to the complex
projective plane CP 2.
3 Relations with the Icosahedron
Emergence of the Icosahedron Let T0 be the tetrahedron with vertex set V =
{x1, x2, x3, x4}. Then, viewed abstractly, the boundary complex of the product
polytope T0 × T0 has vertex set V × V , and faces A × B, where A and B range over all
the subsets of V . The product cell complex for S42 × S42 = (∂T0) × (∂T0) is the
subcomplex consisting of cells A × B, where A and B range over all the proper subsets
of V . We use the notation xij to denote the vertex (xi , xj ) of T0 × T0. For i = j ,
k = l, xij xkl forms an edge of T0 × T0 if and only if it is one of the solid edges of
the icosahedron in Fig. 1. (This picture is a Schlegel diagram obtained by projecting
the boundary of the icosahedron on one of its faces. Thus, there is only one “hidden”
face (namely, x41x42x43) in the picture. What is important for us is the label given to
the vertices.)
Notice that the broken edges in the icosahedron are precisely the edges xij xkl
where {i, j, k, l} is an even permutation of {1, 2, 3, 4}.
To obtain the appropriate triangulation of S2 × S2, we join xii to all vertices for all
i and also introduce the broken edges of the icosahedron. Thus viewed, one sees the
simplicial subdivision (S2 × S2)16 of the cell complex (∂T0) × (∂T0) as a subcomplex
of (∂T ) ∗ (∂I), where T is the tetrahedron with vertex set {xii : 1 ≤ i ≤ 4} and I is
the icosahedron depicted in Fig. 1.
Notice also that the Z2action xij ↔ xji fixes the vertices of T and acts on I as
the antipodal map. Thus, going modulo Z2, we find CP120 as a subcomplex of the
5dimensional simplicial complex S42 ∗ RP62, where S42 is the 4vertex 2sphere given
by the boundary complex of T and RP62 is the (minimal) triangulation of the real
projective plane (with vertices of the same name being identified) given in Fig. 1.
From our nomenclature for the vertices, the inclusion CP120 ⊆ S42 ∗ RP62 is obvious,
as is the fact that (∂T ) ∗ (∂I) is a simplicial subdivision of the boundary complex of
T0 × T0.
i
Finally, note that Δi = {xij : j = i} and Δ = {xji : j = i} are triangles of the
icosahedron, and {Δ1, Δ2, Δ3, Δ4} and {Δ1, Δ2, Δ3, Δ4} are antipodal quadruples
(consisting of triangles) partitioning the vertex set of the icosahedron. It is easy to
see that there are exactly five such pairs in the icosahedron, and the automorphism
group A5 × Z2 of I acts transitively on them. The stabilizer of each such pair is
A4 × Z2, and A4 acts regularly on the vertex set of I. Our choice of nomenclature
for the vertices of I amounts to choosing one such antipodal pair of quadruples. This
is because we have Δi ∩ Δj = ∅ if i = j and = {xij } if i = j . Viewed dually, one
sees Kepler’s regular tetrahedra embedded in the dodecahedron. Namely, the centers
of Δi , 1 ≤ i ≤ 4 (as well as of Δi , 1 ≤ i ≤ 4) are the vertices of a regular tetrahedron
inscribed in the dual dodecahedron.
The 12vertex triangulation (S2 × S2)12 of S2 × S2 is obtained from (S2 × S2)16
by a sequence of bistellar moves (cf. proof of Theorem 5). However, its most elegant
description requires the introduction of the following definition.
Definition 1 Let I1 and I2 be two copies of the icosahedron. A bijection f : V (I1) →
V (I2) is said to be an antimorphism if, for all vertices x, y of I1, we have (a) x and
y are at distance one in I1 if and only if f (x) and f (y) are at distance two in I2, and
(b) x and y are at distance two in I1 if and only if f (x) and f (y) are at distance one
in I2. (It follows that x and y are at distance 3 (antipodal) in I1 if and only if f (x) and
f (y) are at distance 3 (antipodal) in I2.) Here distance refers to the usual graphical
distance on the respective edge graph. In case V (I1) = V (I2) and the identity map is
an antimorphism between I1 and I2, then we say that I1 and I2 are antimorphic. Thus,
the two icosahedra in Fig. 2 below are antimorphic (the map, taking each vertex of
the left icosahedron in Fig. 2 to the vertex of the same name in the right icosahedron,
is an antimorphism).
Another description of (S2 × S2)12 Take an antimorphic pair of icosahedra, say I1
and I2 (with common vertex set V ). It turns out that I1 and I2 have the identical
automorphism group A5 × Z2 (not merely isomorphic; cf. Lemma 9 below). Also,
there is a bijection ϕ from the triangles of I1 to the triangles of I2 such that for
each triangle Δ = abc of I1, ϕ(Δ) = ij k is the only triangle of I2 for which aij ,
bj k and cik are triangles of I2 (cf. Lemma 9). Now, the vertex set of (S2 × S2)12
is V (= V (I1) = V (I2)) and it has two types of facets. (i) For each vertex x, the
neighbors of x in I1 form facets. (ii) For each triangle Δ of I1 and each vertex
y in Δ = ϕ(Δ), (Δ ∪ Δ ) \ {y} is a facet. Thus (S2 × S2)12 has 12 facets of the
first type and 20 × 3 = 60 facets of the second type. From the description, it is
clear that the common automorphism group A5 × Z2 of I1 and I2 is an
automorphism group of (S2 × S2)12. It turns out that its full automorphism group is 2S5
generated by the two automorphisms g = (x12x21x24x42x14x41x43x34x13x31x32x23)
and h = (x12x14x21x24x31)(x13x42x43x32x34). The automorphism g interchanges I1
and I2.
Remark 4 It should be emphasized that the existence of an antimorphic pair of
icosahedra (exploited in the above construction of (S2 × S2)12) is a minor miracle, and
only an empirically verified fact. Its deeper geometric significance, if any, remains to
be understood.
4 A Selfdual CW Decomposition of CP 2
Here we have taken the cell complex ∂T0 × ∂T0, and triangulated it to obtain the
simplicial complex (S2 × S2)16 and finally quotiented this simplicial complex by
Z2 to obtain CP120. This procedure reflects our obsession with simplicial complexes.
However, one may straightaway quotient the cell complex by Z2 to obtain a
(nonregular) CW decomposition of CP 2. This CW complex is selfdual in the sense
that its facevector (10, 24, 31, 24, 10) exhibits a curious palindromic symmetry. We
proceed to describe it in some details. Consider the Z2action on R6 ≡ R3 × R3
given by (x, y) ↔ (y, x). Let η: R6 → R6/Z2 be the quotient map. We know that
η(S2 × S2) = CP 2. We give a CW decomposition W of the space η(∂T0 × ∂T0).
For 0 ≤ i ≤ 4, let W i denote the set of icells in W . For i = 2 the icells in W are
the images (under the map η) of icells in ∂T0 × ∂T0. A 2cell in W is the image of a
2cell F in ∂T0 × ∂T0 which is not of the form E × E for some edge E in ∂T0. More
explicitly
W 0 = V CP120 ,
W 1 = η(E) : E is an edge of ∂T0 × ∂T0 ,
W 2 = η xij xikxil  : 1 ≤ j < k < l ≤ 4, 1 ≤ i ≤ 4
∪ η xi xj  × xkxl  : i < j, k < l and either i < k or i = k and j < l ,
W 3 = η(A) : A is a 3cell of ∂T0 × ∂T0
W 4 = η(B) : B is a 4cell of ∂T0 × ∂T0 .
and
Then, W 1 contains 24 cells, W 2 contains 16 + 15 = 31 cells, W 3 contains 4 ×
6 = 24 cells and W 4 contains 10 cells. Clearly, each 1cell in W is regular (i.e.,
homeomorphic to a closed interval). Since all the 2cells are homeomorphic images
of the corresponding 2cells in ∂T0 × ∂T0, it follows that all the 2cells in W are
regular.
For 0 ≤ i ≤ 4, let Xi = β∈W 0∪···∪W i β. Then ∂α ⊆ Xi−1 if α ∈ W i for i = 3.
Let γ be a 3cell in W . If γ = η(xi xj xk × xi xj ), i < j < k, then γ is
obtained from xi xj xk × xi xj  by identifying xii xjj xij  with xii xjj xji  (by the
identification given by xij ↔ xji ). Thus, γ is a regular 3cell and ∂γ = η(xi xk ×
xi xj ) ∪ η(xj xk × xi xj ) ∪ η(xii xji xki ) ∪ η(xij xjj xkj ). (Now, it is clear why
we do not have to take η(xi xj  × xi xj ) in W 2. In fact, η(xi xj  × xi xj ) is
inside of γ .) Therefore, ∂γ ⊆ X2. Same things are true if γ = η(xi xj xk × xi xk) or
η(xi xj xk × xj xk). On the other hand, if γ = η(F × E), where E is an edge and F
is a 2simplex and E ⊆ F , then γ is homeomorphic to F × E and hence is a regular
3cell. In this case, it follows from the definition of W 2 that ∂γ ⊆ X2. Thus W is a
CW complex.
If σ is a 4cell in W then, either σ = η(xi xj xk × xi xj xk), for some i < j < k
or σ = η(xi xj xk × xi xj xl ), where {i, j, k, l} is an even permutation of {1, 2, 3, 4}.
In the first case, σ is homeomorphic to xii xjj xkkxij xik ∪ xii xjj xkkxij xjk ∪
xii xjj xkkxikxjk and hence σ is a regular 4cell. In the second case, σ is obtained
from xi xj xk × xi xj xl  by identifying xii xjj xij  with xii xjj xji  (by the
identification given by xij ↔ xji ). So, σ is not a regular cell. Thus W 4 contains four regular
4cells and six singular 4cells.
Since each cell in W is the quotient of a cell in S42 × S42, (S2 × S2)16 is a simplicial
subdivision of S42 × S42 and CP120 is the quotient of (S2 × S2)16, it follows that CP120
is a simplicial subdivision of W .
5 Proofs
Definition 2 Let G be a group of simplicial automorphisms of a simplicial complex
X with vertex set V (X). We shall say that the action of G on X is pure if it satisfies:
(a) whenever u, v are distinct vertices from the same Gorbit, uv is a nonedge of X,
and (b) for each Gorbit θ ⊆ V (X) and each α ∈ X, the stabilizer Gα of α in G acts
transitively on θ ∩ V (lkX(α)).
Lemma 7 Let G be a group of simplicial automorphisms of a simplicial complex X.
Let q: V (X) → V (X)/G denote the quotient map, and X/G := {q(α) : α ∈ X}. If the
action of G on X is pure then X/G is a simplicial complex which triangulates X/G
(where the action of G on V (X) is extended to an action of G on X piecewise
linearly, i.e., affinely on the geometric carrier of each simplex). That is, we have
X/G = X/G.
Proof The condition (a) ensures that the quotient map q is oneone on each simplex
of X. The simplicial map q: X → X/G induces a piecewise linear continuous map
q from X onto X/G.
Claim The fibers of q: X → X/G are precisely the Gorbits on simplices of X (that is,
if α, α ∈ X are such that q(α) = q(α ) then there exists g ∈ G such that g(α) = α ).
We prove the claim by induction on k = dim(α) = dim(α ). The claim is trivial for
k = −1. So, assume k ≥ 0, and the claim is true for all smaller dimensions. Choose
a simplex β ⊆ α of dimension k − 1, and let β ⊆ α be such that q(β ) = q(β).
By induction hypothesis, β and β are in the same Gorbit. Therefore, applying a
suitable element of G, we may assume, without loss of generality, that β = β. Let
α = β ∪ {x}, α = β ∪ {x }. Then q(x) = q(x ), i.e., x and x are in the same Gorbit.
Now, by assumption (b), there is a g ∈ Gβ such that g(x) = x . Then g(α) = α . This
proves the claim.
In the presence of condition (a), the claim ensures that the fibers of q are
precisely the Gorbits on points of X. Hence q induces the required homeomorphism
between X/G and X/G.
Up to isomorphism, there are exactly two 6vertex 2spheres, namely, S1 and S2
given in Fig. 3. We need the following lemma to prove Theorem 1.
Lemma 8 Let C be the triangular prism given in Fig. 3(b) (i.e., C is the product of a
2simplex and an edge). Up to isomorphism, there exists a unique 6vertex simplicial
subdivision C of C. The facets (tetrahedra) in C are a1b1b2b3, a1a2b2b3, a1a2a3b3.
Moreover, ∂C is isomorphic to S2 of Fig. 3(a) and determines C uniquely.
Proof Let C be a 6vertex subdivision of C. Then there exists a 3simplex σ in C
which contains the 2simplex b1b2b3. Without loss of generality, we may assume that
σ = a1b1b2b3. Then C is the union of σ and the pyramid P given in Fig. 4. Since we
are not allowed to introduce new vertices, clearly the rectangular base of P must be
triangulated using two triangles, in one of two isomorphic ways, and the remaining
tetrahedra in C must have the apex of P as a vertex and one of these two triangles
as base. Thus, without loss of generality, P = a1a2b2b3 ∪ a1a2a3b3. This proves the
first part.
The last part follows from the fact that the facets of C are the maximal cliques in
the 1skeleton of ∂C.
Proof of Theorem 1 Let X be a 16vertex simplicial subdivision of S42 × S42 satisfying
(i), (ii) and (iii).
For i = j , consider the 2cell xi xj × xi xj . By (iii), xij xji can not be an edge
in X. This implies that xii xjj , xii xji xjj , xii xij xjj ∈ X and xi xj × xi xj = xii xji xjj ∪
xii xij xjj (cf. Fig. 5(a)).
For i, j, k distinct, consider the 2cell xi xj × xi xk . Since X satisfies (iii), both xij
and xji cannot be in lkX(xik). Now, xikxij is an edge in the cell complex S42 × S42 and
hence is an edge in X. Thus, xikxji can not be an edge in X. This implies that xii xjk ,
xii xji xjk , xii xikxjk ∈ X and xi xj × xi xk = xii xji xjk ∪ xii xikxjk (cf. Fig. 5(b)).
Consider the 2cell x1x3 × x2x4. Clearly, x1x3 × x2x4 = x12x32x34 ∪ x12x14x34 or
= x12x32x14 ∪ x32x14x34.
Case 1. x1x3 × x2x4 = x12x32x34 ∪ x12x14x34 (cf. Fig. 5(c)). So, x12x34 ∈ X. Then,
by (ii), x21x43 ∈ X and, by (iii), x12x43, x21x34 ∈ X. This implies that x2x3 × x1x4 =
x21x31x24 ∪ x31x24x34 (cf. Fig. 5(d)). So, x31x24 ∈ X. Then, by (ii), x13x42 ∈ X and,
by (iii), x13x24, x31x42 ∈ X. This implies that x1x2 × x3x4 = x13x23x14 ∪ x23x14x24
(cf. Fig. 5(e)). So, x14x23 ∈ X. Then, by (ii), x41x32 ∈ X and, by (iii), x14x32,
x41x23 ∈ X. These give the 2skeleton of X. Observe that we have already 84 edges
as mentioned in the construction of (S2 × S2)16 and, since X satisfies (iii), all the 36
remaining 2sets are nonedges in X.
Observe that any 3cell in S42 × S42 is the product of a 2simplex and an edge. For
i, j, k distinct, consider the 3cell xi xj xk × xi xj . Since xii xjj , xii xki and xjj xki are
edges, by Lemma 8, xi xj xk × xi xj = xii xij xkj xjj ∪ xii xki xkj xjj ∪ xii xki xji xjj is
the unique subdivision of xi xj xk × xi xj (cf. Fig. 6(a)). Similarly, xi xj × xi xj xk =
xii xji xjkxjj ∪ xii xikxjkxjj ∪ xii xikxij xjj is the unique subdivision of xi xj × xi xj xk
(cf. Fig. 6(b)).
For i, j, k, l distinct, consider the 3cell xi xj xk × xi xl . Here xii xjl and xii xkl are
edges. By interchanging j and k (if required) we may assume that {i, j, k, l} is an
even permutation of {1, 2, 3, 4}. Then xki xjl is an edge and hence, by Lemma 8,
xi xj xk × xi xl = xii xil xkl xjl ∪ xii xki xkl xjl ∪ xii xki xji xjl is the unique subdivision of
xi xj xk × xi xl (cf. Fig. 6(c)). Similarly, for the 3cell xi xl × xi xj xk , we may assume
that {i, j, k, l} is an even permutation of {1, 2, 3, 4}. Then xikxlj is an edge and hence,
by Lemma 8, xi xl × xi xj xk = xii xli xlkxlj ∪ xii xikxlkxlj ∪ xii xikxij xlj is the unique
subdivision of xi xl × xi xj xk (cf. Fig. 6(d)). These give the 3skeleton of X.
For i, j, k distinct, consider the 4cell A = xi xj xk × xi xj xk . The boundary ∂A
of A consists of six 3cells. From above, it follows that S1({xii , xjj , xkk}) ∗ C6 ⊆
X is the subdivision of ∂A, where C6 is the 6cycle C6(xij , xik, xjk, xji , xki , xkj ).
Let D ⊆ X be the subdivision of A. Then, D is a 9vertex 4ball with boundary
∂D = S1({xii , xjj , xkk}) ∗ C6. Clearly, C6 is an induced subcomplex of X. Therefore,
each 4simplex in B must contain xii xjj xkk . Thus, xii xjj xkk is a simplex in D \
∂D. Therefore, lkD(xii xjj xkk) is a cycle and hence = C6. These imply that D =
xii xjj xkk ∗ C6.
Now, consider the 4cell B = xi xj xk × xi xj xl , where i, j, k, l are distinct. By
interchanging i and j (if required) we may assume that {i, j, k, l} is an even
permutation of {1, 2, 3, 4}. The boundary ∂B of B consists of six 3cells. From above,
it follows that the subdivision of ∂B in X is a 9vertex triangulated 3sphere and
1
obtained from S3 ({xii , xjj , xkl }) × C5 by starring the vertex xji in the 3simplex
α := xii xjj xjl xki , where C5 is the 5cycle C5(xij , xil , xjl , xki , xkj ). Since xji xij ,
xji xil , xji xkj and xji xkl are nonedges, it follows that σ := xii xjj xji xjl xki is the
only possible 4simplex containing xji inside B. So, σ ∈ X. Then B = σ ∪ P , where
1
P is a 4cell such that P ∩ σ = α and S3 ({xii , xjj , xkl }) ∗ C5 ⊆ X is the
subdivision of ∂P in X (i.e., P is the 4cell whose geometric carrier is (B \ σ ) ∪ α).
1
Let Q be the simplicial subdivision of P in X. So, ∂Q = S3 ({xii , xjj , xkl }) ∗ C5.
Since C5 is induced in X, it follows that any 4simplex in Q must contain xii xjj xkl .
Since xii xjj xkl ∈ Q \ ∂Q, lkQ(xii xjj xkl ) is a cycle and hence = C5. These imply
that Q = xii xjj xkk ∗ C5. Then B = (xii xjj xkl ∗ C5) ∪ σ¯ .
Now, we have subdivided all the 4cells in S42 × S42. It is routine to check that the
resulting simplicial complex X is identical with the complex (S2 × S2)16 defined in
Sect. 1.
Case 2. x1x3 × x2x4 = x12x32x14 ∪ x32x14x34. By the same method as in
Case 1, one can show that X is uniquely determined and is isomorphic to (S2 ×
S2)16 via the map f given by the transposition (1, 2) on the suffixes, i.e., f ≡
(x11x22)(x13x23)(x14x24)(x31x32)(x41x42). This completes the proof.
Proof of Corollary 2 From Proposition 6, Lemma 7 and Theorem 1, it is immediate
that CP120 triangulates CP 2.
Since the automorphism group A4 = α, β of (S2 × S2)16 commutes with Z2,
it descends to an automorphism group A4 = α¯ , β¯ of CP120. We need to show that
there are no other automorphisms.
It is easy to check that the four vertices xii , 1 ≤ i ≤ 4, are the only ones with
2neighborly links. Therefore, the full automorphism group must fix this set of four
vertices. Since A4 is 2transitive on this 4set, it suffices to show that there is no
nontrivial automorphism γ fixing both x11 and x22. Suppose the contrary. Then γ is a
nontrivial automorphism of lk(x11x22). But lk(x11x22) is the 8vertex triangulated
2sphere given in Fig. 7.
From the picture, it is apparent that lk(x11x22) has only one nontrivial
automorphism, namely (x13, x24)(x14, x23)(x33, x44). Therefore, γ = (x13, x24)(x14, x23)(x33,
x44) and hence γ fixes the 3simplex x11x33x44x34. Then γ must either fix or
interchange the two vertices x13 and x14 in the link of this 3simplex, a contradiction. This
completes the proof.
Proof of Theorem 3 Consider the following sequence of bistellar moves on CP120
(performed one after the other):
(i) x22x33x44 → x23x24x34,
(iii) x11x22x44 → x12x14x24,
(v) x22x34x44 → x13x23x24,
(vii) x12x22x44 → x13x14x24,
(ix) x22x44 → x13x14x23x24.
(ii) x11x33x44 → x13x14x34,
(iv) x14x33x44 → x12x13x34,
(vi) x23x33x44 → x12x24x34,
(viii) x33x44 → x12x13x24x34,
At the end of these moves, we get a 10vertex triangulation K of CP 2.
On K we perform the following sequence of bistellar moves one after another.
(x) x11x24x44 → x12x14x23,
(xii) x11x44 → x12x14x23x34,
(xiv) x44x14 → x34x12x13x23,
(xi) x11x13x44 → x14x23x34,
(xiii) x44x14x24 → x12x13x23,
(xv) x44 → x24x34x12x13x23.
(Note that the last three bistellar moves together is same as the GBM with respect to
(x44 ∗ S3 ({x14, x24, x34}) ∗ S3 ({x12, x13, x23}), S31({x14, x24, x34}) ∗ x12x13x23).) The
1 1
last bistellar move deletes the vertex x44 and hence obtain a 9vertex triangulation L
of CP 2. (Observe that A1 = {x11, x23, x24}, A2 = {x14, x33, x12}, A3 = {x34, x22, x13}
is an amicable partition of L whose layer is of first type (cf. [
4
]).)
Let CP92 be as described in [
11
] with vertex set {1, 2, . . . , 9}. Consider the map
ϕ: L → CP92 given by: ϕ(x11) = 1, ϕ(x23) = 2, ϕ(x24) = 3, ϕ(x34) = 4, ϕ(x22) =
5, ϕ(x13) = 6, ϕ(x14) = 7, ϕ(x33) = 8, ϕ(x12) = 9. It is easy to see that ϕ is an
isomorphism. Thus, CP120 is bistellar equivalent to CP 2.
9
Now, on K we perform the following sequence of bistellar moves:
(xvi) x11x22x33 → x12x13x23,
(xviii) x22x33x13 → x12x14x23,
(xvii) x22x33x24 → x14x23x34,
(xix) x22x33 → x12x14x23x34.
We obtain a 10vertex triangulation M of CP 2. Let K140 be as described in [
7
] with
vertex set {X, Y, Z, 0, 1, . . . , 6}. Consider the map ψ : M → K140 given by ψ (x33) =
X, ψ (x22) = Y , ψ (x44) = Z, ψ (x11) = 0, ψ (x13) = 1, ψ (x12) = 2, ψ (x23) = 3,
ψ (x14) = 4, ψ (x34) = 5, ψ (x24) = 6. It is easy to see that ψ is an isomorphism.
Thus, CP120 is bistellar equivalent to K140. This completes the proof.
Proof of Corollary 4 Part (a) follows from Corollary 2 and Theorem 3.
In [
7
], explicit coordinates for simplices of K140 in the Fubini–Study metric were
given. This shows that the induced plstructure on CP 2 by K140 is the standard one.
Part (b) now follows from Theorem 3.
Lemma 9 Let I1 and I2 be an antimorphic pair of icosahedra. Then we have:
(a) Aut(I1) = Aut(I2) = A5 × Z2.
(b) For each triangle Δ of I1, there is a unique triangle Δ of I2 such that each of the
three triangles of I2 sharing an edge with Δ has its third vertex in Δ. Further,
the map ϕ: Δ → Δ is a bijection from the triangles of I1 to the triangles of I2.
There is a similarly defined bijection ψ from the triangles of I2 to the triangles
of I1, and
(c) Every isomorphism f : I1 → I2 intertwines ϕ and ψ .
(Warning: The maps ϕ and ψ are not induced by any vertextovertex map!)
Proof Recall that I1 and I2 have the same vertex set and the same pairs of
antipodal vertices. Thus, they have the same antipodal map (sending each vertex x to its
antipode x¯ ). Now, the full automorphism group of the icosahedron is generated by
its rotation group A5 and the antipodal map. So, to prove Part (a), it suffices to show
that I1 and I2 share the same rotation group. For each pair x, x¯ of antipodes, Ii has a
i
rotation symmetry αx,x¯ which fixes x and x¯ and rotates the remaining vertices along
the 5cycles lkIi (x) and lkIi (x¯). The rotation group of Ii is generated by these
automorphisms of order five. But, lkI2 (x) (respectively, lkI2 (x¯)) is the graph theoretic
2
complement of the pentagon lkI1 (x¯) (respectively, lkI1 (x)). Therefore, αx,x¯ is the
square of αx1,x¯ . This proves Part (a).
Notice that if f1, f2: I1 → I2 are two antimorphisms, then f1 ◦ f2−1 ∈ Aut(I2) and
f2−1 ◦ f1 ∈ Aut(I1). Thus, the antimorphism is unique up to right multiplication by
elements of Aut(I1) (or left multiplication by elements of Aut(I2)). Therefore, there
is no loss of generality in taking the antimorphic pair of icosahedra as the one given
in Fig. 2.
Since the common automorphism group is transitive on the triangles of I1 (and
of I2), it is enough to look at the triangle Δ = x12x13x14 of I1. From Fig. 2, we
see that the links in I2 of two vertices of Δ have exactly two vertices in
common. Namely, we have V (lkI2 (x12)) ∩ V (lkI2 (x13)) = {x21, x32}, V (lkI2 (x12)) ∩
V (lkI2 (x14)) = {x24, x41}, V (lkI2 (x13)) ∩ V (lkI2 (x14)) = {x31, x43}. Therefore, any
triangle Δ of I2 satisfying the requirement must be contained in the vertex set
{x21, x32, x24, x41, x31, x43}. But one sees that this set of six vertices contains a unique
triangle in I2, namely Δ = x21x31x41. Thus the map ϕ: Δ → Δ is well defined.
Similarly, there is a well defined map ψ from the triangles of I2 to the triangles of I1.
The map ψ ◦ ϕ is the antipodal map on the triangles of I1 to themselves. Similarly,
ϕ ◦ ψ is the antipodal map on triangles of I2. Hence ϕ (as well as ψ ) is a bijection.
This proves Part (b).
To prove Part (c), let f be any isomorphism from I1 to I2. Since I1 and I2 are
antimorphic, it is immediate that f also defines an isomorphism from I2 to I1. Let
Δ be any triangle of I1 and let Δ = ϕ(Δ). By definition, there are three triangles
Δ1, Δ2, Δ3 of I2 each of which shares a vertex with Δ and an edge with Δ . Then
f (Δ) and f (Δ ) are triangles of I2 and I1, respectively. Also, f (Δ1), f (Δ2), f (Δ3)
are three triangles of I1 each of which shares a vertex with f (Δ) and an edge with
f (Δ ). Therefore, by definition of ψ , ψ (f (Δ)) = f (Δ ) = f (ϕ(Δ)).
Proof of Theorem 5 As in the proof of Theorem 1, one may verify that (S2 × S2)16
is a simplicial subdivision of S42 × S42, and hence it triangulates S2 × S2. We apply
the following sequence of bistellar moves to (S2 × S2)16 to create a second 16vertex
triangulation (S2 × S2)16 of S2 × S2:
Since this set of bistellar moves is stable under the automorphism group A4 of
(S2 × S2)16, it follows that (S2 × S2)16 inherits the group A4. Also, both complexes
have lk(x11) = S31({x12, x13, x14}) ∗ S31({x21, x31, x41}). However, while (S2 × S2)16
has both x12x13x14 and x21x31x41 as triangles, we have chosen the bistellar moves
judiciously to ensure that (S2 × S2)16 does not have the triangle x12x13x14. Therefore,
we may apply the following four GBM’s (one after the other) to (S2 × S2)16 to delete
the four vertices xii , 1 ≤ i ≤ 4:
st(x11), D32 {x12, x13, x14} ∗ S31 {x21, x31, x41} ,
st(x22), D32 {x21, x23, x24} ∗ S31 {x12, x32, x42} ,
st(x33), D32 {x31, x32, x34} ∗ S31 {x13, x23, x43} ,
st(x44), D32 {x41, x42, x43} ∗ S31 {x14, x24, x34} .
The resulting complex X is therefore a 12vertex triangulation of S2 × S2. So, to
confirm the first statement of this theorem, it suffices to show that X is isomorphic
to the complex (S2 × S2)12 described in Sect. 3. Indeed, with the antimorphic pair of
icosahedra (and their vertex names) as in Fig. 2, we shall show that we actually have
X = (S2 × S2)12.
Notice that X inherits the automorphism group A4 from (S2 × S2)16, and modulo
this group, the following six are basic facets of X:
x12x14x21x24x31, x12x13x14x21x31, x12x23x31x13x32,
x12x31x34x14x24, x24x31x32x12x21, x24x31x32x12x41.
Each basic facet is in an A4orbit of size 12, yielding a total of 6 × 12 = 72 facets
of X. Since (S2 × S2)12 also has 72 facets and since the group A4 (acting on
subscripts) is a subgroup of the automorphism group A5 × Z2 of (S2 × S2)12, it suffices
to observe that all six basic facets of X listed above are also facets of (S2 × S2)12.
Indeed, the first facet x12x14x21x24x31 is in (S2 × S2)12 since these five vertices are the
neighbors of x23 in I1 (and of x41 in I2). In each of the remaining five basic facets of
X, the first three vertices constitute a triangle Δ of I1 with the last two vertices in the
corresponding triangle Δ = ϕ(Δ) of I2 (cf. Lemma 9). (For instance, Δ = x12x13x14
is a triangle of I1, with corresponding triangle Δ = x21x31x41 of I2. Therefore, the
second basic facet of X is a facet of (S2 × S2)12.) This shows that (S2 × S2)12 = X,
so that (S2 × S2)12 triangulates S2 × S2.
To compute the full automorphism group of (S2 × S2)12, notice that it has exactly
40 triangles of degree 3 (the rest are of degree 5), namely the twenty triangles of
I1 and the twenty triangles of I2. Consider the graph whose vertices are these forty
triangles, two of them being adjacent if and only if they share an edge. This graph
has exactly two connected components, of size 20 each, namely the triangles of I1
and I2. This shows that any automorphism f of (S2 × S2)12 either fixes both I1 and
I2 or interchanges them. So, Aut(I1) = Aut(I2) = A5 × Z2 is a subgroup of index at
most two in the full automorphism group of (S2 × S2)12.
Let f : I1 → I2 be any isomorphism. Since I1 and I2 are antimorphic, it is
immediate that f is also an isomorphism from I2 to I1. Since the five
neighbors in I1 of any vertex are also the neighbors in I2 of the antipodal vertex, it
is immediate that f maps each of the 12 facets of the first kind in (S2 × S2)12
to a facet of the same kind. Also, for any triangle Δ of I1, the construction of
(S2 × S2)12 shows that lk(Δ) = S31(ϕ(Δ)), and also, for any triangle Δ of I2,
lk(Δ ) = S31(ψ (Δ )). Since f intertwines ϕ and ψ (Lemma 9), we also have
lk(f (Δ)) = S31(ψ (f (Δ))) = S31(f (ϕ(Δ))) = f (S31(ϕ(Δ))) = f (lk(Δ)). Similarly,
for any triangle Δ of I2, lk(f (Δ )) = f (lk(Δ )). Thus, f also maps all sixty facets
of the second type in (S2 × S2)12 to facets of the same type. Thus, any
isomorphism between I1 and I2 is also an automorphism of (S2 × S2)12. Therefore, the
full automorphism group G of (S2 × S2)12 has H = A5 × Z2 as an index two
subgroup. Thus, G is of order 240. Indeed, G consists of the 120 common
automorphisms of I1 and I2, and the 120 isomorphisms between I1 and I2. In particular, take
g = (x12x21x24x42x14x41x43x34x13x31x32x23), which is an isomorphism between I1
and I2. Note that g6 is the common antipodal map of I1 and I2, hence it is in the
center of G. Thus, G/ g6 is the extension of A5 by the involution α = g (mod g6).
But A5 has only one nontrivial extension by an involution, namely S5. So, G is an
extension of a central involution by S5. It can not be the split extension S5 × Z2 since
this has no element of order 12. Therefore, G is the unique nonsplit extension 2S5
of Z2 by S5.
Remark 5 If the link of a vertex u in a triangulated 4manifold X is S31({x, y, z}) ∗
S31({a, b, c}) and xyz is not a simplex in X then the GBM (stX(u), D32({x, y, z}) ∗
1
S3 ({a, b, c}) is equivalent to the sequence of the following three bistellar moves:
uab → xyz, ua → cxyz, u → bcxyz. Thus, from the proof of Theorem 5, (S2 ×
S2)12 can be obtained from (S2 × S2)16 by a sequence of bistellar moves only.
Acknowledgements The authors thank the anonymous referee for many useful comments which led to
substantial improvements in the presentation of this paper. The authors are thankful to Siddhartha Gadgil,
Frank Lutz, Wolfgang Kühnel and Alberto Verjovsky for useful conversations and references. The authors
thank Ipsita Datta for her help in the proof of Theorem 3.
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