Classification of superpotentials
CLASSIFICATION OF SUPERPOTENTIALS
A. DANCER AND M. WANG
Abstract. We extend our previous classification [DW4] of superpotentials of “scalar curvature
type” for the cohomogeneity one Ricci-flat equations. We now consider the case not covered in
[DW4], i.e., when some weight vector of the superpotential lies outside (a scaled translate of) the
convex hull of the weight vectors associated with the scalar curvature function of the principal orbit.
In this situation we show that either the isotropy representation has at most 3 irreducible summands
or the first order subsystem associated to the superpotential is of the same form as the Calabi-Yau
condition for submersion type metrics on complex line bundles over a Fano Kähler-Einstein product.
0. Introduction
In this paper we continue the study we began in [DW4] of superpotentials for the cohomogeneity
one Einstein equations. These equations are the ODE system obtained as a reduction of the
Einstein equations by requiring that the Einstein manifold admits an isometric Lie group action
whose principal orbits G/K have codimension one [BB], [EW]. As discussed in [DW3], these
equations can be viewed as a Hamiltonian system with constraint for a suitable Hamiltonian H, in
which the potential term depends on the Einstein constant and the scalar curvature of the principal
orbit, and the kinetic term is essentially the Wheeler-deWitt metric, which is of Lorentz signature.
For any Hamiltonian system with Hamiltonian H and position variable q, a superpotential is a
globally defined function u on configuration space that satisfies the equation
(0.1) H(q, duq) = 0.
From the classical physics viewpoint, u is a C2 (rather than a viscosity) solution of a time-
independent Hamilton-Jacobi equation. The literature for implicitly defined first order partial
differential equations then suggests that such solutions are fairly rare. It is therefore not unrea-
sonable to expect in our case that one can classify (at least under appropriate conditions) those
principal orbits where the associated cohomogeneity one Einstein equations admit a superpotential.
The existence of such a superpotential u in our case leads naturally to a subsystem of equations
of half the dimension of the full Einstein system. One way to see this is via generalised first integrals
which are linear in momenta, described in [DW4]. Schematically, the subsystem may be written as
q̇ = J∇u where J is an endomorphism related to the kinetic term of the Einstein Hamiltonian.
String theorists have exploited the superpotential idea in their search for explicit metrics of special
holonomy (see for example [CGLP1], [CGLP2],[CGLP3], [BGGG] and references in [DW4]). The
point here is that the subsystem defined by the superpotential often (though not always) represents
the condition that the metric has special holonomy. Also, the subsystem can often be integrated
explicitly.
In [DW4], §6, we obtained classification results for superpotentials of the cohomogeneity one
Ricci-flat equations. Besides assuming that G and K are both compact, connected Lie groups such
that the isotropy representation of G/K is multiplicity-free, we also mainly restricted our attention
to superpotentials which are of the same form as the scalar curvature function of G/K, i.e., a finite
sum with constant coefficients of exponential terms. Almost all the known superpotentials are of
this kind.
Date: revised October 24, 2018.
The second author was partly supported by NSERC grant No. OPG0009421.
http://arxiv.org/abs/0704.0210v1
2 A. DANCER AND M. WANG
However, the above classification results were further subject to the technical assumption that
the extremal weights for the superpotential did not lie in the null cone of the Wheeler-de Witt
metric. In [DW4] we gave some examples of superpotentials which do not satisfy this hypothesis.
These included several new examples which do not seem to be associated to special holonomy.
In this paper, therefore, we attempt to solve the classification problem without the non-null
assumption on the extremal weights.
As in [DW4], we use techniques of convex geometry to analyse the two polytopes naturally
associated to the classification problem. The first is (a rescaled translate of) the convex hull
conv(W) of the weight vectors appearing in the scalar curvature function of the principal orbit.
The second is the convex hull conv(C) of the weight vectors in the superpotential. In [DW4] we
showed that the non-null assumption forces these polytopes to be equal, so we could analyse the
existence of superpotentials by looking at the geometry of conv(W).
In the current paper, conv(C) may be strictly bigger than conv(W) because of the existence of
vertices outside conv(W) but lying on the null cone of the Wheeler-de Witt metric. Our strategy
is to consider such a vertex c and project conv(W) onto an affine hyperplane separating c from
conv(W). We can now analyse the existence of superpotentials in terms of the projected polytope.
The analysis becomes considerably more complicated because, whereas in [DW4] we could analyse
the situation by looking at the vertices and edges of conv(W), now, because we have projected onto
a subspace of one lower dimension, we have to consider the 2-dimensional faces of conv(W) also.
We find that in this situation the only polytopes conv(W) arising from principal orbits with more
than three irreducible summands in their isotropy representations are precisely those coming from
principal orbits which are circle bundles over a (homogeneous) Fano product. In the latter case,
the solutions of the subsystem defined by the superpotential correspond to Calabi-Yau metrics, as
discussed in [DW4].
After a review of basic material in §1, we state the main classification theorem of the paper in
§2 and give an outline of the strategy of the proof there.
1. Review and notation
In this section we fix notation for the problem and review the set-up of [DW4].
Let G be a compact Lie group, K ⊂ G be a closed subgroup, and M be a cohomogeneity one
G-manifold of dimension n + 1 with principal orbit type G/K, which is assumed to be connected
and almost effective. A G-invariant metric g on M can be written in the form g = εdt2 + gt where
t is a coordinate transverse to the principal orbits, ε = ±1, and gt is a 1-parameter family of
G-homogeneous Riemannian metrics on G/K. When ε = 1, the metric g is Riemannian, and when
ε = −1, the metric g is spatially homogeneous Lorentzian, i.e., the principal orbits are space-like
hypersurfaces.
We choose an Ad(K)-invariant decomposition g = k ⊕ p where g and k are respectively the Lie
algebras of G and K, and p is identified with the isotropy representation of G/K. Let
(1.1) p = p1 ⊕ · · · ⊕ pr
be a decomposition of p ≈ T(K)(G/K) into irreducible real K-representations. We let di be the
real dimension of pi, and n =
i=1 di be the dimension of G/K (so dimM = n+ 1). We use d for
the vector of dimensions (d1, · · · , dr). We shall assume that the isotropy representation of G/K is
multiplicity free, i.e., all the summands pi in (1.1) are distinct as K-representations. In particular,
if there is a trivial summand it must be 1-dimensional.
We use q = (q1, · · · , qr) to denote exponential coordinates on the space of G-invariant metrics
on G/K. The Hamiltonian H for the cohomogeneity one Einstein equations with principal orbit
G/K is now given by:
H = v−1J + εv ((n− 1)Λ− S) ,
CLASSIFICATION OF SUPERPOTENTIALS 3
where Λ is the Einstein constant, v = 1
ed·q is the relative volume and
(1.2) J(p, p) =
which has signature (1, r − 1). The scalar curvature S of G/K above can be written as
where Aw are nonzero constants and W is a finite collection of vectors w ∈ Z
r ⊂ Rr. The set W
depends only on G/K and its elements will be referred to as weight vectors. These are of three
types
(i) type I: one entry of w is −1, the others are zero,
(ii) type II: one entry is 1, two are -1, the rest are zero,
(iii) type III: one entry is 1, one is -2, the rest are zero.
Notation 1.1. As in [DW4] we use (−1i,−1j , 1k) to denote the type II vector w ∈ W ⊂ Rr with
−1 in places i and j, and 1 in place k. Similarly, (−2i, 1j) will denote the type III vector with −2
in place i and 1 in place j, and (−1i) the type I vector with −1 in place i.
Remark 1.2. We collect below various useful facts from [DW4] and [WZ1]. Also, we shall use
standard terminology from convex geometry, as given, e.g., in [Zi]. In particular, a “face” is not
necessarily 2-dimensional. However, a vertex and an edge are respectively zero and one-dimensional.
The convex hull of a set X in Rr will be denoted by conv(X).
(a) For a type I vector w, the coefficient Aw > 0 while for type II and type III vectors, Aw < 0.
(b) The type I vector with −1 in the ith position is absent fromW iff the corresponding summand
pi is an abelian subalgebra which satisfies [k, pi] = 0 and [pi, pj] ⊂ pj for all j 6= i. If the isotropy
group K is connected, these last conditions imply that pi is 1-dimensional, and the pj , j 6= i, are
irreducible representations of the (compact) analytic group whose Lie algebra is k⊕ pi.
(c) If (1i,−1j ,−1k) occurs in W then its permutations (−1i, 1j ,−1k) and (−1i,−1j , 1k) do also.
(d) If dim pi = 1 then no type III vector with −2 in place i is present in W. If in addition K is
connected, then no type II vector with nonzero entry in place i is present.
(e) If I is a subset of {1, · · · , r}, then each of the equations
i∈I xi = 1 and
i∈I xi = −2
defines a face (possibly empty) of conv(W). In particular, all type III vectors in W are vertices
and (−1i,−1k, 1j) ∈ W is a vertex unless both (−2i, 1j) and (−2k, 1j) lie in W.
(f) For v,w ∈ W (or indeed for any v,w such that
vi or
wi = −1), we have
(1.3) J(v + d,w + d) = 1−
For the remainder of the paper, we shall work in the Ricci-flat Riemannian case, that is, we take
ε = 1 and Λ = 0. As in [DW4], any argument that does not use the sign of Aw would be valid in
the Lorentzian case. We shall also assume that conv(W) is r− 1 dimensional. This is certainly the
case if G is semisimple, as W spans Rr (see the proof of Theorem 3.11 in [DW3]).
The superpotential equation (0.1) now becomes
(1.4) J(∇u,∇u) = ed·q S,
where ∇ denotes the Euclidean gradient in Rr. As in [DW4] we shall look for solutions to Eq.(1.4)
of the form
(1.5) u =
Fc̄ e
4 A. DANCER AND M. WANG
where C is a finite set in Rr, and the Fc̄ are nonzero constants. Now Eq.(1.4) reduces to, for each
ξ ∈ Rr,
(1.6)
ā+c̄=ξ
J(ā, c̄) FāFc̄ =
Aw if ξ = d+ w for some w ∈ W
0 if ξ /∈ d+W.
We shall assume henceforth that r ≥ 2 since the superpotential equation always has a solution
in the r = 1 case, as was noted in [DW4], and J is of Lorentz signature only when r ≥ 2. The
following facts were deduced in [DW4] from Eq.(1.6).
Proposition 1.3. conv(1
(d+W)) ⊂ conv(C).
Proof. If w ∈ W, then Eq.(1.6) implies that d + w = ā + c̄ for some ā, c̄ ∈ C, and hence that
(d+ w) = 1
(ā+ c̄) ∈ conv(C).
Proposition 1.4. If ā, c̄ ∈ C and ā+ c̄ cannot be written as the sum of two non-orthogonal elements
of C distinct from ā, c̄ then either J(ā, c̄) = 0 or ā+ c̄ ∈ d+W.
In particular, if c̄ is a vertex of C, then either J(c̄, c̄) = 0, or 2c̄ = d + w for some w ∈ W and
J(c̄, c̄) F 2c̄ = Aw. In the latter case, J(d+w, d+w) has the same sign as Aw so is > 0 if w is type
I and < 0 if w is type II or III.
As mentioned in the Introduction, for the classification in [DW4] we made the assumption that
all vertices c̄ of C are non-null. Under this assumption, the second assertion of Prop 1.4 implies
that all vertices of C lie in 1
(d+W). Hence conv(C) is contained in conv(1
(d+W)), and by Prop
1.3 they are equal. This meant that in [DW4], subject to the non-null assumption, we could study
the existence of a superpotential in terms of the convex geometry of W.
The aim of the current paper is to drop this assumption. We still have
conv(
(d+W)) ⊂ conv(C),
but can no longer deduce that these sets are equal. The problem is that a vertex c̄ of conv(C) may
lie outside conv(1
(d+W)) if it is null.
In fact, it is clear from the above discussion that conv(1
(d+W)) is strictly contained in conv(C)
if and only if C has a null vertex. For if c̄ is a null vertex of C and 2c̄ = d + w for some w ∈ W,
then Eq.(1.6) fails for ξ = d+ w.
We conclude this section by proving an analogue of Proposition 2.5 in [DW4]. The arguments
below using Prop 1.4 are ones which will recur throughout this paper. Henceforth when we use the
term “orthogonal” we mean orthogonal with respect to J unless otherwise stated.
Theorem 1.5. C lies in the hyperplane {x̄ :
x̄i =
(n−1)} (possibly after subtracting a constant
from the superpotential).
Proof. We can assume 0 /∈ C by subtracting a constant from the superpotential. We shall also use
repeatedly below the fact that as J has signature (1, r− 1) there are no null planes, only null lines.
Denote by Hλ the hyperplane
x̄i = λ, so
(d + W) lies in H 1
(n−1). Suppose there exist
elements of C with
x̄i >
(n − 1). Let λmax denote the greatest value of
x̄i over C. If ãc̃ is
an edge of conv(C) ∩Hλmax , then Prop 1.4 shows that ã, c̃ are null, and that c̃ is orthogonal to the
element of C closest to it on the edge. Hence c̃ is orthogonal to the whole edge. Now J is totally
null on Span{ã, c̃}, so since there are no null planes, ã, c̃ are proportional, which is impossible as
they are both in Hλmax . So C ∩Hλmax is a single point c̃max, which is null.
Next we claim that all elements of C lying in the half-space
x̄i >
(n − 1) must be multiples
of c̃max. If not, let λ∗ be the greatest value such that there is an element of C, not proportional
to c̃max, in Hλ∗ . Let ã be a vertex of conv(C) ∩Hλ∗, not proportional to c̃max. Now, by Prop 1.4,
J(ã, c̃max) = 0, and so ã is not null. Since λ∗ >
(n− 1), we see ã+ ã must be written in another
CLASSIFICATION OF SUPERPOTENTIALS 5
way as a sum of two non-orthogonal elements of C. This sum must be of the form µc̃max + f̃ . But
c̃max is orthogonal to ã and to itself, hence to f̃ , a contradiction establishing our claim.
Similarly, all elements of C lying in
x̄i <
(n − 1) are multiples of an element c̃min, should
they occur. (Note that J is negative definite on H0 and we have assumed 0 /∈ C so λmin 6= 0.)
We denote the sets of elements lying in these open half-spaces by C+ and C− respectively. Note
that, when non-empty, C+ and C− are orthogonal to all elements of C ∩ H 1
(n−1). (For if ã ∈
C ∩ H 1
(n−1) then ã + c̃max cannot be written in another way as a sum of two non-orthogonal
elements of C.) In particular, if c̃max and c̃min are orthogonal, then c̃max is orthogonal to all
of conv(C), which is r-dimensional by assumption. So c̃max is zero, a contradiction. The same
argument implies that C+ and C− are both non-empty.
Let νc̃min and µc̃max be respectively the elements of C− and C+ closest to H 1
(n−1). Suppose that
c̃max+νc̃min = c̃
(1)+ c̃(2) with c̃(i) ∈ C and J(c̃(1), c̃(2)) 6= 0. Non-orthogonality means the c̃(i) cannot
belong to the same side of H 1
(n−1) and by the choice of ν, they cannot belong to opposite sides of
(n−1). Both therefore lie inH 1
(n−1). But by the previous paragraph, J(c̃max+νc̃min, c̃
(1)+c̃(2)) =
0. This means that c̃max + νc̃min is null, which contradicts J(c̃max, c̃min) 6= 0. Hence c̃max + νc̃min
lies in d+W ⊂ Hn−1. Applying the same argument to c̃min+µc̃max, we find that in fact µ = ν = 1,
i.e., C+ = {c̃max} and C− = {c̃min}.
Now C ∩ H 1
(n−1) (and hence its convex hull) is contained in the hyperplanes c̃
max, c̃
min in
(n−1). These hyperplanes are distinct as c̃max is orthogonal to itself but not to c̃min. Hence
d+W ⊂ (C + C) ∩Hn−1 is contained in the union of the point c̃max + c̃min and the codimension 2
subspace c̃⊥max ∩ c̃
min of Hn−1. So conv(d+W) is contained in a codimension 1 subspace of Hn−1,
contradicting our assumption that dimconv(d+W) = r − 1.
Remark 1.6. A notational difficulty arises from the fact that, as seen above, points of C are on
the same footing as points in 1
(d+W) rather than points of W. Accordingly, we shall use letters
c, u, v, ... to denote elements of the hyperplane
ui = −1 (such as elements of W), and c̄, ū, v̄, ...
to denote the associated elements 1
(d+ c), 1
(d+u), 1
(d+v), · · · of the hyperplane
ūi =
(n−1)
(such as elements of C or of 1
(d+W)).
Note that for any convex or indeed affine sum
(j) of vectors ξ(j) in Rr, we have
λjξ(j) =
λj ξ(j).
Since we now know that the set C, like 1
(d +W), lies in H 1
(n−1) := {x̄ :
x̄i =
(n − 1)}, we
will adopt the convention, as in the last paragraph, that when we refer to hyperplanes such as c̄⊥
in the rest of the paper, we mean “affine hyperplanes in H 1
(n−1)”.
2. The classification theorem and the strategy of its proof
We can now state the main theorem of the paper.
Theorem 2.1. Let G be a compact connected Lie group and K a closed connected subgroup such
that the isotropy representation of G/K is the direct sum of r pairwise inequivalent R-irreducible
summands. Assume that dimconv(W) = r−1, where W is the set of weights of the scalar curvature
function of G/K (cf §1). (This holds, for example, if G is semisimple.)
If the cohomogeneity one Ricci-flat equations with G/K as principal orbit admit a superpotential
of form (1.5) where C contains a J-null vertex, then we are in one of the following situations (up
to permutations of the irreducible summands):
(i) W = {(−1)i, (11,−2i) : 2 ≤ i ≤ r}, d1 = 1, C =
(d + {(−11), (11,−2i) : 2 ≤ i ≤ r}) and
r ≥ 2;
6 A. DANCER AND M. WANG
(ii) r ≤ 3.
Remark 2.2. As mentioned before, the situation where C has no null vertex was analysed in [DW4].
Hence, except for the r ≤ 3 case, Theorem 2.1 completes the classification of superpotentials of
scalar curvature type subject to the above assumptions on G and K.
Remark 2.3. The first case of the above theorem is realized by certain circle bundles over a product
of r−1 Fano (homogeneous) Kähler-Einstein manifolds (cf. Example 8.1 in [DW4], and [BB], [WW],
[CGLP3]), and the subsystem of the Ricci-flat equations singled out by the superpotential in these
examples corresponds to the Calabi-Yau condition. For more on the r = 2 case, see the concluding
remarks in §10.
Remark 2.4. Theorem 2.1 remains true if we replace the connectedness of G and K by the
connectedness of G/K and the extra condition on the isotropy representation given by the second
statement in Remark 1.2(d), i.e., if pi is an irreducible summand of dimension 1 in the isotropy
representation of G/K, then [pi, pj] ⊂ k⊕ pj for all j 6= i.
This weaker property does hold in practice. For example, the exceptional Aloff-Wallach space
N1,1 can be written as (SU(3) × Γ)/(U1,1 · ∆Γ), where U1,1 is the set of diagonal matrices of the
form diag(exp(iθ), exp(iθ), exp(−2iθ)) and Γ is the dihedral group with generators
0 1 0
−1 0 0
0 0 1
e2πi/3 0 0
0 e−2πi/3 0
0 0 1
In order to prove Theorem 2.1 we have to analyse the situation when there is a null vertex c̄ ∈ C.
As discussed in §1, conv(C) now strictly includes conv(1
(d + W)) as c̄ is not in conv(1
(d +W)).
Our strategy is to take an affine hyperplane H separating c̄ from conv(1
(d+W)), and consider the
projection ∆c̄ of conv(1
(d+W)) onto H from c̄.
Roughly speaking, whereas in [DW4] we could analyse the situation by looking at the vertices and
edges of conv(1
(d+W)), now, because we have projected onto a subspace of one lower dimension,
we have to consider the 2-dimensional faces of conv(1
(d+W)) also.
This is a natural method of dealing with the situation of a point outside a convex polytope. It
has some relation to the notion of “lit set” introduced in a quite different context by Ginzburg-
Guillemin-Karshon [GGK].
The analysis in the next section will show that the vertices of the projected polytope ∆ can
be divided into three types (Theorem 3.8). We label these types (1A), (1B) and (2). Roughly,
these correspond to vertices orthogonal to c̄, vertices ξ̄ such that the line through c̄ and ξ̄ meets
conv(1
(d+W)) at a vertex, and vertices ξ̄ such that this line meets conv(1
(d+W)) in an edge.
In the remainder of the paper we shall gradually narrow down the possibilities for each type. In
§3 we begin a classification of type (2) vertices. In §4 we are able to deduce that conv(1
(d +W))
lies in the half space J(c̄, ·) ≥ 0. We are able to deduce an orthogonality result for vectors on edges
in conv(1
(d + W)) ∩ c̄⊥. This is analogous to the key result Theorem 3.5 of [DW4] that held (in
the more restrictive situation of that paper) for general edges in conv(1
(d+W)). In §5 we exploit
this result and some estimates to classify the possible configurations of (1A) vertices (i.e. vertices
in c̄⊥), see Theorem 5.18.
In §6 we attack the (1B) vertices, exploiting the fact that adjacent (1B) vertices give rise to a
2-dimensional face of conv(1
(d +W)). This is the most laborious part of the paper, as it involves
a case-by-case analysis of such faces. We show that adjacent (1B) vertices can arise only in a very
small number of situations (Theorem 6.18). In §7 we exploit the listing of 2-dim faces to show that
there is at most one type (2) vertex, except in two special situations (Theorem 7.1). In §8 and 9,
we eliminate more possibilities for adjacent (1B) and type (2) vertices. We find that if r ≥ 4 then
we are either in case (i) of the Theorem or there are no type (2) vertices and no adjacent type (1B)
CLASSIFICATION OF SUPERPOTENTIALS 7
vertices. Using the results of §4, in the latter case we find that all vertices are (1A) except for a
single (1B). Building on the results of §5 for (1A) vertices, we are able to rule out this situation in
§10, see Theorem 10.15 and Corollary 10.16.
3. Projection onto a hyperplane
We first present some results about null vectors in H 1
(n−1).
Remark 3.1. From Eq.(1.3), the set of null vectors in the hyperplane H 1
(n−1) form an ellipsoid
x2i /di = 1}. If c̄ is null, then the hyperplane c̄
⊥ in H 1
(n−1) is the tangent space to this ellipsoid.
So any element x̄ 6= c̄ of c̄⊥ satisfies J(x̄, x̄) < 0.
Lemma 3.2. Let x, y satisfy
yi = −1.
Suppose that J(x̄, x̄) and J(ȳ, ȳ) ≥ 0. Then J(x̄, ȳ) ≥ 0, with equality iff x̄ is null and x̄ = ȳ. In
particular, if x̄, ȳ are distinct null vectors then J(x̄, ȳ) > 0.
Proof. This follows from Eq.(1.3) and Cauchy-Schwartz.
Proposition 3.3. Let H = {x̄ : h(x̄) = λ} be an affine hyperplane, where h is a linear functional
such that conv(1
(d + W)) lies in the open half-space {x̄ : h(x̄) < λ}. Then there is at most one
element of C in the complementary open half-space {x̄ : h(x̄) > λ}. Such an element is a null vertex
of conv(C). Hence any element of C outside conv(1
(d+W)) is a null vertex of C.
Proof. Suppose the points of C with h(x̄) > λ are c̄(1), · · · , c̄(m) with m > 1. Our result is stable
with respect to sufficiently small perturbations of H, so we can assume that h(c̄(1)) > h(c̄(2)) ≥
h(c̄(3)), · · · , h(c̄(m)).
Now c̄(1) + c̄(1) and c̄(1) + c̄(2) cannot be written in any other way as the sum of two elements
of C. Hence, by Prop. 1.4, c̄(1) is null and J(c̄(1), c̄(2)) = 0. The only other way c̄(2) + c̄(2) can be
written is as c̄(1) + c̄ for some c̄ ∈ C. But then c̄ = 2c̄(2) − c̄(1), so J(c̄(1), c̄) = 0, and such sums will
not contribute. Hence J(c̄(2), c̄(2)) = 0, contradicting Lemma 3.2.
Corollary 3.4. For distinct elements c̄, ā of C, the line segment c̄ā meets conv(1
(d+W)).
This gives us some control over the extent to which conv(C) can be bigger than the set conv(1
Lemma 3.5. Let A ⊂ H 1
(n−1) be an affine subspace such that A ∩ conv(
(d +W)) is a face of
conv(1
(d+W)). Suppose there exists c̄ ∈ C ∩A with c̄ /∈ conv(1
(d+W)).
Let x̄ ∈ A. If x̄ = 1
(ā+ ā′) with ā, ā′ ∈ C, then in fact ā, ā′ ∈ A.
Proof. If ā or ā′ equals c̄ this is clear.
We know by Cor 3.4 that if ā, ā′ 6= c̄ then the segments āc̄, ā′c̄′ meet conv(1
(d+W)). So there
exist 0 < s, t ≤ 1 with tā+ (1− t)c̄ and sā′ + (1− s)c̄ in conv(1
(d+W)). Hence
(tā+ (1− t)c̄) +
sā′ + (1− s)c̄
∈ conv(
(d+W)).
As it is an affine combination of x̄, c̄ this point also lies in A, so it lies in A∩ conv(1
(d+W)). Also,
it is a convex linear combination of the points tā+ (1 − t)c̄ and sā′ + (1− s)c̄ of conv(1
(d+W)).
Hence, by our face assumption, both these points lie in A, so ā, ā′ lie in A.
Remark 3.6. The above lemma will be very useful because it means that in all our later calculations
using Prop 1.4 for a face defined by an affine subspace A, we need only consider elements of C lying
in A.
8 A. DANCER AND M. WANG
Proposition 3.7. Let vw be an edge of conv(W) and suppose v̄, w̄ ∈ C.
(i) If there are no points of W in the interior of vw, then J(v̄, w̄) = 0.
(ii) If u = 1
(v+w) is the unique point of W in the interior of vw, J(v̄, w̄) > 0, and u is type II
or III, then Fv̄, Fw̄ are of opposite signs.
Proof. Part(i) is a generalization of Theorem 3.5 in [DW4] and we will be able to apply the proof
of that result after the following argument. Let the edge v̄w̄ of conv(1
(d + W)) be defined by
equations 〈x̄, u(i)〉 = λi : i ∈ I where 〈x̄, u
(i)〉 ≤ λi for i ∈ I and x̄ ∈ conv(
(d + W)). (In
the above, 〈 , 〉 is the Euclidean inner product in Rr.) Note that Span {u(i) : i ∈ I} is the
〈 , 〉-orthogonal complement of the direction of the edge.
Let H be a hyperplane whose intersection with conv(1
(d+W)) is the edge v̄w̄. We can take H
to be defined by the equation 〈x̄,
i∈I biu
(i)〉 =
i∈I biλi where bi are arbitrary positive numbers
summing to 1.
If ā, ā′ are elements of C whose midpoint lies in v̄w̄, then either they are both in H or one of
them, ā say, is on the opposite side of H from conv(1
(d+W)). In the latter case ā is null and the
only element of C on this side of H so, by Prop 1.4, is J-orthogonal to v̄, w̄. Hence, as 1
(ā + ā′)
is an affine combination of v̄, w̄, we see that J(ā, ā′) = 0, and so such sums do not contribute in
Eq.(1.6). We may therefore assume that ā, ā′ are in H. But as this is true for all H of the above
form, the only sums that will contribute are those where ā, ā′ are collinear with v̄w̄.
Now if ā, say, lies outside the line segment v̄w̄, then it is null and J-orthogonal to v̄ or w̄, and
hence to the whole line. So the only sums which contribute are those where ā, ā′ lie on the line
segment v̄w̄. Now the proof of Theorem 3.5 in [DW4] gives (i).
Turning to (ii), note first that the above arguments and Prop 1.4 give (ii) immediately if no
interior points of the edge v̄w̄ lie in C. If there are m interior points in C, we again proceed as in
the proof of Theorem 3.5 in [DW4] and use the notation there. We may assume that Lemma 3.2
(and hence Cor 3.3 and Lemma 3.4) of [DW4] still holds; for the only issue is the statement for
λm+1, but if c
(0) + c(λm+1) cannot be written as c(λj) + c(λk) (0 < λj , λk < m + 1) then what we
want to prove is already true.
Now Lemma 3.4 in [DW4] and our hypothesis J(v̄, w̄) > 0 imply that J00 < 0 and Jλi,λj > 0
except in the three cases listed there. The proof that the elements of C are equi-distributed in v̄w̄
carries over from [DW4] since the midpoint ū is not involved in the arguments.
Suppose next that the points in C ∩ v̄w̄ are equi-distributed. In the special case where m = 1,
we have J(v̄, ū) = 0 = J(ū, w̄), which imply J(ū, ū) = 0. So the midpoint does not contribute to
the equation from c(0) + c(λm+1). If m > 1, we write down the equations arising from c(0) + c(λm+1)
and c(λm−1) + c(λm+1). The formula for Fλj in [DW4] still holds for 1 ≤ j ≤ m, and using this and
the second equation we obtain the analogous formula for Fλm+1 .
Putting all the above information together in the first equation and using Au < 0, we see that
Fm+1λ1 /F
0 is positive if m is even and negative if m is odd. In either case it follows immediately
that F0Fλm+1 < 0, as required.
We shall now set up the basic machinery of the projection of our convex hull onto an affine
hyperplane.
Let c̄ be a null vector in C and let H be an affine hyperplane separating c̄ from conv(1
(d+W)).
Define a map P : conv(1
(d + W)) −→ H by letting P (z̄) be the intersection point of the ray c̄z̄
with H. We denote by ∆ the image of P in H. (P and ∆ of course depend on c̄ and the choice of
H. When considering projections from several null vertices, we will use the vertices as superscripts
to distinguish the cases, e.g., ∆c̄,∆b̄.)
Let us now consider a vertex ξ̄ of ∆. We know that c̄ and ξ̄ are collinear with a subset P−1(ξ̄) of
conv(1
(d+W)). As ξ̄ does not lie in the interior of a positive-dimensional subset of ∆, we see that
CLASSIFICATION OF SUPERPOTENTIALS 9
no point of P−1(ξ̄) lies in the interior of a subset of conv(1
(d +W)) of dimension > 1. So P−1(ξ̄)
is a vertex or an edge of conv(1
(d+W)).
If P−1(ξ̄) is a vertex x̄, then 2x̄ ∈ d +W and in Lemma 3.5 we can take the affine subspace A
to be the line through c̄, ξ̄, x̄. Using this lemma and also Prop 1.4 and Cor 3.4 we see that either
x̄ ∈ C (in which case J(x̄, c̄) = 0), or x̄ /∈ C and x̄ = (ā+ c̄)/2 for some null element ā ∈ C ∩A. We
have therefore deduced
Theorem 3.8. Let ξ̄ be a vertex of ∆. Then exactly one of the following must hold:
(1A) ξ̄ (and hence P−1(ξ̄)) is orthogonal to c̄;
(1B) The line through c̄, ξ̄ meets conv(1
(d + W)) in a unique point x̄, and there exists a null
ā ∈ C such that (ā+ c̄)/2 = x̄;
(2) ξ̄ is not orthogonal to c̄, and c̄ and ξ̄ are collinear with an edge v̄w̄ of conv(1
(d+W)), (and
hence c and ξ are collinear with the corresponding edge vw of conv(W)).
Remark 3.9. If (1B) occurs, then ā = 2x̄− c̄ being null is equivalent to J(x̄, x̄) = J(x̄, c̄), that is,
(3.1)
In particular xi and ci are nonzero for some common index i. We will from now on refer to this
situation by saying that the vectors x and c overlap.
We make a preliminary remark about (1A) vertices.
Lemma 3.10. Suppose that u ∈ W and ū ∈ c̄⊥.
(a) If u = (−2i, 1j) then ci 6= 0.
(b) Suppose that K is connected. If u = (−1i,−1j , 1k), then ci, cj , ck are all nonzero.
Proof. After a suitable permutation, we may let 1, · · · , s be the indices a with ca 6= 0. We need
In case (a) this is impossible if ci = 0 (that is, i /∈ {1, · · · , s}) as then we need dj = 1 = cj and
ca = 0 for a 6= j, contradicting
k=1 ck = −1.
Next, Cauchy-Schwartz on ( ua√
)sa=1, (
)sa=1 shows
≥ 1. In case (b), if, say, ck = 0,
then since 1
≥ 1 and di, dj ≥ 2 (see Remark 1.2(d)) we must have di = dj = 2. The equations
then imply ci = cj = −1 and ca = 0 for a 6= i, j, also giving a contradiction. Similar arguments
rule out ci = 0 or cj = 0.
In the next two sections we shall get stronger results on (1A) vertices. Let us now consider type
(2) vertices.
Theorem 3.11. Consider a type (2) vertex ξ̄ of ∆. So c and ξ are collinear with an edge vw of
conv(W). Suppose there are no points of W in the interior of vw. Then we have
(i) c = 2v − w or
(ii) c = (4v − w)/3.
In (i) the points of C on the line through c̄, ξ̄ are c̄ and w̄. In (ii) they are c̄, w̄ and c̄(1) =
(2v̄ + w̄)/3 = (c̄+ w̄)/2. We need J(c̄(1), w̄) = 0.
Proof. This is very similar to the arguments of §3 in [DW4]. We apply Lemma 3.5 to the line
through v̄, w̄.
(A) We write the elements of C on the line as c̄ = c̄(0), c̄(1), · · · , c̄(m+1) with m ≥ 0. So c̄(m+1) is
either null or is w̄. No other c̄(j) can lie beyond w̄, by Cor 3.4.
10 A. DANCER AND M. WANG
By assumption c̄ = c̄(0) is not orthogonal to the whole line. As c̄ is null, this means c̄ is not
orthogonal to any other point on the line. So c̄(0) + c̄(j) is either 2v̄, 2w̄ or else is a sum of two
other c̄(i). In particular, c̄(0)+ c̄(1) = 2v̄. In fact c̄(0)+ c̄(j) is never 2w̄; for the only possibility is for
c̄(0) + c̄(m+1) = 2w̄, in which case c̄(m+1) is null, and so c̄(m) + c̄(m+1) = 2w̄, contradicting v̄ 6= w̄.
We deduce that for j > 1, we have c̄(0) + c̄(j) = c̄(k) + c̄(p) for some 1 ≤ k, p ≤ j − 1.
(B) Let c̄(m+1) be null. Since the segment c̄(0)c̄(m+1) lies in the interior of the null ellipsoid,
Lemma 3.2 implies that J(c̄(i), c̄(j)) > 0 unless i = j = 0 or m + 1. Arguments very similar to
those in §3 of [DW4] enable us to determine the signs of the Fc̄(j) in (1.5) and show that the
contributions from the pairs summing to c̄(1) + c̄(m+1) cannot cancel. So we have a contradiction
unless c̄(1) + c̄(m+1) = w̄, which can only happen if m = 1, i.e., c̄(0) + c̄(1) = 2v̄, c̄(1) + c̄(2) = 2w̄ and
c̄(0) + c̄(2) = 2c̄(1) (otherwise c̄(0) + c̄(2) cannot cancel). Hence we have
c = (3v − w)/2 ; c(1) = (v + w)/2 ; c(2) = (3w − v)/2.
Writing Fj for Fc̄(j) , we need 2F0F2J(c̄, c̄
(2)) + F 21 J(c̄
(1), c̄(1)) = 0 so that the contributions from
c̄(0) + c̄(2) and c̄(1) + c̄(1) cancel. As J(c̄, c̄(2)) and J(c̄(1), c̄(1)) > 0, we need F0 and F2 to have
opposite signs. Now, as J(c̄, c̄(1)), J(c̄(1), c̄(2)) > 0, we see that Av and Aw have opposite signs. So
we may let w be type I and v be type II or III, as long as the asymmetry between c̄(0) and c̄(2) is
removed. Note that v,w cannot overlap if v is type II, as then Remark 1.2(c) means w is not a
vertex. The possibilities are (up to permutation)
v w c(0) = 1
(3v − w) c(2) = 1
(3w − v)
(1) (−2, 1, 0, · · · ) (−1, 0, · · · ) (−5
, 0, · · · ) (−1
, 0, · · · )
(2) (−2, 1, 0, · · · ) (0,−1, 0, · · · ) (−3, 2, 0, · · · ) (1,−2, 0, · · · )
(3) (−2, 1, 0, · · · ) (0, 0,−1, 0, · · · ) (−3, 3
, · · · ) (1,−1
, 0, · · · )
(4) (1,−1,−1, 0, · · · ) (0, 0, 0,−1, · · · ) (3
, · · · ) (−1
, · · · )
Now, it is clear in (1) and (2) that c̄(0) and c̄(2) can’t both give null vectors. For (3) and (4), we
find that the nullity equations for c̄(0) and c̄(2) have no integral solutions in di (in fact d3 (resp. d4)
must be 5/2).
Therefore in fact c̄(m+1) cannot be null.
(C) Now suppose that c̄(m+1) = w and m > 0. Since J(c̄, v̄) 6= 0, v̄ must lie between c̄(0) and c̄(1).
So J(c̄(0), ·) and J( · , c̄(m+1)) are affine functions on the line, vanishing at c̄(0) and c̄(m) respectively.
Hence J(c̄(0), c̄(i)) (i ≥ 1) and J(c̄(i), c̄(m+1)) (0 ≤ i ≤ m − 1) are the same sign as J(c̄(0), c̄(1)).
It follows that J(c̄(i), ·) is an affine function on the line, taking the same sign as J(c̄(0), c̄(1)) at
c̄(0), c̄(m+1) (for 1 ≤ i ≤ m − 1). Thus J(c̄(i), c̄(j)) is the same sign as J(c̄(0), c̄(1)) except for the
cases
J(c̄(0), c̄(0)) = 0 = J(c̄(m), c̄(m+1)) : sign J(c̄(m+1), c̄(m+1)) = −sign J(c̄(0), c̄(1)).
It then follows that the sign and non-cancellation arguments of (B) (taken from §3 of [DW4]) still
hold, except in the case m = 1.
These give the two cases of the Theorem. If m = 0, we have c(1) = w and c(0) = 2v − w as
c(0) + c(1) = 2v. If m = 1, then c(2) = w, c(0) + c(1) = 2v and c(0) + c(2) = 2c1 (for cancellation).
Hence c(0) = (4v − w)/3, c(1) = (2v + w)/3, as well as J(c̄(1), c̄(2)) = 0.
Remark 3.12. If there are points of W in the interior of vw, we can still conclude that c(0)+c(1) =
2v. Hence c = λv + (1 − λ)w for 1 < λ ≤ 2, since if λ > 2 then c̄(1) is beyond w̄. It must then be
null, and m = 0, so there is no way of getting 2w̄ as a sum of two elements in C.
Lemma 3.13. For case (i) in Theorem 3.11 (i.e., c = 2v −w), either w is type I, or w is type III
and vi = −1, wi = −2 for some index i.
CLASSIFICATION OF SUPERPOTENTIALS 11
Proof. It follows from above that J(w̄, w̄)F 2w̄ = Aw so J(w̄, w̄) is positive if w is type I and negative
if w is type II or III. In the latter case,
> 1, but by nullity, c = 2v − w satisfies
Hence for some i we have |wi| > |ci| = |2vi − wi|. As vi, wi ∈ {−2,−1, 0, 1}, it follows that
vi = −1, wi = −2.
We are now able to characterise the case where c is a type I vector.
Theorem 3.14. If c is a type I vector, say (−1, 0, · · · ) for definiteness, then W is given by
{(−1)i, (11,−2i) : i = 2, · · · , r}.
Remark 3.15. Equivalently, W is as in Ex 8.1 of [DW4], where the hypersurface in the Ricci-flat
manifold is a circle bundle over a product of Kähler-Einstein Fano manifolds. A superpotential was
found for this example in [CGLP3].
Proof. Nullity of c̄ implies d1 = 1, so (−2
1, 1i) /∈ W. Also (−11,−1j , 1k) /∈ W, as then c would be
in conv(W). Let us consider the vertices ξ̄ in ∆. ξ̄ cannot be of type (1A); otherwise ξ1 = −1,
which implies the existence of a type II vector in W with a nonzero first component, contradicting
the above. There can also be no ξ̄ of type (1B) since by Remark 3.9 the vector x̄ satisfies 0 < −x1,
which we ruled out above.
Hence all vertices of ∆ are of type (2), i.e., correspond to edges vw of conv(W) such that
c = λv + (1− λ)w and λ > 1. From this equation it follows that v,w are of the form
v = (−1i), w = (11,−2i)
for some i > 1. As ∆ (being a (r− 2)-dimensional polytope in an (r− 2)-dimensional affine space)
has at least r − 1 vertices, such vectors occur for all i 6= 1.
Now no type II vector can be in W, otherwise v would not be a vertex. Also (1i,−2j) with
i, j 6= 1 cannot be in W, as then (−1j) would not be a vertex. We have already seen (−21, 1i) is
not in W. So W is as claimed.
We shall henceforth exclude this case, i.e. case (i) of Theorem 2.1, from our discussion.
We conclude this section by giving a preliminary listing of the possibilities for c when we have a
type (2) vertex. These are given by cases (i) and (ii) of Theorem 3.11, as well as the possible cases
when there is a point of W in the interior of vw.
For Theorem 3.11(i) the possible v,w, c are:
v w c = 2v − w
(1) (−1, 1,−1, · · · ) (−2, 1, · · · ) (0, 1,−2, · · · )
(2) (−1,−1, 1, · · · ) (−2, 1, · · · ) (0,−3, 2, · · · )
(3) (−1, 0,−1, 1, · · · ) (−2, 1, · · · ) (0,−1,−2, 2, · · · )
(4) (−2, 1, · · · ) (−1, 0, · · · ) (−3, 2, · · · )
(5) (−2, 1, · · · ) (0, 0,−1, · · · ) (−4, 2, 1, · · · )
(6) (−1, 0, · · · ) (0,−1, · · · ) (−2, 1, · · · )
(7) (1,−1,−1, · · · ) (0, 0, 0,−1, · · · ) (2,−2,−2, 1, · · · )
Table 1: c = 2v −w cases
where · · · denotes zeros as usual. To arrive at this list, recall from Lemma 3.13 that w is either
type I or type III with vi = −1, wi = −2 for some i. Note also that if w is type I and v is type II
then v cannot overlap with w as w cannot then be a vertex. Furthermore, the other possibility with
w type I and v type III is excluded as we are assuming in Theorem 3.11 that there are no points
of W in the interior of vw. Finally, the case w = (−2, 1, · · · ), v = (−1, 0, · · · ) can be excluded as
this just gives the example in Theorem 3.14.
In order to list the possibilities under Theorem 3.11(ii), recall that we need J(c̄(1), w̄) = 0 where
c(1) = (2v + w)/3. Equivalently, we need
(3.2) 2J(v̄, w̄) + J(w̄, w̄) = 0.
12 A. DANCER AND M. WANG
This puts constraints on the possibilities for v,w. For instance, w cannot be type I, as for such
vectors J(v̄, w̄) ≥ 0 and J(w̄, w̄) > 0. Also, if w is type II or III, then from the superpotential
equation we need J(w̄, w̄) < 0, so J(v̄, w̄) > 0. If w is type III, say (−2, 1, 0, · · · ), then since d1 ≥ 2,
we have 4
≤ 3, and the above equation gives J(v̄, w̄) ≤ 1
with equality iff d1 = 2, d2 = 1.
By the above remarks and the nullity of c̄, after a moderate amount of routine computations, we
arrive at the following possibilities, up to permutation of entries. In the table we have listed only
the minimum number of components for each vector and all unlisted components are zero. Note
that the entries (12)-(16) can occur only if K is not connected (cf. Remark 5.9).
v w c(1) = (2v +w)/3 c = (4v − w)/3
(1) (0, 0,−2, 1) (−2, 1, 0, 0) (−2
(2) (−2, 0, 1) (−2, 1, 0) (−2, 1
) (−2,−1
(3) (−1, 0, 0, ) (−2, 1, 0) (−4
, 0) (−2
(4) (0, 0,−1) (−2, 1, 0) (−2
(5) (−1, 0, 1,−1) (−2, 1, 0, 0) (−4
) (−2
(6) (−1, 1,−1) (−2, 1, 0) (−4
, 1,−2
) (−2
, 1,−4
(7) (0, 0, 1,−1,−1) (−2, 1, 0, 0, 0) (−2
(8) (0, 1,−1,−1) (−2, 1, 0, 0) (−2
, 1,−2
, 1,−4
(9) (0,−1,−1, 1) (1,−1,−1, 0) (1
,−1,−1, 2
) (−1
,−1,−1, 4
(10) (1,−1, 0,−1) (1,−1,−1, 0) (1,−1,−1
) (1,−1, 1
(11) (1,−2, 0) (1,−1,−1) (1,−5
) (1,−7
(12) (1, 0, 0,−2) (1,−1,−1, 0) (1,−1
) (1, 1
(13) (0, 0, 0,−2, 1) (1,−1,−1, 0, 0) (1
) (−1
(14) (0,−1, 0, 1,−1) (1,−1,−1, 0, 0) (1
,−1,−1
) (−1
,−1, 1
(15) (1, 0, 0,−1,−1) (1,−1,−1, 0, 0) (1,−1
) (1, 1
(16) (0, 0, 0, 1,−1,−1) (1,−1,−1, 0, 0, 0) (1
) (−1
Table 2: c = 1
(4v − w) cases
We will also need a listing of those cases for which vw has interior points lying in conv(W).
v w c
(1) (1,−2, · · · ) (−2, 1, · · · ) (3λ− 2, 1− 3λ, · · · )
(2) (1,−2, · · · ) (−1, 0, · · · ) (2λ − 1, −2λ, · · · )
(3) (−1, 0, · · · ) (1,−2, · · · ) (1− 2λ, 2λ− 2, · · · )
(4) (−2, 1, 0, · · · ) (0, 1,−2, · · · ) (−2λ, 1, 2λ− 2, · · · )
(5) (1,−1,−1, · · · ) (−1, 1,−1, · · · ) (2λ− 1, 1 − 2λ,−1, · · · )
Table 3: Cases with interior points
Recall from Remark 3.12 that 1 < λ ≤ 2 and · · · denote zeros. Note that except in (4) all interior
points which may lie in W actually do.
4. The sign of J(c̄, w̄)
Theorem 4.1. conv(1
(d + W)) lies in the closed half-space J(c̄, ·) ≥ 0, i.e., the same closed
half-space in which the null ellipsoid lies.
Proof. We know that if ξ̄ is a vertex of ∆c̄ then there are three possibilities, given by (1A), (1B)
and (2) of Theorem 3.8. If (1A) occurs, then by definition J(c̄, ξ̄) = 0. If (1B) occurs, let ā be the
null vector in Theorem 3.8. Then by Lemma 3.2, J(c̄, ā) > 0, which in turn implies that J(c̄, ξ̄) > 0.
It is now enough to show that J(c̄, ξ̄) ≥ 0 if ξ̄ is a type (2) vertex of ∆c̄, since it then follows that
∆c̄, and hence conv(1
(d+W)), lies in the half-space J(c̄, ·) ≥ 0.
CLASSIFICATION OF SUPERPOTENTIALS 13
Suppose then that ξ̄ is a type (2) vertex with J(c̄, ξ̄) < 0. By Remark 3.12, c = λv + (1 − λ)w
for some v,w ∈ W with 1 < λ ≤ 2, and both J(c̄, v̄), J(c̄, w̄) < 0. In particular, from Remark 3.1
and Lemma 3.2, J(v̄, v̄), J(w̄, w̄) < 0 since v̄, w̄ lie on the side of c̄⊥ opposite to the null ellipsoid.
0 = 4J(c̄, c̄) = J(d+ λv + (1− λ)w, d+ λv + (1− λ)w)
= J(λ(d + v) + (1− λ)(d + w), λ(d+ v) + (1− λ)(d+ w))
= λ2J(d+ v, d+ v) + 2λ(1− λ)J(d+ v, d + w) + (1− λ)2J(d+ w, d + w).
It follows from the above remarks that J(d+ v, d + w) < 0, that is
One then checks that this condition is only satisfied in the following cases (up to permutation of
indices and interchange of v and w):
(a) v = (−2, 1, 0, · · · ), w = (−2, 0, 1, 0, · · · ) with 1 < d1 < 4;
(b) v = (−2, 1, 0, · · · ), w = (−1, 1,−1, 0, · · · ) with d1 = 2, or (d1, d2) = (3, 2), or d2 = 1;
(c) v = (1,−1,−1, 0, · · · ), w = (1,−1, 0,−1, 0, · · · ) with d1 = 1 or d2 = 1;
(d) v = (1,−1,−1, 0, · · · ), w = (0,−1,−1, 1, 0, · · · ) with d2 = 1 or d3 = 1.
In case (a), c = (−2, λ, 1 − λ, 0, · · · ). The condition d1 < 4 is incompatible with the nullity of c̄.
Interchanging v and w reverses only the role of λ and 1− λ.
A similar argument rules out case (b) with v,w as shown, as here c = (−λ − 1, 1, λ − 1). If
we interchange v and w, then c = (λ − 2, 1,−λ, · · · ). Theorem 3.11 tells us λ = 4/3 or 2, so
c = (−2/3, 1,−4/3, · · · ) or (0, 1,−2, · · · ).
In the former case c(1) := (2v+w)/3 = (−4/3, 1,−2/3, · · · ), so the condition J(w̄, c̄(1)) = 0 gives
8/3d1 + 1/d2 = 1. Thus (d1, d2) = (3, 9) or (4, 3) but in neither case is c̄ null. In the latter case
nullity means 1/d2 + 4/d3 = 1, so J(c̄, v̄) =
(1− 1/d2 − 2/d3) > 0, a contradiction.
In case (c), c = (1,−1,−λ, λ − 1) and if v and w are interchanged, the last two components of
c are interchanged. But c̄ cannot be null if d1 = 1 or d2 = 1. A similar argument works for case
Corollary 4.2. conv(1
(d+W)) ∩ c̄⊥ is a (possibly empty) face of conv(1
((d+W)).
This enables us to adapt Theorem 3.5 of [DW4] to the elements of c̄⊥.
Corollary 4.3. Let vw be an edge of conv(W) and suppose v̄ and w̄ are in c̄⊥. Suppose further
that there are no elements of W in the interior of vw. Then J(v̄, w̄) = 0.
Proof. This is essentially the same as the proof of Theorem 3.5 of [DW1]. As conv(1
(d+W))∩ c̄⊥
is a face of conv(1
(d + W)), Lemma 3.5 shows that for calculations in c̄⊥ we need only consider
elements of C in this hyperplane. Note that by Cor 3.4, no elements of C lie on the opposite side of
c̄⊥ to conv(1
(d+W).
Any vertex of conv(C) ∩ c̄⊥ outside conv(1
(d +W)) ∩ c̄⊥ is, by Prop 1.4, null, so must be c̄ by
Lemma 3.2. Now Cor 3.4 shows that c̄ is the only element of conv(C)∩c̄⊥ outside conv(1
(d+W))∩c̄⊥.
But any sum c̄+ ā with ā ∈ c̄⊥ does not contribute, so in fact we are in the situation of Theorem
3.5 of [DW4].
We introduce the following sets:
Ŝ1 = {i ∈ {1, · · · , r} : ∃ unique w ∈ W with w̄ ∈ c̄
⊥ and wi = −2}
Ŝ≥2 = {i ∈ {1, · · · , r} : ∃ more than one w ∈ W with w̄ ∈ c̄
⊥ and wi = −2}
These are similar to the sets S1, S≥2 of [DW4], but now we require that the vectors w to lie in c̄
It is immediate from Cor 4.3 that di = 4 if i ∈ Ŝ≥2, (cf Prop 4.2 in [DW4]).
14 A. DANCER AND M. WANG
We next prove a useful result about which elements of 1
(d +W) can be orthogonal to c̄. This
will give us information about when (1A) vertices can occur.
Lemma 4.4. Assume that we are not in the situation of Theorem 3.14 (i.e., c is not of type I ).
Let u ∈ W be such that ū ∈ c̄⊥. Then:
(a) there exists i with ci 6= 0 and −2 < ci < 1;
(b) if c ∈ Zr then there is at most one such u, and hence at most one (1A) vertex (wrt c).
Proof. (a) The condition J(ū, c̄) = 0 means
= 1, and nullity of c̄ means
= 1. As
ui ∈ {−2,−1, 0, 1}, if the condition in (a) does not hold, then uici ≤ c
i for all i so we must have
equality for all i. Now ci = ui for all i with ci nonzero. As
ci = −1 and c 6= u (since c /∈ W by
definition), this means c is a type I vector and we are in the situation of Theorem 3.14.
(b) We see from the previous paragraph that we need uici > c
i for some i. If c ∈ Z
r this means
ci = −1 and ui = −2. The orthogonality condition is now
= 1 where uj = 1. As di 6= 1 we
see cj ≥ 0.
If cj = 0 then di = 2. If cj > 0 then di ≥ 3 so
, where the second inequality is due to
the nullity requirement 1
≤ 1. So cj = 1 or 2. Moreover, the latter implies (di, dj) = (3, 6)
and c = (−1i, 2j), which contradicts
ci = −1.
We see that either cj = 1 and (di, dj) = (4, 2) or (3, 3), or cj = 0 and di = 2. Cor 4.3 implies
that if there is more than one such u (say (−2i, 1j) and (−2i, 1k)) for a given i, then di = 4, so
(di, dj , dk) = (4, 2, 2), and (ci, cj , ck) = (−1, 1, 1), contradicting the nullity of c.
It now readily follows that the nullity condition prevents there being more than one u ∈ W
with ū ∈ c̄⊥ except when c = (−1,−1, 1, 0, · · · ) with d = (4, 4, 2, · · · ) or (3, 3, 3, · · · ) and u =
(−2, 0, 1, 0, · · · ), (0,−2, 1, 0, . . .). But in this case if both u occur then c ∈ conv(W), a contradic-
tion.
We shall study (1A) vertices for non-integral c in the next section. The following results will be
useful.
Proposition 4.5. Let v = (−2i, 1j) and w = (−2k, 1l) be elements of W such that v̄, w̄ ∈ c̄⊥.
Suppose that i ∈ Ŝ1 and {i, j} ∩ {k, l} = ∅. Then k ∈ Ŝ≥2 and (di, dk, dl) = (2, 4, 2).
Proof. By Remark 1.2(e) the affine subspace {x̄ : xi+xk = −2, xj +xl = 1}∩ c̄
⊥ meets conv(1
W)) in a face, whose possible elements are v,w, u = (−2k, 1j), y = (−1i, 1j ,−1k) and z = (−1i,−1k, 1l)
(since i ∈ Ŝ1).
As J(v̄, w̄) = 1
, we see from Thm 4.3 that vw is not an edge so z is present in the face. Now
Cor 4.3 on vz implies di = 2. Also, u must be present, otherwise y is present and Cor 4.3 on zw
and yw gives a contradiction. So k ∈ Ŝ≥2, and Cor 4.3 on uw implies dk = 4. Now considering zw
implies dl = 2.
Remark 4.6. This is similar to the proof of Prop 4.6 in [DW4]. But we cannot now deduce that
dj = 1 as the proof of this in [DW4] relied on the existence of t = (−1
i,−1j , 1k), and although we
know this is in W we do not know if t̄ lies in c̄⊥.
Proposition 4.7. If i ∈ Ŝ1 and v = (−2
i, 1j) gives an element of c̄⊥ then w = (−1i,−1j , 1k)
cannot give an element of c̄⊥.
Proof. This is similar to Prop. 4.3 in [DW4]. Since i ∈ Ŝ1, the vectors v̄, w̄ lie on an edge in the face
{x̄ : 2xi + xj = −3} ∩ c̄
⊥ of conv(1
(d+W)), and J(v̄, w̄) = 1
(1− 2
) 6= 0 since di 6= 1.
Corollary 4.8. With v as in Prop 4.7, there are no elements w = (−2j , 1k) with w̄ in c̄⊥.
CLASSIFICATION OF SUPERPOTENTIALS 15
Proof. This is similar to Prop 4.4 in [DW4]. If k = i, then the type I vector u := (−1i) = 1
(2v+w)
lies in W and ū ∈ c̄⊥. By Lemma 5.1 below, u = c, contradicting c /∈ W.
We can therefore take k 6= i. Now v̄, w̄ lie on an edge in the face {x̄ : 3xi +2xj = −4} ∩ c̄
⊥ (this
is a face by Prop 4.7 and the assumption i ∈ Ŝ1). But J(v̄, w̄) =
(1 + 2
) 6= 0.
5. Vectors orthogonal to a null vertex
In this section we analyse the possibilities for 1
(d+W)∩ c̄⊥. This will give us an understanding
of the vertices of type (1A).
We first dispose of the case of type I vectors.
Lemma 5.1. If u is a type I vector and ū ∈ c̄⊥ then c = u, so we are in the situation of Theorem
3.14.
Proof. Up to a permutation we may let u = (−1, 0, · · · ). The orthogonality condition implies
c1 = −d1. But nullity implies
c2i /di = 1, so d1 = 1 and ci = 0 for i > 1. (Note that in particular
u /∈ W.)
We shall therefore assume from now on there are no type I vectors giving points of c̄⊥.
Lemma 5.2. (i) Two type II vectors whose nonzero entries lie in the same set of three indices
cannot both give elements of c̄⊥.
(ii) Two type III vectors (−2i, 1j) and (1i,−2j) cannot both give elements in c̄⊥.
(iii) Three type III vectors whose nonzero entries all lie in the same set of three indices cannot
all give rise to elements in c̄⊥.
Proof. These all follow from Lemma 5.1 by exhibiting an affine combination of the given vectors
which is of type I.
Let u, v ∈ W be such that ū and v̄ ∈ c̄⊥. It follows that λū + (1 − λ)v̄ ∈ c̄⊥ for all λ. Hence
Remark 3.1 shows that for all λ
0 ≥ J(d+ λu+ (1− λ)v, d + λu+ (1− λv))
= J(λ(d+ u) + (1− λ)(d+ v), λ(d + u) + (1− λ)(d+ v))
= λ2(J(d + u, d+ u) + J(d+ v, d+ v)− 2J(d+ u, d+ v)) +
2λ(J(d + u, d+ v)− J(d+ v, d + v)) + J(d+ v, d+ v).
Equality occurs if and only if λu+ (1− λ)v = c, as c̄ is the only null vector in c̄⊥.
Multiplying by −1, using Eq.(1.3), and recalling that the minimum value of a quadratic αλ2 +
βλ+ γ with α > 0 is γ − (β2/4α), we deduce the following result.
Lemma 5.3. If u, v ∈ W and ū, v̄ ∈ c⊥ then
(5.1)
Moreover, equality occurs if and only if c = λu+ (1− λ)v for some λ.
Remark 5.4. By definition, c does not lie in conv(W). So in the case of equality in Eq.(5.1) we
cannot have 0 ≤ λ ≤ 1. This observation will in many cases show that equality cannot occur.
Remark 5.5. The right-hand side of Eq.(5.1) is maximised when
= 1 (i.e., J(ū, v̄) = 0).
In this case Eq.(5.1) just follows from
≥ 1, which is true for any two vectors in c̄⊥. If
J(ū, v̄) 6= 0, we get sharper information.
16 A. DANCER AND M. WANG
Corollary 5.6. Suppose that K is connected. If u, v are type II vectors in W with ū, v̄ ∈ c̄⊥ then
with equality if and only if c = λu+ (1 − λ)v for some λ, in which case all the di = 2 whenever i
is an index such that ui or vi is nonzero.
Proof. Writing X =
and Y =
we see that 1 ≤ X,Y ≤ 3
. The lower bound arises from
ū, v̄ being in c̄⊥, while the upper bound follows from Remark 1.2(d) and the assumption that u, v
are type II vectors.
Now X +Y −XY = 1− (1−X)(1−Y ) is minimised for X,Y in this range if X = Y = 3
, when
it takes the value 3
. The inequality Eq.(5.1) now gives the result.
When K is connected, it follows that any two such type II vectors must overlap. Moreover, if
they have only one common index then we are in the case of equality in Cor 5.6. The nullity of c̄
implies that λ = 1
in this case, contradicting Remark 5.4.
Combining this remark with Cor 5.6 and Lemma 5.2 (i) , we deduce the following result.
Corollary 5.7. Assume K is connected. If u, v are type II vectors in W with ū, v̄ ∈ c̄⊥, then either
u = (−1a,−1b, 1i), v = (−1a,−1b, 1j) or u = (1a,−1b,−1i), v = (1a,−1b,−1j).
Hence the collection of all such type II vectors is of the form, for some fixed a, b:
(i) (−1a,−1b, 1i) : i ∈ I for some set I; or
(ii) (1a,−1b,−1i) : i ∈ I for some set I; or
(iii) (1,−1,−1, 0, · · · ), (1,−1, 0,−1, · · · ), (1, 0,−1,−1, · · · ).
We now investigate type III vectors.
Lemma 5.8. Suppose K is connected. If u is a type II vector and v a type III vector in W with
ū, v̄ ∈ c̄⊥, then
Proof. With the notation of Cor 5.6 we have 1 ≤ X ≤ 3
and 1 ≤ Y ≤ 3. So X + Y − XY =
1− (1 −X)(1 − Y ) ≥ 0, and Eq.(5.1) gives the desired inequality. Also, the case of equality (i.e.,
X = 3
, Y = 3) leads to λ = 2
, again contradicting Remark 5.4.
Remark 5.9. While Cor 5.6 - Lemma 5.8 are stated under the assumption that K is connected,
the actual property we used is that in Remark 2.4. By contrast, the next two results do not require
this property.
Lemma 5.10. Any two type III vectors u, v giving elements of c̄⊥ must overlap.
Proof. Write u = (−2i, 1j) and v = (−2k, 1l). By Cor 4.3, if i, k ∈ Ŝ≥2 then di = dk = 4. Since
J(c̄, ū) = 0 we have (by Cauchy-Schwartz)
Hence
1 + dj
If u and v do not overlap, then the above and the analogous result from considering J(c̄, v̄) = 0,
together with the nullity of c̄, imply that dj = 1 = dl and the only nonzero components of c are
ci = ck = −1, cj = cl =
. But then c is the midpoint of uv, contradicting c /∈ conv(W).
So if u and v do not overlap, we can take i ∈ Ŝ1. Proposition 4.5 shows that k ∈ Ŝ≥2 and
(di, dk, dl) = (2, 4, 2). Hence 2 < X ≤ 3 and Y =
, so X + Y − XY ≥ 0 and
Non-overlap means that equality holds. But then λ = 1/3, contradicting Remark 5.4.
CLASSIFICATION OF SUPERPOTENTIALS 17
Lemma 5.10, together with Lemmas 5.2 and 4.8, implies the following corollary.
Corollary 5.11. The type III vectors associated to elements of 1
(d+W)∩ c̄⊥ are, up to permutation
of indices, either of the form
(a) (−21, 1i), i ∈ I, (with d1 = 4 if |I| ≥ 2), or
(b) (11,−2i), i ∈ I,
for some subset I ⊂ {2, · · · , r}.
Having found the possible configurations for type III vectors in c̄⊥, we start to analyse the type II
vectors for each such configuration. For the rest of this section we will assume that K is connected
(cf Remark 5.9).
Remark 5.12. Lemma 5.8 now shows that in case (a) of Cor 5.11, if |I| ≥ 2, then every type II
vector associated to an element of c̄⊥ must have “-1” in place 1. Similarly, in case (b), if |I| ≥ 3,
then every such type II vector has “1” in place 1. (So if a type II is present then d1 6= 1). If |I| = 2,
the only possible type II vectors with “0” in place 1 are (01,−12,−13, 1i) where i ≥ 4, and all type
II vectors whose first entry is nonzero actually must have first entry equal to 1.
Lemma 5.13. In case (a) of Cor. 5.11 with |I| ≥ 2 there are no type II vectors associated to
elements of c̄⊥.
Proof. Let v = (−21, 1k) and w = (−11, 1i,−1j) give elements of c̄⊥ with k 6= i, j. Consider the
face {x̄ : xi + xk = 1, x1 + xj = −2} ∩ c̄
⊥. Other than v,w the possible elements in this face come
from u = (−11, 1k,−1j) and s = (−21, 1i). As d1 = 4, J(v̄, w̄) 6= 0, so vw is not an edge and u
must be present. But J(ū, w̄) = 1
(1 − 1
) 6= 0 since d1 = 4, giving a contradiction. So k = i
or j for every such v,w.
Hence if such a w exists there are at most two type III vectors. Now if |I| = 2 and the type IIIs
are (−2, 1, 0, · · · ), (−2, 0, 1, · · · ), we cannot have w = (−1, 1,−1, · · · ) or (−1,−1, 1, · · · ) as then a
suitable affine combination of the above vectors give a type I vector. (cf Lemmas 5.1, 5.2). So in
fact no type II vectors give rise to elements of c̄⊥.
Lemma 5.14. The vectors v = (−2, 1, 0, · · · ) and w = (0, 1,−1,−1, 0, · · · ) are not both associated
to elements of c̄⊥, unless (0, 1,−2, 0, · · · ) or (0, 1, 0,−2, 0, · · · ) is also.
Proof. Suppose (0, 1,−2, 0, · · · ), (0, 1, 0,−2, 0, · · · ) are absent. Consider the face {x̄ : x2 = 1, x1 +
x3 + x4 = −2} ∩ c̄
⊥. The other possible elements of this face come from t = (−1, 1,−1, 0, · · · ) and
y = (−1, 1, 0,−1, 0, · · · ). Both these must be present, as J(v̄, w̄) 6= 0. Applying Cor 4.3 to wt, vt
and wy we obtain (d1, d2, d3, d4) = (4, 2, 2, 2).
Now we have equality (for y, t) in Eq.(5.1), as both sides equal 15/16. We find that λ = 1/2,
giving a contradiction again to Remark 5.4.
Combining this with Lemma 5.8 (and using Lemma 4.7) yields:
Corollary 5.15. If there is a unique type III vector u = (−21, 12) with ū in c̄⊥, then the type II
vectors associated to elements of c̄⊥ all have “-1” in place 1. Moreover (−11,−12, 1i) cannot be
present. Also, if (−11, 12,−1i) is present for some i ≥ 3 then (d1, d2) = (4, 2) or (3, 3) and the
index i is unique.
For the last assertion, observe that (−11, 12,−1i) and the type III vector are joined by an edge,
so Cor 4.3 shows the dimensions are as stated. If we have two such type II for i0 and i1 then
Eq.(5.1) implies di0 + di1 ≤ 4. Hence since K is connected, di0 = di1 = 2 and we have equality in
Eq.(5.1) with λ = 1
, giving a contradiction.
Lemma 5.16. Let the type III vectors be as in Cor 5.11(b), i.e., they are (11,−2a), a ∈ I. Assume
that |I| ≥ 2. If we have a type II vector w = (11,−1i,−1j) with w̄ in c̄⊥ then i, j ∈ I
18 A. DANCER AND M. WANG
Proof. Suppose for a contradiction that w = (11,−1i,−1j) is present (so d1 6= 1) and (1
1,−2j)
absent (i.e. j /∈ I). Since |I| ≥ 2, we can consider v = (11,−2k) where k ∈ I (so k 6= j) and k 6= i.
Consider the face {x̄ : x1 = 1, xi + xj + xk = −2} ∩ c̄
⊥. As well as v,w the possible elements of W
in the face giving elements of c̄⊥ are y = (11,−1i,−1k), t = (11,−1j ,−1k) and u = (11,−2i). As
d1 6= 1, vw is not an edge so t is present. Now Cor 4.3 applied to vt and tw gives d1 = dj = 2 and
dk = 4.
Moreover, if i ∈ I then u is present, so the edge wu gives di = 4. Thus we have shown that
da = 4 for all a ∈ I.
Now considering (11,−2a) and (11,−2b) with a, b ∈ I, we see that we have equality in Eq.(5.1)
(both sides equal 3
). In fact c is the average of these two vectors (i.e., λ = 1
), so as in Remark 5.4
we have a contradiction.
Lemma 5.17. Let the type III vectors be as in Cor 5.11(b), i.e., they are (11,−2a), a ∈ I. Assume
that |I| ≥ 3. Then d1 = 1.
Proof. Each pair v,w of type III vectors gives an edge, and if d1 6= 1, then we have J(v̄, w̄) > 0. By
Theorem 4.3 all the midpoint vectors (11,−1a,−1b) are present for a, b ∈ I. Now Prop 3.7 shows
that Fv̄ and Fw̄ have opposite signs, so we have a contradiction if |I| ≥ 3.
Putting together our results so far, we obtain a description of the possibilities for c̄⊥∩ 1
(d+W).
Theorem 5.18. Assume that r ≥ 3 and K is connected, and that we are not in the situation of Thm
3.14. Up to permutation of the irreducible summands, the following are the possible configurations
of vectors in W associated to elements of 1
(d+W) ∩ c̄⊥.
(1) {(−21, 1i), 2 ≤ i ≤ m} for fixed m ≥ 2. There are no type II vectors, and d1 = 4 if m ≥ 3.
(2) {(11,−2i), 2 ≤ i ≤ m} for fixed m ≥ 3 and d1 = 1. There are no type II vectors.
(3)(i) {(11,−22), (11,−23), (−12,−13, 1i), 4 ≤ i ≤ m} with d1 = 1, d2 = d3 = 2.
(ii) {(11,−12,−13), (11,−22), (11,−23), (−12,−13, 1i), 4 ≤ i ≤ m}, d1 6= 1, d2 = d3 = 2.
(4) {(1,−2, 0, 0, · · · ), (1, 0,−2, 0, · · · ), (1,−1,−1, 0, · · · )} with d1 6= 1.
(5) A unique type III (−2, 1, 0, · · · ). Possible type II vectors are
(i) (−1, 1,−1, 0, · · · ) with either (d1, d2) = (4, 2) or (3, 3); or
(ii) {(−11, 13,−1i), 4 ≤ i ≤ m} for fixed m ≤ r and with d1 = 2; or
(iii) {(−11,−13, 1i), 4 ≤ i ≤ m} for fixed m ≤ r and with d1 = 2.
(6) No type III vectors. Possible type II vectors are
(i) {(−11,−12, 1i), 3 ≤ i ≤ m} for fixed m ≤ r, with d1 = d2 = 2 if m ≥ 4; or
(ii) {(11,−12,−1i), 3 ≤ i ≤ m} for fixed m ≤ r, with d1 = d2 = 2 if m ≥ 4; or
(iii) {(11,−12,−13), (11,−12,−14), (11,−13,−14)} with d1 = d2 = d3 = d4 = 2.
Proof. Cor 5.11 gives the possibilities for the type III vectors in c̄⊥. If there are none then Cor 5.7
gives the possibilities in (6). If there is a unique type III vector, then Cor 5.15 and Cor 5.7 give us
the cases listed in (5) (or (1) with m = 2 if there are no type II). If we have two or more type III
vectors with −2 in the same place then Lemma 5.13 shows we are in case (1).
If we have more than two type III vectors with 1 in the same place a, then da = 1 by Lemma
5.17. Remark 5.12 then implies there are no type II vectors and we are in case (2).
If we have exactly two type III vectors with 1 in the same place, e.g., (1,−2, 0, · · · ) and
(1, 0,−2, 0, · · · ), then the proof of Lemma 5.17 shows that if the type II vector (1,−1,−1, 0, · · · )
is absent we must have d1 = 1. On the other hand, if d1 = 1 we are, by Remark 5.12 and the
connectedness of K, in case (2) or (3)(i). If d1 6= 1, then by the above, Remark 5.12, and Cor 5.7,
we are in case (3)(ii) or (4).
The statements about values of the di follow from straightforward applications of Cor 4.3 to the
obvious edges of conv(1
(d+W)) ∩ c̄⊥.
CLASSIFICATION OF SUPERPOTENTIALS 19
Remark 5.19. The possibilities in Theorem 5.18 can be somewhat sharpened. In cases (1), (2),
and (3), m cannot be r; in other words the maximum number of vectors is not allowed. This follows
easily from looking at the system of equations expressing the nullity of c̄, the orthogonality of the
vectors to c̄ and the fact that the entries of c sum up to −1. Similarly, r 6= 3 in (5)(i) and r 6= 4 in
(6)(iii).
When m ≥ 5 in (5)(ii) or (5)(iii), the segment joining two type II vectors is an edge, so Cor 4.3
gives d3 = 2.
6. Adjacent (1B) vertices
We now turn to (1B) vertices. Let ξ̄, ξ̄′ be adjacent (1B) vertices of ∆. Then there exist vertices
x̄, x̄′ of conv(1
(d+W)) such that c̄, ξ̄, x̄ are collinear and c̄, ξ̄′, x̄′ are collinear. Moreover, there exist
null vectors ā, ā′ such that x̄ = (ā+ c̄)/2 and x̄′ = (ā′+ c̄)/2. By Cor 3.4, there must be an element ȳ
of conv(1
(d+W)) on āā′, so P−1(ξ̄ξ̄′) contains the convex hull of x̄, x̄′, ȳ and hence is 2-dimensional.
As ξ̄ξ̄′ is by assumption an edge of ∆, P−1(ξ̄ξ̄′) is a 2-dimensional face of conv(1
(d+W)).
So we need to analyse the 2-dimensional faces of conv(W) containing vertices x, x′ such that
(6.1) x = (a+ c)/2, x′ = (a′ + c)/2, ā, ā′ null,
and such that c lies in the 2-dimensional plane defining this face. The lines through x, c (resp. x′, c)
only meet conv(W) at x (resp. x′).
Most 2-faces of conv(W) are triangular. We list below (up to permutation of components) all the
possible non-triangular faces. For further details regarding how this listing is arrived at, see [DW5].
We emphasize that only the full faces are being listed, i.e., configurations formed by all the possible
elements of W in a given 2-dimensional plane. As the set of weight vectors for a given principal
orbit may be a subset of the full set of possible weight vectors, these full faces may degenerate to
subfaces or even lower-dimensional faces (see Remark 6.2).
Listing convention: In the interest of economy and clarity, we make the convention that when we
list vectors in W belonging to a 2-face we will use the freedom of permuting the summands to
place nonzero components of the vectors first and we will only put down the minimum number of
components necessary to specify the vectors.
Hexagons: There are 3 possibilities.
(H1) This is the face in the plane {x1 + x2 + x3 = −1; xa = 0, for a > 3}. Points of W are
(−2i, 1j), (−1i, 1j ,−1k), (−1i) where i, j, k ∈ {1, 2, 3}. The type III vectors form the vertices of the
hexagon.
(H2) The plane here is {x1 + x2 = −1, x3 + x4 = 0, xi = 0 (i > 4)}. Points of W are vertices
u = (−2, 1, 0, 0), v = (1,−2, 0, 0), y = (−1, 0, 1,−1), y′ = (0,−1, 1,−1),
z = (−1, 0,−1, 1), z′ = (0,−1,−1, 1),
and the interior points
α = (−1, 0, 0, 0), β = (0,−1, 0, 0).
(H3) The plane is {x2 = −1, x1 + x3 + x4 = 0, xi = 0 (i > 4)}. Points of W are the vertices
u = (−1,−1, 1, 0), v = (0,−1, 1,−1), w = (1,−1, 0,−1),
x = (1,−1,−1, 0), y = (0,−1,−1, 1), z = (−1,−1, 0, 1)
and the centre
t = (0,−1, 0, 0).
Square: (S) with midpoint t = (0,−1, 0, 0, 0) and vertices
v = (−1,−1, 1, 0, 0), u = (0,−1, 0, 1,−1),
s = (0,−1, 0,−1, 1), w = (1,−1,−1, 0, 0).
20 A. DANCER AND M. WANG
Trapezia: We have vertices v, u, s, w, t with 2v − s = 2u − w and t = 1
(s + w), i.e., these are
symmetric trapezia. Below we list the possible v, u, s, w.
v u s w
(T1) (−2, 1, 0, 0) (−2, 0, 1, 0) (0, 0,−2, 1) (0,−2, 0, 1)
(T2) (−2, 0, 1, 0) (−2, 1, 0, 0) (0,−1, 1,−1) (0, 1,−1,−1)
(T3) (−1,−1, 0, 1) (0,−1,−1, 1) (−2, 1, 0, 0) (0, 1,−2, 0)
(T4) (0, 0, 1,−1,−1) (1, 0, 0,−1,−1) (−2, 1, 0, 0, 0) (0, 1,−2, 0, 0)
(T5) (−1, 0, 0, 1,−1) (0, 0,−1, 1,−1) (−2, 1, 0, 0, 0) (0, 1,−2, 0, 0)
(T6) (1,−1,−1, 0, 0) (1,−1, 0,−1, 0) (0, 0,−1, 1,−1) (0, 0, 1,−1,−1)
Table 4: Possible trapezoidal faces
Note that the configuration with vertices (−1,−1, 1, 0, 0), (−1,−1, 0, 1, 0), (0, 0, 1,−1,−1), and
(0, 0,−1, 1,−1) is equivalent to (T6) under the composition of a permutation and a J-isometric
involution.
Parallelograms: We have vertices v, u, s, w with v − u = s− w.
v u s w
(P1) (−2, 1, 0, 0) (−1, 0,−1, 1) (−2, 0, 1, 0) (−1,−1, 0, 1)
(P2) (−2, 1, 0, 0, 0) (−2, 0, 1, 0, 0) (0, 1, 0,−1,−1) (0, 0, 1,−1,−1)
(P3) (−2, 1, 0, 0, 0) (−2, 0, 1, 0, 0) (0, 0,−1,−1, 1) (0,−1, 0,−1, 1)
(P4) (−2, 1, 0, 0) (−1, 0, 1,−1) (−1,−1, 0, 1) (0,−2, 1, 0)
(P5) (−2, 1, 0, 0, 0) (−1, 0, 0, 1,−1) (−1, 0, 1,−1, 0) (0,−1, 1, 0,−1)
(P6) (−2, 1, 0, 0, 0) (−1, 0, 0,−1, 1) (−1, 0,−1, 1, 0) (0,−1,−1, 0, 1)
(P7) (−2, 1, 0, 0, 0) (0, 1,−1,−1, 0) (−1, 0, 0, 1,−1) (1, 0,−1, 0,−1)
(P8) (1,−1,−1, 0, 0, 0) (0, 0, 0, 1,−1,−1) (1,−1, 0,−1, 0, 0) (0, 0, 1, 0,−1,−1)
(P9) (1,−1,−1, 0, 0, 0) (1,−1, 0, 0,−1, 0) (0, 0,−1, 1, 0,−1) (0, 0, 0, 1,−1,−1)
(P10) (1,−1,−1, 0, 0, 0) (1, 0, 0, 0,−1,−1) (0,−1,−1, 1, 0, 0) (0, 0, 0, 1,−1,−1)
(P11) (0, 0, 1,−1,−1, 0) (0,−1, 0,−1, 0, 1) (1, 0, 0, 0,−1,−1) (1,−1,−1, 0, 0, 0)
(P12) (1, 0,−1, 0,−1) (1,−1,−1, 0, 0) (0, 0,−1, 1,−1) (0,−1,−1, 1, 0)
(P13) (−1, 0,−1, 0, 1) (0, 0,−1,−1, 1) (0,−1,−1, 1, 0) (1,−1,−1, 0, 0)
(P14) (−1, 0, 1, 0,−1) (−1,−1, 1, 0, 0) (0, 0,−1, 1,−1) (0,−1,−1, 1, 0)
(P15) (−1, 0, 1, 0,−1) (−1,−1, 1, 0, 0) (0, 0, 1,−1,−1) (0,−1, 1,−1, 0)
(P16) (−2, 1, 0, 0) (0, 1,−2, 0) (−1, 0, 1,−1) (1, 0,−1,−1)
(P17) (−2, 1, 0, 0) (0, 1, 0,−2) (−2, 0, 1, 0) (0, 0, 1,−2)
Table 5: Possible parallelogram faces
Remark 6.1. (P1), (P2), (P3), and (P17) are actually rectangles. (P16) also includes the midpoints
y = (u+ v)/2 = (−1, 1,−1, 0) and z = (s+ w)/2 = (0, 0, 0,−1). The rectangle (P17) also includes
the midpoints y = (u+ v)/2 = (−1, 1, 0,−1) and z = (s+ w)/2 = (−1, 0, 1,−1).
Remark 6.2. We must also consider subshapes of the above. Each symmetric trapezium contains
two parallelograms. The two rectangles with midpoints (P17), (P16) will contain asymmetric
trapezia. (P17) also contains parallelograms and squares. (For (P16), note that s is present iff w is.)
Furthermore, there are numerous subshapes of the hexagons. The regular hexagon (H3) contains
rectangles with midpoint (by omitting opposite pairs of vertices). Besides triangles, the hexagon
(H2) contains pentagons, rectangles and squares (with midpoints), and kite-shaped quadrilaterals
(e.g. y′uz′v). For (H1) see the discussion before Theorem 6.12. Finally, the triangle with midpoints
of all sides (where the vertices are the three type III vectors with 1 in the same place) contains a
trapezium (by omitting one vertex) and hence parallelograms.
Remark 6.3. We also note for future reference that there are examples where we can have four
or more coplanar elements of W but the plane cannot be a face. These examples are not of course
CLASSIFICATION OF SUPERPOTENTIALS 21
relevant to the case of adjacent (1B) vertices, but some will be relevant when we consider multiple
vertices of type (2). The examples which we will need in that context are the following three
trapezia
v u s w
(T ∗1) (0, 1,−1,−1) (1, 0,−1,−1) (−2, 1, 0, 0) (1,−2, 0, 0)
(T ∗2) (0,−1, 1,−1) (1,−1, 0,−1) (−2, 1, 0, 0) (0, 1,−2, 0)
(T ∗3) (1,−1,−1, 0) (1,−1, 0,−1) (−1, 0,−1, 1) (−1, 0, 1,−1)
Table 6: Further trapezia
In (T*2),(T*3), as in (T1)-(T7), we have 2v − s = 2u− w. In these examples t = 1
(s+w) may
also be present. In (T*1) we have s− w = 3(v − u), and the vectors t = (2s + w)/3 = (−1, 0, 0, 0)
and r = (s+ 2w)/3 = (0,−1, 0, 0) will also be present.
As an example, we explain why the trapezium (T*2) can never be a face. As u is present
in W, so are u′ = (−1, 1, 0,−1) and u′′ = (−1,−1, 0, 1). Now (2u′ + u′′)/3 = (2s + u)/3 =
(−1, 1
, 0,−1
) is in the plane, but u′ is not, so this plane cannot give a face. Similar arguments
involving (1, 0,−1,−1), (−1, 0, 0, 0) (resp. (−1, 0,−1, 1), (−1, 0, 1,−1)) show (T*1) (resp. (T*3))
cannot be faces.
These arguments also show several parallelograms cannot be faces, but these will not be relevant
for our purposes.
We now begin to classify the possible 2-faces which arise from adjacent (1B) vertices. We shall
repeatedly use Prop 1.4, Cor 3.4, and Lemma 3.5. Let E denote the affine 2-plane determined by
the 2-face being studied.
Theorem 6.4. Suppose we have adjacent (1B) vertices corresponding to a parallelogram face vusw
of conv(W). So we have ū = (ā+c̄)/2 and w̄ = (ā′+c̄)/2 for null ā, ā′. Suppose the vertices v, u, s, w
are the only elements of W in the face. Then u,w are adjacent vertices of the parallelogram, and
either
(i) C ∩ E = {c̄, ā, ā′, ē} where ē is null with v = (a+ e)/2 and s = (a′ + e)/2; or
(ii) v̄, s̄ ∈ C and J(ā, v̄) = J(ā′, s̄) = J(s̄, v̄) = 0.
Moreover, if none of v, u, s, w is type I, then (i) cannot occur.
Proof. We may introduce coordinates in the 2-plane E using the sides sv and sw to define the
coordinate axes. In this way we can speak of “left” or “right”, “up” or “down”. If we extend
the sides of the parallelogram to infinite lines, these lines divide the part of the plane outside the
parallelogram into 8 regions, and c̄ must be in the interior of one such region.
We first observe that if c̄ is in one of the four regions which only meet the parallelogram at a
vertex, then āā′ does not meet the parallelogram, contradicting Lemma 3.4.
(A) Let c̄ then lie in a region which meets the parallelogram in an edge. Without loss of generality
we may assume the edge is uw. By Cor 3.4, all elements of C ∩ E lie on or between the rays from
c̄ through ā, ā′. Hence, by Lemma 3.2, J(b̄, c̄) > 0 for all b̄ ∈ C \ {c̄}. If b̄ is a rightmost element
of (C ∩E) \ {c̄}, then as b̄+ c̄ cannot be written in another way as a sum of two elements of C, we
deduce from Prop 1.4 that b̄+ c̄ ∈ d+W. So b̄ is either ā or ā′. All other elements of C ∩ E lie to
the left of āā′. Note also that a rightmost element of (C ∩ E) \ {c̄, ā, ā′} satisfies b+ c = a+ a′, 2v
or 2s.
(B) Next let ē = 2v̄ − ā. Observe that as well as v̄ = (ā + ē)/2, we have s̄ = (ā′ + ē)/2, since
2v̄ − ā = 2(v̄ − ū) + c̄ = 2(s̄ − w̄) + c̄ = 2s̄− ā′.
If ē ∈ C, then it must be null, and the same argument as above shows that no elements of
(C ∩ E) \ {ē} lie to the left of ā, ā′, so we are in case (i). Now, Lemma 3.2 shows J(h̄, k̄) > 0 for
all h̄ 6= k̄ ∈ C ∩E. If v, u, s, w are all type II/III, we see that Fc̄, Fē are of one sign and Fā, Fā′ the
other sign. But now the contributions from ā+ ā′ and c̄+ ē in the superpotential equation cannot
cancel.
22 A. DANCER AND M. WANG
If ē /∈ C then, as in the argument before Theorem 3.8, s̄, v̄ ∈ C and we are in case (ii). Prop 3.7
shows v̄, s̄ are orthogonal. Moreover, note that the remark at the end of (A) shows that v + c or
s+ c is left of a+ a′.
Lemma 6.5. In case (ii) of Theorem 6.4, we have J(v̄, v̄) = J(s̄, s̄).
Proof. As c̄ and ā = 2ū− c̄ are both null, and similarly c̄ and ā′ = 2w̄− c̄ are both null, we deduce
(cf Remark 3.9)
(6.2) J(ū, ū) = J(ū, c̄) : J(w̄, w̄) = J(w̄, c̄).
We also have
(6.3) 2J(ū, v̄) = J(c̄, v̄) : 2J(w̄, s̄) = J(c̄, s̄)
from the orthogonality conditions on ā, v̄ and ā′, s̄.
Now J(s̄, s̄)− J(v̄, v̄) = J(s̄, s̄)− J(w̄− ū− s̄, w̄− ū− s̄), which, on expanding out and using the
second relations of Eqs.(6.2),(6.3), becomes J(2ū− c̄, w̄ − s̄)− J(ū, ū). Now
J(2ū− c̄, w̄ − s̄)− J(ū, ū) = J(2ū− c̄, ū− v̄)− J(ū, ū) = J(2ū− c̄, ū)− J(ū, ū) = J(ū− c̄, ū) = 0.
We have used the first relations of Eqs.(6.3), (6.2) in the second and fourth equalities.
Remark 6.6. We must also consider the case when the midpoint of one side or a pair of opposite
sides of the parallelogram face is in W. This can happen for (P16) and (P17). Note that v, u, s, w
are type II/III in these cases.
In fact, the argument of Theorem 6.4 is still valid if one or both of the midpoints of vu, sw is in
W and c lies in the region to the right of uw (or the left of vs).
Keeping c in the region to the right of uw, we now need to consider the case where one or both
of the midpoints of vs, uw is in W. The conclusions (in 6.4(ii)) still hold except that we no longer
have J(v̄, s̄) = 0.
However, we have to make slight modifications to the arguments as 1
(ā + ā′) may be in C ∩ E.
If ē ∈ C, then, as ā + ā′ is not in d + W, the usual sign argument shows that the terms in the
superpotential equation summing up to ā + ā′ do not cancel, which is a contradiction. So ē /∈ C
and our previous arguments hold except for the use of Prop 3.7.
Note that we also have to consider the possibility that a, a′, and e lie on the line through vs.
But now the midpoint of uw must be present and C ∩E = {c̄, ā, ā′, 1
(ā+ ā′)}, with v + s = a+ a′.
The usual sign argument then forces the midpoints of uw and vs to be present and of type I. Hence
this special configuration cannot occur in (P16) or (P17).
Lastly, since the proof of Lemma 6.5 makes no mention of midpoints, it remains valid if midpoints
are present.
The conditions of Theorem 6.4 and Lemma 6.5, together with the nullity of ā, ā′, c̄, put very strong
constraints on vusw and the dimensions. In fact, one can check that these constraints cannot be
satisfied for any of our parallelograms (including those of Remark 6.2) with one exception. This is
the rectangle yy′z′z in (H2) with c = (−2, 1, 0, · · · ) and 1
, which will be dealt with in
Lemma 8.5. We now give an example of how to apply the above conditions in a specific case.
Example 6.7. Consider parallelogram (P8). The equation of the 2-plane E containing the paral-
lelogram is
(6.4) x2 = −x1, x5 = x6, x2 + x5 = −1, x1 + · · · + x6 = −1
and xi = 0 for i > 6. As all vertices are type II/III, we must be in case (ii) of Theorem 6.4.
(A) Take c to face the side uw. Note that vs and uw have equation x1 = 1, x1 = 0 respectively,
so c1 < 0. Also, the remarks at the end of parts (A) and (B) in the proof of Theorem 6.4 shows
that c1 > −
, as v + c or s+ c is left of a+ a′ so 1 + c1 > −2c1
CLASSIFICATION OF SUPERPOTENTIALS 23
The condition J(v̄, s̄) = 0 implies d1 = d2 = 2 and Lemma 6.5 implies d3 = d4. Eqs.(6.2) and
(6.3) give four linear equations in ci. Now d3 = d4 and Eq.(6.3) show c3 = c4, so the equations for
the plane give c = (1
− c4, −
+ c4, c4, c4,−
− c4, −
− c4). Next d1 = d2 = 2 and Eq.(6.3) show
c4 = 3d4/(2d4 + 2) and c1 = (1− 2d4)/(2d4 + 2).
But the condition −1
< c1 < 0 now implies d3 = d4 = 1, and it follows that c cannot be null.
(B) The argument if c faces vs is very similar. We have d3 = d4 and d5 = d6 = 2, and the
orthogonality equations imply c3 = c4. So c has the same form as in the second paragraph of (A)
above. We find c4 = −3d4/(2d4+2)) and c1 = (1+4d4)/(2d4 +2). But we now have the inequality
1 < c1 <
, so again d3 = d4 = 1, violating nullity.
(C) If c faces vu or sw then we need J(s̄, w̄) = 0 (resp. J(v̄, ū) = 0), which is impossible.
Example 6.8. The example of the square (S) with midpoint can be treated in essentially the same
way as the parallelograms. By symmetry, we may assume that c lies in the region that intersects
uw. However, because 1
(a + a′) may now be the midpoint and hence in W, the configuration
of Theorem 6.4(i) can occur, even though all vertices are type II. We have C ∩ E = {c̄, ā, ā′, ē}
with a = (−1,−1, 1, 1,−1, · · · ), a′ = (1,−1,−1,−1, 1, . . .), c = (1,−1,−1, 1,−1, · · · ), and e =
(−1,−1, 1,−1, 1, · · · ) with nullity condition
= 1. We will be able to rule this case out in
§7. On the other hand, the configuration of Theorem 6.4(ii) cannot occur, as one easily checks.
Next assume that adjacent (1B) vertices in ∆c̄ determine a trapezium vusw as shown in the
diagram below:
II III
VVIVII
where t is the midpoint of sw and vu is parallel to sw. We assume that v, u, s, w ∈ W but our
conclusions hold whether or not t lies in W. We will now derive constraints on the 2-face and E∩C
resulting from having c lie in one of the regions shown above. For theoretical considerations, we
need only treat the cases where c lies in regions I to VI. In practice, for an asymmetric trapezium,
we must consider c lying in the remaining regions as well. In the following we will adopt the
convention that ā, ā′ always denote null vectors in C.
(I) c in region I: This is impossible because then s̄ = 1
(c̄ + ā) and w̄ = 1
(c̄ + ā′) for some ā, ā′,
and so āā′ would not intersect conv(1
(d+W), a contradiction to Cor 3.4.
(II) c in region II: Then v̄ = 1
(c̄+ ā), ū = 1
(c̄+ ā′) for some ā, ā′. We get a contradiction to Cor
3.4 if ā, ā′ lie below the line sw. They also cannot lie on the line sw since the argument in (A) in
the proof of Theorem 6.4 and Cor 3.4 imply that C ∩ E = {c̄, ā, ā′}, and the terms corresponding
to s̄, w̄ in the superpotential equation would be unaccounted for.
24 A. DANCER AND M. WANG
Let e = 2s − a, e′ = 2w − a′. These points lie in region VI, and since we have a trapezium,
e 6= e′. We may now apply Theorem 3.8 to ā and ā′ to obtain the possibilities:
(i) s̄, w̄ ∈ C; J(ā, s) = 0 = J(ā′, w̄),
(ii) s̄ ∈ C, J(ā, s̄) = 0; w /∈ C, ē′ ∈ C is null, J(ē′, s̄) = 0,
(iii) w̄ ∈ C, J(w̄, ā′) = 0; s̄ /∈ C, ē ∈ C is null, J(ē, w̄) = 0.
Note that the last condition in (ii) (resp. (iii)) results from applying Theorem 3.8 to ē′ (resp. ē).
(III) c in region III: We have v̄ = 1
(c̄+ ā), w̄ = 1
(c̄+ ā′) for some ā, ā′ lying respectively in regions
VIII and VI (in view of Cor 3.4). Applying Theorem 3.8 we obtain the possibilities:
(i) s̄ ∈ C, J(ā, s̄) = 0 = J(ā′, s̄),
(ii) s̄ /∈ C, 2s̄ = ā+ ā′ (which implies c+ s = v + w).
(IV) c in region IV: We have ū = 1
(c̄+ ā), w̄ = 1
(c̄+ ā′) for some ā, ā′ ∈ C ∩E.
If a lies in region IX, then Cor 3.4 implies that ā′ lies in region VI. Applying Theorem 3.8 to ā
and ā′ we obtain the possibilities:
(i) s̄ ∈ C, J(ā, s̄) = 0 = J(ā′, s̄),
(ii) 2s = a+ a′, i.e., c+ s = u+ w.
If a lies on the line sv, then we may apply Theorem 3.8 to ā′. We cannot have 2s̄ = ā′ + ē′ with
ē′ ∈ C and null, otherwise āē′ would not intersect conv(1
(d+W)). So we have
(iii) s̄ ∈ C and J(s̄, ā′) = 0.
If a lies in region II, then ā′ lies in region VI. Let ē = 2v̄ − ā and ē′ = 2s̄ − ā′. As we have a
trapezium, ē 6= ē′. Now e lies in region VII or VIII while e′ lies in region VIII or IX, so by Cor 3.4
ē and ē′ cannot both lie in C and hence be null. Theorem 3.8 now gives the possibilities:
(iv) v̄, s̄ ∈ C, J(ā, v̄) = 0 = J(ā′, s̄), (and by Prop 3.7 J(v̄, s̄) = 0),
(v) v̄ ∈ C, J(ā, v̄) = 0, ē′ ∈ C is null, and J(ē′, v̄) = 0,
(vi) s̄ ∈ C, J(ā′, s̄) = 0, ē ∈ C is null, and J(ē, s̄) = 0.
(V) c in region V: We have ū = 1
(c̄+ ā′), s̄ = 1
(c̄+ ā) for some ā, ā′ lying respectively in regions
VIII and II (by Cor 3.4). Theorem 3.8 now gives the possibilities:
(i) v̄ ∈ C, J(ā, v̄) = 0 = J(ā′, v̄),
(ii) v̄ /∈ C, 2v̄ = ā+ ā′ (which implies c+ v = u+ s).
(VI) c in region VI: We have s̄ = 1
(c̄ + ā), w̄ = 1
(c̄ + ā′) for some ā, ā′ lying respectively in
regions VIII and IV (by Cor 3.4). (To rule out ā, ā′ lying in the line vu, we proceed as in case (II),
except that when t ∈ W, we conclude instead that C ∩E = {c̄, ā, ā′, 1
(ā+ ā′)}. One can still check
that v̄, ū cannot be both accounted for.) Now let ē = 2v̄ − ā and ē′ = 2ū − ā′. Again, having a
trapezium means ē 6= ē′ and Theorem 3.8 now gives the possibilities:
(i) ū, v̄ ∈ C, J(ā, v̄) = 0 = J(ā′, ū), (and J(ū, v̄) = 0 by Prop 3.7),
(ii) v̄ ∈ C, J(ā, v̄) = 0, ū /∈ C, ē′ ∈ C is null,
(iii) ū ∈ C, J(ā′, ū) = 0, v̄ /∈ C, ē ∈ C is null.
Remark 6.9. We mention a useful inequality which holds in (II) and (VI) above, as well as in
parallelogram faces with the same configuration (cf Example 6.7(A)).
Let us consider (II), where we choose in E coordinates such that the first coordinate axis is
parallel to s̄w̄ (assumed to be horizontal) and the second coordinate axis is arbitrary, with the
second coordinate increasing as we go up. As in (A) in the proof of Theorem 6.4, all points in
(C ∩ E) \ {c̄, ā, ā′} must lie below the line āā′. Let b̄ be a point among these with largest second
coordinate. Since we have seen above that either s̄ or w̄ lies in C∩E, we have s2 ≤ b2. Furthermore,
as b̄ + c̄ cannot lie in d +W it must be balanced by sums of elements in C ∩ E, with the limiting
configuration given by ā + ā′. So we have 1
(b2 + c2) ≤ a2 = a
2 = 2v2 − c2. Combining the two
inequalities we get 3c2 ≤ 4v2 − s2.
Equality in the above holds iff b̄ lies in s̄w̄ and b̄ + c̄ = ā + ā′. In particular, b̄ is unique, so in
II(i), the inequality above is strict.
CLASSIFICATION OF SUPERPOTENTIALS 25
Note that we only need v̄ū and s̄w̄ to be parallel and the presence or absence of t in W is
immaterial. Hence in Theorem 6.4(ii) we also have an analogous strict inequality, which we have
already used, e.g., in (B) of Example 6.7. (For a parallelogram, there may be midpoints on the
pair of non-horizontal sides lying in 1
(d+W), but 1
(b̄+ c̄) can never equal these midpoints, so we
still get the inequality we want.)
For the configuration in (VI), we still have an analogous inequality, but since 1
(ā + ā′) ∈ C, we
lose uniqueness of b̄ and hence the strict inequality.
We will also have occasion to apply the above analysis to appropriate trapezoidal regions in
hexagon (H3).
The method described above together with Remark 6.9 can now be used to rule out the trapezia
(T1)-(T6) as well as those mentioned in Remark 6.2.
Example 6.10. For the trapezium (T3), the vectors v, u, s, w are given in Table 4, and lie in the
2-plane {x1 + x2 + x3 + x4 = −1, x2 +2x4 = 1}. vu is given by x4 = 1 while sw is given by x4 = 0.
sv is given by x3 = 0 and wu is given by x1 = 0. The vector c that we are looking for has the
form (−c3 + c4 − 2, 1 − 2c4, c3, c4). Since the trapezium is symmetric, an explicit symmetry being
induced by interchanging x1 and x3, we need only consider c lying in regions II-VI.
(A) If c lies in region III, then c1 > 0, c4 > 1. Since a = 2v − c, we obtain a = (c3 − c4,−3 +
2c4,−c3, 2 − c4). Similarly, a
′ = (c3 − c4 + 2, 1 + 2c4,−4 − c3,−c4). If we are in case (ii), then
c = v + w − s = (1,−1,−2, 1), which violates c4 > 1. So we must be in case (i).
It follows from J(ā, s̄) = 0 = J(ā′, s̄) that d1+3 = −2c3+4c4 and J(w̄, s̄) = J(v̄, s̄). The second
equality implies that d1 = d2. Using this together with the first equality and the null condition
for ā′ (in the form J(w̄, w̄) = J(w̄, c̄), see Remark 3.9) we get c4 = d1(d1 − 1)/(4d1 + 2d3). Since
c4 > 1, we have d1(d1 − 5) > 2d3, so d1 > 5. But by Remark 3.1, J(s̄, s̄) < 0, which gives d1 < 5
(since d1 = d2), a contradiction.
(B) Let c lie in region IV, so that c1 > 0, 0 < c4 < 1. We obtain a = (2 + c3 − c4, 2c4 −
3,−2 − c3, 2 − c4) and a
′ = (2 + c3 − c4, 1 + 2c4,−4 − c3,−c4). We claim that a3 > 0, so that
a lies in region IX. To see this, we solve for c3, c4 using the null conditions J(ū, ū) = J(c̄, ū) and
J(w̄, w̄) = J(c̄, w̄) for ā, ā′ respectively. We obtain c4 = (d2d3 + 2d3d4 − d2d4)/(d3(d2 + 3d4)) and
a3 = −2− c3 = (d2d3 + 2d3d4 − d2d4)/(d2(d2 + 3d4)). Since c4 > 0 we obtain our claim.
Since a lies in region IX, we first check if (ii) holds. In this case, c = (2,−1,−3, 1) which
contradicts c4 < 1. The equations in (i) together imply the contradiction 0 = −4/d2.
(C) Suppose c lies in region V, so that c1 > 0, c4 < 0. We obtain a = (c3−c4−2, 1+2c4,−c3,−c4)
and a′ = (c3−c4+2, 2c4−3,−2−c3, 2−c4). If (ii) holds then c = (−1, 1,−1, 0) and this contradicts
c1 > 0. Hence (i) must hold.
By Remark 3.9, the null condition for ā is J(s̄, s̄) = J(s̄, c̄), which is c3
− ( 1
)c4 = 0. The
two equations in (i) imply J(ū, v̄) = J(s̄, v̄) and J(v̄, v̄) < 0, which in turn give d1 = 2. Using
this, the null condition for ā, and J(ā′, v̄) = 0 we obtain c4 =
and c3 = −
d2(d2+1)
. But
c1 = c4 − c3 − 2 > 0, which simplifies to 1 > d2(d2 + 1), a contradiction.
(D) Let c lie now in region II. Then c1 < 0, c3 < 0, 1 < c4 ≤
where the last upper
bound comes from the inequality in Remark 6.9. We obtain a = (c3 − c4, 2c4 − 3,−c3, 2 − c4)
and a′ = (2 + c3 − c4, 2c4 − 3,−2− c3, 2− c4). The null conditions for ā, ā
′ then give
c3 = −
2d1d4 + d1d2 − d2d4
(d1 + d3)(d2 + 2d4)− d2d4
, c4 =
(d1 + d3)(d2 + 2d4)
(d1 + d3)(d2 + 2d4)− d2d4
Suppose we are in case (i). The two equations and the above values of c3, c4 combine to give
(d1 − d3)((d1 + d3)(d2 + 2d4) − d2d4) = 0. However, the upper bound c4 ≤
translates into
(d1 + d3)(d2 + 2d4) ≥ 4d2d4. So the second factor is positive and we have d1 = d3. Putting this
26 A. DANCER AND M. WANG
information into the equation J(ā, s̄) = 0, we get
d1d2d4(d2 + 15) = 2d
1d2(d2 + 1)− 6d1d
2 + 2d4(2d
1d2 + 2d
1 + d
By Remark 3.1 we also have J(s̄, s̄) < 0, i.e., 1 < 4
, so either d2 = 1 or d1 < 8. Substituting
these values into the equation above and using c4 ≤
we obtain in each instance a contradiction.
If we are in case (ii), then by adding the equations J(ā, s̄) = 0 and 2J(w̄, s̄) − J(ā′, s̄) = 0
(equivalent to J(ē′, s̄) = 0), we obtain 1 = 2
. Hence (d1, d2) = (4, 2) or (3, 3). One then
checks that these values are incompatible with the null condition for ē′, J(ā, s̄) = 0, and the bound
c3 < 0.
An analogous argument works to eliminate case (iii), where we now need the bound c4 ≤
instead.
(E) Lastly suppose c lies in region VI, so c1, c3 < 0 and −
≤ c4 < 0, where the lower bound for
c4 results from Remark 6.9. We have a = (c3 − c4 − 2, 1 + 2c4,−c3,−c4) and a
′ = (2 + c3 − c4, 1 +
2c4,−4− c3,−c4). Using the null conditions for ā, ā
′, we obtain
c1 = −
2(d2 + d3)
d1 + d2 + d3
, c2 =
d1 + 5d2 + d3
d1 + d2 + d3
, c3 =
−2(d1 + d2)
d1 + d2 + d3
, c4 =
d1 + d2 + d3
If we are in case (i), J(ū, v̄) = 0 gives d2 = d4 = 2. The other two equations and the above
values of c3, c4 then give 3(d1 + d3 + 2)(d1 + d3 − 4) = 4(3d1 + 3d3 − 2). The lower bound −
becomes d1 + d2 + d3 ≥ 6d2. Using this inequality in the above Diophantine relation leads to a
contradiction. (Alternatively, observe the relation is a quadratic in d1 + d3 with no rational roots).
For case (ii), using the two equations and the above values for c3, c4, we arrive at the relation
(d1 + d2 + d3)((d1 − 5)d2d4 + d1d4 + d2d3 + 2d3d4) = 2d2(d1d2 + 2d1d4 − d2d4 + d2d3 + 2d3d4).
Using the lower bound −1
≤ c4 in the above relation we see that d1 ≤ 3. By direct substitution, we
further obtain d1 6= 3. Finally, if d1 = 2, the null condition for c̄ gives 1 >
c21 and so d2 + d3 ≤ 4.
The lower bound on c4 now implies d2 = 1. Since c2 > 1, the null condition for c̄ is violated.
Case (iii) reduces to case (ii) upon interchanging the first and third summands. Therefore, the
trapezium (T3) has been eliminated.
We discuss next the hexagons (H1)-(H3). As the three cases are similar, we will focus on (H3)
and refer to the following (schematic) diagram:
IVIV′
VII′ VII
Example 6.11. The hexagon (H3) lies in the 2-plane given by {x2 = −1, x1 + x3 + x4 = 0}. So c
has the form (−c3 − c4,−1, c3, c4). The lines vw and zy are given respectively by x1 + x3 = 1 and
CLASSIFICATION OF SUPERPOTENTIALS 27
x1 + x3 = −1. Similarly, the lines uv and yx are given by x3 = 1 and x3 = −1 respectively. The
lines uz and wx are given by x1 = −1 and x1 = 1 respectively.
Interchanging x1 and x3 induces the reflection about the perpendicular bisector of vw, while
(x1, x2, x3, x4) 7→ (−x3, x2,−x1,−x4) induces the reflection about ux. These symmetries reduce
our consideration to those c lying in regions I-VI. Moreover, (H3) is actually a regular hexagon.
The symmetry (x1, x2, x3, x4) 7→ (−x4, x2,−x3,−x1) induces the reflection about zw, which swaps
region II with region IV and region I with region VI. Finally, the symmetry (x1, x2, x3, x4) 7→
(−x3, x2,−x4,−x1) induces the rotation in E about t taking x to w, and maps region V to region
III. Therefore, we need only consider c lying in regions I, II, and V.
In the discussion below we again adopt the convention that ā, ā′ always denote null vectors in C.
If c lies in region I, then ū = 1
(c̄+ ā), x̄ = 1
(c̄+ ā′) for some ā, ā′, and we immediately see that
āā′ cannot meet conv(1
(d+W)), a contradiction to Cor 3.4.
c lying in region II:
We have c1, c3, < 1 and c1 + c3 > 1. The assumption of adjacent (1B) vertices means that
v̄ = 1
(c̄+ ā) and w̄ = 1
(c̄+ ā′) for some ā, ā′ ∈ E ∩ C. Hence a = (c3 + c4,−1, 2− c3,−2− c4) and
a′ = (2 + c3 + c4,−1,−c3,−2− c4). One checks easily that ā
′ lies in region IV and ā lies in region
IV′. Moreover, the null conditions for these vectors yield
d3 + d4
d1 + d3 + d4
, c3 =
d1 + d4
d1 + d3 + d4
, c4 = −
d1 + d3 + 2d4
d1 + d3 + d4
Let e := 2u − a and e′ := 2x − a′. These lie respectively in regions VII′ and VII. We can now
apply Theorem 3.8 to ā and ā′ to obtain the following possibilities:
(i) ū, x̄ ∈ C and J(ā, ū) = 0 = J(ā′, x̄);
(ii) ū ∈ C, J(ā, ū) = 0, x̄ /∈ C, ē′ ∈ C is null;
(iii) x̄ ∈ C, J(ā′, x̄) = 0, ū /∈ C, ē ∈ C is null;
(iv) ū, x̄ /∈ C, ē, ē′ are both null.
We can eliminate (i)-(iii) by noting that the two equations in each case together with the values
of c3, c4 above imply that 1 =
. Using this relation (and the values of c3, c4) in the null
condition for c̄ then leads to a contradiction.
For case (iv) we can again apply Theorem 3.8 to the null vertices ē and ē′. The conditions
J(ē, z̄) = 0 and J(ē′, ȳ) = 0 lead, as above, to 1 = 1
and 1 = 1
respectively.
Using this in the null condition for c̄ again leads to a contradiction. Hence z̄, ȳ /∈ C and q̄ := 2z̄− ē
and q̄′ := 2ȳ− ē′ are null vectors in E ∩C. In fact we now find that q = q′, so caeqe′a′ is a hexagon
circumscribing (H3).
Let us consider the pair of null vertices c̄, q̄. We apply the argument in (A) of the proof of Theorem
6.4 to the wedge with vertex c̄ bounded by the rays c̄ā and c̄ā′. All elements of (C ∩ E) \ {ā, ā′, c̄}
lie below the line āā′. Let b̄ be a highest (with respect to x1 + x3) element among these. Since
ē ∈ C, b1+ b3 > −1 and so c̄+ b̄ cannot equal 2ū, 2t̄, 2x̄. Hence c̄+ b̄ = ā+ ā
′, and we compute that
b1 + b3 =
d1+d3−2d4
d1+d3+d4
. The analogous argument applied to the wedge bounded by the rays q̄ē and q̄ē′
gives a lowest element b̄′ of (C ∩E) \ {q̄, ē, ē′} satisfying b̄′ + q̄ = ē+ ē′ and b′1 + b
2d4−d1−d3
d1+d3+d4
avoid a contradiction, we must have d1 + d3 ≥ 2d4.
We can repeat the above argument with the null vertex pairs {ē, ā′} and {ē′, ā}, obtaining the
inequalities d3 + d4 ≥ 2d1 and d1 + d4 ≥ 2d3 respectively. The three inequalities then imply that
in fact d1 = d3 = d4 and c = (
,−1, 2
). Furthermore, C ∩ E = {ā, ā′, c̄, ē, ē′, t̄, q̄} and the null
condition for c̄ gives (d1, d2) = (3, 9) or (4, 3).
By looking at the terms in the superpotential equation corresponding to the vertices (all of
type II), we find that the coefficients Fc̄, Fē, Fē′ have the same sign, which is opposite to that of
Fā, Fā′ , Fq̄. Next we note that the only ways to write d+(
,−1, 1
) (resp. d+(−1
,−1, 2
as a sum of element of C are t̄+ c̄ = ā+ ā′ (resp. t̄+ ā = c̄+ ē). The superpotential equation then
28 A. DANCER AND M. WANG
gives FāFā′J(ā, ā
′)+Ft̄Fc̄J(t̄, c̄) = 0 and Fc̄FēJ(c̄, ē)+Ft̄FāJ(t̄, ā) = 0. Since J(ā, ā
′), J(c̄, ē), J(t̄, c̄)
and J(t̄, ā) are all positive, the above equations and facts imply that Fc̄ and Fā have the same sign,
a contradiction.
So c cannot lie in region II.
c lying in region V:
We have c3 < −1 < −c4 < 1 < c1. The adjacent (1B) vertices assumption implies that
w̄ = 1
(c̄+ā) and ȳ = 1
(c̄+ā′) for some ā, ā′ ∈ C∩E. It follows that a = (2+c3+c4,−1,−c3,−2−c4)
and a′ = (c3 + c4,−1,−2− c3, 2− c4). The null conditions on these vectors give
(2d1 + d4)(d3 + d4)
d4(d1 + d3 + d4)
, c3 = −
2d3 + d4
d1 + d3 + d4
, c4 =
d3 − d1
d1 + d3 + d4
Since a3 = −c3 > 1, a lies above the line uv. Also, a
1 = c3 + c4 = −c1 < −1, so a
′ lies below the
line uz. We can therefore apply Theorem 3.8 to ā and ā′ to get the following possibilities:
(i) ū ∈ C, J(ā, ū) = 0 = J(ā′, ū),
(ii) ū /∈ C, e := 2u− a, e′ := 2u− a′ lie in C ∩ E and are null.
If (i) occurs, then the two orthogonality conditions imply that d1 = d3, so c4 = 0, c1 = −c3 =
1 + d1
. Substituting these values of ci into J(ā
′, ū) = 0 gives 1 = 2
. But the null condition
for c̄ is 1 = 1
(1 + d1
)2 > 1
= 1, which is a contradiction.
Hence (ii) must occur. Note that if the above diagram is rotated so that the lines x1 + x3 = κ
(for arbitrary constants κ) are horizontal, then the lines x1 − x3 = κ would be vertical. u is the
only point in the hexagon lying on x1 − x3 = −2. Observe that a1 − a3 = a
1 − a
3 ≥ −2, otherwise
āā′ would not intersect conv(1
(d+W)), which contradicts Cor 3.4. If, however, a1−a3 > −2, then
ēē′ would not intersect conv(1
(d+W)). So in fact a = e′, e = a′ and u all lie on x1 − x3 = −2. In
other words, the hexagon is circumscribed by the triangle caa′ with intersections at w, u and y.
It follows easily from the above that c = (2,−1,−2, 0), d1 = d3 = d4, and the null condition for
c̄ is 1 = 8
. Also, we have C ∩E = {c̄, ā, ā′, t̄}. Since w, u, y are type II, by Lemma 3.2, we see
that the signs of Fā, Fc̄, and Fā′ in the superpotential equation cannot be chosen compatibly. We
have thus shown that the hexagon (H3) cannot occur.
The hexagon (H2) is not regular, but has reflection symmetry about uv and the perpendicular
bisector of yy′. It can be eliminated by similar arguments, but we now have to consider c lying in
regions III and IV as well. The hexagon (H1) can also be eliminated by the above methods. Here
the hexagon is invariant under the symmetric group permuting the coordinates x1, x2, x3. Together
with Cor 3.4, this fact reduces our consideration to those c lying in three of the regions formed by
extending the sides of the hexagon.
As mentioned in Remark 6.2, we also need to rule out subshapes of the hexagons. For (H2) and
(H3) the methods used above can also be applied to rule out all the sub-parallelograms and trapezia
except the rectangle yy′z′z of (H2) (see Lemma 8.5 and the discussion immediately before Ex 6.7).
All sub-triangles will be dealt with at the end of this section. (There is a triangle with midpoint
in (H2) but that can be dealt with by similar methods.) For (H2) this leaves the pentagon yy′vz′z
and the kite y′uz′v, both of which can still be eliminated using the above methods.
The possible subshapes of (H1) are rather numerous. However, if r ≥ 4 we will be able to
eliminate all of them in Lemma 8.6. Without this assumption, the above methods can be used to
eliminate those subshapes which do not contain all three type I vectors. Of course the following
discussion will handle the sub-triangles.
Lastly, we consider triangular faces.
Theorem 6.12. Suppose we have adjacent (1B) vertices in ∆c̄ corresponding to a triangular face
x̄x̄′x̄′′ of conv(1
(d +W)). Let E be the affine 2-plane determined by the triangular face. So there
are null vectors ā, ā′ in C ∩ E such that x = 1
(a+ c), x′ = 1
(a′ + c).
CLASSIFICATION OF SUPERPOTENTIALS 29
Suppose the vertices of the triangle are the only elements of W in the face. Then we are in one
of the following two situations:
(i) C ∩ E = {c̄, ā, ā′, x̄′′}, with c+ x′′ = a+ a′ and J(x̄′′, ā) = J(x̄′′, ā′) = 0;
(ii) C ∩ E = {c̄, ā, ā′} where 1
(a + a′) = x′′, one of x, x′, x′′ is type I, and the others are either
both type I or both type II/III.
Proof.
PPPPPPPPPPPPP
(i) ❝✟
❵❵❵❵❵❵❵❵
(A) We may introduce coordinates in E so that x̄x̄′ is vertical and to the right of c̄. As āā′ must
meet conv(1
(d + W)), we see x̄′′ is on or to the right of āā′. Let b̄ be any leftmost point of
(C ∩E) \ {c̄}. As in Theorem 6.4, we see that b̄+ c̄ ∈ d+W, so all elements of C ∩E except c̄, ā, ā′
are to the right of āā′.
(B) Considering āx̄′′ and ā′x̄′′ we see (using Theorem 3.8 and Cor 3.4) that either
(1) x̄′′ ∈ C and J(x̄′′, ā) = 0 = J(x̄′′, ā′), or
(2) x̄′′ /∈ C and x′′ = 1
(a+ a′).
In case (1), (x̄′′)⊥ ∩E is the line through āā′. By Prop. 3.3 and Cor. 3.4, observe that all elements
of (C∩E)\{x̄′′} are left of x̄′′. Let b̄ be a rightmost element of (C∩E)\{x̄′′}. So either J(b̄, x̄′′) = 0
or b̄+ x̄′′ ∈ d+W. Since b̄ is not to the left of āā′, the second alternative cannot hold and so b̄ must
lie on āā′. Combining this with our results in (A), we see C ∩ E is as in (i). Also, as J(ā, ā′) > 0
and ā+ ā′ /∈ d+W, we see a+ a′ must equal c+ x′′.
In case (2), by Cor 3.4 there are no elements of C ∩E right of āā′. Hence C ∩E is as in (ii). Now
J(b̄, ē) > 0 for all b̄ 6= ē in C ∩ E, so the last statement of (ii) follows.
Remark 6.13. We must also consider the case when some midpoints of the sides of our trian-
gular face lie in W. (This could happen if two vertices were (1,−1,−1, · · · ), (−1, 1,−1, · · · ) or
(1,−2, · · · ), (1, 0,−2, · · · ) or (1,−2, · · · ), (−1, 0, · · · ).) Let us denote the midpoints of xx′, xx′′ and
x′x′′ respectively by z, y, t.
If z is absent, the arguments of (A) in the proof of Theorem 6.12 still hold, so we have the
alternatives (1),(2) in (B). If (1) holds then, choosing b̄ as above, if b is right of aa′, we have
(b + x′′) ∈ W. This gives a contradiction since 1
(b + x′′) cannot be y or t as b 6= x, x′. Now
C ∩ E = {c̄, ā, ā′, x̄′′}, and as c + x′′ /∈ 2W it must equal a+ a′. It follows that the midpoints y, t
cannot arise. If instead (2) holds, then C ∩ E = {c̄, ā, ā′} and again no midpoints can be present.
Suppose now the midpoint z of xx′ is present. The argument of (A) shows that to account for
(ā+ ā′) ∈ C, and all elements of (C ∩E) \ {c̄, ā, ā′, (ā+ ā′)/2} are right of āā′. We still have the
alternatives (1) and (2), but (2) immediately gives a contradiction.
In (1) we see as before there are no elements of C ∩ E lying to the right of āā′, so C ∩ E =
{c̄, x̄′′, ā, ā′, (ā+ ā′)/2}. Note that J(ā, (ā+ ā′)/2) and J(ā′, (ā+ ā′)/2) > 0.
If c+x′′ = a+ a′, we find after some algebra that ā+(ā+ ā′)/2 6= 2ȳ and also cannot be written
as a different sum of elements of C, giving a contradiction.
If c̄ + x̄′′ 6= ā + ā′ then one sees that c̄ + x̄′′ /∈ d +W, and by relabelling x and x′, a and a′ we
may assume that c̄ + x̄′′ = ā + 1
(ā + ā′) and also ā + ā′ = 2ȳ and ā′ + 1
(ā + ā′) = 2t̄ = x̄′ + x̄′′.
These relations imply a = x′, a contradiction. So no triangle with any midpoints present can arise.
30 A. DANCER AND M. WANG
Remark 6.14. There are also triangular faces with two points of W in the interior of an edge.
This can only happen if two vertices are (−2, 1, 0, · · · ) and (1,−2, 0, · · · ) (up to permutation). The
other sides of the triangle now have no interior points in W unless the triangle is contained in the
hexagon (H1). We can again modify the proof of Theorem 6.12 to treat this situation.
If the interior points z, w lie on xx′, then (2ā + ā′)/3, (ā + 2ā′)/3 must be in C, and all points
of C ∩ E except for these two and c̄, ā, ā′ lie to the right of āā′. By Prop 3.3, alternative (1) must
now hold. The usual argument shows x̄′′ is the only element of C ∩ E on the right of āā′. Now
again J(ā, 1
(2ā′ + ā)) > 0, J(ā′, 1
(ā+ 2ā′)) > 0, and the sums a+ (2a′ + a)/3 and a′ + (a+ 2a′)/3
cannot give points in 2W. Since they also cannot both be cancelled by c+x′′ in the superpotential
equation, we have a contradiction.
The other possibility for two interior points is, after relabelling the vertices if necessary, when
z = (2x+ x′′)/3 and w = (2x′′ + x)/3. As usual all elements of C ∩ E except for c̄, ā, ā′ are on the
right of āā′. Alternative (1) must hold, or else we cannot account for z, w. The usual argument
shows either x̄′′ is the only element of C ∩ E right of āā′, or z ∈ C is the rightmost element of
(C ∩ E) \ {x̄′′} (so (z + x′′)/2 = w). In the former case we cannot get both z and w, as (c+ x′′)/2
can’t equal both z and w. In the latter, considering āz̄ shows J(ā, z̄) = 0. But as J(ā, x̄′′) = 0, this
means ā is orthogonal to x̄ and hence to c̄, a contradiction.
So no triangle with points of W in the interior of an edge can arise (except possibly for a
subtriangle of (H1)).
Nullity of c̄, ā, ā′ and the conditions in Theorem 6.12(i),(ii) again put severe constraints on x, x′, x′
and the dimensions. The possible triangles for case (i) are as follows, where (Tr11)-(Tr22) occur
only if K is not connected, and we have also listed the vectors c, a, a′ for future reference. Further
details of how the following listing is arrived at can be found in [DW5].
x′′ x x′
(Tr1) (−2, 1, 0, 0, 0) (0, 0,−2, 1, 0) (0, 0,−2, 0, 1)
(Tr2) (−2, 1, 0, 0) (0, 1,−2, 0) (0, 1,−1,−1)
(Tr3) (0, 0, 0,−2, 1) (−2, 1, 0, 0, 0) (0, 1,−2, 0, 0)
(Tr4) (−2, 1, 0, 0, 0, 0) (0, 0,−2, 1, 0, 0) (0, 0, 0, 1,−1,−1)
(Tr5) (−2, 1, 0, 0, 0) (0, 1,−1, 0,−1) (0, 1,−1,−1, 0)
(Tr6) (−2, 1, 0, 0, 0, 0) (0, 0, 1,−1,−1, 0) (0, 0, 1,−1, 0,−1)
(Tr7) (−2, 1, 0, 0, 0, 0) (0, 0, 1,−1,−1, 0) (0, 0,−1,−1, 0, 1)
(Tr8) (−2, 1, 0, 0, 0, 0) (0, 0,−1,−1, 1, 0) (0, 0,−1,−1, 0, 1)
(Tr9) (−2, 1, 0, 0, 0, 0, 0) (0, 0, 1,−1,−1, 0, 0) (0, 0, 1, 0, 0,−1,−1)
(Tr10) (−2, 1, 0, 0, 0, 0, 0) (0, 0,−1, 1,−1, 0, 0) (0, 0,−1, 0, 0, 1,−1)
(Tr11) (0, 0, 0, 1,−1,−1) (−2, 1, 0, 0, 0, 0) (−2, 0, 1, 0, 0, 0)
(Tr12) (0, 1, 0,−1,−1) (−2, 1, 0, 0, 0) (−1, 1,−1, 0, 0)
(Tr13) (0, 0, 0,−1,−1, 1) (−2, 1, 0, 0, 0, 0) (0, 1,−2, 0, 0, 0)
(Tr14) (0, 0, 0,−1,−1, 1) (0, 1,−1,−1, 0, 0, 0) (−2, 1, 0, 0, 0, 0, 0)
(Tr15) (0, 0, 0, 1,−1,−1, 0) (1,−1,−1, 0, 0, 0, 0) (1,−1, 0, 0, 0, 0,−1)
(Tr16) (0, 0, 0, 1,−1,−1, 0) (1,−1,−1, 0, 0, 0, 0) (0,−1, 1, 0, 0, 0, 1)
(Tr17) (0, 0, 0, 1,−1,−1, 0) (1,−1,−1, 0, 0, 0, 0) (0,−1,−1, 0, 0, 0, 1)
(Tr18) (0, 0, 0, 1,−1,−1, 0) (1,−1,−1, 0, 0, 0, 0, 0) (1, 0, 0, 0, 0, 0,−1,−1)
(Tr19) (0, 0, 0, 1,−1,−1, 0) (1,−1,−1, 0, 0, 0, 0, 0) (0,−1, 0, 0, 0, 0,−1, 1)
(Tr20) (−1,−1, 1, 0, 0, 0) (0, 0, 1,−1,−1, 0) (0, 0, 1,−1, 0,−1)
(Tr21) (−1, 1,−1, 0, 0, 0) (0, 0,−1, 1,−1, 0) (0, 0,−1, 1, 0,−1)
(Tr22) (−1, 1,−1, 0, 0, 0) (0, 0,−1,−1, 1, 0) (0, 0,−1,−1, 0, 1)
CLASSIFICATION OF SUPERPOTENTIALS 31
3c 3a 3a′
(Tr1) (2,−1,−8, 2, 2) (−2, 1,−4, 4,−2) (−2, 1,−4,−2, 4)
(Tr2) (2, 3,−6,−2) (−2, 3,−6, 2) (−2, 3, 0,−4)
(Tr3) (−4, 4,−4, 2,−1) (−8, 2, 4,−2, 1) (4, 2,−8,−2, 1)
(Tr4) (2,−1,−4, 4,−2,−2) (−2, 1,−8, 2, 2, 2) (−2, 1, 4, 2,−4,−4)
(Tr5) (2, 3,−4,−2,−2) (−2, 3,−2, 2,−4) (−2, 3,−2,−4, 2)
(Tr6) (2,−1, 4,−4,−2,−2) (−2, 1, 2,−2,−4, 2) (−2, 1, 2,−2, 2,−4)
(Tr7) (2,−1, 0,−4,−2, 2) (−2, 1, 6,−2,−4,−2) (−2, 1,−6,−2, 2, 4)
(Tr8) (2,−1,−4,−4, 2, 2) (−2, 1,−2,−2, 4,−2) (−2, 1,−2,−2,−2, 4)
(Tr9) (2,−1, 4,−2,−2,−2,−2) (−2, 1, 2,−4,−4, 2, 2) (−2, 1, 2, 2, 2,−4,−4)
(Tr10) (2,−1,−4, 2,−2, 2,−2) (−2, 1,−2, 4,−4,−2, 2) (−2, 1,−2,−2, 2, 4,−4)
(Tr11) (−8, 2, 2,−1, 1, 1) (−4, 4,−2, 1,−1,−1) (−4,−2, 4, 1,−1,−1)
(Tr12) (−6, 3,−2, 1, 1) (−6, 3, 2,−1,−1) (0, 3,−4,−1,−1)
(Tr13) (−4, 4,−4, 1, 1,−1) (−8, 2, 4,−1,−1, 1) (4, 2,−8,−1,−1, 1)
(Tr14) (−4, 4,−2,−2, 1, 1,−1) (4, 2,−4,−4,−1,−1, 1) (−8, 2, 2, 2,−1,−1, 1)
(Tr15) (4,−4,−2,−1, 1, 1,−2) (2,−2,−4, 1,−1,−1, 2) (2,−2, 2, 1,−1,−1,−4)
(Tr16) (2,−4, 0,−1, 1, 1,−2) (4,−2,−6, 1,−1,−1, 2) (−2,−2, 6, 1,−1,−1,−4)
(Tr17) (2,−4,−4,−1, 1, 1, 2) (4,−2,−2, 1,−1,−1,−2) (−2,−2,−2, 1,−1,−1, 4)
(Tr18) (4,−2,−2,−1, 1, 1,−2,−2) (2,−4,−4, 1,−1,−1, 2, 2) (2, 2, 2, 1,−1,−1,−4,−4)
(Tr19) (2,−4,−2,−1, 1, 1,−2, 2) (4,−2,−4, 1,−1,−1, 2,−2) (−2,−2, 2, 1,−1,−1,−4, 4)
(Tr20) (1, 1, 3,−4,−2,−2) (−1,−1, 3,−2,−4, 2) (−1,−1, 3,−2, 2,−4)
(Tr21) (1,−1,−3, 4,−2,−2) (−1, 1,−3, 2,−4, 2) (−1, 1,−3, 2, 2,−4)
(Tr22) (1,−1,−3,−4, 2, 2) (−1, 1,−3,−2, 4,−2) (−1, 1,−3,−2,−2, 4)
Remark 6.15. In making the above table, it is useful to observe from the nullity and orthogonality
conditions that x′′ cannot be type I, and that if x′′ is type III, say, (−2i, 1j), then xi = x
i iff xj = x
The possibilities for Theorem 6.12(ii) are as follows (up to permutation of x, x′, x′′ and the
corresponding permutation of c, a, a′):
x′′ x x′
(Tr23) (−1, 0, 0, 0, 0) (0,−2, 1, 0, 0) (0, 0, 0,−2, 1)
(Tr24) (−1, 0, 0, 0) (0, 1,−2, 0) (0,−1,−1, 1)
(Tr25) (−1, 0, 0, 0, 0, 0) (0, 1,−2, 0, 0, 0) (0, 0, 0, 1,−1,−1)
(Tr26) (−1, 0, 0, 0, 0) (0, 1,−1,−1, 0) (0,−1,−1, 0, 1)
(Tr27) (−1, 0, 0, 0, 0, 0, 0) (0, 1,−1,−1, 0, 0, 0) (0, 0, 0, 0, 1,−1,−1)
(Tr28) (−1, 0, 0) (0,−1, 0) (0, 0,−1)
c a a′
(Tr23) (1,−2, 1,−2, 1) (−1,−2, 1, 2,−1) (−1, 2,−1,−2, 1)
(Tr24) (1, 0,−3, 1) (−1, 2,−1,−1) (−1,−2, 1, 1)
(Tr25) (1, 1,−2, 1,−1,−1) (−1, 1,−2,−1, 1, 1) (−1,−1, 2, 1,−1,−1)
(Tr26) (1, 0,−2,−1, 1) (−1, 2, 0,−1,−1) (−1,−2, 0, 1, 1)
(Tr27) (1, 1,−1,−1, 1,−1,−1) (−1, 1,−1,−1,−1, 1, 1) (−1,−1, 1, 1, 1,−1,−1)
(Tr28) (1,−1,−1) (−1,−1, 1) (−1, 1,−1)
Remark 6.16. In drawing up the above listing, recall from Theorem 6.12 that one of the vectors,
without loss of generality x′′, is of type I. We write x′′ = (−1, 0, 0, · · · ). It now easily follows from
nullity and the relations between x, x′, x′′ and c, a, a′ that x1 = x
Also, observe that as x′′ is a vertex of W, no type II vector may have a nonzero entry in the
first position.
32 A. DANCER AND M. WANG
In contrast to the earlier listing of non-triangular faces, the above lists result from examining
all triangular faces, including ones which arise from other faces because certain vertices are absent
from W.
The restrictions on the dimensions of the corresponding summands are as follows:
(Tr1) (2, 1, 16, 4, 4, · · · )
(Tr2) (2, 3, 12, 4, · · · )
(Tr3) (16, 4, 16, 2, 1, · · · )
(Tr4) (2, 1, 16, 4, d5 , d6, · · · ),
(Tr5) (2, 3, 6, 6, 6, · · · )
(Tr6) (2, 1, d3, d4, 4, 4, · · · ),
(Tr7) (2, 1, 12, 3, 12, 12, · · · )
(Tr8) (2, 1, d3, d4, 4, 4, · · · ),
(Tr9) (2, 1, 4, d4 , d5, d6, d7, · · · ),
(Tr10) (2, 1, 4, d4 , d5, d6, d7, · · · ),
(Tr11) (16, 4, 4, 1, 1, 1, · · · )
(Tr12) (12, 3, 4, 1, 1, · · · )
(Tr13) (16, 4, 16, 1, 1, 1, · · · )
(Tr14) (16, 4, d3, d4, 1, 1, 1, · · · )
(Tr15) (d1, d2, 4, 1, 1, 1, 4, · · · ),
(Tr16) (12, 3, 12, 1, 1, 1, 12, · · · )
(Tr17) (4, d2, d3, 1, 1, 1, 4, · · · ),
(Tr18) (4, d2, d3, 1, 1, 1, d7 , d8, · · · ),
(Tr19) (d1, 4, d3, 1, 1, 1, d7 , d8, · · · ),
(Tr20-22) (1, 1, 3, 6, 6, 6, · · · ), (1, 2, 2, 8, 8, 8, · · · ), or (2, 1, 2, 8, 8, 8, · · · )
(Tr23) 1
(Tr24) d3 = 2d2 :
(Tr25) 1
(Tr26) d2 = d3 :
(Tr27)
(Tr28) 1
Note that (Tr28) is a subtriangle of (H1), (Tr2) is a subtriangle of a triangle with midpoints of
all sides in W, and (Tr12) is a subtriangle of a triangle with the midpoint of one side.
Let us now illustrate by an example how one arrives at the above tables.
Example 6.17. One possible triangle has vertices V1 = (0, 0, 0,−2, 1), V2 = (−2, 1, 0, 0, 0), V3 =
(0, 1,−2, 0, 0) with the midpoint V4 = (−1, 1,−1, 0, 0) of V2V3 in W. The triangle has a symmetry
given by interchanging the first and third entries. It therefore suffices to consider V1, V2, V4 as pos-
sibilities for x′′. Of course, by Remark 6.13 the full triangle cannot occur. The possible subtriangles
xx′x′′ are V2V3V1, V2V4V1, V4V1V2, V3V1V2, and V3V1V4. Now 3c = 2x+2x
′−x′′, 3a = 4x−2x′+x′′,
and 3a′ = −2x+ 4x′ + x′′ can be used to compute these vectors in each case. For V2V4V1 one gets
3c = (−6, 4,−2, 2,−1), 3a = (−6, 2, 2,−2, 1) and so c̄ and ā cannot be both null. Similarly, for the
last three possibilities, ā′ and c̄ cannot be both null. That leaves the first case, which gives (Tr3).
The condition J(ā, x̄′′) = 0 is 3 = 4
, which implies (d4, d5) = (2, 1). Putting this into the null
conditions for c̄, ā, ā′ gives the equations
CLASSIFICATION OF SUPERPOTENTIALS 33
The last two equations imply that d1 = d3 and the first two equations give d1 = 4d2. These in turn
give (d1, d2, d3) = (16, 4, 16), as in the tables above.
Putting all the results in this section together we obtain
Theorem 6.18. If we have two adjacent (1B) vertices, then the associated 2-face of W is given
by a triangle in the list (Tr1) − (Tr27), the square with midpoint (S), a proper subshape of the
hexagonal face (H1) containing all three type I vectors, or the sub-rectangle yy′zz′ of (H2).
We note for future reference the following properties of the c vector of the non-triangular faces
appearing in the above theorem: for (S), all nonzero entries have the same absolute value, and
there are only 3 (resp. 2) nonzero entries for the subfaces of (H1) (resp. (H2)).
7. More than one type (2) vertex
In this section we shall now show there is at most one type (2) vertex in ∆c̄, except in the
situation of Theorem 3.14 and one other possible case.
Suppose we have two type (2) vertices of V . Then we have elements v,w, v′, w′ of W with c, v, w
collinear and c, v′, w′ collinear. So we have four coplanar elements v,w, v′, w′ of W where vw and
v′w′ are edges. Moreover, the edges vw and v′w′ meet at c outside conv(W). Hence vwv′w′ do not
form a parallelogram or a triangle.
From our listing of polygons in §6 and considering their sub-polygons we see that the possibilities
for further analysis are the following:
• Trapezia (T1)-(T6): We must have c = 2v− s = 2u−w. Also, we note for future reference
that sw is always an edge of conv(W) in (T3) and (T5), regardless of whether or not the
whole trapezium is a face, since sw can be cut out by {x2 = 1, x1 + x3 = −2} (cf 1.2(e)).
• Hexagons (H1)-(H3)
• Rectangle with midpoints (P17): While the rectangle itself cannot occur, we need to con-
sider the trapezia obtained by omitting one vertex, so that the edges are a side of the rec-
tangle and the segment joining the remaining vertex to the opposite midpoint. As above,
note that the longer of the two parallel sides of the trapezium is always an edge of conv(W).
• Parallelogram with midpoints (P16): This case is similar to (P17). The sub-polygons
to consider are the trapezia obtained by omitting one vertex of the parallelogram. By
symmetry, we are reduced to omitting either u or w. But since s occurs, w cannot be
omitted.
• Triangle with midpoints of all sides: We need to consider the trapezia obtained by omitting
a vertex. By symmetry all three trapezia are equivalent. This triangle is always a face of
conv(W) as it is cut out by {x2 = 1, x1 + x3 + x4 = −2} (cf 1.2(e)).
• Trapezia (T*1),(T*2), (T*3): By Rmk 6.3 these cannot be faces of conv(W), so cannot
come from adjacent type (2) vertices of V . For (T*1), besides the full trapezium, we need
to consider the two trapezia obtained by omitting either s or w. By symmetry these are
equivalent.
For (T1),(T2),(T4),(T6),(T*2),(T*3) we must have c = 2v − s = 2u − w, and so Lemma 3.13
applied to vs gives a contradiction. The same argument works for (P16), as up to permutations,
c = 2v − s = 2y − w. For (T1*), since Theorem 3.11 rules out c = (3v − s)/2 = (3u − w)/2
(corresponding to the full trapezium), the only other possible c is 2v − s = 2u − r, and again
Lemma 3.13 rules this out.
For (T3),(T5) and the trapezium coming from the triangle with midpoints, we need more infor-
mation from the superpotential equation. Since J(c̄, s̄), J(c̄, w̄) > 0, while Av, Au < 0, Fs̄, Fw̄ must
have the same sign, which must be opposite to that of Fc̄. Since stw is always an edge by earlier
remarks, the nullity of c̄ implies J(s̄, w̄) > 0, contradicting Prop 3.7(ii).
34 A. DANCER AND M. WANG
Essentially the same argument works for (P17), as up to permutations c = 2s − v = 2z − u =
(−2,−1, 2, 0).
For (H3) most quadruples cannot give pairs of edges. For we observe that u (resp. v,w) is present
iff x (resp. y, z) is. Thus, if u is missing, so is x, and v,w, y, z must all be present (otherwise we do
not have a 2-dimensional polygon). But we now get a rectangle, which is not admissible. Hence all
vertices are present and by symmetry we may assume that one of our edges is uz or uv. From this
we quickly find that the two possible c (up to permutations) are (−1,−1,−1, 2) = 2y − x = 2z − u
and (1,−1, 1,−2) = 2v − u = 2w − x. Both cases are ruled out by Lemma 3.13.
For (H2), observe that y (resp. y′) is present iff z (resp. z′) is. As these four vectors cannot all be
absent (otherwise we do not have a 2-dim polygon), by the symmetries of (H2), we can assume y′ is
present. If v is present, then all possibilities are eliminated by Theorem 3.11. (Note that although
2α − y′ = 2z − z′ it is impossible for αy′ and zz′ to both be edges.) On the other hand, if v is
absent, then y′z′ is an edge. Since the polygon cannot be a parallelogram or a triangle, it follows
that u is present and the polygon is a pentagon. In this case, the only possibility compatible with
Theorem 3.11 is c = (0,−1, 2,−2) = 2y−u = 3
y′− 1
z′. (This is not a priori ruled out by Theorem
3.11 as y′z′ has an interior point β).
To discuss (H1), we write
u = (−2, 1, 0), p = (0, 1,−2), v = (1, 0,−2), w = (0,−2, 1), s = (1,−2, 0), q = (−2, 0, 1)
for the vertices,
x = (−1, 1,−1), y = (−1,−1, 1), z = (1,−1,−1),
for midpoints of the longer sides, and
α = (−1, 0, 0), β = (0, 0,−1), γ = (0,−1, 0)
for the interior points, with the understanding that the rest of the components of the above vectors
are zero.
As before we consider pairs of vectors which can form edges of an admissible polygon. We then
compute the possibilities for c and apply Theorem 3.11. This will eliminate most possibilities. (For
many quadruples of points we can see, as in (H3), that they cannot all be vertices.) So up to
permutations, the remaining possibilities are as follows.
If no type II is present:
(1) c = 2α− u = 2β − p = (0,−1, 0, · · · )
(2) c = 2u− v = 2q − s = (−5, 2, 2, · · · )
(3) c = 2u− p = 2q − γ = (−4, 1, 2, · · · )
(4) c = 2q − u = 2α− p = (−2,−1, 2, · · · )
If all type II are present:
(5) c = 2u− x = 2q − y = (−3, 1, 1, · · · )
(6) c = 2u− y = 2x− z = (−3, 3,−1, · · · )
(7) c = (3y − z)/2 = 2q − u = (−2,−1, 2, · · · )
(8) c = (3p − u)/2 = (3v − s)/2 = (1, 1,−3, · · · )
Again, we cannot immediately rule out (7) and (8) using 3.11 because of the presence of interior
points. However for (8) we easily see using the arguments of 3.11 that the elements of C on the line
through v, s are c̄, c̄1 = (v̄+ s̄)/2 = z̄ and c̄2 = (3s̄− v̄)/2. Now as s, v are type III we need Fc̄ and
Fc̄2 to have the same sign, which is the opposite sign to Fc̄1 . But the superpotential equation now
gives a contradiction to the fact that Az < 0.
In (2)-(7) Lemma 3.13 applied respectively to uv, up, qu, ux, uy, qu gives a contradiction. Note
that case (4) only occurs when K is not connected, as the vectors β, γ are absent (cf 1.2(b)). We
are left with (1), which is precisely the situation of Theorem 3.14. We have therefore proved
Theorem 7.1. Apart from the situation of Theorem 3.14, the only other possible case where we
can have more than one type (2) vertex is, up to permutation of summands, when two type (2)
CLASSIFICATION OF SUPERPOTENTIALS 35
vertices are adjacent and the 2-plane determined by them and c̄ intersects conv(1
(d +W)) in the
pentagon with vertices uyy′zz′ contained in the hexagon (H2).
We will be able to rule this case out in §8.
8. Adjacent (1B) vertices revisited
We now return to our classification of when adjacent (1B) vertices can occur.
The idea is as follows: each of the configurations of §6 involves, as well as the null vector c̄, two
new null vectors ā, ā′. Hence the arguments of earlier sections also apply to ā, ā′. That is, we may
consider the associated polytopes ∆ā and ∆ā
The following lemma is useful when applied to ∆c̄,∆ā and ∆ā
Lemma 8.1. Suppose we have a (1B) vertex with exactly k adjacent (1B) vertices. Then
r ≤ #((1A) vertices) + #((2) vertices) + k + 2.
Suppose we have a (1B) vertex with no adjacent (1B) vertices. Then
r ≤ #((1A) vertices) + #((2) vertices) + 2.
If there are no (1B) vertices then
r ≤ #((1A) vertices) + #((2) vertices) + 1.
Proof. By our assumption that dim conv(W) = r− 1, it follows that ∆c̄ is a polytope of dimension
r − 2. Any vertex in it has at least r − 2 adjacent vertices. So for a (1B) vertex, the first two
statements follow immediately. If there are no (1B) vertices the third inequality follows because
∆c̄ has at least r − 1 vertices.
Lemma 8.2. Configurations (Tr1) - (Tr22) cannot arise from adjacent (1B) vertices.
Proof. The strategy is to count the number of type (1A), (2) and adjacent (1B) vertices in ∆ā or
∆c̄ and apply Lemma 8.1 to get a contradiction.
(i) We first observe that for these configurations c and a have at least four nonzero entries (at
least five except for (Tr2)), so they cannot be collinear with an edge vw with points of W in the
interior of vw (see Table 3 in §3). So if ∆c̄ or ∆ā has a type (2) vertex, by Theorem 3.11, c or a
must equal 2v − w or (4v − w)/3. It is easy to check that this is impossible except for c in (Tr3),
using the forms of c in Tables 1, 2 in §3.
(ii) Next we consider (1A) vertices. For (Tr1)-(Tr10) we have |ai/di| ≤
for all i. For Tr(1),
Tr(6),Tr(8) there are three i where equality holds. In these cases one of the associated di equals
1. Moreover for (Tr1) and (Tr8) two of these ai/di equal 1/3 and the third is −1/3, wheras for
(Tr6) it is the other way round. For (Tr2)-(Tr5), (Tr7) and (Tr9)-(Tr10) there are only two i where
equality holds. Further, for (Tr2) and (Tr5) |ci/di| ≤
for all i, with equality for just two i, and
here ci/di =
. It follows that for (Tr1), (Tr8) there are at most two (1A) vertices in ∆ā, while
for (Tr2)-(Tr5), (Tr7) and (Tr9)-(Tr10) there is at most one. In the case of (Tr6) there are at most
three (1A) vertices in general but at most two if K is connected. For (Tr2) and (Tr5) there are no
(1A) vertices in ∆c̄.
By means of similar considerations, we find that there is one (1A) vertex (corresponding to x̄′′)
in ∆ā for (Tr12), (Tr13),(Tr14) (Tr16), (Tr18), (Tr19) and at most two (1A) vertices for (Tr11),
(Tr17), (Tr20) and the d = (1, 1, 3, 6, 6, 6, · · · ) case of (Tr21), (Tr22). For (Tr15) there are at most
four (1A) vertices in ∆ā, and for the remaining cases of (Tr21) and (Tr22) there are at most r − 4
(1A) vertices (r − 6 of those correspond to (−23, 1j) where j > 6).
(iii) Finally, consider the (1B) vertex ξ̄ in ∆ā corresponding to x̄ in each of the triangles. In order
for there to be an adjacent (1B) vertex, ā must be (up to permutation) of the form of the null vector
c̄ in the 2-faces in Theorem 6.18. Now observe that for examples (Tr1), (Tr3), (Tr4), (Tr6)- (Tr20)
36 A. DANCER AND M. WANG
the null vector a does not appear in the list of possible c. Hence ξ̄ has no adjacent (1B) vertices.
From above, type (2) vertices cannot occur, so combining the bounds for (1A) vertices in ∆ā with
Lemma 8.1 gives an upper bound for r less than the minimum required by each configuration, a
contradiction.
(iv) Let us now consider (Tr21) and (Tr22). The vector a of (Tr21) has the same form as c in
(Tr22) and vice versa. An adjacent 2-face containing aξc of (Tr21) can only be a triangle of type
(Tr22) containing 1
(1,−1,−3,−2,−2, 4, 0, · · · ). Thus ξ has at most one adjacent (1B) vertex, and
we get the bound r− 2 ≤ 2+1+0 in the d = (1, 1, 3, 6, 6, 6, · · · ) subcase and r− 2 ≤ (r− 4)+1+0
in the other two subcases, a contradiction.
(v) For the remaining two triangles (Tr5) and (Tr2) we consider ∆c̄ instead.
For (Tr5), observe that c determines the plane xx′x′′ and does not occur as a possible null vector
for any other configurations. So we have at most one adjacent pair of (1B) vertices in ∆c̄. From
above, there are no type (1A) or (2) vertices. But r ≥ 5, giving a contradiction.
For (Tr2), consider the vertex ξ̄′ of ∆c̄ corresponding to x′. If there is a (1B) vertex adjacent
to it, we have a 2-dim face including c, x′. By Theorem 6.18 the only one is the face including x,
so there is at most one (1B) vertex of ∆c̄ adjacent to ξ̄′. Also, type (1A) and (2) vertices cannot
occur, so r ≤ 3, a contradiction.
Lemma 8.3. Configurations (Tr23)-(Tr27) cannot arise from adjacent (1B) vertices.
Proof. Note first that all entries of c, a, a′ are integers, so Lemma 4.4 shows in each case there is at
most one (1A) vertex, and for (Tr23),(Tr24) one checks that there are no (1A) vertices in ∆c̄. Note
also that for all these configurations, as x′′ is a vertex, there are no type II vectors with nonzero
entry in place 1.
Observe as in Lemma 8.2 that there are no type (2) vertices in ∆c̄,∆ā or ∆ā
. (For (Tr24), we
need to rule out the possibility that c has the form (4) in Table 3 in §3 with λ = 3/2. This follows
since the interior point in that case would be a type II vector with nonzero entry in place 1.)
So in all cases if we have a (1B) vertex with exactly k (1B) vertices adjacent to it, then by
Lemma 8.1 we have r ≤ k + 3. For (Tr23), (Tr24) (using ∆c̄), we have r ≤ k + 2. We will work
with ∆c̄ below.
First consider (Tr23) and look for (1B) vertices adjacent to ξ̄ where ξ̄ corresponds to the vertex
x. We need a 2-face including c, x. By Theorem 6.18 such a face must be of type (Tr23), and
having fixed c and a, the only freedom lies in assigning 1 in the third null vector to the first or fifth
place. So k ≤ 2, which gives r ≤ 4, a contradiction.
Similarly, for (Tr24), since a type (H1) face cannot contain c and x, we need only consider faces
of type (Tr24), for which there are again two possibilities. However, as mentioned above, in one of
these possibilities the vector “x′” has a 1 in place 1 and hence cannot occur. So there is at most
one (1B) vertex adjacent to ξ, and we deduce r ≤ 3, a contradiction.
For (Tr25),(Tr27) we similarly deduce that the only 2-face containing x, c is itself because as
above we cannot have any type II vectors with nonzero entry in place 1. So r ≤ 4, a contradiction.
Finally, for (Tr26) the above argument still works since (Tr24) has been ruled out (the vectors
a, a′ of (Tr24) are of the same form as c, a of (Tr26), so a priori (Tr24) could be an adjacent
2-face).
Lemma 8.4. Configuration (S) (square with midpoint) cannot arise from adjacent (1B) vertices.
Proof. We refer to §6 for the expressions for the vertices vusw of the square. The null vertex c̄
corresponds to (1,−1,−1, 1,−1, 0, · · · ) and the 2-dimensional face is cut out by x2 = −1, x1+x3 =
0 = x4 + x5, and xk = 0, for k > 5. Lemma 4.4(b) shows that there is at most one (1A) vertex
in ∆c̄. As r ≥ 5 and all the nonzero entries of c have the same absolute value, it follows that
there are no type (2) vertices. Let ξ̄ denote the vertex of ∆c̄ so that ξ is collinear with u and
a = 2u − c = (−1,−1, 1, 1,−1, 0, · · · ). A (1B) vertex adjacent to ξ̄ gives a 2-dimensional face
CLASSIFICATION OF SUPERPOTENTIALS 37
including c̄, ū. By what we have analysed so far about 2-faces given by adjacent (1B) vertices, this
face must again be a face of type (S), and the only possibilities are itself or the face obtained from
this by swapping indices 2 and 5. Hence there are at most two (1B) vertices adjacent to ξ̄, and at
vertex ξ̄, we have 3 ≤ r − 2 ≤ 1 + 2. Thus r = 5 is the remaining possibility, in which case ξ̄ has
exactly two adjacent (1B) vertices and one adjacent (1A) vertex.
Let us denote by ξ̄′ the (1B) vertex such that ξ′ is collinear with w and a′ := 2w − c =
(1,−1,−1,−1, 1). Let η̄ denote the other (1B) vertex adjacent to ξ̄. Then the 2-face deter-
mined by c, ξ, η is cut out by x5 = −1, x1 + x3 = 0 = x2 + x4. The ray cη intersects conv(W)
at z = (1, 0,−1, 0,−1) and b := 2z − c = (1, 1,−1,−1,−1) corresponds to a null vertex. Similarly,
there is a (1B) vertex η̄′ (besides ξ̄) adjacent to ξ̄′, and the corresponding 2-face (also of type (S)) is
cut out by x3 = −1, x1+x2 = 0 = x4+x5. The ray cη
′ intersects conv(W) at z′ = (0, 0,−1, 1,−1).
The vector b′ := 2z′ − c = (−1, 1,−1, 1,−1) corresponds to a null vertex.
Let us examine the (1A) vertex in ∆c̄ more closely. Let y ∈ W such that J(ȳ, c̄) = 0. As r = 5,
the null condition for c̄ implies that di ≥ 2 with at most one equal to 2. Also, for some j ∈ {2, 3, 5}
(i.e., j is an index for which the corresponding entry of c is −1) we must have yj = −2, so y is
type III. Let i be the index such that yi = 1. Then i ∈ {1, 4} (i.e., i is an index for which the
corresponding entry of c is 1), and the orthogonality condition implies (di, dj) = (2, 4) or (3, 3).
There are thus six possibilities for y, but only one can actually occur.
With the possible exception of the existence of the (1A) vertex, the above arguments apply
equally to the projected polytopes ∆ā, ∆b̄, ∆ā
, and ∆b̄
as the entries of a, b, a′ and b′ are just
permutations of those of c. We claim that whichever possibility for y occurs in ∆c̄, there is another
projected polytope with no (1A) vertex. Applying the above arguments to this polytope would
result in the contradiction r − 2 ≤ 2 and complete our proof.
We can use ∆ā for the contradiction if d1 = 2 or if any of (d1, d2), (d3, d4), (d1, d5) = (3, 3). If
d4 = 2 or if (d2, d4) = (3, 3) we can use ∆
ā′ instead. Finally, if (d1, d3) = (3, 3) we can use ∆
b̄′ and
if (d4, d5) = (3, 3) we can use ∆
For example, when (d1, d3) = (3, 3) (so y = (1
1,−23)), the null condition for c̄ implies that d2, d4
in particular cannot equal 2 or 3. In order to have a (1A) vertex in ∆b̄
, we must have a type III
vector (1i,−2j) with i ∈ {2, 4}. But this requires one of d2, d4 to be 2 or 3.
When (d4, d5) = (3, 3), then d1, d2 cannot be 2 or 3. But in ∆
b̄ a (1A) vertex corresponds to
(1i,−2j) with i ∈ {1, 2}, which implies that one of d1, d2 is 2 or 3.
The remaining cases are handled similarly.
Lemma 8.5. The subrectangle yy′zz′ of (H2) cannot arise from adjacent (1B) vertices.
Proof. Recall c = (−2, 1, 0, · · · ), so by Lemma 4.4 there are no (1A) vertices of ∆c̄. Moreover,
using Tables 1-3 in §3, one may check that there are no type (2) vertices either. (Note that type
II vectors other than y, z with a nonzero entry in place 1 cannot occur as then the subrectangle
cannot be a face. Similarly the line through c, α, β and (1,−2, 0, · · · ) will not give a type (2) vertex
as this line cannot be an edge.)
Let η̄ denote the vertex of ∆c̄ collinear with c and y. Any (1B) vertex adjacent to η̄ will give rise
to a face containing c and y, which cannot be of type (H1), and must therefore be of type (H2),
since we have eliminated all other possibilities. In fact, it must be the face we started with.
So there is just one (1B) vertex adjacent to η̄, and from above there are no (1A) or (2) vertices.
As r ≥ 4 for (H2), this contradicts Lemma 8.1.
Lemma 8.6. If r ≥ 4, configuration (H1) or subshapes cannot arise from adjacent (1B) vertices.
Proof. We first note some special properties of W. Since (H1) is a face, there can be no type II
vectors in W with nonzero entry in a place ∈ {1, 2, 3} and in a place /∈ {1, 2, 3}. Also, if (−2i, 1k)
with i ∈ {1, 2, 3}, k /∈ {1, 2, 3}, then (−1k) must be absent, which has strong implications, as noted
in Remark 1.2(b).
38 A. DANCER AND M. WANG
Let ξ̄ be a (1B) vertex in the plane. We have at most one (1B) vertex adjacent to ξ̄, as the
associated face must again be of type (H1) and is now determined by c̄, ξ̄. It also readily follows
that c cannot be collinear with an edge of W not in the face (assuming as usual we are not in the
situation of Theorem 3.14).
Now the special properties of W in the first paragraph imply that (1A) vertices in ∆c̄ can
correspond only to type III vectors in W which overlap with c. A straight-forward check using the
null condition for c̄ shows that the possible type IIIs have form (−2i, 1k) with i ∈ {1, 2, 3}, k /∈
{1, 2, 3} and ci/di = −1/2. It follows that di = 2 or 3 and hence, by nullity, the index i is unique.
So there are at most r−3 (1A) vertices. By Lemma 8.1, all r−3 (1A) vertices must occur. Applying
Cor 4.3 we conclude that r ≤ 4 (as di 6= 4 and r > 4 forces i ∈ Ŝ≥2).
We will now improve this estimate to r ≤ 3. Let the vertices of (H1) be as in §7. If r = 4 then
a (1A) vertex does exist and we can take it to come from t = (−2, 0, 0, 1) with (−14) absent. It
follows that besides t the only other possible members of W lying outside the 2-plane containing
(H1) are (0,−2, 0, 1) and (0, 0,−2, 1). As noted just before Theorem 6.12 we may assume the type
I vectors α, β, γ are all present. (If K is connected, d4 = 1 and so this last fact follows without
having first to eliminate those subshapes not containing one of the type I vectors.)
As noted above, d1 = 2 or 3, and c1 = −1 or −
respectively.
First consider c1 = −1, so d1 = 2. Now c = (−1, c2,−c2, 0), and by swapping the 2, 3 coordinates
if necessary, we may take c2 > 0. Observe u, q are absent, as if u is present or if u is absent but q
is present, then u (resp. q) gives a (1B) vertex, which contradicts nullity as the associated a would
have a1 = −3. Now the type II vectors x, y, z are absent, as if one is present they all are, and we
have a type (2) vertex. We deduce α gives a (1B) vertex so a = (−1,−c2, c2, 0). The other (1B)
vertex cannot correspond to w since β, γ are present. It also cannot be given by v, s as this violates
nullity, so must correspond to p or β.
If it is p, we have a′ = (1, 2− c2, c2−4, 0). Now Remark 3.9 implies c2 = (d3+4d2)/(d3 +2d2) so
1 < c2 < 2. But now no entry of a
′ equals −1 or −3
. We can now check that there are no (1A) or
(2) vertices with respect to a′, so there is at most one vertex of ∆ā
adjacent to p, a contradiction.
If it is β then p, v must be absent. Now a′ = (1,−c2, c2 − 2) and Remark 3.9 implies c2 = 1.
Hence c = (−1, 1,−1), a = (−1,−1, 1), a′ = (1,−1,−1), and nullity implies 1
. It is easy
to check by considering the vertices of ∆ā
that w, s must also be absent, so W just contains the
three type I vectors, t and possibly one or both of (0,−2, 0, 1), (0, 0,−2, 1). But we can check that,
if present, these three latter vectors give respectively vertices with respect to a, a′ which cannot
satisfy any of the conditions (1A), (1B) or (2). So in fact we have r = 3.
Similar arguments rule out the case c1 = −
Lemmas 8.2-8.6 give the following improvement of Theorem 6.18.
Theorem 8.7. It is impossible to have adjacent (1B) vertices except possibly when r = 3, in
which case conv(1
(d+W)) is a proper subface of (H1) containing all three type I vectors (e.g., the
tri-warped example (Tr28)).
We are now in a position to strengthen Theorem 7.1 by eliminating the remaining case of the
pentagon.
Theorem 8.8. Let c̄ be a null vertex of conv(C) such that ∆c̄ contains more than one type (2)
vertex. Then we are in the situation of Theorem 3.14.
Proof. We just have to eliminate the case of the pentagon uyy′zz′ in Theorem 7.1. Recall r ≥ 4 for
this configuration, and c is (0,−1, 2,−2, · · · ).
Using the nullity of c̄ we check that the only elements of W which can give an element of c̄⊥ are
(−22, 1i) where i > 4 and we have d2 = 2. Note that (1
1,−22) cannot be present as then y′z′ is
not an edge. By Cor 4.3, at most one such vector can arise. So there is at most one (1A) vertex,
which occurs only if r ≥ 5.
CLASSIFICATION OF SUPERPOTENTIALS 39
If we can show there are no (1B) vertices, then we are done because if we look at the adjacent
vertices of the type (2) vertex associated to y (in the pentagon), besides one (1A) possibility, the
other possibility is the type (2) vertex associated to y′ (by Theorem 7.1). As there must be at least
r− 2 adjacent vertices, we deduce r− 2 ≤ 2, so r = 4. But now, from above there is no (1A) vertex
so in fact we get r − 2 ≤ 1, a contradiction.
We now use Remark 3.9 to make a list of the possible x ∈ W associated to (1B) vertices
of ∆c̄. These are (0, 1,−1,−1, 0, · · · ), (0,−2, 1, 0, · · · ), (13,−1i,−1j), (−14, 1i,−1j), (13,−14,−1i),
(−13,−14, 1i) and (13,−2i) where i, j 6= 2, 3, 4. Note that type II vectors with nonzero entries in
places 2, k,m cannot occur except for y′, z′ as then y′z′ is not an edge. For each x in this list, we
consider the projected polytope ∆ā, where a = 2x − c. By looking at the form of a, we see from
Theorem 7.1 that there is at most one type (2) vertex in ∆ā. Also, the nonzero components of a
are either ≥ 1 or ≤ −2. By Lemma 4.4(a), there are no vertices of type (1A) in ∆ā. Since r ≥ 4,
by Theorem 8.7, the type (1B) vertex in ∆ā corresponding to x̄ has no adjacent (1B) vertices. So
we have a contradiction to Theorem 8.1.
The above result together with Theorem 8.7 and Lemma 8.1 gives us lower bounds on the number
of (1A) vertices.
Theorem 8.9. Let c̄ be a null vertex of conv(C) and ∆c̄ be the corresponding projected polytope.
Suppose further that c is not type I, i.e., we are not in the case of Theorem 3.14.
(i) If there are no (1B) vertices in ∆c̄, then there are at least r − 2 type (1A) vertices.
(ii) If either there is a type (2) vertex or r ≥ 4, then there are at least r − 3 type (1A) vertices
in ∆c̄. Hence there are at least r − 3 elements of 1
(d+W) orthogonal to c̄.
9. Type (2) vertices
In this section we consider again type (2) vertices of ∆c̄. In view of Theorem 8.8, it remains to
deal with the case of a unique type (2) vertex in ∆c̄. By Theorem 8.7 there are no adjacent (1B)
vertices in this situation.
Let c be collinear with an edge vw of conv(W). We first consider the situation where there are
no interior points of vw lying in W. By Theorem 3.11, we have the two possibilities c = 2v − w
and c = (4v − w)/3. Moreover, a preliminary listing of the cases appears in Tables 1 and 2 of §3.
Case (i): c = 2v − w
We have to analyse cases (1)-(7) in Table 1 of §3. The idea is to determine the number of (1A)
and (1B) vertices using respectively Lemma 4.4 and Remark 3.9, and then get a contradiction
(sometimes using Theorem 8.9). Note that J(w̄, w̄) < 0 for (1)-(3).
In (1), (2) and (4)-(7), Lemma 4.4 shows that there are no elements of 1
(d+W) orthogonal to
c̄ (recall c /∈ W), so Theorem 8.9 shows that r ≤ 3. This already gives a contradiction in case (7).
(Note that when r = 3 and w is type I, since w is a vertex there are no type II vectors in W.)
In (1) the only x ∈ W that could satisfy Eq.(3.1) and give a (1B) vertex with respect to c̄ is
(1,−1,−1). But the associated a = 2x− c is (2,−3, 0) and it easily follows that ā, c̄ cannot both be
null. For (2), the possible x ∈ W which correspond to (1B) vertices are (1,−2, 0) and (1,−1,−1)
respectively. In each case we find the nullity of c̄ and Remark 3.9 imply J(w̄, w̄) > 0, a contradiction
to F 2w̄ J(w̄, w̄) = Aw < 0 (as w is type III).
In (5) with r = 3, one checks that the only possible x ∈ W corresponding to a (1B) vertex is
(0, 1,−2). Let us consider the distribution of points of W in the plane x1 + x2 + x3 = −1. The
point (0, 1,−2), if present, would lie on one side of the line vw while the point (−1, 0, 0) lies on
the other side. Now (−1, 0, 0) must lie in W as otherwise v cannot be present by Remark 1.2(b).
So since vw is an edge by assumption, (0, 1,−2) cannot lie in W, which gives a contradiction to
Theorem 8.9(i).
Hence in (1),(2),(5) Theorem 8.9 shows r ≤ 2, which is a contradiction.
40 A. DANCER AND M. WANG
In case (4) the nullity of c̄ translates into 1 = 9/d1 + 4/d2. Hence d2 6= 1, so if K is connected
(0,−1, 0, · · · ) is present and w is not a vertex, which is a contradiction. If r = 3 and K is not
connected, by Remark 1.2(b), (1,−2, 0) and (0,−2, 1) must be absent, and, from Remark 3.9, the
possibilities for x ∈ W associated to the (1B) vertex are x = (−2, 0, 1) and y = (0, 1,−2). In
the first case, conv(W) is the triangle with vertices v,w, x and a = 2x − c = (−1,−2, 2). Now
J(ā, w̄) > 0, contradicting the superpotential equation. In the second case, a = (3, 0,−4) with
J(ā, ȳ) > 0, and aw intersects conv(W) in an edge. By Theorem 3.11, t = (1, 0,−2) ∈ W and
conv(W) is a parallelogram with vertices v, y, w, t. Moreover, Remark 3.12 implies that a and w
are the only elements of C in aw. But then the midpoint (0, 0,−1) of wt is unaccounted for in the
superpotential equation.
For (6) with r = 3, there should be at least two vertices in ∆c̄. But we find there are no (1B)
vertices, a contradiction. So r = 2, and we are in the situation of the double warped product
Example 8.2 of [DW4].
In case (3), Lemma 4.4 shows 1
(d + W) ∩ c̄⊥ has at most one element. Hence ∆c̄, which has
dimension ≥ 2 since r ≥ 4, must contain at least one (1B) vertex. By Theorem 8.7, such a (1B)
vertex has at most 2 adjacent vertices. It follows that r = 4 and (1,−2, 0, 0) corresponds to the
(1A) vertex; also d2 = 2. Also, since (−1, 0, 0, 0) ∈ W, (0,−1, 0, 0) cannot be a vertex of conv(W).
But now routine computations using Eq.(3.1) show there are no (1B) vertices, a contradiction.
So the only possible case if K is connected is that giving Example 8.2 of [DW4]. If K is
disconnected there is the further possibility of (4) with r = 2, i.e., W = {(−2, 1), (−1, 0)}. This
is discussed in the third paragraph of Example 8.3 of [DW4]. An example in the inhomogeneous
setting is treated there and in [DW2]. An example where the hypersurface is a homogeneous space
G/K is discussed in the concluding remarks at the end of section §10.
Case (ii): c = (4v − w)/3
For clarity of exposition let us assume K is connected, using the assumption as indicated in
Remark 5.9. We examine the cases (1)-(11) in Table 2 of §3.
Some of these cases can be immediately eliminated. In (3), Eq.(3.2) implies (d1, d2) = (3, 3) or
(4, 1). In neither case is c̄ null. In (11) Eq.(3.2) and J(v̄, w̄) > 0 imply (d1, d2, d3) = (2, 4, 4), (2, 5, 2)
or (3, 3, 3), and again c̄ is not null. In (4) and (6) Eq.(3.2) implies (d1, d2) = (2, 1) and (d1, d2) =
(3, 9) or (4, 3) respectively. In neither case does the nullity condition have an integral solution in
Further cases can be eliminated by finding the possible (1A) vertices (using Lemmas 3.10 and
4.4) for the given value of c and using Theorem 8.9. In particular, we get a contradiction whenever
r ≥ 4 and there are no (1A) vertices.
In (1), Eq.(3.2) implies (d1, d2) = (2, 1), and nullity of c̄ implies
= 3. But we now find
that 1
(d+W) ∩ c̄⊥ is empty, giving a contradiction as r ≥ 4.
In (5) Eq.(3.2) and nullity imply (d1, d2) = (3, 3) and {d3, d4} = {3, 8}. One can now check that
(d+W) ∩ c̄⊥ is empty, which is a contradiction as r ≥ 4.
In (7), Eq.(3.2) implies (d1, d2) = (2, 1) and nullity implies
. One can now check
that the only possible elements ū orthogonal to c̄ correspond to u = (1, 0, 0,−2, · · · ) if d4 = 4 and
(1, 0, 0, 0,−2, · · · ) if d5 = 4. The nullity condition means that at most one of these can occur. This
is a contradiction as r ≥ 5.
In (8) Eq.(3.2) gives (d1, d2) = (2, 3) and nullity of c̄ gives
. Again one can check
that 1
(d+W) ∩ c̄⊥ is empty, a contradiction as r ≥ 4.
In (9), Eq.(3.2) and the nullity of c̄ give (d1, d4) = (2, 16) and {d2, d3} = {2, 3}. The only u which
can give ū ∈ c̄⊥ are (0,−2, 0, 0, 1i , · · · ) if d2 = 2 or (0, 0,−2, 0, 1
i , · · · ) if d3 = 2, where i ≥ 5. In
each case, i is unique since d2 (resp. d3) 6= 4. Since r ≥ 4, Theorem 8.9 now implies r = 4. But now
these u are not present (as i ≥ 5). Therefore there are actually no (1A) vertices, a contradiction
to r = 4.
CLASSIFICATION OF SUPERPOTENTIALS 41
In (10) we have (d3, d4) = (2, 16) and {d1, d2} = {2, 3}. The only u which can give an element
of c̄⊥ is (0,−2, 0, 0, 1i , · · · ) (for i unique and ≥ 5) if d2 = 2. The final argument in (9) now applies
equally here.
Finally, we can eliminate (2) by an analysis of both the (1A) and (1B) vertices. First, Eq.(3.2)
and nullity of c̄ force (d1, d2, d3) = (6, 1, 8). Next we check that
(d+W) ∩ c̄⊥ is empty, so r = 3.
Using Remark 3.9 we then find there can be no (1B) vertices, giving a contradiction.
So case (ii) cannot occur if K is connected.
Remark 9.1. Case (ii) is the only part of this section that relies on the connectedness of K.
In fact, the analysis of the cases where w is type III does not use this assumption. If K is not
connected, using the same methods and with more computation we obtain the following additional
possibilities (all of which are associated to a w of type II).
v w c = (4v − w)/3 d r
(9∗) (0,−1,−1, 1) (1,−1,−1, 0) (−1
,−1,−1, 4
) (1, 2, 6, 8) 4, 5
(1, 6, 2, 8) 4, 5
(10∗) (1,−1, 0,−1) (1,−1,−1, 0) (1,−1, 1
) (3, 3, 1, 8) 4
(6, 2, 1, 8) 4, 5
(14) (0,−1, 0, 1,−1) (1,−1,−1, 0, 0) (−1
,−1, 1
) (1, 3, 1, d4 , d5) 5
In (9*) and (10*) there is always a (1B) vertex in ∆c̄, and r = 4 or 5 according to whether the
cardinality of c̄⊥ ∩ 1
(d+W) is 1 or 2. The dimensions d4, d5 in (14) must satisfy
(i.e.,
{d4, d5} = {5, 20}, {6, 12}, {8, 8}) and again there is always a (1B) vertex in ∆
Interior points
Finally, we must consider the cases, listed in Table 3, §3, when there may be points of W in the
interior of vw. As in the earlier cases, we analyse the possible (1A) and (1B) vertices for these c.
For (1) and (2), as 1 < λ ≤ 2, the nonzero entries of c are either < −2 or > 1. Hence by Lemma
4.4 there are no (1A) vertices. By Theorem 8.9 we have r = 2 or 3.
In case (3) the nullity of c̄ implies that a vector u ∈ W not collinear with vw and with ū ∈ c̄⊥
must be of the form (−2, 0, 1j). So 2λ− 1 = d1
and c = (−d1
,−1 + d1
, 0, · · · ). From the range for
λ and the nullity of c̄, we have d1 = 3, d2 = 1. But d2 6= 1 since w ∈ W. So again there are no
(1A) vertices and by Theorem 8.9 we have r ≤ 3.
In case (4), a straight-forward preliminary analysis reduces the possibilities of u ∈ W such that
ū ∈ c̄⊥ to the choices u = (−2, 0, 1, · · · ), (0,−2, 1, · · · ) or (−1,−1, 1, · · · ). Note that c1 < −2 and
c2 = 1, so by Lemma 4.4(a) we see c3 < 1, i.e. λ <
. Now the second vector cannot occur because
the orthogonality equation and λ ≤ 2 imply that d3 = 1, contradicting the presence of w. Since
the three vectors are collinear, if two satisfy the orthogonality equation then all do. So there is at
most one (1A) vertex and so r ≤ 4 by Theorem 8.9. This can be improved to r ≤ 3 as follows. If
the third vector (−1,−1, 1, · · · ) occurs then the orthogonality relation, the bound on λ, and nullity
imply that d1 = 5, d3 = 2 and λ =
2 + 1
. Now the nullity equation may be written as a
quadratic in 2+ 1
with no rational root. If the first vector (−2, 0, 1, · · · ) occurs then orthogonality
implies λ =
d1(d3+2)
4d3+2d1
, and the bound on λ gives 6
> 1. Nullity implies d1 ≥ 5 and d1 > d
We can deduce d3 = 2 and λ =
, and one can check that nullity fails.
For case (5), again a straight-forward analysis of the orthogonality condition with the help of
the nullity of c̄ gives the following u ∈ W as possibilities such that J(c̄, ū) = 0:
(a) (−23, 1i), i ≥ 4 and d3 = 2,
(b) (1,−2, 0, · · · ),
(c) (−22, 1i), i ≥ 4 and d2 = 3,
(d) (0,−2, 1, 0, · · · ).
42 A. DANCER AND M. WANG
Note (1, 0,−2, · · · ) cannot be in W as then vw is not an edge.
It follows from Cor 4.3 that among (a) only one vector can occur and among (b), (c), (d) also
only one vector can occur. (The orthogonality conditions of (b) and (d) are incompatible with
1 < λ ≤ 2.) So c̄⊥∩ 1
(d+W) contains at most two elements. If it has two elements, one must then
come from (a) and the other from (b)-(d). Together they give an edge of c̄⊥ ∩ conv(1
(d + W))
with no interior points in 1
(d +W). Using Cor 4.3 and the null condition, we find that all these
two-element cases cannot occur. Hence r ≤ 4.
If r = 4 then the possible u with J(c̄, ū) = 0 are given by (a)-(d) with i = 4. Now, we can show
using techniques similar to those of Theorem 3.11 that c∗ = (1− 2λ, 2λ− 1,−1, 0) also gives a null
element of C. The possible vectors orthogonal to this element come from (a) and the vectors (b∗),
(c∗), (d∗) obtained from (b), (c), (d) by swapping places 1 and 2.
If (a) does not give an element in c̄⊥ ∩ 1
(d + W), it is straightforward to show, using the
orthogonality and nullity conditions for c̄ and c∗ together, that the (1A) vertices for c̄ and c∗ are
given by (b) and (b∗) respectively. Also we must have c = (4
,−1, 0) and d = (4, 4, 9, d4).
We need a (1B) vertex outside x4 = 0. From Remark 3.9, the only possible (1B) vertices for c̄
correspond to (1, 0, 0,−2) and (1,−1, 0,−1). In particular there can be no vertices, and hence no
elements of W, with x4 > 0. Hence (−1,−1, 0, 1) and therefore (1,−1, 0,−1) are not in W. So
the (1B) vertices for c̄ and c∗ are given by (1, 0, 0,−2) and (0, 1, 0,−2) respectively. Now the line
joining the corresponding null vectors a, a∗ misses conv(W), a contradiction.
The remaining case is when (a) gives the element in c̄⊥ ∩ 1
(d+W) and in c∗
(d+W). Now
for vw to be an edge we need (−14) absent, so by Remark 1.2(b) the only possible members of W
lying outside {x4 = 0} are the three type IIIs with x4 = 1. In particular, all vectors in conv(W)
have x4 ≥ 0. As (−1,−1, 1, 0) ∈ W, there must be (1B) vertices lying in {x4 = 0} for both c̄ and c∗.
We then find that the only possibilities for such a (1B) vertex are given by (b), (b∗) respectively.
It follows that d1 = d2, but now nullity is violated.
We conclude that there are no (1A) vertices, so r ≤ 3.
Theorem 9.2. Let c̄ be a null vertex in C such that ∆c̄ contains a type (2) vertex corresponding
to an edge vw of conv(W). Suppose we are not in the situation of Theorem 3.14.
(i) If there are no points of W in the interior of vw, then either we are in the situation of
Example 8.2 of [DW4] or K is not connected and we are in one of cases in the table of Remark 9.1
or in the situation of the third paragraph of Example 8.3 of [DW4].
(ii) If there are interior points of vw in W then r ≤ 3.
For further remarks about the r = 2 case see the concluding remarks at the end of §10.
10. Completing the classification
Throughout this section we will assume that K is connected (and we are not in the situation of
Theorem 3.14). Theorems 8.7 and 9.2 then tell us that if r ≥ 4 there are no type (2) vertices and no
adjacent (1B) vertices in ∆c̄, for any null vector c̄ ∈ C. Since all (1B) vertices lie in the half-space
{J(c̄, ·) > 0} bounded by the hyperplane c̄⊥ containing the (1A) vertices, we must therefore have
in each ∆c̄ exactly one (1B) vertex, with the remaining vertices all of type (1A). So if r ≥ 4 the
only remaining task is to analyse such a situation.
As dim∆c̄ = r − 2, we see dim(∆c̄ ∩ c̄⊥) = r − 3. In particular, there must be at least r − 2
elements of W giving elements of c̄⊥ ∩ 1
(d+W).
Theorem 5.18 lists the possible configurations of such elements when r ≥ 3. The above discussion,
together with Remark 5.19, shows that in cases (1), (2), (3) we can take m = r− 1, in cases (5)(ii),
(5)(iii), (6)(i), (6)(ii) we can take m = r, and in (6)(iii) we have r = 5.
Finally, since the vectors in (4) are collinear, so that dim(∆ ∩ c̄⊥) = 1, we have r = 3 or 4.
If r = 3, it follows that c̄ and the edge in conv(1
(d + W)) determined by the vectors in (4) are
CLASSIFICATION OF SUPERPOTENTIALS 43
collinear. This contradicts the orthogonality condition for the configuration of vectors in (4) and
we conclude that r = 4.
For each configuration we can consider the possible vectors u ∈ W giving the (1B) vertex. Besides
the nullity condition on c̄ and the condition that c̄ should be orthogonal to the (1A) vectors, we
have a further relation coming from the null condition in Remark 3.9. In most cases, routine (but
occasionally tedious) computations show that these relations have no solution.
As a result, we obtain the following possibilities for u (up to obvious permutations):
case (1A) vectors possible (1B) vector
(1) (−21, 1i), 2 ≤ i ≤ r − 1 (−1r), (12,−2r), (−11,−12, 1r)
(−11, 12,−13), (−11, 12,−1r)
(2) (11,−2i), 2 ≤ i ≤ r − 1 (−12), (−12,−13, 1r)
(3) (11,−22); (11,−23); possibly(11,−12,−13);
(−12,−13, 1i), 4 ≤ i ≤ r − 1 (14,−2r), (11,−2r), (−1r)
(4) (11,−22), (11,−23), (11,−12,−13) (−14), (11,−24), (11,−12,−14),
(5)(i) (−21, 12), (−11, 12,−13) (−11), (−14)
(5)(ii) (−21, 12); (−11, 13,−1i), 4 ≤ i ≤ r (13,−24), (−11, 12,−14), (13,−14,−15)
(5)(iii) (−21, 12); (−11,−13, 1i), 4 ≤ i ≤ r (−22, 14), (−11,−12, 14), (−13, 14,−15)
(6)(i) (−11,−12, 1i), 3 ≤ i ≤ r (−11,−13, 14)
(6)(ii) (11,−12,−1i), 3 ≤ i ≤ r (11,−13,−14), (−12, 13,−14), (11,−23)
(6)(iii) (11,−12,−1i), i = 3, 4; (11,−13,−14) (−15)
Table 7: Unique 1B cases
Remark 10.1. The possibilities for the (1B) vertex in cases (5)(ii) and (5)(iii) only apply to the
r ≥ 5 situation. When r = 4, the two cases become the same if we switch the third and fourth
summands and the possibilities are discussed in Lemma 10.14 below.
Remark 10.2. Note that u such as (−12), (−13) in (4), (−14) in (5)(ii) or (−1i) with i > 2 in
(6)(ii) cannot arise because they will not be vertices, due to the presence of the type II vectors in
Remark 10.3. The dimensions must satisfy certain constraints in each case. Some such constraints
were stated in Theorem 5.18 and Remark 5.19. We also have constraints coming from the nullity
conditions for c̄ and ā. These typically involve the requirement that some expression in the di is a
perfect square. The following is a summary of general constraints in each case:
case (1): d1 = 4; case (2): d1 = 1; case (3): d2 = d3 = 2; case (4): d2 + d3 ≤ 4d1/(d1 − 1) and
d2, d3,≥ 2; case (5)(i): (d1, d2) = (4, 2), (3, 3); case (5)(ii,iii): d1 = 2 and if r ≥ 5 also d3 = 2; case
(6)(i, ii): d1 = d2 = 2; case (6)(iii): d1 = d2 = d3 = d4 = 2, d5 = 25.
Our strategy now is reminiscent of that in §8. We have a (1B) vertex corresponding to ū and a
second null vector ā satisfying a = 2u − c. Now we may apply our arguments to ā, and conclude
that the vectors in ā⊥ ∩ 1
(d+W) are also of the form given in the above table, up to permutation.
The resulting constraints will allow us to finish our classification.
In some cases we can actually show that ā⊥∩ 1
(d+W) is empty and we have a contradiction. A
simple example when this happens is case 6(iii), where we now have c = (6
,−1) and
(ai/di) = (−
). Other cases are treated in Lemma 10.4 below.
Next we shall show that cases (6)(i)(ii) cannot arise (cf Lemma 10.6), so in all remaining cases
there must be at least one type III vector w with w̄ in ā⊥. We now use our explicit formulae
for c and a to derive inequalities on the entries of a and find when there can be such a type III
vector orthogonal to ā. For each such instance we then check whether ā⊥ ∩ 1
(d + W) forms a
configuration equivalent to one of those in Table 4. This turns out to be possible in only two
44 A. DANCER AND M. WANG
situations (cf Lemmas 10.7, 10.11 and Lemma 10.9). These have such distinctive features that W
can be completely determined and judicious applications of Prop 3.7 lead to contradictions. This
yields our main classification theorem.
Lemma 10.4. The following cases cannot arise:
case (4) with u = (1,−1, 0,−1),
case (6)(ii) with u = (1, 0,−2, · · · ),
case (4) with u = (0, 0, 0,−1) except for the case c = (4
,−1) with d = (4, 2, 2, 9).
Proof. (α) For case (4) with u = (1,−1, 0,−1) we find that the nullity and orthogonality conditions
and Remark 3.9 leave us with the following possibilities:
d c (ai/di)
(2, 5, 3, 20) (1,−5
, 0) (1
(2, 6, 2, 12) (1,−3
, 0) (1
(3, 4, 2, 12) (1,−4
, 0) (1
(5, 3, 2, 15) (1,−6
, 0) (1
It is easy to see that we can never have
= 1 for w ∈ W, so ā⊥ ∩ 1
(d+W) is empty and we
have a contradiction.
(β) Similarly, nullity, orthogonality and Remark 3.9 give:
d c (ai/di)
(2, 2, 225, d4 , . . . , dr) (
,− d4
, . . . ,− dr
) (109
, . . . , 1
(2, 2, 98, d4 , . . . , dr) (
,− d4
, . . .− dr
) (23
, . . . , 1
(2, 2, 36, d4 , . . . dr)) (
,−1,−d4
, . . . ,−dr
, . . . 1
Moreover n equals 962, 226, 50 respectively. It is easy to see that we can never have
for w ∈ W, so we have a contradiction.
(γ) For case (4) with u = (0, 0, 0,−1) we find that the nullity and orthogonality conditions and
Remark 3.9 leave us with the following possibilities, up to swapping places 2 and 3:
d c (ai/di)
(2, 2, 4, 25) (6
,−1) (−3
(2, 3, 3, 25) (6
,−1) (−3
(3, 2, 3, 121) (15
,−1) (− 5
(2m− 2, 2, 2,m2) (
2(m−1)
, 1−m
, 1−m
,−1) (− 1
, m−1
, m−1
It is now straightforward to see that we cannot have w ∈ W with
= 1, except in two cases
(both associated to the last entry of the table). One is the case stated in the Lemma. The other
occurs if m = 2, so a = (−1, 1
,−1) which is orthogonal to (−1, 0, 1,−1), (−1, 1, 0,−1). But as
(0, 0, 0,−1) is a vertex, neither of these vectors can be in W. So ā⊥∩ 1
(d+W) is still empty, giving
the desired contradiction.
As discussed above, we now turn to showing that case (6) cannot occur. The following remark
is useful in finding when type III vectors can give elements of ā⊥ ∩ 1
(d+W).
Lemma 10.5. If w = (−2i, 1j) and w̄ ∈ ā⊥, then ai
< 0 (assuming we are not in the situation of
Theorem 3.14).
Proof. We need
(10.1)
= 1 +
CLASSIFICATION OF SUPERPOTENTIALS 45
so if ai
≥ 0 then
≥ 1. Hence
≥ dj ≥ 1. As ā is null this means a = (−1
j) and we
are in the situation of Theorem 3.14.
Lemma 10.6. Configurations of type (6) cannot arise.
Proof. Recall that we have dealt with (6)(iii) and we have d1 = d2 = 2.
Case (6)(i): We have u = (−1, 0,−1, 1, · · · ), and from the nullity and orthogonality conditions we
deduce that
(ci/di) =
2(m+ 1)
2(m− 1)
m2 − 1
, · · · ,
m2 − 1
where 1
2(m+1)
and n− 1 = m2 for some positive integer m. We have, therefore,
(ai/di) =
2(m+ 1)
2(m− 1)
m2 − 1
m2 − 1
m2 − 1
, · · · ,
m2 − 1
Let us estimate the size of the entries in (ai/di). First observe that, as d3, d4 > 2(m + 1) from
above, we have
4(m+ 1) < d3 + d4 ≤ n− d1 − d2 = n− 4 = m
2 − 3
so we deduce m ≥ 6. Hence we have 3
≤ |a1
| < 1
≤ |a2
| < 1
, |a3
| ≤ 6
, |ai
| ≤ 1
for i ≥ 5. Also
note that a4
. Finally, a4
> 0, else we would have 2n − 4 = 2m2 − 2 ≤ d4, which
is impossible.
Consider now a type III vector w = (−2i, 1j) with w̄ ∈ ā⊥. By Remark 10.5, we need i = 1, 3 or
≥ 5. If i = 1 then, by Eq.(10.1),we have
∈ (0, 1
]. So we must have j = 4 and 1
contradicting our above remarks. If i = 3 then
, which is impossible. Similarly, if i ≥ 5
, which is impossible.
Hence there are no such type III vectors, so we are in case (6) with respect to ā. We cannot be
in 6(ii) as then the null vector has exactly one positive entry (see below), but a has two positive
entries. For 6(i), the null vector has exactly two negative entries. Now a has r− 2 negative entries,
so r = 4. But for 6(i) the negative entries have modulus < 2, while a3 = −2−
m2−1 , a contradiction.
Case 6(ii): Here there are two possibilities.
Subcase (α): u = (0,−1, 1,−1, · · · ). Then, as above, the null condition for c̄ gives
(ci/di) =
(m− 1)(m+ 2)
2(m+ 1)2
2(m+ 1)
(m+ 1)2
, · · · ,
(m+ 1)2
where 1
2(m+1)
and n− 1 = m2. So the vector (ai/di) is given by
(m− 1)(m+ 2)
2(m+ 1)2
2(m+ 1)
(m+ 1)2
(m+ 1)2
(m+ 1)2
, · · · ,
(m+ 1)2
As before, we have m ≥ 6. We deduce |a1
| ≤ 1
≤ |a2
| < 1
, |a3
| ≤ 8
, |a4
| ≤ 1
, |ai
| ≤ 1
i ≥ 5. Also a4
< 0, else d4 ≥ 2(m+ 1)
2 > 2n, which is impossible.
We look for vectors w = (−2i, 1j) with w̄ ∈ ā⊥. By Lemma 10.5 we have i = 1, 2 or 4. If i = 1
= m+3
(m+1)2
, so j = 3, but now Eq.(10.1) contradicts 2
. A similar argument works if
i = 2, while if i = 4, Eq.(10.1) implies
, a contradiction.
So 6(ii) must hold for ā, as we have already ruled out 6(i). But now we need a to have exactly
one positive entry, which has modulus < 2. So r = 4 and this positive entry is a3, but we have
a3 > 2, a contradiction.
46 A. DANCER AND M. WANG
Subcase (β): u = (1, 0,−1,−1, · · · ). We similarly have
(ci/di) =
(m− 2)(m+ 1)
2(m− 1)2
2(m− 1)
(m− 1)2
, · · · ,
(m− 1)2
(ai/di) =
m2 − 3m+ 4
2(m− 1)2
2(m− 1)
(m− 1)2
(m− 1)2
(m− 1)2
, · · · ,
(m− 1)2
where n− 1 = m2 and 1
= m+1
2(m−1)2 . The last two equations easily imply that m ≥ 6, and so
| < 1
for all i.
A type III (−2i, 1j) giving an element of ā⊥ must have i = 3 or 4, by Lemma 10.5. In both cases
we find from Eq.(10.1) that
, which is impossible. So 6(ii) holds for a, which is impossible
as a has at least two positive entries.
Lemma 10.7. The only possible example in case (4) is when c =
, u = (0, 0, 0,−1),
and d = (4, 2, 2, 9). ∆ā is then in case (1) with a = (−4
,−1) and ā⊥ ∩ 1
(d +W) consists of
(−2, 1, 0, 0), (−2, 0, 1, 0).
Proof. By Lemma 10.4 we just have to eliminate the possibility u = (1, 0, 0,−2). Now
(ai/di) =
8− 2d1 + d4
d1(4 + 2d1 + d4)
(d1 − 1)(2d1 + d4)
2d1(4 + 2d1 + d4)
(d1 − 1)(2d1 + d4)
2d1(4 + 2d1 + d4)
2(d1 − 1)
d1(2d1 + d4 + 4)
The null condition is
−(d1 − 1) d
4 − 4(d
1 − d1 − 1) d
4 + 4d1(3d
1 + d1 + 8)d4 + 16d
1(d1 + 2)
2 = 0.
For w = (−2i, 1j) with w̄ ∈ ā⊥ we need, by Lemma 10.5, i = 1 or 4. If i = 1, then for j = 2
or 3, Eq.(10.1) can be rewritten as 2d21 + d1d4 + 2d1 = −5d4 − 32, which is absurd. For j = 4 it
can be rewritten as −1 − 2
= 14+2d4−2d1
d1(2d1+d4+4)
. So the right hand side is < −1, which on clearing
denominators is easily seen to be false.
If i = 4, then for j = 1 Eq.(10.1) becomes 4
= 1. So (d1, d4) = (2, 8), (3, 6) or (5, 5), all of
which violate the null condition. For j = 2, 3 we obtain from Eq.(10.1) the equation 4d4(d1 − 1) =
(2d1 + d4 + 4)(d1d4 + d4 − 8d1), which can only have solutions if d4 ≤ 9. On the other hand, the
null condition has no integer solutions if d4 ≤ 9.
So no such type III exists, contradicting Lemma 10.6.
Lemma 10.8. Configurations of type (5)(i) cannot occur.
Proof. It is useful to note that the null condition for c̄ implies that d3 ≤ 4 when (d1, d2) = (4, 2)
and d3 ≤ 3 when (d1, d2) = (3, 3). One further finds the following possibilities:
u d c (ai/di)
(−1, 0, 0, 0) (3, 3, 1, 2) (−1, 1,−1
) (−1
(3, 3, 2, 1) (−1, 1,−2
) (−1
(4, 2, 1, 3) (−1, 1,−1
) (−1
(4, 2, 2, 2) (−1, 1,−1
) (−1
(4, 2, 3, 1) (−1, 1,−3
), (−1
(0, 0, 0,−1) (3, 3, 2, 121) (− 9
,−1) ( 3
(4, 2, 2, 25) (−4
,−1) (1
One easily checks that ā⊥ ∩ 1
(d +W) is empty in the last two cases, and consists only of type
II vectors in the third to fifth cases, giving a contradiction to Lemma 10.6. For the first two cases,
note that ā⊥ ∩ 1
(d + W) contains 1
(d + (−1,−1, 1, 0)) since by hypothesis for 5(i) (−1, 1,−1, 0)
is in W. Hence (1),(2) cannot hold with respect to ā. Also the vector d of dimensions rules out
(3),(4) and (5), so we have a contradiction.
CLASSIFICATION OF SUPERPOTENTIALS 47
Lemma 10.9. The only possible example for case (3) is when c =
, u =
(0, 0, 0, 0,−1), d = (2, 2, 2, 2, 9). Then a = (−2
,−1) and ∆ā is again in case (3).
Proof. (A) Let u = (−1r). The null condition for c̄ gives 4dr = (δ+d1+2)
2, where δ = d4+· · ·+dr−1.
In particular, dr is a square. Also, (ai/di) =
1− 1√
1− 1√
, −1√
, · · · , −1√
If dr = 4, then we find there are no type III vectors in ā
⊥, a contradiction. So dr ≥ 9 and
we have |ai
| ≤ 1
for i = 1, 4, · · · , r − 1, ≤ 1
for i = r and < 1
for i = 2, 3. Lemma 10.5 shows
i 6= 2, 3. From Eq.(10.1) and the above estimates, we first get i 6= r, and for the remaining values
of i, we have
> 0, so that j = 2 or 3. Also, dr = 9 (and hence d1 + d4 + · · · + dr−1 = 4), and
(ai/di) = (−
, · · · ,−1
). Upon applying Theorem 5.18 to ∆ā together with Theorem
8.9 and the above Lemmas, we deduce that we are in case (3)(ii) with r = 5 and d1 = d4 = 2,
giving the example in the statement of the Lemma.
(B) Next let u = (1, 0, · · · ,−2). Now (ai/di) = (
− α, 1−α
, 1−α
,−α, · · · ,−α,
(n−2−dr)α−5
where, as a consequence of the null condition for c̄, we have
(10.2) α =
n− 2−m
: m2 = dr(n− 1).
Next we get the identity (n− 2)2 −m2 = (n− 1)(d1 + 1+ δ) + 1 =
m2(d1+1+δ)
+1, where δ is as
given in (A) above. We deduce m <
n−3(n− 2), and hence α <
d1+1+δ
n−dr−3 ≤ min(
So ai
is positive for i ≤ 3 and negative for 3 < i ≤ r−1. Note also that (n−2−dr)α <
2(n−2−dr)
n−3−dr =
1 + 1
n−3−dr
. In particular, ar
< − 5
As usual, we look for (−2i, 1j) giving an element of ā⊥. By Lemma 10.5, i 6= 1, 2, 3. If 4 ≤ i ≤ r−1
then Eq.(10.1) says
= 1 − 2α > 0, so j = 1, 2 or 3. If j = 1 we obtain α = 1 − 2
. Comparing
this with Eq.(10.2) shows d1 = 3 and α =
, but now ā is not null. If j = 2 or 3, we obtain α = 1
we deduce from Eq.(10.2) that d1 ≤ 5, and again one can check that all possibilities violate nullity.
So all type III (−2i, 1j) have i = r. Therefore we must be in case (1) or (5) with respect to ā,
and dr = 4 or 2 respectively.
If j = 1, 2, 3 then
> 0. Now Eq.(10.1) combined with the estimate above for ar
show that
> ar > −
, so dr > 5, a contradiction.
If 4 ≤ j ≤ r − 1, then in the case dr = 4, we find Eq.(10.1) gives α =
n−4 . Combining with
Eq.(10.2) we get n = 10, which is incompatible with dr = 4 and r ≥ 5. If dr = 2 we find similarly
that α = 4
n−3 and m satisfies 3m
2 − 8m− 4 = 0; but this has no integral roots.
So u = (1, 0, · · · − 2) cannot occur.
(C) For u = (14,−2r), we have (ai/di) = (−α,
, 1−α
− α,−α, · · · ,−α,
(n−2−dr)α−5)
) and
Eq.(10.2) still holds. The arguments of case (B) carry over to this case, on swapping indices
1, 4.
Lemma 10.10. Case (2) cannot occur.
Proof. (A) Consider u = (−12). Now (ai/di) = (
− 1, −1
, · · · , 1
(2− n+1−dr
Nullity implies d2 ≥ 3, so −1 <
and |ai
| ≤ 1
for 2 ≤ i ≤ r − 1. Also we have
dr(n− 1) = m
2, and for this choice of u we have m = n+1− 2d2, so
= dr−m
which is positive if
m < 0 and negative if m > 0. By Lemma 10.5 and the fact that d1 = 1, we only have to consider
(−2i, 1j) with i = 2 or r.
If i = 2, then Eq.(10.1) says
= 1 − 2
> 0 so j ≥ 3. If 3 ≤ j ≤ r − 1 then Eq.(10.1) shows
d2 = 3, so m = n− 5 and dr =
(n−5)2
n−1 = n− 9 +
n−1 . As n ≥ 7 we must then have (dr, n) = (9, 17)
or (2, 9). Imposing the nullity condition on ā shows there are only three possibilities, corresponding
48 A. DANCER AND M. WANG
to d = (1, 3, 2, 2, 9), (1, 3, 4, 9), (1, 3, 3, 2). In the first two there is only one type III in ā⊥, as d1 = 1
and d2 6= 4, so we are in case (5) with respect to ā, contradicting the fact that d2 6= 2. In the last
case we must be in case (4) with respect to ā, but now (0,−1, 1,−1) is present, so u is not a vertex.
If j = r then Eq.(10.1) becomes d2 =
3dr−n−1
dr−2 which is less than 3, a contradiction. (We cannot
have dr = 2 and 3dr = n+ 1 as n ≥ 7.)
Hence all such (−2i, 1j) have i = r and we are in case (1) or (5) with respect to ā. For case (1)
we need r− 2 of the
(j < r) equal. This can only happen for our a if d2 = 3, which is ruled out
as in the previous paragraph. For case (5) we have dr = 2 and
= 3 − n−1
. The possibilities on
the left-hand side are 2
− 1,− 1
respectively. On using our relations for m,n, dr we find that
only the third possibility can occur, and d2 = 3. The argument in the previous paragraph again
eliminates this case.
(B) Consider u = (0,−1,−1, · · · , 1). Now (ai/di) = (2β−1, β−
, β− 2
, β, · · · , β,
4−(n+1−dr)β
where again from the null condition of c̄ we have
(10.3) β :=
(1− c1) =
n+ 1−m
3 + dr(
n+ dr + 1
: dr(n− 1) = m
2 (m > 0).
Now 0 < β < 1
(the case β = 1
leads to r = 4, d2 = d3 = 2 and a = (0,−1,−1, 1) which violates
nullity). So for w̄ ∈ ā⊥ we just have to consider w = (−2i, 1j) with i = 2, r (as d1 = 1 we can’t
have i = 1; also by symmetry the case i = 3 is treated the same way as i = 2).
If i = 2 then Eq.(10.1) says
= 2β + 1 − 4
. If 4 ≤ j ≤ r − 1 we get β = 4
− 1; the only
possibility consistent with our bounds on β is d2 = 3, β =
and it is straightforward to check this
is incompatible with the null condition for ā.
If i = 2 and j = 1 then Eq.(10.1) implies d2 = 2 and again one checks that nullity for c̄ fails.
If j = 3 Eq.(10.1) says β = 4
− 1, so as β > 0 either d2 = 2 and β = 1 −
or d2 = 3
and β = 1
. In the former case the bound β < 1
shows d3 = 3 and β =
, and now nullity
for ā fails. In the latter case the bound β > 0 shows d3 > 6. Substituting this into the quadratic
which must vanish for nullity of c̄, we see δ = d4 + · · · + dr−1 is < 4. Checking the resulting short
list of cases yields no examples where nullity holds. If j = r then Eqs.(10.1) and Eq.(10.3) imply
= 1+ 1
and one check that the possible (d2, dr) yield no examples where nullity of c̄ holds.
So all such type III have i = r, and case (1) or (5) holds for ā. For case (1), then, as in (A), r−2
of the
(j ≤ r− 1) must be equal. So either r = 5 and d2 = d3 with β = 1−
or r = 4 and one
of the preceding equalities holds. If β = 1 − 2
holds, then the bounds on β show d2 = 3, β =
and as usual nullity for ā fails. If d2 = d3 holds, then using our formulae for β and substituting
into the null condition for ā gives a quadratic with no integer roots.
If case (5) holds, then dr = 2. Now Eq.(10.1) gives
= 5 − (n − 1)β. If j = 1, j = 2, or
4 ≤ j ≤ r − 1 we get β = 6
5+(2/d2)
respectively. (As usual, the case j = 3 is treated in just
the same way as j = 2.) Now using the equations in Eq.(10.3) relating n,m in each case gives a
quadratic with no real roots.
Lemma 10.11. For case (1) the only possibility is when c = (−4
,−1) with u = (0, 0, 0,−1)
and d = (4, 2, 2, 9). ∆ā is then in case (4).
Proof. (A) Consider u = (0, · · · , 0,−1). From the null condition for c̄ we see that dr = k
2, n− 1 =
(k + 1)2 for some positive integer k and (ai/di) = (
(1 − 1
),− 1
, · · · ,− 1
). Note that since
d1 = 4, n > 5 and so k 6= 1.
We must consider solutions of Eq.(10.1). By Lemma 10.5, i 6= 1. If i = r we have
= 1 − 2
The resulting equation has no solution in integer k > 1 for any choice of j. If 2 ≤ i ≤ r − 1, we
= 1 − 2
. We only obtain a solution k > 1 if j = 1; in this case k = 3, so n = 17, dr = 9
CLASSIFICATION OF SUPERPOTENTIALS 49
and we see r = 4 with {d2, d3} = {2, 2} or {3, 1}. The former case is that in the statement of the
Lemma. In the latter case we can have just one type III and one type II in ā⊥ (since d2 or d3 is
1, one potential type III is missing), so we must be in case (5) with respect to ā; but no di is 2, a
contradiction.
(B) Consider u = (0, 1, 0, · · · , 0,−2). Now (ai/di) = (
(1− β), 2
− β,−β, · · · ,−β,
(n−dr−2)β−5
where β = dr+6d2
d2(2n−dr−4) . The nullity condition for c̄ implies d2 ≥ 3, d2 > δ and dr > 2d2 + 4, where
δ now denotes d3 + · · · + dr−1. We can then deduce that β < 1, 0 <
, 0 < a2
particular β < 2
. By Lemma 10.5, we must consider elements of ā⊥ coming from vectors (−2i, 1j)
with i ≥ 3.
If 3 ≤ i ≤ r−1, Eq.(10.1) says
= 1−2β. As β < 1, this immediately rules out 3 ≤ j ≤ r−1. If
j = 1 we get β = 1
. Combining this with our formula above for β we get 2d2(d2+δ−7)+(d2−3)dr =
0. The only possibilities are d2 = 3, δ = 4 which violates the null condition, or d = (4, 4, 2, 8) which
violates the condition that dr(n − 1) should be a square. If j = 2, we get β = 1 −
. Since we
saw above that β < 2
we get d2 = 3 and β =
, which is ruled out as above. If j = r, Eq.(10.1)
implies β = dr+5
2+d2+δ+2dr
. Comparing this with the formula for β above leads to a contradiction.
The remaining possibility is for i = r. So we are in case (1) or (5) with respect to ā, and dr = 4
or 2 respectively. But dr > 2d2 + 4, so this is impossible.
(C) Let u = (−1,−1, 0, · · · , 0, 1). Now (ai/di) = (−
β,− 2
−β,−β, · · · ,−β,
1+(n−dr−2)β
) where
dr(d2−4)
2d2(2n+dr−4) , so 0 < β <
(noting that the nullity condition for c̄ implies d2 ≥ 5).
We look for vectors (−2i, 1j) giving elements of ā⊥. Now Lemma 10.5 rules out i = r, while if
3 ≤ i ≤ r − 1 we need
= 1− 2β > 2
. So j = r, and Eq.(10.1) yields β = dr−1
n+dr−2 . Equating this
to the expression above for β gives an equation which may be rearranged so that it says a sum of
positive terms is zero, which is absurd.
If i = 1 then Eq.(10.1) says
= 1−β. Clearly this can only possibly hold if j = r. The equation
then gives β = dr−1
n−2 , and equating this with the earlier expression for β leads, as in the previous
paragraph, to a contradiction.
So the only possibility is i = 2, and we are therefore in case (1) or (5) with respect to ā. But
d2 ≥ 5 so this is impossible.
(D) Let u = (−1, 1,−1, 0, · · · ). Now (ai/di) = (−
− β,− 2
− β,−β, . . . ,−β,
(n−dr−2)β−1
and β = 1
−2( 1
). An analysis of the nullity condition for c̄ shows that it can only be satisfied
, so d2, d3 ≥ 5 and 0 < β <
Let us now consider solutions to Eq.(10.1). If i = r, we have
= 1− 2
2(n−dr−2)β
. If j 6= 2,
this equation implies that the positive quantity (1 +
2(n−dr−2)
)β (or (1
2(n−dr−2)
)β if j = 1)
equals a nonpositive quantity (recall dr > 1 as i = r). If j = 2, we get that it equals
But d2 ≥ 5 so dr = 2 or 3, and in each case we find the nullity condition for c̄ is violated.
If i = 1, Eq.(10.1) says
= 1 − β > 9
, so j = 2 or r. But for j = 2 we get d2 = 2, which is
impossible as we know d2 ≥ 5, so in fact j = r.
If i = 2, Eq.(10.1) is
= 1 + 4
− 2β. We cannot then have j = 1, 3 or 4 ≤ j ≤ r − 1 as they
lead to β > 2
, > 1, > 1 respectively. So we must have j = r.
If i = 3, we see
= 1 − 4
− 2β. If j = 1, 2 or 4 ≤ j ≤ r − 1, we see in all cases (using our
bounds on d2, d3) that β >
, a contradiction. Hence again j = r.
If 4 ≤ i ≤ r − 1, then
= 1 − 2β > 4
so j = 2 or r. If j = 2 we obtain β = 1 − 2
contradicting our earlier inequality for β; so again we have j = r.
We have shown that any (−2i, 1j) giving an element of ā⊥ has j = r, so we are in case (3), (4)
or (5) with respect to ā. It cannot be case (3) as we know from Lemma 10.9 that then each di is
50 A. DANCER AND M. WANG
2 or 9, and we have d1 = 4. If we are in case (4), then Lemma 10.7 tells us that d = (4, 2, 2, 9).
Moreover, as (−2, 0, 0, 1), (0,−2, 0, 1) are the elements of ā⊥ ∩ 1
(d+W), we must have β = 4
; but
now β > 1
, a contradiction. If it is case (5), then we have di = 2 for some i, which we can take
to be 4. Now ā must be orthogonal to vectors associated to (−14, 15,−1k) or (−14,−15, 1k), and
either case is incompatible with our expressions for ai/di.
(E) Consider u = (−1, 1, 0, · · · , 0,−1). Now (ai/di) = (−
− β,−β, · · · ,−β,
((n−dr−2)β−3)
and β =
8d2−dr(d2−4)
2d2(2n−dr−4) . It is easy to check that β <
. Also, the nullity condition for c̄ implies
d2 ≥ 3 and (
− 1)dr + 8 > 0; hence β > 0.
The analysis is similar to that in (D). If i = r then Eq.(10.1) implies that a positive quantity
times β equals a positive linear combination of reciprocals of di, minus 1. This sum of reciprocals
is therefore > 1, which gives us upper bounds on dr. The only case where Eq.(10.1) and the null
condition can hold is if j = 2 and d2 = 7, dr = 4, d3 + · · ·+ dr−1 = 11
If i = 1 then Eq.(10.1) says
= 1− β > 0, so j = 2 or r. But j = 2 implies d2 = 2, which from
above cannot hold, so j = r.
Now Lemma 10.5 rules out i = 2. If 3 ≤ i ≤ r − 1, we have
= 1 − 2β. If j = 1 then we get
β = 2
, which cannot hold. If j = 2 then β = 1 − 2
, and as β < 2
we deduce d2 = 3 and β =
which violates the null condition for c̄. If 3 ≤ j ≤ r − 1, then β = 1, which is impossible. So we
have j = r.
So in all cases we have j = r, except in the exceptional case discussed above where we can have
i = r and j = 2. But our list (1)-(6) of possible configurations in ā⊥ shows that if the (i, j) = (r, 2)
case occurs then no other type III can be in ā⊥. So we are in case (5), which is impossible as
dr = 4 6= 2 for this example. Hence the exceptional case cannot arise.
We see therefore that j = r in all cases. So, as in (D), we must be in case (3), (4) or (5) with
respect to ā. As before, the fact that d1 = 4 rules out case (3). For case (4) we need the dk to be
4, 2, 2, 9 and dr to be 4 (as the (−2
i, 1j) in ā⊥ have j = r), but this contradicts d1 = 4.
So we are in case (5). Now the orthogonality condition for the family of type II vectors leads to
β > 2
, which is impossible.
Lemma 10.12. Case (5)(ii) cannot occur if r ≥ 5.
Proof. (A) Consider u = (0, 0, 1,−2, 0, · · · ). We have
(ai/di) = (1− β, 1− 2β,
n−2d2−3
n−d2−2 − (
n−6−3d2
n−d2−2 )β,
2(d2+2)β
n−d2−2 −
n−d2−2 −
2(d2+2)β
n−d2−2 −
n−d2−2 , · · · )
where all terms from the fifth onwards are equal and where
β := 1 + c1
8(n−d2−2)+d4(n+d2)
2d4(n+d2+2)
The nullity condition for c̄ implies d4 > 52, so n > 56 and we deduce 0 < β <
. Hence
> 0. It is also easy to show that ai
> 0 for i ≥ 5 and a3
So if (−2i, 1j) gives an element of ā⊥ we need i = 2 or 4. As d4 > 52, Lemmas 10.6 - 10.11 show
that case (5) must hold with respect to ā. In particular di = 2, so we cannot have i = 4. Hence
i = 2 and d2 = 2. Now Eq.(10.1) implies β =
, 2n−5
3n−4 ,
4(n−2) +
d4(n−2) or
4(n−2) , depending on
whether j = 1, 3, 4 or ≥ 5. In all cases this contradicts the bound β < 3
and n > 56.
(B) Consider u = (−1, 1, 0,−1, 0, · · · ). We have (ai/di) =
(−β, 2
+ 1− 2β, − d2+1
n−d2−2 − (
n−6−3d2
n−d2−2 )β,
2(d2+2)β
n−d2−2 −
n−d2−2 −
2(d2+2)β
n−d2−2 −
n−d2−2 , · · · )
where all terms from the fifth onwards are equal and
β := 1 + c1
= n+d2−2
2(n+d2+2)
+ n−2
d2(n+d2+2)
+ n−d2−2
d4(n+d2+2)
The nullity condition for c̄ implies d2 ≥ 9 and d4 ≥ 4. It is now easy to check that
< β < 31
and that ai
> 0 for i ≥ 5.
As in (A), case (5) must hold with respect to ā. Now if (−2i, 1j) gives an element of ā⊥ we need
di = 2. This, combined with Lemma 10.5, means i = 1 or 3.
CLASSIFICATION OF SUPERPOTENTIALS 51
If i = 1, Eq.(10.1) immediately shows j cannot be 2. Moreover, if j = 3 or ≥ 5, Eq.(10.1) yields a
value for β that violates the nullity condition for c̄. If j = 4 we obtain β = n−1
+ n−d2−2
. As we are
in case (5) with respect to ā, we need to consider the elements in ā⊥∩ 1
(d+W) corresponding to type
II vectors. Their number and pattern, as stipulated by Theorem 5.18, together with orthogonality
to ā, imply further linear relations among the components of (ai/di) and small upper bounds for
r (usually of the form r = 5, 6). In all cases these additional constraints can be shown to be
incompatible with the above values of β.
As an illustration of the above method, note that our type III vector is (−2, 0, 0, 1, 0, · · · ). If ∆ā
is in case (5)(iii), Theorem 5.18 says that the possible type IIs must have a −1 in place 1 and a 0 in
place 4. Since r ≥ 5, the remaining −1 must be in a place whose corresponding dimension is 2. As
d3 = 2 we can have (−1, ∗,−1, 0, ∗, · · · ) where ∗ indicates a possible location of the 1 in the type II.
The other possibility is for −1 to be in place k for some k ≥ 5. After a permutation we can assume
k = 5, and d5 = 2 must hold. The type II is then of the form (−1, ∗, ∗, 0,−1, ∗, · · · ) where ∗ again
indicates possible positions for the 1. In the first case, the orthogonality conditions imply a2
which gives β = n−1
+ n−d2−2
. Comparing with the value of β from Eq.(10.1), we get d2 = d4.
Using this in the first value of β in (B) gives a contradiction after some manipulation. In the second
case, the argument we just gave implies that we can only have r = 5 and the orthogonality condition
implies a2
, which gives β = n−1
n+d2+2
2(n−d2−2)
d2(n+d2+2)
. After a short computation, one sees that
the two values of β are again incompatible. If ∆ā is in case (5)(ii), the argument is essentially the
same, as we only have to switch the places of the second −1 and the 1 in the type IIs.
Let us now take i = 3. If j = 1, Eq.(10.1) implies β = 3d2+4−n
5d2+10−n . If the denominator is negative,
then β > 1, which is a contradiction. If it is positive we find that this is incompatible with the
inequality β > n+d2−2
2(n+d2+2)
which comes from the displayed expression for β above.
If j = 2, we get β = d2+1
2(d2+2)
+ n−d2−2
2d2(d2+2)
. As above, we can rule this out by considering the vectors
in ā⊥ ∩ 1
(d+W) associated to type II vectors. A similar argument works for j ≥ 5, where we find
β = n−2d2−3
2(n−2d2−4) , and for j = 4, where we have β =
n−d2−2
d4(n−2d2−4) +
n−2d2−3
2(n−2d2−4) .
(C) Next let u = (0, 0, 1,−1,−1, 0, · · · ). We have (ai/di) = (1 − β, 1 − 2β,
n−2d2−3
n−d2−2 −
(n−6−3d2
n−d2−2 )β,
2(d2+2)
n−d2−2β −
n−d2−2 −
2(d2+2)
n−d2−2β −
n−d2−2 −
2(d2+2)
n−d2−2β −
n−d2−2 , · · · ) where all
terms from the sixth on are equal and
β := 1 + c1
= n+d2
2(n+d2+2)
n−d2−2
(n+d2+2)
The nullity condition for c̄ implies d4 and d5 ≥ 13. It is now easy to check that 0 < β <
> 0. As in (A), we find also that ai
> 0 for i ≥ 6, and that a3
As in (A) again, case (5) must hold with respect to ā, so if (−2i, 1j) gives an element of ā⊥ we
need di = 2. This, combined with Lemma 10.5, means i = 2. In this situation in all cases Eq.(10.1)
gives a value of β incompatible with our bounds on n and β.
Lemma 10.13. Case (5)(iii) cannot arise if r ≥ 5.
Proof. This is similar to the proof of the previous Lemma so we will be brief.
(A) Consider u = (0,−2, 0, 1, 0, · · · ). Now (ai/di) is given by
(1− β,− 4
+ 1− 2β,
(n−3)+(n+d2−2)(β−1)
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 , · · · ,
2d2β−(d2+1)
n−d2−2 )
where
β := 1 + c1
(d2+4d4)(n−2−d2−d4)+4d24
2d2d4(2n−d2−4) .
The nullity condition for c̄ implies d4 ≥ 8, d2 ≥ 27 and d2 > 2d4, and it readily follows that
< β < 1
. In particular, a1
As before, we see that case (5) holds with respect to ā, so for ā⊥ we have to consider type III
vectors (−2i, 1j) where di = 2. So we need only consider i = 3 or i ≥ 5.
52 A. DANCER AND M. WANG
In either situation, we proceed as in part (B) of the proof of Lemma 10.12, and obtain inconsis-
tencies in the equations involving β or contradiction to the bounds on β or the dimensions.
(B) Let u = (0, 0,−1, 1,−1, 0, · · · ). The nullity condition on c̄, which has a symmetry in d4 and
d5, now implies d4, d5 ≥ 46 and > 28d2. Now (ai/di) is given by
(1− β, 1− 2β, (n+d2−2
n−d2−2)β −
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 , · · · )
where all terms from the sixth on are equal and
β := 1 + c1
n+d2−2 + (
)(n−d2−2
n+d2−2).
We deduce that 1
< β < 3
and so 1
> 0. ∆ā must be in case (5), and for (−2i, 1j)
associated to elements of ā⊥ ∩ 1
(d+W), we have di = 2 and we need only consider i = 2, 3 or ≥ 6.
If i = 2, Eq.(10.1) and the upper bound on β imply
. As d2 = 2, one checks that this
never holds.
If i ≥ 6, Eq.(10.1) implies
. The bound d4, d5 > 28d2 can be used to show that this never
happens.
If i = 3, Eq.(10.1) and the expression for β above are seen to be incompatible if we use the
bounds on d4, d5 and β.
(C) Consider u = (−1,−1, 0, 1, 0, · · · ). The nullity condition on c̄ gives d4 ≥ 4, d2 ≥ 5. The
vector (ai/di) is
(−β, 1− 2β − 2
(n+d2−2)β−(d2+1)
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 ,
2d2β−(d2+1)
n−d2−2 , · · · )
where all terms from the fifth on are equal and
β := 1 + c1
d24+(d2+d4)(n−d2−d4−2)
d2d4(3n−d2−6) .
Now as 1
, we see that 1
< β < 1
Again ∆ā is in case (5) and we consider vectors (−2i, 1j) associated to elements of ā⊥∩ 1
(d+W),
where we must have di = 2, so i 6= 2, 4.
If i = 1, Eq.(10.1) becomes
= 1 − 2β > 0. This immediately means j 6= 2, 5, · · · , r. If j = 3,
the value of β from Eq.(10.1) and the above expression for β lead to d4 ≤ 10/3. For j = 4, we
obtain a contradiction by the method of part (B) in the proof of Lemma 10.12.
If i ≥ 5, Eq.(10.1) says
4d2β+(n−3d2−4)
n−d2−2 . If i = 3, Eq.(10.1) say
2(d2+1)
n−d2−2 +
2(n+d2−2)β
n−d2−2 .
In both situations, we can again apply the method of part (B) in the proof of Lemma 10.12 to
obtain contradictions.
The last case to consider is case (5)(ii) with r = 4, which is the same as case (5)(iii) with r = 4
if we interchange the third and fourth summands.
Lemma 10.14. No configurations for case (5)(ii) with r = 4 can occur.
Proof. When r = 4 we no longer have d3 = 2, but the nullity condition for c̄ implies that
Hence either {d3, d4} is one of {3, 3}, {3, 4}, {3, 5}, {3, 6}, {4, 4} or one of d3 or d4 is 2. Using this
together with the nullity conditions for ā, c̄ and the orthogonality conditions, we see that we only
need to consider u = (0,−2, 1, 0), (0, 0, 1,−2), (−1, 1, 0,−1) and (−1,−1, 1, 0).
(A) Let u = (0,−2, 1, 0). From the nullity condition for c̄ we deduce that d4 = 2, d2 ≥ 13, and
d3 ≥ 3. Now
(ai/di) = (1− β, 2β − 1−
2d2β−(d2+1)
(2d2+d3+2)β−(d2+1)
n−d2−2 )
where β := 1 + c1
d2+4d3+2d
d2d3(d2+2d3+4)
. We find that 11
< β < 1
and so a1
The above facts imply that ∆ā is again in case (5)(ii), and for (−2i, 1j) associated to an element
of ā⊥ we must have i = 4. If j = 1, 2, Eq.(10.1) leads to a contradiction to the above dimension
restrictions. The case j = 3 can be eliminated using the method of part (B) in the proof of Lemma
10.12.
(B) Next let u = (0, 0, 1,−2). The nullity condition for c̄ implies that d3 = 2 and d4 > 32d2+14 ≥
46. Now (ai/di) is given by
CLASSIFICATION OF SUPERPOTENTIALS 53
(1− β, 1− 2β, d4−d2+1
+ (2d2+2−d4
)β, − 4
2(d2+2)β−(d2+1)
where β := 1 + c1
d24+2d2d4−16
2d4(2d2+d4+6)
. One easily sees that 1
< β < 1, so that 0 < a1
. Since
d4 > 2d2 + 2 we obtain 0 <
Therefore, ∆ā is in case (5)(ii) and for (−2i, 1j) associated to an element of ā⊥ we must have
(by Lemma 10.5) i = 2 and so d2 = 2. Putting this value of d2 into the nullity condition for c̄ gives
a cubic equation in d4 with no integral roots, a contradiction.
(C) Consider now u = (−1, 1, 0,−1). From the nullity condition for c̄ we deduce that d2 ≥
4, d3 = 2, d4 ≥ 3 and 4d2 > 3d4. Also,
(ai/di) = (−β,
+ 1− 2β,
(2d2+2−d4)β−(d2+1)
, − 2
2(d2+2)β−(d2+1)
where β := 1 + c1
2d2+2d4+d
d2d4(2d2+d4+6)
. It follows that 1
< β < 3
and a2
Now we see that ∆ā is either in case (1) or (4) or (5)(ii). In the first two instances, by Lemmas
10.11, 10.7 d is a permutation of (4, 2, 2, 9). Since 3d4 < 4d2 we have d2 = 9, d4 = 4. But then the
null condition for c̄ is violated. So we are in case (5)(ii). For (−2i, 1j) associated to an element of
ā⊥, as di = 2, we have i = 1 or 3.
If i = 1, then Eq.(10.1) becomes
= 1 − 2β < 0, so j = 3, 4. When j = 3 the value of β given
above together with Eq.(10.1) imply that d4 = 3 or 4. But then the null condition for c̄ is violated.
For j = 4 we may use the argument of part (B) of the proof of Lemma 10.12.
If i = 3, using β > 1
in Eq.(10.1), we see that j = 2, 4. In either case, applying our bounds for
the dimensions in Eq.(10.1) lead to contradictions.
(D) Let u = (−1,−1, 1, 0). The null condition for c̄ implies that d4 = 2, d3 ≥ 3, d2 ≥ 5. With
β := 1 + c1
, we have
(ai/di) = (−β, 1 − 2β −
2d2β−(d2+1)
(2d2+d3+2)β−(d2+1)
One computes that β = 1
2d2+d
3+2d3
d2(3d
3+2d2d3+6d3)
, and from the dimension bounds one gets 5
≤ β <
. ∆ā cannot be in case (1) or (4), otherwise as d2 ≥ 5, we must have d2 = 9, d3 = 4, and the null
condition for c̄ is violated. So ∆ā is in case (5)(ii). For (−2i, 1j) associated to an element of ā⊥,
we must then have i = 1, 4.
If i = 1, then Eq.(10.1) is
= 1 − 2β > 0, so j = 3 or 4. In either situation, we may apply
the argument of part (B) of the proof of Lemma 10.12 to get a contradiction. If i = 4, Eq.(10.1)
together with the dimension bounds above show first that we can only have j = 3. In that case, a
more detailed look at Eq.(10.1) leads to a contradiction.
We can summarise our discussions thus far by
Theorem 10.15. Let r ≥ 4 and K be connected. Suppose that we are not in the situation of
Theorem 3.14. Assume that c̄ ∈ C is a null vector such that ∆c̄ has the property that there is
a unique vertex of type (1B) and all other vertices are of type (1A). Then the only possibilities
are given by Lemmas 10.7/10.11 and 10.9, up to interchanging ā and c̄ and a permutation of the
irreducible summands.
We will now sharpen the above Theorem using Proposition 3.7.
Corollary 10.16. Let r ≥ 4. Assume that K is connected and we are not in the situation of
Theorem 3.14. Then the possibilities given by Lemmas 10.7 and 10.9 (and hence Lemma 10.11)
cannot occur.
Proof. We will discuss the r = 4 case (i.e. that in Lemmas 10.7, 10.11) in detail and leave the
details of the r = 5 case (from Lemma 10.9) to the reader, as the arguments are very similar.
In the r = 4 case, first observe that C has exactly two null vectors, c̄ and ā in the notation of
Lemma 10.7, as the entries of a, c are determined by the vector d of dimensions. Hence, by Prop.
3.3, these are the only elements of C outside conv(1
(d+W)).
54 A. DANCER AND M. WANG
Since (−14) is a vertex of W, all type II vectors in W must be zero in place 4. As (1,−1,−1, 0)
is associated to an element of c̄⊥, it, together with (−1, 1,−1, 0), (−1,−1, 1, 0) are the only type II
vectors in W.
Next we analyse vectors in W and see if they are associated to elements of C, this last property
being important for applying Prop 3.7. Recall that (1,−2, 0, 0), (1, 0,−2, 0), (−2, 1, 0, 0), (−2, 0, 1, 0)
must be in W. The first two give elements of c̄⊥, the last two give elements of ā⊥.
First consider v = (1,−2, 0, 0). Now v̄ is a vertex of conv(1
(d + W)). By the superpotential
equation, d+ v = 2v̄ can be written as c̄(α)+ c̄(β), with c̄(α), c̄(β) ∈ C. Since v̄ is a vertex, every such
expression must involve ā or c̄, unless it is the trivial expression v̄ + v̄ and v̄ ∈ C. By computing
2v−a, 2v−c we find that these cannot lie in conv(W) (it is enough to exhibit one component < −2
or > 1). Thus v̄ ∈ C. Now an analogous argument shows that if w = (0,−2, 1, 0) is in W then
w̄ also lies in C. But vw is an edge of conv(W) with no interior points in W. So Prop 3.7 gives
= 4J(v̄, w̄) = 0, a contradiction to d2 = 2. Hence w /∈ W. Similarly we see (0,−2, 0, 1) /∈ W.
Next consider z = (−1,−1, 1, 0). By Remark 1.2(e) and the above, z is a vertex of conv(W). As
above we can show that z̄ ∈ C. Now v, z are the only elements of the face {x1+2x2 = −3}∩conv(W)
(cf. proof of Prop. 4.3 in [DW4]). So applying Prop 3.7 to vz we obtain 0 = 4J(v̄, z̄) = 1
contradiction.
To handle the r = 5 case (from Lemma 10.9), first observe that null elements of C must have
entries 2
in two of the places 1, · · · , 4 and −2
in the other two places. For a, c as in Lemma 10.9,
we can take (1,−2, 0, 0, 0), (1, 0,−2, 0, 0), (0, 1, 0,−2, 0), (−2, 1, 0, 0, 0) to lie in W. The argument
above to show that v̄ is in C still works for such type III vectors v. As above, we can use Prop 3.7
to show the other type III vectors (−2i, 1j) (i ≤ 4) do not lie in W; hence the ā, c̄ of Lemma 10.9
are the only null elements of C.
Let z = (−1, 1,−1, 0, 0) (it lies in W since (1,−1,−1, 0, 0) is associated to an element of c̄⊥) and
v = (1, 0,−2, 0, 0). As above we find z̄ is in C, and the arguments of Prop 4.3 in [DW4] show vz is
an edge of conv(W). A contradiction results as above from applying Prop 3.7 to vz.
The discussion at the beginning of this section now tells us that if K is connected and r ≥ 4 the
only case when we have a superpotential of the kind under discussion is that of Theorem 3.14. The
proof of Theorem 2.1 is now complete.
Concluding remarks.
1. When r = 2, then c is collinear with the elements of W. In other words, the projected
polytope ∆c̄ reduces to a single vertex, which must be of type (2). The possible elements of W are
(−2, 1), (−1, 0), (0,−1), (1,−2). If W has just two elements then Theorem 9.2 tells us we are either
in the situation of Theorem 3.14 (the Bérard Bergery examples), or in Example 8.2 or the third case
of Example 8.3 in [DW4]. In fact one can show that this last possibility can be realised in the class
of homogeneous hypersurfaces exactly when (d1, d2) = (8, 18). An example for these dimensions is
provided by G = SU(2)9 ⋉ Sym(9) (where Sym(9) acts on SU(2)9 by permutation) and K is the
product of the diagonal U(1) in SU(2)9 with Sym(9). The arguments of [DW2] show that this in
fact gives an example where the cohomogeneity one Ricci-flat equations are fully integrable.
If W has three elements, we may adapt the proof of Theorem 3.11 to derive a contradiction.
Here the essential point is that whenever we had to check that a sum of two elements of C does not
lie in d+W, such a fact remains true because the interior point of vw is the midpoint.
If W contains all four possible elements, then k ⊂ g is a maximal subalgebra (with respect to
inclusion). We suspect that this case also does not occur. In any event, it is of less interest because
the only way to obtain a complete cohomogeneity one example is by adding a Z/2-quotient of the
principal orbit as special orbit.
2. The only parts of this paper which depend on K being connected (or slightly more generally,
on the condition in Remark 2.4) are parts of §5, Case (ii) of §9, and all of §10. To remove this
CLASSIFICATION OF SUPERPOTENTIALS 55
condition, the main task would be generalizing Theorem 5.18 by getting a better handle on the
type II vectors associated to (1A) vertices (cf Lemmas (5.6)-(5.8)).
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Jesus College, Oxford University, Oxford, OX1 3DW, United Kingdom
E-mail address: dancer@maths.ox.ac.uk
Department of Mathematics and Statistics, McMaster University, Hamilton, Ontario, L8S 4K1,
Canada
E-mail address: wang@mcmaster.ca
http://www.maths.ox.ac.uk/
http://www.math.mcmaster.ca/mckenzie
	0. Introduction
	1. Review and notation
	2. The classification theorem and the strategy of its proof
	3. Projection onto a hyperplane
	4. The sign of J(, )
	5. Vectors orthogonal to a null vertex
	6. Adjacent (1B) vertices
	7. More than one type (2) vertex
	8. Adjacent (1B) vertices revisited
	9. Type (2) vertices
	10. Completing the classification
	References
