Perfect Cuboids and the Box Variety
Ernst Kani
Queen’s UniversityQuébec-Maine Conference, Laval University
27 Sept 2014

Outline
1. Introduction
2. Early History
3. New Ideas (using Arithmetic Geometry)
4. The Bombieri-Lang Conjecture
5. Further Results
6. Diagonal Quotient Surfaces
7. Modular Correspondences
8. Mazur’s Question

1. Introduction
I Consider a rectangular box (also known as cuboid):

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wazxb c

I By Pythagoras we have the following relations:(1)a2 + b 2 = x 2(2)b2 + c 2 = y 2(3)

a2 + c 2 = z 2(4)

a2 + b 2 + c 2 = w 2

1. Introduction – 2I A rational cuboid is a solution of (1) – (3) witha, b, c, x, y , z ∈ Q+ .
I A perfect cuboid is a solution of (1) – (4) with a, . . . , w ∈ Q+ .

• Problem 1: (Sanderson, 1740, Euler 1770) Find(parametric families of) rational cuboids.
• Open Problem 2: Are there any perfect cuboids?
• Open Problem 3: Are there at most finitely many perfectcuboids?

2. Early History• P. Halcke, 1719: observed that (a, b, c) = (44, 240, 117)
defines a rational cuboid. (This is the smallest one!)• Sanderson, 1740, Euler 1770: Using pythagorean triples,Sanderson and Euler gave in their respective texts “Elementsof Algebra” a systematic method for constructing rational
cuboids. These are now called Euler cuboids; cf. Euler,Elements of Algebra, II, Art. 238 (p. 443).
• A. Martin, 1894: In L’intermediaire des mathématiciens vol. 1(1894), p. 214, Artemas Martin (Washington) posed Problem
2 as Question 361. A solution was offered by Brocard (1895),
which was criticized by Tannery (1896), as is mentioned in L.
Dickson’s History of Number Theory, vol. II, ch. XVII.

2. Early History - 2• H. Olsen, 1916: Problem 254 of the American Mathematical
Monthly 23 (1916) (proposed by H. Olsen, Chicago) requires
you to find all perfect cuboids.
V. Spunar, 1917 submits a “solution” (which is published)that no such cuboids exist. (This solution is mentioned but is
not criticized by Dickson, although it should be.)• M. Kraitchik, 1954: establishes some congruence conditions
for the sides a, b, c of a perfect cuboid.
• J. Leech, 1977: studies cuboids for which a, b, c and only 3 ofx, y , z, w are rational (suggested by M. Gardner, 1970).
• I. Korec, 1992: Using his earlier results and a computer search,
Korec shows that for a perfect cuboid max(a, b, c) > 4 × 109 .

3. New Ideas (using Arithmetic Geometry)
I Question: Is there a geometric reason why it is easy to findlots of rational cuboids but difficult to find perfect ones?
• A. Bremner, 1988: studies the projective surface V ⊂ P5defined by equations (1) – (3). Thus, the rational points
V (Q) of V (with abc 6= 0) correspond to rational cuboids.
He constructs another surface W which is birationallyequivalent to V , and classifies all curves of degree ≤ 3 on W .
He mentions that “the surface W has a superabundance ofrational curves lying upon it.”• F. Beukers, van Geemen, < 2000: In their (unpublished)
preprint, they show that V is birationally equivalent to an
elliptic surface V 0 fibered over P1 . The sections of thisfibration give rise to infinitely many rational curves on V 0 andhence on V .

3. New Ideas (using Arithmetic Geometry) - 2• R. van Luijk, 2000: In his (unpublished) thesis, [vL] studies
the box variety (the name is due to [FS]) B ⊂ P6 defined byequations (1) – (4), whose rational points B(Q) (withabc 6= 0) correspond to perfect cuboids.
He proves that B ⊗ C is a normal surface with 48 singularities,
and that its desingularization B̃ is of general type.
He also finds a set L of 92 curves of genus ≤ 1 on B ⊗ C :– 24 curves/Q isomorphic to P1 (the components of abc = 0,)– 8 curves/Q(i), isomorphicto P1 ,√– 60 elliptic curves/Q( 2, i), none defined over Q.
• Thus: The geometry of the underlying surfaces V and B isradically different.

4. The Bombieri-Lang Conjecture
I Question: What special diophantine properties do varieties ofgeneral type have?
• Theorem of Faltings, 1983: If C /K is a curve of general type(⇔ gC ≥ 2) over a number field K , then C (K ) is finite.
• Bombieri-Lang Conjecture: If X /K is a variety of general type,then X (K ) is not Zariski-dense in X . (K a number field.)• Lang Conjecture (LC): If X /K is a variety of general type,then there is a proper closed subset E (X ) ⊂ X such thatU(X ) := X \ E (X ) has finitely many K 0 -rational points forevery number field K 0 /K .
• Remark: If (LC) is true for a surface X /K , then we must
have: E (X ) = union of all genus 0 and 1 curves on X ⊗ K .
Since E (X ) is supposed to be a closed set, this implies:

4. The Bombieri-Lang Conjecture - 2• Geometric Lang Conjecture (GLC): A surface X /K of general
type contains at most finitely many curves of genus ≤ 1.
• Theorem of Bogomolov, 1977: (GLC) is true for a smoothsurface X /C, provided that c12 (X ) > c2 (X ).
• Difficulties: 1) How can we determine the exceptional set
E (X )? Even in the situation of Bogomolov, there is noalgorithm for determining E (X ).
2) The desingularization B̃C of the box variety does not satisfy
Bogomolov’s hypothesis. (Here c12 (B̃C ) = 16, c2 (B̃C ) = 80.)

5. Further Results- all unpublished, but available on arXiv [math.AG].
• M. Stoll, D. Testa, 2010: study the box variety B and its
desingularization B̃ in detail. For example:– they compute all the geometric invariants of B̃C = B̃ ⊗ C,
– they determine Aut(B̃C ) (so | Aut(B̃C )| = 1536 = 29 · 3),– they prove that the Picard group Pic(B̃C ) = NS(B̃C ) ' Z64
and is generated by 140 curves: the 92 curves found by [vL],
and the 48 exceptional curves which resolve the 48
singularities.
• Question/Conjecture[ST]: Is E (BC ) = L? Is every curve ofgenus ≤ 1 on BC one of the 92 curves found by [vL]?
• Note: Conjecture[ST] + Lang’s Conjecture (LC) (+ [vL])
imply that Problem 3 has a positive answer: there are only
finitely many perfect cuboids.

5. Further Results - 2• A. Beauville, 2013: gives a more intrinsic construction of thebox variety BC . This implies:
BC is a diagonal quotient surface!
• E. Freitag, R. Salvati Manni, 2013: give an analytic and a
modular description of BC , Aut(BC ) and of L.
1) They construct BC as an (explicit) quotient of H∗ × H∗ ;2) They show that (an open part of) B ⊗ Q(i) has a modularinterpretation;3) All α ∈ Aut(BC ) are modular (i.e., induced by Γ(1) × Γ(1));4) All curves in L are modular (or cuspidal).
In particular: B ⊗ Q(i) is a (generalized) modular diagonalquotient surface.

6. Diagonal Quotient Surfaces
I Let: X be a smooth, projective curve over a field K ,G ≤ Aut(X ) a finite group of automorphisms of X ,π : X → X := G \X , the quotient map,
Y := X × X , the product surface,∆G = {(g , g ) : g ∈ G } ≤ G × G , the diagonal subgroup,
ZG := ∆G \Y , the diagonal quotient surface defined by G ,φ = φG : Y → ZG , the associated quotient map,ψ = ψG : ZG → Y := X × X , the induced map.
I Remarks: 1) We have that ψ ◦ φ = π × π :φψπ × π : Y = X × X → ZG → Y = X × X .
Thus, φ and ψ are finite of degree deg(φ) = deg(ψ) = |G |.
2) In general, ZG has finitely many quotient singularities. The
structure of ZG and of its desingularization Z̃G can be workedout; cf. K.-Schanz, 1997.

6. Diagonal Quotient Surfaces - 2• Application: We apply this to:
X := X (8) = Γ(8)\H∗ , the modular curve of level 8,
G = Γ(4)/Γ(8) ' (Z/2Z)3 , so
X = X (4) ' P1 .
Note: gX = 5, and X and G are defined over Q.
• Theorem 1: (∼[FS]) Let ZG /Q be the diagonal quotient
surface of X = X (8)/Q and G = Γ(4)/Γ(8). Then(5)B ⊗ Q(i) ' ZG ⊗ Q(i).

• Key Idea: [FS] use the fact that there is an isomorphism
X (8) ' Proj(A),where A := C[ϑ00 (z), ϑ10 (z), ϑ01 (z), ϑ00 (2z), ϑ10 (2z)]) is thegraded ring generated by the classical theta-functions ϑa,b (ofweight 12 ).

6. Diagonal Quotient Surfaces - 3
I Recall that the theta functions ϑa,b are defined byϑa,b (z) =∞X2e πi((n+a/2) z+b(n+a/2)) .

n=−∞and satisfy the classical theta-relations
ϑ200 (z) = ϑ200 (2z) + ϑ210 (2z)ϑ201 (z) = ϑ200 (2z) − ϑ210 (2z)ϑ210 (z) = 2ϑ00 (2z)ϑ10 (2z).
[FS] also work out the action of the generators
T = 10 41 , T 0 = 14 01 , R = 85 138
of G = Γ(4)/Γ(8) on the basis ϑ00 (z), ϑ10 (z), . . . , ϑ10 (2z).
This allows them them to compute the ring of invariants ofA ⊗ A with respect to ∆G . (They use instead a group∆(4, 8), but this leads to the same ring of invariants.)

7. Modular Correspondences
I Question: What is special about a product of modular curves?
I Partial answer: It comes equipped with a rich supply ofcurves, the modular correspondences. (→ Hecke operators.)• Construction: Let α ∈ GL+2 (Q) ∩ M2 (Z), and letT (α) = {(z, α(z)) : z ∈ H∗ } ⊂ H∗ × H∗
be its graph. Then for any Γ = Γ(N), its imageTΓ (α) ⊂ X (N) × X (N) = Γ(N)\H∗ × Γ(N)\H∗
is an irreducible (algebraic) curve on X (N) × X (N).
• Example: For Γ = Γ(1), each modular correspondence is of1 0
the form TΓ (αn ), for some n ≥ 1, where αn = 0 n .
Moreover, TΓ (αn ) ∼ X0 (n).

7. Modular Correspondences - 2• Theorem 2: ([FS]) Via the isomorphism (5), each curve in L
is either the image of modular correspondence T (α) (withdet(α) = 1) or the image of cuspidal curve {c} × H∗ orH∗ × {c}, where c ∈ H∗ \ H is a cusp.
The following is a partial converse:• Theorem 3: If C is curve of genus ≤ 1 on BC ' ZG ⊗ C which
is modular (i.e., C = φ(TΓ(8) (α)), for some α), then C ∈ L.
• Thus: The Conjecture [ST] (i.e., E (B) = L) ⇔(*) every curve in E (B) is modular or cuspidal.

8. Mazur’s Question• Mazur, 1978: To what extent are the isogeny classes of
elliptic curves E /Q determined by their mod N Galois
representations?
• Frey (1985), Darmon (1994): Formulated conjectures whichmake this question much more precise.
• Remark: To study isomorphisms of mod N Galois
representations, one is naturally led to consider the modular
diagonal quotient surfaceZN = ZX (N),GN , where GN = Γ(1)/(±Γ(N)),
because the points of ZN “classify” such isomorphisms.
• Conjecture (1995): If N = p is a prime with p ≥ 23, thenevery curve C in E (Zp ) is modular.
• Note: This conjecture + (LC) ⇒ Darmon’s Conjecture.

8. Mazur’s Question - 2• Theorem 4 (Bakker/Tsimerman, 2013): The above conjectureis true for p >> 0.
• Remark: Unfortunately, their proof does not give any estimate
on how large p has to be.
I But suppose: that their methods could be refined to give aproof of the above conjecture. Then it might be possible toalso prove a similar statement for ZG ⊗ C = BC , i.e. to provecondition (*). Then we would have:I (LC) ⇒ #(perfect cuboids) < ∞ (Problem 3)