1Curves of Genus 2 anda Conjecture of Gauss
1. Introduction
Let E1 and E2 be two elliptic curves over K = K.
Question: Is there a (smooth, irreducible) genus 2 curve C on
the product surface E1 × E2?
Equivalent Question: Is there a curve C such that its Jacobian
JC is isomorphic to E1 × E2?
Definition: The pair (E1, E2) is called irreducible if such a curve
exists, and is called reducible if no such curve exists.
Problem 1: Classify the reducible pairs (E1, E2).
Remarks: 1) This problem was studied by:Hayashida(1965), Hayashida/Nishi(1965) → partial results
Ibukiyama/Katsura/Oort (1986) If E1, E2 are supersingular,then (E1, E2) is reducible ⇔ char(K) = 2 or 3.
2) If E1 is not isogenous to E2, then (E1, E2) is reducible.
Assume henceforth: E1 ∼ E2 and E1 is not supersingular.

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Basic Observation: The irreducibility of (E1, E2) depends only
on the nature of the quadratic form qE1,E2 on Hom(E1, E2)
which is defined byqE1,E2 (f ) := deg(f ) for f ∈ Hom(E1, E2) ' Zr .
Here r = 2 if E1 has CM and otherwise r = 1.
Notes: 1) Thus, by choosing a basis of Hom(E1, E2), the map
qE1,E2 defines an equivalence class of positive definite quadratic
forms in r ≤ 2 variables.
2) Conversely, it can be shown that every positive definite
quadratic form q in r ≤ 2 variables is equivalent to qE1,E2 ,for some pair (E1, E2) of elliptic curves.
Due to this observation, we can split Problem 1 into two subproblems:Problem 1a: Classify the “exceptional” quadratic forms q suchthat q ∼ qE1,E2 , where (E1, E2) is a reducible pair (i.e. there
is no genus 2 curve on E1 × E2).
Problem 1b: For each exceptional quadratic form q, classify thepairs (E1, E2) of elliptic curves with q(E1,E2) ∼ q.
Note: While Problem 1b is relatively simple, Problem 1a is quitedifficult, for it is closely connected to a Conjecture of Gauss.

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2. Main Results
Theorem 1 (Non-CM Case). (a) The form q(x) = dx2 is exceptional if and only if d 6≡ 3(4) and 4d is an idoneal (or suitableor convenient) number.
Thus, if Gauss’s Conjecture (or if GRH) is true, then q is
exceptional ⇔ either d = 1 or d is one of the 20 known idoneal
numbers d ≡ 2, 4, 6 (mod 8) ⇔ d ∈ L :={1, 2, 4, 6, 10, 12, 18, 22, 28, 30, 42, 58, 60, 70, 78, 102, 130, 190, 210, 330, 462}.

(b) Given an integer d ≥ 1 (exceptional or not), the pairs(E1, E2) with qE1,E2 ∼ dx2 are parametrized by the (non-CM)
points of the modular curve X0(d).
Remark: It follows from the work of Weinberger (1973) that
there are either 21 or 22 exceptional q’s.
Theorem 2 (CM Case). If r = 2, then there are precisely 15
exceptional forms, and these come from 46 (distinct) pairs ofCM-curves (E1, E2).
Remarks: 1) If we restrict attention to those CM-curves forwhich End(Ei) is a maximal order, then there are only 4 pairsof curves/forms, as was proven by Hayashida and Nishi (1965).
2) The proof of Theorem 2 uses Theorem 1 and Weinberger’s“at most one more” theorem, but does not involve any unproven hypotheses.43. A Conjecture of Gauss
Definition: An idoneal√ number is an integer d ≥ 1 such thatCl(−4d) = Pic(Z[ −d]) is an elementary abelian 2-group.
Note that by Gauss’s genus theory we have thatd is idoneal⇔c(x2 + dy 2) = 1,

where c(f ) denotes the number of equivalence classes of forms
in the genus of f . (Watson calls c(f ) the class number of f .)Notes: 1) The term idoneal was introduced by Euler (1772), whoused such numbers to find large primes. It is a non-trivial
fact that his definition is equivalent to the one above; cf. Cox’s
book, Primes of the form x2 + ny 2, p. 61.
Conjecture (Gauss, DA, Art. 303): d is an idoneal number
if and only if it is one of the 65 numbers found by Euler. In
particular, if d is idoneal, then d ≤ 1848.
– the fact that d is bounded was proved by Chowla (1934) byextending Heilbronn’s method (to show that h(−d) → ∞).
– in the 1930’s, the conjecture was studied by Dickson and hisstudents (e.g. N. Hall), who obtained useful partial results.
– Swift (1948): conjecture is true for d ≤ 107 (computations
were carried out using Lehmer’s linear congruence machine)– Weinberger (1973) proved:1) there is at most one fundamental counterexample (this requires Lehmer’s computations for d < 2.1 × 1011)
2) GRH (Generalized Riemann Hypothesis) ⇒ there are nocounterexamples, i.e. the conjecture is true.

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4. Gauss’s Problem: Generalizations
Note: As will be explained below, Problem 1a for non-CM curves
naturally leads to classifying idoneal numbers, i.e. to Gauss’sConjecture. For CM curves, however, we need a generalizationof this to ternary forms. One such generalization is:
Watson’s Problem: Classify the positive definite forms q withc(q) = 1.
In 1965-80, Watson studied this problem for r ≥ 3 variables(and stated that the case r = 2 is impossible):1) There exist only finitely many classes of positive definite
primitive forms with c(q) = 1 (and none for r ≥ 11).
2) For r = 3, ∃ precisely 790 classes of such forms.
Here is another generalization:
Problem 2. Classify the positive definite quadratic forms q inr ≥ 2 variables which satisfy the property:(1)q 0 → 1,for all q 0 ∈ gen(q),where gen(q) is the genus of q (i.e. the set of forms which are
genus-equivalent to q) and q 0 → 1 means that q represents 1.
Remarks: 1) Clearly, if q → 1 and c(q) = 1 ⇒ (1) holds. Thus,
the solutions of Problem 2 include those solutions q of Watson’s
Problem for which q → 1.
2) If r = 2, then Problem 2 is essentially equivalent to Gauss’s
Problem (or Conjecture) and to Watson’s Problem (becauseq → 1 ⇔ q ∼ 1∆).65. The Refined Humbert InvariantAim: Translate the existence of genus 2 curves into a problem
about quadratic forms.
Let A be an abelian surface (dim(A) = 2),NS(A) = Div(A)/ ≡ its Néron-Severi group.
Observation: If C ⊂ A is a (smooth) curve of genus 2, thenC 2 = 2 and so its class θC = cl(C) ∈ NS(A) is a principal
polarization on A.
The converse is false: not every θ ∈ P(A) := {principal polarizations on A} comes from an irreducible genus 2 curve.
Definition: The refined Humbert invariant of a principally polarized abelian surface (A, θ) is the (positive definite) quadratic
form qθ on NS(A, θ) := NS(A)/Zθ defined by(2)qθ (D) = (D.θ)2 − 2D2,for D ∈ Div(A).

Remark: In [ECAS] (1994) I showed how qθ is related to (andrefines) the classical Humbert invariant ∆(A, θ) ∈ N.
Key Lemma: Let θ ∈ P(A). Then θ = cl(C), for some (smooth)genus 2 curve C on A ⇔ qθ (D) 6= 1, ∀D ∈ Div(A).
Proof (Sketch) (⇐) If not, then by a theorem of Weil(1957),
θ = cl(D), where D = E1 + E2, and the Ei’s are elliptic
curves with (E1.E2) = 1. But then qθ (Ei) = 1, contradiction.
(⇒) If θ = cl(C) but qθ (D) = 1, then by [ECAS] we have thatD ≡ E1 and θ − D ≡ E2, where the Ei are elliptic curves.
Thus θ ≡ E1 + E2 6≡ C (by Riemann-Roch), contradiction.

7Consequence: The existence (or non-existence) of genus 2 curves
C on A can be translated to a problem about the quadratic
form qA associated to the intersection pairing on NS(A), i.e.
1 2qA(D) = D ,2

for all D ∈ NS(A).

Corollary: If A is an abelian surface, then there is no smoothgenus 2 curve on A if and only if(3) (qA)θ represents 1, for every θ ∈ NS(A) with qA(θ) = 1.
Note: If A = E1 × E2, thenqA ∼ xy ⊥ (−qE1,E2 ),where xy is the quadratic form defined by the hyperbolic planeand qE1,E2 is (as above) the quadratic form defined by thedegree map.
Definition: A positive definite quadratic form q is called exceptional if the form Q := xy ⊥ (−q) satisfies (3), i.e.
Qθ → 1 for all θ with Q(θ) = 1.
Here, following Watson, “q → 1” means “q represents 1”, andQθ is defined by replacing (the role of) qA in (2) by Q.
Note: By the above Corollary, this definition is consistent withthe previous use of the term “exceptional” (which was defined
only for the quadratic form qE1,E2 since its definition used ageometric property of E1 × E2).

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6. Exceptional Forms: the Case r = 1
Proposition 1: Let q(z) = dz 2, where d > 0, and put Q(x, y, z) =xy − dz 2. Then:(a) If d ≡ 3 (mod 4), then ∃θ with Q(θ) = 1 such that Qθ isnot primitive. In particular, Qθ 6→ 1, so q is not exceptional.
(b) If d 6≡ 3 (mod 4), then{Qθ : Q(θ) = 1} = gen(1−16d)
is the principal genus of discriminant −16d. Thus q is exceptional ⇔ c(1−16d) = 1 ⇔ 4d is an idoneal number.
Proof. Preprint [Jacobians] = Jacobians isomorphic to . . .
Corollary: The form dz 2 is exceptional ⇔d ∈ L∗ := {d ≥ 1 : c(1−16d) = 1 and d 6≡ 3(4)}.
Remarks: 1) It is easy to see (Gauss) that L ⊂ L∗, and that
equality holds if Gauss’s Conjecture is true.
If, however, there is a d∗ ∈ L∗ \ L, then d∗ ≡ 2, 4, 6 (mod 8)
and by Hall (1940) d∗ is squarefree. Thus −4d∗ is a fundamental discriminant, and then by Weinberger it is the unique(fundamental) counterexample to Gauss’s Conjecture. ThusL∗ = L ∪ {d∗} in this case.
2) This proves the first part of Theorem 1. The second part isessentially trivial, for if E1 has no CM, thenqE1,E2 ∼ dx2 ⇔ ∃h : E1 → E2, Ker(h) cyclic of degree d⇔ (h : E1 → E2) ∈ X0(d)(K).

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7. Exceptional Forms: the Case r = 2
Let q = (a, b, c) be a positive definite binary quadratic form, i.e.
q(x, y) = ax2 + bxy + cy 2,
d = b2 − 4ac its discriminantQ(x, y, z, w) = xy − q(z, w)
1q (x, y, z) = x2 + 4q(y, z)
Proposition 2. (a) If d ≡ 0 (mod 4) and q → a, where a ≡3 (mod 4), then there is a θ with Q(θ) = 1 such that Qθ is notprimitive. In particular, q is not exceptional.
(b) If d ≡ 1 (mod 4) or if q 6→ a, for any a ≡ 3 (mod 4), then{Qθ : Q(θ) = 1} ⊂ gen(1q )Thus, if c(1q ) = 1, then q is exceptional.
Main Theorem. If q is as in Proposition 2(b), then TFAE:(i) q is exceptional;(ii) 1q satisfies property (1) of Problem 2;(iii) c(1q ) = 1;(iv) q ∈ L := {k(1, 1, 1) : k = 1, 2, 4, 6, 10}∪ {k(1, 0, 1) : k = 1, 2, 6}∪ {(1, 1, 2), (1, 1, 4)}∪ {2(1, 1, c) : c = 3, 9}∪ {2(1, 0, c) : c = 2, 5}∪ {2(2, 0, 3)}.
Proof (Sketch). (iii) ⇒ (ii) ⇒ (i): trivial (by Proposition 2(b)).

10(i) ⇒ (iv): If q is exceptional, then using Proposition 1(b), one
proves that q satisfies:(i0) q → n, n < |d| ⇒ n ∈ L∗.
Using Weinberger’s result, this can be sharpened to(i00) q → n, n < |d| ⇒ n ∈ L.
Indeed, if q = (a, b, c) is (wlog) reduced, then by (i0) we havea, c, a + b + c ∈ L∗. But if c ∈ L∗ \ L = {d∗}, then a + b + c >/ L∗, contradiction. Thus, a, c ≤ 462, soc = d∗, so a + b + c ∈|d| ≤ 4 · 4622 < 106 < d∗, and hence (ii00) holds.
We therefore have only finitely many d’s to consider, and by asomewhat tedious argument (using (ii00)) we obtain that q ∈ L.
(iv) ⇒ (iii) For each q ∈ L, apply the mass formula of Eisenstein/Smith/Brandt to the ternary form 1q . This has the form
M (1q ) =
−kd0 Y6 · 2ν

p|δ

11− 2pYp|kd0

 0   d 1−4k 2d0 1
1+1+p ppp

where k = cont(q), d0 = kd2 , δ = gcd(4k 2, d0), etc. andX
1M (1q ) =.
| Aut(f )|f ∈gen(1q )/∼11For each q ∈ L one calculates that M (1q ) = | Aut(q)|= | Aut(1,q )|and so it follows that c(1q ) = 1.

118. References[ECAS] Elliptic curves on abelian surfaces. Manusc. math. 84(1994), 199–223.
[Jacobians] Jacobians isomorphic to a product of two elliptic
curves. Preprint, 39pp.