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.
Basic Observation: The irreducibility depends only on the nature of the quadratic form
qE1,E2 (f ) = deg(f ) on Hom(E1, E2) ' Zr .
Here r = 2 if E1 has CM and otherwise r = 1.

2Notes: 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.
By using deep results in number theory (due to Chowla andHeilbronn), it is possible to prove:
Theorem 1: There exist only finitely many equivalence classesof positive definite quadratic forms q in r ≤ 2 variables suchthat q ∼ qE1,E2 , for some reducible pair (E1, E2).
Problem 1a: Classify the (finitely many) “exceptional” quadratic
forms of Theorem 1.
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.

3

2. A Conjecture of Gauss√
Recall: If K = Q( −d) is an imaginary quadratic field withhK = 1, then d ≤ 163 (provided that d is squarefree).
– this was “conjectured” by Gauss (1801)– the fact that d is bounded was proved by Heilbronn (1934)– the conjecture was proved by Heegner (1952), Stark (1967),. . .
However: the above conjecture is only a portion of what he actually conjectured in Article 303 of the Disquisitiones Arithmeticae. Translated to number fields, his conjecture is:√If K = Q( −d) is an imaginary quadratic field whoseclass group Cl(OK ) is an elementary abelian 2-group, thend ≤ 5460.
– the fact that d is bounded was proved by Chowla (1934) byextending Heilbronn’s method.
– 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 counterexample (this requires Lehmer’s
computations that the conjecture is true for d < 2.1 × 1011)
2) GRH (Generalized Riemann Hypothesis) ⇒ there are nocounterexamples, i.e. the conjecture is true.
Note: Chowla is the only person who mentions that this is (essentially) a conjecture of Gauss.

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Conjecture of Gauss: If q is a primitive, positive definite binary quadratic form of discriminant ∆(q) = −4D, thenc(q) = 1 ⇔ D is one of the 65 idoneal numbers of Euler.
Here (cf. Watson), c(q) is the class number of the form q, i.e.
c(q) = #(equivalence classes of forms in the genus of q).
Remarks: Watson studied in 1965-80 the “c(q) = 1” problem forr ≥ 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.
Theorem 2 (Non-CM Case). If r = 1, then there are either 21
or 22 exceptional forms q(x) = dx2. If Gauss’s Conjecture (orif GRH) is true, then q is exceptional ⇔ either d = 1 or d isone of the 20 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}.

Moreover, to each such d belongs an infinite family of pairs(E1, E2); these are parametrized by the (non-CM) points ofthe modular curve X0(d).
Theorem 3 (CM Case). If r = 2, then there are precisely 15
exceptional forms, and these come from 46 (distinct) pairs ofCM-curves (E1, E2).
Note: If we restrict attention to those CM-curves for which End(Ei)is a maximal order, then there are only 4 pairs of curves/forms,
as was proved by Hayashida and Nishi (1965).

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3. The Refined Humbert Invariant
Aim: 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(1)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 ellipticcurves 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.

6Consequence: 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(2) (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 (2), 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 (1) 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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4. Gauss’s Problem: A Generalization
Note: As was mentioned above, one such generalization was studied (and solved for r = 3) by Watson:
Classify the positive definite forms q with c(q) = 1.
Here is another generalization:
Problem 2. Classify the positive definite quadratic forms q inr ≥ 2 variables which satisfy the property:(3)q 0 → 1,for all q 0 ∈ gen(q),where gen(q) denotes the genus of q, i.e. the set of forms which
are genus-equivalent to q.
Remarks: 1) Clearly, if q → 1 and c(q) = 1 ⇒ (3) holds. Thus,
the solutions of Problem 2 include the solutions q of Watson’s
Problem with 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∆).

85. 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.
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) By Gauss we know 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 2. 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).

96. 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 (3) 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), oneproves 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, so
c = 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

1
1− 2p

Yp|kd0

 0   d 1−4k 2d0 11+1+p p
ppwhere k = cont(q), d0 = kd2 , δ = gcd(4k 2, d0), etc. andX
1M (1q ) =.
| Aut(f )|f ∈gen(1q )/∼1For each q ∈ L one calculates that M (1q ) = | Aut(q)|, and soc(1q ) = 1.

11This proves the Main Theorem and hence also the first part ofTheorem 3. For the second part, use:
Proposition 3. Let E1 ∼ E2 be two elliptic curves with CM by
the imaginary quadratic field k, so End(Ei) is an order in kof discriminant Di = fi2dk , where dk is the discriminant of k.
Thendisc(qE1,E2 ) = −lcm(D1, D2),lcm(D1, D2)
cont(qE1,E2 ) =.
gcd(D1, D2)
2Remark: From the above we see that for a given binary formq, there are only finitely many pairs (E1, E2) of elliptic curves
such that qE1,E2 ∼ q. These can be found precisely by using anexplicit formula for qE1,E2 in terms of the ideals “defining” theEi relative to a common curve E (i.e. such that Ei = E/Ii).

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7. Connection with Moduli Spaces
Let A2 denote the moduli space of princ. pol. abelian surfaces,
M2 the moduli space of smooth genus 2 curves;
M2 ⊂ A2 via the Torelli map: C 7→ (JC , θC ).
Definition: Let q be a (positive definite) quadratic form. The
generalized Humbert variety associated to q is the subset of A2
defined byH(q) = {(A, θ) ∈ A2(K) : q(A,θ) → q}
Here q(A,θ) is the refined Humbert invariant of (A, θ), and“q(A,θ) → q” means that q(A,θ) primitively represents the formq.
Examples: 1) The classical Humbert surface of discrim. ∆ is(4)H∆ = H(∆x2),

as was explained in [ECAS]. Thus, by the Key Lemma we have(5)H 1 = A2 \ M 2 ,which (for K = C) is a theorem of Biermann(1886) and Humbert(1901).
2) If m and n are distinct positive integers, then[Hm ∩ Hn =H(q),where the union runs over the (finitely many) equivalence classesof positive definite binary quadratic forms q with q → m andq → n.

13Theorem 5: (a) Let q ∈ gen(1−16d) ∪ gen(4 · 1−4d) be a positive definite binary quadratic form of discriminant −16d inthe principal genus. If q 6→ 1, then H(q) is a curve lying
completely in M2, and every C ∈ H(q) has the property that
JC ' E1 × E2, for some elliptic curves E1 and E2.
(b) Conversely, if C is a curve with JC ' E1 × E2, for some
elliptic curves E1 and E1, then C ∈ H(q), for some binary
form q as in part (a).
(c) If q is in part (a), then there are morphisms µs : X0(d) →H(q) which are either of degree 2 or 4. Thus, H(q) ∼ X0(d)+or to a degree 2 quotient thereof by an Atkin-Lehner involution.
Moreover, the latter case occurs if and only if q is an ambiguousform.
Remark: In [Jacobians], a curve which appears in[T (d) =H(d),q∈gen(1−16d )∪gen(4·1−4d )is called a curve of type d. Thus, Theorem 2 ⇔Theorem 20: There is no curve of type d ⇔ d ∈ L∗.
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.