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Humbert surfacesDefinitionsAlgebraic models
Computing Humbert surfacesDegree formulaPower seriesLinear algebraRuntime analysisExample
Humbert intersectionsQuaternion ordersComputing Shimura curvesExample
The Siegel upper half plane
DefinitionThe Siegel upper half plane of degree g is
Hg = {τ ∈ Matg×g(C) | tτ = τ , Im (τ) > 0} .
I Each τ ∈ Hg corresponds to a PPAV Aτ/C with periodmatrix (τ Ig) ∈ Matg×2g(C).
I Aτ∼= Aτ ′ ⇔ ∃M =
(a bc d
)∈ Sp2g(Z) such that
τ ′ = M · τ := (aτ + b)(cτ + d)−1.
I Ag = Sp2g(Z)\Hg is a moduli space for dimension g PPAV’s.
I dimAg = 12g(g + 1). In particular, dimA2 = 3 and A2 is
called the Siegel modular threefold.
Extra endomorphisms
Let A be a PPAS (g = 2). Then End(A) is an order inEnd(A)⊗Q which isomorphic to one of the following algebras:
(0) quartic CM field
(1) indefinite quaternion algebra over Q(2) real quadratic field
(3) Q
The irreducible components of the corresponding moduli spaces inA2 which have “extra endomorphisms” are known as
(0) CM points
(1) Shimura curves
(2) Humbert surfaces
Humbert’s equation
Humbert showed that any
(τ1 τ2
τ2 τ3
)∈ A2 satisfying the equation
kτ1 + `τ2 − τ3 = 0
defines a Humbert surface H∆ of discriminant ∆ = 4k + ` > 0.
Example
H1 = Sp4(Z)\{(
τ1 τ3
τ3 τ3
)}= Sp4(Z)\
{(τ1 00 τ3
)}, the set of
abelian varieties which split as a product of elliptic curves.
Task: Find “useful” algebraic models for H∆.
Algebraic models
I Torelli says that the map C 7→ Jac(C) defines a birationalmap between M2, the moduli space of genus 2 curves and A2
(In fact M2∼= A2 −H1).
I The function field of A2 (and hence M2) is C(j1, j2, j3)where ji are the absolute Igusa invariants.
I There exists an irreducible polynomial H∆(j1, j2, j3) whosezero set is the Humbert surface of discriminant ∆.
Unfortunately, working with ji is impractical (enormous degrees,giant coefficients).Solution: add some level structure.
Algebraic modelsRunge’s model
Runge uses level Γ∗(2, 4) structure, with four theta functions:
θ
[a
(0, 0)
](2τ), a ∈ Z2/2Z2
where
θ
[m′
m′′
](τ) =
∑x∈Z2
e2πi
(12(x+m′
2)·τ ·t(x+m′
2)+(x+m′
2)·t(m′′
2))
are classical theta functions of half integral characteristicsdetermined by values m′,m′′ ∈ Z2/2Z2.
Algebraic modelsRosenhain model
We use level Γ(2)-structure with three functions
λ1(τ) =(
θ[ 0 00 0 ] θ[ 0 0
1 0 ]θ[ 0 0
1 1 ] θ[ 0 00 1 ]
)2
,
λ2(τ) =(
θ[ 0 01 0 ] θ[ 1 1
0 0 ]θ[ 0 0
0 1 ] θ[ 1 11 1 ]
)2
,
λ3(τ) =(
θ[ 0 00 0 ] θ[ 1 1
0 0 ]θ[ 0 0
1 1 ] θ[ 1 11 1 ]
)2
called Rosenhain invariants. These generate the function field ofA2(2) = Γ(2)\H2.
Relation to genus 2 curves
I Given a genus 2 curve C : y2 =∏6
i=1(x− ui) we can sendthree of the ui to 0, 1,∞ via a fractional linear transformationto get an isomorphic curve with a Rosenhain model
y2 = x(x− 1)(x− t1)(x− t2)(x− t3) .
The ti are called Rosenhain invariants.
I (0, 1,∞, t1, t2, t3) determines an ordering of the Weierstrasspoints and a level 2 structure on Jac(C) (∈ A2(2)).
I Let M2(2) denote the moduli space of genus 2 curvestogether with a full level 2 structure. Points of M2(2) aregiven by triples (t1, t2, t3) where ti 6= tj , 0, 1 for all i, j.
I The forgetful morphism M2(2) →M2 is a Galois covering ofdegree 720 = |S6| where S6 acts on the Weierstrass 6-tuple bypermutations, followed by renormalising the first three to(0, 1,∞).
Runge’s method
Let φ : A′ → A2 be a finite cover of A2. Then
φ−1H∆ =⋃finite
H(i)∆ .
Given functions {fi(τ)}i=1,...,n generating the function field of A′,
compute H(i)∆ (f1, . . . , fn) as follows:
1. Calculate the degree of the Humbert components H(i)∆ (given
by a formula).
2. Compute power series representations of the fi(τ) restrictedto H∆ ⊂ H2.
3. Solve H(i)∆ (f1, . . . , fn) = 0 in the power series ring (truncated
series with large precision) using linear algebra.
We shall consider the level 2 covering A2(2) → A2.
Step 1 - degree formula
Fortunately much arithmetic-geometric information is known aboutHumbert surfaces (van der Geer ’82). The number of Humbertcomponents in A2(2) is
m(∆) =
10 if ∆ ≡ 1 mod 815 if ∆ ≡ 0 mod 46 if ∆ ≡ 5 mod 8
(see Besser ’98).
The degree of any Humbert component H(i)∆ in A2(2) is given by a
recursive formula
a∆ =∑x>0
v(∆/x2)m(∆/x2) deg(H
(i)∆/x2
)where
v(x) =
1/2 if x = 11 if x ≥ 2, x ≡ 0, 1 mod 4.
0 otherwise
Moreover, a∆ is the coefficient of a certain modular form of weight5/2 for the group Γ0(4), which fortunately has a more elementarydescription due to a formula of Siegel:
a∆ − 24∑x∈Z
σ1
(∆− x2
4
)=
{12∆− 2 if ∆ = �
0 otherwise
Here are the degrees for small discriminants:∆ 1 4 5 8 9 12 13 16 17 20 21 24
deg(H(i)∆ ) 2 4 8 8 24 16 40 32 48 32 80 48
actual deg 1 2 8 8 16 16 40 24 48 32 80 48
Remarks
I When ∆ ≡ 0 (mod 4) we have
mRunge(∆) = 4×mRosenhain(∆)
⇒ degRunge(∆) =14× degRosenhain(∆) .
I In reality (after computing these equations) the actual degrees
of H(i)n2 (λj) are less than what the formula produces. For
example
H1 : {ei − ej = 0, i 6= j} ∪ {ei = 0} ∪ {ei − 1 = 0} .
Step 2 - power series
Write ∆ = 4k + ` where ` is either 0 or 1, and k is uniquelydetermined. The Humbert surface of discriminant ∆ can bedefined by the set
H∆ = Sp4(Z)\{(
τ1 τ2
τ2 kτ1 + `τ2
)∈ H2
}.
Restrict θ[
a bc d
]to H∆ to get a Laurent series
θ[
a bc d
](τ) =
∑(x1,x2)∈Z2
eπi(x1c+x2d)r(2x1+a)2+k(2x2+b)2q2(2x1+a)(2x2+b)+`(2x2+b)2
where r = e2πiτ1/8 and q = e2πiτ2/8.
Unfortunately q has negative exponents. Substitute r = pq to get∑(x1,x2)∈Z2
(−1)x1c+x2dp(2x1+a)2+k(2x2+b)2q(2x1+a+2x2+b)2+(k+`−1)(2x2+b)2
which is a power series with integer coefficients.Using this representation one can compute the restriction ofλ1, λ2, λ3 to a Humbert surface as elements of Z[[p, q]]/(pN , qN )fairly easily.
Step 3 - linear algebra
Let d = deg(H(i)∆ ). To find the algebraic relation H
(i)∆ :
I Compute all monomials of degree ≤ d in the variablese1, e2, e3.
I Substitute ei = λi(p, q) ∈ Z[[p, q]]/(pN , qN ) in eachmonomial.
I Use linear algebra to find linear dependencies between thepower series monomials pmqn (compute null space of a bigmatrix).
I With high enough precision there will be exactly one linearrelation between the monomials ei. This produces the
polynomial relation H(i)∆ (e1, e2, e3) = 0 which defines a
Humbert component.
I Once one component has been determined, the others caneasily be found by looking at the Rosenhain (S6) orbit of acomponent.
These other components will turn out to be useful when we look atShimura curves.
Runtime analysis
I There are:I
(d+33
)= O(d3) monomials to be evaluated
I O(N2) coefficients of evaluated power series expressions ofprecision N .
I Runtime cost is dominated by the nullspace calculation:O(d6N2) ≥ O(d9) to find a unique solution.
I Symmetries of the equation (arising from the fixed group ofthe humbert component) can be exploited to reduce thematrix size by a constant factor, giving a speedup by aconstant factor.
I Not overly efficient, but least it’s only a one time calculation..
Example
We calculate a component of H5:
λ1 = 1 + 16p4q8 + O(p12q12)λ2 = 1 + 4q4 + 8q8 − 8p4q4 − 24p4q8 + 4p8q8 + 48p8q8 + O(p12q12)λ3 = 1 + 4q4 + 8q8 + 8p4q4 + 40p4q8 + 4p8q8 + 48p8q8 + O(p12q12)
Using power series with precision 65, we compute the Humbertcomponent
e22e
23 − 2e
22e
33 + e
22e
43 + 2e1e2e
33 − 2e1e2e
43 − 2e1e
22e3 − 2e1e
22e
23 + 4e1e
22e
33 + 2e1e
32e3
−2e1e32e
33 + e
21e
43 − 2e
21e2e
33 + e
21e
22 + 4e
21e
22e3 − 4e
21e
22e
23 − 2e
21e
32 − 2e
21e
32e3
+4e21e
32e
23 + e
21e
42 − 2e
21e
42e3 + e
21e
42e
23 − 2e
31e
33 − 2e
31e2e3 + 4e
31e2e
23 + 2e
31e2e
33
−2e31e
22e
23 + 2e
31e
32e3 − 2e
31e
32e
23 + e
41e
23 − 2e
41e2e
23 + e
41e
22e
23
Application: Computing Shimura Curves
Quaternion ordersbare essentials
Let R be an order in an indefinite Q-quaternion algebra A.
I R is a QM-order if R = End(X) for some abelian surface X.
I Any x ∈ A satisfies x2 − tx + n = 0 where t, n are thereduced trace, norm respectively.
I ∆(x) = t(x)2 − 4n(x) defines a discriminant form
∆(x, y) =12(∆(x + y)−∆(x)−∆(y)) .
I The discriminant d(x1, . . . , x4) of a module generated byx1, . . . , x4 is defined to be the positive square root of
d(x1, . . . , x4)2 = −det(t(xixj)) .
Theorem (Runge ’99)
1. Any QM-order can be written as R = Z + Zα + Zβ + Zαβsuch that the discriminant matrix
S∆ =(
∆(α) ∆(α, β)∆(α, β) ∆(β)
)is positive definite. The discriminant of R equals det(S∆)/4.
2. A change of basis corresponds to changing the discriminantmatrix to tgS∆g for some g ∈ GL2(Z). ⇒ can assumediscriminant matrix is reduced.
3. If two orders have the same discriminant matrix which isprimitive (gcd(entries)=1) then the corresponding Shimuracurves are isomorphic.
Theorem (Hashimoto ’95, Runge ’99)
Let O = Z[ω] be a quadratic order of discriminant ∆. Let S∆ be adiscriminant matrix of a QM order R. The following areequivalent:
1. ∆ is primitively represented by S∆.
2. There exists an embedding O ↪→ R such that R ∩Q(ω) = O.
3. A Shimura curve C with QM order R is contained in H∆.
If we work in a finite cover, we have
C(h) ⊂ H(i)∆(α) ∩H
(j)∆(β)
if and only if we can write
tgS∆g =(
∆(α) ∗∗ ∆(β)
)for some g ∈ GL2(Z).
Example
H5 ∩H8 contains four Shimura curves CS :
Discriminant matrix QM-order discriminantS = det(S)/4(
5 00 8
)10(
5 22 8
)9(
5 44 8
)∼
(5 11 5
)6(
5 66 8
)∼
(1 00 4
)4
I This intersection was first computed by Hashimoto andMurabayashi (1995).
Computing intersections
I For intersections of Humbert components H∆i(e1, e2, e3) wecan find plane affine models simply by taking resultants withrespect to e1.
I As we are working with coordinates in M2(2) = A2(2)−H1,we will not be able to compute any Shimura curves in H1.
I S6 acts on Humbert components, hence acts on theirintersections producing isomorphic curves.
I Take one curve from each S6-orbit. Each of theseintersections is a component of a Shimura curve CS for somediscriminant matrix S.
Example
In our H5 ∩H8 example, there are three non-equivalentintersections:
C1 : a genus 1 curveC2 : a genus 3 hyperelliptic curveC3 : a genus 3 non-hyperelliptic curve
and the Shimura curves in M2(2) are
C( 5 00 8 ), C( 5 2
2 8 ) and C( 5 11 5 )
so there is a one-one correspondence between the Ci and the CS ,to be determined.
Look at other Humbert intersections with “related” discriminants.Write D(a, b) for the set of discriminant matrices of QM-orders ofShimura curves in Ha ∩Hb. We have
D(5, 5) = { ( 5 11 5 ), ( 4 2
2 5 )}D(4, 5) = { ( 4 0
0 5 ), ( 4 22 5 )} ∪ {( 1 0
0 4 )}D(5, 9) = {( 5 1
1 9 ) , ( 5 22 8 ), ( 4 0
0 5 )} ∪ {( 1 00 4 )}
D(5, 8) = {( 5 00 8 ) , ( 5 2
2 8 ), ( 5 11 5 )} ∪ {( 1 0
0 4 )}
I Since D(4, 5) ∩ D(5, 5) = {( 4 22 5 )} we can identify the
corresponding curve. Hence we also know ( 5 11 5 ) (and ( 4 0
0 5 )).I Similarly ( 5 2
2 8 ) can be matched byD(5, 8) ∩ D(5, 9) = {( 5 2
2 8 )}.
In the end we find that:
C( 5 22 8 ) : the genus 1 curve
C( 5 11 5 ) : the genus 3 hyperelliptic curve
C( 5 00 8 ) : the genus 3 non-hyperelliptic curve
Thanks for listening!
