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Motivation The Hilbert Modular Group Resolution of the Cusps Signatures Class Numbers, Continued Fractions, and the Hilbert Modular Group Jordan Schettler University of California, Santa Barbara 11/8/2013

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Page 1: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Class Numbers, Continued Fractions, and theHilbert Modular Group

Jordan Schettler

University of California, Santa Barbara

11/8/2013

Page 2: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Outline

1 Motivation

2 The Hilbert Modular Group

3 Resolution of the Cusps

4 Signatures

Page 3: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Motivation

Page 4: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

‘Minus’ Continued Fractions

Let α ∈ R\Q. Then ∃unique continued fraction expansion

α = [[b0; b1,b2, . . .]] := b0 −1

b1 −1

b2 − · · ·

where b0 ∈ Z and b1,b2, . . . ∈ Z>1.

(b0,b1, . . .) is eventually periodic⇔ α is algebraic of degree 2.

Note: [[2; 2,2, . . .]] = 1, so we must have bk ≥ 3 for∞ly many k

Page 5: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

“An Amusing Connection”

Let ` > 3 be a prime such that ` ≡ 3 (mod 4). Then√` = [[b0; b1, . . . ,bm]]

where m = minimal period is even, bm = 2b0 and

(b1,b2, . . . ,bm−1) = (bm−1,bm−2, . . . ,b1).

Theorem (Hirzebruch)

If Q(√`) has class number 1, the class number of Q(

√−`) is

h(−`) =13

m∑k=1

(bk − 3).

Page 6: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The Hilbert Modular Group

Page 7: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Notation

H = {z ∈ C : =(z) > 0}

GL+2 (R) acts on H via Möbius transformations:(

a bc d

)· z =

az + bcz + d

The action induces an isomorphism between the group ofbiholomorphic maps H → H and the group

PL+2 (R) = GL+

2 (R)/{(

a 00 a

): a ∈ R×

}

Page 8: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Notation

Fix an integer n ≥ 1, and consider Hn = H×H× · · · × H︸ ︷︷ ︸n times

.

(PL+2 (R))n acts on Hn component-wise.

∃SES1→ (PL+

2 (R))n → An → Sn → 1

where An = group of biholomorphic maps Hn → Hn andSn = a symmetric group.

Page 9: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Notation

F = number field of degree n over Q

Assume F is totally real: ∃n distinct embeddings

F ↪→ R : x 7→ x (j) for j = 1, . . . ,n

GL+2 (F ) = {A ∈ GL2(F ) : det(A)(j) > 0,∀j = 1, . . . ,n}

We use the embeddings to view

PL+2 (F ) ⊂ (PL+

2 (R))n

Page 10: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Notation

OF = ring of integers of F

We define the Hilbert modular group

G = SL2(OF )/{±1} ⊂ PL+2 (F )

G is a discrete, irreducible subgroup of (PL+2 (R))n

More generally, let Γ denote any discrete, irreduciblesubgroup of (PL+

2 (R))n such that

[G : Γ] :=[G : G ∩ Γ]

[Γ : G ∩ Γ]<∞

Page 11: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Notation

With coordinates zj = xj + iyj on Hn, ∃Gauss-Bonnet form:

ω =(−1)n

(2π)n ·dx1 ∧ dy1

y21

∧ · · · ∧ dxn ∧ dyn

y2n

Theorem (Siegel)

Hn/Γ

∫ω = [G : Γ] · 2ζF (−1) ∈ Q

where ζF (s) is the Dedekind zeta function of F .

Page 12: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Define the isotropy group at z ∈ Hn by

Γz = {γ ∈ Γ : γ · z = z}

Let am(Γ) = # of Γ-orbits of points z with |Γz | = m

Every Γz is finite cyclic, and∑

m≥2 am(Γ) <∞.

TheoremHn/Γ is a non-compact complex analytic space with finitelymany “quotient” singularities and Euler characteristic

χ(Hn/Γ) =

Hn/Γ

∫ω +

∑m≥2

am(Γ)m − 1

m

Page 13: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The n = 1 Case

If n = 1, then F = Q and G = SL2(Z)/{±1}

∃biholomorphism j : H/G→ C

1 = χ(C) = χ(H/G) = 2ζ(−1) +12

+23

= 1

Page 14: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The n = 1 Case Continued

For n = 1 we could take Γ = Γ0(N), a congruence subgroup oflevel N

non-compact Riemann surface H/Γ = Y0(N)

compact Riemann surface X0(N) = Y0(N) ∪ {cusps} where

{cusps} = ({∞} ∪Q)/Γ0(N)

Page 15: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Back to General Case

If Γ ⊆ G, we can compactify Hn/Γ by adding “cusps” P1(K )/Γwhere we view

P1(K ) ⊂ ({∞} ∪ R)n = ∂Hn

Theorem∃bijection {cusps of Hn/G} ↔ class group C of F :

orbit of [α, β] ∈ P1(K ) with α, β ∈ OF 7→ ideal class of (α, β)

Page 16: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

From now on take n = 2, so F = Q(√

d) for a squarefree d > 1.

OF =

Z[√

d ] if d ≡ 2,3 (mod 4)

Z[

1+√

d2

]if d ≡ 1 (mod 4)

O×F = {±1} × εZ

H2/G is a Hilbert modular surface.

The # of cusps = the class number h(d) of Q(√

d).

Page 17: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The number of quotient singularities of H2/G is related to theclass numbers of imaginary quadratic fields.

Theorem (Prestel)

For d > 6 and (d ,6) = 1,

a2(G) =

10h(−d) if d ≡ 3 (mod 8)4h(−d) if d ≡ 7 (mod 8)h(−d) if d ≡ 1 (mod 4)

a3(G) = h(−3d), am(G) = 0 for m > 3.

Page 18: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Example

Suppose d ≡ 1 (mod 12). Then

2ζF (−1) =1

15

∑1≤b<

√d

b odd

σ1

(d − b4

4

)

where σ1(m) = sum of divisors of m.

Thus

30χ(H2/G) = 2∑

1≤b<√

db odd

σ1

(d − b4

4

)+ 15h(−d) + 20h(−3d)

Page 19: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Resolution of the Cusps

Page 20: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Consider a cusp of H2/G with representative x ∈ P1(F ).

Translate the cusp to infinity:

ρx =∞ = (∞,∞)

for some ρ ∈ (PL+2 (R))2.

∃deleted closed neighborhoods for∞, x :

W (r) = {(x1 + iy1, x2 + iy2) ∈ H2 : y1y2 ≥ r}

U(r) = ρ−1W (r)

Page 21: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

We can choose r � 0 so that

U(r)/G = U(r)/Gx ≈W (r)/ρGxρ−1 ⊆ H2/ρGxρ

−1

There is a SES

1→ M → ρGxρ−1 → V → 1

where M is a fractional ideal and V = {u2 : u ∈ O×F }.

The narrow ideal class of M is uniquely determined by the cusp(indep. of ρ), and we may choose ρ such that

ρGxρ−1 = G(M,V ) = {( v m

0 1 ) : v ∈ V ,m ∈ M}

Page 22: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The quotient space H2/G(M,V ) is a complex manifold.

We can compactify

H2/G(M,V ) = H2/G(M,V ) ∪ {∞}

to obtain a complex analytic space with a singularity at∞.

We now show how M,V are determined and how to resolve thesingularity at∞.

Page 23: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The narrow class group C+ = fractional ideals modulo strictequivalence: a ∼ b⇔ a = λb for some totally positive λ ∈ F

For a fractional ideal a of F , a 7→ a−2 induces a homomorphism

Sq : C → C+

where C+ is the narrow class group of F

Hence to each cusp corresponding to an ideal class a, we havean associated narrow ideal class C = Sq(a).

Page 24: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Every narrow ideal class C ∈ C+ contains an ideal of the form

M = Z + wZ

with w ∈ K and w > 1 > w ′ > 0 (w ′ = Galois conjugate).

This implies w has a purely periodic continued fraction:

w = [[b0; b1, . . . ,bm−1]]

where m = minimal period. (Note: all bk ≥ 2 and bj > 2 some j)

Page 25: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The cycle ((b0,b1, . . . ,bm−1)) (defined up to cyclic permutation)depends only on C ∈ C+.

We define bk by extension using periodicity for all k ∈ Z.

For each k ∈ Z take Rk = C2 with coordinates (uk , vk ).

∃biholomorphism

ϕk : R′k → R′′k+1 : (uk , vk ) 7→ (ubkk vk ,1/uk )

where R′k = Rk\{uk = 0} and R′′k+1 = Rk+1\{vk+1 = 0}.

Page 26: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Take the disjoint union ∪kRk and identify R′k with R′′k+1 via ϕk .

This gives a complex manifold Y of dimension 2 with chartsψk : Rk → C2 given by coordinates (uk , vk )

∃curves Sk in Y given by uk+1 = 0 in Rk+1 (and vk = 0 in Rk ).

By construction, Sk · Sk+1 = 1 while Sk · Sj = 0 for k < j + 1,and we can compute the self-intersections:

Sk · Sk = −bk

Page 27: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

M acts freely on C2 via λ · (z1, z2) = (z1 + λ, z2 + λ′)

Note that

Y −⋃k∈Z

Sk = {(u0, v0) : u0 6= 0 6= v0},

so the map

2πiz1 = w log(u0) + log(v0)

2πiz2 = w ′log(u0) + log(v0)

induces a biholomorphism

Φ: Y −⋃k∈Z

Sk −→ C2/M

Page 28: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Define A0 = 1 and inductively Ak+1 = w−1k+1Ak where

wk+1 = [[bk+1,bk+2, . . . ,bk+m]]

Then Am generates the group U+ of totally positive units, andAcm = Ac

m generates V = (O×F )2 where c = [U+ : V ] ∈ {1,2}.

The group V acts on Y + = Φ−1(H2/M) ∪⋃

k∈Z Sk :

(Acm)n sends (uk , vk ) in the k th coordinate system to the pointwith the same coordinates in the (k + ncm)th coordinatesystem.

Under the action, Sk is mapped by (Acm)n to the curve Sk+ncm

Page 29: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Y + is an open submanifold of Y with a free and properlydiscontinuous action of V .

Y ((b0, . . . ,bcm−1)) = Y +/V is a complex manifold

∃cycle of curves S0,S1, . . . ,Scm−1 with intersection matrix

−b0 1 0 · · · 0 11 −b1 1 0 · · · 00 1 −b2 1 0 · · ·· · · · · · · · · · · · · · · · · ·0 · · · 0 1 −bcm−2 11 0 · · · 0 1 −bcm−1

or(−b0 2

2 −b1

)for cm = 2, or (−b0 + 2) for cm = 1.

Page 30: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

The Resolution

∃holomorphic map

σ : Y ((b0, . . . ,bcm−1))→ H2/G(M,V )

such that:

σ−1(∞) =cm−1⋃k=0

Sk

and

Y ((b0, . . . ,bcm−1))−cm−1⋃k=0

Sk → H2/G(M,V )

is a biholomorphism.

Page 31: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Signatures

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Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Definition

Let M be a complex surface. ∃symmetric bilinear form

β : H2(M,R)× H2(M,R)→ R

given by the intersection of homology classes.

We define the signature of M by

sign(M) = b+ − b−

where b+ (resp. b−) is the # of positive (resp. negative)eigenvalues of a matrix representing β.

Page 33: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Two Geometric Formulas

Let M be a connected, complex surface.

Theorem (Adjunction Formula)For a nonsingular, compact curve S on M,

χ(S) = −K · S − S · S

where K is a canonical divisor on M.

Theorem (Signature Formula)

If M is a compact manifold with no boundary,

sign(M) =13

(K · K − 2χ(M))

where K is a canonical divisor on M.

Page 34: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

∀a ∈ C we associate a compact manifold with boundary Xaobtained by resolving the singularity (as above) of

W (r)/G(M,V ) ≈ (U(r)/G) ∪ {x} ⊆ H2/G

where r � 0 and x is the cusp corresponding to a.

We define the signature deviation invariant

δ(Xa) =13

(K · K − 2χ(Xa))− sign(Xa)

where K is a canonical divisor on Xa.

Page 35: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Computing δ(Xa)

Xa is constructed by blowing up the cusp x into a cycle of cmnonsingular curves S0, . . . ,Scm−1.

The intersection matrix (as in the previous section) gives

sign(Xa) = −cm

In fact, Xa is homotopy equivalent to⋃cm−1

k=0 Sk , so

χ(Xa) = 1− 1 + cm = cm

Page 36: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Computing δ(Xa)

For each k the adjunction formula gives

2 = χ(Sk ) = −K · Sk − Sk · Sk = −K · Sk + bk ,

so

−K =cm−1∑k=0

Sk ,

and

K · K = −cm−1∑k=0

bk + 2cm.

Page 37: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Computing δ(Xa)

Therefore

δ(Xa) =13

(−c

m−1∑k=0

bk + 2cm − 2cm

)− (−cm) = −c

3

m−1∑k=0

(bk − 3)

Suppose F = Q(√

d) has no units of negative norm. (c = 2)

Theorem (Curt Meyer)

If ζ(s,C) = partial zeta function of C = Sq(a),

ζ(0,C) = −ζ(0,C∗) =16

m−1∑k=0

(bk − 3)

(=−12c

δ(Xa)

)

where a−2 = C ∪ C∗.

Page 38: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Assume F = Q(√`) where ` > 3 is a prime with ` ≡ 3 (mod 4).

∃unique, character ψ on C+ which is non-trivial and real-valued.

Meyer’s theorem implies

sign(H2/G) =∑a∈C

δ(Xa) = −4∑C∈C+

ψ(C)ζ(0,C)

= −4L(0, ψ) = −4h(−`)h(−1)

2= −2h(−`)

Page 39: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

If, additionally, F = Q(√`) has class number 1, then C = {a}

where a = [OF ] is the trivial ideal class.

C = Sq(a) is the trivial narrow ideal class, so we may choose

M = OF = Z + (d√`e+

√`)Z

with √` = [[b0; b1, . . . ,bm]]

where m = minimal period, b0 = d√`e, bm = 2b0, and whence

d√`e+

√` = [[2b0; b1, . . . ,bm−1]].

Page 40: Class Numbers, Continued Fractions, and the Hilbert ...web.math.ucsb.edu/~jcs/HilbertModularSurfaces.pdf · MotivationThe Hilbert Modular GroupResolution of the CuspsSignatures We

Motivation The Hilbert Modular Group Resolution of the Cusps Signatures

Thus we recover the “amusing connection”

−2h(−`) = sign(H2/G) = δ(Xa) = −23

(2b0 +

m−1∑k=1

(bk − 3)

),

or more simply

h(−`) =13

m∑k=1

(bk − 3)

Note: We did NOT need ANY signatures or surfaces to derivethe formula for the class number.