Lecture 2.THE ARAKELOV CLASS GROUP

Ha Tran

ICTP–CIMPA summer school 2016HCM University of Science– Saigon University

1 / 26

We have studied:• Number field, the ring of integers.• Fractional ideals: J/α for some 0 6= α ∈ OF

and J ⊂ OF is an ideal.• The class group ClF = IdF/PrincF and class

number hF = #ClF .• The Φ map:

Φ = (σ1, · · · , σr1, σr1+1, · · · , σr1+r2).Φ(I ) is a lattice in FR.

• The L map: L(x) = (log |σ(f )|)σ, ∀x ∈ F×.Λ = L(O×F ) is a lattice in H = ....T 0 = H/Λ is a real torus of dim. r1 + r2 − 1.

• Ideal lattices: (I , q), where ...I : factional ideal; u =∈ (R>0)r1+r2. Then(I , qu) is an ideal lattice.

• Many famous lattices arise from ideal lattices. 2 / 26

T 0 and Λ = L(O×F )

Denote by

H =

(xσ) ∈ ⊕σR :∑σ real

xσ + 2∑

σ complex

xσ = 0

and

3 / 26

T 0 and Λ = L(O×F )

Denote by

H =

(xσ) ∈ ⊕σR :∑σ real

xσ + 2∑

σ complex

xσ = 0

and

Λ = L(O×F ).

3 / 26

T 0 and Λ = L(O×F )

Denote by

H =

(xσ) ∈ ⊕σR :∑σ real

xσ + 2∑

σ complex

xσ = 0

and

Λ = L(O×F ).

Λ is a lattice contained in the vector space H .

3 / 26

T 0 and Λ = L(O×F )

Denote by

H =

(xσ) ∈ ⊕σR :∑σ real

xσ + 2∑

σ complex

xσ = 0

and

Λ = L(O×F ).

Λ is a lattice contained in the vector space H .

3 / 26

T 0 and Λ = L(O×F )Denote by

H =

(xσ) ∈ ⊕σR :∑σ real

xσ + 2∑

σ complex

xσ = 0

and

Λ = L(O×F ).

Λ is a lattice contained in the vector space H .

Let T 0 = H/Λ. Then T 0 is a compact real torus ofdimension r1 + r2 − 1 (Dirichlet).

3 / 26

T 0 and Λ = L(O×F )

Ex: Let F = Q(√−2). Then r1 = 0, r2 = 1 and

T 0 =? dim(T 0) =?

4 / 26

T 0 and Λ = L(O×F )

Ex: Let F = Q(√

2). Then r1 = 2, r2 = 0 and

H = {(x , y) ∈ R2 : x + y = 0} ' R.

T 0 =? dim(T 0) =?

5 / 26

T 0 and Λ = L(O×F )Ex: Let F = Q(

√2). Then r1 = 2, r2 = 0 and

H = {(x , y) ∈ R2 : x + y = 0} ' R.

6 / 26

T 0 and Λ = L(O×F )

Ex: Let F be a totally real cubic field. Thenr1 = 3, r2 = 0 and

H = {(x , y , z) ∈ R3 : x + y + z = 0} ' R2.

T 0 =? dim(T 0) =?

Ex: F = Q( 4√

2)?

7 / 26

T 0 and Λ = L(O×F )

Ex: Let F be a totally real cubic field. Thenr1 = 3, r2 = 0 and

H = {(x , y , z) ∈ R3 : x + y + z = 0} ' R2.

T 0 =? dim(T 0) =?Ex: F = Q( 4

√2)?

7 / 26

What we study today?

The Arakelov class group Pic0F ,

• Pic0F tells you the class number hF , theregulator RF ,...

• (Main Theorem) There is a bijection

Pic0Fψ→ {Isometry classes of ideal lattices of covol.

√|∆F |}.

8 / 26

Content

1 Arakelov divisorsWhat are Arakalov divisors?Principal Arakelov divisorsDegreeThe Hermitian line bundle

2 The Arakelov class group Pic0F

3 The structure of Pic0F

4 Main theorem

9 / 26

What are Arakalov divisors?

Arakelov divisors of a number field are analogous todivisors on an algebraic curve.

Algebraic curveDivisorD =

∑P points

nPP

nP ∈ Z.

Number field FArakelov divisorD =

∑p primes

npp +∑

σ xσσ

σ infinite primes of F .np ∈ Z but xσ ∈ R.

10 / 26

What are Arakalov divisors?

Arakelov divisors of a number field are analogous todivisors on an algebraic curve.

Algebraic curveDivisorD =

∑P points

nPP

nP ∈ Z.

Number field FArakelov divisorD =

∑p primes

npp +∑

σ xσσ

σ infinite primes of F .np ∈ Z but xσ ∈ R.

10 / 26

What are Arakalov divisors?

Arakelov divisors of a number field are analogous todivisors on an algebraic curve.

Algebraic curveDivisorD =

∑P points

nPP

nP ∈ Z.

Number field FArakelov divisorD =

∑p primes

npp +∑

σ xσσ

σ infinite primes of F .np ∈ Z but xσ ∈ R.

10 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

• F = Q• F = Q(i), Q(

√2).

11 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

Ex 4: F = Q, OF = Z,p1 = 2Z, p2 = 5Z: 2 prime ideals;σ : Q→ C, q 7−→ q the infinite prime;D = p1 − 3p2 + πσ is an Arakelov divisor.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

• F = Q• F = Q(i), Q(

√2).

11 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

• F = Q• F = Q(i), Q(

√2).

11 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

• F = Q• F = Q(i), Q(

√2).

11 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

Ex 5: DivF =?

• F = Q• F = Q(i), Q(

√2).

11 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

Ex 5: DivF =?

• F = Q

• F = Q(i), Q(√

2).

11 / 26

What are Arakalov divisors?DefinitionAn Arakelov divisor is of the form

D =∑

p primes

npp +∑σ

xσσ

σ infinite primes of F ; np ∈ Z but xσ ∈ R.

• The set of all Arakelov divisors of F is anadditive group denoted by DivF ' ⊕pZ×⊕σR.

• D1 + D2 =?, the neutral of DivF , −D =?

Ex 5: DivF =?

• F = Q• F = Q(i), Q(

√2).

11 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.• h0(D).

• ...

12 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.• h0(D).

• ...

12 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.• h0(D).

• ...

12 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.• h0(D).

• ...

12 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.

• h0(D).

• ...

12 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.• h0(D).

• ...

12 / 26

Analogies

Algebraic curve

• Divisor.

• Principal divisor.

• Picard group.

• Canonical divisor κ.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− (g − 1).

• h0(D).

• ...

Number field F

• Arakelov divisor.

• Principal Arakelovdivisor.

• Arakelov class group.

• The inverse different.

• Riemann–Rochh0(D)− h0(κ−D) =deg(D)− 1

2 log |∆|.• h0(D).

• ...

12 / 26

Principal Arakelov divisors

• For f ∈ F×, the principal Arakelov divisor

(f ) =∑

p primes

npp +∑σ

xσσ

where np = ordp(f ) and xσ = − log |σ(f )|,∀σ.

Ex 6: f = −1. Then (f ) =?Ex 7: F = Q(

√2), f = 1−

√2 ∈ Q(

√2)× : (f ) =?

g = 3−√

2 ∈ Q(√

2)× : (g) =?Ex 8: Let F = Q(i) and f = 2 + i ∈ F×. Then(f ) =?

13 / 26

Principal Arakelov divisors

• For f ∈ F×, the principal Arakelov divisor

(f ) =∑

p primes

npp +∑σ

xσσ

where np = ordp(f ) and xσ = − log |σ(f )|,∀σ.

Ex 6: f = −1. Then (f ) =?

Ex 7: F = Q(√

2), f = 1−√

2 ∈ Q(√

2)× : (f ) =?g = 3−

√2 ∈ Q(

√2)× : (g) =?

Ex 8: Let F = Q(i) and f = 2 + i ∈ F×. Then(f ) =?

13 / 26

Principal Arakelov divisors

• For f ∈ F×, the principal Arakelov divisor

(f ) =∑

p primes

npp +∑σ

xσσ

where np = ordp(f ) and xσ = − log |σ(f )|,∀σ.

Ex 6: f = −1. Then (f ) =?Ex 7: F = Q(

√2), f = 1−

√2 ∈ Q(

√2)× : (f ) =?

g = 3−√

2 ∈ Q(√

2)× : (g) =?

Ex 8: Let F = Q(i) and f = 2 + i ∈ F×. Then(f ) =?

13 / 26

Principal Arakelov divisors

• For f ∈ F×, the principal Arakelov divisor

(f ) =∑

p primes

npp +∑σ

xσσ

where np = ordp(f ) and xσ = − log |σ(f )|,∀σ.

Ex 6: f = −1. Then (f ) =?Ex 7: F = Q(

√2), f = 1−

√2 ∈ Q(

√2)× : (f ) =?

g = 3−√

2 ∈ Q(√

2)× : (g) =?Ex 8: Let F = Q(i) and f = 2 + i ∈ F×. Then(f ) =?

13 / 26

Degree

deg(p) = log(N(p)) where N(p) = #OF/p,

deg(σ) =

{1 if σ real2 if σ complex

• The degree of D is defined bydeg(D) :=

∑p np logN(p) +

∑σ deg(σ)xσ.

Let f ∈ F×. Compute deg(f ) ifEx 6: f = −1?Ex 7: F = Q(

√2), f = 1−

√2? , f = 3−

√2?

Ex 8:F = Q(i) and f = 2 + i?

• The set of all Arakelov divisors of degree 0form a group, denoted by Div 0F (⊃ PrincF ).

14 / 26

Degree

deg(p) = log(N(p)) where N(p) = #OF/p,

deg(σ) =

{1 if σ real2 if σ complex

• The degree of D is defined bydeg(D) :=

∑p np logN(p) +

∑σ deg(σ)xσ.

Let f ∈ F×. Compute deg(f ) ifEx 6: f = −1?Ex 7: F = Q(

√2), f = 1−

√2? , f = 3−

√2?

Ex 8:F = Q(i) and f = 2 + i?

• The set of all Arakelov divisors of degree 0form a group, denoted by Div 0F (⊃ PrincF ).

14 / 26

Degree

deg(p) = log(N(p)) where N(p) = #OF/p,

deg(σ) =

{1 if σ real2 if σ complex

• The degree of D is defined bydeg(D) :=

∑p np logN(p) +

∑σ deg(σ)xσ.

Let f ∈ F×. Compute deg(f ) ifEx 6: f = −1?

Ex 7: F = Q(√

2), f = 1−√

2? , f = 3−√

2?Ex 8:F = Q(i) and f = 2 + i?

• The set of all Arakelov divisors of degree 0form a group, denoted by Div 0F (⊃ PrincF ).

14 / 26

Degree

deg(p) = log(N(p)) where N(p) = #OF/p,

deg(σ) =

{1 if σ real2 if σ complex

• The degree of D is defined bydeg(D) :=

∑p np logN(p) +

∑σ deg(σ)xσ.

Let f ∈ F×. Compute deg(f ) ifEx 6: f = −1?Ex 7: F = Q(

√2), f = 1−

√2? , f = 3−

√2?

Ex 8:F = Q(i) and f = 2 + i?

• The set of all Arakelov divisors of degree 0form a group, denoted by Div 0F (⊃ PrincF ).

14 / 26

Degree

deg(p) = log(N(p)) where N(p) = #OF/p,

deg(σ) =

{1 if σ real2 if σ complex

• The degree of D is defined bydeg(D) :=

∑p np logN(p) +

∑σ deg(σ)xσ.

Let f ∈ F×. Compute deg(f ) ifEx 6: f = −1?Ex 7: F = Q(

√2), f = 1−

√2? , f = 3−

√2?

Ex 8:F = Q(i) and f = 2 + i?

• The set of all Arakelov divisors of degree 0form a group, denoted by Div 0F (⊃ PrincF ).

14 / 26

Degree

deg(p) = log(N(p)) where N(p) = #OF/p,

deg(σ) =

{1 if σ real2 if σ complex

• The degree of D is defined bydeg(D) :=

∑p np logN(p) +

∑σ deg(σ)xσ.

Let f ∈ F×. Compute deg(f ) ifEx 6: f = −1?Ex 7: F = Q(

√2), f = 1−

√2? , f = 3−

√2?

Ex 8:F = Q(i) and f = 2 + i?

• The set of all Arakelov divisors of degree 0form a group, denoted by Div 0F (⊃ PrincF ). 14 / 26

The Hermitian line bundle

• Let D =∑

p primes npp +∑

σ xσσ. Denote

I :=∏p

p−np and u := (e−xσ)σ ∈ FR.

Then (I , u) is called the Hermitian line bundleassociated to D. We can identity D = (I , u).

Ex: The Hermitian line bundle ass. to• the zero divisor D = 0?• the principal divisor D = (f ) ?• D1 + D2 =? if D1 = (I1, u1),D2 = (I2, u2)?• −D =? if D = (I , u).

15 / 26

The Hermitian line bundle

• Let D =∑

p primes npp +∑

σ xσσ. Denote

I :=∏p

p−np and u := (e−xσ)σ ∈ FR.

Then (I , u) is called the Hermitian line bundleassociated to D. We can identity D = (I , u).Ex: The Hermitian line bundle ass. to• the zero divisor D = 0?• the principal divisor D = (f ) ?• D1 + D2 =? if D1 = (I1, u1),D2 = (I2, u2)?• −D =? if D = (I , u).

15 / 26

What is the Arakelov classgroup Pic0

F?

It is an analogue of the Picard group of an algebraiccurve.

DefinitionThe Arakelov class group Pic0F is the quotient ofDiv 0F by its subgroup of principal divisors.

Ex 1: F = Q, Pic0F =?Ex 2: F = Q(

√−1), Pic0F =?

Ex 3: F = Q(√

2), Pic0F =?

16 / 26

What is the Arakelov classgroup Pic0

F?

It is an analogue of the Picard group of an algebraiccurve.

DefinitionThe Arakelov class group Pic0F is the quotient ofDiv 0F by its subgroup of principal divisors.

Ex 1: F = Q, Pic0F =?Ex 2: F = Q(

√−1), Pic0F =?

Ex 3: F = Q(√

2), Pic0F =?

16 / 26

The structure of Pic0F

Consider the maps

φ1 : T 0 −→ Pic0F

(xσ)σ + Λ 7−→ class of (OF , u) where u = (exσ)σ,

andφ2 : Pic0F −→ ClF

class of (I , u) 7−→ class of I

17 / 26

The structure of Pic0F

Consider the maps

φ1 : T 0 −→ Pic0F

(xσ)σ + Λ 7−→ class of (OF , u) where u = (exσ)σ,

andφ2 : Pic0F −→ ClF

class of (I , u) 7−→ class of I

17 / 26

The structure of Pic0F

PropositionThe following sequence is exact.

0 −→ T 0 φ1−→ Pic0Fφ2−→ ClF −→ 0.

Remark

• T 0 is a compact topological group and#ClF <∞ ⇒ Pic0F is a compact topo. gp.

• The compactness of Pic0F ⇒ the Dirichlet unittheorem and the finiteness of the class group.

• D,D ′ ∈ Pic0F on the same connectedcomponent, then there exists unique u ∈ T 0 stD − D ′ = (OF , u).

18 / 26

The structure of Pic0F

PropositionThe following sequence is exact.

0 −→ T 0 φ1−→ Pic0Fφ2−→ ClF −→ 0.

Remark

• T 0 is a compact topological group and#ClF <∞ ⇒ Pic0F is a compact topo. gp.

• The compactness of Pic0F ⇒ the Dirichlet unittheorem and the finiteness of the class group.

• D,D ′ ∈ Pic0F on the same connectedcomponent, then there exists unique u ∈ T 0 stD − D ′ = (OF , u).

18 / 26

The structure of Pic0F

vol(T 0) =√n2−r2/2RF with RF the regulator of F .

The number of connected components of Pic0F isthe class number hF .

19 / 26

The structure of Pic0F

vol(T 0) =√n2−r2/2RF with RF the regulator of F .

The number of connected components of Pic0F isthe class number hF .F = Q then Pic0F = 0.

19 / 26

The structure of Pic0F

vol(T 0) =√n2−r2/2RF with RF the regulator of F .

The number of connected components of Pic0F isthe class number hF .

r1 = 0, r2 = 1 so T 0 is a point.

19 / 26

The structure of Pic0F

vol(T 0) =√n2−r2/2RF with RF the regulator of F .

The number of connected components of Pic0F isthe class number hF .

r1 = 2, r2 = 0 so T 0 is a circle.

19 / 26

The structure of Pic0F

vol(T 0) =√n2−r2/2RF with RF the regulator of F .

The number of connected components of Pic0F isthe class number hF .

r1 = 3, r2 = 1 so T 0 is a real torus in R3.

19 / 26

The structure of Pic0F

vol(T 0) =√n2−r2/2RF with RF the regulator of F .

The number of connected components of Pic0F isthe class number hF .

Note: Buchmann’s algorithm to find the regulatorand class number of the number field.

19 / 26

The structure of Pic0F

Let D = (I , u). z ∈ I ,Φ(z) = (σ(z))σ ∈ FR,uz := (uσ · σ(z))σ ∈ FR.We define

qu(x , y) := 〈ux , uy〉 for any x , y ∈ I .

(the scalar product defined on FR)

Proposition(I , qu) is an ideal lattice.

Proof. Ex.We called (I , qu) the ideal lattice associated to D.In particular, ‖x‖2u = qu(x , x) =?

20 / 26

The structure of Pic0F

Let D = (I , u). z ∈ I ,Φ(z) = (σ(z))σ ∈ FR,uz := (uσ · σ(z))σ ∈ FR.We define

qu(x , y) := 〈ux , uy〉 for any x , y ∈ I .

(the scalar product defined on FR)

Proposition(I , qu) is an ideal lattice.

Proof. Ex.

We called (I , qu) the ideal lattice associated to D.In particular, ‖x‖2u = qu(x , x) =?

20 / 26

The structure of Pic0F

Let D = (I , u). z ∈ I ,Φ(z) = (σ(z))σ ∈ FR,uz := (uσ · σ(z))σ ∈ FR.We define

qu(x , y) := 〈ux , uy〉 for any x , y ∈ I .

(the scalar product defined on FR)

Proposition(I , qu) is an ideal lattice.

Proof. Ex.We called (I , qu) the ideal lattice associated to D.

In particular, ‖x‖2u = qu(x , x) =?

20 / 26

The structure of Pic0F

Let D = (I , u). z ∈ I ,Φ(z) = (σ(z))σ ∈ FR,uz := (uσ · σ(z))σ ∈ FR.We define

qu(x , y) := 〈ux , uy〉 for any x , y ∈ I .

(the scalar product defined on FR)

Proposition(I , qu) is an ideal lattice.

Proof. Ex.We called (I , qu) the ideal lattice associated to D.In particular, ‖x‖2u = qu(x , x) =?

20 / 26

Main theorem

TheoremLet F be a number field of discriminant ∆F . Thereis a bijection

Pic0Fψ→ {Isometry classes of ideal lattices of covol.

√|∆F |}

class of D = (I , u) 7−→ class of (I , qu).

21 / 26

Main theoremProof.ψ is injectiveψ is surjective

• (I , qu) ' (I ′, qu′).• ∃f ∈ F ∗ st I ′ = fI andqu′(fx , fx) = qu(x , x),∀x ∈ I .Hence ‖u′fx‖ = ‖ux‖ for all x ∈ I .

• Extend qu and qu′ to I ⊗ R = FR.⇒ ‖u′fx‖ = ‖ux‖,∀x ∈ FR.

• For each σ, let eσ ∈ FR : σ(eσ) = 1 whileσ′(eσ) = 0 for all σ′ 6= σ.

• Substituting eσ with x ⇒ |σ(f )u′σ| = |uσ|,∀σ⇒ |f | = u′/u.

22 / 26

Main theoremProof. ψ is injective:

• (I , qu) ' (I ′, qu′).

• ∃f ∈ F ∗ st I ′ = fI andqu′(fx , fx) = qu(x , x),∀x ∈ I .Hence ‖u′fx‖ = ‖ux‖ for all x ∈ I .

• Extend qu and qu′ to I ⊗ R = FR.⇒ ‖u′fx‖ = ‖ux‖,∀x ∈ FR.

• For each σ, let eσ ∈ FR : σ(eσ) = 1 whileσ′(eσ) = 0 for all σ′ 6= σ.

• Substituting eσ with x ⇒ |σ(f )u′σ| = |uσ|,∀σ⇒ |f | = u′/u.

22 / 26

Main theoremProof. ψ is injective: Assume ψ(D) = ψ(D ′) forsome D = (I , u),D ′ = (I ′, u′) ∈ Pic0Fwe have to show that

D ′ ≡ D in Pic0F

⇔ D ′ − D = (f ) for some f ∈ F ∗.

• (I , qu) ' (I ′, qu′).• ∃f ∈ F ∗ st I ′ = fI andqu′(fx , fx) = qu(x , x),∀x ∈ I .Hence ‖u′fx‖ = ‖ux‖ for all x ∈ I .

• Extend qu and qu′ to I ⊗ R = FR.⇒ ‖u′fx‖ = ‖ux‖,∀x ∈ FR.

• For each σ, let eσ ∈ FR : σ(eσ) = 1 whileσ′(eσ) = 0 for all σ′ 6= σ.

• Substituting eσ with x ⇒ |σ(f )u′σ| = |uσ|,∀σ⇒ |f | = u′/u.

22 / 26

Main theoremProof. ψ is injective: Assume ψ(D) = ψ(D ′) forsome D = (I , u),D ′ = (I ′, u′) ∈ Pic0F

• (I , qu) ' (I ′, qu′).• ∃f ∈ F ∗ st I ′ = fI andqu′(fx , fx) = qu(x , x),∀x ∈ I .Hence ‖u′fx‖ = ‖ux‖ for all x ∈ I .

• Extend qu and qu′ to I ⊗ R = FR.⇒ ‖u′fx‖ = ‖ux‖,∀x ∈ FR.

• For each σ, let eσ ∈ FR : σ(eσ) = 1 whileσ′(eσ) = 0 for all σ′ 6= σ.

• Substituting eσ with x ⇒ |σ(f )u′σ| = |uσ|,∀σ⇒ |f | = u′/u.

⇒ D ′ − D = (f ).22 / 26

Main theoremProof. ψ is injective: Assume ψ(D) = ψ(D ′) forsome D = (I , u),D ′ = (I ′, u′) ∈ Pic0F• (I , qu) ' (I ′, qu′).• ∃f ∈ F ∗ st I ′ = fI andqu′(fx , fx) = qu(x , x),∀x ∈ I .Hence ‖u′fx‖ = ‖ux‖ for all x ∈ I .

• Extend qu and qu′ to I ⊗ R = FR.⇒ ‖u′fx‖ = ‖ux‖,∀x ∈ FR.

• For each σ, let eσ ∈ FR : σ(eσ) = 1 whileσ′(eσ) = 0 for all σ′ 6= σ.

• Substituting eσ with x ⇒ |σ(f )u′σ| = |uσ|,∀σ⇒ |f | = u′/u.

⇒ D ′ − D = (f ).22 / 26

Main theoremProof.ψ is injective: doneψ is surjective

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

23 / 26

Main theoremProof. ψ is surjective:

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

23 / 26

Main theoremProof. ψ is surjective: Let (I , q) be an ideal lattice.We have to show that

(I , q) ' ψ(D) for some D = (J , u) ∈ Pic0F

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

23 / 26

Main theoremProof. ψ is surjective: Let (I , q) be an ideal lattice.We have to show that

(I , q) ' ψ(D) for some D = (J , u) ∈ Pic0F

⇔ (I , q) ' (J , qu) for some D = (J , u) ∈ Pic0F .

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

23 / 26

Main theoremProof. ψ is surjective: Let (I , q) be an ideal lattice.We have to show that

(I , q) ' ψ(D) for some D = (J , u) ∈ Pic0F

⇔ (I , q) ' (J , qu) for some D = (J , u) ∈ Pic0F .

Here we let J = I and

construct u

and then construct qu using q st (I , q) ' (I , qu).

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

23 / 26

Main theoremProof. ψ is surjective: Let (I , q) be an ideal lattice.

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

⇒ (I , q) ' (I , qu) for some D = (I , u) ∈ Pic0F .23 / 26

Main theoremProof. ψ is surjective: Let (I , q) be an ideal lattice.

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

⇒ (I , q) ' (I , qu) for some D = (I , u) ∈ Pic0F .23 / 26

Main theoremProof. ψ is surjective: Let (I , q) be an ideal lattice.

• Extend q to FR.• u =

∑σ q(eσ, eσ)1/2eσ ∈ F ∗R, D = (I , u).

• e2σ = eσ, q is Hermitian and eσeτ = 0,

⇒ q(eσ, eτ) = q(e2σ, eτ) = q(eσ, eσeτ) = 0,∀σ 6= τ.

• For all x , y ∈ FR,

qu(x , y) = 〈ux , uy〉 =∑σ

u2σxσyσ

=∑σ

q(eσ, eσ)xσyσ = q(x , y).

⇒ (I , q) ' (I , qu).23 / 26

Oh, no : ( : ( : (

Show at least 2 points that have been lacked

in the proof!

Prove these points!

24 / 26

Oh, no : ( : ( : (

Show at least 2 points that have been lacked

in the proof!

Prove these points!

Your exercise : (.

24 / 26

Oh, no : ( : ( : (

Show at least 2 points that have been lacked

in the proof!

Prove these points!

Your exercise : (.

There will be a gift for this :).

24 / 26

Recap• Arakelov divisors (I , u).

• The degree, norm.

• Principal Arakelov divisors.

• The Hermitian line bundle.

• The Arakelov class group Pic0F = Div 0F/PrincF .

• The structure of the Arakelov class group.

0 −→ T 0 φ1−→ Pic0Fφ2−→ ClF −→ 0 is exact.

• There is a bijection

Pic0Fψ→ {Isometry classes of ideal lattices of covol.

√|∆F |}

class of D = (I , u) 7−→ class of (I , qu).25 / 26

ReferencesEva Bayer-Fluckiger. Lattices and number fields.

In Algebraic geometry: Hirzebruch 70 (Warsaw, 1998),volume 241 of Contemp. Math., pages 69–84. Amer.Math. Soc., Providence, RI, 1999.

Hendrik W. Lenstra, Jr. Lattices.

In Algorithmic number theory: lattices, number fields,curves and cryptography, volume 44 of Math. Sci. Res.Inst. Publ., pages 127–181. Cambridge Univ. Press,Cambridge, 2008.

Rene Schoof. Computing Arakelov class groups.

In Algorithmic number theory: lattices, number fields,curves and cryptography, volume 44 of Math. Sci. Res.Inst. Publ., pages 447–495. Cambridge Univ. Press,Cambridge, 2008. 26 / 26