# Lattices and modular forms

Document Sample

```					Lattices and modular forms.

Gabriele Nebe

Lehrstuhl D fur Mathematik
¨

EPF Lausanne, 20.7.2009
Lattices
Deﬁnition.
A lattice L in Euclidean n-space (Rn , (, )) is the Z-span of an R-basis
B = (b1 , . . . , bn ) of Rn
n
L = b1 , . . . , bn   Z   ={         ai bi | ai ∈ Z}.
i=1

Ln := {L ≤ Rn | L is lattice } the set of all lattices in Rn .

1   0
G      =
0   1

1   1
H      =
1   2
Invariants of lattices.
Gram matrix.
Gram(L) = {g Gram(B)g tr | g ∈ GLn (Z)} where

Gram(B) = ((bi , bj )) = BB tr ∈ Rn×n
sym

is the Gram matrix of B.
det(L) = det(Gram(B)) = det(BB tr ) the determinant of L is the
square of the volume of the fundamental parallelotop of B.
min(L) = min{( , ) | 0 = ∈ L} the minimum of L.
The density of the associated sphere packing is then
proportional to the Hermite function

min(L)
γ(L) :=
det(L)1/n

γn := max{γ(L) | L ∈ Ln } is called the Hermite constant.

n    1    2      3       4        5      6      7     8    24
γn   1   1.15   1.26    1.41     1.52   1.67   1.81   2    4
Orthogonal decomposition.

Deﬁnition.
Let L1 ≤ Rn1 and L2 ≤ Rn2 be lattices. Then L1 ⊥ L2 ≤ Rn1 ⊥ Rn2 is
called the orthogonal sum of L1 and L2 . A lattice is orthogonally
indecomposable if it cannot be written as orthogonal sum of proper
sublattices.
If Gi ∈ Gram(Li ) are Gram matrices of Li , then the block diagonal
matrix diag(G1 , G2 ) is a Gram matrix of L1 ⊥ L2 , but not all Gram
matrices of L1 ⊥ L2 are block diagonal.
Theorem (M. Kneser).
Every lattice L has a unique orthogonal decomposition
L = L1 ⊥ . . . ⊥ Ls with indecomposable lattices Li .
Construction of orthogonal decomposition.

Proof. Call x ∈ L indecomposable, if x = y + z for y , z ∈ L − {0},
(y, z) = 0.
Then any 0 = x ∈ L is sum of indecomposables, because if x is not
itself indecomposable then x = y + z with (y , z) = 0 and hence
0 < (y , y) < (x, x), 0 < (z, z) < (x, x).
So this decomposition process terminates.
In particular L is generated by indecomposable vectors.
Two indecomposable vectors y, z ∈ L are called connected, if there
are indecomposable vectors x0 = y , x1 , . . . , xt = z in L, such that
(xi , xi+1 ) = 0 for all i. This yields an equivalence relation on the set of
indecomposable vectors in L with ﬁnitely many classes K1 , . . . , Ks .
If Li := Ki Z then L = L1 ⊥ . . . ⊥ Ls is the unique orthogonal
decomposition of L in indecomposable sublattices.
The dual lattice.

Dual lattice.
Let L = b1 , . . . , bn   Z   ≤ Rn be a lattice. Then the dual lattice

L# := {x ∈ Rn | (x, ) ∈ Z∀ ∈ L}
∗            ∗
is again a lattice in Rn and the dual basis B ∗ = (b1 , . . . , bn ) with
∗                                     #
(bi , bj ) = δij is a lattice basis for L .

Integral lattices.
L is called integral, if L ⊂ L# or equivalently Gram(B) ∈ Zn×n .
L is called even, Q( ) := 1 ( , ) ∈ Z for all ∈ L.
2
Even lattices are integral and an integral lattice is even if
(bi , bi ) ∈ 2Z for all i = 1, . . . , n.
L is called unimodular if L = L# .

The square lattice Z2 is unimodular but not even.
Grammatrix and dual basis.

Grammatrix and dual lattice.
n              ∗
bi =    j=1 (bi , bj )bj   and Gram(B ∗ ) = Gram(B)−1
If L is integral, then Gram(B) is the relation matrix for the ﬁnite
∗           ∗
abelian group L# /L = b1 + L, . . . , bn + L . In particular
#
|L /L| = det(L) and

L# /L ∼ Z/d1 Z × . . . × Z/ds Z
=

where (d1 , . . . , ds ) are the nontrivial elementary divisors of
Gram(B).
An integral lattice L is unimodular, if det(L) = 1.
Orthogonal lattices and pure sublattices.

Deﬁnition.
If L ≤ M are lattices then

L⊥,M := L⊥ := {m ∈ M | ( , m) = 0∀ ∈ L}

is called the orthogonal lattice of L in M.
A sublattice L ≤ M is pure if L = {m ∈ M | m ∈ L R } = M ∩ RL.
For L ≤ M the orthogonal lattice L⊥ is a pure sublattice of M and
L = (L⊥ )⊥ iff L is pure in M.
Theorem.
Let M = M # be a unimodular lattice and L ≤ M a pure sublattice.
Then det(L) = det(L⊥ ), more precisely L# /L ∼ (L⊥ )# /L⊥ as ﬁnite
=
abelian groups.
Equivalence and automorphism groups.

Equivalence.
The orthogonal group
On (R) = {g ∈ GLn (R) | (vg, wg) = (v , w) for all v , w ∈ Rn } acts on
Ln preserving all invariants that can be deduced from the Gram
matrices like integrality, minimum, determinant, density etc..
Lattices in the same On (R)-orbit are called isometric.

Automorphism group.
The automorphism group of L is

Aut(L) = {σ ∈ On (R) | σ(L) = L}
∼ {g ∈ GLn (Z) | g Gram(B)g tr = Gram(B)}
=

Aut(L) is a ﬁnite group and can be calculated efﬁciently, if the ﬁnite
set of vectors { ∈ L | Q( ) ≤ maxn Q(bi )} can be stored.
i=1
Theta-series of lattices.
The theta series θL =    ∈L   q Q( ) .
Assume that L is an even lattice and let La := { ∈ L | Q( ) = a}.
∞
Then La is a ﬁnite Aut(L)-set and θL = a=0 |La |q a .
√                                               2 0
L = 2Z2 the square lattice with Gram matrix                    :
0 2
θL = 1 + 4q 1 + 4q 2 + 4q 4 + 8q 5 + 4q 8 + 4q 9 + 8q 10 + . . .
Aut(L) ∼ D8 (the symmetry group of a square)
=
Example: the hexagonal lattice.

The hexagonal lattice.
√      √
2 1
Basis B = ((1, 1), ( 1+2 3 , 1−2 3 )), Gram(B) =
1 2
2
det(L) = 3, min(L) = 2, γ(L) =   √
3
∼ 1.1547 (density .91)
7
θL = 1 + 6q + 6q 3 + 6q 4 + 12q + 6q 9 + 6q 12 + 12q 13 + 6q 16 + . . .
Aut(L) ∼ D12 (the symmetry group of a regular hexagon)
=
Example: the E8 -lattice.

Let (e1 , . . . , e8 ) be an orthonormal basis of R8 and consider
L := Z8 = e1 , . . . , e8 Z = L# .
Let D8 := { ∈ L | ( , ) ∈ 2Z} be the even sublattice of L.
θD8 = 1 + 112q + 1136q 2 + 3136q 3 + 9328q 4 + 14112q 5 + . . .
∼
Then D# /D8 = e1 + D8 , v + D8 = C2 × C2 , where v = 1
8
ei .
8                                                   2   i=1
8
(v , v ) =   4   = 2 and E8 = D8 , v is an even unimodular lattice.
θE8 = θD8 + θv +D8 = 1 + 240q + 2160q 2 + 6720q 3 + 17520q 4 +
30240q 5 + . . . = 1 + 240(q + 9q 2 + 28q 3 + 73q 4 + 126q 5 + . . .)
D8#

E8                L        E8

D8
Theta series as holomorphic functions.

In the following we will consider even lattices L and the associated
n
integral quadratic form Q : L → Z, → 1 ( , ) = 1 j=1 2 .
2         2       j

Theorem.
Deﬁne q(z) := exp(2πiz) and H := {z ∈ C | (z) > 0} the upper half
plane. The function
∞
θL : H → C, z → θL (z) =         exp(2πiz)Q( ) =         |La |q(z)a
∈L                     a=0

is a holomorphic function on the upper half plane H.
It satisﬁes θL (z) = θL (z + 1).
The theta series of the dual lattice.

Poisson summation formula.
For any well behaved function f : Rn → C and any lattice L ∈ Ln

det(L)1/2         f (x) =           ˆ(y )
f
x∈L             y ∈L#

where ˆ(y ) =
f         Rn
f (x) exp(−2πi(x, y ))dx is the Fourier transform of f .

Theorem.
z n/2
Let L ∈ Ln . Then θL ( −1 ) =
z         i        det(L)−1/2 θL# (z).
Proof.
z n/2
Proof of θL ( −1 ) =
z                 i     det(L)−1/2 θL# (z).
Both sides are holomorphic functions on H, so it sufﬁces to prove the
identity for z = it and t ∈ R>0 .
The Fourier transform of
−2π                √n
f (x) = exp(       Q(x)) is ˆ(y) = t exp(−2πtQ(y)).
f
t
Hence Poisson summation yields

−1                                         ˆ(y) = det(L)−1/2 t n/2 θL# (it).
θL (      )=         f (x) = det(L)−1/2           f
it
x∈L                         y∈L#

Poisson summation:

det(L)1/2         f (x) =           ˆ(y)
f
x∈L             y ∈L#
The space of modular forms.
The group of biholomorphic mappings of the upper half plane
H := {z ∈ C | (z) > 0} is the group of Mobius transformations
¨

az + b             a     b
z → A(z) :=           , A=                   ∈ SL2 (R).
cz + d             c     d

For all k ∈ Z this yields an action |k of SL2 (R) on the space of
meromorphic functions f : H → C deﬁned by

az + b
f |k A(z) := (cz + d)−k f (          ).
cz + d

Deﬁnition.
A holomorphic function f : H → C is called a modular form of weight
k , f ∈ Mk , if
f |k A = f for all A ∈ SL2 (Z)
and f is holomorphic at i∞.
0
f is called cuspform, f ∈ Mk , if additionally limt→∞ f (it) = 0.
Fourier expansion.
az+b
Remember: f |k A(z) := (cz + d)−k f ( cz+d ).

1   1                0     1
SL2 (Z) = T :=                  , S :=
0   1                −1    0
1
where S acts on H by z → − z and T by z → z + 1.
Theorem.
A holomorphic function f : H → C is a modular form of weight k , if
f (z) = f (z + 1) and f ( −1 ) = (−z)k f (z) and f is holomorphic at i∞.
z

Theorem.
Let f ∈ Mk for some k . Then f (z) = f |k T (z) = f (z + 1) and hence f
has a Fourier expansion
∞                       ∞
f (z) =         cn exp(2πiz)n =         cn q(z)n
n=0                     n=0
The graded ring of modular forms.

az+b
Remember: f |k A(z) := (cz + d)−k f ( cz+d ).
Since |k is multiplicative

(f |k A)(g|m A) = (fg)|k +m A

for all A ∈ SL2 (R) the space of all modular forms is a graded ring
∞
M :=         Mk .
k=0

Theorem.
Mk = {0} if k is odd.
Proof: Let A = −I2 ∈ SL2 (Z) and f ∈ Mk . Then
f |k A(z) = (−1)k f (z) = f (z) for all z ∈ H and hence f = 0 if k is odd.
Eisenstein series.

Theorem.
Let k ≥ 4 be even. Then the Eisenstein series

Gk (z) :=                       (mz + n)−k
(m,n)∈Z2 −{(0,0)}

is a modular form of weight k .
Proof. Gk is holomorphic on H because k ≥ 4.
Clearly Gk (z + 1) = Gk (z).
To obtain Gk ( −1 ) we calculate
z

1            k          z        k             1     k
=                   = zk
−m(1/z) + n              nz − m                 nz − m

and conclude that Gk ( −1 ) = z k Gk (z).
z
Normalized Eisenstein series and ∆.

Normalized Eisenstein series.
1
Ek (z) :=   2ζ(k) Gk (z)   has Fourier expansion

∞
2k
Ek (z) = 1 −                σk−1 (m)q m
Bk
m=1

where σs (m) =        d|m   d s . In particular
∞
E4 (z) = 1 + 240            m=1   σ3 (m)q m
∞
E6 (z) = 1 − 504            m=1   σ5 (m)q m

The ∆ function.
1     3          2         0
∆(z) :=     1728 (E4 (z)   − E6 (z)) ∈ M12 has Fourier expansion

∆(z) = q − 24q 2 + 252q 3 − 1472q 4 + . . .
Fundamental domain.
A fundamental domain for the action of SL2 (Z) on H by T : z → z + 1
1
and S : z → − z is

F := {z ∈ C | |z| ≥ 1 and   (z) ∈ [−1/2, 1/2) and |z| > 1 if   (z) > 0}.

i

−1       −1/2             1/2      1
The order formula.
Let f : H → C be holomorphic and p ∈ H.
∞
Then f (z) = n=0 bn (z − p)n for some bn ∈ C.
vp (f ) := min{n | bn = 0} is called the order of f at p.
∞
Write f (z) = n=0 cn q n . Then vi∞ (f ) := min{n | cn = 0}.
So vi∞ (f ) > 0 if f is a cuspform.

Order formula.
Let 0 = f ∈ Mk . Then

1          1                                   k
vi∞ (f ) +     vi (f ) + vω (f ) +              vp (f ) =      .
2          3                                   12
i,ω=p∈F

√
where ω = −1+i 3 .
2
Proof in the case that f has no zeros on the boundary C of F using
the argument principle:

1        f (z)
dz =         vp (f ).
2πi   C    f (z)
p∈F
f (z)
Calculation of       C f (z) dz.
Assume no zeros on C and choose (A), (E) large enough such
that f has no zeros inside F ∩ { (z) ≥ (A)}
B       C        D         E       A
=           +        +        +        +
C       A           B        C         D       E
A                         E

B               Ci        D

−1       −1/2                     1/2       1
B      E
Now   A
=−    D
since T (AB) = (ED) and f is invariant under T .
Moreover S(BC) = (DC) and f (Sz) = z k f (z) so
C             D f (z)
Calculation of (                B   +         C ) f (z) dz.

Remember S(BC) = (DC) and f (Sz) = z k f (z) so

f (Sz)   k  f (z)
= +
f (Sz)  z   f (z)

and hence
C       D                         C                                  C
f (z)                      f (z) f (Sz)                  k         k
(       +       )          dz =            (          −        )dz = −         dz = 2πi .
B       C        f (z)            B         f (z)   f (Sz)           B   z         12
A
Computation of E ff (z) dz:
(z)

Substituting q = exp(2πiz) transforms (EA) to a circle in which 0 is
the only possible zero of f . so
A
f (z)
dz = −2πivi∞ (f ).
E        f (z)
Conclusion of proof.
Order formula.
1          1                                  k
vi∞ (f ) +     vi (f ) + vω (f ) +             vp (f ) =      .
2          3                                  12
i,ω=p∈F

1     f (z)
p∈F vp (f ) = 2πi C f (z) dz =
1    B f (z)        C f (z)      D f (z)             E f (z)            A f (z)
2πi ( A f (z) dz + B f (z) dz + C f (z) dz        +   D f (z)
dz   +     E f (z)
dz)
1             k
2πi (0 + 2πi 12 − 2πivi∞ (f )).
k
so vi∞ (f ) + p∈F vp (f ) = 12 .
A               E

B      C        D
The order formula.

Hence we have proved the order formula.
Order formula.
Let 0 = f ∈ Mk . Then

1          1                                  k
vi∞ (f ) +     vi (f ) + vω (f ) +             vp (f ) =      .
2          3                                  12
i,ω=p∈F

Theorem.
(a) Mk = {0} for k odd, for k < 0, and for k = 2.
(b) M0 = C, Mk = CEk for k = 4, 6, 8, 10.
(c) Mk−12 ∼ Mk via multiplication by ∆ := 1728 (E4 − E6 ).
= 0                                1    3   2

Proof: (a) follows from the order formula since vp (f ) ∈ Z≥0 .
(b) follows from (c) and the fact that the C-linear mapping Mk → C,
∞
f = n=0 cn q n → c0 has kernel Mk . 0
Proof of (c).

Remember: vi∞ (f ) + 1 vi (f ) + 1 vω (f ) +
2           3             i,ω=p∈F   vp (f ) =   k
12 .

0
Need to show f → ∆f , Mk−12 → Mk is an isomorphism.

0
Now ∆ ∈ M12 and vi∞ (∆) = 1. So the map is well deﬁned and
injective. Moreover vp (∆) = 0 for all p ∈ H.
0             g
For g ∈ Mk the function ∆ ∈ Mk−12 is holomorphic since vi∞ (g) ≥ 1
and hence a preimage of f .

Corollary.
M = C[E4 , E6 ].
∞
M := k=0 M4k = C[E4 , ∆].
Even unimodular lattices have dimension 8d.
Theorem.
Let L = L# ∈ Ln be even. Then n ∈ 8Z.
Proof. Assume not. Replacing L by L ⊥ L or L ⊥ L ⊥ L ⊥ L, if
necessary, we may assume that n = 4 + 8m. Then by Poisson
summation
−1      z
θL (Sz) = θL (      ) = ( )n/2 θL (z) = −z n/2 θL (z)
z       i
and since θL is invariant under T , we hence get

θL ((TS)(z)) = −z n/2 θL (z)

where (TS)(z) = −1 + 1 =
z
z−1
z .   (TS)2 (z) =   −z
z−1   +1=    −1
z−1 .   Since
(TS)3 = 1 we calculate
1
θL (z) = θL ((TS)3 z) = θL ((TS)(TS)2 z) = −( z−1 )n/2 θL ((TS)2 z)
1 n/2 z−1 n/2                1 n/2
= ( z−1 ) ( z ) θL ((TS)z) = ( z ) θL ((TS)z) = −θL (z)

Theta series of even unimodular lattices are modular
forms

Theorem.
If L = L# ∈ Ln is even, then θL (z) ∈ Mk with k = n .2
In particular the weight of θL is half of the dimension and hence a
multiple of 4 so θL ∈ M .
Proof. θL (z) = θL (z + 1) because L is even.
From the Poisson summation formula we get

−1   z    n/2
θL      =            det L−1/2 θL# (z) = z n/2 θL# (z)
z    i
since n is a multiple of 8 and det(L) = 1.
Theta series of certain lattices.

Corollary.
Let L be an even unimodular lattice of dimension n.
∞
If n = 8 then θL = θE8 = E4 = 1 + 240     m=1   σ3 (m)q m .
If n = 16 then
2
θL = θE8 ⊥E8 = E4 = 1 + 480q + 61920q 2 + 1050240q 3 + . . ..
For n = 24 let c1 = |L1 | be the number of roots in L.
Then θL = 1 + c1 q + (196560 − c1 )q 2 + . . . .
Let L be an even unimodular lattice of dimension 80 with
minimum 8. Then | Min(L)| = 1 250 172 000.
Extremal modular forms.
∞
M =         M4k = C[E4 , ∆]
k=0

E4 = θE8 = 1 + 240q + . . . ∈ M4 ,         ∆ = 0 + q + . . . ∈ M12 .
Basis of M4k :
k
E4 =             1+ 240q+ ∗q 2 +           ...
k −3
E4 ∆ =                 q+ ∗q 2 +           ...
E4 −6 ∆2 =
k
q2+             ...
.
.
.
E4 −3a ∆a =
k
...                qa+          ...
n        k
where a =    24   =   3   .
Deﬁnition.
This space contains a unique form
(k)            (k )
f (k) := 1 + 0q + 0q 2 + . . . + 0q a + fa+1 q a+1 + fa+2 q a+2 + . . .

f (k) is called the extremal modular form of weight 4k .
Extremal even unimodular lattices.

Theorem (Siegel).
(k )                    (k)
fa+1 > 0 for all k and fa+2 < 0 for large k (k ≥ 5200).

Corollary.
Let L be an n-dimensional even unimodular lattice. Then
n
min(L) ≤ 2 + 2      .
24
Lattices achieving this bound are called extremal.

Extremal even unimodular lattices L≤ Rn
n      8    16    24        32     48          56   72   80
min(L)    2     2    4         4      6            6    8   8
number of
extremal    1    2      1      ≥ 106    ≥3     many      ?    ≥2
lattices

```
DOCUMENT INFO
Categories:
Stats:
 views: 34 posted: 9/4/2009 language: English pages: 32
How are you planning on using Docstoc?