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Theta functions and weighted theta functionsof Euclidean lattices, with some applications

Noam D. Elkies1

March, 2009

0. Introduction and overview

[. . . ]

1. Lattices in Rn: basic terminology, notations, and examples

By “Euclidean space” of dimensionn we mean a real vector space of dimensionn,equipped with a positive-definite inner product〈·, ·〉. We usually call such a space“Rn” even when there is no distinguished choice of coordinates.

A lattice in Rn is a discrete co-compact subgroupL ⊂ Rn, that is, a discrete sub-group such that the quotientRn/L is compact (and thus necessarily homeomorphicwith then-torus(R/Z)n). As an abstract groupL is thus isomorphic with the freeabelian groupZn of rankn. ThereforeL is determined by the images, call themv1, . . . , vn, of the standard generators ofZn under a group isomorphismZn ∼→ L.We say thevi generate, or aregeneratorsof, L: each vector inL can be written as∑n

i=1 aivi for someuniqueintegersa1, . . . , an. Vectorsv1, . . . , vn ∈ Rn generatea lattice if and only if they constitute anR-linear basis forRn, and thenL is theZ-span of this basis. For instance, theZ-span of the standard orthonormal basise1, . . . , en of Rn is the latticeZn. This more concrete definition is better suited forexplicit computation, but less canonical because most lattices have no canonicalchoice of generators even up to isometries ofRn.

An equivalent approach definesL as a free abelian group of rankn with a positive-definite symmetric pairing; the Euclidean space is then recovered asL ⊗Z R,which inherits a symmetric bilinear pairing fromL. With this approach we must becareful about the definition of a “positive-definite pairing” so that the extension toL⊗Z R remains positive-definite. For most of the lattices we consider, the pairingtakes values inQ, and then the usual definition for inner products suffices:

v ∈ L, v 6= 0 =⇒ 〈v, v〉 > 0. (1)

1Department of Mathematics, Harvard University, Cambridge, MA 02138 USA; e-mail:(elkies@math.harvard.edu). Supported in part by NSF grant DMS-0501029.

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But for a generalR-valued pairing, (1) guarantees only that the pairing onL⊗Z R

is positive semidefinite, not necessarily positive definite: a standard counterexam-ple isL = Z2 with

〈v, w〉 = (v1 − tv2)(w1 − tw2)

for some irrationalt, when the nonzero vectorx = (t, 1) ∈ L ⊗Z R satisfies〈x, x〉 = 0. For this general case we say the pairing〈·, ·〉 is “positive-definite” ifit has a positive-definite Gram matrixA. TheGram matrixof 〈·, ·〉 with respect togeneratorsv1, . . . , vn of L is the symmetric matrix with(i, j) entryAij = 〈vi, vj〉.The Gram matrix depends on the choice of generators, but ifw1, . . . , wn are anyother generators then eachwk =

∑ni=1 bikvi for some integersbik forming a matrix

B of determinant±1, and the Gram matrix with respect tow1, . . . , wn is BTAB,which is positive-definite if and only ifA is. If v = m1v1 + · · · +mnvn for somem = (m1, . . . ,mn) ∈ Zn then〈v, v〉 is the value atm of thequadratic form

(m,AmT) =n∑

i=1

n∑

j=1

Aijmimj =n∑

i=1

n∑

j=1

〈vi, vj〉mimj , (2)

a homogeneous polynomial of degree2 in n variables.

We next recall some further invariants2 of L and the corresponding properties ofthe Gram matrix. Thediscriminantof L is

discL = (Vol(Rn/L))2 = detA; (3)

in particular,detA is independent of the choice of generators, which we can alsoverify directly: if detB = ±1 thendetA = detBTAB. The volume

√discL of

the torusRn/L is known as thecovolumeof L. A lattice of discriminant1 is saidto beunimodular. Thedual latticeL∗ is defined by

L∗ = {v∗ ∈ Rn | ∀v ∈ L, 〈v, v′〉 ∈ Z}. (4)

If L is theZ-span ofv1, . . . , vn with Gram matrixA, thenL∗ is theZ-span of thedual basisv∗1 , . . . , v

∗n with Gram matrix(AT)−1; in particular

discL∗ = (discL)−1. (5)

We sayL is integral if 〈v, v′〉 ∈ Z for all v, v′ ∈ L; equivalently, ifL ⊆ L∗. In thiscaseL∗/L is a finite group with#(L∗/L) = disc(L). In particular,L = L∗ if andonly if L is integral and unimodular; we naturally say such a lattice isself-dual.

2These are “invariant” in the sense that they do not depend on achoice of generators, nor onother extrinsic features such as an embedding inR

n. The Gram matrix depends on the choice ofgenerators, and is thus not an invariant: different Gram matrices may give rise to the same lattice.

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The basic example of a lattice isZ with the pairing〈x, y〉 = xy; this lattice is self-dual, and is the unique unimodular lattice in the1-dimensional Euclidean spaceR.More generally, for every realD > 0 there is a unique lattice of discriminantDin R, namelyD1/2Z, or equivalentlyZ with the pairing〈x, y〉 = Dxy instead;this lattice is integral if and only ifD ∈ Z.

We next give some constructions of new lattices from old. A subgroupL′ of finiteindex in a latticeL in Rn is itself a lattice inRn whose dual containsL∗. Com-paring covolumes, we see that[L′∗ : L∗] = [L : L′]; in fact much more is true: theinner product onRn induces a perfect pairing(L′∗/L∗) × (L/L′) → Q/Z on thequotient subgroups, so in particular these subgroups are isomorphic, albeit not ingeneral canonically isomorphic.

If v1, . . . , vn generateL andw1, . . . , wn generateL′, then [L : L′] = |detB|,whereB is the integer matrix formed by the coefficientsbik of the expansionswk =

∑ni=1 bikvi. If A is the Gram matrix ofL with respect to thevi, then as

beforedetBTAB is the Gram matrix ofL′ with respect to thewk. Therefore

discL′ = [L : L′]2 discL, (6)

an identity that can also be obtained from the first equality in (3) because

Vol(Rn/L′) = [L : L′] Vol(Rn/L).

If L is integral then so isL′. In the other direction, ifL′ is integral, letG be anyset of generators ofL/L′, andG an arbitrary lift ofG to a subset ofL; thenL isintegral if and only ifL ⊂ L′∗ and〈g, g′〉 ∈ Z for all g, g′ ∈ G.

If n = n1 + n2 andL1 andL2 are lattices inRn1 andRn2 respectively, then thedirect sum

L = L1 ⊕ L2 := {(v1, v2) | v1 ∈ L1, v2 ∈ L2} (7)

is a lattice inRn of discriminantdisc(L1) disc(L2); if Ai (i = 1, 2) is a Grammatrix forLi thenL has a block-diagonal Gram matrix with blocksA1, A2. Thedual ofL1 ⊕ L2 is L∗

1 ⊕ L∗2. The direct sumL is integral if and only ifL1 and

L2 are integral, and self-dual if and only ifL1 andL2 are self-dual. For example,Z2 = Z⊕Z ⊂ R2 is self-dual; iterating the construction yields the self-dual latticeZn ⊂ Rn for eachn = 1, 2, 3, . . ..

Of course oncen > 1 this self-dual lattice is no longer unique, because we canobtain uncountably many others by applying an orthogonal linear transformationto Rn; but the resulting lattices are isomorphic. It is known thatfor n ≤ 7 everyself-dual lattice is isomorphic withZn (see Proposition 7 below for one approachto this result), and for everyn there are only finitely many isomorphism classes.

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But for largen the number of isomorphism classes grows rapidly, exceeding(cn)n2

for some positivec. We conclude this section by constructing the self-dual latticeE8 ⊂ R8 and showing thatE8 6∼= Z8.

Forn = 1, 2, 3, . . ., define a latticeDn ∈ Rn by

Dn ={

(x1, . . . , xn) ∈ Zn :

n∑

i=1

xi ≡ 0 mod 2}

, (8)

a sublattice3 of Zn of index2. An explicit Z-basis consists ofei + ei+1 for 0 <i < n together with2e1. The dual latticeD∗

n is the union ofZn and the translateof Zn by the half-lattice vector

h := (1, 1, . . . , 1)/2 =1

2

n∑

i=1

ei. (9)

The4-element quotient groupD∗n/Dn is cyclic if n is odd, and of exponent2 when

n is even. In the latter case,2h ∈ Dn, so

D+n := Dn ∪ (Dn + h) (10)

is a lattice, which is unimodular because[D+n : Dn] = 2 = [Zn : Dn]. An explicit

basis consists ofei + ei+1 for 0 < i < n− 1 together with2e1 andh. This latticeis integral, and thus self-dual, if and only if4|n, because〈h, h〉 = n/4. Havingclaimed that forn < 8 every unimodular lattice inRn is isomorphic withZn,we should haveD+

4∼= Z4, and indeedh, h − e1, h − e2, h − e3 are orthonormal

generators forD+4 . But if n 6= 4 thenD+

n 6∼= Zn, becauseD+n contains no vectorsx

with 〈x, x〉 = 1. In particular,D+8 is a self-dual lattice not isomorphic withZ8.

The latticeD+8 is usually calledE8; it has some remarkable properties (some of

which we shall see later) not shared byD+n for n 6= 8. It is known (see for instance

[CS2, p.49 and Table 16.7]) that for eachn ≤ 13 every self-dual lattice inRn isisomorphic with one ofZn, Zn−8 ⊕ E8 (for n ≥ 8), or Zn−12 ⊕D+

12 (for n = 12or n = 13). Forn = 8 this means every self-dual lattice inR8 is isomorphic witheitherZ8 orE8; see Exercise 3.5 below (and Exercise 3.6 for the classification forall n ≤ 15).

3Warning: since we regard any latticeL as living in a Euclidean spaceRn = L ⊗ R, anysubgroupL′ is also a lattice in some Euclidean space, but possibly a proper subspace ofRn —indeedL′ is a lattice inRn if and only if the index is[L : L′] is finite. But we callL′ a “sublattice”whether or not it is of finite index; thus a “sublattice of a lattice inR

n” is not necessarily a “latticein R

n” in our sense.

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Exercises

1.1 i) Every latticeL in Rn has, for each integerm > 0, a sublatticemL. How arethe Gram matrix, discriminant, and dual lattice ofmL related to those ofL?ii) For any latticeL in Rn and any positiveα ∈ R, defineL〈α〉 to be the latticeobtained fromL by multiplying the pairing byα. For example,mL ∼= L〈m2〉.How are the Gram matrix, discriminant, and dual lattice ofL〈α〉 related to thoseof L?

1.2 i) Verify that the4-element groupD∗n/Dn is cyclic if and only ifn is odd, and

thatDn is irreducible (not the direct sum of two sublattices of positive dimension)except forD2

∼= Z〈2〉 ⊕ Z〈2〉(∼= Z2〈2〉). [For the irreducibility ofDn for n ≥ 3you might consider the vectorsx ∈ Dn with 〈x, x〉 = 2, which is the smallestvalue of〈x, x〉 for nonzerox ∈ Dn.]ii) For all n = 1, 2, 3, . . ., each of the latticesZn andDn contains2(n2−n) vectorsx with 〈x, x〉 = 2, all equivalent under automorphisms of the lattice.iii) The same is true ofD+

n for eachn ≡ 0 mod 4, except thatD+8 = E8 has240

such vectors, not112. Can you find an automorphism ofE8 mappinge1 + e2 to h?[From the existence of such an automorphism we readily deduce thatAut(E8) actstransitively on all240 vectorsx ∈ E8 such that〈x, x〉 = 2; we later give severalother approaches to this result, each of which also lets us calculate the number ofautomorphisms.]

1.3 i) The latticeAn in then-dimensional Euclidean space

{

(x0, x1, . . . , xn) ∈ Rn+1 :n∑

i=0

xi = 0}

is defined by

An ={

(x0, x1, . . . , xn) ∈ Zn+1 :n∑

i=0

xi = 0}

. (11)

Describe the latticeA2 ⊂ R2 geometrically. Show that eachAn is theZ-span ofthe independent vectorsei − ei−1 (i = 1, . . . , n). Use the Gram matrix for thesegenerators to finddisc(An) = n + 1. In particulardisc(A3) = disc(D3); showthat in factA3

∼= D3 by finding an explicit isometry.ii) Prove that the groupA∗

n/An of ordern + 1 is cyclic, generated by the coset ofthe vector

−e0 +1

n+ 1

n∑

i=0

xi ∈ A∗n.

Thus for each positive factord of n + 1 there exists a unique latticeA+dn ⊆ A∗

n

that containsAn with indexd. Thusdisc(A+dn ) = (n+ 1)/d2. What is the dual of

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A+dn ? Show thatA+d

n is integral if and only ifd2|n + 1, and self-dual if and onlyif d2 = n + 1. In particular we should haveA+2

3∼= Z3; prove this by finding an

explicit isomorphism. Show on the other hand thatA+38 6∼= Z8 because there is no

x ∈ Z+38 such that〈x, x〉 = 1. Thus we should haveA+3

8∼= E8 ; can you find an

explicit isometry?

1.4 Recall that a square matrixA is called “tridiagonal” ifAij = 0 for all i, j suchthat |i − j| > 1. Thus vectorsv1, . . . , vn have a tridiagonal matrix if and only ifany two vectors whose indices differ by more than1 are orthogonal. An exampleis our generatorsvi = ei − ei−1 of An. Show more generally that a latticeL has atridiagonal Gram matrix whose entries are all integers of absolute value at most2if and only ifL is the direct sum of lattices each isomorphic with eitherZ or someAn. Find a tridiagonal Gram matrix forE8 each of whose entries is one of0, 1, 2, 4,with 4 occurring only once.4

1.5 i) The integral latticeA+27 of discriminant(7+1)/22 = 2 has the special name

E7. Show thatE7 contains no vectorsx such that〈x, x〉 = 1, and126 vectorsxsuch that〈x, x〉 = 2.ii) Find a self-dual latticeL ⊃ E7⊕E7, and prove that it is the unique such lattice.(Hint: start by showing thatL must be contained in the dual ofE7 ⊕ E7.) ProvethatL contains no vectorx such that〈x, x〉 = 1, and thus thatL is not isomorphicwith any of the self-dual latticesZ14, Z6⊕E8, andZ2⊕D+

12 that we already knowin 14-dimensional space.

We can now account for all the self-dual lattices tabulated in [CS2, p.49] forn <16: they areZn, Zn−8 ⊕ E8 (for n ≥ 8), Zn−12 ⊕ D+

12 (for n ≥ 12), the latticeL of exercise 1.5ii forn = 14, andZ ⊕ L andA+4

15 for n = 15. As before, wecan check thatA+4

15 is distinct from the other four self-dual lattices we know inthis dimension by verifying that it contains no vectorsx with 〈x, x〉 = 1. We latergive one approach to proving the completeness of that list ofself-dual lattices forn < 16, see Proposition 7 and Exercise 3.6 below.

4Iwaniec [Iw, p.176] exhibits a tridiagonal Gram matrix forE8 with much larger entries, namely2, 2, 4, 4, 20, 12, 4, 2 on the main diagonal and1, 1, 3, 5, 3, 1, 1 next to it. Curiously this alreadyappeared in [H2,§7], presented as a quadratic polynomial inx1, . . . , x8, though with the coefficients2, 2, 4, 4 misprinted as1, 1, 2, 2. Perhaps Iwaniec presented this form only to silently correct Hecke’serror, but that still leaves the mystery of how the tridiagonal form was obtained in the first place, sincea considerably simpler one was available; even if we requireeach off-diagonal entry to be odd, wequickly find the example with diagonal2, 2, 2, 2, 2, 6, 2, 4 and off-diagonal1, 1, 1, 1, 1, 1, 3, 1 bycomputer search. The same form appears (in matrix notation,with the correct coefficients) in [Sch,p.520], where it is attributed to Minkowski. Indeed Minkowski [Min, p.77] gives this tridiagonalmatrix, as well as the Gram matrix for what we now call theE8 root system (with some sign changes),with Aii = 2, Aij = 1 when|i − j| = 1, and all otherAij zero except forA25 = A52 = −1.

We shall meet tridiagonal Gram matrices at least once later (Exercise 5.2).

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2. Theta functions of lattices

By thenormof a vectorx in a lattice or inner-product space we mean〈x, x〉 (andnot the length〈x, x〉1/2). For any latticeL ⊂ Rn and eachk ∈ R, define

Nk(L) = #{v ∈ L | 〈v, v〉 = k}. (12)

The set is finite; indeed for anyk0 ∈ R the sum∑

k≤k0

Nk(L) = #{v ∈ L | 〈v, v〉 ≤ k0} (13)

is finite becauseL is discrete and the subset{〈x, x〉 ≤ k0} in Rn is compact (aclosed ball ifk0 ≥ 0, empty ifk0 < 0). TheNk(L) are important invariants ofL.For example, in the context of sphere packing one is interested in theminimal(nonzero) normof L, which is the smallestk > 0 for whichNk(L) > 0, and thekissing numberof L, which is the value ofNk(L) for thatk.5 The theta functionor theta seriesΘL is a generating function that encodes these invariantsNk(L):

ΘL(q) :=∑

v∈L

q〈v,v〉/2 = 1 +∑

k>0

N2k(L)qk. (14)

Note thatN0(L) = 1; the factors of2 are convenient as we soon see. The sum

in (14) converges absolutely if0 ≤ q < 1, because the sum (13) isOL(kn/20 ) as

k0 → ∞ (indeed it is asymptotic tocLkn/20 , wherecL is disc(L)−1/2 times the

volumeπn/2/Γ((n/2) + 1) of a unit sphere inRn. This makes it easy to justifythe following derivation of the product formula for the theta function of a directsum:

ΘL1⊕L2(q) =

∑

(v1,v2)∈L1⊕L2

q〈v1+v2,v1+v2〉/2

=∑

(v1,v2)∈L1⊕L2

q(〈v1,v1〉+〈v2,v2〉)/2

=∑

v1∈L1

q〈v1,v1〉/2∑

v2∈L2

q〈v2,v2〉/2

= ΘL1(q) ΘL2

(q). (15)

For example,

ΘZn(q) = ΘZ(q)n =

( ∞∑

m=−∞qm2/2

)n

. (16)

5The minimal norm is(2r)2 wherer is the radius of the largest ball inRn whose translates byLhave disjoint interiors. These translates constitute the sphere packing associated withL. Each ofthem is tangent toκ others, whereκ is the kissing number.

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A more remarkable identity relates the theta functions of a lattice and its dual:

Proposition 1 (functional equation for theta series)For any latticeL in Rn we have

ΘL∗(e−2πt) = disc(L)1/2t−n/2ΘL(e−2π/t). (17)

for all t > 0.

Already the first example, withn = 1 andL = L∗ = Z, is surprising and impor-tant: ∞

∑

m=−∞e−πm2t = t−1/2

∞∑

m=−∞e−πm2/t. (18)

A famous application is Riemann’s proof of the analytic continuation and func-tional equation of the zeta functionζ(s) =

∑∞m=1m

−s: multiply ΘZ(e−2πt) − 1

by ts/2 dt/t and integrate termwise over0 < t <∞ to find

∫ ∞

0(ΘZ(e−2πt) − 1) ts/2dt

t= 2

∞∑

m=1

∫ ∞

0e−πm2t ts/2dt

t

= 2

∞∑

m=1

(πm2)−s/2Γ(s/2)

= 2π−s/2Γ(s/2)ζ(s) = ξ(s) (19)

for Re(s) > 0; then split the integral as∫∞0 =

∫ 10 +

∫∞1 and apply (18) to

∫ 10 to

obtain

ξ(s) +1

s+

1

1 − s=

1

2

∫ ∞

1(ΘZ(e−2πt) − 1)(ts/2 + t(1−s)/2)

dt

t(20)

for 0 < Re(s) < 1. To recover the analytic continuation ofξ(s) to all of C (withsimple poles ats = 0 ands = 1), observe that the integral in (20) is an analyticfunction ofs on all ofC, becauseΘZ(e−2πt)− 1 decays exponentially ast→ ∞;the functional equationξ(s) = ξ(1 − s) then follows from the symmetry of theintegral unders↔ 1 − s.

Applying the same definite integral toΘL for a general latticeL in Rn, but usingts instead ofts/2, we obtain

∫ ∞

0(ΘL(e−2πt) − 1) ts

dt

t= π−sΓ(s)ζL(s) = ξL(s), (21)

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whereζL is the zeta function ofL, defined by

ζL(s) :=∑

v∈Lv 6=0

〈v, v〉−s, (22)

andξL is defined by the last equality in (21). Transforming the integral as before,and using the functional equation (17) relatingΘL with ΘL∗ , we obtain the identity

ξL∗

(n

2− s)

= disc(L)1/2ξL(s). (23)

See Exercises 2.3 and 2.5ii below.

For a rather more frivolous application of (18) (and one admittedly unrelated to ourmain topic), differentiate both sides of (18) with respect to t and sett = 1 to find6

d

dtΘZ(e−2πt)

∣

∣

∣

∣

t=1

= −1

4ΘZ(e−2π),

whence

1 + 2∞∑

m=1

e−πm2t = ΘZ(e−2π) = 8π∞∑

m=1

m2e−πm2t.

Therefore8π < eπ + 2, with a rather small difference

eπ + 2 − 8π ≈ (32π − 2)e−3π = 0.00795+ (24)

becausee−3π is tiny. Subtracting from this the error22 − 7π = 0.00885+ in thefamiliar approximationπ ≈ 22/7 yields the striking

eπ − π = 19.999099979+,

which has even featured in anxkcd comic [Mu]. For a variation on this theme seeExercise 2.4 below.

Returning to our main thread, we review the proof of the functional equation forΘL because we shall re-use the technique several times later. The key tool is thePoisson summation formula inRn, which relates the sums of a Schwartz functionoverL and of its Fourier transform overL∗. Here aSchwartz functionis aC∞

functionf : Rn → C such thatf and all its partial derivatives decay aso(〈x, x〉k)for all k as〈x, x〉 → ∞. TheFourier transformf : Rn → C is defined by

f(y) =

∫

x∈Rn

f(x) e2πi〈x,y〉 dµ(x), (25)

and is a Schwartz function iff is.6The next identity was certainly known to Riemann, if not to Poisson himself: an equivalent

form appears in the last displayed equation before formula (2) in [Ed, p.17], where Edwards recitesRiemann’s derivation of a series representation forξ( 1

2+ it).

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Theorem 2 (Poisson summation in Rn)LetL be any lattice inRn. Then

∑

x∈L

f(x) = (discL)−1/2∑

y∈L∗

f(y) (26)

for all Schwartz functionsf : Rn → C.

Note that they = 0 term in (26) is

(discL)−1/2f(0) =1

Vol(Rn/L)

∫

x∈Rn

f(x) dµ(x). (27)

After multiplying (26) bydisc(L)1/2, we can thus interpret the left-hand side as aRiemann sum approximating the integralf(0). The termsf(y) for nonzeroy ∈ L∗

then measure the discrepancy between this Riemann sum and the integral.

Proof of Poisson summation: DefineF : Rn → C by

F (z) =∑

x∈L

f(x+ z). (28)

Becausef is Schwartz, the sum converges absolutely to aC∞ function, whosevalue atz = 0 is the left-hand side of (26) SinceF (z) = F (x+ z) for all z ∈ Rn

andx ∈ L, the function descends to aC∞ function onRn/L, and thus has aFourier expansion

F (z) =∑

y∈L∗

F (−y) e2πi〈y,z〉, (29)

where

F (y) =1

Vol(Rn/L)

∫

z∈Rn/LF (z) e2πi〈z,y〉 dµ(z). (30)

Note that the vectorsy ∈ L∗ are exactly those for whiche2πi〈x,y〉 is well-definedon Rn/L. Now choose a fundamental domainR for Rn/L; for instance, letv1, . . . , vn be generators ofL and setR = {a1v1 + · · · + anvn : 0 ≤ ai < 1}.Then we have

Vol(Rn/L)F (y) =

∫

z∈RF (z) e2πi〈y,z〉 dµ(z)

=

∫

z∈R

∑

x∈L

f(x+ z) e2πi〈y,z〉 dµ(z)

=∑

x∈L

∫

z∈R−xf(z) e2πi〈y,z〉 dµ(z)

=

∫

z∈Rn

f(z) e2πi〈y,z〉 dµ(z) = f(y), (31)

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where we used in the last step the fact thatRn is the disjoint union of the translatesR− x of R by lattice vectors. Thus (29) becomes

F (z) =1

Vol(Rn/L)

∑

y∈L∗

f(−y) e2πi〈y,z〉. (32)

Takingz = 0 we obtain (26), Q.E.D.

The functional equation (17) is then the special casef(x) = exp(−π〈x, x〉/t)of (26).

Proof of the functional equation (17) for theta series: Letf(x) = exp(−π〈x, x〉/t)in (26). We claim thatf(y) = tn/2 exp(−π〈y, y〉t). Choosing any orthogonalcoordinates(x1, . . . , xn) for Rn, we see that the integral (25) definingf(y) factorsas

n∏

j=1

∫ ∞

−∞e−πx2

j/te2πixjyj dxj ,

which reduces our claim to the casen = 1, which is the familiar definite integral∫ ∞

−∞e−πx2/t e2πixy dx = t1/2e−πty2

.

Using thesef andf in the Poisson summation formula (26) we deduce the func-tional equation (17), Q.E.D.

Exercises

2.1 How are the theta series and zeta function ofL〈α〉 related withΘL andζL?

2.2 Having given in (16) a formula forΘZn in terms ofΘZ, we find formulas forthe theta functions ofDn, D∗

n, andD+n in terms ofΘZ and two further functions7

ΦZ,ΨZ, defined for0 ≤ q < 1 by

ΦZ(q) =∞∑

m=−∞(−1)mqm2/2 = 1 − 2q1/2 + 2q4/2 − 2q9/2 + − · · · , (33)

ΨZ(q) =∞∑

m=−∞q(m+ 1

2)2/2 = 2

(

q1/8 + q9/8 + q25/8 + q49/8 · · ·)

. (34)

Thus

Φ Z(q) = ΘZ(q) − 2ΘZ(q4), ΨZ(q) = ΘZ(q1/4) − ΘZ(q). (35)

7The use ofΦZ, ΨZ for these functions is an ad hoc notation. Usually these would be written asthe value atu = 1 of Jacobi’s theta functionsϑ4, ϑ2; likewiseΘZ is Jacobi’sϑ3 evaluated atu = 1.

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Prove that forn = 1, 2, 3, . . .

ΘDn(q) =1

2

(

ΘZ(q)n + ΦZ(q)n)

, ΘD∗n(q) = ΘZ(q)n + ΨZ(q)n, (36)

and forn even

ΘD+n(q) =

1

2

(

ΘZ(q)n + ΨZ(q)n + ΦZ(q)n)

, (37)

In particular, sinceD+n∼= Z4 we have Jacobi’s identity

ΘZ(q)4 = ΨZ(q)4 + ΦZ(q)4. (38)

2.3 Complete the derivation of the functional equation (23)relatingξL andξL∗ ,showing along the way thatξL is holomorphic except for simple poles ats = 0 ands = n/2. What are the residues at these poles?

2.4 Show thatA∗2∼= A2〈1/3〉. (The geometric description ofA2 from Exercise 1.3i

should help.) Use this and the functional equation forΘA2 to prove

e2π/√

3 > 8√

3π − 6.

The two sides are not nearly as close as in (24), despitee−2π/√

3 being even smallerthane−π; why?

2.5 [Sphere packing bounds from Poisson summation]

2.6 We can get further use of the formula (32) by evaluating both sides atz /∈ L.Here are two examples for the casen = 1:i) Taking z = 1/2, L = L∗ = Z, andf(x) = exp(−πx2/t), obtain a functionalequation relatingΨZ andΦZ. Check that this equation is consistent with the resultof applying theΘZ functional equation (18) to the formulas (35) forΨZ andΦZ interms ofΘZ. Verify also that your equation, and the formulas (36,37), are consis-tent with (17) forL = Dn andL = D+

n .ii) Let χ12 be the Dirichlet character mod12 defined byχ12(1) = χ12(−1) = 1,χ12(5) = χ12(−5) = −1 (andχ12(m) = 0 if 2|m or 3|m). Show that iff is aSchwartz function onR then

∞∑

m=−∞χ12(m)f(m) = 12−1/2

∞∑

n=−∞χ12(n)f(n/12).

Letting f(x) = e−πx2t, obtain an identity analogous to (18), and deduce a func-tional equation for the DirichletL-seriesL(s, χ12) =

∑∞m=1 χ12(m)m−s.

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2.7 The Schwartz condition is much more restrictive than is needed to justify Pois-son summation (though Schwartz functions suffice for all ourapplications to lat-tices and theta series).8 Show for instance that our derivation of (26) is valid alsofor n = 1 andf(x) = e−|x|, f(y) = 2/((2πy)2 + 1), and use this to evaluatein closed form

∑∞n=1 1/(n2 + c2) for c > 0. Verify that your answer approaches

ζ(2) = π2/6 asc→ 0.

3. Theta functions of self-dual lattices as modular forms for Γ+

We next considerΘL as a function of a complex variable. For general latticesLwecannot make sense ofΘL(q) as a function ofq in a neighborhood ofq = 0 in C,because the exponents〈v, v〉/2 in (14) need not be integers. However, the changeof variablesq = e−2πt suggested by the functional equation (17) yields a functionof t that extends to a holomorphic function on the half-planeRe(t) > 0. Thatfunctional equation then extends to this half-plane, either by analytic continuationor by using the same proof.

Suppose now thatL is self-dual. ThenΘL = ΘL∗ , so the functional equation (14)relates the values att and1/t of the same function. Also, each exponent〈v, v〉/2 isin 1

2Z becauseL is integral. ThusΘL(e−2πt) is also invariant under the imaginarytranslationt 7→ t+ 2i. Combining this invariance with the functional equation wethen obtain further identities, one for each fractional linear transformation gener-ated byt 7→ 1/t andt 7→ t+ 2i.

We make the coefficients of these transformations integral using the further changeof variablet = iτ . Thenτ is in the Poincare upper half-plane

H = {τ ∈ C | Im(τ) > 0},

andq = e2πiτ . (39)

Our transformationst 7→ 1/t andt 7→ t+ 2i then become

S : τ 7→ −1/τ, T 2 : τ 7→ τ + 2

acting onH. We use the notationT 2 because we later need also the translationT : τ 7→ τ + 1.

Recall that the group of orientation-preserving isometries of H with respect tothe hyperbolic metric|dτ |/ Im(τ) consists of the fractional linear transformations

8For example, Poisson summation holds if there existsδ > 0 for which bothf(x) andf(x) areO(〈x, x〉−(n/2)−δ) as〈x, x〉 → ∞ [SW, Ch.VII, Cor. 2.6]; this includes the examplef(x) = e−|x|

in Exercise 2.6 withn = 1.

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γ : τ 7→ (aτ + b)/(cτ + d) with a, b, c, d ∈ R and ad − bc > 0, uniquelydetermined up to scaling(a, b, c, d) 7→ (λa, λb, λc, λd). We may regardγ as theprojective linear transformation of the(τ : 1)-line that takes

(

τ1

)

to(

a bc d

)(

τ1

)

. Thus

we composeγ’s by multiplying the corresponding2 × 2 matrices(

a bc d

)

. Requiringad−bc = 1 determinesλ up to sign; we thus identify our group of fractional lineartransformation ofH with PSL2(R) = SL2(R)/{±1}.

Now the mapsS : τ 7→ −1/τ andT : τ 7→ τ + 2 become the integer matricesS =

(0 −11 0

)

andT2 =(1 20 1

)

of determinant1. These generate a subgroup〈S, T 2〉of the full modular group

Γ := SL2(Z)/{±1} = PSL2(Z)

BecauseT2 is congruent mod2 to the identity matrixI, everyγ ∈ 〈S, T 2〉 iscongruent to eitherI or

(

0 11 0

)

mod 2. (The reduction of(

a bc d

)

∈ PSL2(Z) mod2

is well-defined because−(

a bc d

)

≡(

a bc d

)

mod 2.) It turns out that this necessarycondition is also sufficient:〈S, T 2〉 is the group, call itΓ+, of all PSL2(Z) matricescongruent mod2 to eitherI or

(0 11 0

)

. See Exercise 3.1 below for one approach tothe identification of〈S, T 2〉 with Γ+.

We are thus led to define for any latticeL in Rn the function

θL(τ) := ΘL(e2πiτ ) = 1 +∑

k>0Nk(L) 6=0

Nk(L)eπikτ , (40)

onH. If L is self-dual thenθL satisfies the functional equations

θL(τ + 2) = θL(τ), θL(−1/τ) = (τ/i)n/2θL(τ), (41)

where “(τ/i)n/2” is thenth power of the principal square root ofτ/i (that is, thesquare root with positive real part). Iterating these functional equations yields foreachγ ∈ 〈S, T 2〉 = Γ+ an identity

θL(γ(τ)) = ǫnc,d(cτ + d)n/2θL(τ) (42)

for someǫc,d ∈ C∗ with ǫ8c,d = 1.9 A holomorphic functionφ : H → C satisfying

φ(γ(τ)) = ǫnc,d(cτ + d)n/2φ(τ) for all γ = ±(a bc d

)

∈ Γ+ is a “weakly modularform of weightn/2 for Γ+” (more fully, for Γ+ and theǫnc,d; these factors are saidto form a “multiplier system of weightn/2” [Iw, 2.6] — a multiplier system being a

9We write ǫc,d rather thanǫγ because if someγ′ has the same(c, d) asγ thenγ′ = T 2kγ forsomek ∈ Z, soθL(γ(τ)) = θL(γ′(τ)) soγ andΓ′ have the sameǫ; also,γ determines(c, d) onlyup to sign, and changing(c, d) to (−c,−d) multipliesǫc,d by±i.

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system of factors in an identity such as (42) that is consistent with θL(γ1(γ2(τ)) =θL((γ1γ2)(τ)) for all choices ofγ1, γ2.) We shall soon see thatθL also satisfies theadditional condition that make it modular, not just weakly modular; this will let uscite results from the theory of modular forms that for eachn confineθL to an affinevector space of dimension⌊n/8⌋. To motivate this additional condition we mustfirst review a few further facts concerning the action ofΓ andΓ+ onH.

The relevant results forΓ are very well known:

Theorem 3 i) Γ = PSL2(Z) is generated byS andT .ii) The action ofΓ onH has a fundamental domain

F = {τ ∈ H : |Re(τ)| ≤ 1/2, |τ | ≥ 1}. (43)

The second assertion means that everyΓ-orbit has a representative inF , which isunique if it is in the interior ofF . Thus the images ofF underΓ coverH anddo not overlap except on the boundaries; see Figure 1 for a picture of part of thistiling of H. See [Se, VII, Theorems 1 and 2] for an exposition that nicelyprovesboth parts of Theorem 3 together. WhileF is not compact, it is closed inH and asequence{zj} with no accumulation point inF must approach the “cusp”i∞ inthe sense thatIm zj → ∞.

R

t

i

T−1F F TF

T−1SF SF TSF

−2 −1 0 1 2

Figure 1: The fundamental domainF for Γ and some of its nearby images

The corresponding facts forΓ+ are:

Theorem 4 i) Γ+ is generated byS andT 2.ii) The action ofΓ+ onH has a fundamental domain

F+ = {τ ∈ H : |Re(τ)| ≤ 1, |τ | ≥ 1}. (44)

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This can be proved by adapting the argument of [Se, VII] for Theorem 3. Alterna-tively it can be derived from Theorem 3 as follows. For part (i), see Exercise 3.1 be-low. For part (ii), we first show that[Γ : Γ+] = 3. (This, as well as part (i), is notedby Serre in the concluding “Complements” section of [Se, VII].) We saw alreadythat reduction mod2 gives a well-defined homomorphismΓ → SL2(Z/2Z). Thehomomorphism is readily seen to be surjective; for example,check that the imagesof S andT generateSL2(Z/2Z). SinceSL2(Z/2Z) has order6, the preimageΓ+

of the2-element subgroup{1,(

0 11 0

)

} has index6/2 = 3 as claimed. We then obtaina fundamental domain forΓ+ by forming the union of the images ofF under thecoset representatives1, T, TS. The fundamental domainF+ is then obtained byapplyingT−2 ∈ Γ+ to the right half ofTF ∪ TSF . See Figure 2.

R

t

i

F

−1 0 1

Figure 2: The fundamental domainF+ dissected into three images ofF

We see thatF+ has two cusps, at∞ and±1. For a general finite-index subgroupΓ′ ⊆ Γ, we can similarly get a fundamental domain by combining[Γ : Γ′] imagesof F , and the cusps may be identified with the orbits of the action of Γ′ on Q ∪{∞} = P1(Q).

Now our theta functionθL not only satisfies the identity (42) that makes it weaklymodular, but also remains bounded asIm τ → ∞ — that is, asτ approaches thecuspi∞ — becauseq → 0 asτ → i∞. For eachγ ∈ Γ+ it follows via (42) that(Im τ)−n/2θL(γ(τ)) remains bounded asτ → i(∞), as long asτ remains inF .Changing variables we see that(Im τ)n/2θL(τ) remains bounded asτ → a/c =γ(i∞), as long asτ remains inγF . This does not constrain the growth ofθL(τ)asτ approaches the cusp1 or its images underΓ+. But we can prove directly:

Lemma 5 Let L be a lattice inRn. Then for anyt0 > 0 the functionτ 7→(Im τ)n/2θL(τ) is bounded on the strip{τ ∈ H : Im τ ≤ t0}.

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Proof: If Im τ = t then

|tn/2θL(τ)| = tn/2|ΘL(e2πiτ )| ≤ tn/2|ΘL(e−2πt)|,

because each of the termseπi〈v,v〉τ in the sum definingΘL(e−2πt) has absolutevalue equal to the corresponding terme−π〈v,v〉t in ΘL(e−2πt). Hence by the func-tional equation (17) we have

|tn/2θL(τ)| ≤ disc(L)1/2ΘL(e−2π/t),

and ΘL(e−2π/t) ≤ ΘL(e−2π/t0), again because the inequality holds termwise.This gives the upper bounddisc(L)1/2ΘL(e−2π/t0) on |(Im τ)n/2θL(τ)|, Q.E.D.

This suggests the following definitions: fix a finite-index subgroupΓ′ of Γ andsome multiplier system{εc,d} (for all (c, d) occurring as the bottom row of someγ ∈ Γ′). A holomorphic functionφ : H → C satisfying

φ(γ(τ)) = εc,d(cτ + d)n/2φ(τ)

for all γ = ±(a bc d

)

∈ Γ′ is aweakly modular form of weightn/2 for Γ′ and theεc,d;if moreover|(Im τ)n/2φ(τ)| is bounded in stripsIm τ ≤ t0 thenφ is a modularform of weightn/2 for Γ′ and theεc,d. The boundedness condition is equivalentto one growth condition for each cusp ofH/Γ′. Thus Lemma 5, combined withthe preceding analysis, states thatif L is self-dual thenθL is a modular form ofweightn/2 for Γ+ and the multiplier system{ǫnc,d}.

GivenΓ′, m, and theεc,d, these modular forms constitute a vector space overC,denoted byMn/2(Γ

′, {εc,d}), or simplyMn/2(Γ′) if the εc,d are known. This is

a finite-dimensional vector space, and much is known about the coefficients of itselements. We shall see that for any latticeL ⊂ Rn with rational inner productsthe theta functionθL is inMn/2(Γ

′, {εc,d}) for someΓ′ andεc,d. This can be usedto obtain very precise statements on theNk(L), that is, on the radial distributionof lattice vectors. We shall also study the angular distribution by generalizingθL

to “weighted theta functions” and showing and show that they, too, are modularforms. For the rest of this section and the next we illustratethis by using thestructure ofMn/2(Γ

′) for Γ′ = Γ+ andΓ′ = Γ to study self-dual lattices via theirvectors’ radial distribution.

If {ε(1)c,d} and{ε(2)

c,d} are multiplier systems of weightsn1/2 andn2/2 for Γ′, then

{ε(1)c,dε

(2)c,d} is a multiplier system of weight(n1+n2)/2, and the product of modular

forms inMn1/2(Γ′, {ε(1)

c,d}) andMn2/2(Γ′, {ε(2)

c,d}) is inM(n1+n2)/2(Γ′, {ε(1)

c,dε(2)c,d}).

This happens in our setting whereΓ′ = Γ+ and we use for each weightn/2 the

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multipliers ǫnc,d of (42). (Note that this is consistent with the identity (15): thechange of variableq = e2πiτ transforms that identity toθL1

θL2= θL1⊕L2

, equat-ing a modular form of weight(n1 + n2)/2 with the product of forms of weightsn1/2 andn2/2.) This gives the direct sum

M+ :=∞⊕

n=0

Mn/2(Γ+) (45)

the structure of a graded algebra. For generalΓ′ such algebras can be quite com-plicated, but ourM+ is known to have the following simple description:

Theorem 6 The algebraM+ is freely generated overC by the modular formsθZ

of weight1/2 andθE8of weight4; equivalently, byθ

Zand10

∆+ :=1

16(θ8

Z − θE8) = q1/2 − 8q + 28q3/2 − 64q2 + − · · · . (46)

Thus eachMn/2(Γ+) (n ≥ 0) has dimension1 + ⌊n/8⌋ and basis

{θn−8mZ

∆m+ : m = 0, 1, 2, . . . , ⌊n/8⌋}. (47)

This theorem can be proved starting from the description ofF+ in much the sameway that Serre obtains the generators of the algebra of modular forms forΓ ([Se,VII, Theorem 4]; Theorem 11 below).

Corollary. LetL be a self-dual lattice inRn. Then

θL = θnZ +

⌊n/8⌋∑

m=1

cmθn−8mZ

∆m+ (48)

for some constantscm (m = 1, 2, . . . , ⌊n/8⌋).Proof: By Theorem 6 we haveθL =

∑⌊n/8⌋m=0 cmθ

n−8mZ

∆m+ for somecm (m =

0, 1, 2, . . . , ⌊n/8⌋); so we need only showc0 = 1. But c0 is the value of the sum atq = 0, which isN0(L) = 1 as desired.

In particular, the coefficientsNk(L) of θL for k = 1, 2, . . . , ⌊n/8⌋ determineθL,because we can use them to calculate thecm iteratively. We can already use this toprove our earlier claim:

10The notation “∆+” for this form is not standard; we use it here in analogy to theweight-12 form∆ which plays a similar role in the next section.

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Proposition 7 If n < 8 then every self-dual lattice inRn is isomorphic withZn.

Proof: Here⌊n/8⌋ = 0 so our Corollary determines the theta series completely:θL = θ

Zn . Comparing coefficients, we deduceNk(L) = Nk(Zn) for all k. In

particular,N1(L) = 2n, becauseN1(Zn) = 2n, as may be seen either directly or

by expanding (16) in powers ofq1/2. ThusN1(L) = 2n, soL containsn pairs±vi

(1 ≤ i ≤ n) of vectors with〈vi, vi〉 = 1. For i 6= j we then have|〈vi, vj〉| < 1 byCauchy-Schwarz; sinceL is integral, it follows that〈vi, vj〉 = 0. That is, thevi areorthonormal. ThereforeL contains theirZ-span, call itL0, which is isomorphicwith Zn. But thenL = L∗ ⊆ L∗

0 = L0, soL = L0∼= Zn, Q.E.D.

See Exercises 3.5 and 3.6 for the classification of self-dualL ⊂ Rn with n = 8 and9 ≤ n ≤ 15; Exercise 3.7 for an application of the casen = 2 to prove Fermat’stwo-squares theorem, which determinesNk(Z

2) whenk is an odd prime; and Ex-ercise 3.8 for a different construction ofθ

Z2 that yields a formula forNk(Z2) for

all k.

We conclude this section with a warning: we used the fact thatθL = θZn to prove

L ∼= Z2, but in general it is possible for latticesL andL′ to have the same theta se-ries (equivalently: to satisfyNk(L) = Nk(L

′) for all k) without being isomorphic.We say more about this later, and give an example with self-dual lattices inR16.

Exercises

3.1 Verify that the matricesS =(0 −11 0

)

andT =(1 10 1

)

satisfyS2 = (ST)3 = −I,and thus that their imagesS, T in PSL2(Z) = Γ satisfyS2 = (ST )3 = 1. Usethis, and the fact thatΓ = 〈S, T 〉, to prove thatΓ+ = 〈S, T 2〉. [Write anyγ ∈ Γ+

asg1g2g3 . . . gr where eachgj ∈ {S, T, T−1}, and eliminate any occurrences ofSS, T−1T , or TT−1 in the product. Then, working from one side (say from theleft), combine pairs withgj = gj+1 = T±1 to obtain a product of factors in{S, T 2, T−2}. If an impasse arises, use(ST )3 = I to replaceTST or T−1ST−1

by ST−1S or STS respectively; note thatT−1ST = T−2TST andTST−1 =T 2T−1ST−1. This process either terminates or proves thatγ /∈ Γ+ by writing γ asthe product of an element of〈S, T 2〉 with eitherT or TS.]

3.2 Define aq-series

η =∞∑

n=1

χ12(n)qn2/24

= q1/24(1 − q − q2 + q5 + q7 − q12 − q15 + q22 + −− + · · · ), (49)

whereχ12 is the even Dirichlet character mod12 introduced in Exercise 2.5(ii).Use the identity proved in that Exercise to show thatη satisfies, for eachγ =

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±(

a bc d

)

∈ Γ, a functional equation

η(γ(τ)) = εc,d(cτ + d)1/2η(τ) (50)

for someεc,d ∈ C∗ with ε24c,d = 1. We thus say thatη is a modular form of weight

1/2 for Γ and theεc,d (not just weakly modular, becauseη is clearly bounded atthe one cuspq → 0).

3.3 i) LetL be an integral lattice inRn that contains a vectorv of norm1. Showthat everyx ∈ L can then be written uniquely asx′ + nv for somen ∈ Z andx′ ∈ L with 〈x′, v〉 = 0, and thus that such vectorsx′ constitute a latticeL′ in theorthogonal complement ofRv, whenceL ∼= L′ ⊕ Z.ii) More generally, letL be an integral lattice inRn, andL1 ⊂ L a lattice in somesubspaceV of dimensionn1. If L1 is self-dual, show thatL = L1 ⊕ L′ whereL′ = L ∩ V ⊥. (Hint: givenx ∈ L, use the homomorphismL1 → R, v 7→ 〈x, v〉.Part (i) is the special caseL1 = Zv.)

3.4 Either by iterating the construction in part (i) of the previous exercise, or usingpart (ii) directly, prove that every integral latticeL in Rn can be uniquely writtenasL′ ⊕ L1 whereL1

∼= Zn1 for some integern1 ∈ [0, n] andL′ is a lattice inRn−n1 that contains no vectors of norm1. (This gives an alternative conclusion ofthe proof that forn < 8 every self-dual lattice inRn is isomorphic withZn; heren1 = n andL′ is the zero lattice.)

3.5 Show that ifL is a self-dual lattice inR8 then eitherL ∼= Z8 or θL = θE8.

[In particular, in the latter caseN2(L) = N2(E8) = 240. If you know about rootsystems then this is all you need to deduceL ∼= E8, becauseE8 is the only rootlattice of rank at most8 with as many as240 vectors of norm2. ThusL has asublatticeL1 of finite index isomorphic withE8, and then[L : L1] = 1 becausedisc(E8) = 1, soL = L1.]

3.6 More generally, show that if0 < n < 16 andL is a self-dual lattice inR8

without vectors of norm1 thenn ≥ 8 andθL = θnZ− 2nθn−8

Z∆+. In particular,

N2(L) = 2n(23 − n). [Using this plus the classification of root systems we canshow thatE8 is the only such lattice withn < 12. For9 ≤ n ≤ 15, it follows fromProposition 7 and Exercise 3.3ii that the root system cannotcontainE8; this leavesonly the root systemsD12 for n = 12, E2

7 for n = 14, andA15 or A4 ⊕ D11 forn = 15. In each case we then know a sublattice of finite index inL. Because thissublattice must have square discriminant,A4 ⊕ D11 cannot occur. In each of theremaining cases the root lattice determinesL uniquely as we saw in Exercises 1.2,1.3, and 1.5 above.]

3.7 (Fermat’s two-squares theorem) Letp be an odd prime. Fermat’s two-squarestheorem asserts thatp can be represented as a sumx2

1+x22 of squares of two integers

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if and only if p ≡ 1 mod 4, in which case the representation is unique up to theobvious transformationsx1 ↔ −x1, x2 ↔ −x2, andx1 ↔ x2. Equivalently: ifp ≡ −1 mod 4 thenNp(Z

2) = 0; and ifp ≡ +1 mod 4 thenNp(Z2) = 8 and the

dihedral group of symmetries ofZ2 acts simply transitively on the8 vectors with〈x, x〉 = p. We give a version of a standard proof of this theorem that concludesby invoking the casen = 2 of Proposition 5.

If p = x21 + x2

2 thenx2 6= 0 so we may letr = x1/x2 ∈ Z/pZ, so thatr2 = −1.By Legendre’s criterion,−1 is a square inZ/pZ if and only if (−1)(p−1)/2 ≡+1 mod p. This already proves the casep ≡ −1 mod 4. If p ≡ +1 mod 4 thenthere are two square roots of−1 in Z/pZ; let r be one of them. The vectors(x1, x2) ∈ Z2 such thatx1 ≡ rx2 mod p constitute a lattice, call itLp. Wewant x ∈ Lp such that〈x, x〉 = p. Prove thatLp〈1/p〉 is integral. Check that[Z2 : Lp] = p, and thus thatdisc(Lp) = p2. Conclude thatLp〈1/p〉 is self-dual.ThusLp〈1/p〉 ∼= Z2 by Proposition 5. In particularLp〈1/p〉 contains four unitvectors. Use this to complete the proof of the two-squares theorem.

3.8 (The coefficientsNk(Z2) of θ

Z2) 11 Recall that thehyperbolic cosineis de-

fined bycosh z = cos(iz) = (ez + e−z)/2, and thehyperbolic secantis sech z =1/ cosh(z) = sec(iz) = 2/(ez + e−z). It is a known application of contour inte-gration that fort > 0 (or evenRe(t) > 0) the Fourier transform ofsech(πtx) ist−1 sech(πt−1y). Use this to prove that

s2(τ) :=

∞∑

m=−∞sech(mπiτ) =

∞∑

m=−∞sec(mπτ) (52)

is a modular form of weight1 for Γ+. Sinces2(τ) → 1 asτ → i∞ we must have

11The formulas2(τ) = θZ(τ)2 = θZ2(τ) proved in this Exercise is over a century old; the proof

via modular forms is also not new but its origins are harder totrack down. Dickson [Di, p.235]attributes the identity

(1 + 2q + 2q4 + 2q9 + · · · )2 = 1 + 4q

1 − q− 4

q3

1 − q3+ 4

q5

1 − q5− · · · (51)

to Jacobi, in a “letter to Legendre, Sept. 9, 1828”; the right-hand side is obtained from the formula(51) for s2(τ) by interchanging the sums overm andd. For Jacobi and others, including Dicksonhimself [KD], the identity arises in the theory of elliptic functions and the proof does not explicitlyuse the fact thats2 andθ2

Z are modular. Much more recently, the development in [SS, Ch.10, §3.1,pages 297–304] closely follows the modular approach of our Exercise; this approach is also notedin [Coh, p.106, Remark (6)] though without deducing the formula for Nk(Z2). But this too wasprobably known much earlier.

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s2(τ) = θZ(τ)2 = θZ2(τ). Then expands2(τ) in powers ofq:

s2(τ) = 1 + 4∞∑

m=1

qm/2

1 + qm= 1 + 4

∞∑

m=1

∞∑

d=1

χ4(d)qdm/2, (53)

whereχ4 is the Dirichlet character mod 4 defined byχ4(±1) = ±1 andχ4(0) =χ4(2) = 0. Deduce the formula for the numberNk(Z

2) of representations ofan integerk > 0 as the sum of two squares:Nk(Z

2) = 0 if there is a primep ≡ 3 mod 4 such that thep-valuationvp(k) is odd; otherwise,

Nk(Z2) = 4

∏

p≡1 mod 4

(1 + vp(k)). (54)

Recover Fermat’s two-squares theorem as the special case thatk is an odd prime.

4. The characteristic coset and shadow of a self-dual lattice

For any integral latticeL the identity

〈v + w, v + w〉 = 〈v, v〉 + 2〈v, w〉 + 〈w,w〉 (55)

yields a homomorphism

L→ Z/2Z, v 7→ 〈v, v〉 mod 2. (56)

Suppose further thatdiscL is odd. Then〈·, ·〉 mod 2 gives an isomorphism fromL/2L to LHom(L,Z/2Z); in particular there is a unique cosetC of 2L in L,called theharacteristic coset[Se, Ch. V], that maps to the homomorphism (56).Thusw ∈ C if and only if 〈v, w〉 ≡ 〈v, v〉 mod 2 for all v ∈ L. Scaling by1/2 yields a coset ofL in 1

2L called theshadowof L; we denote the shadow bys(L). For example, the latticeZ has characteristic cosetC = 2Z + 1 and shadows(Z) = 1

2C = Z + 12 , and likewise forZ〈α〉 for any odd integerα > 0. The

characteristic coset and shadow are additive: ifL1 andL2 are integral latticesof odd discriminant with characteristic cosetsC1 andC2, then the integral latticeL1 ⊕L2 has odd discriminant and its characteristic coset and shadow areC1 ⊕C2

ands(L1) ⊕ s(L2) respectively. For example, the characteristic vectors ofZn arethe vectors all of whose coordinates are odd.

Now the vectors in any coset of2L in L have the same norm mod4, but for thecharacteristic coset we actually get congruence mod8. Thus ifw,w′ ∈ C we maywritew′ = w + 2v for somev ∈ L, so

〈w′, w′〉 = 〈w + 2v, w + 2v〉 = 〈w,w〉 + 4(〈v, w〉 + 〈v, v〉) ≡ 〈w,w〉 mod 8.

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Thus we have a lattice invariantσ(L) mod 8 which is the common residue mod8of the norms of all characteristic vectors. (Equivalently,all shadow vectors havequarter-integral norm congruent to14σ(L) mod 2Z.) For example,σ(Z〈α〉) = αbecause every odd square is congruent to1 mod 8. Again we have additivity: iflatticesL1, L2 have odd discriminant thenσ(L1 ⊕ L2) = σ(L1) ⊕ σ(L2). Forexample,σ(Zn) = n. The self-dual latticesD+

n (4|n) also have characteristicvectors of norm≡ n mod 8 (see Exercise 4.1). In fact it turns out thatσ(L) = nfor any self-dual latticeL ⊂ Rn. This can be proved algebraically, for exampleby extendingσ to indefinite quadratic forms of discriminant±1 and studying thearithmetic of such forms as in [Se, Ch.I–V]. In our positive-definite setting thecharacteristic coset and shadow arise naturally when we study the transformationof θL under arbitraryγ ∈ Γ (not justγ ∈ Γ+). In the course of this study we shallgive a proof ofσ(L) = n for self-dualL ⊂ Rn using theta functions.

We already knowθL(γ(τ)) for all γ ∈ Γ+. We shall determineθL(T (τ)) andθL(TS(τ)); this will suffice to obtainθL(γ(τ)) for all γ ∈ Γ becauseT andTSare representatives of the two nontrivial cosets ofΓ+ in Γ.

The action ofT : τ 7→ τ + 1 is easy: this map takeseπiτ to −eπiτ , so for anyintegral latticeL we have simply

θL(T (τ)) = θL(τ + 1) = 1 +∞∑

k=1

(−1)kNk(L)eπikτ .

The formula forθL(TS(τ)) can be expressed in terms of the theta series for theshadow ofL; as suggested by our notationΨZ in Exercise 2.2, we define

ΨL(q) :=∑

v∈s(L)

q〈v,v〉/2 = 1 +∞∑

k=1

N2k(s(L))qk, (57)

ψL(τ) := ΨL(e2πiτ ) =∑

k>0Nk(s(L)) 6=0

Nk(s(L))eπikτ , (58)

whereNk(s(L)) = #{v ∈ s(L) | 〈v, v〉 = k}. (59)

Then we have:

Proposition 8 For any self-dual latticeL in Rn we have

θL(TS(τ)) = (τ/i)n/2ψL(τ) (60)

for all τ ∈ H.

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Proof: Letw be a characteristic vector. Then

θL(T (τ)) = θL(τ + 1) =∑

v∈L

(−1)〈v,v〉eπi〈v,v〉τ =∑

v∈L

eπi(〈v,v〉τ+〈v,w〉) (61)

because(−1)c = eπic for any integerc. We now apply Poisson summation to thesum. The Fourier transform ofexpπi(〈x, x〉τ+〈x,w〉) is the integral overx ∈ Rn

of exp(πi(〈x, x〉τ+〈x,w + 2y〉)), which is to say the value aty+ w2 of the Fourier

transform ofexpπi〈x, x〉τ , which we already know is(τ/i)−n/2 exp(−πi〈x, x〉/τ).Poisson summation then gives

θL(T (τ)) = (τ/i)−n/2∑

v∈L

e−πi〈v+ w2

,v+ w2〉/τ = (τ/i)−n/2

∑

v∈s(L)

e−πi〈c,c〉/τ

(62)which is(τ/i)−n/2ψL(S(τ)); replacingτ by Sτ we recover (60), Q.E.D.

Corollary. We haveσ(L) = n for any self-dual latticeL in Rn.

Proof: BecauseψL(t+ 1) = eπiσ(L)/4ψL(t), the claimσ(L) = n is equivalent to

ψL(t+ 1) = eπin/4ψL(t). (63)

Using Proposition 8, together withS2 = (ST )3 = 1 andθL(T 2τ) = θL(τ), wecalculate

(

t+ 1

i

)n/2

ψL(t+ 1) = θL(TST (t)) = θL(ST−1S(t))

= (T−1S(t)/i)n/2θL(T−1S(t)) =

(

i(t+ 1)

t

)n/2

θL(TS(t))

(in which we usedS2 = (ST )3 = 1 and the invariance ofθL underT 2, and againwriten/2 power to meannth power of principal square root). Comparing with (60)yields the desired identity (63), Q.E.D.

See Exercise 4.4 for an alternative approach, and Exercise 4.3 for an application tothe determination ofσ(L) for latticesL ⊂ Rn of (odd) prime discriminant.

The identity (60) can be used to study the norm distribution of unimodular latticesand their shadows. For instance, we have seen that the characteristic coset ofZn

consists of the vectors all of whose coordinates are odd, whence it contains2n

vectors of normn but no vectors of norm less thann. We showed in [E1] that thislatter property characterizes theZn lattice:

Theorem 9 [E1] LetL ⊂ Rn be a self-dual lattice with no characteristic vectorw such that〈w,w〉 < n. ThenL ∼= Zn.

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Proof:12 By Theorem 6 we can write

θL =

⌊n/8⌋∑

m=0

cmθn−8mZ

θmE8

(64)

for some constantscm. We shall see thatψE8= θE8

(Exercise 4.4 below, and atgreater length in the next section). Hence by (60) we have

ψL =

⌊n/8⌋∑

m=0

cmψn−8mZ

θmE8. (65)

We haveψZ

= 2q1/4 +O(q9/4) asq → 0, while θE8= 1 +O(q2). By hypothesis

ψL = O(qn/4). Hencecm = 0 for m > 0, andθL = c0θnZ

. SinceN0(L) = 1it follows that c0 = 1, soL has the same theta function asZn. In particularN1(L) = 2n. It follows as in Exercises 3.3 and 3.4 thatL ∼= Zn, Q.E.D.

This characterization ofZn answered a question that arose in the geometry of4-manifolds (where self-dual lattices can arise via the intersection pairing on thesecond homology group). See also [E2] for the determinationof all self-dualL ⊂ Rn whose characteristic coset has minimal normn − 8, and [CS1, RS] forthe use of (60) to obtain upper bounds on the minimal nonzero norm ofL.

Exercises

4.1 Prove directly thatσ(D+n ) = n for all n ≡ 0 mod 4 by showing that(n/2)ei is

a characteristic vector ofD+n for any i = 1, 2, . . . , n. Find a characteristic vector

of the self-dual latticeA+415 (see Exercises 1.3 and 1.5 above), and check that its

norm is≡ 7 mod 8.

4.2 i) If L is an integral lattice of odd discriminant, andL1 ⊂ L is a sublattice ofodd index, thenσ(L1) = σ(L).ii) Let L ⊂ Rn be an integral lattice of odd discriminant, andL1 ⊂ L a lattice ofodd discriminant in some subspaceV ⊂ Rn. ThenL′ := L ∩ V ⊥ also has odddiscriminant andσ(L′) = σ(L) − σ(L1). [Show thatL1 ⊕ L′ has (finite and) oddindex inL.]

4.3 LetL ⊂ Rn be an integral lattice whose discriminant is an odd primep.SupposeL∗ contains a vectorv∗ such thatp〈v∗, v∗〉 ≡ −1 mod p. Prove that

12The proof in [E1] uses an alternative, and probably preferable, approach to the key fact that sucha lattice must haveθL = θZn , which avoids the explicit determination ofMn/2(Γ+) in Theorem 6or indeed any explicit mention of modular forms. Instead we use that fact thatθZ vanishes at thecuspτ = ±1 but nowhere inH, deducing in effect thatθL/θZn is in M0(Γ+) and thus constant.

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σ(L) = n + 1 − p. (Apply 4.2i to a suitable lattice containingL ⊕ Z〈p〉.) Whathappens whenL∗ does not contain suchv∗? [In general ifL is an integral latticewith discL = p, andv∗ ∈ L∗ is any dual vector not inL, thenp〈v∗, v∗〉 is an inte-ger not divisible byp, and all choices ofv∗ yield integers with the same quadraticcharacter modp; thusp〈v∗, v∗〉 ≡ −1 mod p is one of only two possibilities.]

4.4 Show that0 is a characteristic vector ofE8, again consistent withσ(L) = n.Use Theorem 6 to derive (63), and thus the fact thatσ(L) = n, for all self-duallatticesL from the special casesL = Z andL = E8.

4.5 Generalize formula (60) to integral lattices of odd discriminant.

4.6 LetL ⊂ Rn be an integral lattice whose characteristic coset has minimalnormn − 8m0. Show that the coefficientcm of θn−8m

Z∆m

+ in the expansion (48)of θL vanishes for eachm > m0, while cm0 is (−1)m0212m0−nNn−8m0(C). Inparticular(−1)mcm0 > 0.

4.7 i) [E2] Use this to prove without using root systems that if L ⊂ Rn is a uni-modular lattice for somen < 12 such thatN1(L) = 0 thenn = 8 andθL = θE8

.(Use the formula forθL obtained in Exercise 3.6 and the fact thatNn−8m0(C) isan integer which is even ifn− 8m0 6= 0.)

ii) Deduce further that a nonzero integral lattice with no vectors of norm1 or 2must have rank at least23, and ifL ⊂ R23 is an integral lattice with no vectors ofnorm1 or 2 then

θL = θ23Z − 46θ15

Z ∆+ = 1 + 4600q3/2 + 93150q2 + 953856q5/2 + · · · .

[It is known that there is a unique such latticeL ⊂ R23, the “odd Leech lattice”or “shorter Leech lattice”; it can be obtained from the LeechlatticeΛ24 ⊂ R24

by choosingv0 ∈ Λ24 of norm4 (all suchv0 are equivalent underAut(Λ24)), andprojecting allv ∈ Λ24 such that2|〈v0, v〉 to the23-dimensional space(Rv0)⊥.]

5. Even self-dual lattices and their theta functions

An important special case is an integral latticeL (with no restriction ondisc(L))for which the homomorphism (56) is zero; that is, lattices such that〈v, v〉 ∈ 2Z forall v ∈ L. Such a lattice is said to beevenan integral lattice that is not even is saidto beodd(it contains vectors of odd norm). By (55),L is automatically integral if〈v, v〉 ∈ 2Z for all v ∈ L. More commonly we check thatL is even by verifyingthat it is integral and has generators of even norm; equivalently, that it has a Grammatrix with integer entries and even diagonal entries. As with integral lattices, thedirect sum of even lattices is even, as is any sublattice of aneven lattice.

A nonzero vector in an even lattice must have norm at least2. A norm-2 vector in

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an integral latticeL is called aroot. A lattice generated by its roots is known as aroot lattice; a root lattice is even because it is integral and generated by vectors ofnorm2. Such lattices are surprisingly ubiquitous, playing crucial roles in the theoryof Lie groups and algebras, in the study of singularities in algebraic geometry,and elsewhere; they are also important building blocks of even lattices and otherintegral lattices. The direct sum of root lattices is again aroot lattice, and theirreducible root lattices are precisely the latticesAn (n ≥ 1), Dn (n ≥ 4), andEn (n = 6, 7, 8). We have already encountered all of these exceptE6, which maybe defined as the sublattice ofE8 orthogonal to any isometric copy ofA2 in E8

(all are equivalent underAut(E8)). For instance, having constructedE8 asD+8 we

may choose the copy spanned bye1 − e2 ande2 − e3, and thenE6 is the sublatticeconsisting of vectors whose first three coordinates are equal. See Exercise 5.1 forthe discriminants and roots of the latticesAn, Dn, En. For any integral latticeLthe “root (sub)lattice ofL” is the sublattice generated by the roots ofL.

If the integral latticeL has odd discriminant thenL is even if and only if0 is acharacteristic vector, which is to sayL is its own shadow. In this caseσ(L) = 0,whence we have:

Theorem 10 (Schoeneberg [Sch, p.520])If Rn contains an even self-dual latticethenn ≡ 0 mod 8.

Proof: We have seen thatσ(L) = n for every self-dualL ⊂ Rn. If L is also eventhenσ(L) = 0 son vanishes inZ/8Z. Q.E.D.

Theorem 10 can be used as the starting point for proving congruences mod8 onthe ranks of even lattices of other discriminants; see for example 5.4i–iii (discrim-inant 2) and 5.7i–ii (even latticesL such asD4 for which L∗/L ∼= (Z/2Z)n/2).Exercise 5.2 gives a more offbeat application of Theorem 10 to Diophantine equa-tions.

A self-dual lattice is said to be ofType Ior Type IIdepending on whether it is oddor even. The necessary condition8|n on the existence of Type II lattice inRn

is also sufficient: if8|n thenRn contains the Type II latticeEn/88 (direct sum of

n/8 copies ofE8). For n = 8 this is the only such lattice. Forn = 16 thereis another Type II lattice, namelyD+

16, which is not isomorphic withE28 because

E28 is generated by its roots andD+

16 is not. We shall soon see that these arethe only Type II lattices inR16 up to isomorphism. InR24 there are24, as firstshown with great effort by Niemeier [Ni]. The classificationwas considerablysimplified by Venkov [Ve], using the weighted theta functions we introduce in thenext section. The Niemeier lattices are ubiquitous in many other classificationproblems for lattices of rank up to24. See Exercise 5.4iv for an example of one

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such use.

If L is a Type II lattice then its theta functionθL is invariant underT , not justT 2, becauseq never appears raised to half-integral powers, only integral ones. Be-causeθL still transforms underS by Proposition 1, we conclude that it transformsunder the full modular groupΓ generated byS andT . Moreover, since8|n theS transformation simplifies toθL(S(τ)) = τn/2θL(τ), whence we have for allγ = ±

(

a bc d

)

∈ Γ the identity

θL(γ(τ)) = (cτ + d)n/2θL(τ). (66)

In other words,the theta function of a Type II latticeL ∈ Rn is a modular form ofweightn/2 for Γ with trivial multiplier system.

While the weight ofθL is a multiple of4, the transformation(cτ + d)kφ(τ) =φ(γ(τ)) is well-defined for all evenk (becauseτ determines±(cτ +d)); as beforewe sayφ is amodular form of weightk for Γ if it is a holomorphic function onHthat satisfiesφ(γ(τ)) = (cτ + d)kφ(τ) for all γ ∈ Γ andφ remains bounded asτ → i∞. We shall later need such forms also fork ≡ 2 mod 4, so we considerthem together. Again the modular forms constitute a graded algebra, which we call

M :=⊕

k≥0k even

Mk(Γ); (67)

and again the algebra turns out to be freely generated. Here these generators arethe normalized Eisentein series13

E4 = 1 + 240∞∑

m=1

m3qm

1 − qm= 1 + 240q + 2160q2 + 6720q3 + · · · , (68)

E6 = 1 − 504∞∑

m=1

m5qm

1 − qm= 1 − 504q − 16632q2 − 122976q3 − · · · , (69)

of weights4 and6 respectively. The analogue of Theorem 6 thus reads:

Theorem 11 The algebraM is freely generated overC by the modular formsE4

andE6. In other words, eachMn(Γ) (n = 0, 2, 4, . . .) has basis

{Ea4E

b6 : a, b ≥ 0, 4a+ 6b = n}. (70)

13Warning: there is an unfortunate but unavoidable notational collision here between the latticeE8 and the Eisenstein seriesEk. Even worse, the usual indexing of Eisenstein series makesθE8

equalE4, notE8 which is the theta function of each of the Type II lattices inR16. We always use a

sanserifE for the Eisenstein series — which also suggests the mnemonicthat thisE, being thin, hashalf the weight.

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HencedimMn(Γ) is the number of possible(a, b) in (70), which is⌊n/24⌋ ifn ≡ 2 mod 12 and1 + ⌊n/24⌋ for other evenn ≥ 0:

n dimMn n dimMn n dimMn · · ·0 1 12 2 24 3 · · ·2 0 14 1 26 2 · · ·4 1 16 2 28 3 · · ·6 1 18 2 30 3 · · ·8 1 20 2 32 3 · · ·10 1 22 2 34 3 · · ·

(71)

See [Se, VII, Theorem 4] for a classic exposition of the proof.

Corollary. LetL be an even self-dual lattice inRn. Then

θL = En/84 +

⌊n/24⌋∑

m=1

cmEn−24m4 ∆m (72)

for some constantscm (m = 1, 2, . . . , ⌊n/8⌋), where

∆ :=E3

4 − E26

123= 1−24q+252q2−1472q3+4830q4−6048q5−16744q6 · · · (73)

Proof: Because8|n any element ofMn/2(Γ) is a polynomial inE4 andE26. We may

thus writeθL =∑⌊n/24⌋

m=0 cmθn−8mZ

∆m for somecm (m = 0, 1, 2, . . . , ⌊n/24⌋),and as with (48) we use evaluate atq = 0 to obtainc0 = N0(L) = 1 as desired.

Again it follows that the coefficientsN2k(L) of θL for k = 1, 2, . . . , ⌊n/24⌋ deter-mineθL, because we can use them to calculate thecm iteratively.

In particular, whenn < 24 there are no undetermined coefficients, soθL =

θn/8E8

= 1 + 30nq + O(q2), whenceL has30n roots. Forn = 8 we alreadyknow thatN2(L) = 240, and observed that this forcesL ∼= E8. The powerseries expansion (68) then determinesN2d(E8) for all integersd > 0, namelyN2d(E8) = 240

∑

d|n d3 [Sch, p.521]. This lattice yields a sphere packing ofR8

that is densest in its dimension among lattice packings (this was proved in 1935by Blichfeldt [Bl], who also obtained the corresponding results forE6 andE7; indimensionsn = 1, 2, 3, 4, 5 the densest lattice packings come from the root latticesA1, A2, A3, D4, D5). TheE8 packing is also expected to be the densest among allpackings of identical spheres inR8, and known to be within a factor1 + 10−14 ofthe maximal density ([CK], extending the computations of [CE] in the course ofgiving a new proof of its optimality among lattice packings).

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Forn = 16 we haveN2(L) = 480, and the only root systems of rank at most16that have480 roots areE2

8 andD16. In the former case the root lattice already hasdiscriminant1, soL ∼= E2

8 ; in the latter case,D16 ⊂ L ⊂ D∗16 with each inclusion

having index2; there are three such lattices, of which one is the odd lattice Z16

and the other two are isomorphic withD+16, soL ∼= D+

16. SincedimM4(Γ) = 1 italso follows thatθL is the normalized Eisenstein seriesE8, which yields the closedform

N2d(E28) = N2d(D

+216 ) = 480

∑

d|nd7 (74)

for the theta function coefficients.

The existence of two inequivalent latticesE28 andD+

16 that cannot be distinguishedby their theta functions has a nice consequence in differential geometry: as Mil-nor [Mil] observed, the toriR16/E2

8 andR16/D+16 are non-isometric but “isospec-

tral”, in that their Laplacians have the same eigenvalues with the same multiplic-ities. In the language of [Ka], we cannot hear the differencebetween these two16-dimensional drums. In general, for any latticeL ⊂ Rn the Laplacian on thetorusRn/L has an eigenbasis consisting of functionsx 7→ e2πi〈v∗,x〉 for v∗ ∈ L∗;the corresponding eigenvalue is−(2π)2〈v∗, v∗〉, so the toriRn/L1 andRn/L2 areisospectral if and only if14 θL∗

1= θL∗

2, or equivalently (by (17))θL1 = θL2 . Thus

if L1 6∼= L2 we obtain non-isomorphic isospectral tori. Subsequently exampleswith n < 16 were discovered; it is now known thatθL uniquely determinesL forlatticesL ⊂ Rn if n ≤ 3, but already forn = 4 (and thus for eachn ≥ 4, usingdirect sums withZ〈α〉n−4) there is a two-dimensional family of non-isomorphicpairs{L1, L2} with the same theta functions.

Oncen > 16 we need at least one coefficient in (72). The first example isn = 24,when (72) givesθL = θ3

E8− (720−N2(L))∆. For example, ifL has no roots then

θL = E34 − 720∆ = 1 + 196560q2 + 16773120q3 + 398034000q4 + · · · . (75)

In particular, the minimal norm of a Type II lattice inR24 is at most4, and ifit is that large then the kissing number is196560. Leech [Le] costructed such alatticeΛ24, and Conway [Con], while studying his sporadic simple groupCo1 =Aut(Λ24)/{±1}, showed thatΛ24 is the unique Type II lattice of minimal norm4.As with E8, the Leech lattice yields a sphere packing expected to be thedensestin its dimension, now known to be the densest among lattice packing (proved in

14Clearly if Nk(L∗1) = Nk(L∗

2) for all k thenθL∗1

= θL∗2. To show the converse, note that if

Nk(L∗1) = Nk(L∗

2) doesnot hold for all k then there is a least counterexamplek0 because the setof lattice norms is discrete. But theneπk0t(θL∗

1(it) − θL∗

2(it)) → (Nk0

(L∗1) − Nk0

(L∗2)) 6= 0 as

t → ∞, soθL∗16= θL∗

2.

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2004 [CK], and the only example of a lattice packing ofRn that has been provedoptimal for anyn > 8); it is known to be within a factor1 + ǫ of the maximaldensity for any24-dimensional sphere packing, here withǫ = 1.65 · 10−30 ([CK],again extending the computations of [CE]).

Forn = 32 andn = 40 there is again one coefficientc1 to choose. IfN2(L) = 0thenc1 = −30n and we findN4(L) = 90n(211− 5n), so again the minimal normis 4 and the kissing number is4. Already forn = 32 it is known [Ki] that there areliterally millions of such lattices, and thus millions of isospectral32-dimensionaltori. Each of these lattices has kissing number146880, and we shall see that thereare enumerative properties beyond the theta coefficientsNk(L) shared by all theselattices’ configurations of short vectors. Similarly forn = 40, with an even largernumber of lattices, each of kissing number39600.

For n = 48 there are two unknown coefficientsc1, c2; the q andq2 terms inθL

vanish if and only if(c1, c2) = (−1440, 125280), giving

θL = E64 − 1440E

34∆ + 125280∆2

= 1 + 52416000q3 + 39007332000q4 + · · · . (76)

Thus any such lattice has minimal norm6 and kissing number52416000. Severalsuch lattices have been constructed (see [CS2] for more information), and the as-sociated sphere packings are the densest known inR48. We shall see that theirconfigurations of short vectors must satisfy many combinatorial constraints. Onemight hope to use these constraints to fully describe such lattices, but so far nei-ther a full classification nor a large lower bound on the number of lattices has beenachieved.

For anyn ≡ 0 mod 8, we can likewise use (72) to determine the theta seriesof any Type II lattice inRn with no nonzero vectors of norm2m or less wherem = ⌊n/24⌋. The resulting modular form is sometimes called the “extremal thetafunction” in Mn/2(Γ). One might hope that theqm+1 coefficient could vanish aswell, so that the minimal norm might exceed2m + 2; but Siegel [Si] showed thatthis cannot happen. See Exercise 5.8 for one approach to thisresult, which Siegelproved using the Lagrange-Burmann theorem. It follows thata Type II latticeL ⊂ Rn has minimal norm at most2⌊n/24⌋ + 2, and for any such latticeθL

is the extremal theta function inMn/2(Γ). SuchL is calledextremal. Pastn = 8,extremal lattices yield particularly good sphere packingswhenn = 24m, when(2⌊n/24⌋+2)/n has a local maximum. Unfortunately none are known in this casepastm = 2; the existence of any such lattice, and in particular them = 3 case (anextremal Type II lattice inR72), is a long-standing open question. They are knownfor all othern ≤ 88 (subject to8|n as always) and for a few other values. It is

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known that for very largen (above roughly41000) there are no extremal Type IIlattices because the extremal theta series has negative theqm+2 coefficient [MOS].

Exercises

5.1 i) We have already seen thatDn has2(n2 − n) roots (Exercise 1.2ii), and thatE7 andE8 have respectively126 and240 roots. Find the72 roots ofE6, and then2 + n roots ofAn.ii) We already know thatdisc(An) = n + 1 (Exercise 1.3i) anddisc(Dn) = 4(because[Zn : Dn] = 2). Show thatE6 has generatorsei − ei−1 (i = 4, 5, 6, 7),e7 + e8, andh = 1

2

∑ni=1 ei, and use this to compute thatdisc(E6) = 3.

iii) More generally we can define for eachn = 2, 3, . . . , 8 a sublatticeEn of E8

by requiring the first9 − n coordinates to be equal. This agrees with the usualEn for n = 6, 7, 8. Identify the latticesE5, E4, E3 with the root latticesD5, A4,A1 ⊕ A2 respectively, and show thatE2 is a lattice with Gram matrix

(

2 11 4

)

(not aroot lattice); in each casedisc(En) = 9 − n.

5.2 [Another application of Theorem 10]15 For positive integersa, b, c, d, letM bethe tridiagonal matrix

M =

2a 1 0 01 2b 1 00 1 2c 10 0 1 2d

.

Show thatM is positive definite (hint: the special casea = b = c = d = 2should be familiar), and is thus the Gram matrix of an even lattice L. Note thatM mod p has rank at least3 for any primep, and deduce thatL∗/L is cyclic. UseTheorem 10 to conclude thatdet(M) cannot be a square.

[This could also have been done using Proposition 7 (Z4 is the only self-dual latticein R4), but the approach here generalizes ton×nmatrices for all evenn 6≡ 0 mod8. Only the casen ≡ 4 mod 8 is of interest: ifn ≡ 2 mod 4 thendet(M) ≡3 mod 4 so det(M) can never be a square. Square values can be attained when8|n; for example,M could be a tridiagonal Gram matrix for the latticeA8 ofdiscriminant9 = 32.

The casen = 4 can also be solved more laboriously by rewritingdet(M) = x2 as

15I do not know the original source of this problem. Henri Cohenposted it May 25, 1990 to theNMBRTHRY mailing list, writing that it is “linked to some problems of algebraic topology” and wasposed to him about 15 years earlier. Cohen no longer recalledthe poser’s name, but was sure that heor she knew how to solve the problem. (In later communicationCohen dated the event to a graduatecourse taught in 1971, but still did not know who originated the problem.) I posted the solutionoutlined here to the mailing list, and Atkin posted a solution using quadratic reciprocity. Henri alsoreported that H. Lenstra solved it, but did not specify the method.

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(4ab− 1)(4cd− 1) = x2 + 4ad. We always havedet(M) ≡ 1 mod 4; numericalcomputation suggests that all non-square positive integers congruent to1 mod 4occur asdet(M) for some choice ofa, b, c, d, with the exception of105 and2961.If a, b, c, d are not required to be positive thendet(M) = 1 can be attained bytakinga = b = c = 0 to obtain a Gram matrix of II2,2 (the indefinite self-dual evenlattice of signature(2, 2)).]

5.3 [Uniqueness ofE8 (again) and# Aut(E8)] We use theta functions to provethatE8 is the unique Type II lattice inR8 and count its automorphisms. LetLbe a Type II lattice inR8. ThusθL = E4. Consider the distribution of vectors ofnorm at most4 among the28 cosets of2L in L. The zero coset contains only0;if r, r′ are roots in the same coset thenr′ = ±r. Thus zero and the roots accountfor 1 + (240/2) = 121 cosets, leaving only28 − 121 = 135. Show that each cosetcontains at most16 vectors of norm4, and if there are as many as16 vectors thenthey are±2ei for some orthonormal basis{ei}8

i=1 of R8. But theq2 coefficientof E4 is exactly2160 = 16 ·135. Hence each of our135 leftover cosets contains16vectors of that form. Since they are congruent mod2L we deduce thatL containseach of the norm2 vectorsei ± ej , whenceL ⊃ D8. We have seen already thatthis implies thatL is one of the two lattices other thanZ8 contained betweenD8

andD∗8, whenceL ∼= D+

8 = E8 . Show thatAutD8 is the hyperoctahedral groupof order288!, and deduce thatE8 has135 · 278! = 696729600 automorphisms.

5.4 [Bimodular even lattices] SupposeL ⊂ Rn is an even lattice of discriminant2.(Integral lattices of discriminant2 are sometimes called “bimodular”.) Letv∗ ∈L∗ − L andς = 2〈v∗, v∗〉.i) Show thatς is an integer whose residue mod4 does not depend on the choiceof v∗. We may thus writeς = ς(L) ∈ Z/4Z.ii) Show thatς is odd. [ElseL∗ would be an integral lattice of discriminant1/2.]Check thatς(A1) = 1 andς(E7) = 3.iii) Let L′ = E7 or A1 according asς(L) = 1 or ς(L) = 3. Construct an evenunimodular latticeL containingL⊕ L′ with index2. Conclude thatn ≡ 1 mod 8if ς(L) = 1 while n ≡ 7 mod 8 if ς(L) = 3.iv) If n = 1 then clearlyL ∼= Z〈2〉 = A1. Show that ifn = 7 thenL ∼= E7; ifn = 9 thenL ∼= A1 ⊕ E8; while if n = 15 then eitherL ∼= E7 ⊕ E8 or L is theunique even lattice containingA1 ⊕D14 with index2.[You’ll use several times the existence of a bijection between copies ofA1 andE7 in E8, namely the orthogonal slice. Forn = 17 such lattices correspond toNiemeier latticesL containingE7; there are four, with root systemsD10 ⊕ E2

7 ,A17 ⊕ E7, E3

8 , andD16 ⊕ E8. The first of these yieldsL with root sublatticeD10 ⊕E7, which has index2 in L; the other three yield the latticesA+3

17 ,A1 ⊕E28 ,

andA1 ⊕ D+216 respectively. Forn = 23 we need a Niemeier lattice with a root,

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and there are31 distinct possibilities. Borcherds [Bo] enumerated121 lattices forn = 25 using his analysis of the even Lorentzian lattice II25,1. For n ≥ 31 anenumeration is likelye hopeless: by the mass formula there are millions of themfor eachn = 32 ± 1, and many more for even largern.]

5.5 [The theta function of a bimodular even lattice, and the Kohnen space] LetL ⊂ Rn be an even lattice of discriminant2, and this time setψL = θL∗ − θL, thetheta function of the nontrivial coset ofL in L∗. Thus by the functional equation

ψL(τ) = 21/2(τ/i)n/2θL(S(τ)) − θL(τ).

Also θL(T (τ)) = θL(τ) as usual, andψL(T (τ)) = iς(L)ψL(τ).i) Use these transformation rules and the identity(ST )3 = 1 to give an analyticproof thatn ≡ 1 or 7 mod 8 according asς(L) = 1 or ς(L) = 3.ii) Show thatθL andψL are modular forms of weightn/2 for Γ(2) = ker(Γ →SL2(Z/2Z)).iii) [ E7 via “Construction A”] Define a latticeL such thatA7

1 ⊂ L ⊂ A∗17 as

follows: identifyA∗17/A7

1 with (Z/2Z)7, and letL be the preimage of the subspaceof (Z/2Z)7 consisting of zero and the cyclic permutations of(0, 0, 1, 0, 1, 1, 1).Check that this is indeed a subspace; it is known in coding theory as the dualHamming code. Deduce thatL is a lattice, and verify that it is even and bimodular.Thus by the previous ExerciseL ∼= E7 (can you find an explicit isomorphism?).Show that its theta function isθ7

Z+ 7θ3

Zψ4

Z, and use this to compute the first few

coefficients ofθE7.

5.6 [Uniqueness ofE7 (again) and# Aut(E7)] Now taken = 7. ThenθL =1 + 126q + 756q2 +O(q3). Again we consider the distribution of vectors of normat most4 among the cosets of2L in L. There are27 cosets. The two cosets thatconstitute2L∗ contain the zero vector and no other vector of norm less than6. Ofthe remaining27 − 2 = 126 cosets,N2(L)/2 = 63 contain only a pair of roots,leaving63 others. As in Exercise 5.2, each of these remaining cosets isrepresentedby vectors±2ei for some orthonormalei; but this time we cannot have a fullorthonormal frame because thenL would contain2Z7 with finite index, whichis impossible because2Z7 has square discriminant andL does not. Thus there areat most6 pairs, and sinceN4(L) = 756 = 2 · 6 · 63 each of the cosets must haveexactly6. ThusL ⊃ D6. Construct a mapL/D6 → D∗

6/D6 by sendingv ∈ Lto the class of the homomorphismD6 → Z, w 7→ 〈v, w〉. SinceL/D6 is infinitecyclic andD∗

6/D6 has exponent2, the kernel of the map has index1 or 2 in L/D6.Compare discriminants to show that the index is2 andL containsD6⊕A1. Finallycheck that there are only two even integral lattices containingD6⊕A1 with index2,which are related byAut(D6 ⊕ A1). This shows that all bimodular evenL ⊂ R7

are isomorphic. Since we already know such a latticeE7 we conclude thatL ∼= E7,

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and moreover that# Aut(E7) = 63# Aut(D6) = 63 · 266! = 2903040.

5.7 [2-modular lattices and their theta functions]

5.8 [Extremal theta series] Letf(q) andg(q) be power series of the form

f(q) = 1 +O(q), g(q) = q +O(q2).

For a nonnegative integerk and anya0, a1, . . . , ak there exists a unique homoge-neous polynomialP (·, ·) of degreek such thatP (f, g) =

∑ki=0 aiq

i + O(qk+1).[Construct this polynomial iteratively starting from the leading coefficienta0, aswe did for(f, g) = (θ8

Z,∆+) or (E3

4,∆).] In particular, takinga0 = 1 andai = 0for i = 1, 2, . . . , k, we findP such thatP (f, g) = 1+Cqk+1 +O(qk+2). We shallshow thatC is k/(k + 1) times the1/q coefficient off ′/g(k + 1). In particular,C > 0 for all k > 0 if f and1/g have positive coefficients. This is the case for(f, g) = (E3

4,∆), using the expansion (68) ofE4 and the infinite product formula∆ = q

∏∞n=1(1 − qn)24.

Divide both sides ofP (f, g) = 1 + Cqk+1 + O(qk+2) by gk to obtainp(f/g) =1/gk+Cq+O(q2) wherep is the degree-k univariate polynomialp(X) = P (X, 1).Sinceg andg/f are both of the formq +O(q2), we can find a (unique) power se-riesF (z) = z + b2z

2 + b3z3 + · · · such thatg = F (g/f). (This power series is

holomorphic in a neighborhood ofz = 0 if f andg have positive radii of conver-gence, but we needF only as a formal power series.) Takingz = g/f , we find1/F (z)k = p(1/z)−Cz+O(z2), in which the right-hand side is the beginning ofthe Laurent expansion of1/F (z)k aboutz = 0. Therefore−C is thez coefficientof that expansion. But thenC is the residue of−(1/F (z)k) dz/z2 atz = 0.

6. Weighted theta functions

We next introduceweighted theta functionsΘL,P , θL,P of a latticeL ⊂ Rn. Thesegeneralize the theta functionsΘL, θL: they are generating functions that encode notjust the numberN2k(L) of lattice vectors of each norm2k but also their distributionon the sphere〈x, x〉 = 2k.

Weighted theta functions are defined as follows. TheweightP is aharmonic poly-nomialon Rn, that is, a homogeneous polynomial whose Laplacian vanishes (weshall give a fuller description of such polynomials soon). Then

ΘL,P (q) :=∑

v∈L

P (v) q〈v,v〉/2 =∑

k≥0

N2k(L,P ) qk, (77)

θL,P (τ) :=∑

v∈L

P (v) eπi〈v,v〉τ =∑

k≥0

Nk(L,P ) eπikτ = ΘL,P (e2πiτ ), (78)

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where we defineNk(L,P ) :=

∑

v∈L〈v,v〉=k

P (v). (79)

Let d = deg(P ). If d = 0 thenP is a constant, soΘL,P andθL,P reduce to scalarmultiples ofΘL andθL. If d is odd thenΘL,P = θL,P = 0 because thev and−vterms cancel. For evend > 0 we get a nontrivial generalization ofΘL andθL; inthis caseN0(L,P ) = 0 so the sums overk in (77, 78) may be taken overk > 0.

The definitions ofΘL,P andθL,P make sense for any polynomialP , harmonic ornot. We require thatP be harmonic so that we can generalize Proposition 1 (thefunctional equation (17)) to weighted theta functions. Again we shall prove thefunctional equation using Poisson summation; here the relevant functions onRn

aref(x) = P (x) e−π〈x,x〉t. (80)

Theorem 12 Suppose thatt > 0 andP is a harmonic polynomial onRn of de-greed, and define a functionf : Rn → R by (80). Then the Fourier transformof f is

f(y) = idt−( n2+d)P (y) e−π〈y,y〉/t. (81)

This will yield:

Proposition 13 (functional equation for weighted theta series)For any latticeL in Rn, and any harmonic polynomialf of degreed, we have

ΘL∗,P (e−2πt) = id disc(L)1/2t−(n/2)−dΘL(e−2π/t) (82)

for all t > 0.

Proof (modulo Theorem 12): Apply Poisson (26) toL and the function (80), anduse the formula (81) for the Fourier transform of this function. Q.E.D.

Note that sinceΘL,P andΘL∗,P vanish identically for odddwe can write the factor

id as(−1)d/2.

To prove Theorem 12, and then to use Proposition 13 to study lattices, we nextreview some key properties of harmonic polynomials.

Let P be theC-vector space of polynomials onRn, andPd (d = 0, 1, 2, . . .) itssubspace of homogeneous polynomials of degreed, so thatP =

⊕∞d=0 Pd. The

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Laplacianis the differential operator16

∆ =n∑

j=1

∂2

∂x2j

: C∞(Rn) → C∞(Rn), P → P, Pd → Pd−2. (83)

Herex1, . . . , xn are any orthonormal coordinates onRn, andPd is taken to be{0}for d < 0. The space of harmonic polynomials of degreed is then

P0d := ker(∆ : Pd → Pd−2); (84)

this is the degree-d homogeneous part of

P0 := ker(∆ : P → P). (85)

Examples.P00 andP0

1 are the spaces of constant and linear functions respectively,of dimensions1 andn. If n = 1 thenP0

d = {0} for all d > 1. If n = 2 thenP0d

is 2-dimensional for eachd > 0; see Exercise 6.3 below for an explicit basis. Foranyn, A quadratic polynomialP =

∑

1≤j≤k≤n ajkxjxk is harmonic if and onlyif∑n

j=1 ajj = 0, because∆P is the constant polynomial2∑n

j=1 ajj .

We shall see that∆ : Pd → Pd−2 is surjective, whence

dimP0d = dim(Pd) − dim(Pd−2) =

(

n+ d− 1

d

)

−(

n+ d− 3

d

)

. (86)

Indeed we shall give a more precise result using two further operators onC∞(Rn)and on its subspaceP. The first is17

E := x · ∇ =

n∑

j=1

xj∂

∂xj. (87)

Euler proved that ifP ∈ C∞(Rn) is homogeneous of degreed thenEP = d · P ;in particularPd is thed-eigenspace ofE|P . The second operator is multiplicationby the norm:

F := 〈x, x〉 =

n∑

j=1

x2j , P 7→ 〈x, x〉P. (88)

16The use of∆ for both this operator and the modular formη24 = qQ∞

n=1(1 − qn)24 may beunfortunate, but should not cause confusion because the two∆’s never appear together outside thisfootnote. The alternative notationL for the Laplacian would be much worse when we regularly useL for a lattice.

17Fortunately this operator will never appear together with an Eisenstein seriesE2k. . .

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ClearlyF injects eachPd into Pd+2. ThusP0d = ker(F∆ : Pd → Pd); that is,

P0d is the zero eigenspace of the operatorF∆ onPd. We next show that the other

eigenspaces areFkP0d−2k for k = 1, 2, . . . , ⌊d/2⌋, and thatPd is the direct sum of

these eigenspaces, from which the surjectivity of∆ : Pd → Pd−2 will follow as acorollary.

We begin with by finding the commutators of∆,E,F. Recall that thecommutatorof any two operatorsA,B on some vector space is

[A,B] = AB −BA = −[B,A]. (89)

For example,[xj, xk] = [∂/∂xj, ∂/∂xk] = 0 for all j, k, while [∂/∂xj, xk] = δjk(Kronecker delta).

Lemma 14 (Commutation relations for ∆,E,F). We have

[∆,F] = 4E + 2n, [E,∆] = −2∆, [E,F] = +2F. (90)

Proof: These are direct computations using the pairwise commutators of the oper-atorsxj and∂/∂xk. For example,[∂2/∂x2

j , x2k] = 0 unlessj = k, and then we

calculate

∂2

∂x2j

◦ x2j =

∂

∂xj◦(

xj∂

∂xj+[ ∂

∂xj, xj

]

)

◦ xj =( ∂

∂xj◦ xj

)2+

∂

∂xj◦ xj

=(

xj∂

∂xj+1)2

+xj∂

∂xj+1 =

(

xj∂

∂xj

)2+3xj

∂

∂xj+2 = x2

j

∂2

∂x2j

+4xj∂

∂xj+2,

whence[ ∂2

∂x2j

, x2k

]

= δjk

(

4xj∂

∂xj+ 2)

.

Summing overj, k = 1, . . . , n yields [∆,F] = 4E + 2n as claimed. We completethe proof of Lemma 14 by verifying the remaining two identities in (90) via asimilar but easier calculation (see Exercise 6.1 below), Q.E.D.

For a further check on the formulas for[E,∆] and[E,F], note that they are consis-tent with the action of∆,E,F on thePd: if P ∈ Pd then∆P ∈ Pd−2 yields

[E,∆]P = E∆P − ∆EP = (d− 2)∆P − ∆(d · P ) = −2∆P,

consistent with[E,∆] = −2∆, and likewise for the third part[E,F] = +2F of (90).

See the further Remarks at the end of this section for the interpretation of thecommutation relations (90) (and Lemmas 17,18(ii) below) interms ofsl2 and otherLie algebras and groups.

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Now supposeP ∈ Pd is in theλ-eigenspace ofF∆ for someλ. Then〈x, x〉P =FP is in the(λ+ 4d+ 2n)-eigenspace ofF∆ acting onPd+2, because

F∆FP = F(F∆ + [∆,F])P = F(F∆ + 4E + 2n)P = F(λ+ 4d+ 2n)P.

By induction onk = 0, 1, 2, . . . it follows thatFkP is an eigenvector ofF∆|Pd+2k

with eigenvalue

λ+

k−1∑

j=0

4(d+ 2j) + 2n = λ+ k(

4(d+ k − 1) + 2n)

.

Replacingd by d− 2k and takingλ = 0, we see that ifP ∈ P0d−2k thenFkP is an

eigenvector ofF∆|Pdwith eigenvalue

λd(k) := k (4(d− k − 1) + 2n) . (91)

We next prove that this accounts for all the eigenspaces ofF∆|Pd.

Lemma 15 Fix d ≥ 0 . For integersk, k′ such that0 ≤ k < k′ ≤ d/2 we haveλd(k) < λd(k

′).

Proof: By induction it is enough to check this fork′ = k + 1. We compute

λd(k + 1) − λd(k) = 2n+ 4(d− 2k′) ≥ 2n > 0,

Q.E.D.

Corollary. The sum of the subspacesFkP0d−2k of Pd overk = 0, 1, . . . , ⌊d/2⌋ is

direct.

Proof: SinceFkP0d−2k is a subspace of theλd(k) eigenspace ofF∆, it is enough to

prove that theλd(k) are pairwise distinct. By Lemma 15, they are strictly increas-ing, Q.E.D.

Proposition 16 For k = 0, 1, . . . , ⌊d/2⌋, letPkd = FkP0

d−2k. Then:i) The map∆ : Pd → Pd−2 is surjective.

ii) Pd =⊕⌊d/2⌋

k=0 Pkd = P0

d ⊕ FPd−2, andP =⊕∞

k=0 FkP0.iii) Pk

d is the entireλd(k) eigenspace ofF∆|Pd, andF∆|Pd

has no eigenvaluesother than theλd(k) for k = 0, 1, . . . , ⌊d/2⌋.iv) dimP0

d = dim(Pd) − dim(Pd−2) as claimed in (86).

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Proof: The sum in (ii) is direct by the previous Corollary. We provethat it equalsPd by comparing dimensions. SinceF is injective we havedimPk

d = dimP0d−2k;

moreoverdimP0

d−2k ≥ dimPd−2k − dimPd−2k−2,

with equality if and only if∆ : Pd−2k → Pd−2k−2 is surjectve, becauseP0d−2k =

ker(∆ : Pd−2k → Pd−2k−2). Hencedim⊕⌊d/2⌋

k=0 Pkd is

⌊d/2⌋∑

k=0

dimPkd =

⌊d/2⌋∑

k=0

dimP0d−2k ≥

⌊d/2⌋∑

k=0

dimPd−2k − dimPd−2k−2, (92)

and the last sum telescopes todimPd. Thus equality holds in the last step of (92)anddim

⊕⌊d/2⌋k=0 Pk

d = dimPd. The first of these proves (i) (by takingk = 0). Thesecond yields

Pd =

⌊d/2⌋⊕

k=0

Pkd , (93)

as claimed in (ii); taking the direct sum overd yields P =⊕∞

k=0 FkP0, alsoclaimed in (ii). to complete the proof of (ii) we compare the decompositions (93)of Pd andPd−2 and note thatPk

d = FPk−1d−2 for eachk > 0. Claim (iii) follows

because the decomposition (93) diagonalizesF∆|Pd. Finally (iv) is the casek = 0

of equality in (92), Q.E.D.

Remark: Part (ii) impliesP0d ∩FPd−2 = {0}, and thus thatP0 contains no nonzero

multiple of 〈x, x〉. Proving this was set as problem B-5 on the 2005 Putnam exam[KAL, p.736], which was the hardest of the 12 problems that year, solved by onlyfive of the top 200 scorers [KAL, p.741]. The solution printedin [KAL, p.742]uses some of the ingredients used here to prove Proposition 16.18

We next characterize the functionsf(x) = P (x) e−π〈x,x〉t of (80) using the opera-tors∆,E,F. For t ∈ C define an operator

Gt : C∞(Rn) → C∞(Rn), g 7→ e−πt〈x,x〉g (94)

18Suppose〈x, x〉|P and∆P = 0. Write P =P

d≥0 Pd with eachPd ∈ Pd. Then〈x, x〉|Pd and∆Pd = 0 for eachd, and we may choosed so thatPd 6= 0. Let m be the largest integer such thatPd = 〈x, x〉mQ for some polynomialQ; by assumptionm > 0. In our notations, then,Q ∈ Pd−2m

with ∆FmQ = 0 andQ /∈ FP. Using in effect the formula for[∆, F] and Euler’s description ofE,

compute that∆F

mQ = Fm−1[F∆ + 2m(n + 2(d − m − 1))]Q

for all Q ∈ Pd−2m. Thus∆FmQ = 0 impliesF∆Q = −2m(n + 2(d − m − 1))Q, and the factor

n + 2(d − m − 1) is positive (becaused ≥ 2m ≥ m + 1), soQ ∈ FP, contradiction.

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that multiplies everyC∞ function by the Gaussiane−πt〈x,x〉; these operators con-stitute a one-parameter group:GtGt′ = Gt+t′ for all t, t′. We are then interestedin f = GtP for P ∈ P in the intersection of the kernel of∆ with an eigenspaceof E. If P ∈ Pd then

d · f = Gt(d · P ) = GtEP = (GtEG−t)GtEP = (GtEG−t)f, (95)

sof is in thed-eigenspace ofGtEG−t; likewisef ∈ kerGt∆G−t. Since our one-parameter group{Gt} has infinitesimal generator−πF, we expect that conjugationbyGt will take ∆,E to some linear combination of∆,E,F. Indeed we find:

Lemma 17 (Conjugation of ∆,E,F by Gt) The operatorsGt commute withF,and we have

GtEG−t = E + 2πtF, Gt∆G−t = ∆ + πt(4E + 2n) + (2πt)2F. (96)

Proof: As with Lemma 14, this comes down to an exercise in differential calculus.Here we start from the fact thatGt commutes with eachxj whileGt(∂/∂xj)G−t =2πtxj, whence the first formula in (96) quickly follows, whileGtF = FGt isimmediate. A somewhat longer computation (see Exercise 6.2below) establishesthe second formula, Q.E.D.

Corollary. The operators∆,E,F act onGtP, and the subspaceGtP0d is the inter-

section ofker(∆ + πt(4E + 2n) + (2πt)2F) with thed-eigenspace ofE + 2πtF inGtP.

We next relate the Fourier transform of a Schwartz functionf with the Fouriertransforms of its images under∆,E,F, and use this to prove Theorem 12.

Lemma 18 (Conjugation of ∆,E,F by the Fourier transform). Let f : Rn →C be any Schwartz function. Then:i) For eachj = 1, . . . , n, the Fourier transform ofxjf is (2πi)−1∂f/∂yj , and theFourier transform of∂f/∂xj is−2πiyj f .ii) The Fourier transforms of∆f , (2E + n)f , andFf are respectively−(2π)2Ff ,−(2E + n)f , and−(2π)−2∆f .

Proof: Again a calculus exercise, this time with definite integrals. The formulafor the Fourier transform of∂f/∂xj is obtained by integrating by parts with re-spect toxj . The Fourier transform ofxjf can be obtained from this using Fourierinversion, or directly by differentiation with respect toyj of the integral (25) thatdefinesf(y). We then obtain (ii) by iterating the formulas in (i) to find the Fourier

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transform of∂2f/∂x2j , xj∂f/∂xj, or x2

jf , and summing overj. The case ofEfcan be explained by writing the operator2E+n as

∑nj=1 xj(∂/∂xj)+(∂/∂xj)◦xj ,

Q.E.D.

We first use this to show that iff ∈ GtP thenf ∈ G1/tP, that is, thatf is some

polynomial multiplied bye−π〈y,y〉/t; more precisely:

Proposition 19 Let t ∈ C with Re(t) > 0. If f = GtP for someP ∈ Pd thenf = G1/tP for someP =

∑dd′=0 Pd′ with eachPd′ ∈ Pd′ andPd = idt−( n

2+d)P .

As beforet−( n2+d) denotes the−(n+ 2d) power of the principal square root oft.

Proof: We use induction ond. The base cased = 0 is the fact that the Fourier trans-form of e−πt〈x,x〉 is t−n/2e−π〈y,y〉/t, which we showed already. Suppose we haveestablished the claim forP ∈ Pd. By linearity and the fact thatPd+1 is spanned byits subspacesxjPd, it is enough to prove the Proposition withP replaced byxjP .By part (i) of Lemma 18, the Fourier transform ofGtxjP = xjGtP is

1

2πi

∂

∂yj(G1/tP ) =

1

2πiG1/t

( ∂P

∂yj− 2π

tyjP

)

. (97)

By the inductive hypothesisP has degreed and leading partPd = idt−( n2+d)P .

Therefore the right-hand side of (97) has degreed+ 1 and leading part

−2πt−1

2πiPd =

i

tPd = id+1t−( n

2+d+1)yjP.

This completes the induction step and the proof, Q.E.D.

Proof of Theorem 12: SupposeP ∈ P0d andf(x) = P (x) e−π〈x,x〉t = GtP. By the

Corollary to Lemma 17,

(∆ + πt(4E + 2n) + (2πt)2F)f = 0, (E + 2πtF)f = d · f. (98)

Taking the Fourier transform and applying Lemma 18(ii), we deduce

(−(2π)2F − πt(4E + 2n) − t2∆)f = 0, −(E + n+t

2π∆)f = d · f . (99)

Eliminating ∆f , we findd · f = (E + 2πt F)f ; that is, f is in thed-eigenspace

of E + 2πt−1F. By Proposition 19 we knowf = G1/tP for someP ∈ P. By

Lemma 17, then,P is in thed-eigenspace ofE; that is,P ∈ Pd. By Proposition 19we conclude thatP = idt−( n

2+d)P , Q.E.D.

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Remark: Multiplying the first equation in (99) by−t−2, we recover the first equa-tion of (98) witht replaced byt−1; this lets us show without Euler’s theorem thatthe Fourier transform takesGtP0 to G1/tP0. See the further Remarks at the endof this section for a sketch of an alternative approach to Theorem 12, using theconnection withsl2 andSL2 and avoiding Proposition 19.

A natural application of weighted theta functions is to the question of equidistribu-tion of lattice points in spherical shells. TheNk(L) lattice vectorsv ∈ L on thesphere{x ∈ Rn | 〈x, x〉 = k} yield a configuration, call it

Sk(L) := k−1/2{v ∈ L | 〈v, v〉 = k}, (100)

of Nk(L) vectors on the unit sphereΣ ⊂ Rn. As k → ∞ through the latticenorms, are theSk asymptotically equidistributed onΣ? Recall that a sequence{Cm}∞m=1 of nonempty finite subsets ofΣ areasymptotically equidistributedif,for every continuous functionϕ : Σ → C, the average ofϕ overCm approachesthe average ofϕ overΣ asm→ ∞:

∫

x∈Σϕ(x) dνx = lim

m→∞1

#Cm

∑

x∈Cm

ϕ(x). (101)

Heredν is the invariant measure onΣ such that∫

x∈Σ dµx = 1; for instance wemay define

∫

x∈Σϕ(x) dνx = tn/2

∫

x∈Rne−πt〈x,x〉 ϕ

(

x

〈x, x〉1/2

)

dµx (102)

for any t > 0. The coefficients of weighted theta functionsΘL,P give us the sumin (101) whenP is a harmonic polynomial. Using the decomposition (93) fromProposition 16(ii), we show that these are enough to test equidistribution:

Proposition 20 P0|Σ is dense inC(Σ); that is, for every continuousϕ : Σ → C

and anyǫ > 0 there existsP ∈ P0 such that

∀x ∈ Σ : |P (x) − ϕ(x)| < ǫ. (103)

Proof: By the Stone–Weierstrass theorem there existsP ∈ P satisfying (103). Itis thus enough to prove that for everyP ∈ P there existsQ ∈ P0 such thatP = QonΣ. Applying Proposition 16(ii) to each homogeneous part ofP we write

P =

⌊deg(P )/2⌋∑

k=0

〈x, x〉kQk (104)

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for someQk ∈ P0. Since〈x, x〉 = 1 on Σ, the polynomialQ =∑

k Qk ∈ P0

agrees withP onΣ, Q.E.D.

Theorem 21 A sequence{Cm}∞m=1 of nonempty finite subsets ofΣ is asymptoti-cally equidistributed if and only if

limm→∞

1

#Cm

∑

x∈Cm

P (x) = 0 (105)

for all harmonic polynomialsP of positive degree.

Proof: For the “only if” direction, assume (101) holds for allϕ ∈ C(Σ), and takeϕ = P |Σ. We claim (101) is then equivalent to (105), i.e., that

∫

x∈Σ P (x)dνx = 0.

Equivalently (see Exercise 6.5), we claim∫

x∈Rn P (x) e−πt〈x,x〉 dµx = 0. But

the integral is the value aty = 0 of the Fourier transform ofP (x) e−πt〈x,x〉. Weobtained this Fourier transform in Theorem 12; it vanishes at y = 0 as claimed,becauseP (0) = 0 for P of positive degree.

Since both sides of (101) are linear, we have thus proved the converse implicationfor functionsϕ that are the restriction toΣ of any finite linear combination ofharmonic polynomials of positive degree. We can drop the condition of positivedegree, because (101) holds automatically forϕ = 1: its left-hand side equals1,and the right-hand side reduces tolimm→∞ 1. Thus (105) implies (101) for allϕof the formP |Σ with P ∈ P0. By Proposition 20, every continuousϕ : Σ → C

can be uniformly approximated by suchP |Σ. Hence (101) holds for allϕ ∈ C(Σ)by the following standard argument. Changingϕ to P moves both

∫

x∈Σ ϕ(x) dνx

and each average(#Cm)−1∑

x∈Cmϕ(x) by at mostǫ. For large enoughm, the

average ofP overCm is within ǫ of∫

x∈Σ P (x) dνx. Therefore the average andintegral ofϕ are within3ǫ of each other. Sinceǫ is arbitrary, we are done. Q.E.D.

[. . . ]

FurtherRemarks:19 The commutation relations in Lemma 14 are tantamount to anisomorphism of Lie algebras fromsl2 to the span of{∆,E + n

2 ,F} that takes thestandard basis(X,H, Y ) =

((

0 10 0

)

,(

1 00 −1

)

,(

0 01 0

))

of sl2 to ( 14π∆,−(E+ n

2 ),−πF).

The decompositionPd =⊕⌊d/2⌋

k=0 Pkd in Proposition 16ii then says in effect that

P = ⊕∞d=0P0

d ⊗ Vn2+d, where for any realm > 0 we writeVm for the infinite-

dimensional irreducible representation ofsl2 with basis{Y kv}∞k=0 whereXv = 0andHv = −mv.

19The reader anxious to reach the application of weighted theta series to the study of extremallattices etc. will likely want to skim or skip these remarks,and the last exercise for this section, atleast on first reading(s).

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Moreover, we noted already thatGt = exp(−πtF), and it is known that the Fouriertransform ise−πin/4 exp πi

2 (πF − 14π∆).20 Thus Lemmas 17 and 18, which give

the action on∆,E,F of conjugation byGt and the Fourier transform, correspondto the action onsl2 of conjugation by the elementsetY =

(1 0t 1

)

ande−πi(X+Y )/2 =

−(0 ii 0

)

of SL2.

In Proposition 19, we obtainP by applying toP the operatorsGt, then the Fouriertransform, thenG−1/t; up to the constant factore−πin/4, this corresponds to theproduct

(

1 0−1/t 1

) (

0 −i−i 0

) (

1 0t 1

)

= −i(

t 10 −1/t

)

,

in SL2, which can be written as

−i(

t 10 −1/t

)(

1 1/t0 1

)

=

(

t/i 00 (it)−1

)(

1 i/t0 1

)

.

Now if P ∈ P0d thenP is fixed by the one-parameter subgroup

(

1 ∗0 1

)

of SL2 gener-ated byX. ThusP is a multiple of the image ofP under a diagonal matrix inSL2.Such diagonal matrices constitute the one-parameter groupgenerated byH. SinceP is an eigenvector ofH, we deduce thatP is proportional toP . More precisely,sinceHP = −(n

2 + d)P we havediag(eβ, e−β)P = eβHP = e−β( n2+d)P ; taking

eβ = t/i (with Im(β) = −π/2),21 and restoring the factore−πin/4, we recover

P = e−πin/4(t/i)−( n2+d)P = idt−( n

2+d)P,

which is Theorem 12.

The differential operators∂2/∂xj∂xk, xj∂/∂xk + 12δjk, andxjxk generate a Lie

algebra isomorphic withsp2n, which contains the span of∆,E + n2 ,F as the sub-

algebra invariant under the orthogonal group ofRn. See the final exercise for this

20This has a memorable physical interpretation: running the Schrodinger equation on a quantumharmonic oscillator for1/4 of its classical period applies a multiple of the Fourier transform to thewave function. The distribution of factors ofπ in this formula is the reason we chose the homomor-phism fromsl2 that maps

`

0 10 0

´

and`

0 01 0

´

to 14π

∆ and−πF, rather than12∆ and− 1

2F which seems

more natural at first.Alternatively, we could use Lemma 18 to identify the Fouriertransform with conjugation by

±i`

0 11 0

´

.21We must specify the path because in general a representationof sl2 lifts to a representation not

of SL2 but of its universal cover. A more honest treatment would either carefully check that we areworking in a contractible patch ofSL2(C), or compute the constant factorid of (81) in some otherway.

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section. Multiplying∂2/∂xj∂xk andxjxk by i (to make all the generators skew-Hermitian) yields infinitesimal generators of the Lie algebra of Weil’s projectiverepresentation [Wei] ofSp2n(R) onL2(Rn).

Exercises

6.0 Verify directly that Theorem 21 holds forn = 1.

6.1 Check the commutation relations[E,∆] = −2∆ and[E,F] = +2F of (90).

6.2 Verify the second formula in (96).

6.3 (Harmonic polynomials onR2) Letn = 2, and identifyR2 with C as usual byx1 + ix2 = z. Show that for eachd ≥ 1 the polynomialszd and zd constitute abasis forP0

d . [Hint: recall that

∆ =( ∂

∂x1+ i

∂

∂x2

)( ∂

∂x1− i

∂

∂x2

)

. ]

Deduce that the casen = 2 of Theorem 21 is tantamount to Weyl’s characteriza-tion [Wey] of asymptotic equidistribution inR/Z, or equivalently in the unit circleΣ1

∼= {z ∈ C : |z| = 1}: a sequence of finite nonempty subsetsCm ⊂ Σ1 isasymptotically equidistributed inΣ1 if and only if for eachd = 1, 2, 3, . . . we have(#Cm)−1

∑

z∈Cmzd → 0 ask → ∞.22

6.4 (a weighted theta function twisted by a Dirichlet character) Show that iff is aSchwartz function onR then

∞∑

m=−∞χ4(m)f(m) =

1

2i

∞∑

n=−∞χ4(n)f(n/4),

whereχ4 is the Dirichlet character mod 4 introduced in Exercise 3.7.Sinceχ4 isodd, we would get only the trivial identity0 = 0 using the even functionf(x) =e−πtx2

. Howeverf(x) = xe−πtx2yields a new identity. Use this identity to prove

that

∞∑

n=1

nχ4(n)qn2/8 = q1/8(1 − 3q + 5q3 − 7q6 + 9q10 − 11q15 + − · · · ) (106)

is a modular form of weight3/2 for Γ and the8th roots of unityε3c,d, with εc,d asin Exercise 2.5; deduce that this modular form isη3.

22Weyl also shows that that this is equivalent to the conditionthat#(I ∩ Cm)/#Cm → |I|/|Σ|for all arcsI ⊆ Σ (where| · | denotes the length). See also [Ko, Ch.3]. Analogous statements can bemade on the unit sphere inRn for anyn, but we do not need those versions of equidistribution here.

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6.5 Prove that there are positive constantsc0, c2, c4, . . . (depending onn) such thatif ϕ is the restriction toΣ of a homogeneous polynomialP of even degreed then

∫

x∈Σϕ(x) dνx = cdt

(n+d)/2

∫

x∈Rne−πt〈x,x〉 P (x) dµx. (107)

for all t > 0. We defined the integral overΣ so thatc0 = 1; can you evaluate theothercd explicitly? (For oddd, (107) holds for allcd because both sides vanish. . . )

6.6 Define an inner product〈·, ·〉 on the space of Schwartz functions onRn by

〈f, g〉 :=

∫

x∈Rn

f(x) g(x)dµx. (108)

Recall that an operatorA on this space isself-adjointif 〈Af, g〉 = 〈f,Ag〉 for allSchwartz functionsf, g. Prove that the operators∆ andF are self-adjoint. [Thisis immediate forF, and uses integration by parts for∆. Warning: E is not self-adjoint; indeed the fact that∆ andF are self-adjoint implies that4E + 2n = [∆,F]is anti-self-adjoint.] Use this and the Corollary to Lemma 17 to prove thatP0

d

andP0d′ are orthogonal ford 6= d′, that is, that

∫

x∈Σ P (x)Q(x) dνx = 0 if PandQ are harmonic polynomials of different degrees. (We showed the special cased = 0 during the proof of Theorem 21.) Recover the orthogonal decompositionL2(Σ, dνx) =

⊕∞d=0 P0

d using this and Proposition 20.

6.7 (Failure of equidistribution forn = 2) Equidistribution ofSkm(L) can fail forn = 2 even ifNkm(L) → ∞, and already for the familiar latticeL = Z2. Tostudy this question we must use the intimate connection between the Diophantineequation〈x, x〉 = x2

1 +x22 = k and arithmetic in the Gaussian integers{x1 + ix2 |

x1, x2 ∈ Z} that was kept in the background of Exercises 3.7 and 3.8.

(i) Let p be a prime such thatp ≡ 1 mod 4, and{km} a sequence of integers withpm|km. Then ifNkm 6= 0 thenNkm > 4m; in particularNkm → ∞. Show that inthis case{Skm} is asymptotically equidistributed on the unit circle.

(ii) Prove that there existkm such thatNkm → ∞ but{Skm} is notasymptoticallyequidistributed on the unit circle. [Use the following special case of Hecke’s theo-rem [H1] on equidistribution of prime ideals in number fields: for all ǫ there existintegersx1, x2 with 0 < x2 < ǫx1 such thatx2

1 + x22 is prime.]

6.8 (Identification of a Lie algebra of differential operators with sp2n) Let V bethe2n-dimensionalR-vector space of differential operators onRn with basisxj

and∂/∂xj (1 ≤ j ≤ n); and letg be the space of dimension2n2 + n with basis∂2/∂xj∂xk, xj∂/∂xk + 1

2δjk, andxjxk (1 ≤ j, k ≤ n).i) Check that ifA,B ∈ g then [A,B] ∈ g, while if A ∈ g and v ∈ V then[A, v] ∈ V . Thusg is a Lie algebra andV is a representation ofg.

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ii) The map[·, ·] : V × V → R is a perfect alternating pairing onV . Use theJacobi identity[[A,B], C] + [[B,C], A] + [[C,A], B] = 0 to prove that[gv, v′] +[v, gv′] = 0 for all g ∈ g andv, v′ ∈ V . Thusg is contained in the Lie algebraof the symplectic group for our pairing; sincedim g = dim sp2n, this gives anisomomorphismg

∼→ sp2n.

7. Extremal lattices and spherical designs

8. Further directions

Lattices with arbitrary discriminants; periodic weights

Higher θ functions; coding analogues

Murphy’s Law

Acknowledgements

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