STRUCTURE THEORY OF UNIMODULAR LATTICES

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1 STRUCTURE THEORY OF UNIMODULAR LATTICES TONY FENG 1 Unimodular Lattices 11 Definitions Let E be a lattice, by which we mean a free abelian group equipped with a symmetric bilinear form, : E E Z Definition 11 We say E is unimodular if one of the two equivalent conditions hold: (1) ( Extrinsic ) For a basis e i of E, we have det(a ij := e i, e j ) = ±1 (2) ( Intrinsic ) The map E E given by x x, is an isomorphism To see that these equivalent, note that the matrix (a ij ) expresses e i, in terms of the dual basis (e i ) of E, where e i (e j ) = δ ij Note that the intersection matrix (a ij ) is well-defined only up to the action of GL(n, Z) given by b (a ij ) = b t (a ij )b We let ULat be the monoid of unimodular lattices under 12 Invariants If we are going to give any sort of classification of unimodular lattices, we then we have to write down some interesting invariants of them Rank We define r(e) = rank Z E Index We can consider V = E R The classification of real symmetric bilinear forms tells us that this is determined by the signature (p, q), and we define τ(e) = p q Discriminant Since det GL n Z = {±1}, the discriminant det(a ij ) of E is well-defined and equal to ±1 by definition Note however, that this is already determined by the rank and index in the unimodular case Type If x, x 0 (mod 2) for each x E then we say that E is even, or of Type II Otherwise we say that E is odd, or of Type I Note that in terms of an intersection matrix (a ij ), even is equivalent to a ii 0 (mod 2) for all i Wu class Consider E = E Z Z/2, a vector space over F 2 Since the map x x, x Date: April 19,

2 2 TONY FENG is linear for x E, the unimodular assumption implies that x, x = v 2, x for some v 2 E If we let ṽ 2 denote any lift of v 2, then ṽ 2, ṽ 2 is well-defined modulo 8 Indeed, any other lift is of the form ṽ 2 + 2v, but ṽ 2 + 2v, ṽ 2 + 2v = ṽ 2, ṽ 2 + 4( ṽ 2, v + v, v ) and ṽ 2, v + v, v is even by definition of v 2 We then define σ(e) := v 2, v 2 Z/8 Remark 12 While this last definition may be natural from the experience of topology, it feels hacky to me A more systematic approach would be to consider E Z Q p for each prime p, and then to use a classification of quadratic forms over p-adic fields This is a little complicated to state, but we just remark that at odd primes p 2 one would find no interesting invariants, and at p = 2 one would recover the datum of σ (but packaged slightly differently) 13 Examples We denote by I + the rank 1 lattice with intersection matrix (1) We denote by I the rank 1 lattice with intersection matrix ( 1) We denote by U the hyperbolic lattice with intersection matrix ( ) We denote by D n the index-2 sublattice of Z n consisting of (x i ) with x i 0 (mod 2) We denote by E 4k the lattice E 4k = D 4k + (D 4k + ( 12,, 12 ) ) with the form inherited from R n (We need a multiple of 4 for the form to be integral) Note that this is even if k = 2m is even Remark 13 For us the only aspect of the E 8 lattice that will matter is its invariants There are other ways of describing it, for instance as the intersection matrix for the lattice of ( 1)-curves in the Picard group of the blowup of P 2 at 8 points Here is a summary of the invariants of the lattices which are important to us Lattice r τ D Type σ I I 1 I I -1 si + + ti s + t s t ( 1) t I s t U II 0 E 8m 8m 8m 1 II 0 2 Structure Theory 21 Indefinite lattices The two main results are: Theorem 21 Let E be indefinite of type I Then E = si + ti Theorem 22 Let E be indefinite of type II If τ(e) 0 then E = pu qe 8

3 STRUCTURE THEORY OF UNIMODULAR LATTICES 3 Remark 23 If τ(e) < 0 then E = p( U) q( E 8 ), where the negative lattice is obtained by negating the Cartan matrix These results can be reformualted as saying that a unimodular indefinite lattice is completely determined by its rank, index, and type There is no significance to E 8 other than that together with U all ranks and indices can be achieved 22 Consequences Corollary 24 If E, E ULat have the same rank and index then E I + = E I + or E I = E I Proof By adding either I + or I, we can ensure that E and E are indefinite and odd, and then apply Theorem 21 (Strictly speaking, this is false if E = E = 0, but that case is also fine) The following results are most elegantly phrased in terms of the Grothendieck K-group K(ULat) We recall that the Grothendieck group is a construction taking in any abelian monoid and producing an abelian group, essentially by formally adjoining inverses The elements of K(ULat) can be thought of as formal differences of lattices, and two lattices E and E become isomorphic in K(ULat) if they were stably isomorphic, ie if there exists some lattice F such that E F = E F The K-group K(ULat) is universal with respect to maps from the monoid of lattices to groups Theorem 25 K(ULat) is a free abelian group on generators I + and I This follows from Theorem 21 by adding I + or I Alternatively, we have just seen that any two lattices with the same rank and index are stably isomorphic Since rank and index are always congruent mod 2, this also shows that the map (r, τ) induces an isomorphism K(ULat) = {(a, b) Z 2 a b (mod 2)} This completely characterizes lattices up to stable isomorphism Theorem 26 We have σ(e) = τ(e) (mod 8) for all lattices E Proof By inspection this holds for si + + ti Corollary 27 If E is even (type II), then τ(e) 0 (mod 8) Corollary 28 If E is definite and even (type II), then r(e) 0 (mod 8) Proof Definite implies that τ(e) = ±r(e) Let E be a unimodular latice 3 Some Lemmas Lemma 31 If F E, then F is a summand if and only F is unimodular Proof If F is a summand then obviously F needs to be unimodular, since the discriminant of E is a multiple of that for E Conversely, suoppose F is unimodular We need to produce a splitting E F, but we always have E F and the unimodularity implies E = E and F = F Corollary 32 If x E has x, x = ±1 then E = x x

4 4 TONY FENG Lemma 33 If x is indivisible, then there exists y E such that x, y = 1 Proof Under the isomorphism E = E we have x x, Since x is indivisible, so is the functional x,, so it takes some y E to 1 The major input is: 4 Proof of Theorem 21 Theorem 41 If E is an indefinite unimodular lattice then there exists x E such that x, x = 0 Proof We will not give a full proof For rank 4 the argument is by direct analysis; for n 5 perhaps the nicest way to see it is by the Hasse-Minkowski Theorem, which says that if a quadratic form over Q represents λ in Q p for all p and also in R, then it represents λ over Q Obviously there is a solution over R Over Q p, at least for p 2 it suffices to find a non-zero solution in F p, which can then be lifted by Hensel s Lemma One argues that: if a, b, c are units in Z/p, then there is a non-zero solution to ax 2 + by 2 + cz 2 0 (mod p) This can be obtained by counting: ax 2 takes roughly p/2 values, as do the other terms The need for five squares comes from the fact that among five elements of Z p, some three will have the same p-adic valuation modulo 2 Now, suppose that E is odd and definite (and unimodular) Then there is some x E such that x, x = 0, which we can assume to be indivisible By Lemma 33 there is then y E such that x, y = 1 By Lemma 31, we have E = x, y F Since E is odd, the intersection matrix for x, y is of the form ( ) m + 1 By replacing x by mx y and y by (m + 1)x y, we have the new intersection matrix ( ) This shows that E = I + I F You might be tempted to conclude the result by induction, but be careful: we do not know that F is odd and indefinite However, if F is non-zero then we can always guarantee this to be the case by taking F I + or F I 5 Proof of Theorem 22 Lemma 51 Let E be indefinite and even Then there is F ULat such that E = U F Proof Arguing as before, there exists an indivisible x E such that x, x = 0, and y E such that x, y = 1, and a splitting The intersection matrix of x, y is E = x, y F ( ) m

5 STRUCTURE THEORY OF UNIMODULAR LATTICES 5 By the change of variables y y mx, we get the intersection matrix ( ) which is U Lemma 52 Let F 1, F 2 ULat be even Suppose I + I F 1 = I+ I F 2 Then U F 1 = U F2 Granting this for now, let s see how it finishes off the Theorem We want to show that an even indefinite unimodular lattice is determined by its rank and index, so suppose E 1 and E 2 are two lattices with the same rank and index Then E 1 = U F1 and E 2 = U F2 Then I + I F 1 = I+ I F 2 by Theorem 21, and so we are done by Lemma 52 To prove the classification E = pu qe 8, simply note that we can recover r = r(e) τ(e) 2 and q = τ(e)/8 Proof of Lemma 52 Let W = I + I and E i = W F i Consider the index 2 sublattices Ei 0 = {x E i : x, x 0 (mod 2)} We have evidently Ei 0 = W 0 F i where W i is the index 2 sublattice of W consisting of points (x, y) with x y (mod 2) We now consider also the dual lattice (Ei 0) = (W 0 ) F i, noting that (W 0 ) = {(x, y) x, y 1 Z; x y Z} 2 Then (W 0 ) /W 0 = Z/2 Z/2 has three subgroups of index 2 One is W, and the other is isomorphic to U, as can be seen directly: a basis for (W 0 ) /W 0 is (1/2, 1/2) and (1/2, 1/2), so the other two subgroups are (1/2, 1/2), (1, 1) and (1, 1), (1/2, 1/2) Now, any isomorphism W F 1 = W F2 induces an isomorphism (W 0 ) F 1 = (W 0 ) F 2, hence also an isomorphism of the rank 2 sublattices (up to some ordering) 6 The Siegel Mass formula The definite unimodula lattices do not admit such nice classification theorems However, there is a Siegel mass formula which describes how many different isomorphism classes there are For simplicity, we consider the even case The automorphism group of a positivedefinite lattice is finite, since it is the intersection of a compact group with a discrete group For a lattice E, let g E be the size ofthe automorphism group of E Let C n be the set of isomorphism classes of even unimodular lattices with rank n, and set M n = E C n 1/g E Theorem 61 We have where M n = B 2k 8k t e t 1 = m=0 4k 1 j=1 B j 4j B m t m m! The right hand side grows extremely quickly, which tells us that there are many different isomorphism classes

6 6 TONY FENG n M n The growth of M n is superexponential In particular, this suggests the impossibility of giving a classification of definite unimodular lattices: it would require too many invariants Example 62 In practice the mass formula is useful for showing that we have accounted for all lattices of a given rank For instance, for n = 8 it gives , which is the size of Aut(E 8 ) Similarly, for n = 16 we get that the only lattices up to isomorphism are E 16 and E 8 E 8 References [1] Serre, A Course in Arithmetic Springer-Verlag Graduate Texts in Mathematics, 1973

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