DISCRETE SUBGROUPS, LATTICES, AND UNITS.

DISCRETE SUBGROUPS, LATTICES, AND UNITS. IAN KIMING 1. Discrete subgroups of real vector spaces and lattices. Definitions: A lattice in a real vector space V of dimension d is a subgroup of form: Zv 1 +... + Zv d where v 1,..., v d is a basis of V. A subset B of the real vector space R n is called discrete if the intersection between B and any bounded subset of R n is finite (here we envision R n as being equipped with the usual metric coming from the usual measure v of length of vectors v R n ). Proposition 1. (i). If Λ is a lattice in R n then Λ is discrete in R n. (ii). Suppose that Λ be a discrete subgroup of the real vector space R n, i.e., Λ is a subgroup of R n that is discrete as a subset of R n. Then Λ is a lattice in the subspace V of R n that it generates as vector space. In particular, Λ is a free abelian group of rank n. Proof. (i). Write Λ = Zv 1 +... Zv n with v 1,..., v n a basis of R n. The matrix M having the v i s as columns is then invertible. If x is a lattice point we can write: ) x = M ( a1 with integers a i and where we view x as a column vector. We obtain: ) ( a1. a n. a n = M 1 x and consequently (with just a rough estimate), a i c n x where c is the maximum of the numerical values of entries of M 1. Thus, if we require a lattice point x to be in some bounded subset of R n there are only finitely many possibilities for the a i, and hence only finitely many such lattice points. (ii). Let V be the subspace of R n that is generated by Λ (i.e., V is the subspace consisting of all R-linear combinations of elements of Λ). Let d := dim V so that d n. Now, Λ must contain a basis of V as vector space; suppose that v 1,..., v d Λ is a basis of V. It is easily seen that Λ is discrete as a subset of V. 1

2 IAN KIMING Consider then the subgroup Λ 0 := Zv 1 +... + Zv d of Λ. Then Λ 0 = Z d as the v i are R-linearly independent, and hence in particular Z-linearly independent. Since V is generated over R by the v i we see that any element v V can be written: ( ) v = a 1 v 1 +... a d v d + x 0, with a i R, 0 a i < 1, x 0 Λ 0. As Λ V we see in particular that any element of Λ can be written in the form ( ). We deduce that any coset of Λ/Λ 0 has a representative x of form ( ) x = a 1 v 1 +... a d v d, with a i R, 0 a i < 1. But if x Λ has shape ( ) then x i a i v i i v i. But then x is in the intersection Λ {v V v i v i } which is a finite set because Λ is discrete in V. We conclude that Λ/Λ 0 is finite. If we put a := [Λ : Λ 0 ] we have then that a Λ Λ 0. By exercise 24, page 44 of [1], we have that a Λ is a free abelian group of rank d. Since x ax is clearly an isomorphism of Λ onto a Λ, we have that Λ is a free abelian group of rank d. But since Λ contains Λ 0 which is free of rank d, we conclude that Λ is free of rank d. If now x 1,..., x d is a Z-basis of Λ then the v i are Z-linear combinations of the x i. We conclude that the x i generate V as a real vector space, and hence that Λ is a lattice in V. 2. The logarithmic map on units in algebraic number fields. In this section we work with the following setup: K is an algebraic number field of degree n over Q. Denote by U K the abelian group of units, that is, multiplicatively invertible elements of the ring O K of algebraic integers in K. We denote by r and s the number of real embeddings K R, and the number of pairs of complex conjugate embeddings K C, respectively. Thus, r + 2s = n. Let σ 1,..., σ r denote the r real embeddings K R, and let (τ 1, τ 1 ),..., (τ s, τ s ) denote the s pairs of complex conjugate embeddings K C. We then define the logarithmic map l: U K R r+s as follows: l(u) := (log σ 1 (u),..., log σ r (u), log τ 1 (u) 2,..., log τ s (u) 2 ). Notice that this definition is independent of whether we use τ j or τ j from a pair (τ j, τ j ) of complex conjugate embeddings in the definition. Proposition 2. (i). The logarithmic map l is a homomorphism of abelian groups with image l(u K ) contained in the hyperplane H := {(x 1,..., x r+s ) x 1 +... + x r+s = 0}. (ii). If B is a bounded subset of R r+s then l 1 (B) is a finite subset of U K.

DISCRETE SUBGROUPS, LATTICES, AND UNITS. 3 Proof. (i). That l is a homomorphism, i.e., that l(u 1 u 2 ) = l(u 1 ) + l(u 2 ) is immediately clear. Let u U K. We know that then N K/Q (u) = ±1; but this means: r s 1 = N K/Q (u) = σ i (u) τ j (u) τ j (u) = i=1 r σ i (u) i=1 as τ j (u) = τ j (u) for j = 1,..., s. So, r 0 = log σ i (u) + i=1 which just means that l(u) H. j=1 s τ j (u) 2 j=1 s log τ j (u) 2 (ii). Let u U K. Observe that any bound on l(u) implies a bound on the numerical values of the coordinates of l(u) and hence a bound on the numerical value of all conjugates σ i (u) and τ j (u), τ j (u) of u. This again implies a bound on the numerical values of the coordinates of φ(u) where φ: K R n is the map that we introduced earlier (cf. [1], p. 133): φ(u) := (σ 1 (u),..., σ r (u), Re(τ 1 (u)), Im(τ 1 (u)),..., Re(τ s (u)), Im(τ s (u))). We have shown that if l(u) lies in a given bounded subset B of R r+s then φ(u) lies in a certain bounded subset of R n (that could be specified explicitly in terms of B). But since we already know, cf. the theorem 36, p. 134, of [1], that φ(o K ) is a lattice in R n, and since U K O K, we conclude by Proposition 1 (i) that there are only finitely many possibilities for u U K if l(u) is required to lie in a given bounded subset B. Corollary 1. The logarithmic map l maps U K to a lattice in some subspace of the hyperplane H := {(x 1,..., x r+s ) x 1 +... + x r+s = 0}. The kernel of l consists of the roots of unity in K. Proof. Proposition 2 implies that l(u K ) is a discrete subgroup of H. The first claim then follows from Proposition 1 (ii). Let u U K. If u is a root of unity then so is every conjugate of u and so every conjugate of u has numerical value 1. We see then that l(u) = 0. On the other hand, consider the set: j=1 W := {u U K l(u) = 0}. We certainly have that W is contained in l 1 of the unit ball of R r+s ; proposition 2 (ii) then implies that W is finite. But since l is a homomorphism U K R r+s of abelian groups we have u i W for all i Z if u W. So if u W there are necessarily 2 distinct integers i and j such that u i = u j. Then u i j = 1 and u is a root of unity.

4 IAN KIMING Notice that the previous proof gives in particular a new proof of the fact that there are only finitely many roots of unity in K. Corollary 2. For the abelian group U K we have: U K = W F where W is the finite subgroup of roots of unity in K and F is a free abelian group of rank r + s 1. Proof. The image l(u K ) under the logarithmic map is a lattice in a subspace of R r+s of dimension r+s 1; in particular this image is free of some rank d r+s 1. Choose units u 1,..., u d U K such that l(u 1 ),..., l(u d ) is a Z-basis of the free abelian group l(u K ). Since the kernel of l is W it is now clear that U K is generated as abelian group by W and the u i. Letting F denote the subgroup of U K generated by the u i, we have that F is free of rank d with Z-basis u 1,..., u d : For any relation u a1 1 ua d d = 1 with a i Z implies a 1 l(u 1 ) +... + a d l(u d ) = 0 and thus a 1 =... a d = 0. In particular, the only element of finite order in F is 1 so we have W F = {1}. But then U K = W F follows. The following lemma is an application of Minkowski s lemma on convex bodies. We will need it for the proof of Dirichlet s unit theorem. Lemma 1. Fix any integer k with 1 k r + s. Then given any nonzero element α O K there is a nonzero β O K such that ( ) s 2 N K/Q (β) disc(k), π and such that: and: σ i (β) < σ i (α) if i k, τ j (β) < τ j (α) if r + j k. Proof. Since α 0 we can, and will, choose positive real numbers c i, i = 1,..., r+s such that: c i < σ i (α) for i = 1,..., r, i k, c r+j < τ j (α) 2 for j = 1,..., s, r + j k, and furthermore: ( ) s 2 c 1 c r+s = disc(k). π Consider now the subset E of R n defined by the inequalities: x 1 c 1,..., x r c r, x 2 r+1 + x 2 r+2 c r+1,..., x 2 n 1 + x 2 n c r+s.

DISCRETE SUBGROUPS, LATTICES, AND UNITS. 5 Then E is compact and easily seen to be centrally symmetric and convex with volume: vol(e) = (2c 1 ) (2c r ) (πc r+1 ) (πc r+s ) = 2 r π s i c i = 2 r+s disc(k) = 2 r+2s vol(r n /φ(o K )) = 2 n vol(r n /φ(o K )) where again φ is the embedding K R n discussed earlier, and where we used theorem 36 of [1]. Now Minkowski s lemma implies the existence of a nonzero point in the intersection between E and the lattice φ(o K ). Writing this lattice point as φ(β) with β O K, we have β 0. Since β E we find: N K/Q (β) ( ) s 2 c i = disc(k), π i and also that β has the remaining desired properties: σ i (β) c i < σ i (α) for i = 1,..., r, i k, τ j (β) c r+j < τ j (α) for j = 1,..., s, r + j k. Lemma 2. Fix any integer k with 1 k r + s. Then there is u U K such that if l(u) = (y 1,..., y r+s ) then y i < 0 for i k. Proof. Choosing any nonzero element α 1 O K and applying Lemma 1 repeatedly we find a sequence α 1, α 2,... of nonzero elements of O K such that, for all m N, and: σ i (α m+1 ) < σ i (α m ) if i k, τ j (α m+1 ) < τ j (α m ) if r + j k, and such that all numbers N K/Q (α m ) are bounded by a certain constant. Since (α m ) = N K/Q (α m ), a previous argument shows that there are only finitely many distinct ideals among the principal ideals (α m ) (consider the prime factorizations of the ideals). Consequently, there exist m, t N with m < t and such that (α m ) = (α t ). But then α t = u α m with a unit u, and we find (using t > m): log σ i (u) = log σ i (α t ) log σ i (α m ) < 0 for i = 1,..., r, i k, log τ j (u) 2 = 2 log τ j (α t ) 2 log τ j (α m ) < 0 for j = 1,..., s, r + j k. As l(u) = (log σ 1 (u),..., log σ r (u), log τ 1 (u) 2,..., log τ s (u) 2 ), the unit u has the desired properties.

6 IAN KIMING References [1] D. A. Marcus: Number fields. Springer-Verlag 1995 (Corr. 3rd printing). Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK- 2100 Copenhagen Ø, Denmark. E-mail address: kiming@math.ku.dk