A Note on the Special Unitary Group of a Division Algebra

A NOTE ON THE SPECIAL UNITARY GROUP OF A DIVISION ALGEBRA 1. Introduction Let K be a number field, and let D be a division algebra with center K, with an involution of the second kind, τ . Let U (1, D) be the unitary group of D, that is, the set of elements in D∗ such that dτ (d) = 1. Let SU (1, D) be the special unitary group, that is, the set of elements of U (1, D) with reduced norm 1. An old theorem of Wang [7] shows that for any central division algebra over a number field, SL(1, D) is the commutator subgroup of D∗. It is an open question (see [4, p.536]) whether the group SU (1, D) equals the group [U (1, D), U (1, D)] generated by unitary commutators. We show in this note that, by contrast, if K is a function field in one variable over a number field, and if D is an algebra with center K and with an involution of the second kind, the group SU (1, D) modulo [U (1, D), U (1, D)] can be infinite in general. More precisely, we prove : Theorem 1.1. Let n ≥ 3, and let ζ be a primitive n-th root of one. Then, there exists a division algebra D of index n with center Q(ζ)(x) which has an involution of the second kind such that the corresponding group SU (1, D)/[U (1, D), U (1, D)] is infinite. Our algebra will be the symbol algebra D √= (a, x; ζ, K, n) where K = Q(ζ)(x) and a ∈ Q is such that [Q(ζ)( n a) : Q(ζ)] = n. This is the K-algebra generated by two symbols r and s subject to the relations rn = a, sn = x, and sr = ζrs. If we write L for the K subalgebra of D generated by r, it is clear that L is just the field √ Q(ζ, n a)(x). The Galois group L/K is generated by σ that sends r to ζr: note that conjugation of L by s has the same effect as σ on L. An easy computation shows that xn is the smallest power of x that is a norm from L to K, so standard results from cyclic algebras ( [3, Chap. 15.1] for instance) show that D is indeed a division algebra. It is well known that D has a valuation on it that extends the x-adic valuation on K. This valuation will be crucial in proving our theorem. 2. The valuation on D We recall here how the x-adic valuation is defined on D. Recall first how the x-adic (discrete) valuation is defined on any field P function i E(x) over a field E: it is defined on polynomials f = i ai x (ai ∈ E) by v(f ) = min{i | ai 6= 0}, and on quotients of polynomials f /g by v(f /g) = v(f ) − v(g). The value group ΓL is Z, while the residue L is E. This definition gives valuations on all three fields Q(ζ + ζ −1 )(x), K, and L, all of which we will refer to as v. √ These fields have residues (respectively) Q(ζ + ζ −1 ), Q(ζ) and Q(ζ, n a) with respect to v. It is standard that the valuation v on Q(ζ + ζ −1 )(x) extends uniquely to K, a fact that will be crucial to us. With v as above, we define a function, also denoted v, from D∗ to (1/n)Z as follows: first, note that each d in D∗ can be uniquely written as d = l0 + l1 s + · · · + ln−1 sn−1 , for li ∈ L. (We will call each expression of the form li si , i = 0, 1, · · · n − 1, a monomial.) Define v(s) = 1/n, and v(li si ) as v(li ) + iv(s). Note that the n values v(li si ); 0 ≤ i < n are all distinct, since they lie in different cosets of Z in (1/n)Z. Thus, exactly one of these n monomials has the least value among them, and we define v(d) to be the value of this monomial. It is easy to check that v indeed gives a valuation on D. We find ΓD = (1/n)Z, so √ n ΓD /ΓK = Z/nZ. Also, the residue D contains the field Q(ζ, a). The fundamental inequality ( [5, p.21]) [D : K] ≥ [ΓD /ΓK ][D : K] shows √ n that D = L = Q(ζ, a). Note that since D is valued, the valuation v (restricted to K) extends uniquely from K to D ( [6]). 3. Computation of SU (1, D) and [U (1, D), U (1, D)] Write k for the field Q(ζ + ζ −1 )(x), and τ for the nontrivial automorphism of K/k that sends ζ to ζ −1 . Note that since a and x belong to the field k, we may define an involution on D that extends the automorphism of K/k by the rule τ (f ri sj ) = τ (f )ζ ij ri sj for any f ∈ K (so τ (r) = r, τ (s) = s; see [2, Lemma 7].) Proof of the theorem. Let d be in U (1, D), so dτ (d) = 1. Since v and v ◦ τ are two valuations on D that coincide on k, and since v extends uniquely from k to K, and then uniquely from K to D, we must have v ◦ τ = v. Thus, we find 2v(d) = 0, that is, d must be a unit. Then, for any d and e in U (1, D), we take residues to find √ −1 ded−1 e−1 = ded e−1 . However, D = L = Q(ζ)( n a) is commutative, so d and e commute, so ded−1 e−1 = 1. Note that we have a natural inclusion of L in the v-units of L; we identify L with its image in L. Under this identification, for any l ∈ L ⊆ L, l = l. Since the commutator of two elements in U (1, D) has residue 1, it suffices to find infinitely many elements in SU (1, D) ∩ L to show that SU (1, D) modulo [U (1, D), U (1, D)] is infinite. Write L1 and L2 (respectively) for the subfields Q(ζ + ζ −1 )(r) and Q(ζ) of L; note that L2 is the residue field of K. Then the involution τ on D acts as the nontrivial automorphism of L/L1 , so for any l ∈ L, lτ (l) is the norm map from L to L1 . The automorphism σ of L/K restricts to an automorphism (also denoted by σ) of L/L2 , and it is standard that the reduced norm of l viewed as an element of D is just the norm of l from L to K ( [3, Chap. 16.2] for instance), and hence the norm of l from L to L2 . We thus need to find infinitely many l ∈ L such that NL/L1 (l) = NL/L2 (l) = 1. Now, the set S1 = {l ∈ L : NL/L1 (l) = 1} is indexed by the L1 (1) points of the torus T1 = RL/L Gm (see [4], § 2.1). Similarly, the set 1 S2 = {l ∈ L : NL/L2 (l) = 1} is indexed by the L2 points of the torus (1) T2 = RL/L Gm . To show that S1 ∩S2 is infinite, we switch to a common 2 field by noting that the groups T1 (L1 ) and T2 (L2 ) are just the k0 points of the groups (RL1 /k0 T1 ) and (RL2 /k0 T2 ) respectively, where k0 = Q(ζ + ζ −1 ). Thus, it suffices to check that (RL1 /k0 T1 ∩RL2 /k0 T2 )(k0 ) is infinite, and for this, it is sufficient to check that (RL1 /k0 T1 ∩ RL2 /k0 T2 )0 (k0 ) is infinite. As both RL1 /k0 T1 and RL2 /k0 T2 are k0 -tori, the connected component (RL1 /k0 T1 ∩ RL2 /k0 T2 )0 is a k0 -torus as well, since it is a connected commutative group defined over k0 consisting of semisimple elements. So, its k0 points are Zariski dense in its Q points by a theorem of Grothendieck (see p.120 of [1]). Hence, it suffices to check that there are infinitely many Q points in (RL1 /k0 T1 ∩ RL2 /k0 T2 )0 . But for this, it clearly suffices to check that there are infinitely many Q points in (RL1 /k0 T1 ∩ RL2 /k0 T2 ). Write any l ∈ L as l = P X + (ζ − ζ −1 )Y where X, Y ∈ L1 . Then, Pn−1 i X = i=0 xi ri and Y = n−1 i=0 yi r where xi , yi ∈ k0 . Consider the equations NL/L1 (l) = 1 and NL/L2 (l) = 1. Rewrite these in terms of powers of r, invoking the actions of σ and τ and using the fact that rn = a. The first equation now involves the 2n variables xi , yi and has coefficients in L1 . Equating the coefficients of ri (i = 0, . . . , n − 1) on both sides, we get n equations in the variables xi , yi with coefficients in k0 . Similarly, the second equation involves the variables xi , yi and has coefficients in L2 . Using the fact that (ζ − ζ −1 )2 ∈ k0 and equating the coefficients of 1 and ζ − ζ −1 on both sides, we get two equations in the variables xi , yi with coefficients in k0 . As n ≥ 3, we have n + 2 < 2n and these equations have infinitely many common solutions over Q. This proves the theorem. 4. Concrete illustration for n = 3 We illustrate the theorem for n = 3 by concretely constructing infinitely many elements in SU (1, D)/[U √ (1, D), U (1, D)]. We take a = 2 for simplicity. Write l = a + b −3, where a and b are in L1 . Then s2 − 3 NL/L1 (l) = a2 + 3b2 = 1 has a parametrized set of solutions a = 2 , s +3 2s , for s ∈ L1 . Write s = t0 + t1 r + t2 r2 for ti ∈ Q and b = 2 s +3 substitute in a and b above. Then compute NL/L2 (l), noting that σ(s) = (t0 + ωt1 r + ωt2 r2 ). We solve for the ti so that NL/L2 (l) = 1. We claim that if we take t0 = 1 and t1 = 0, then for arbitrary t2 = t, NL/L2 (l) = 1. Indeed, l = u/v where u = 2ω + t2 r − 2tω 2 r2 , v = 2 + t2 r + tr2 . Then, an easy computation, using r3 = 2, shows that NL/L2 (u) = (2ω+t2 r−2tω 2 r2 )(2ω+t2 ωr−2tωr2 )(2ω+t2 ω 2 r−2tr2 ) = −8t3 +2t6 . Similarly, NL/L2 (v) = (2+t2 r+tr2 )(2+t2 ωr+tω 2 r2 )(2+t2 ω 2 r+tωr2 ) = −8t3 +2t6 . Thus, we have an infinite set of solutions and we are done. (Actually, the parametric solution above was first obtained using MathematicaT M . The program gives other parametric solutions as well, for instance, t0 = 0, t1 = − 2t12 .)