A Metabebelian Group Admitting Integral Polynomial Exponents

Transcription

1 A Metabebelian Group Admitting Integral Polynomial Exponents Dennis Spellman Philadelphia, PA U.S.A August 2013 Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

2 Let r be a cardinal - here viewed as an ordinal not in bijective correspondence with any prior ordinal. Let F be a group free of rank r with free basis { a ξ+1 0 ξ < r }. Let be a short exact sequence of groups. 1 R F ρ H 1 Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

3 Let ZH be the integral group-ring and let M be a free rank r left ZHmodule with basis { t ξ+1 0 ξ < r }. The set of matrices of the type [ ] h m (h H, m M) forms a group under multiplication. If R is the derived group of R, a classical result of Magnus asserts that we get a faithful representation of F /R in the above group of matrices via the assignment [ ] a ξ+1 R ρ(aξ+1 ) t ξ (0 ξ < r). Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

4 Applying this in the case R F, the derived group of F, gives a faithful representation of the free rank r metabelian group F /F via the assignment [ ] a ξ+1 F ρ(aξ+1 ) t ξ (0 ξ < r). Let G F /F, a ξ+1 F α ξ+1 (0 ξ < r) and α ξ+1 ρ(a ξ+1 ) (0 ξ < r). Also let A ρ(f ) F /F. So A is a multiplicative free rank r abelian group. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

5 Translating into our new notation: We get a faithful representation of G via [ ] αξ+1 t α ξ+1 ξ+1 (0 ξ < r). The commutator subgroup of the overgroup [ ] h m (h A, m M) [ ] 1 m consists of matrices of the form (m M) and G itself is represented by those matrices (and only those) in the image of G. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

6 [ ] n [ ] 1 m 1 n m Ifn is an integer, then it is easy to see that. Suppose h A\{1} and n 0 is an integer. An easy induction on n shows [ ] n 1 n h m hn h k m k0 The integral group ring ZA is an integral domain. Letting K be its field of fractions we may write the above as [ ] n h m [ h n h n 1 h 1 m where the (1,2)-component of the above is viewed as lying in the r-dimensional vector space over K with basis { t ξ+1 0 ξ < r }. ] Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

7 On the other hand [ ] n h m ([ h ] n ) 1 m [ h n h n 1 h 1 m ] 1 [ h n h n hn 1 [ h n h n 1 h 1 m ] h 1 m ] Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

8 So, for arbitrary integers n, [ ] n [ h m h n h n 1 h 1 m ]. [ ] h m Now we extend the group to a group admitting exponents from the integral polynomial ring. First extend A to B A Z Z[θ] which we write multiplicatively with basis still denoted { α ξ+1 0 ξ < r }. The group ring Z[θ]B is an integral domain. We let Φ be its field of fractions. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

9 [ ] h m We embed our group into the matrices where h B and m lies in the r-dimensional vector space over Φ with basis { t ξ+1 0 ξ < r }. We define exponentiation from Z[θ] via [ ] f (θ) 1 m [ ] 1 f (θ) m and if h 1 [ ] [ f (θ) h m h f (θ) hf (θ) 1 h 1 ] m. Then we let G Z[θ] denote the Z[θ] -subgroup, generated as a Z[θ]-group, by the image of G. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

10 Precisely, G Z[θ] is the union of the chain (identifying G with its faithful image) G 0 G 1 G 2 G n where G 0 G and G n+1 is the group generated by g f (θ) as g varies over G n and f (θ) varies over Z[θ]. G n is the level-n subgroup of G Z[θ]. The analogies with Lyndon s free exponential group F Z[θ] are striking. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

11 If we define the sequence of derived ideals δ Z[θ] n (F Z[θ] ) recursively by δ Z[θ] 0 (F Z[θ] ) F Z[θ] δ Z[θ] n+1 (F Z[θ] ) [δ Z[θ] n (F Z[θ] ), δ Z[θ] n (F Z[θ] )] Z[θ] to be the Z[θ]-normal closure of the derived group of δ Z[θ] n (F Z[θ] ), then we conjecture that we have a faithful representation of F Z[θ] / δ Z[θ] 2 (F Z[θ] ). Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

12 The utility of Lyndon s group is that the integral substitutions θ k yield a discriminating family of retractions F Z[θ] F. We will see that the same is true (we illustrate with rank r 2) when relativized to the metabelian variety, i.e., we get a discriminating family of retractions (identifying G with its faithful image) G Z[θ] G induced by the integral substitutions θ k. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

13 The argument is tricky since for level n (with n 2) the denominator of the (1,2)-entry can contain a factor of the form h f (θ) 1; if k is an integral root of f (θ), the substitution θ k leads to the indeterminate form 0/0. We now show how to remedy this inconvenience. Here we illustrate with the case of rank r 2 but a similar argument will hold for any countable rank. We will just show here that the retractions are well-defined and ask you to trust that we really get a discriminating family. Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

14 Consider [ ] the subgroup of GL 2 (R) consisting of those matrices of the form a b where a > 0. We introduce real exponents by [ ] λ 1 b [ ] 1 λ b and if a 1 [ ] [ λ a b ] aλ 1 a 1 b. a λ Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

15 Suppose G has rank 2. We may take A to be any rank 2 free abelian subgroup of the positive reals under multiplication. Say A e, e e. The field extension Q(e, e e ) is countable so we may chose t R \ Q(e, e e ) and easily infer that {1, t} is linearly independent over Q(e, e e ). We then identify G with the group of real matrices generated by and extend to G Z[θ]. α 1 [ ] e 1 and α 2 [ ] e e t Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

16 An induction on the length of a Z[θ]-word representing an element γ(θ) G Z[θ] shows that for every integer k lim γ(θ) exists and lies in G. θ k Thus our quandary with the indeterminate form 0/0 is resolved! Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16

17 Dennis Spellman (Institute) A Metabebelian Group Admitting Integral Polynomial ExponentsAugust / 16