LENGTH OF COMMUTATORS IN

CLASSICAL-LIKE GROUPS OVER RINGS

Alexei Stepanov Nikolai Vavilov

Chair of Mathematics II Sankt-Petersburg Electrotechnical University 197342, Sankt-Petersburg, Russia and Department of Mathematics and Mechanics Sankt-Petersburg State University 198904, Petrodvorets, Russia

Introduction

1. Notation

(1.1) Let R be the center of a ring A and s ∈ R. Denote by A_{s} the localization of A in the multiplicative subset {s^{n} | n ∈ ℕ}.

2. Classical-like groups

(2.1) Let Φ be a root system and G = G(Φ,_{−}) be a group valued functor on a category ℛ of rings (a ring does not necessarily have a unit element). For each α ∈ Φ let X_{α} : ℛ → 𝒜b be a functor which takes a ring A to an additive subgroup of A. Suppose that we are given a family of natural transformations t_{α} : X_{α} → G. If A is a ring and ξ ∈ A, then, by definition, ξ ∈ X_{α}(A). By abuse of notation, we shall write t_{α}(ξ) instead of t_{α}(A)(ξ). These elements of G(A) will be called transvections.

Let E be the subfunctor of G such that E(A) is generated by all transvections in G(A). If I is a two-sided ideal in A, then we denote by E(A, I) the normal closure of the image of E(I) in E(A).

For a ring A and a subset M ⊆ A denote by T (A, M) the set of all transvections t_{α}(ξ), α ∈ Φ, ξ ∈ M. If M = A, then we set T (A) = T (A, A). Let H be a group and X a subset of H. We denote by X^{h} a subset of H consisting of all elements x_{1} ⋯ x_{j}, where j ⩽ h and each x_{i} is in X. For h = 0 we set X^{0} = {1}. In particular, T^{h}(A, M) denotes the subset of G(A) consisting of all elements which can be written as a product of h transvections of the form t_{α}(ξ), α ∈ Φ, ξ ∈ M.

Key words and phrases. Chevalley group, General Quadratic group, Elementary subgroup, Word length, Dimension of maximal spectrum.

The work on this article was partially supported by the INTAS 93–436 and the SFB 343 an der Universität Bielefeld

Suppose that the functors G and E satisfy the following properties.

2.2 G preserves monomorphisms and kernels. Let I be a two-sided ideal of a ring A ∈ ℛ. Denote by G(A, I) the image of G(I) in G(A). Then it follows that G(A, I) ≅ G(A) and G(A)/ G(A, I) = G(A/I).

2.3 Given G there is a positive integer ℓ_{2} = ℓ_{2}(G) such that if I is an ideal in A and A

is semilocal or I ⩽ Rad A, then G(A, I) ⊆ T^{ℓ_{2}}(A). 2.4 There is ℓ_{1} ∈ ℕ depending only on G such that given p ∈ ℕ there are m, n ∈ ℕ

such that for any central elements s, t ∈ A each commutator of transvections from T^{1}(A_{st}, \frac{t^{m}}{s} A) and T^{1}(A_{st}, \frac{s^{n}}{t} A) belongs to T^{ℓ_{1}}(A_{st}, (st)^{p}A).

2.5. Definition. A bunch consisting of a root system Φ, a group valued functor G, and natural transformations t_{α} : X_{α} → G (α ∈ Φ) satisfying the properties 2.2–2.4 will be called a classical-like group functor. Usually we shall write G or G(Φ,_{−}) instead of (Φ, G, t_{α}).

3. The main result

3.1. Theorem. Let G be one of the group functors listed above of rank n ⩾ 2 and let E be its elementary group subfunctor. Let further R be a ring with dim Specm R = d < ∞ and A a quasi-finite R-algebra (if G ≠ SGQ_{2n} we assume that A = R). Then there exists a natural number ℓ depending only on G and d such that each commutator [a, b] of elements a ∈ E(A) and b ∈ G(A) is a product of at most ℓ elementary transvections.

(3.2) The proof goes as follows. Since G(A, Rad A) has bounded word length we may assume that Rad A = 0. Take elements a ∈ G(A) and b ∈ E(A). As in the proof of the main result of [SS] we can choose elements s_{0}, …, s_{d} ∈ R generating R such that for each i the image of a in G(A_{s_{i}}) is a product of ℓ_{2} transvections. Decompose b as a product of b^{(i)} ∈ E(A, s_{i}^{m}A) where m ∈ ℕ is large enough. By induction arguments using the commutator formula (???) it suffices to prove that the length of [a, b^{(i)}] is bounded. By Lemma 3.4 G(A, s_{i}A) injects into G(A_{s_{i}}). Therefore it suffices to show that the image of [a, b^{(i)}] in G(A_{s_{i}}) is a product of a bounded number of transvection from the image of E(A, s_{i}A). Using induction and the same commutator formula as above we come up to the following proposition (Corollary 3.10).

Let s ∈ R, ξ ∈ A, and α ∈ Φ. Then there exists a positive integer m such that given c in the image of G(s^{m}A) in G(A_{s}) the commutator [x_{α}(\frac{ξ}{s}), b] is a product of at most ℓ_{3}(G) transvections from the image of E(A, s^{m′}A).

To prove this one can use either the decomposition of transvections as in [SS] or localization and patching as in [B]. The resulting bound for the length is much better if one uses the decomposition of transvections. But it is not known yet in the full generality. Therefore we proceed with localization and patching. The proof is based on the same ideas as we have just described.

First, we reduce to the case of a ring with trivial radical and prove the injectivity statement for such rings.

3.3. Lemma. Let A be a ring. Suppose that for any ā ∈ E(A/ Rad A) and b̄ ∈ G(A/ Rad A) the length of the commutator [ā, b̄] is at most h for some integer h. Then the length of a commutator [a, b], where a ∈ E(A) and b ∈ G(A), is bounded by h + ℓ_{2}.

Proof. By the Gauss decomposition any element of G(A, Rad A) can be written as a product of ℓ_{2} transvections. By the condition of the lemma a commutator [a, b] (a ∈ E(A), b ∈ G(A)) is a product of an element of length ⩽ h and an element from G(A, Rad A). Hence it length ⩽ h + ℓ_{2}.

3.4. Lemma. Let s be an element from the center of A. If Rad A = 0, then sA injects into A_{s} and, hence, G(A, sA) injects into G(A_{s}).

Proof. Suppose that sξ ∈ sA goes to 0 in A_{s}. Then there is m ∈ ℕ such that s^{m}sξ = 0. It follows that (sξA)^{m+1} = 0. Hence by the condition of the lemma sξA = 0. Thus sξ = 0.

Now we start the proof of Theorem 3.1 from the end of the plan above.

The following group theoretic lemma will be used in the sequel several times to compute the length of the commutator [a, b_{1} ⋯ b_{h}] proceeding from the length of each [a, b_{i}].

3.5. Lemma. Let h, ℓ′ ∈ ℕ and f a function ℕ → ℕ defined by the formula f(k) = (ℓ′ + 1)^{k} − 1. Let further X(k), Y (k) and Z(k) (k ∈ ℕ) be families of subsets in H satisfying the following properties.

(i) X(k + 1) ⊆ X(k), Y (k + 1) ⊆ Y (k), and Z(k + 1) ⊆ Z(k).

(ii) Z(k) ⊆ X(k).

(iii) Each Z(k) is closed under taking the inverse. Suppose that

(iv) given p ∈ ℕ there are m′, n′ ∈ ℕ such that for any x ∈ X(m′) and y ∈ Y (n′) their

commutator [x, y] belongs to Z^{ℓ′}(p). Then given p, h ∈ ℕ there are m, n ∈ ℕ such that for any a ∈ X(m) and b ∈ Y^{h}(n)

their commutator [a, b] belongs to Z^{f(h)}(p).

Proof. We proceed by induction on h. The result for h = 0 follows from condition (i). For h > 0 choose m′ and n′ satisfying condition (iv). According to (i) we may assume m′ ⩾ p. By the induction hypothesis with m′ instead of p and condition (i) there are m ⩾ m′ and n ⩾ n′ such that [a, c] ∈ Z^{f(h−1)}(m′) for any a ∈ X(m) and c ∈ Y (n). Take y ∈ Y (n) and c ∈ Y^{h−1}(n) such that b = yc and put [a, c] = z_{1} ⋯ z_{f(h−1)}, where each z_{i} is an element of Z(m′) ⊆ Z(p) ∩ X(m′). Then

[a, b] = [a, yc] = [a, y][a, c]^{y^{−1}} = [a, y] ∏_{i=1}^{f(h−1)}[y, z_{i}]z_{i} ∈ Z^{ℓ′}(p) ∏_{i=1}^{f(h−1)} Z^{ℓ′+1}(p)

([y, z_{i}] = [z_{i}, y]^{−1} ∈ Z^{ℓ′+1}(p) by (iii)). Thus [a, b] ∈ Z^{f(h)}(p), where f(h) = ℓ′ + (ℓ′ + 1)f(h − 1) = (ℓ′ + 1)^{h} − 1.

Proof. Let ξ, η ∈ A and α, β ∈ Φ. We have to prove that [x_{α}(ξ \frac{t^{m}}{s^{h}}), x_{β}(η \frac{s^{n}}{t^{k}})] belongs to T^{ℓ_{1}}(A_{st}, (st)^{p}A). If β ≠ −α this follows from commutator formula (???). Otherwise ???

3.6. Corollary. Let h ∈ ℕ and ℓ_{3}(h) = (ℓ_{1}+1)^{h}−1. Then, given s, t ∈ R and p ∈ ℕ there are m, n ∈ ℕ such that for any a ∈ T^{1}(A_{st}, \frac{t^{m}}{s} A), and b ∈ T^{h}(A_{st}, \frac{s^{n}}{t} A) their commutator [a, b] belongs to T^{ℓ_{3}(h)}(A_{st}, (st)^{p}A).

Proof. Use Lemma 3.5 with X(m) = T^{1}(A_{st}, \frac{t^{m}}{s} A), Y (n) = T^{1}(A_{st}, \frac{s^{n}}{t} A), and Z(p) = T^{1}(A_{st}, (st)^{p}A).

3.7. Lemma. Let Λ be a semi-local ring, φ : B → Λ a ring homomorphism, and r an element from the center of B. Then there is a positive integer ℓ_{2} = ℓ_{2}(G) such that for any a ∈ G(B, rB) its image in G(Λ) belongs to T^{ℓ_{2}}(Λ, rφ(B)).

Proof. The proof follows immediately from the Gauss decomposition (see [???]) and the definition of G(rA).

3.8. Lemma. Let r ∈ R and a ∈ G(rA). Then there exist t_{0}, …, t_{d} ∈ R generating R such that for each k = 0, …, d the image of a in G(A_{t_{k}}) belongs to T^{ℓ_{2}}(A_{t_{k}}, \frac{r}{t_{k}} A).

Proof. The proof is the same as for Lemma 1.3 of [SS]. (Should we give this proof here ???)

3.9. Lemma. Let s, t_{0}, …, t_{d} ∈ R, q ∈ ℕ, Y (n) = ⋃_{i=0}^{d} T^{1}(A_{s}, \frac{t_{i}^{n}}{s} A), and Z(p) =

T^{1}(A_{s}, t_{i}^{p}s^{q}A). Denote by X(m) the set of all elements x ∈ T^{∞}(A_{s}, s^{m}A) whose image in G(A_{st_{i}}) lies in T^{ℓ_{2}}(A_{st_{i}}, \frac{s^{m}}{t_{i}} A) for each i = 0, …, d. Define a function ℓ_{4} : ℕ → ℕ by the formula ℓ_{4}(h) = (ℓ_{1} + 1)^{ℓ_{2}h} − 1. Then given h, p ∈ ℕ there are m, n ∈ ℕ such that for any ā ∈ X(m) and b̄ ∈ Y^{h}(n) their commutator [ā, b̄] belongs to Z^{ℓ_{4}(h)}(p).

Proof. Clearly, if m ⩾ p, then X(m), Y (n), and Z(p) satisfy conditions (i), (ii), and (iii) of Lemma 3.5. We shall verify 3.5(iv) with ℓ′ = ℓ_{3}(ℓ_{2}) = (ℓ_{1} + 1)^{ℓ_{2}} − 1.

By Lemma 3.4 G(A_{s}, t_{i}A_{s}) injects into G(A_{st_{i}}) for each i = 0, …, d. Since Z(p) is a descending sequence, we may assume that p ⩾ q and p ⩾ 1. Using Lemma 3.6 choose m, n ∈ ℕ such that for any i = 0, …, d, a ∈ T^{1}(A_{st_{i}}, \frac{t_{i}^{m}}{s} A), and b ∈ T^{ℓ_{2}}(A_{st_{i}}, \frac{s^{n}}{t_{i}} A) the commutator [a, b] belongs to T^{ℓ_{3}(ℓ_{2})}(A_{st_{i}}, (st_{i})^{p}A). Clearly we may assume that n ⩾ p. Take an arbitrary elements ā ∈ X(m) and b̄ ∈ Y (n). By definition of Y (n) there is i such that b̄ ∈ T^{1}(A_{s}, \frac{t_{i}^{n}}{s} A). Let a and b be the images in G(A_{st_{i}}) of ā and b̄ respectively. By the choice of m and n the image of [ā, b̄] belongs to T^{ℓ_{3}(ℓ_{2})}(A_{st_{i}}, (st_{i})^{p}A). On the other hand, [ā, b̄] ∈ G(A_{s}, t_{i}^{p}A_{s}) because n ⩾ p. By the injectivity assertion we get [ā, b̄] ∈ T^{ℓ_{3}(ℓ_{2})}(A_{s}, (st_{i})^{p}A). Since p ⩾ q this implies condition 3.5(iv). Thus we can apply Lemma 3.5 to complete the proof.

3.10. Corollary. Let s ∈ R and q ∈ ℕ. Let b ∈ G(A) be such that its image in G(A_{s}) is a product of h transvections. Then there is m ∈ ℕ such that for any a ∈ G(s^{m}A) the commutator [a, b] is a product of at most ℓ_{4}(h) = (ℓ_{1} + 1)^{(d+1)hℓ_{2}} − 1 transvections from E(s^{q}A).

Proof. For the given q and p = 1 choose m, n ∈ ℕ satisfying the conditions of Lemma 3.9. Denote by ā and b̄ the images of a and b in G(A_{s}). By Lemma 3.8 there is a unimodular row (t_{0}, …, t_{d}) ∈ R^{d+1} such that ā belongs to T^{ℓ_{2}}(A_{t_{k}}, s^{m}A) for each k = 0, …, d. Since the row (t_{0}^{n}, …, t_{d}^{n}) is unimodular as well, in the notation of Lemma 3.9 b̄ belongs to Y^{h(d+1)}(n). Applying the lemma we get [ā, b̄] ∈ T^{ℓ_{4}(h(d+1))}(A_{s}, s^{q}A).

Clearly [a, b] ∈ G(A, sA). By Lemma 3.4 [ā, b̄] has a unique preimage in G(A, sA) which completes the proof.

Proof of Theorem 3.1. Assume first that Rad A is trivial. Let ℓ_{1} = ℓ_{1}(G) and ℓ_{2} = ℓ_{2}(G). Let a ∈ E(A) and b ∈ G(A). By Lemma 3.8 we can choose a unimodular row (s_{0}, …, s_{d}) ∈ R^{d+1} such that for each k = 0, …, d the image of b in G(A_{s_{k}}) lies in T^{ℓ_{2}}(A_{s_{k}}, \frac{1}{s_{k}} A). Denote by X(m) = X the set of all elements in G(A) whose images in G(A_{s_{k}}) lie in

T^{ℓ_{2}}(A_{s_{k}}, \frac{1}{s_{k}} A) for each k = 0, …, d. Let Y (n) = ⋃_{k=0}^{d} E(A, s_{k}^{n}A) and Z(p) = T^{1}(A). Then

Corollary 3.10 garantees condition (iv) of Lemma 3.5 whereas conditions 3.5(i)–(iii) hold by the definition of X, Y, and Z. Since for any n the row (s_{0}^{n}, …, s_{d}^{n}) is unimodular, the groups E(A, s_{k}^{n}A) generate E(A), hence a ∈ Y^{d+1}(n) for any n. Now applying Lemma 3.5 with ℓ′ = ℓ_{4}(ℓ_{2}) = (ℓ_{1} + 1)^{(d+1)ℓ_{2}^{2}} − 1 we see that the commutator [a, b] has length ℓ ⩽ f(d + 1) = (ℓ_{1} + 1)^{(d+1)^{2}ℓ_{2}^{2}} − 1.

If Rad A is not trivial, by Lemma 3.3 we have

ℓ ⩽ (ℓ_{1} + 1)^{(d+1)^{2}ℓ_{2}^{2}} − 1 + ℓ_{2}.