Subgroups of a Chevalley group between two different rings

Alexei STEPANOV (St.Petersburg)

Notation R ⊆ A are commutative rings with 1. Φ is a reduced irreducible root system. We shall assume that Φ ≠ A_{1}, C_{2}, G_{2}. G = G(Φ, ) is a Chevalley group scheme of type Φ. E = E(Φ, ) is its elementary subgroup functor. For an elements a, b of a group: a^{b} = b^{−1}ab and [a, b] = a^{−1}b^{−1}ab. Definition of standard sandwich classification (SSC).

Fix a triple (Φ, R, A). The SSC holds for subgroups between E(R) and G(A) if given a subgroup H between E(R) and G(A) there exists a unique subring P between R and A such that

E(P) ⩽ H ⩽ N(P)

where N(P) denotes the normalizer of E(P) in G(A). Known results. SSC holds:

0. A = R and any Φ. In this case SSC follows from normality of E(R) in G(R) (Taddei, 1986).

1. Φ = A_{n}, A is the field of fractions of a Dedekind domain R (R.A.Shmidt, 1979).

2. Any Φ, A is the field of fractions of a principle ideal domain R (Nuzhin, Yakushevich, 2000).

3. Any Φ, A is an algebraic extension of a field R (Nuzhin, 1983). SSC does not hold

4. All roots in Φ have the same length (i.e. Φ = A_{n}, D_{n}, E_{6}, E_{7}, E_{8}), arbitrary R, A is the affine algebra of G(Φ, R) (follows easily from Gordeev’s theorem published in 1998)

5. Φ = A_{n}, A is not quasi algebraic over R (for example A = R[t]; Stepanov 2003).

Definition. A ring A is called quasi transcendental over its subring R if there exists a diagram

R ⟶ P ⟶ A
↓  ↓
F ⟶ F [t]

where F is a field, horizontal arrows are natural inclusions, and vertical ones are surjective. Otherwise, A is called quasi algebraic over R. Remark. A is quasi algebraic over R iff any finitely generated R-subalgebra P ⊆ A is quasi finite over R. Theorem (Stepanov 2003) Suppose that R is a finitely generated algebra over a field or over ℤ, and A is a domain. Then A is quasi algebraic over R iff either A is an integral extension of R or dim R ⩽ 1 and A is contained in the algebraic closure of the field of fractions of R.

Main results

Theorem 1. Suppose that all roots in Φ has the same length, R = F is a field and A = F [t]. Let ¯·: G(F [t]) → G(F [t])/Center. Then there exists a nontrivial element y ∈ \overline{G(F [t])} such that the subgroup generated by \overline{E(F)} and y is the free product of \overline{E(F)} and the cyclic subgroup ⟨y⟩. Corollary. Suppose that all roots in Φ has the same length. If A is not quasi algebraic over R then SSC does not hold for subgroups between E(R) and G(A). Theorem 2. Let Φ = B_{n} or C_{n} and \frac{1}{2} ∈ R. Then SSC for subgroups between E(R) and G(A) holds for any ring A. The same is true if Φ = C_{n} and 2 = 0.

Outline of the proof of Theorem 2

Lemma 1. For a matrix g ∈ G(A) the following conditions are equivalent.

(1) g ∈ N(R).

(2) E(R)^{g} ⩽ G(R).

(3) g′_{ij}g_{kℓ} ∈ R for any i, j, k, ℓ ∈ {1, …, n}.

(4) g normalizes G(R).

(5) E(R)^{g} ⩽ N(R).

Moreover,

– [N(R), N(R)] ⩽ G(R),

– E(R) is a characteristic subgroup of N(R),

– N(R) is self normalized,

– if R is Noetherian, then by theorem of Hazrat and Vavilov N(R)/E(R)

is nilpotent by Abelian. Definition. Let H be a subgroup of G(A), containing E(R). Define the subring associated with H to be the largest subring P of A such that E(P) ⩽ H. We denote it by P_{H}.

Clearly, R ⊆ P_{H} and P_{H} = {ξ ∈ A | ∃α ∈ Φ : x_{α}(ξ) ∈ H}. Definition. Let h belong to a torus of G(R). Recall that h is called small if it is annihilated by all long roots. Clearly, if all roots in Φ have the same length, then there are no non-central small semisimple elements. For Φ = B_{n}, C_{n}, F_{4} such an element exists over any ring provided 2 ≠ 0. Lemma 2. Let h ∈ G(R) be a small semisimple element, α a long root, and g ∈ G(A). Then X_{α}^{h^{g}} commutes with X_{α}. Lemma 3. Suppose that g ∈ H commutes with a long root subgroup X_{α} for some α ∈ Φ. Then g ∈ N(P_{H}). Proof of Theorem 2. Let α be a long root and g ∈ H. Take two arbitrary elements a, b from E(P_{H}) and consider the element c = x_{α}(1)^{h^{agb}} ∈ H. By Lemma 2 this element commutes with X_{α} and by Lemma 3 c ∈ N(P). Rewrite c in the form

c = b^{−1}(g^{−1}(a^{−1}h^{−1}a)g(bx_{α}(1)b^{−1})g^{−1}(a^{−1}ha)g)b

Since b ∈ E(P), the element bcb^{−1} is in N(P). Fix a and let b vary. Since the subgroup generated by bx_{α}(1)b^{−1} is normal in E(P), it must coincide with E(P). Thus, E(P)^{g^{−1}(a^{−1}h^{−1}a)g} ∈ N(P), and by Lemma 1 (a^{−1}h^{−1}a)^{g} ∈ N(P).

Again, the elements a^{−1}h^{−1}a generate the normal subgroup of E(P) as a ranges over E(P). But the minimal normal subgroup of E(P) containing h must be equal to the relative elementary subgroup E(P, 2P). Therefore, E(P, 2P)^{g} ∈ N(P). Since 2 is invertible in P, we have E(P, 2P) = E(P) and by Lemma 1 g ∈ N(P).

Outline of the proof of Theorem 2

Assumption. All roots in Φ have the same length. Theorem (Gordeev). Let K be an infinite field. Then given noncentral elements a_{1}, …, a_{ℓ} ∈ G(K) and nonzero integers m_{1}, …, m_{ℓ} there exists g ∈ G(Φ, K) such that g^{m_{1}}a_{1} ⋯ g^{m_{ℓ}}a_{ℓ} ≠ 1. Lemma 1. Fix an ordering on Φ. Let α be the maximal root, K a field, a, b ∈ G(K), and ba^{−1} ∈ BwB. Then the following conditions are equivalent.

(1) X_{α}^{a} and X_{α}^{b} are opposite (i.e. if c ∈ G(K) is such that (X_{α}^{a})^{c} = X_{β} and (X_{α}^{b})^{c} = X_{γ}, then γ = −β).

(2) w(α) = −α

(3) For any x ∈ X_{α}^{a} \ {1} and y ∈ X_{α}^{b} \ {1} the commutator [x, y] does not commute with x.

Lemma 2. Let α ∈ Φ be a root. There exists an element g ∈ G(F [t]) such that for any noncentral element a in G(F) the root subgroups X_{α} and X_{α}^{gag^{−1}} are opposite. Proof. For g ∈ G(F [t]) denote by C(g) the set of all elements a ∈ G(F) such that

[[X_{α}, X_{α}^{gag^{−1}}], X_{α}] = {1}.

Clearly C(g) is Zariski closed in G(F). We shall construct a sequence of elements g_{0} = 1, g_{1}, …, g_{m} ∈ G(F [t]) such that C(g_{0}) ⫌ C(g_{1}) ⫌ ⋯ ⫌ C(g_{m}) = Center is a proper chain of Zariski closed subsets in G(F).

Suppose that we have constructed g_{0}, …, g_{k} satisfying the above conditions. Clearly, we can choose M ∈ ℕ such that for any a ∉ C(g_{k})

[[X_{α}, X_{α}^{g_{k}ag_{k}^{−1}}], X_{α}] ≢ {1} mod t^{M}F [t]

We shall choose g_{k+1} ≡ g_{k} mod t^{M}F [t]. Take a ∈ C(g_{k}). Obviously, the set X of all g ∈ G(F (t)) such that

[[X_{α}, X_{α}^{(g_{k}g)a(g_{k}g)^{−1}}], X_{α}] ≠ {1}

is open in G(F (t)). By Gordeev’s theorem, this set is nonempty. Now, since the congruence subgroup G(F [t], t^{M}F [t]) is Zariski dense in G(F (t)), it has nonempty intersection with X. Put g_{k+1} = g_{k}g. Since g_{k+1} ≡ g_{k} mod t^{M}F [t], by the choice of M we have C(g_{k+1}) ⊆ C(g_{k}). But a ∈ C(g_{k}) \ C(g_{k+1}), therefore C(g_{k+1}) ≠ C(g_{k}), which completes the proof.

Proof of Theorem 2. Take N ∈ ℕ big enough and put y = x_{α}(t^{N})^{g}, where α is the maximal root and g is from the previous lemma. Consider the product

d = c_{1}y^{m_{1}}c_{2} ⋯ y^{m_{ℓ}},

where c_{i} ∈ G(Φ, F) \ Center and each m_{i} is not divisible by char F (otherwise, y^{m_{i}} = x_{α}(m_{i}t^{N})^{g} = 1). We have to prove that d ≠ 1. Conjugating d by g^{−1} we get

(gc_{1}g^{−1})x_{α}(m_{1}t^{N})(gc_{2}g^{−1})x_{α}(m_{2}t^{N}) ⋯ (gc_{ℓ}g^{−1})x_{α}(m_{ℓ}t^{N}) = 1

Let ac_{k}a^{−1} = b′_{k}w_{k}b′′_{k} be the Bruhat decomposition of ac_{k}a^{−1} with b′_{k}, b′′_{k} lying in the unipotent radical U of B. Since X_{α} commutes with U, we can rewrite the equation above in the form

(⋆)  b_{1}w_{1}x_{α}(m_{1}t^{N})b_{2}w_{2}x_{α}(m_{2}t^{N}) ⋯ b_{ℓ}w_{ℓ}x_{α}(m_{ℓ}t^{N}) = (b′′_{ℓ})^{−1}

where b_{k} = b′′_{k−1}b′_{k} and b_{1} = b′_{1}. Recall that by the choice of a we have w_{k}(α) = −α. Consider the action of the left hand side of (⋆) on the root element e_{α} of the corresponding Lie algebra.

1. e_{α}^{b_{1}} = e_{α}.

2. e_{α}^{w_{1}} = f(t)e_{−α}, where f ∈ F (t).

3. (f(t)e_{−α})^{x_{α}(m_{1}t^{N})} = f(t)(e_{−α} + m_{1}t^{N}h_{α} − m_{1}^{2}t^{2N}e_{α}). Thus, the degree of the coefficient at e_{α} increases by 2N + deg f

whereas other coefficients increase at most by N + something. Since N > something − deg f, other summands has no effect on the degree of the coefficient at e_{α}. Therefore, the degree of the image of e_{α} under the action of the left hand side of (⋆) is huge (at least nonzero). At the same time the right hand side of (⋆) leaves e_{α} invariant. This shows that the equality d = 1 is impossible, and we are done.