Siberian Mathematical Journal, Vol. 50, No. 1, pp. 66–76, 2009 c 2009 Kabanov A. N. and Roman kov V. A. Original Russian Text Copyright
STRICTLY NONTAME PRIMITIVE ELEMENTS OF THE FREE METABELIAN LIE ALGEBRA OF RANK 3 A. N. Kabanov and V. A. Roman kov
UDC 512.54
Abstract: We prove that the free metabelian Lie algebra M3 of rank 3 over an arbitrary field K admits strictly nontame primitive elements. Keywords: Lie algebra, automorphism, tame automorphism, primitive element, free algebra, free derivation
1. Introduction Free algebras. Let K be a field of an arbitrary characteristic; and let Xn = {x1 , . . . , xn }, n ≥ 2, be a free finite set. Assume that x1 < x2 < · · · < xn . Define some free algebras with the set of free generators Xn . Denote by Pn = K[Xn ] the polynomial algebra over K in commuting variables in Xn . Denote by An = KXn the free associative algebra over K with the set of free generators Xn . It is well known that Xn ⊆ An freely generates the Lie algebra Ln = K(Xn ) with respect to the operation [u, v] = uv − vu, u, v ∈ An . Given a Lie algebra T , we put T = [T, T ], T = [T , T ]. Define Mn = Ln /Ln as the free metabelian Lie algebra Mn = K((Xn )) freely generated by Xn . Automorphisms. We write ϕ for an arbitrary automorphism (in general, an endomorphism) ϕ = (f1 , . . . , fn ) of the free algebra Fn with the set of free generators Xn , where ϕ(xi ) = fi , i = 1, . . . , n. If Fn is a Lie algebra then the automorphism (endomorphism) ϕ ∈ Aut Fn (ϕ ∈ End Fn ) is called an IAautomorphism (IA-endomorphism) provided that ϕ induces the identical mapping on Fn /Fn . We have the subgroup IAut Fn ≤ Aut Fn and the subsemigroup IAEnd Fn ≤ End Fn consisting of the automorphisms (respectively, endomorphisms) of the shape pointed above. Tame automorphisms. Let Fn be as above. An automorphism ϕ ∈ Aut Fn is called elementary if ϕ is of the shape ϕ = (x1 , . . . , xj−1 , αxj + u(x1 , . . . , xj−1 , xj+1 , . . . , xn ), xj+1 , . . . , xn ), where α ∈ K ∗ , and the element u(x1 , . . . , xj−1 , xj+1 , . . . , xn ) does not depend on xj . An automorphism ψ is called tame if ψ belongs to the subgroup TAut Fn ≤ Aut Fn generated by the elementary automorphisms; otherwise, ψ is called nontame or wild. Some available results. In 1964 P. M. Cohn [1] proved that all automorphisms of the free Lie algebra Ln are tame. It is well known that the automorphisms of the polynomial algebra P2 and the free associative algebra A2 are tame over every field K [2–5]. In 1972 M. Nagata [6] conjectured wildness of the afterwards famous automorphism 2 σ = x1 + x21 − x2 x3 x3 , x2 + 2 x21 − x2 x3 x1 + x21 − x2 x3 x3 , x3 of P3 if K is of characteristic 0. Using the Poisson brackets, degree bounding, and weak reduction, I. P. Shestakov and U. U. Umirbaev [7–9] proved that Nagata’s automorphism σ is indeed wild. The authors were supported by the Russian Foundation for Basic Research (Grant 07.01.00392). Omsk. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 50, No. 1, pp. 82–95, January–February, 2009. Original article submitted August 27, 2007. 66
c 2009 Springer Science+Business Media, Inc. 0037-4466/09/5001–0066
In the case of the free associative algebra A3 , the existence problem for the wild automorphisms is known as Cohn’s problem. U. U. Umirbaev in [10] proved that the so-called Anick automorphism δ = (x1 + x3 (x1 x3 − x3 x2 ), x2 + (x1 x3 − x3 x2 )x3 , x3 ) of A3 = KX3 over K of characteristic 0 is wild. Alongside the tame automorphisms, in the case of the free metabelian Lie algebras Mn (n ≥ 2) the natural subgroup Inn Mn is distinguished of the so-called inner automorphisms. The group Inn Mn is an analog to conjugations in groups. An automorphism μ ∈ Inn Mn is given by μ = μ(u) = (x1 + [x1 , u], x2 + [x2 , u], . . . , xn + [xn , u]), where u ∈ Mn . It is easy to verify that μ(u1 )μ(u2 ) = μ(u1 + u2 ), whence Inn Mn (Mn , +). V. A. Artamonov proved that Aut M2 = TAut M2 · Inn M2 [11]. It is easy to see that TAut M2 is isomorphic to GL2 (K), and every automorphism ϕ ∈ TAut M2 is in one-to-one correspondence with a nondegenerate linear change of X2 . Thus, Aut M2 is isomorphic to GL2 (K) · (M2 , +), and Aut M2 obviously contains wild automorphisms. Y. Bahturin and S. Nabiyev [12] similarly established an existence of wild automorphisms in Aut Mn for all n ≥ 3. However, the inner automorphisms in Inn Mn may be essentially considered as tame in virtue of their simplicity. From this point of view, the existence problem appears naturally for the automorphisms in Aut Mn (n ≥ 3) which do not belong to the subgroup generated by the tame and inner automorphisms. Call these automorphisms beyond the group strictly nontame. In the recent article [13] V. A. Roman kov established existence of strictly nontame automorphisms in Aut M3 for every field K. He pointed out how to construct these automorphisms extensively. Furthermore, a series of much stronger assertions on the structure of Aut M3 was established in [13]. From the Roman kov announced assertion on the generators of Aut Mn with n ≥ 4 [14] it follows that there are no strongly nontame automorphisms in case n ≥ 4. Primitive elements. An element g of Fn is called primitive if g may be included into some set Yn of free generators in Fn . A primitive element g ∈ Fn is called tame provided that g is the image ϕ(x1 ) = g under some tame automorphism ϕ ∈ TAut Fn ; otherwise, g is called nontame or wild. A primitive element g of the free metabelian algebra Mn is called strictly nontame if g is not the image ϕ(x1 ) = g for any automorphism ϕ in the subgroup generated by the tame and inner automorphisms. The aim of the present article is to prove that M3 possesses strictly nontame primitive elements for every field K. From the above-mentioned Roman kov result [14] it follows that there are no strictly nontame primitive elements in Mn in case n ≥ 4. Fox derivatives. Denote by KGn the group algebra of the free group Gn of rank n with the set of free generators Xn over a field K. In [15], Fox gave a detailed description of differential calculus in the group ring ZGn . This description is obviously extended to every group algebra KGn . Since every free associative algebra An = KXn is naturally embedded into KGn and the partial Fox derivatives on KGn transfer An ≤ KGn into itself; therefore the majority of technical assertions about these derivatives remain true for An . Recall that the (partial) free derivatives of An , induced by the corresponding derivatives of the group algebra KGn , are exactly the mappings d/dxj : An → An ,
1 ≤ j ≤ n,
satisfying the following conditions for α, β ∈ K and u, v ∈ An : dxi /dxj = δij
(Kronecker’s delta);
d(αu + βv)/dxj = αdu/dxj + βdv/dxj ; d(uv)/dxj = du/dxj v ε + udv/dxj , where ε : An → K is the specialization homomorphism, defined by xεj = 1, j = 1, . . . , n. 67
By definition, dα/dxj = 0 for all α ∈ K, j = 1, . . . , n. Direct calculation shows that d[xi , xj ]/dxi = 1 − xj . The following is known as the main identity for the Fox derivatives: n
du/dxj (xj − 1) = u − uε .
j=1
Since Ln is a Lie subalgebra in An , we may also consider the derivatives d/dxj : Ln → An ,
j = 1, . . . , n.
Let π : An → Pn be the standard epimorphism. Define the derivatives ∂/∂xj : An → Pn , j = 1, . . . , n, putting ∂u/∂xj = (du/dxj )π for every u ∈ An . Since ∂u/∂xj = 0 for every u ∈ Ln and j = 1, . . . , n (which is easy), the derivatives ∂/∂xj : Mn → Pn ,
j = 1, . . . , n,
are also defined. The normal form of an element of a free metabelian Lie algebra. The following identities hold in every free metabelian Lie algebra Mn over an arbitrary field: [u, v] = −[v, u]; [u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0 (the Jacobi identity); [[u, v], [w, t]] = 0; [z1 , [z2 , [. . . , [zk , [x, y]] . . . ]]] = [zσ(1) , [zσ(2) , [. . . , [zσ(k) , [x, y]] . . . ]]] for every σ ∈ Sk . Hence, the subalgebra Mn2 = Mn of the free metabelian Lie algebra Mn of rank n ≥ 2 may be considered to be a module over the polynomial algebra Pn as follows: Each v ∈ Pn may be uniquely presented as αk1 ,...,kn (x1 − 1)k1 . . . (xn − 1)kn , (1) v =α+ (k1 ,...,kn )∈N n
where α, αk1 ,...,kn ∈ K. Given u ∈ Mn2 and v ∈ Pn of the shape (1), put uv = αu + αk1 ,...,kn [xn , [. . . , [x1 , u] . . . ]], (k1 ,...,kn )∈N n
where xj is repeated kj times. It is easy to see that ∂uv /∂xj = v∂u/∂xj ,
j = 1, . . . , n.
Define the right-normed commutator [xi1 , . . . , xim ], putting [xi1 , . . . , xim ] = [xi1 , [. . . , [xim−1 , xim ] . . . ]]. We have [xi1 ] = xi1 when m = 1. In what follows, we only consider the elements with the right-ordered parentheses. By [16], a base for the free metabelian Lie algebra Mn consists of all commutators of the type [xi1 , . . . , xik ], where xij ∈ Xn , xik > xik−1 , xik−1 ≤ · · · ≤ xi1 . We say that an element of Mn is written in normal form if it is presented in the shape of a linear combination of some basis elements. 68
2. Preliminary Results In what follows, we denote xi − 1 by bi for brevity. Let λ be a mapping that assigns the Jacobi matrix Jϕ = ∂xϕ to each endomorphism ϕ ∈ IAEnd Mn . /∂x j i Lemma 1. The mapping λ is a homomorphism. ψ Proof. Let ϕ, ψ ∈ IAEnd Mn and xϕ i = xi + ui , xi = xi + vi , where ui = ui (x1 , . . . , xn ), vi = vi (x1 , . . . , xn ) ∈ Mn2 . Prove that Jϕ Jψ = Jϕψ . We have
Jϕ = E + (∂ui /∂xj ), On the one hand,
xϕψ i
Jψ = E + (∂vi /∂xj ).
= xi + vi + ui (x1 + v1 , . . . , xn + vn ). Hence, it suffices to show that ∂wi /∂xj = ∂ui /∂xj +
n
∂ui /∂xk · ∂vk /∂xj ,
k=1
where wi = ui (x1 + v1 , . . . , xn + vn ). Assume that ui = [xm , xl ]s , where s ∈ Pn , m < l ≤ n. Then n
∂ui /∂xk · ∂vk /∂xj = −bl s · ∂vm /∂xj + bm s · ∂vl /∂xj ,
k=1
and wi = ui −
bl s vm
+
vlbm s .
Therefore, ∂wi /∂xj = ∂ui /∂xj − bl s · ∂vm /∂xj + bm s · ∂vl /∂xj .
Since an arbitrary element ui is a linear combination of the elements of the shape [xm , xl ]s , from the equality above we obtain Jϕ Jψ = Jϕψ as required. Lemma 2. The kernel of λ is trivial. Proof. Assume that there is a nonidentical endomorphism ϕ ∈ IAEnd Mn possessing the identical ϕ 2 Jacobi matrix Jϕ . Put xϕ i = xi + ui , where ui ∈ Mn . Then ∂xi /∂xj = δij implies ∂ui /∂xj = 0. In the normal form of the derivative ui , the elements of the type [x1 , xm ]s are only nonzero under the variable x1 , where m > 1, s ∈ Pn . From the normal form of the monomial [xi1 , xi2 , . . . , xik ] (ik > ik−1 , ik−1 ≤ ik−2 ≤ · · · ≤ i1 ), it follows that the derivatives of [x1 , xm1 ]s1 and [x1 , xm2 ]s2 with different m1 and m2 cannot be canceled. Hence, there are no the elements of the shape [x1 , xm ]s in ui . The lack of other elements in ui may be proved by induction with the help of a derivative by the corresponding variable. The lemma is proved. Corollary 1. Let ϕ, ψ ∈ IAEnd M and J(ϕψ) = J(ϕ)J(ψ) = E. Then ϕ, ψ ∈ IAut M and ϕ = ψ −1 . Define the stabilizer Stab(b1 , . . . , bn )T = {A ∈ GLn (Pn ) | A · (b1 , . . . , bn )T = (b1 , . . . , bn )T }, where T denotes transposition. Lemma 3. Let A ∈ Stab(b1 , . . . , bn )T . Then there is an automorphism ϕ ∈ IAut M with the Jacobi matrix Jϕ equal to A. Proof. Let A = E + (aij ). Since A · (b1 , . . . , bn )T = (b1 , . . . , bn )T , we have ai1 b1 + · · · + ain bn = 0 for every i. By induction, we prove an existence of ui ∈ Mn2 such that ∂ui /∂xj = aij , j = 1, . . . , n. If n = 1 then a11 b1 = 0, whence a11 = 0. In this case u1 may be taken equal 0. Let n > 1. Write aij in the shape aij = cij bn + dij , where dij does not contain bn . Then n−1 j=1
cij bn bj + ain bn +
n−1
dij bj = 0.
j=1
69
n−1 Under the specialization xn → 1, xi → xi , i = n, we obtain j=1 dij bj = 0. By the induction 2 such that ∂vi /∂xj = dij , j = 1, . . . , n − 1. Now, hypothesis, there is vi ∈ Mn−1 n−1
cij bn bj + ain bn = 0.
j=1
Consider wi =
n−1
[xn , xj ]cij .
j=1
The derivatives of wi are of the shape ∂wi /∂xj = cij bn (1 ≤ j ≤ n − 1). Hence, n−1
∂wi /∂xj · bj + ain bn = 0.
j=1
On the other hand, it is known that n−1
∂wi /∂xj · bj + ∂wi /∂xn · bn = 0,
j=1
whence ain = ∂wi /∂xn . We may take ui = vi + wi . The so-constructed map ϕ is an endomorphism. The same argument may be used for the matrix A−1 , and some endomorphism ψ may be obtained. Since J(ϕψ) = E, we have ϕ = ψ −1 ∈ IAut M by Corollary 1. The lemma is proved. Consider the conjugation of the Jacobi matrix Jϕ of some automorphism ϕ ∈ IAut M3 :
1 0 b1 0 1 b2 0 0 b3
−1
Jϕ
1 0 0
0 1 0
b1 b2 b3
A(ϕ)
=
∗ ∗
0 0 , 1
∂u1 /∂x2 − b1 b−1 ∂u3 /∂x2 3 A(ϕ) = E + . (2) ∂u2 /∂x2 − b2 b−1 3 ∂u3 /∂x2 Denote by μ the homomorphism that assigns the matrix A(ϕ) ∈ GL2 P3 + b−1 3 P2 to each ϕ ∈ IAut M3 . Lemma 4. The kernel of μ : IAut M3 → GL2 P3 + b−1 3 P2 consists of inner homomorphisms. Proof. Letting A(ϕ) be the identity matrix, we get where
∂u1 /∂x1 − b1 b−1 3 ∂u3 /∂x1 ∂u2 /∂x1 − b2 b−1 3 ∂u3 /∂x1
∂ui /∂xj − bi b−1 3 ∂u3 /∂xj = 0,
i, j = 1, 2,
whence b3 ∂ui /∂xj = bi ∂u3 /∂xj . Hence, there are some polynomials vj ∈ P3 (j = 1, 2) such that ∂ui /∂xj = bi vj for i = 1, 2, 3. Therefore, there exists w ∈ M32 such that ui = wbi , i = 1, 2, 3, as required. Since an arbitrary c ∈ P3 is uniquely represented as c = b23 c2 + b3 c1 + c0 with c2 ∈ P3 and c1 , c0 ∈ P2 , A(ϕ) is uniquely represented as A(ϕ) = E + b23 A2 + b3 A1 + A0 + b−1 3 A−1 , where A2 ∈ M2 (P3 ) and A1 , A0 , A−1 ∈ M2 (P2 ). 70
(3)
Consider the automorphism ψ ∈ IAut M3 that fixes x1 and x2 and maps x3 into x3 +[x1 , x2 ] (since the Jacobian of this mapping is equal to 1; therefore, ψ is an automorphism). Applying the homomorphism μ to ψ, we obtain −1 b1 b3 b2 −b1 b−1 b1 3 A(ψ) = E + . −1 b2 b−1 3 b2 −b2 b3 b1 Putting b1 b2 −b21 , (4) X= b22 −b2 b1 we have A(ψ) = E + b−1 3 X. The annihilator of A ∈ M2 (P2 ) is Ann A = {B ∈ M2 (P2 ) | BA = AB = 0}. Lemma 5. Let X ∈ M2 (P2 ) be of the shape (4). Then Ann X = αX, where α ∈ P2 . Proof. Let a1 a2 A= ∈ M2 (P2 ) a3 a4 and AX = XA = 0. Computing AX and equating its elements to 0, we get a1 b1 + a2 b2 = 0,
a3 b1 + a4 b2 = 0,
whence a1 = α1 b2 , a2 = −α1 b1 , a3 = α2 b2 , a4 = −α2 b1 , where α1 , α2 ∈ P2 . Analogously, we deduce from XA = 0 that a1 = α3 b1 , a3 = α3 b2 , a2 = α4 b1 , a4 = α4 b2 , where α3 , α4 ∈ P2 . Since a1 = α1 b2 = α3 b1 , we have α1 = αb1 and α3 = αb2 , where α ∈ P2 . Also, α2 = α3 and α4 = −α1 . Thus, αb1 b2 −αb21 = αX. A= αb22 −αb2 b1 The lemma is proved. Take an arbitrary automorphism ϕ ∈ IAut M3 and compute A(ϕψ). Putting A = A(ϕ) and taking into account (3), we arrive at −1 −1 −2 A E + b−1 3 X = A + b3 X + b3 A2 X + A1 X + b3 A0 X + b3 A−1 X. Since A(ϕψ) should be of the shape (2), there is no b−2 3 in the representation of A(ϕψ), whence A−1 X = 0. Analogously, computing A(ψϕ), we get XA−1 = 0. Hence, A−1 ∈ Ann X, and A−1 = αX, α ∈ P2 , by Lemma 5. There is b−2 3 XA0 X in the representation of A(ψϕψ). Therefore, XA0 X = 0. From the representation of A(ϕψψ) it follows that A0 X 2 = 0, whence A0 X = βX, β ∈ P2 . Analogously, A(ψψϕ) gives X 2 A0 = 0, and XA0 = γX, γ ∈ P2 . Applying the same argument to A(ψϕψψ) and A(ψψϕψ), we infer (XA1 X)X = X(XA1 X) = 0, whence XA1 X = δX, δ ∈ P2 . Define the mapping 1+β α , ϕ ∈ IAut M3 . ρ:ϕ→ δ 1+γ Lemma 6. The mapping ρ : IAut M3 → GL2 (P2 ) is a homomorphism. Proof. Let ϕ, ψ ∈ IAut M3 and μ(ϕ) = A, μ(ψ) = B. Then, using the decompositions of the shape (3) of A and B, we obtain (AB)−1 = A−1 B0 + A0 B−1 + A−1 + B−1 = α1 XB0 + α2 A0 X + α1 X + α2 X = α1 γ2 X + α2 β1 X + α1 X + α2 X. Analogously, in view of X 2 = 0, we have (AB)0 X = (A0 + B0 + A1 B−1 + A−1 B1 + A0 B0 )X = (β1 + β2 + α1 δ2 + β1 β2 )X, X(AB)0 = (γ1 + γ2 + α2 δ1 + γ1 γ2 )X, X(AB)1 X = X(A1 + B1 + A2 B−1 + A−1 B2 + A1 B0 + A0 B1 )X = (δ1 + δ2 + β2 δ1 + γ1 δ2 )X. Therefore, ρ(ϕψ) = ρ(ϕ)ρ(ψ) as required. Find some formulas to compute the coefficients α, β, γ, and δ. Denote by uij the derivative ∂ui /∂xj of ui ∈ M3 ; and by ukij , the component corresponding to bk3 in the expansion uij = b23 u2ij + b3 u1ij + u0ij . 71
Lemma 7. Let ϕ = (x1 + u1 , x2 + u2 , x3 + u3 ) ∈ IAut M3 , ui ∈ M32 , i = 1, 2, 3, and 1+β α ρ(ϕ) = , δ 1+γ where α, β, γ, δ ∈ P2 . Then the elements of ρ(ϕ) may be calculated by the following formulas: 0 −1 α = −u031 b−1 2 = u32 b1 ;
(5)
β = −u131 b1 − u132 b2 ;
(6)
−1 0 0 γ = u011 − u021 b1 b−1 2 = u22 − u12 b2 b1 ;
(7)
δ = u023 b1 − u013 b2 .
(8)
Proof. Represent A(ϕ) as in (3). From (2) we see that −b1 u031 −b1 u032 . A−1 = −b2 u031 −b2 u032 0 0 −1 0 −1 Hence, αb2 = −u031 , αb α = −u31 b2 = u32 b1 . 10 = u32 ,1 and From (2), A0 = uij − bi u3j , i, j = 1, 2. Computing A0 X and equating the components of A0 X to the corresponding components of βX, we get 0 u11 − b1 u131 b1 + u012 − b1 u132 b2 = βb1 , u021 − b2 u131 b1 + u022 − b2 u132 b2 = βb2 .
Since u11 b1 + u12 b2 + u13 b3 = 0, we have u011 b1 + u012 b2 = 0. Analogously, u021 b1 + u022 b2 = 0. Hence, −u131 b21 − u132 b1 b2 = βb1 , −u131 b1 b2 − u132 b21 = βb2 , and β = −u131 b1 − u132 b2 . Analogously, from XA0 = γX we obtain u011 b2 − u021 b1 = γb2 , Therefore,
u012 b2 − u022 b1 = γb1 .
−1 0 0 γ = u011 − u021 b1 b−1 2 = u22 − u12 b2 b1 .
From (2), A1 = u1ij − b1 u23j , i, j = 1, 2. Since XA1 X = δX, we have XA1 = δE, whence δ = b1 b2 u111 − b1 u231 − b21 u121 − b2 u231 + b22 u112 − b1 u232 − b1 b2 u122 − b2 u232 and
δ = b1 b2 u111 − u122 − b21 u121 + b22 u112 = b2 b1 u111 + b2 u112 − b1 b1 u121 + b2 u122 .
From (6), u11 b1 + u12 b2 = −u13 b3 . Then δ = u023 b1 − u013 b2 . The lemma is proved. Hence, the matrix of the shape
1 + Δ2 Δ22
P2 1 + Δ2
with Δ2 = id(b1 , b2 ) is the image of an IA-automorphism under ρ. Since ρ is a homomorphism, the determinant of this matrix is equal to 1. Indeed, if ϕ ∈ IAut M3 and ρ(ϕ) = A then ρ(ϕ−1 ) = A−1 . Consequently, det A ∈ K, and det A = 1, which follows from the shape of A.
2 . Denote by SE2 P2 , Δ2 , Δ2 = SE
2 Denote the set of all these matrices by SL2 P2 , Δ2 , Δ22 = SL 2
2 consisting of all tame matrices of the above-mentioned shape. the subgroup of SL 72
Lemma 8. If char K = 2 then ρ is surjective. Proof. Let
A=
1 + a1 b21 a31 + b1 b2 k + b22 a32
a2 1 + a4
2 , ∈ SL
where a1 , a4 ∈ Δ2 , a2 , a31 , a32 ∈ P2 , and k ∈ K. Considering that char K = 2, we take the automorphism σ = (x1 + [x1 , x3 ]k/2 − [x2 , x3 ]k/2 , x2 + [x1 , x3 ]k/2 − [x2 , x3 ]k/2 , x3 ). Direct computation shows that det J(σ) = 1. Therefore, it is an automorphism. Using (5)–(8), we obtain 1 0 = t21 (δσ ), ρ(σ) = δσ 1 where δσ = b21 − 2b1 b2 + b22 k/2. Multiplying A on the transvection t21 (δσ ), we dispose of b1 b2 k and obtain a2 1 + a1
2 , ∈ SL B= b21 a31 + b22 a32 1 + a4 where a1 ∈ Δ2 , a31 , a32 ∈ P2 . The matrix B may be represented as B = E + B1 + b2 B2 ,
2 , we have E + B1 ∈ SL
2 . Indeed, det B = det(E + B1 ) + where B1 does not depend on b2 . Since B ∈ SL d = 1, where d = b2 d , d ∈ P2 . However, det(E + B1 ) depends only on b1 . Hence, d = 0 and det B1 = 1. Therefore, B(E + B1 )−1 = E + b2 B3 , where B3 = B2 (E + B1 )−1 . The matrix E + B1 is of the shape 1 + b1 c 1 c2 , b21 c3 1 + b1 c4 where c1 , c2 , c3 , c4 ∈ P2 . A preimage for this matrix is the automorphism ϕ1 = (x1 , x2 + [x1 , x2 ]c4 + [x1 , x3 ]c3 , x3 + [x1 , x2 ]c2 + [x1 , x3 ]c1 ). This ϕ1 is an automorphism, because det J(ϕ1 ) = 1. The equality ρ(ϕ1 ) = E + B1 may be verified by direct computation using (5)–(8). It suffices to show that E + b2 B3 possesses a preimage in IAut M3 . The matrix E + b2 B3 is of the shape 1 + b2 d1 b2 d2 , b22 d3 1 + b2 d4 where d1 , d2 , d3 , d4 ∈ P2 . By analogy to the case (E + B1 )−1 , the automorphism ϕ2 = (x1 , x2 + [x1 , x2 ]d4 + [x1 , x3 ]d3 , x3 + [x1 , x2 ]d2 + [x1 , x3 ]d1 ) is a preimage of (E + b2 B3 ). Thus, A = ρ(ϕ2 ϕ1 σ −1 ) as required. 73
2 . Lemma 9. The image of every tame IA-automorphism under ρ lies in SE Proof. Let ψ ∈ Aut M3 be an arbitrary elementary automorphism. Then ψ is of the shape 3 ψ = x1 , . . . , xi−1 , αj xj + u, xi+1 , . . . , x3 , j=1
where αj ∈ K, αi = 0, and u ∈
M32
does not depend on xi . Write ψ in another form: ψ = τ ϕ, where 3 τ = x1 , . . . , xi−1 , αj xj , xi+1 , . . . , x3 j=1
and ϕ is an elementary IA-automorphism. Let ξ ∈ T IAut M3 . Then m ξ= τi ϕi , i=1
where τi are some linear automorphisms and ϕi are some elementary IA-automorphisms. We have m m −1 −1 τi ϕi = τi τ m . . . τ2−1 ϕ1 τ2 . . . τm . . . τm ϕm−1 τm ϕm . i=1
i=1
Thus, ξ=
m
τi
i=1
where ϕλi i = λ−1 i ϕi λi ,
λi =
m
ϕλi i ,
i=1
m
τj (i < m),
λm = id .
j=i+1
Since ξ ∈ IAut M3 ; therefore,
m
τi = id
and ξ =
i=1
ρ(ϕτ )
m
ϕλi i .
i=1
2 for an arbitrary elementary ϕ ∈ IAut M3 and an arbitrary ∈ SE
It suffices to prove that linear automorphism τ . Assume that τ is given by x → Ax, where x = (x1 , x2 , x3 )T , A = (αij ), i, j = 1, 2, 3. Let τ −1 : x → Bx, where B = A−1 = (βij ), i, j = 1, 2, 3. Consider the case ϕ = (x1 , x2 , x3 + u), where u ∈ M32 does not contain x3 , u = [x1 , x2 ]s and s ∈ P2 . Now, τ
s
s
xϕ i = xi + αi3 w
(i = 1, 2, 3),
s
where w = [x1 , x2 ]M33 +[x1 , x3 ]M32 +[x2 , x3 ]M31 , Mij are the minors of B, and s = τ −1 (s). Therefore, ∂w/∂x1 = −(M33 b2 + M32 b3 )s , ∂w/∂x2 = (M33 b1 − M31 b3 )s . Using the expansion s = s2 b23 + s1 b3 + s0 with s1 , s0 ∈ P2 and (5)–(8), we obtain α = −α33 M33 s0 , β = α33 (M32 b1 + M31 b2 )s0 , Since B = A−1 , we have If α33
γ = M33 (α23 b1 − α13 b2 )s0 ,
δ = −(M32 b1 + M31 b2 )2 s0 .
M31 = α13 , M32 = −α23 , M33 = α33 .
2 ; otherwise = 0 then ρ(ϕτ ) = t21 (δ) ∈ SE 1 + kα α τ
2 , = t21 (−k)t12 (α)t21 (k) ∈ SE ρ(ϕ ) = −k 2 α 1 − kα
−1 where k = α33 (α23 b1 − α13 b2 ). The cases ϕ = (x1 , x2 + u, x3 ) and ϕ = (x1 + u, x2 , x3 ) may be considered analogously.
74
Lemma 10. There are strictly nontame automorphisms in IAut M3 . Proof. It is known that SL2 (P2 ) = SE2 (P2 ) [1]; i.e., there are nontame matrices in SL2 (P2 ). Let A ∈ SL2 (P2 ) be a nontame matrix. Then A is of the shape k1 + a11 k2 + a12 A= , k3 + a21 k4 + a22 where aij ∈ Δ2 . Let k2 = 0. Multiply A on the transvection t21 (λ1 ), where λ1 = (1 − k1 )/k2 . Then 1 + a11 + λ1 a12 k2 + a12 A t21 (λ1 ) = = A1 . k3 + λ1 k4 + a21 + λ1 a22 k4 + a22 Multiply A1 on t21 (λ2 ), where λ2 = −k3 − λ1 k4 . We get 1 + a11 + λ1 a12 t21 (λ2 )A1 = λ2 a11 + λ1 λ2 a12 + a21 + λ1 a22
k2 + a12 k4 + λ2 k2 + λ2 a12 + a22
.
Since A ∈ SL2 (P2 ), we have det A = k1 k4 − k2 k3 = 1; whence k4 + λ2 k2 = k4 − k3 k2 − k4 (1 − k1 ) = 1.
2 , and A1 is nontame, since A1 is obtained from A by some multiplications on Therefore, A1 ∈ SL transvections. If k2 = 0 then k1 = 0 and k4 = 0; otherwise, det A = 1. In this case, an analogous result may be obtained by multiplication on t21 (−k3 /k4 ) and the diagonal matrix 1/k1 0 = t12 (1/k1 )t21 (1 − k1 )t12 (−1)t21 (1 − 1/k1 ). 0 k1
2 \SE
2 If char K = 2 then, by Lemma 8, there is an automorphism in IAut M3 with the image in SL under ρ. Since Inn M3 ≤ ker ρ, this automorphism is strictly nontame by Lemma 9.
2 or Im ρ contains all If char K = 2 then from the proof of Lemma 8 we see that either Im ρ = SL
matrices from SL2 , but the matrix of the shape 1 + Δ2 P2 , (9) b21 P2 + b22 P2 + kb1 b2 1 + Δ2
2 then take a nontame matrix in SL
2 \ Im ρ of the shape (9). where k ∈ K ∗ = K\{0}. If Im ρ = SL Multiplying this matrix on t21 (−kb1 b2 ), we get a nontame matrix in Im ρ. The lemma is proved. Theorem. Let ϕ ∈ IAut M3 be a nontame automorphism. Then g = ϕ(x3 ) ∈ M3 is primitive strictly nontame. Proof. Assume that g is primitive and tame, i.e., there is a tame automorphism ψ = (g1 , g2 , g3 = g). Without loss of generality, we may assume that the set of free generators {g1 , g2 , g} is IA-tame. Indeed, let 3 3 g1 = α1j xj + v1 , g2 = α2j xj + v2 , j=1
j=1
where αij ∈ K, v1 , v2 ∈ M32 . Write ψ as ψ : X → DX + V, where
X=
x1 x2 x3
,
D=
α11 α21 0
α12 α22 0
α13 α23 1
,
V =
v1 v2 u3
. 75
Since ψ is an automorphism, the matrix D is invertible; the inverse automorphism should be of the shape ψ −1 : X → D−1 X + W, where W = (w1 , w2 , w3 )T , wi ∈ M32 . Acting on G = (g1 , g2 , g3 )T by the automorphism θ : X → D−1 X, T we obtain Gθ = X + V , where V = v1θ , v2θ , v3θ . Thus, we may consider the tame IA-automorphism ψ = ψθ = (g1 , g2 , g) instead of ψ (by the linearity of θ).
2 . From (5) and (6) it follows that the first line of B = ρ(ψ ) depends The image ψ under ρ lies in SE only on u3 . Hence, the first lines of B and A = ρ(ϕ) are coincided. Thus, if A = (aij ), i, j = 1, 2, then a11 a12 B= , b21 b22 where b21 ∈ Δ22 , b22 ∈ 1 + Δ2 . Therefore, B = CA, where 1 0 −1 , C = BA = c 1 c2
c1 , c2 ∈ P2 .
2 then B ∈
2 . This
2 , we have c2 = 1. Hence, C is a transvection, and if A ∈ / SE / SE Since B ∈ SE contradiction proves that g = ϕ(x3 ) is primitive and nontame. The proof of the fact that g is strictly nontame is practically the same since Inn M3 ≤ ker ρ. References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
15. 16.
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[email protected];
[email protected];
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