The group of automorphisms of the Lie algebra of derivations of a field of rational functions
aa r X i v : . [ m a t h . R A ] A p r The group of automorphisms of the Lie algebra of derivationsof a field of rational functions
V. V. Bavula
Abstract
We prove that the group of automorphisms of the Lie algebra Der K ( Q n ) of derivationsof the field of rational functions Q n = K ( x , . . . , x n ) over a field of characteristic zero iscanonically isomorphic to the group of automorphisms of the K -algebra Q n . Key Words: Group of automorphisms, monomorphism, Lie algebra, automorphism, locallynilpotent derivation, the field of rational functions in n variables.Mathematics subject classification 2010: 17B40, 17B20, 17B66, 17B65, 17B30. In this paper, module means a left module, K is a field of characteristic zero and K ∗ is its groupof units, and the following notation is fixed: • P n := K [ x , . . . , x n ] = L α ∈ N n Kx α is a polynomial algebra over K where x α := x α · · · x α n n and Q n := K ( x , . . . , x n ) is the field of rational functions, • G n := Aut K − alg ( P n ) and Q n := Aut K − alg ( Q n ); • ∂ := ∂∂x , . . . , ∂ n := ∂∂x n are the partial derivatives ( K -linear derivations) of P n , • D n := Der K ( P n ) = L ni =1 P n ∂ i ⊆ E n := Der K ( Q n ) = L ni =1 Q n ∂ i are the Lie algebras of K -derivations of P n and Q n respectively where [ ∂, δ ] := ∂δ − δ∂ , • G n := Aut Lie ( D n ) and E n := Aut Lie ( E n ), • δ := ad( ∂ ) , . . . , δ n := ad( ∂ n ) are the inner derivations of the Lie algebras D n and E n wheread( a )( b ) := [ a, b ], • D n := L ni =1 K∂ i , • H n := L ni =1 KH i where H := x ∂ , . . . , H n := x n ∂ n , • for each natural number n ≥ u n := K∂ + P ∂ + · · · + P n − ∂ n is the Lie algebra oftriangular polynomial derivations (it is a Lie subalgebra of D n ) and Aut Lie ( u n ) is its groupof automorphisms. Theorem 1.1 [4] G n = G n . The aim of the paper is to prove the following theorem.
Theorem 1.2 E n = Q n . tructure of the proof . (i) Q n ⊆ E n via the group monomorphism (Lemma 2.3 and (3)) Q n → E n , σ σ : ∂ σ ( ∂ ) := σ∂σ − . (ii) Let σ ∈ E n . Then ∂ ′ := σ ( ∂ ) , . . . , ∂ ′ n := σ ( ∂ n ) are commuting derivations of Q n such that E n = L ni =1 Q n ∂ ′ i (Lemma 2.12.(2)) and(iii) T ni =1 ker Q n ( ∂ ′ i ) = K (Lemma 2.12.(1)).(iv)(crux) There exist elements x ′ , . . . , x ′ n ∈ Q n such that ∂ ′ i ( x ′ j ) = δ ij for i, j = 1 , . . . , n (Lemma 2.12.(3)).(v) σ ( x α ∂ i ) = x ′ α ∂ ′ i for all α ∈ N n and i = 1 , . . . , n (Lemma 2.12.(6)).(vi) The K -algebra homomorphism σ ′ : Q n → Q n , x i x ′ i , i = 1 , . . . , n is an automorphismsuch that σ ′ ( q∂ i ) = σ ′ ( q ) ∂ ′ i for all q ∈ Q n and i = 1 , . . . , n .(vii) Fix E n ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } (Proposition 2.9.(1)). Hence, σ = σ ′ ∈ Q n , by (v)and (vi), i.e. E n = Q n . (cid:3) The groups of automorphisms of the Lie algebras D n and u n . Theorem 1.3 (Theorem 5.3, [3]) Aut
Lie ( u n ) ≃ T n ⋉ (UAut K ( P n ) n ⋊ ( F ′ n × E n )) where T n is analgebraic n -dimensional torus, UAut K ( P n ) n is an explicit factor group of the group UAut K ( P n ) of unitriangular polynomial automorphisms, F ′ n and E n are explicit groups that are isomorphicrespectively to the groups I and J n − where I := (1 + t K [[ t ]] , · ) ≃ K N and J := ( tK [[ t ]] , +) ≃ K N . Comparing the groups G n , E n and Aut Lie ( u n ) we see that the group UAut K ( P n ) n of polynomialautomorphisms is a tiny part of the group Aut Lie ( u n ) but in contrast G n = G n and E n = Q n . Theorem 1.4 [1] Every monomorphism of the Lie algebra u n is an automorphism. Not every epimorphism of the Lie algebra u n is an automorphism. Moreover, there are count-ably many distinct ideals { I iω n − | i ≥ } such that I = { } ⊂ I ω n − ⊂ I ω n − ⊂ · · · ⊂ I iω n − ⊂ · · · and the Lie algebras u n /I iω n − and u n are isomorphic (Theorem 5.1.(1), [2]). Conjecture , [4].
Every homomorphism of the Lie algebra D n is an automorphism. The groups of automorphisms of the
Witt W n ( n ≥
2) and the
Virasoro
Vir Lie algebras werefound in [5].
This section can be seen as a proof of Theorem 1.2. The proof is split into several statements thatreflect ‘Structure of the proof of Theorem 1.2’ given in the Introduction.Let G be a Lie algebra and H be its Lie subalgebra. The centralizer C G ( H ) := { x ∈ G | [ x, H ] =0 } of H in G is a Lie subalgebra of G . In particular, Z ( G ) := C G ( G ) is the centre of the Lie algebra G . The normalizer N G ( H ) := { x ∈ G | [ x, H ] ⊆ H} of H in G is a Lie subalgebra of G , it is thelargest Lie subalgebra of G that contains H as an ideal.Let V be a vector space over K . A K -linear map δ : V → V is called a locally nilpotent map if V = S i ≥ ker( δ i ) or, equivalently, for every v ∈ V , δ i ( v ) = 0 for all i ≫
1. When δ is a locallynilpotent map in V we also say that δ acts locally nilpotently on V . Every nilpotent linear map2 , that is δ n = 0 for some n ≥
1, is a locally nilpotent map but not vice versa, in general. Let G be a Lie algebra. Each element a ∈ G determines the derivation of the Lie algebra G by the rulead( a ) : G → G , b [ a, b ], which is called the inner derivation associated with a . The set Inn( G ) ofall the inner derivations of the Lie algebra G is a Lie subalgebra of the Lie algebra (End K ( G ) , [ · , · ])where [ f, g ] := f g − gf . There is the short exact sequence of Lie algebras0 → Z ( G ) → G ad → Inn( G ) → , that is Inn( G ) ≃ G /Z ( G ) where Z ( G ) is the centre of the Lie algebra G and ad([ a, b ]) = [ad( a ) , ad( b )]for all elements a, b ∈ G . An element a ∈ G is called a locally nilpotent element (respectively, a nilpotent element ) if so is the inner derivation ad( a ) of the Lie algebra G . The Lie algebra E n . Since E n = n M i =1 Q n ∂ i = n M i =1 Q n H i (1)every element ∂ ∈ E n is a unique sum ∂ = P ni =1 a i ∂ i = P ni =1 b i H i where a i = x i b i ∈ Q n . Thefield Q n is the union S = f ∈ P n P n,f where P n,f is the localization of P n at the powers of f . Thealgebra Q n is a localization of P n,f . Hence D n,f := Der K ( P n,f ) = L ni =1 P n,f ∂ i ⊆ E n and E n = [ = f ∈ P n D n,f .Q n is an E n -module . The field Q n is a (left) E n -module: E n × Q n → Q n , ( ∂, q ) ∂ ∗ q . Inmore detail, if ∂ = P ni =1 a i ∂ i where a i ∈ Q n then ∂ ∗ q = n X i =1 a i ∂q∂x i . The E n -module Q n is not a simple module since K is an E n -submodule of Q n , and n \ i =1 ker Q n ( ∂ i ) = K. (2) Lemma 2.1
The E n -module Q n /K is simple with End E n ( Q n /K ) = K id where id is the identitymap.Proof . We have to show that for each non-scalar rational function, say pq − ∈ Q n , the E n -submodule M of Q n /K it generates coincides with the E n -module Q n /K . By (2), a i = ∂ i ∗ ( pq − ) = 0 for some i . Then for all elements u ∈ Q n , ua − i ∂ i ∗ ( pq − + K ) = u + K . So, Q n /K is asimple E n -module. Let f ∈ End E n ( Q n /K ). Then applying f to the equalities ∂ i ∗ ( x + K ) = δ i for i = 1 , . . . , n , we obtain the equalities ∂ i ∗ f ( x + K ) = δ i for i = 1 , . . . , n. Hence, f ( x + K ) ∈ T ni =2 ker Q n /K ( ∂ i ) ∩ ker Q n /K ( ∂ i ) = ( K ( x ) /K ) ∩ ker Q n /K ( ∂ i ) = K ( x + K ).So, f ( x + K ) = λ ( x + K ) and so f = λ id, by the simplicity of the E n -module Q n /K . (cid:3) The Cartan subalgebra H n of E n . A nilpotent Lie subalgebra C of a Lie algebra G is calleda Cartan subalgebra of G if it coincides with its normalizer. We use often the following obviousobservation: An abelian Lie subalgebra that coincides with its centralizer is a maximal abelian Liesubalgebra . Lemma 2.2 H n is a Cartan subalgebra of E n . . H n = C E n ( H n ) is a maximal abelian Lie subalgebra of E n .Proof . 2. Clearly, H n ⊆ C E n ( H n ). Let ∂ = P ni =1 a i H i ∈ C E n ( H n ) where a i ∈ Q n . Then all a i ∈ ∩ ni =1 ker Q n ( H i ) = ∩ ni =1 ker Q n ( ∂ i ) = K , by (2), and so ∂ ∈ H n . Therefore, H n = C E n ( H n ) isa maximal abelian Lie subalgebra of E n .1. By statement 2, we have to show that H n = N := N E n ( H n ). Let ∂ = P ni =1 a i H i ∈ N , wehave to show that all a i ∈ K . By statement 2, for all j = 1 , . . . , n , H n ∋ [ H j , ∂ ] = P ni =1 H j ( a i ) H i ,and so H j ( a i ) ∈ K for all i and j . This condition holds if all a i ∈ K , i.e. ∂ ∈ H n . Suppose that a i K for some i , we seek a contradiction. Then necessarily, a i K ( x , . . . , b x j , . . . , x n ) for some j . Since Q n = K ( x , . . . , b x j , . . . , x n )( x j ), the result follows from the following claim. Claim: If a ∈ K ( x ) \ K then H ( a ) K . The field K ( x ) is a subfield of the series field K (( x )) := { P i> −∞ λ i x i | λ i ∈ K } . Since H ( P i> −∞ λ i x i ) = P i> −∞ iλ i x i , the Claim is obvious.Then, by the Claim, H j ( a i ) K , a contradiction. (cid:3) Lemma 2.3 [5] Let R be a commutative ring such that there exists a derivation ∂ ∈ Der( R ) such that r∂ = 0 for all nonzero elements r ∈ R (eg, R = P n , Q n and δ = ∂ ). Then the grouphomomorphism Aut( R ) → Aut
Lie (Der( R )) , σ σ : δ σ ( δ ) := σδσ − , is a monomorphism. The Q n -module E n . The Lie algebra E n is a Q n -module, Q n × E n → E n , ( σ, ∂ ) σ ( ∂ ) := σ∂σ − . By Lemma 2.3, the Q n -module E n is faithful and the map Q n → E n , σ σ : ∂ σ ( ∂ ) = σ∂σ − , (3)is a group monomorphism. We identify the group Q n with its image in E n , Q n ⊆ E n . Everyautomorphism σ ∈ Q n is uniquely determined by the elements x ′ := σ ( x ) , . . . , x ′ n := σ ( x n ) . Let M n ( Q n ) be the algebra of n × n matrices over Q n . The matrix J ( σ ) := ( J ( σ ) ij ) ∈ M n ( Q n ),where J ( σ ) ij = ∂x ′ j ∂x i , is called the Jacobian matrix of the automorphism (endomorphism) σ andits determinant J ( σ ) := det J ( σ ) is called the Jacobian of σ . So, the j ’th column of J ( σ ) is the gradient grad x ′ j := ( ∂x ′ j ∂x , . . . , ∂x ′ j ∂x n ) T of the polynomial x ′ j . Then the derivations ∂ ′ := σ∂ σ − , . . . , ∂ ′ n := σ∂ n σ − are the partial derivatives of Q n with respect to the variables x ′ , . . . , x ′ n , ∂ ′ = ∂∂x ′ , . . . , ∂ ′ n = ∂∂x ′ n . (4)Every derivation ∂ ∈ E n is a unique sum ∂ = P ni =1 a i ∂ i where a i = ∂ ∗ x i ∈ Q n . Let ∂ :=( ∂ , . . . , ∂ n ) T and ∂ ′ := ( ∂ ′ , . . . , ∂ ′ n ) T where T stands for the transposition. Then ∂ ′ = J ( σ ) − ∂, i . e . ∂ ′ i = n X j =1 ( J ( σ ) − ) ij ∂ j for i = 1 , . . . , n. (5)In more detail, if ∂ ′ = A∂ where A = ( a ij ) ∈ M n ( Q n ), i.e. ∂ i = P nj =1 a ij ∂ j . Then for all i, j = 1 , . . . , n , δ ij = ∂ ′ i ∗ x ′ j = n X k =1 a ik ∂x ′ j ∂x k δ ij is the Kronecker delta function. The equalities above can be written in the matrix formas AJ ( σ ) = 1 where 1 is the identity matrix. Therefore, A = J ( σ ) − . The maximal abelian Lie subalgebra D n of E n . Suppose that a group G acts on a set S . For a nonempty subset T of S , St G ( T ) := { g ∈ G | gT = T } is the stabilizer of the set T in G and Fix G ( T ) := { g ∈ G | gt = t for all t ∈ T } is the fixator of the set T in G . Clearly, Fix G ( T ) isa normal subgroup of St G ( T ). Lemma 2.4 C E n ( D n ) = D n and so D n is a maximal abelian Lie subalgebra of E n .2. Fix Q n ( D n ) = Fix Q n ( ∂ , . . . , ∂ n ) = Sh n .3. Fix Q n = ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } .4. Cen E n ( D n + H n ) = 0 .Proof . 1. Statement 1 follows from (2): Clearly, D n ⊆ C E n ( D n ). Let ∂ = P a i ∂ i ∈ C E n ( D n )where a i ∈ Q n . Then all elements a i ∈ T ni =1 ker Q n ∂ i = K , by (2), and so ∂ ∈ D n . So, C E n ( D n ) = D n and as a result D n is a maximal abelian Lie subalgebra of E n .2. Let σ ∈ Fix Q n ( D n ) and J ( σ ) = ( J ij ). By (5), ∂ = J ( σ ) ∂ , and so, for all i, j = 1 , . . . , n , δ ij = ∂ i ∗ x j = J ij , i.e. J ( σ ) = 1, or equivalently, by (2), x ′ = x + λ , . . . , x ′ n = x n + λ n for some scalars λ i ∈ K , and so σ ∈ Sh n (since x ′ i − x i ∈ T nj =1 ker Q n ( ∂ j ) = K for i = 1 , . . . , n ).3. Let σ ∈ Fix Q n = ( ∂ , . . . , ∂ n , H , . . . , H n ). Then σ ∈ Fix Q n ( ∂ , . . . , ∂ n ) = Sh n , by statement2. So, σ ( x ) = x + λ , . . . , σ ( x n ) = x n + λ n where λ i ∈ K . Then x i ∂ i = σ ( x i ∂ i ) = ( x i + λ i ) ∂ i for i = 1 , . . . , n , and so λ = · · · = λ n = 0. This means that σ = e . So, Fix Q n =( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } .4. Statement 4 follows from statement 1 and Lemma 2.2. (cid:3) Lemma 2.5
Let A be a K -algebra, Der K ( A ) be the Lie algebra of K -derivations of A and D ( A ) be the ring of differential operators on A . If the algebra D ( A ) is simple and generated by A and Der K ( A ) then the D ( A ) -module A is simple.Proof . Let a be a nonzero D ( A )-submodule of A . So, a is an ideal of A such that ∂ ( a ) ⊆ a forall ∂ ∈ Der K ( A ). The algebra D := D ( A ) is generated by A and D . So, D a ⊆ a D and a D ⊆ D a ,i.e. D a = a D is a nonzero ideal of the simple algebra D . Hence, 1 ∈ D a and so 1 = P i a i d i forsome elements d i ∈ D and a i ∈ a ⊆ D . Then1 = 1 ∗ X i a i d i ∗ ∈ a , hence a = A , i.e. A is a simple D ( A )-module. (cid:3) Theorem 2.6 E n is a simple Lie algebra.2. Z ( E n ) = { } .3. [ E n , E n ] = E n .Proof . 1. (i) n = 1, i.e. E = K ( x ) ∂ is a simple Lie algebra : We split the proof into severalsteps.(a) D := K [ x ] ∂ and W := K [ x, x − ] ∂ are simple Lie subalgebras of E (easy).(b) For all λ ∈ K , W ( λ ) := K [ x, ( x − λ ) − ] is a simple Lie subalgebra of E , by applying the K -automorphism s λ : x x − λ of the K -algebra Q to W , i.e. s λ ( W ) = W ( λ ).(c) For any nonempty subset I ⊂ K , W ( I ) := W ( I ) K := K [ x, ( x − λ ) − | λ ∈ I ] ∂ is a simpleLie subalgebra of E : Let a be a nonzero ideal of W ( I ) and 0 = a∂ ∈ a . Then either a∂ ∈ D or5 = [ p∂, a∂ ] ∈ D ∩ a for some p ∈ P . Since D ⊆ W ( λ ) for all λ ∈ I and W ( λ ) are simple Liealgebra, a ∩ W ( λ ) = W ( λ ). Hence a = W ( I ) since W ( I ) = [ λ ∈ I W ( λ ) , i.e. W ( I ) is a simple Lie algebra.(d) If K is an algebraically closed field then E is a simple Lie algebra since E = W ( K ).The algebra E is the union S = f ∈ P W [ f − ] of the Lie algebras W [ f − ] := P ,f ∂ where P ,f is the localization of P at the powers of the element f . Let a be the ideal of E generated by anonzero element a = pq − ∂ for some pq − ∈ Q . Clearly, a ∈ W [( f q ) − ] for all nonzero elements f ∈ P and E = S = f ∈ P W [( f g ) − ]. So, to finish the proof of (i) it suffices to show that all thealgebras W [ f − ] are simple.(e) A := W [ f − ] is a simple Lie algebra for all = f ∈ P : Let K ′ := K ( ν , . . . , ν s ) be thesubfield of the algebraic closure K of K generated by the roots ν , . . . , ν s of the polynomial f and G = Gal( K ′ /K ) be the Galois group of the finite Galois field extension K ′ /K (since char( K ) = 0).Let K ′ = ⊕ di =1 Kθ i for some elements θ i ∈ K ′ and θ = 1. By (c), A ′ := K ′ [ x, f − ] ∂ = W ( ν , . . . , ν s ) K ′ is a simple Lie K ′ -algebra. Let a ∈ A \{ } , a and d a ′ be the ideals in A and A ′ respectivelythat are generated by the element a . Then a ′ = A ′ , by (c). Notice that A ′ = P di =1 θ i A and for a ′ = P di =1 θ i a i , b = P di =1 θ i b i ∈ A ′ where a i , b i ∈ A , [ a ′ , b ] = P di =1 θ i θ j [ a i , b j ]. Moreover, everyelement in A ′ = a ′ is a linear combination of several commutators in A ′ (where c = P di =1 θ k c k ∈ A ′ and c k ∈ A ), [ a, [ a ′ , . . . [ b, c ] . . . ] = X θ i · · · θ j θ k [ a, [ a i , . . . [ b j , c k ] . . . ] . (6)The symmetrization map Sym : K ′ → K , λ
7→ | G | − P g ∈ G g ( λ ), is a surjection such thatSym( µ ) = µ for all µ ∈ K . Clearly, K ′ ( x ) /K ( x ) is a Galois field extension with the Galoisgroup G where the elements of G act trivially on the element x . So, the symmetrization map Symcan be extended to the surjection K ′ ( x ) → K ( x ) by the same rule, and then to the surjection A ′ → A , f ∂ Sym( f ) ∂ .Each element e ∈ A ⊆ A ′ , can be expressed as a finite sum of elements in (6). Then applyingSym, we see that e is a linear combination of elements (commutators) from a , i.e. A is a simpleLie algebra.(ii) E n is a simple Lie algebra for n ≥
2: Let a ∈ E n \{ } and a = ( a ) be the ideal in E n generated by the element a = P ni =1 a i ∂ i where a i ∈ Q n .(a) a ∩ D n = 0: If a ∈ D n then there is nothing to prove. Suppose that a D n .(a1) Suppose that a i ∈ K ( x i ) for all i . Then a i K [ x i ] for some i (since a D n ), and so a ∋ [ H i , a ] = H i ( a i ) ∂ i ∈ K ( x i ) ∂ i \{ } . By (i), ∂ ∈ a ∩ D n .(a2) Suppose that a i K ( x i ) for some i . Then ∂ j ( a i ) = 0 for some j = i . Let q ∈ P n bethe common denominator of the fractions a , . . . , a n , that is a = p q − , . . . , a n = p n q − for someelements p i ∈ P n . For all n ≥ D n ∩ a ∋ [ q n ∂ j , a ] = q n ∂ j ( a i ) ∂ i + X k = i ( . . . ) ∂ k = 0 . (b) a = D n since D n is a simple Lie algebra, [4].(c) a ⊇ K ( x i ) ∂ i for i = 1 , . . . , n : In view of symmetry it suffices to prove that a ⊇ K ( x ) ∂ .Notice that for all u ∈ Q n and i = 2 , . . . , n , a ∋ [ u∂ , x ∂ i ] = u∂ i − x ∂ i ( u ) ∂ . a + Q n ∂ = E n . The field of rational functions Q n = Q n ( K ) can be seen as the field ofrational functions Q n ( K ) = Q n − ( K ′ ) where K ′ = K ( x ). Then E ′ n − := Der K ′ ( Q n − ( K ′ )) = n M i =2 Q n − ( K ′ ) ∂ i = n M i =2 Q n ∂ i . By Lemma 2.5, the E ′ n − -module Q ′ n − /K ′ = Q n /K ( x ) is simple. The Lie algebra E ′ n − is aLie subalgebra of E n , and E n can be seen as a left E ′ n − -module with respect to the adjointaction. The ideal a of E n is an E ′ n − -submodule of E n . The Lie algebra K ( x ) ∂ is simpleand a ∩ K ( x ) ∂ is a nonzero ideal of it (by (b)). Therefore, K ( x ) ∂ ⊆ a . The E ′ n − -module E n / a = ( a + Q n ∂ ) / a ≃ Q n ∂ / a ∩ Q n ∂ is an epimorphic image of the simple E ′ n − -module Q n /K ( x ) via ϕ : Q n /K ( x ) → Q n ∂ / a ∩ Q n ∂ , u + K ( x ) u∂ + a ∩ Q n ∂ , with 0 = ( P n + K ( x )) /K ( x ) ⊆ ker( ϕ ). Therefore, Q n ∂ = a ∩ Q n ∂ ⊆ a , and so E n = a + Q n ∂ = a . So, E n is a simple Lie algebra.2 and 3. Statements 2 and 3 follow from statement 1. (cid:3) Lemma 2.7
For all nonzero elements q ∈ Q n and i = 1 , . . . , n , C E n ( qP n ∂ i ) = { } .Proof . Let c ∈ C E n ( qP n ∂ i ). Then for all elements p ∈ P n ,0 = [ c, qp∂ i ] = c ( p ) · q∂ i + p [ c, q∂ i ] = c ( p ) · q∂ i . Then c ( p ) = 0 for all p ∈ P n , and so c = 0. (cid:3) Proposition 2.8 [4]
Fix G n ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } . Let d , . . . , d n be a commuting linear maps acting in a vector space E . Let Nil E ( d , . . . , d n ) := { e ∈ E | d ji e = 0 for all i = 1 , . . . , n and some j = j ( e ) } . Let Nil E n ( D n ) := Nil E n ( δ , . . . , δ n ).Clearly, Nil E n ( D n ) = D n is a Lie subalgebra of E n . Proposition 2.9 Fix E n ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } .2. Fix E n ( ∂ , . . . , ∂ n ) = Sh n .Proof . 1. Let σ ∈ F := Fix E n ( ∂ , . . . , ∂ n , H , . . . , H n ). We have to show that σ = e . Then σ − ∈ F and σ ± (Nil E n ( D n )) ⊆ Nil E n ( D n ), i.e. σ ( D n ) = D n since Nil E n ( D n ) = D n . So, σ | D n ∈ Fix G n ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } (Proposition 2.8), i.e. σ ( ∂ ) = ∂ for all ∂ ∈ D n . Let0 = δ ∈ E n . Then δ = q − ∂ for some 0 = q ∈ P n and ∂ ∈ D n . Now, [ q p∂ i , δ ] = ∂ ′ ∈ D n for all p ∈ P n . Applying σ to the equality yields the equality [ q p∂ i , σ ( δ )] = ∂ ′ . By taking the difference,we obtain σ ( δ ) − δ ∈ C E n ( q P n ∂ i ) = { } , by Lemma 2.7, hence σ = e .2. Clearly, Sh n ⊆ F := Fix E n ( ∂ , . . . , ∂ n ). Let σ ∈ F and H ′ i := σ ( H i ) , . . . , H ′ n := σ ( H n ).Applying the automorphism σ to the commutation relations [ ∂ i , H j ] = δ ij ∂ i gives the relations[ ∂ i , H ′ j ] = δ ij ∂ i . By taking the difference, we see that [ ∂ i , H ′ j − H j ] = 0 for all i and j . Therefore, H ′ i = H i + d i for some elements d i ∈ C E n ( D n ) = D n (Lemma 2.4.(1)), and so d i = P nj =1 λ ij ∂ j forsome elements λ ij ∈ K . The elements H ′ , . . . , H ′ n commute, hence[ H j , d i ] = [ H i , d j ] for all i, j, or equivalently, λ ij ∂ j = λ ji ∂ i for all i, j. This means that λ ij = 0 for all i = j , i.e. H ′ i = H i + λ ii ∂ i = ( x i + λ ii ) ∂ i = s λ ( H i )7here s λ ∈ Sh n , s λ ( x i ) = x i + λ ii for all i . Then s − λ σ ∈ Fix E n ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } (statement 1), and so σ = s λ ∈ Sh n . (cid:3) The automorphism ν . Let ν be the K -automorphism of Q n given by the rule ν ( x i ) = x − i for i = 1 , . . . , n . Then ν ( ∂ i ) = − x i H i , ν ( H i ) = − H i , ν ( x i H i ) = − ∂ i , i = 1 , . . . , n. (7)By (7), the elements X := x H , . . . , X n := x n H n commute and the next lemma follows fromLemma 2.4 and Proposition 2.9 since X n := ν ( D n ) = L ni =1 KX i . Lemma 2.10 C E n ( X n ) = X n is a maximal abelian Lie subalgebra of E n .2. Fix Q n ( X , . . . , X n ) = Fix E n ( X , . . . , X n ) = Sh n .3. Fix Q n ( X , . . . , X n , H , . . . , H n ) = Fix E n ( X , . . . , X n , H , . . . , H n ) = { e } . The following lemma is well-known and it is easy to prove.
Lemma 2.11
Let ∂ be a locally nilpotent derivation of a commutative K -algebra A such that ∂ ( x ) = 1 for some element x ∈ A . Then A = A ∂ [ x ] is a polynomial algebra over the ring A ∂ := ker( ∂ ) of constants of the derivation ∂ in the variable x . The next lemma is the core of the proof of Theorem 1.2.
Lemma 2.12
Let σ ∈ E n , ∂ ′ := σ ( ∂ ) , . . . , ∂ ′ n := σ ( ∂ n ) and δ ′ := ad( ∂ ′ ) , . . . , δ ′ n := ad( ∂ ′ n ) . Then1. ∂ ′ , . . . , ∂ ′ n are commuting derivations of Q n such that T ni =1 ker Q n ( ∂ ′ i ) = K .2. E n = L ni =1 Q n ∂ ′ i .3. For each i = 1 , . . . , n , σ ( x i ∂ i ) = x ′ i ∂ ′ i for some elements x ′ i ∈ Q n . The elements x ′ , . . . , x ′ n are algebraically independent and ∂ ′ i ( x ′ j ) = δ ij for i, j = 1 , . . . , n .4. Nil Q n ( ∂ ′ , . . . , ∂ ′ n ) = P ′ n where P ′ n := K [ x ′ , . . . , x ′ n ] .5. Nil E n ( δ ′ , . . . , δ ′ n ) = L ni =1 P ′ n ∂ ′ i .6. σ ( x α ∂ i ) = x ′ α ∂ ′ i for all α ∈ N n and i = 1 , . . . , n .7. σ ′ : Q n → Q n , x i x ′ i , i = 1 , . . . , n is a K -algebra homomorphism (statement 3) such that σ ′ ( a∂ i ) = σ ′ ( a ) σ ( ∂ i ) .8. The K -algebra homomorphism σ ′ is an automorphism.Proof . 1. The elements ∂ , . . . , ∂ n are commuting derivations, hence so are ∂ ′ , . . . , ∂ ′ n . Let λ ∈ T ni =1 ker Q n ( ∂ ′ i ). Then λ∂ ′ ∈ C E n ( ∂ ′ , . . . , ∂ ′ n ) = σ ( C E n ( ∂ , . . . , ∂ n )) = σ ( C E n ( D n )) = σ ( D n ) = σ ( n M i =1 K∂ i ) = n M i =1 K∂ ′ i , since C E n ( D n ) = D n , Lemma 2.4.(1). Then λ ∈ K since otherwise the infinite dimensional space L i ≥ Kλ i ∂ ′ would be a subspace of the finite dimensional space σ ( D n ).2. It suffices to show that the elements ∂ ′ , . . . , ∂ ′ n of the n -dimensional (left) vector space E n over the field Q n are Q n -linearly independent (the key reason for that is statement 1). Let V = P ni =1 Q n ∂ ′ i . Suppose that m := dim Q n ( V ) < n , we seek a contradiction. Up to order, let ∂ ′ , . . . , ∂ ′ m be a Q n -basis of V . Then ∂ m +1 = P mi =1 a i ∂ ′ i for some elements a i ∈ Q n . By applying δ ′ j ( j = 1 , . . . , n ), we see that 0 = P mi =1 ∂ ′ j ( a ) ∂ ′ i , and so a i ∈ T ni =1 ker Q n ( ∂ ′ j ) = K , by statement 1.This means that the elements ∂ ′ , . . . , ∂ ′ m are K -linearly dependent, a contradiction.8. Let H ′ i := σ ( x i ∂ i ) for i = 1 , . . . , n . By statement 2, H ′ i = P ni =1 a ij ∂ ′ j for some elements a ij ∈ Q n . Applying the automorphism σ to the relations δ ij ∂ j = [ ∂ j , H i ] yields the relations δ ij ∂ ′ i = n X i =1 ∂ ′ j ( a ik ) ∂ ′ k . Let x ′ i := a ii . Then ∂ ′ j ( x ′ i ) = δ ji and ∂ ′ j ( a ik ) = 0 for all k = i . By statement 1, a ik ∈ K for all i = k . Now, H ′ i := x ′ i ∂ ′ i + X j = i a ij ∂ ′ j . The elements H ′ , . . . , H ′ n commute, hence for all i = j , 0 = [ H ′ i , H ′ j ] = − a ji ∂ ′ i + a ij ∂ ′ j , and so a ij = 0. Therefore, H ′ i = x ′ i ∂ ′ i .The equalities ∂ ′ i ( x ′ j ) = δ ij imply that the elements x ′ , . . . , x ′ n ∈ Q n are algebraically inde-pendent over K : Suppose that f ( x ′ , . . . , x ′ n ) = 0 for some nonzero polynomial f ( t , . . . , x n ) ∈ K [ t , . . . , x n ]. We can assume that the (total) degree deg( f ) is the least possible. Clearly, f K ,hence ∂f∂x i = 0 for some i and deg( ∂f∂x i ) < deg( f ), but ∂f∂x i ( x ′ , . . . , x ′ n ) = ∂ i ( f ( x ′ , . . . , x ′ n )) = ∂ i (0) = 0, a contradiction.4. Let D ′ n = P ni =1 K∂ ′ i and N = Nil Q n ( D ′ n ). By statement 3 and Lemma 2.11, N = N D ′ n [ x ′ , . . . , x ′ n ] = K [ x ′ , . . . , x ′ n ]since K ⊆ N D ′ n ⊆ Q D ′ n n = K (by statement 1).5. Let ∂ = P ni =1 a i ∂ ′ i ∈ N := Nil E n ( δ ′ , . . . , δ ′ n ) where a i ∈ Q n (statement 2). For all α ∈ N n , δ ′ α ( ∂ ) = n X i =1 ∂ ′ α ( a i ) ∂ ′ i where δ ′ α := Q ni =1 δ ′ α i i , δ ′ i = ad( ∂ ′ i ) and ∂ ′ α := Q ni =1 ∂ ′ α i i . So, δ ′ α ( a i ) = 0 iff ∂ ′ α ( a i ) = 0 for i = 1 , . . . , n (statement 2). Now, statement 5 follows from statement 4.6. First, let us show that, by induction on | α | , that σ ( x α ∂ i ) − x ′ α ∂ ′ i ∈ Cen E n ( D ′ n ) = D ′ n (Lemma 2.4.(1)). The initial case when | α | = 0 is obvious. So, let | α | >
0. Then[ ∂ ′ j , σ ( x α ∂ i ) − x ′ α ∂ ′ i ] = σ ([ ∂ j , x α ∂ i ]) − α j x ′ α − e j ∂ ′ i = σ ( α j x α − e j ∂ i ) − α j x ′ α − e j ∂ ′ i = α j x ′ α − e j ∂ ′ i − α j x ′ α − e j ∂ ′ i = 0 . Therefore, σ ( x α ∂ i ) = x ′ α ∂ ′ i + P λ ij ∂ ′ j for some scalars λ ij = λ ij ( α ) ∈ K . Notice that σ ( H i ) = σ ( x i ∂ i ) = x ′ i ∂ ′ i := H ′ i , by the definition of the elements x ′ i . Since | α | > α j = 0 for some j . Applying the automorphism σ to the equalities ( α j − δ ij ) x α ∂ i = [ H j , x α ∂ i ] we have (we may assume that x α ∂ i = H i )( α j − δ ij )( x ′ α ∂ ′ i + n X k =1 λ ik ∂ ′ k ) = σ (( α j − δ ij ) x α ∂ i ) = σ ([ H j , x α ∂ i ]) = [ H ′ j , x ′ α ∂ ′ i + n X k =1 λ ik ∂ ′ k ]= ( α j − δ ij ) x ′ α ∂ ′ i − λ ij ∂ ′ j , and so ( α j − δ ij + 1) λ ij = 0 and ( α j − δ ij ) λ ik = 0 for all k = j . This means that all λ is = 0.7. By statement 3, σ ′ is a K -algebra homomorphism such that im( σ ′ ) = Q ′ n := K ( x ′ , . . . , x ′ n ).By statement 3, for all elements a ∈ Q n , ∂ ′ i σ ′ ( a ) = σ ′ ∂ i ( a )since ∂ ′ i acts as ∂∂x ′ i on Q ′ n . 9et a = pq − = 0 where p, q ∈ P n . Then, for all r ∈ q P n , [ a∂ i , r∂ i ] = ( a∂ i ( r ) − ∂ i ( a ) r ) ∂ i ∈ P n ∂ i . By applying σ , we have the equality[ σ ( a∂ i ) , σ ′ ( r ) ∂ ′ i ] = σ ′ ( a∂ i ( r ) − ∂ i ( a ) r ) ∂ ′ i . On the other hand,[ σ ′ ( a ) ∂ ′ i , σ ′ ( r ) ∂ ′ i ] = ( σ ′ ( a ) ∂ ′ i σ ′ ( r ) − ∂ ′ i σ ′ ( a ) σ ′ ( r )) ∂ ′ i = ( σ ′ ( a ) σ ′ ∂ i ( r ) − σ ′ ∂ i ( a ) σ ′ ( r )) ∂ ′ i = σ ′ ( a∂ i ( r ) − ∂ i ( a ) r ) ∂ ′ i . Hence, σ ( a∂ i ) − σ ′ ( a ) ∂ ′ i ∈ C E n ( σ ′ ( q P n ) ∂ ′ i ) = C E n ( σ ( q P n ∂ i )) = σ ( C E n ( q P n ∂ i )) = σ ( C E n ( q P n ∂ i )) = 0 , by Lemma 2.7. Therefore, σ ( a∂ i ) = σ ′ ( a ) σ ( ∂ i ).8. Since σ ( Q n ∂ i ) = σ ′ ( Q n ) ∂ ′ i for all i = 1 , . . . , n (statement 7), we must have σ ′ ( Q n ) = Q n ,by statement 2, and so σ ′ ∈ Q n . (cid:3) Proof of Theorem 1.2 . Let σ ∈ E n . By Corollary 2.12.(8), we have the automorphism σ ′ ∈ Q n such that, by Lemma 2.12.(3,6), σ ′− σ ∈ Fix E n ( ∂ , . . . , ∂ n , H , . . . , H n ) = { e } (Proposition2.9). Therefore, σ = σ ′ and so E n = Q n . (cid:3) Acknowledgements
The work is partly supported by the Royal Society and EPSRC.
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C. R. Acad. Sci. Paris, Ser. I , (2012) no. 11-12, 553–556. (Arxiv:math.AG:1205.0797).[2] V. V. Bavula, Lie algebras of triangular polynomial derivations and an isomorphism criterion fortheir Lie factor algebras, Izvestiya: Mathematics , (2013), in print. (Arxiv:math.RA:1204.4908).[3] V. V. Bavula, The groups of automorphisms of the Lie algebras of triangular polynomial derivations,Arxiv:math.AG/1204.4910.[4] V. V. Bavula, The group of automorphisms of the Lie algebra of derivations of a polynomial algebra.Arxiv:math.RA:1304.6524.[5] V. V. Bavula, The groups of automorphisms of the Witt W n and Virasoro Lie algebras.Arxiv:math.RA:1304.6578.Department of Pure MathematicsUniversity of SheffieldHicks BuildingSheffield S3 7RHUKemail: v.bavula@sheffield.ac.ukand Virasoro Lie algebras.Arxiv:math.RA:1304.6578.Department of Pure MathematicsUniversity of SheffieldHicks BuildingSheffield S3 7RHUKemail: v.bavula@sheffield.ac.uk