An Envelope for Left Alternative Algebras

Transcription

1 International Journal of Algebra, Vol. 7, 2013, no. 10, HIKARI Ltd, An Envelope for Left Alternative Algebras Josef Rukavicka Department of Mathematics Faculty of Electrical Engineering Czech Technical University in Prague Copyright c 2013 Josef Rukavicka. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Let A be the free non-associative algebra and let T be the T -ideal generated by the identity (x, y, z)+(y, x, z). Given an ideal J T, then C J = A/J is a left alternative algebra. We construct an ideal Γ J and we define a universal enveloping algebra of C J as A/Γ J. We introduce a linear map ω : C J A/Γ J, such that ω((a, b, c)) = (a, b, c) (b, a, c). As a conjecture we state that ω is injective. The injection of C J into A/Γ J is similar to the injection of a Lie algebra into an associative algebra by [a, b] =ab ba; moreover we show how to construct a spanning set of A/Γ J from a basis of C J and we define a universal property analogously like in the case of Lie algebras. Mathematics Subject Classification: 17A50, 17D15, 17A30 Keywords: Non-associative algebras, Enveloping algebras, Left alternative algebras, Lie algebras 1 Introduction There is a well known construction of a Lie algebra, [2]: given an associative algebra A asc, a Lie algebra may be obtained from A asc by defining a new bilinear multiplication [x, y] =xy yx on the underlying vector space of A asc. The algebra obtained in this way is usually denoted as A asc. Actually the famous PBW (Poincaré Birkhoff Witt) theorem implies that any

2 456 Josef Rukavicka Lie algebra I is a subalgebra of A asc for some unital associative algebra A asc, [2]. The corresponding associative algebra A asc is called universal enveloping algebra for the Lie algebra I. The universal enveloping algebras possess a universal property. The PBW theorem has been extended to Malcev algebras, [3] and then to Bol algebras, [4]. We apply a similar approach like in the case of Lie algebras to construct universal enveloping algebras for left alternative algebras. A left alternative algebra is embedded into its envelope by means of the linear map ω which sends an associator (a, b, c)to(a, b, c) (b, a, c) (note the similarity with [x, y] = xy yx). The paper is organized as follows. In the second section a special basis B of the free non-associative algebra A is constructed; B contains maximal number of associators in the sense that given an associator (a, b, c) A than either (a, b, c) B or (a, b, c) is equal to a linear combination of associators from B; it means (a, b, c) = ijk r ijk (a i,b j,c k ) where (a i,b j,c k ) B and r ijk are scalars. This feature turns out to be essential for several proofs in the paper. The third section defines the T -ideal T generated by a defining identity (x, y, z) +(y, x, z) and a linear map π : A A with the kernel denoted Π, such that π((a, b, c)) = (π(a), π(b), π(c)) (π(b), π(a), π(c)) where (a, b, c) B. Bases of vector spaces π(a) and Π are constructed. Next it is proved that Π T and π(t ) T. In the fourth section we consider a left alternative algebra C J = A/J where J T is an ideal in A. We introduce an ideal Γ J generated by the subspace π(j). A linear map ω : C J A/Γ J based on the linear map π is presented. As a conjecture we state that π(a) Γ J = π(j), what implies immediately that ω is injective. We define a universal enveloping algebra of C J as A/Γ J. It is showed how to build up a set Υ from a basis of C J in such a way that {a+γ J a Υ} spans the vector space A/Γ J. Although the construction of Υ is analogous to the construction of a basis of the universal enveloping algebra from a basis of a Lie algebra, [2], unfortunately the elements {a +Γ J a Υ} are not linearly independent. Thus it remains as an open question how build up a basis of A/Γ J from a basis of C J. Finally we prove that A/Γ J possesses a universal property as follows: Given two left alternative algebras C J and C I with injections ω and ω into their universal enveloping algebras, respectively, and a homomorphism σ : C J C I. Then there is a unique homomorphism μ : A/Γ J A/Γ I such that

3 An envelope for left alternative algebras 457 μ ω = ω σ. 2 Associator basis of the free non-associative algebra In the paper all algebras and vector spaces are considered over a field K with characteristic 0. Let A be the free non-associative algebra on a set of generators X = {x 1,x 2,...x k }. The elements of A will be called polynomials. It is well known that A = i>0 A i where A i is a homogeneous component of A spanned by all words of the length i. Let (a, b, c) =(ab)c a(bc) denote the associator in A where a, b, c A. Let B = i>0 B i denote a set of polynomials defined as follows where B i = B A i. B 1 = X (a, b, c) B where a, b, c B ax B where a B and x X Proposition 2.1. The set B spans the vector space A. Proof. The proposition obviously holds for A 1 and A 2. We suppose it holds for A i where i<nand we prove that it holds for A n where n>2. Let w A n be a word or an associator that is not in B. Three cases may occur: 1. w =(w 1,w 2,w 3 ), then replace w 1,w 2,w 3 by linear combination of elements of B and multiply out. The result follows from the fact that (a, b, c) B if a, b, c B. 2. w = w 1 x where x X, then replace w 1 by linear combination of elements of B and multiply out. The result follows from the fact that ax B if a B and x X. 3. w = w 1 w 2 where w 2 X, then w 2 = w 3 w 4. Thus w = w 1 w 2 = w 1 (w 3 w 4 )=(w 1 w 3 )w 4 (w 1,w 3,w 4 ). The associator (w 1,w 3,w 4 ) turns into the case 1 and (w 1 w 3 )w 4 turns into the case 2 or 3 with the difference that we decreased the length of the last multiplicand. By iterating the process we eventually achieve the case 2, where the last multiplicand is from X.

4 458 Josef Rukavicka Let S = i>0 S i denote the set of all non-associative words on X where S i contains the words of the length i. It is known that the set S is a basis of A. Proposition 2.2. The set B forms a basis of the vector space A. Proof. Consider a map τ : B S defined by τ(x) =x where x X τ(ab) =τ(a)τ(b) where ab B τ((a, b, c)) = τ(a)(τ(b)τ(c)) where (a, b, c) B. Since the polynomials of the form a(bc) cannot appear in B, it is easy to see the map τ is an injection. The fact that B i spans A i implies that τ is in fact a bijection. This proves that B is a basis of A. Corollary 2.3. Given (a, b, c) A such that (a, b, c) B. Then (a, b, c) = ijk α ijk (a i,b j,c k ) where (a i,b j,c k ) B and α ijk is a non-zero scalar. In this sense B contains maximal number of associators. 3 A linear map π : A A Definition 3.1. Let π : A A be a linear map defined on the basis B as follows. The kernel of π will be denoted Π= i Π i where Π i =Π A i. π(x) =x where x X π((a, b, c)) = (π(a),π(b),π(c)) (π(b),π(a),π(c)) where (a, b, c) B π(ax) =π(a)π(x) =π(a)x where ax B, x X. Remark 3.2. Note that due to Corollary 2.3 it holds π((a, b, c)) = (π(a),π(b),π(c)) (π(b),π(a),π(c)) for any a, b, c A. For a B, π(a) 0, we define sets Ḃa, B a B as follows: Ḃ a = {b b B,π(a) =π(b)} B a = {b b B,π(a) = π(b)} It is easy to verify that π(a) = b Ḃ a b b B a b where π(a) 0.

5 An envelope for left alternative algebras 459 Example 3.3. Let X = {x, y, z, t, u, v} and let a =((x, y, z)t, u, v) B, then π(a) =(π((x, y, z)t), π(u), π(v)) (π(u), π((x, y, z)t), π(v)) = ((x, y, z)t, u, v) ((y, x, z)t, u, v) (u, (x, y, z)t, v)+(u, (y, x, z)t, v). And it holds that Ḃ a = {((x, y, z)t, u, v), (u, (y, x, z)t, v)} and B a = {((y, x, z)t, u, v), (u, (x, y, z)t, v)}. Let B be ordered. And let L = i L i where L i B i be a set defined as follows: L 1 = X (a, b, c) L where a, b, c L and a>b(with regard to the order of B) ax L where a L and x X Definition 3.4. We define a multiple as in [1], p Given polynomials g, f A. We call g a multiple of f if there is a sequence p =(f 0,f 1,...,f r ), r 0, such that: f 0 = f,f r = g, f i A f i = h i f i 1 or f i 1 h i with h i A. Proposition 3.5. A set π(l) ={π(a) a L} forms a basis of the vector space π(a). Proof. A set π(b) ={π(a) a B} clearly spans π(a). The fact that the set π(l) spans π(a) follows from that (a, b, c)+(b, a, c) Π: given p B having a multiple of the form (a, b, c), if we replace (a, b, c) by(b, a, c) then π(p) only changes a sign, hence it is enough to include in L only associators where a<b. To prove the linear independence of elements of π(l), just realize that for any a, b L,a b it holds (Ḃa B a ) (Ḃb B b )=. Corollary 3.6. The set {a +Π a L} is a basis of A/Π. Let T = i T i denote a T -ideal in A generated by the defining identity (x, y, z)+(y, x, z) where T i = T A i. Definition 3.7. We define: Δ ={a + αb a B \ L, b L, α { 1, 1},π(a) 0,π(a + αb) =0} Δ ={a a B,π(a) =0} Δ= Δ Δ Remark 3.8. To see that the definition of Δ makes sense, note that any a B \ L, π(a) 0is a multiple of at least one associator (c 1,c 2,c 3 ) B, thus b L exists and is uniquely determined as well as the scalar α (less formally said b arises from a when we order all associators (c 1,c 2,c 3 ) in a, so that c 1 >c 2 ).

6 460 Josef Rukavicka Proposition 3.9. The set Δ forms a basis of Π. Proof. Given any a + αb Δ ora Δ where a B \ L, b L, α { 1, 1}, then a appears in no other polynomial from Δ \{a + αb, a}. It follows that elements of Δ are linearly independent. The facts that π(δ) = 0, Δ + L = B, and span(δ L) =A imply that Δ forms a basis of Π. Corollary Π T Proof. It is easy to see that for the basis Δ of Π it holds Δ T. Example To see that Π is a proper subset of T : x(x, x, x) T \ Π, since π(x(x, x, x)) = π( (x, x, x)x +(x, x, xx) (x, xx, x)+(xx, x, x)) = 2(x, xx, x)+2(xx, x, x). On the other hand note that every polynomial of the form (a, b, c) +(b, a, c) where a, b, c A lies in Π due to Corollary 2.3. Remark It can be proved that π(t ) T and π(t ) T 0. For example let f = y(y, (y, x, x),x) y((x, y, x),y,x), then f T and π(f) T. On the other hand let f = t(x, y, z)+t(y, x, z), then π(f) 0and f,π(f) T. We omit the laborious proof. 4 Envelope of left alternative algebras Let J be an ideal in A such that T J and let C J = A/J, then C J is a left alternative algebra. Let Γ J be the ideal generated by the subspace π(j). Let ω : C J A/Γ J be a linear map defined as ω(b) =π(b) +Γ J A/Γ J where b + J C J = A/J. The linear map ω is well defined, since π(j) Γ J. Conjecture 4.1. π(a) Γ J = π(j) Remark 4.2. The above conjecture has been confirmed by a computer for J being the free left alternative algebra on 2 and 4 generators for J i where i 6 and i 4, respectively. Corollary 4.3. The linear map ω : C J A/Γ J is injective. Proof. Note that a linear map ˆω : C J A/π(J), ˆω(b) =π(b)+π(j) A/π(J), b + J C J = A/J is clearly injective since Π T J. The injection of ω then follows from the previous conjecture. Definition 4.4. We call A/Γ J a universal enveloping algebra of a left alternative algebra C J.

7 An envelope for left alternative algebras 461 Let C J L be a set such that the set {a + J a C J } forms a basis of C J = A/J. Next we require that for any (a, b, c),ax C J it follows that a, b, c, x C J, x X. It is easy to see such C J exists, since L B and L itself satisfies this condition. Let Υ ={π(a) a C J }. Recall that B is ordered, and that C J L B. We define a set Ϋ: (π(a),π(b),π(c)) Ϋ where a, b, c C J, a b (a, b, c) Ϋ where a, b, c Ϋ ax Ϋ where a Ϋ, x X Let Υ = Υ Ϋ and let Ψ be a set defined as follows: π(a)x Ψ where a C J, x X (π(a),π(b),π(c)) Ψ where a, b, c C J, a>b Proposition 4.5. The set {a +Γ J a Υ} spans the vector space A/Γ J. Proof. Obviously Υ generates the algebra A/Γ J since C 1 Υ. Hence the it is clear that span({a +Γ J a Υ Ψ}) =A/Γ J (note that the set Υ Ψ is constructed in an analogous way like the basis B on generators X, with difference that the generators are from Υ; less formally we can say that the set Υ Ψ is an associator basis on generators Υ, we don t consider multiples of elements from Ψ since A/Γ J is an algebra). Hence it is enough to prove that {a +Γ J a Ψ} span({a +Γ J a Υ}). π(a)x Ψ, a C J, x X Either ax C J, then π(ax) =π(a)x or p = ax i α i b i T, b i C J, α i K, then π(p) =π(a)x i α i π(b i ) π(t ) Γ J where π(b i ) Υ; in consequence π(a)x +Γ J span({a +Γ J a Υ}). (π(a),π(b),π(c)) Ψ, a, b, c C J, a>b Either (a, b, c) C J, then π((a, b, c)) = (π(a),π(b),π(c)) (π(b),π(a),π(c)) and (π(b),π(a),π(c)) Ϋ, π((a, b, c)) Υ. Or p =(a, b, c) i α i b i T, b i C J, α i K, then π(p) =(π(a),π(b),π(c)) (π(b),π(a),π(c)) i α i π(b i ) π(t ) Γ J where π(b i ) Υ, (π(b),π(a),π(c)) Ϋ; in consequence (π(a),π(b),π(c)) + Γ J span({a +Γ J a Υ}). Finally we show the universal enveloping algebra possesses a universal property:

8 462 Josef Rukavicka Proposition 4.6. Given two left alternative algebras C J and C I with injections ω and ω into their universal enveloping algebras, respectively, and a homomorphism σ : C J C I. Then there is a unique homomorphism μ : A/Γ J A/Γ I such that μ ω = ω σ. Proof. It follows easily from the observation that the set Λ = {π(a) +Γ J a + J C J } generates A/Γ J, hence any homomorphism defined on elements of Λ extends to a unique homomorphism of universal enveloping algebras. References [1] L. Gerritzen: Tree polynomials and non-associative Gröbner bases. Journal of Symbolic Computation 41 (2006), no. 3-4, [2] N. Jacobson: Lie Algebras New York: Dover, [3] J.M. Pérez-Izquierdo, I.P. Shestakov: An envelope for Malcev algebras J. Algebra 272 (2004) [4] J.M. Pérez-Izquierdo: An envelope for Bol algebras J. Algebra 284 (2005) [5] R. D. Schafer: An Introduction to Nonassociative Algebras, New York: Dover, [6] K. A. Zhevlakov, A. M. Slinko, I.P. Shestakov, A. I. Shirshov: Rings that are nearly associative, Moscow, Nauka, 1978 (in Russian); English translation: Academic Press, N.Y Received: May 1, 2013