Free Malcev algebra of rank three
ABSTRACT We find a basis of the free Malcev algebra on three free generators over a
field of characteristic zero. The specialty and semiprimity of this algebra are
proved. In addition, we prove the decomposability of this algebra into
subdirect sum of the free Lie algebra rank three and the free algebra of rank
three of variety of Malcev algebras generated by a simple seven-dimensional
Malcev algebra.
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- 01/1982; Academic Press., ISBN: 0-12-779850-1
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arXiv:1103.6011v1 [math.RA] 30 Mar 2011
Free Malcev algebra of rank three.
A. I. Kornev∗
March 31, 2011
Abstract
We find a basis of the free Malcev algebra on three free generators over
a field of characteristic zero. The specialty and semiprimity of this algebra
are proved. In addition, we prove the decomposability of this algebra into
subdirect sum of the free Lie algebra rank three and the free algebra of rank
three of variety of Malcev algebras generated by a simple seven-dimensional
Malcev algebra.
The problem of finding of a basis of a free algebra is important for different varieties.
For free Malcev algebras this problem is posed by Shirshov in [1, the problem 1.160].
For alternative algebras with three generators similar problem is solved in
Recall that the Malcev algebra is called special if it is a subalgebra of a commutator
algebra A−for some alternative algebra A. The question of the speciality of a Malcev
algebras was posed by Malcev in [3]. In this paper we find the basis of the free
Malcev algebra with three free generators, and prove the specialty of this algebra.
In addition, we prove decomposition of this algebra into subdirect sum of free Lie
algebra of rank three and a the free algebra of rank three of the variety of Malcev
algebras generated by a simple seven-dimensional Malcev algebra.
Shestakov [4] in 1976 proved that a free Maltsev algebra of n > 8 generators over
commutative ring Φ is not semiprime provided 7! ?= 0 in Φ. Filippov [5] in 1979 then
proved that in fact a free Maltsev Φ−algebra of n > 4 generators is not semiprime
if 6Φ ?= 0. We prove that the free Malcev of rank three is semiprime.
For brevity, we omit the brackets in the terms of the following type (...((x1x2)x3)...)xn.
In addition, the products of the form axx...x we denote as axn.
[2].
∗Supported by FAPESP grant No. 2008/57680-5.
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Linear algebra M over a field F, which satisfies following identities
x2= 0,
J(x,y,xz) = J(x,y,z)x,
where J(x,y,z) = xyz + zxy + yzx, is called a Malcev algebra. In what follows,
the characteristic of F is assumed to be zero. Let Rabe the operator of right mul-
tiplication by element a of the algebra M and R(M) is the algebra generated by
all Rawhere a is the element of the algebra M. We will adhere to following notations:
La,b= 1/2(RaRb+ RbRa),
[Ra,Rb] = RaRb− RbRa.
(a,b,c) = (ab)c − a(bc)
In addition, by ∆i
we denote the center of the algebra M and by G(a,b,c,d) - the function defined in
[7] with equality
a(b) we denote the operator defined in [6, Chapter 1, §4], by Z(M)
G(a,b,c,d) = J(ab,c,d) − bJ(a,c,d) − J(b,c,d)a.
Let X = {x,y,z} and M be the free Malcev algebra with the set of the free genera-
tors X. For brevity, the expressions of the form G(...G(G(t,x,y,z),x,y,z)...),x,y,z)
will be denoted as tGn.
By Alt[X] denote the free alternative algebra generated by the set of free generators
X and by Ass[X] denote the free associative algebra, generated by the set of free
generators X. Furthermore, for a,b ∈ Alt[X] we denote a ◦ b =1
a ∈ Alt[X] denote by R+
xR+
defined as follows: xL+
is the alternative algebra, then by B−denote the commutator algebra of the algebra
B.
The main result:
Theorem. Let M be the free Malcev algebra with free generators X = {x,y,z}
Let U = {J(x,y,z)GkLl
the set of the vectors U∪Ux∪Uy ∪Uz ∪Uxy ∪Uxz ∪Uyz forms the basis of the
space J(M,M,M). Besides M is special.
The algebras M and R(M) satisfy the following identities:
2(ba + ab) and for
athe operator in the algebra Alt[X], defined as follows:
a= x ◦ a for any x ∈ Alt[X]. And L+
a,b= R+
a,b, is the operator in the algebra Alt[X],
a◦bfor any x ∈ Alt[X] and a,b ∈ Alt[X]. If B
aR+
b−R+
x,xLm
y,yLn
z,zLp
x,yLq
x,zLr
y,z| k,l,m,n,p,q,r ∈ N ∪ {0}}. Then
(ab)(cd) = acbd + dacb + bdac + cbda,(1)
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3J(wa,b,c) = J(a,b,c)w − J(b,c,w)a − 2J(c,w,a)b + 2J(w,a,b)c,(2)
G(t,a,b,c) =2
3(J(t,b,c)a + J(a,t,c)b + J(a,b,t)c − J(a,b,c)t),
G(t,a,b,c) = 2(J(ta,b,c) + J(t,a,bc)),
(3)
(4)
J(J(a,b,c),a,b) = 3(ab)J(a,b,c),(5)
J(c,ba2k−1,b) = J(c,a,b)a2k−2b,(k ≥ 1),(6)
uLa,bt = utLa,b+ uLat,b− uLa,tb. (7)
The identities (1), (2), (3), (4) and (5) are proved in [7], (6) is the identity (5) of
[8, §1] and identity (7) is the identity (12) of [8, §1] rewritten in our notation .
Moreover, from the identity (3) it is clear that the function G is a skew-symmetric
for any two arguments.
Lemma 1. The algebras M and R(M) satisfy the following identities
(ta)J(a,b,c) = −1
2J(a,t,c)ab − J(b,c,ta)a −1
J(a,b,tac) = −1
2J(a,t,c)[Ra,Rb] + J(a,b,t)La,c,
J(a,b,c)Lk
2J(t,a,b)ac −3
2J(a,t,cb)a(8)
(9)
b,ba = J(a,b,c)aLk
b,b, (10)
J(a,b,c)Lk
a,aLl
b,b= J(a,b,c)Ll
b,bLk
a,a,(11)
Proof. Applying the operator ∆1
b(h) to the identity bJ(a,b,c) = J(a,b,cb) obtain
hJ(a,b,c) = J(a,h,c)b + J(a,h,cb) + J(a,b,ch).
From the identity (2):
hJ(a,b,c) =1
3J(a,h,c)b + J(a,h,cb) +1
3J(a,b,h)c −2
3J(b,c,h)a +1
3hJ(a,b,c).
That is
hJ(a,b,c) =1
2J(a,h,c)b − J(b,c,t)a +1
2J(a,b,h)c +3
2J(a,h,cb).
Replacing now h by ta obtain the identity (8).
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Twice applying to the first assertion of Lemma to the identity (8), obtain identity
(ta)J(a,b,c) = −1
2J(a,t,c)ab −1
6J(a,b,t)ca −1
6J(a,t,c)ba −2
3J(t,b,c)aa+
+2
3J(a,b,c)ta −1
2J(a,b,t)ac.
From the identity (2) obtain
J(a,b,tac) =2
3J(ta,c,a)b −2
3J(b,ta,c)a −1
3J(a,b,ta)c +1
3J(c,a,b)(ta) =
= −2
3J(t,c,a)ab +4
2
9J(a,b,c)ta +1
9J(a,b,t)ca +4
3J(a,b,t)ac −1
9J(a,t,c)ba −2
9J(t,b,c)aa+
3(ta)J(a,b,c)
Applying now the previous identity, we obtain (9).
The identity (10) shall prove by induction on k. From the identity (7), replacing a by
b, t by a and u by J(a,b,c), we have J(a,b,c)Lb,ba = J(a,b,c)aLb,b−2J(a,b,c)Lb,ab.
From the identity (5): J(J(a,b,c),b,ab) = −3J(a,b,c)(ab)b, J(J(a,b,c),b,ab) =
−J(J(a,b,bc),a,b) = 3J(a,b,bc)(ab) = 3J(a,b,c)b(ab). That is, −3J(a,b,c)(ab)b =
3J(a,b,c)b(ab). Thus, J(a,b,c)Lb,ab= 0. That is, J(a,b,c)bba = J(a,b,c)abb. And
J(a,b,c)Lb,ba = J(a,b,c)aLb,b. Now we have
J(a,b,c)Lk+1
b,ba = J(a,b,cLk
b,b)Lb,ba = J(a,b,cLk
b,b)aLb,b=
= J(a,b,c)Lk
b,baLb,b= J(a,b,c)aLk+1
b,b.
The identiy (11). We apply (6) and identity (10).
J(a,b,c)Lk
a,aLl
b,b= −J(ba2k+1,b,cb2l−1) = −J(a,b,cb2l−1)Lk
a,ab =
= −J(a,b,cb2l−1)bLk
a,a= J(a,b,c)Ll
b,bLk
a,a.
?
Lemma 2. For an arbitrary polynomial f of degree n from the subalgebra
generated by the elements a and b of Malcev algebra M, the following equalities are
true:
1) J(a,b,J(a,f,c)) + (−1)nJ(a,f,J(a,b,c)) = 0,
2) J(a,b,J(f,b,c)) + (−1)nJ(f,b,J(a,b,c)) = 0,
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3) J(a,b,fc) = (−1)nfJ(a,b,c).
Proof. We shall prove the conjunction of all three statements by induction on n.
For n = 1 all identities are obvious. Suppose, they are correct when n = k. Prove
them for n = k+1. Since M is binary Lie algebra, then we can assume that f = f1a
or f = f1b.
1) If f = f1a,the proof is obvious. Let f = f1b. Applying operator ∆1
identity J(a,cb,c) = J(a,b,c)c obtain
c(f1) to the
J(a,f1b,c) = J(a,f1,cb) + J(a,b,c)f1
Next, using this equation and the induction hypothesis, obtain
J(a,b,J(a,f1b,c))+(−1)k+1J(a,f1b,J(a,b,c)) = J(a,b,J(a,f1,cb))+J(a,b,J(a,b,c)f1)+
+(−1)k+1J(a,f1,J(a,b,c)b) + (−1)k+1J(a,b,J(a,b,c))f1= 0
2) The proof is similar to 1).
3) Let f = f1a.
J(a,b,f1ac) = J(a,b,J(f1,a,c) − cf1a − acf1) = −J(a,b,J(a,f1,c))+
+(−1)k+1(f1J(a,b,c)a+J(a,b,c)af1) = −J(a,b,J(a,f1,c))+(−1)k+1(J(J(a,b,c),a,f1)−
−af1J(a,b,c)) = (−1)k+1f1aJ(a,b,c)
If f = f1b, then the arguments are similar and used the equality 2). ?
Corollary 1. Under the conditions of Lemma 2 the equalities are true :
J(a,f,c)b − J(b,f,c)a =3(−1)n+ 1
2
fJ(a,b,c)(12)
J(a,fb,c) = −J(a,f,c)b + (−1)nfJ(a,b,c)(13)
Proof.
J(a,b,fc) =
obtain the first equality. Second equality follows from the first after the application
of identity (2) to J(a,fb,c). ?
From Lemma 2: J(a,b,fc) = (−1)nfJ(a,b,c). From the identity (2):
2
3J(f,c,a)b −2
3J(b,f,c)a −1
3J(a,b,f)c +1
3J(c,a,b)f. Combining, we
Proposition 1. Let u ∈ J(M,M,M). There are αifrom F for which
u =
?
i
αiJ(x,y,z)xi1xi2...xiki,xi,j∈ X,ki∈ N ∪ {0}.
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