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Springerplus. 2016; 5(1): 833.
Published online 2016 June 22. doi:  10.1186/s40064-016-2545-2
PMCID: PMC4917519

Ideals and primitive elements of some relatively free Lie algebras

Abstract

Let F be a free Lie algebra of finite rank over a field K. We prove that if an ideal v~ of the algebra F/γm+1(F) contains a primitive element u~ then the element v~ is primitive. We also show that, in the Lie algebra F/γ3(F) there exists an element v¯ such that the ideal v¯ contains a primitive element u¯ but, u¯ and v¯ are not conjugate by means of an inner automorphism.

Keywords: Primitive element, Free Lie algebra, Ideal, Lower central series, Free nilpotent

Background

Let F be a free Lie algebra of finite rank n, with n ≥ 2, freely generated by the set {x1, …, xn} over a field K. By F and F we denote the subalgebras [FF] and [FF] of F respectively. An ideal V in the free Lie algebra F is called a verbal ideal if for any g(x1, …, xn) ∈ V and any h1, …, hn ∈ F the Lie polynomial g(h1, …, hn) belongs to V. Let V be a non-trivial verbal ideal of F. An element of F is said to be primitive if it can be included in a free generating set of F. Similarly an element of the relatively free Lie algebra F / V is called primitive if it is extendible to a free generating set of F / V.

Let LF/F be the free metabelian Lie algebra. Write x¯i=xi+F,i=1,2,,n. Thus, the set x¯1,,x¯n is a free generating set for L (Bahturin 1987). For g ∈ L, let g be the ideal generated by g and let h be a primitive element of L. It is known that if h ∈ 〈g then g is a primitive element in L (Chirkov and Shevelin 2001). In fact there is an inner automorphism θ of L such that θ(h) = g. For each v ∈ L the linear operator

adv:L ⟶ L

defined by

adv(w) = [wv],  w ∈ L

is a derivation of L and ad2v = 0 because L = {0}. Hence the linear mapping

exp(adv) = 1 + adv

is well defined and it is an inner automorphism of L. In Chirkov and Shevelin (2001) proved that for g ∈ L if a primitive element h of L belongs to the ideal g then h and g are conjugate by means of an inner automorphism of L. This result was obtained by Evans (1994) for free metabelian groups. Does a similar result, as in L, holds for the Lie algebras F/γm+1(F) and F/γ3(F)? In the group case this question was answered by Timoshenko (1997). In the present paper we answer this question. We obtain an affirmative answer for the Lie algebra F/γm+1(F). In contrast to the case of free metabelian Lie algebras and free Lie algebras of the form F/γm+1(F), for the Lie algebra F/γ3(F) we prove that the question has a negative answer. Our main results are similar to the result of Timoshenko (1997) in the case of groups but there are some essential differences.

Preliminaries

Let F be the free Lie algebra generated by a set X = {x1, …, xn} over a field K of characterisitic zeroU(F) be the universal enveloping algebra of F and Δ its augmentation ideal, that is, the kernel of the natural homomorphism σ:U(F) ⟶ K defined by σ(xi) = 0, 1 ≤ i ≤ n. For a given subalgebra R of F we denote by ΔR the left ideal of U(F) generated by the subalgebra R. In the case where R is an ideal of F, ΔR becomes a two-sided ideal of U(F). In fact ΔR is the kernel of the natural homomorphism U(F) ⟶ U(F/R). For any element u of F we denote by u the ideal of F generated by the element u.

Fox (1953) gave a detailed account of the differential calculus in a free group ring. We introduce here free derivations xi:UFUF,1in such that xixj=δijKronecker delta,uvxi=uxiσv+uvxi. It is an obvious consequence of the definitions that xi1=0 . The ideal Δ is a free left U(F)-module with a free basis X and the mappings xi are projections to the corresponding free cyclic direct summands. Thus any element f ∈ Δ can be uniquely written in the form

f=i=1nfxixi.

For any elements g1, …, gn of U(F) we can always find an element f of U(F) such that fxi=gi,1in.

Let f be the column vector fx1,,fxnT, where T indicates transpose.

For any Lie algebra G,  the lower central series

Gγ1(G) ⊇ γ2(G) ⊇  ⋯  ⊇ γk(G) ⊇  ⋯ 

is defined inductively by γ2(G) = [GG], γk(G) = [γk-1(G), G], k ⩾ 2. We usually write G for γ2(G).

Let R be an ideal of F. If u is an element of F, then we denote the images of u under the natural homomorphisms as follows: by u^ in F / R,  by u¯ in F/R and by u~ in F/γm+1(R),  where m ≥ 1.

In (Umirbaev 1993), Umirbaev has defined the right derivatives in the algebras F/R and F/γm+1(R). We give a summary here referring to (Umirbaev 1993).

Let

ρ:[U(F)n]T → [U(F/R)n]T

be the natural componentwise homomorphism, i.e.,

ρf1,,fnT=f1^,,fn^T.

where f1^,,fn^T is the transpose of the vector f1^,,fn^.

Consider the composition mapping

ρ:FUFnTρUF/RnT.
1

This mapping induces the mappings

¯:F/RUF/RnT,~:F/γm+1RUF/RnT.

Since the kernel of the mapping ~ is R/γm+1(R) (see Umirbaev 1993 for details) then it induces the mapping ¯¯:HUF/RnT, where HF/γm+1(R)/R/γm+1(R).

For any element f of F the components ¯f¯xi,~f~xi and ¯¯f¯¯xi of the vectors

¯f¯=¯f¯x1,,¯f¯xnT,~f~=~f~x1,,~f~xnT,¯¯=¯¯f¯¯x1,,¯¯f¯¯xn

are called the partial derivatives of f¯,f~ and f¯¯ respectively. Here we use left derivatives instead of right derivatives.

For each u ∈ R/γm+1(R) the derivation adu: F/γm+1(R) ⟶ F/γm+1(R) is nilpotent and (adu) = 0,  because γm+1( R/γm+1(R)) = {0}.

Hence the linear mapping

expadu=1+adu1!+ad2u2!++admum!

is well defined and it is an inner automorphism of F/γm+1(R), m ≥ 1,  that is, since [[w,u],,um+1-times]=0,

expadvw=w+w,u1!+w,u,u2!++[[w,u],,um-times]m!.

We need the following technical lemmas. The first lemma is an immediate consequence of the definitions.

Lemma 1

LetJ be an arbitrary ideal ofU(F) andu ∈ Δ. Thenu ∈ JΔ if and only ifuxiJ for eachi, 1 ≤ i ≤ n.

The next lemma can be found in Yunus (1984).

Lemma 2

LetR be an ideal ofF andu ∈ F. Thenu ∈ ΔRΔ if and only ifu ∈ R.

Main results

Let F be the free Lie algebra generated by a set X = {x1, …, xn}, n ≥ 2,  over a field K of characteristic zero and let R be a non-trivial verbal ideal of F.

For an element f of F the vector fx1,,fxn is called unimodular, if there exist a1, …, an ∈ U(F) such that

a1fx1++anfxn=1.

Umirbaev (1993) has proved a criterion of primitiveness for a system of elements in a finitely generated free Lie algebra of the form F/γm+1(R), where m ≥ 1 and RF. Umirbaev’s criterion for the primitivity of an element of the algebra F/γm+1(R) is stated below.

Proposition 3

Let RF. An elementu~ ofF/γm+1(R) is primitive if and only if the vector~u~x1,,~u~xn is unimodular inU(F/R).

We are going to consider the case RF.

Proposition 4

An elementf¯ of the free metabelian Lie algebraF/F is primitive if and only if the imagef~ is primitive in the free nilpotent-by-abelian Lie algebraF/γm+1(F),  wheref ∈ Fm ≥ 2.

Proof

Suppose that the element f¯ of F/F is primitive. If we put m = 1 in Proposition 3 we have that the vector ¯f¯x1,,¯f¯xn is unimodular in U(F/F),  that is, there exist a1,…,an ∈ U(F/F) such that i=1nai¯f¯xi=1.

Let HF/γm+1(F)/F/γm+1(F). We calculate the derivative ¯f¯xi by using the natural homomorphism θ:F/γm+1(F) → H the isomorphism φ:H → F/F and the chain rule for derivatives:

¯f¯xi=¯φf¯¯xi=¯φf¯¯f¯¯.¯¯f¯¯xi=¯φf¯¯f¯¯.¯¯θf~xi=¯φf¯¯f¯¯.¯¯θf~f~.~f~xi.

It is clear that ¯φf¯¯f¯¯,¯¯θf~f~UF/F. Therefore from the equality i=1nai¯f¯xi=1 we get i=1nbi~f~xi=1, where bi=ai¯φf¯¯f¯¯.¯¯θf~f~. Hence by Proposition 3, f~ is primitive in F/γm+1(F).

Now suppose that f~ is primitive element of the algebra F/γm+1(F). By definition it can be extended to a free generating set Y=f~=f~1,,f~n of F/γm+1(F). Clearly Y is linearly independent modulo (F/γm+1(F)). Therefore the image θ(Y) of Y in the algebra H is linearly independent modulo H. As a simple application of theorem 4.2.4.9 of Bahturin (1987) we see that θ(Y) freely generates the algebra H. Hence the image of θ(Y) under the isomorphism φ:H → F/F generates the algebra F/F. That is, the algebra F/F is freely generated by the set φθY=f¯=f¯1,,f¯n. Thus, f¯ is a primitive element of the algebra F/F.

As a consequence of the result of Chirkov and Shevelin (2001), we obtain the following proposition. Although its proof is given in Ersalan and Esmerligil (2014), our proof is more explicit. The idea of the proof is similar to the idea of the proof of Proposition 2 of the paper by Timoshenko (1997) for groups.

Proposition 5

Letu~ be a primitive element of the algebraF/γm+1(F) and letv~F/γm+1F whereuv ∈ Fm ≥ 1. Ifu~v~ thenv~ is also primitive.

Proof

Let u~,v~ ∈ F/γm+1(F). Assume that u~ is primitive and that it is contained in the ideal v~ of F/γm+1(F). By Proposition 4, u¯ is primitive. In the view of Proposition 4, it sufficies to the prove that the element v¯ of F/F is primitive.

Since

u~=u+γm+1Fv+γm+1F,m1

we have

u ∈ 〈v〉(modF), 

that is,

u¯=u+Fv+F.

From the result of Chirkov and Shevelin (2001), we obtain that the elements u¯ and v¯ are conjugate by means of an inner automorphism. Therefore v¯ is primitive. Hence the result follows.

The mapping ^:FF/R can be extended to the mapping ^:UFUF/R for which we preserve the same notation.

The following lemma will play a crucial role in proving our main result.

Lemma 6

LetR be a verbal ideal ofFr ∈ R and letv ∈ F. ThenrR ∈ 〈v〉 + R if and only if there exist an elementα^UF/R and an elementβi^Δv^, such thatr^xi=α^v^xi+βi^,where i = 1, …, n andΔv^ is the ideal generated by the elementv^ in the algebraU(F/R).

Proof

Let r be an element of the ideal R, v ∈ F and r¯v¯. Then r ∈ 〈v〉(modR),  where v is the ideal of F generated by v. Any element of the ideal v can be written as linear combinations of commutators of F depending on the element v. Applying the Jacobi identitiy and the anticommutativity, these commutators can be rewritten as linear combinations of commutators of the form

[[…[[vxi1], xi2], …], xik],  xi1, …xik ∈ {x1, …, xn},  k ≥ 0.
2

If r ≡ v(modR) then clearly r^xi=v^xi,i=1,n.

Now assume that the element r is written as a linear combination of elements of the form (2). Without loss of generality we may assume that

r ≡ [[…[vxi1], …], xik](modR),  k ≥ 1, 

By straightforward calculations we see that the form of the derivatives rxi are

rxi=αvxi+βi+ΔR,

where α ∈ U(F/R), βi ∈ Δvi = 1, …, n.

Therefore

r^xi=α^v^xi+βi^.

Let now r^xi=α^v^xi+βi^, where α^UF/R,βi^Δv^,i=1,n.

The kernel of the natural homomorphism ^:UFUF/R is ΔR and hence

rxi+ΔR=αvxi+βi+ΔR.

Then there exists an element g of ΔR such xir-αv=βi+g, where βi ∈ Δv that is, xir-αvΔv+ΔR. By Lemma 1 we have rαv ∈ ΔvΔ + ΔRΔ. Hence the element rαv of F can be written as rαvhz where h ∈ ΔvΔ, z ∈ ΔRΔ. By Lemma 2 we get h ∈ 〈v and z ∈ R. Hence rRαvhR. This completes the proof.

In contrast to the case of free metabelian Lie algebras we can show that there exists an element v¯ of the algebra F/γ3(F) such that the ideal v¯ of F/γ3(F) contains a primitive element u¯, but u¯ and v¯ are not conjugate by means of an inner automorphism.

Theorem 7

There is an elementv¯ in the algebraF/γ3(F) such that the idealv¯ ofF/γ3(F) contains the elementx1¯ , but the elementsv andx1 are not conjugate moduloγ3(F) by means of an inner automorphism.

Proof

We consider the element v¯=x1+x1,x2,x2,x1,x2+γ3F of F/γ3(F) which is an analogue of the element given in Fox (1953) for groups. Let w = [[[[x1x2], x2], x1], x2].

We have

wx1=-x2·x1,x2,x2+x2·x1·x1,x2,x2x1,wx2=x1,x2,x2,x1+x2·x1·x1,x2,x2x2.

Now consider the images w^xi under the homomorphism

^:UFUF/γ3F,i=1,2.

Then

w^x1=x2^·x1^·x1,x2,x2^x1,w^x2=x2^·x1^·x1,x2,x2^x2,w^xk=0,k>2.

Clearly w^xiΔx1^=Δv^. In the above equalities if we set α^=0 and βi^=x2^·x1^·x1,x2,x2^xi,i=1,2, then we see that

w^xi=α^·v^xi+βi^,i=1,2.

By Lemma 6 wγ3(F) ∈ 〈vγ3(F). Therefore we have

x1γ3(F)vwγ3(F) ∈ 〈v〉 + γ3(F).

Now we are going to verify that the element w can not be written in the form [x1u] in the algebra F/γ3(F).

Assume that the rank of F equal to 2, u ∈ γ3(F) and

w = [x1u].
3

Let us calculate the derivative x1 of both sides of (3). We have

-x2·x1,x2,x2+x2·x1·x2·x2=-u+x1·ux1.

Taking the image under the homomorphism ^:UFUF/γ3F we get

x2^·x1^·x2^·x2^=x1^u^x1.
4

It is well known that the set x1^,x2^,x1^,x2^ is a basis of F/γ3(F). Therefore by Poincare–Birkhoff–Witt’s theorem the algebra U(F/γ3(F)) is a free K- module generated 1 and the all ordered monomials of the form

x1^,x2^r·x1^s·x2^k,r0,s0,k0,r,s,k0,0,0.

Thus every element of U(F/γ3(F)) can be uniquely written as

αrskr,s,k0x1^,x2^r·x1^s·x2^k,αijkK.
5

Let us express each side of (4) in the form (5):

x2^·x1^·x2^·x2^=x2^,x1^x2^·x2^+x1^·x2^·x2^·x2^,x1^u^x1=x1^αijki,j,k0x1^,x2^r·x1^s·x2^k.

Then

x1^·αijki,j,k0x1^,x2^i·x1^j·x2^k=x2^,x1^·x2^2+x1^·x2^3.
6

We note that in the algebra F/γ3(F) we have x1^,x2^,xi^=0,i ∈ {1, 2}. That is, the elements x1^,x2^ and xi^ commute in the algebra U(F/γ3(F)).

So from (6) we get

i,j0αij2x1^,x2^i·x1^j+1=x2^,x1^
7

and

i,j0αij3x1^,x2^i·x1^j+1=x1^.

Using (7) we obtain

j0α1j2x1^j+1=-1,

which is impossible. This contradiction proves the theorem.

Conclusions

In this work we found a relation between the generator of a one-generated ideal of a relatively free Lie algebra and a primitive element which is contained in this ideal. One can expects to adopt our results for ideals of some relatively free Lie algebras which have more than one generator.

Authors’ contributions

All authors participated in the design of this work and performed equally. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Contributor Information

Naime Ekici, rt.ude.uc@iciken.

Zerrin Esmerligil, rt.ude.uc@nirreze.

Dilek Ersalan, rt.ude.uc@hakelid.

References

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  • Evans MJ. Presentations of the free metabelian group of rank 2. Can Math Bull. 1994;37(4):468–472. doi: 10.4153/CMB-1994-068-9. [Cross Ref]
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