Enter Your Search:Search tips Search criteria Articles Journal titles Advanced

Springerplus. 2016; 5(1): 833.
Published online 2016 June 22.
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

defined by

adv(w) = [wv],  w ∈ L

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

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 $∂∂xi:UF⟶UF,1≤i≤n$ such that $∂∂xixj=δijKronecker delta,∂uv∂xi=∂u∂xiσv+u∂v∂xi$. 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

$f=∑i=1n∂f∂xixi.$

For any elements g1, …, gn of U(F) we can always find an element f of U(F) such that $∂f∂xi=gi,1≤i≤n.$

Let f be the column vector $∂f∂x1,…,∂f∂xnT$, 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

$ρ∘∂:F→∂UFnT→ρUF/RnT.$
1

This mapping induces the mappings

$∂¯:F/R′→UF/RnT,∂~:F/γm+1R→UF/RnT.$

Since the kernel of the mapping $∂~$ is R/γm+1(R) (see Umirbaev 1993 for details) then it induces the mapping $∂¯¯:H→UF/RnT,$ 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 $[[w,u],…,u⏟m+1-times]=0,$

$expadvw=w+w,u1!+w,u,u2!+⋯+[[w,u],…,u⏞m-times]m!.$

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 if$∂u∂xi∈J$ 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 $∂f∂x1,…,∂f∂xn$ is called unimodular, if there exist a1, …, an ∈ U(F) such that

$a1∂f∂x1+⋯+an∂f∂xn=1.$

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 element$u~$ 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 element$f¯$ of the free metabelian Lie algebraF/F is primitive if and only if the image$f~$ 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

Let$u~$ be a primitive element of the algebraF/γm+1(F) and let$v~∈F/γm+1F′$ whereuv ∈ Fm ≥ 1. If$u~∈v~$ then$v~$ 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

$u~=u+γm+1F′∈v+γm+1F′,m≥1$

we have

u ∈ 〈v〉(modF),

that is,

$u¯=u+F″∈v+F″.$

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 $^:F→F/R$ can be extended to the mapping $^:UF→UF/R$ 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 element$α^∈UF/R$ and an element$βi^∈Δv^,$ such that$∂r^∂xi=α^∂v^∂xi+βi^,$where i = 1, …, n and$Δv^$ is the ideal generated by the element$v^$ 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 $∂r∂xi$ are

$∂r∂xi=α∂v∂xi+βi+ΔR,$

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 $^:UF⟶UF/R$ is ΔR and hence

$∂r∂xi+ΔR=α∂v∂xi+βi+ΔR.$

Then there exists an element g of ΔR such $∂∂xir-αv=βi+g,$ where βi ∈ Δv that is, $∂∂xir-αv∈Δv+ΔR.$ 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 element$v¯$ in the algebraF/γ3(F) such that the ideal$v¯$ ofF/γ3(F) contains the element$x1¯$ , 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

$∂w∂x1=-x2·x1,x2,x2+x2·x1·∂x1,x2,x2∂x1,∂w∂x2=x1,x2,x2,x1+x2·x1·∂x1,x2,x2∂x2.$

Now consider the images $∂w^∂xi$ under the homomorphism

$^:UF⟶UF/γ3F,i=1,2.$

Then

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

Clearly $∂w^∂xi∈Δx1^=Δv^.$ 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

$-x2·x1,x2,x2+x2·x1·x2·x2=-u+x1·∂u∂x1.$

Taking the image under the homomorphism $^:UF⟶UF/γ3F$ 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

$x1^,x2^r·x1^s·x2^k,r≥0,s≥0,k≥0,r,s,k≠0,0,0.$

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

$∑αrskr,s,k≥0x1^,x2^r·x1^s·x2^k,αijk∈K.$
5

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

$x2^·x1^·x2^·x2^=x2^,x1^x2^·x2^+x1^·x2^·x2^·x2^,x1^∂u^∂x1=x1^∑αijki,j,k≥0x1^,x2^r·x1^s·x2^k.$

Then

$x1^·∑αijki,j,k≥0x1^,x2^i·x1^j·x2^k=x2^,x1^·x2^2+x1^·x2^3.$
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

$∑i,j≥0αij2x1^,x2^i·x1^j+1=x2^,x1^$
7

and

$∑i,j≥0αij3x1^,x2^i·x1^j+1=x1^.$

Using (7) we obtain

$∑j≥0α1j2x1^j+1=-1,$

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

• Bahturin YuA. Identical relations in Lie algebras. Utrecht: VNU Science Press BV; 1987.
• Chirkov IV, Shevelin MA. Ideals of free metabelian Lie algebras and primitive elements. Sib Math J. 2001;42(3):610–612. doi: 10.1023/A:1010443714483.
• Ersalan D, Esmerligil Z. Primitive elements and preimage of primitive sets of free Lie algebras. Int J Pure App Math. 2014;95(4):535–541.
• 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.
• Fox RH. Free differential calculus I. Derivations in free group rings. Ann Math. 1953;57(2):547–560. doi: 10.2307/1969736.
• Timoshenko EI. Primitive elements of the free groups of the varieties 𝔘𝔑n$UNn$. Math Notes. 1997;61(6):739–743. doi: 10.1007/BF02361216.
• Umirbaev UU. Partial derivatives and endomorphisms of some relatively free Lie algebras. Sibirskii Mat Zh. 1993;34(6):179–188.
• Yunus IA. On the Fox problem for Lie algebras. Uspekhi Math Nauk. 1984;39:251–252.

Articles from SpringerPlus are provided here courtesy of Springer-Verlag

 PubMed Central Canada is a service of the Canadian Institutes of Health Research (CIHR) working in partnership with the National Research Council's national science library in cooperation with the National Center for Biotechnology Information at the U.S. National Library of Medicine(NCBI/NLM). It includes content provided to the PubMed Central International archive by participating publishers.