Soluble Product of Parafree Lie Algebras and Its Residual Properties

Parafree groups arose from G. Baumslag’s works [2,3]. G. Baumslag has obtained some interesting results about parafree groups [4,5]. In 1978, H. Baur has translated the notion of parafree groups to parafree Lie algebras [6]. He has defined parafree Lie algebras as in the group case and he proved the existence of a non-free parafree Lie algebra [5]. Further, N. Ekici and Z. Velioğlu have worked on unions [9] and direct limit [10] of parafree Lie algebras. In [12], Z. Velioğlu has investigated a residual property of metabelian parafree Lie algebras. G. Baumslag’s and Baur’s works have given a start for studies in the theory of parafree Lie algebras. Although there are many questions about that algebras have remained unanswered. The aim of this work is to define soluble product of parafree Lie algebras and to investigate some residual properties of that product as a generalized of the study in [12].

Preliminaries

Let K be a field of characteristic zero and L be a Lie algebra over K. The lower central series of L, $L = γ_{1} (L) \supseteq γ_{2} (L) \supseteq ... \supseteq γ_{n} (L) \supseteq ...$ L = {\gamma _1}\left(L \right) \supseteq {\gamma _2}\left(L \right) \supseteq... \supseteq {\gamma _n}\left(L \right) \supseteq... is defined inductively by $γ_{2} (L) = [L, L], γ_{n + 1} (L) = [γ_{n} (L), L], n \geq 2 .$ {\gamma _2}(L) = [L,L],\,{\gamma _{n + 1}}(L) = [{\gamma _n}(L),L],\,n \ge 2. If n is the smallest integer satisfying γ_n(L) = 0, then L is called nilpotent of class n.

The derived series of L, $L = δ^{0} (L) \supseteq δ^{1} (L) \supseteq \dots \supseteq δ^{n} (L) \supseteq \dots$ L = {\delta ^0}(L) \supseteq {\delta ^1}(L) \supseteq \ldots \supseteq {\delta ^n}(L) \supseteq \ldots is defined inductively by $δ^{0} (L) = L, δ^{n + 1} (L) = [δ^{n} (L), δ^{n} (L)] .$ {\delta ^0}(L) = L,\,{\delta ^{n + 1}}(L) = [{\delta ^n}(L),{\delta ^n}(L)].

If k is the smallest integer satisfying δ^k (L) ≠ 0, then L is called solvable of class k.

Definition 1

A Lie algebra is called Hopfian if it satisfies the following equivalent conditions:i)

It is not isomorphic to any of its proper quotients.

ii)

Every surjective endomorphism of it is an automorphism.

Definition 2

Let P be a property of Lie algebras that invariant under isomorphisms and let L be any Lie algebra. L is called residually-P, if there exists an ideal I of L such that for all 0 ≠ x ∈ L, x ∉ I and the quotient algebra L/I has the property P.

Equivalently,

L is residually-P if and only if I is an ideal of L such that L/I has property P and ∩ I = {0}

For example, let L be a Lie algebra, if $\cap_{n = 1}^{\infty} δ^{n} (L) = {0}$ \bigcap\nolimits_{n = 1}^\infty {\delta ^n}(L) = \{0\} , then L is called residually soluble, if $\cap_{n = 1}^{\infty} γ_{n} (L) = {0}$ \bigcap\nolimits_{n = 1}^\infty {\gamma _n}(L) = \{0\} , then L is called residually nilpotent.

Definition 3

Let F be the free Lie algebra freely generated by X. A Lie algebra P is called parafree over a set X if, i)

P is residually nilpotent, and

ii)

For every n ≥ 1, P/γ_n (P) ≅ F/γ_n (F), i.e. P has the same lower central sequence as F.

The cardinality of X is called rank of L.

Soluble Product of Parafree Lie Algebras

In this section, we want to define soluble product of parafree Lie algebras. For more details about the structure of products of algebras, the reader may consult [1].

Definition 4

Let $\prod_{α \in I}^{*} L_{α}$ \prod\nolimits_{\alpha \in I}^ * {L_\alpha}and ⊕ L_α be the free product and the direct sum of Lie algebras {L_α|α ∈ I}, respectively. For e ∈ L_αand α ∈ I consider an epimorphism as follows $\begin{matrix} χ : \prod_{α \in I}^{*} L_{α} \to \oplus L_{α}, \\ χ (e) = e . \end{matrix}$ \matrix{{\chi :\prod\nolimits_{\alpha \in I}^ * {L_\alpha} \to \mathop \oplus \nolimits_ {L_\alpha},} \cr {\chi (e) = e.}}

The kernel of the epimorphism χ is called cartesian subalgebra of $\prod_{α \in I}^{*} L_{α}$ \prod\nolimits_{\alpha \in I}^ * {L_\alpha} .

Definition 5

Let L₁ * L₂be the free product of Lie algebras L₁and L₂and D be the cartesian subalgebra of L₁ * L₂. The n-th solvable product of L₁and L₂is denoted as $L_{1} *_{sol}^{n} L_{2}$ {L_1} * _{sol}^n{L_2}and defined as $L_{1} *_{sol}^{n} L_{2} = (L_{1} * L_{2}) / (δ^{n} (L_{1} * L_{2}) \cap D) .$ {L_1} * _{sol}^n{L_2} = ({L_1} * {L_2})/({\delta ^n}({L_1} * {L_2}) \cap D).

Theorem 6

Let P₁and P₂be two parafree Lie algebras. Then P₁ * P₂is parafree.

Proof

The proof can be found in [6].

Theorem 7

Let P₁and P₂be two parafree Lie algebras. Then $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2}is parafree.

Proof

Put P = P₁ * P₂, by the Theorem 6, P is parafree. Firstly we want to show that $P_{1} *_{sol}^{n} P_{2} = P / (δ^{n} (P) \cap D)$ {P_1} * _{sol}^n{P_2} = P/({\delta ^n}(P) \cap D) is residually nilpotent. Let $u \in \cap_{k = 1}^{\infty} γ_{k} (P / (δ^{n} (P) \cap D))$ u \in \bigcap\nolimits_{k = 1}^\infty {\gamma _k}(P/({\delta ^n}(P) \cap D)) . Then for all positive integers k, $u \in γ_{k} (P / (δ^{n} (P) \cap D)) = (γ_{k} (P) + (δ^{n} (P) \cap D)) / (δ^{n} (P) \cap D) .$ u \in {\gamma _k}(P/({\delta ^n}(P) \cap D)) = ({\gamma _k}(P) + ({\delta ^n}(P) \cap D))/({\delta ^n}(P) \cap D). Let a ∈ γ_k (P) + (δⁿ (P) ∩ D), so u = a + (δⁿ (P) ∩ D). It is clear that $a \in \cap_{k = 1}^{\infty} γ_{k} (P) + (δ^{n} (P) \cap D) .$ a \in \bigcap\nolimits_{k = 1}^\infty {\gamma _k}(P) + ({\delta ^n}(P) \cap D). Since P is residually nilpotent, then a ∈ δⁿ(P) ∩ D and so u = δⁿ (P) ∪ D. Therefore $\cap_{k = 1}^{\infty} γ_{k} (P / (δ^{n} (P) \cap D)) = {0},$ \bigcap\nolimits_{k = 1}^\infty {\gamma _k}(P/({\delta ^n}(P) \cap D)) = \{0\}, i.e., $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} is residually nilpotent. Now we want to show that $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} has the same lower central sequence as a free Lie algebra. Consider that $(P / δ^{n} (P) \cap D) / γ_{k} (P / δ^{n} (P) \cap D) ≅ (P / δ^{n} (P) \cap D) / (γ_{k} (P) / (δ^{n} (P) \cap D)) ≅ P / γ_{k} (P) .$ (P/{\delta ^n}(P) \cap D) / {\gamma _k}(P/{\delta ^n}(P) \cap D) \cong (P/{\delta ^n}(P) \cap D) / ({\gamma _k}(P)/({\delta ^n}(P) \cap D)) \cong P/{\gamma _k}(P). Since P is parafree, there is a free Lie algebra F such that $P / γ_{k} (P) ≅ F / γ_{k} (F) .$ P/{\gamma _k}(P) \cong F/{\gamma _k}(F). Hence, $P_{1} *_{sol}^{n} P_{2} / γ_{k} (P_{1} *_{sol}^{n} P_{2}) ≅ F / γ_{k} (F),$ {P_1} * _{sol}^n{P_2}/{\gamma _k}({P_1} * _{sol}^n{P_2}) \cong F/{\gamma _k}(F), i.e., $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} is parafree.

Residual Properties of Soluble Product of Parafree Lie Algebras

In this section, we want to investigate some residual properties of soluble product of parafree Lie algebras.

Note that if P₁ and P₂ are two parafree solvable Lie algebras of class n, then δⁿ (P₁ * P₂) ∩ D = δⁿ (P₁ * P₂). Hence $P_{1} *_{sol}^{n} P_{2} = (P_{1} * P_{2}) / δ^{n} (P_{1} * P_{2}) .$ {P_1} * _{sol}^n{P_2} = ({P_1} * {P_2})/{\delta ^n}({P_1} * {P_2}). By the Theorem 7, $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} is a parafree Lie algebra. Since P₁ and P₂ are soluble, then $\cap_{n = 1}^{\infty} δ^{n} (P_{1} *_{sol}^{n} P_{2}) = {0},$ \bigcap\nolimits_{n = 1}^\infty {\delta ^n}({P_1} * _{sol}^n{P_2}) = \{0\}, i.e. $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} is residually soluble. On the other hand in the proof of Theorem 7, we have shown that $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} is residually nilpotent. Now we want to investigate another residual property of $P_{1} *_{sol}^{n} P_{2}$ {P_1} * _{sol}^n{P_2} .

Lemma 8

If a set Y freely generates a free Lie algebra F, modulo γ₂(F). Then for n = 2,3,... Y freely generates F, modulo γ_n(F).

Proof

The proof can be found in [8].

Theorem 9

A parafree Lie algebra is residually parafree of finite rank.

Proof

The proof can be found in [6].

Theorem 10

Let P₁and P₂be two parafree solvable Lie algebras of class n that each one generated by more than one element and $M = P_{1} *_{sol}^{n} P_{2}$ M = {P_1} * _{sol}^n{P_2} . Then M is residually parafree of rank two.

Proof

Since P₁ and P₂ are soluble and generated by more than one element, then M is also soluble and generated by more than one element. By Theorem 7, M is a parafree Lie algebra.

Now we want to show that M is residually parafree. By the Theorem 9, M has a finite rank n (n ≥ 2). Let 0 ≠ a ∈ M. Since M is parafree, for some k ≥ 2, a ∉ γ_k(M). It is clear that M/γ_k(M) is free soluble and nilpotent.

Let N be a subalgebra of M such that a ∉ N and γ_k(M) ⊆ N. Then the quotient algebra (M/γ_k(M))/(N/γ_k(M)) is free of rank 2. It is well know that free Lie algebras are projective. Therefore for an ideal N/γ_k(M) of M/γ_k(M), there exists a subalgebra L/γ_k(M) of M/γ_k(M) such that $\begin{matrix} M / γ_{k} (M) ≅ ((N / γ_{k} (M)) \oplus (L / γ_{k} (M)), \\ (N / γ_{k} (M)) \cap (L / γ_{k} (M)) = 0 \end{matrix}$ \matrix{{M/{\gamma _k}(M) \cong ((N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M)),} \cr {(N/{\gamma _k}(M)) \cap (L/{\gamma _k}(M)) = 0}} and $L / γ_{k} (M) ≅ (M / γ_{k} (M)) / (N / γ_{k} (M)) .$ L/{\gamma _k}(M) \cong (M/{\gamma _k}(M)) / (N/{\gamma _k}(M)). Now consider M/γ₂(M) : $M / γ_{2} (M) ≅ (M / γ_{k} (M)) / γ_{2} (M / γ_{k} (M)) .$ M/{\gamma _2}(M) \cong (M/{\gamma _k}(M)) / {\gamma _2}(M/{\gamma _k}(M)). Since M/γ_k(M) ≅ ((N/γ_k(M)) ⊕ (L/γ_k(M)), then $(M / γ_{k} (M)) / γ_{2} (M / γ_{k} (M)) = ((N / γ_{k} (M)) \oplus (L / γ_{k} (M)) / γ_{2} ((N / γ_{k} (M)) \oplus (L / γ_{k} (M))) .$ (M/{\gamma _k}(M)) / {\gamma _2}(M/{\gamma _k}(M)) = ((N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M)) / {\gamma _2}((N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M))). Now we want to investigate γ₂((N/γ_k(M)) ⊕ (L/γ_k(M))): $\begin{array}{l} γ_{2} ((N / γ_{k} (M)) \oplus (L / γ_{k} (M)) = [(N / γ_{k} (M)) \oplus (L / γ_{k} (M)), (N / γ_{k} (M)) \oplus (L / γ_{k} (M))] \\ = ([N, N] \oplus [N, L] \oplus [L, L]) / γ_{k} (M) = (γ_{2} (N) \oplus [N, L] + γ_{2} (L)) / γ_{k} (M) . \end{array}$ \matrix{{{\gamma _2}((N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M)) = [(N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M)),(N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M))]} \hfill \cr {= ([N,N] \oplus [N,L] \oplus [L,L]) / {\gamma _k}(M) = ({\gamma _2}(N) \oplus [N,L] + {\gamma _2}(L)) / {\gamma _k}(M).} \hfill}

Therefore, $\begin{array}{l} ((N / γ_{k} (M)) \oplus (L / γ_{k} (M)) / γ_{2} ((N / γ_{k} (M)) \oplus (L / γ_{k} (M)) \\ = (N / γ_{k} (M)) \oplus (L / γ_{k} (M)) / ((γ_{2} (N) \oplus [N, L] \oplus γ_{2} (L)) / γ_{k} (M)) . \\ = ((N / γ_{k} (M)) / (γ_{2} (N) \oplus [N, L]) / (γ_{k} (M))) \oplus ((L / γ_{k} (M)) / (γ_{2} (L) / γ_{k} (M))) \\ = ((N / γ_{k} (M)) / (γ_{2} (N) \oplus [N, L] \oplus γ_{k} (M)) / γ_{k} (M))) \oplus ((L / γ_{k} (M)) / ((γ_{2} (L) + γ_{k} (M) / γ_{k} (M))) \\ = (N / (γ_{2} (N) \oplus [N, L] + γ_{k} (M))) \oplus (L / (γ_{2} (L) + γ_{k} (M)) . \end{array}$ \matrix{{((N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M)) / {\gamma _2}((N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M))} \hfill \cr {= (N/{\gamma _k}(M)) \oplus (L/{\gamma _k}(M)) / (({\gamma _2}(N) \oplus [N,L] \oplus {\gamma _2}(L))/{\gamma _k}(M)).} \hfill \cr {= ((N/{\gamma _k}(M)) / ({\gamma _2}(N) \oplus [N,L])/({\gamma _k}(M))) \oplus ((L/{\gamma _k}(M)) / ({\gamma _2}(L)/{\gamma _k}(M)))} \hfill \cr {= ((N/{\gamma _k}(M)) / ({\gamma _2}(N) \oplus [N,L] \oplus {\gamma _k}(M))/{\gamma _k}(M))) \oplus ((L/{\gamma _k}(M)) / (({\gamma _2}(L) + {\gamma _k}(M)/{\gamma _k}(M)))} \hfill \cr {= (N/({\gamma _2}(N) \oplus [N,L] + {\gamma _k}(M))) \oplus (L/({\gamma _2}(L) + {\gamma _k}(M)).} \hfill} Hence, we have $M / γ_{2} (M) = (N / (γ_{2} (N) \oplus [N, L] + γ_{k} (M))) \oplus (L / (γ_{2} (L) + γ_{k} (M))) .$ M/{\gamma _2}(M) = (N/({\gamma _2}(N) \oplus [N,L] + {\gamma _k}(M))) \oplus (L/({\gamma _2}(L) + {\gamma _k}(M))).

It is easy to see that the quotient algebra M/γ₂(M) is free abelian of rank n. On the other hand, due to the choice of L, the quotient algebra L/(γ₂(L) + γ_k(M)) is free abelian of rank 2. Hence the algebra N/(γ₂(N) ⊕ [N,L] + γ_k(M)) is free abelian of rank n-2.

Let a₁,...,a_n−2 ∈ N and a_n−1,a_n∈ L, such that elements a₁,...,a_n−2 freely generate N, modulo γ₂(N) ⊕ [N,L]+γ_k(M) and elements a_n−1,a_n freely generate L, modulo γ₂(L)+γ_k(M). By the Hopfianicity [11], elements a₁,...,a_n freely generates M, modulo γ₂(M). Let T be an ideal generated by a₁,...,a_n−2, then again by the Hopfianicity, we have N = T + γ_k(M).

Now consider $J / T = \cap_{n = 1}^{\infty} γ_{n} (M / T) .$ J/T = \bigcap\nolimits_{n = 1}^\infty {\gamma _n}(M/T).

We want to show that a ∉ J and M/J is parafree. Since a ∉ N, it is clear that a ∉ J. Therefore it remains to show that M/J is parafree. By the definition of J, the algebra M/J is residually nilpotent. By the Lemma 8, if the elements a₁,...,a_n freely generate M, modulo γ₂(M), then they freely generate M, modulo γ_k(M). Then the algebra M/(T + γ_k(M)) is free soluble of rank 2. We chose the ideal J as the smallest ideal of M containing T such that M/J residually nilpotent. Therefore J ⊆ (T + γ_k(M)). Hence we have $M / (T + γ_{k} (M)) = (M / J) / (T + γ_{k} (M) / J) = (M / J) / γ_{k} (M / J) .$ M/(T + {\gamma _k}(M)) = (M/J) / (T + {\gamma _k}(M)/J) = (M/J) / {\gamma _k}(M/J). Since M/(T + γ_k(M)) is free, so is (M/J)/γ_k(M/J). Therefore M/J is parafree and M is residually parafree.

eISSN:: 2444-8656
Language:: English

Publication timeframe:: Volume Open
Journal Subjects:: Life Sciences, other, Mathematics, Applied Mathematics, General Mathematics, Physics

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Soluble Product of Parafree Lie Algebras and Its Residual Properties

Published Online: Oct 25, 2019

Page range: 509 - 514

Received: Mar 25, 2019

Accepted: May 07, 2019

DOI: https://doi.org/10.2478/amns.2019.2.00036

KeywordsParafree Lie Algebras, Soluble product, Residual properties

© 2020 Zehra Velioğlu, published by Sciendo

This work is licensed under the Creative Commons Attribution 4.0 International License.

Keywords
Parafree Lie Algebras, Soluble product, Residual properties