Equivalence of definitions of 2-Engel Lie ring

This article gives a proof/explanation of the equivalence of multiple definitions for the term 2-Engel Lie ring
View a complete list of pages giving proofs of equivalence of definitions

Statement

The following are equivalent for a Lie ring:

No.	Shorthand	Condition on $L$
1	2-Engel identity	for any $a,b\in L$, we have $[a,[a,b]]=0$ (Note that if $a=b$, this would follow automatically, so we can restrict attention to the case $a\neq b$).
2	triple Lie bracket is alternating	The function $(x,y,z)\mapsto [x,[y,z]]$ is an alternating function in all pairs of variables.
3	2-locally class at most two	any subring of $L$ generated by a subset of size at most two is a Lie ring of nilpotency class two, i.e., any such subring has class at most two.
4	cyclic symmetry of Lie bracket	for any $x,y,z\in L$, we have $[x,[y,z]]=[y,[z,x]]=[z,[x,y]]$. Note that $x,y,z$ are possibly equal, possibly distinct.
5	union of abelian ideals	There is a collection $A_{i},i\in I$ (for some indexing set $I$) of abelian ideals of $L$ such that $L=\bigcup _{i\in I}A_{i}$. Equivalently, every element of $L$ is contained in an abelian ideal.

Facts used

1. Polarization trick
2. Alternating function condition is transitive
3. Alternating and skew-symmetric in pairs with a common variable implies alternating in all three variables
4. Symmetric or skew-symmetric function condition needs to be checked only on a generating set for the symmetric group

Proof

Slick version of equivalence of (1), (2), and (4)

Define $f:L\times L\times L\to L$ by:

$f(x,y,z)=[x,[y,z]]$

We know already that $f$ is alternating in the second and third variable. We now see that:

• The equivalence of (1) and (2) follows from Fact (2).
• By Fact (4) and the observation that $f$ is alternating in the second and third variable, and the fact that the 3-cycle and a transposition generate symmetric group:S3, we obtain that condition (4) (cyclic symmetry) is equivalent to being skew-symmetric in all pairs of variables. By Facts (1) and (3), this is equivalent to being alternating in all pairs of variables, which is condition (2).

Since this slick version of equivalence may not be very clear, explicit proofs also follow.

(1) implies (2)

Given: A Lie ring $L$. $[a,[a,b]]=0$ for all $a,b\in L$.

To prove: Any triple Lie product that involves a repeated variable is zero. In particular, for any $x,y\in L$, we have $[x,[x,y]]=0,[x,[y,x]]=0$, and $[y,[x,x]]=0$.

Proof: Setting $a=x$ and $b=y$, we get $[x,[x,y]]=0$, so also $[x,[y,x]]=0$ by the skew symmetry of the Lie bracket. By the alternating condition on the Lie bracket, we have $[y,[x,x]]=0$.

(2) implies (3)

Given: A Lie ring $L$ such that any triple Lie product with a repeated variable in $L$ is zero.

To prove: For any $x,y\in L$, the subring generated by $x,y$ has nilpotency class at most two.

Proof: The subring generated by $x,y$ is the additive subgroup generated by all iterated Lie products of $x,y$. It thus suffices to show that all triple Lie products involving only $x,y$ equal zero. Note that a triple product involving only $x,y$ must necessarily involve a repeated variabl, hence by the given, must be zero, completing the proof.

(3) implies (4)

Given: $L$ is a Lie ring such that the triple Lie product $(x,y,z)\mapsto [x,[y,z]]$ is alternating in all three variables.

To prove: For any $x,y,z$, $[x,[y,z]]=[y,[z,x]]$ (note that the rest of cyclic symmetry follows by repeated application).

Proof (simple): By Fact (1), alternation in the first two variables gives that:

$[x,[y,z]]+[y,[x,z]]=0$

Alternation in the second and third variable gives:

$[y,[x,z]]+[y,[z,x]]=0$

Combining, we get:

$[x,[y,z]]=[y,[z,x]]$

Proof (sophisticated): Since the function is alternating in all variables, we obtain that for any permutation of the set $\{1,2,3\}$, we have:

$[x_{\sigma (1)},[x_{\sigma (2)},x_{\sigma (3)}]]=\operatorname {sgn} (\sigma )[x_{1},[x_{2},x_{3}]]$

Apply this to $\sigma =(1,2,3)$, the 3-cycle in symmetric group:S3, to get:

$[x_{2},[x_{3},x_{1}]]=[x_{1},[x_{2},x_{3}]]$

Now write $x=x_{1},y=x_{2},z=x_{3}$.

(4) implies (1)

Given: A Lie ring $L$ such that $[x,[y,z]]=[y,[z,x]]=[z,[x,y]]$ for all $x,y,z\in L$

To prove: $[a,[a,b]]=0$ for all $a,b\in L$

Proof (simple): Set $x=b,y=z=a$. The first equality yields:

$[b,[a,a]]=[a,[a,b]]$

The first expression is zero, hence so is the second expression.

Proof (sophisticated): We are trying to reverse the polarization trick here. In general, reversing the polarization trick requires a torsion-free condition: skew-symmetric does not imply alternating. In this case, however, we already have genuine alternation in the pair of the second and third variable, and this is sufficient.

Equivalent conditions (1)-(4) imply (5)

Given: A Lie ring satisfying equivalent conditions (1)-(4). An element $z\in L$

To prove: The ideal of $L$ generated by $z$ is abelian.

Proof: This is somewhat tricky. The first step is to show that the ideal generated by $z$ is generated, as an additive group, by $z$ and $\{[l,z]\mid l\in L\}$. Then, we note that the bracket of any two elements in this generating set is zero, hence the ideal is abelian.

Step no.	Assertion/construction	Explanation
1	For any $x,y\in L$, the element $[x,[y,z]]$ in the form $[l,z]$ for some $l\in L$.	By condition (4), it equals the element $[z,[x,y]]$, which by skew symmetry (twice) is $[[y,x],z]$.
2	Any iterated Lie product that involves $z$ can be rewritten in the form $[l,z]$ for some $l\in L$.	Using skew symmetry, the bracket is of the form $[x_{1},[x_{2},\dots ,[x_{n},z]\dots ]]$. Use Step (1) on the inside repeatedly to keep shortening the length of the bracket, until it has length one.
3	The ideal generated by $z$ is generated, as an additive group, by $z$ and $\{[l,z]\mid l\in L\}$.	Follows from Step (2), plus the observation that the elements described in Step (2), along with $z$ itself, generate $I$ as an additive group.
4	The ideal generated by $z$ has a generating set of elements whose pairwise brackets are zero	The pairwise brackets possible are of the form $[z,z]$ (which is zero by alternation), $[z,[l,z]]$ (which is zero by version (2) of the 2-Engel definition), and $[[x,z],[y,z]]$. For the last type of expression, first use skew symmetry to rewrite it as $-[[z,x],[y,z]]$. Then, use cyclic symmetry (condition (4)) in $[z,x],z,y$ to rewrite as $-[y,[z,[z,x]]]$. This is zero by the 2-Engel condition between $z$ and $x$.
5	The result now follows from Steps (3) and (4).

(5) implies (1)

Given: A Lie ring $L$ such that for every $x\in L$, there is an abelian ideal $I$ of $L$ containing $x$.

To prove: $[a,[a,b]]=0$ for all $a,b\in L$.

Proof: Set $x=a$ and consider the abelian ideal $I$. Clearly, both $a$ and $[a,b]$ are in $I$. Since $I$ is abelian, we must have $[a,[a,b]]=0$.