Equivalence of definitions of 2-Engel Lie ring

Statement
The following are equivalent for a Lie ring:

Facts used

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

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.

(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 $$[a,[a,b]] = 0$$.