Nilpotency class three is 3-local for Lie rings
This article gives a proof/explanation of the equivalence of multiple definitions for the term Lie ring of nilpotency class three
View a complete list of pages giving proofs of equivalence of definitions
The following are equivalent for a Lie ring :
- The nilpotency class of is at most three, i.e., we have for all (where we allow some or all of to be equal).
- The 3-local nilpotency class of is at most three, i.e., the subring generated by any three elements of is nilpotent of class at most three.
- Polarization trick: In particular, this states that alternating implies skew-symmetric.
(1) implies (2)
This is immediate.
(2) implies (1)
The proof is incomplete.
Given: A Lie ring of 3-local nilpotency class three
To prove: for all
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||The map is alternating and multi-linear, and hence skew-symmetric, in all pairs of inputs. In particular, this means that its sign is reversed under any odd permutation of the inputs and is preserved under any even permutation of the inputs.||Fact (1)||3-local class three||For any pair of inputs, if we set them to be equal, then we have a situation of a Lie bracket of length four involving three elements, which must be zero. Hence, we have the alternating condition. Note that the alternating implies skew-symmetric deduction is a special case of Fact (1).|
|2||For all , we have .||Step (1)||By Step (1), we have . By linearity in and the Jacobi identity in , the sum is zero, hence we get the result.|
|3||For all , we have||Step (2)||Use the Jacobi identity to get . Now, the middle term is the negative of , which by the alternating condition we know to be the negative of . So, the middle term equals . Thus, the Jacobi identity expression gives . Combine with the conclusion of Step (2) to obtain that .|
|4||For all , we have||Step (3)||Interchange the roles of with , and with , in Step (3).|
|5||For all , we have||Steps (1), (3), (4)||Combine Steps (3) and (4) with the observation that, by Step (1), because the underlying permutation is even.|
|6||For all , we have||Use the skew symmetry of the Lie bracket|
|7||For all , we have||Steps (5), (6)||Step-combination direct|
|8||For all , we have||Steps (3), (7)||Step-combination direct|
|9||For all , we have||Steps (2), (8)||Step-combination direct|