We estimate Gröbner-Shirshov bases for the Lie algebra A n given arbitrary orders of generators (nodes of a Dynkin graph). Previously, the Gröbner-Shirshov basis was computed in [1] for the particular case where nodes of the Dynkin graph are ordered successively.
REGULAR WORDSWe generalize results of [1]. In what follows X is a finite linearly ordered set (of letters); L(X) is a Lie algebra freely generated by X over any field of characteristic zero; Words(X) is a set of all associative words in X (including the empty word 1).We extend the linear ordering on X to a least order on the set of all non-empty associative words as follows:if y z then xy xz; moreover, if y = z then ya zb for any associative words x, y, z, a, b (x, a, and b may be empty). A word and its proper initial subword under are incomparable.We extend the linear ordering on X to a linear order ≤ on Words(X) as follows: for any words y and z in Words(X), the inequality y ≤ z is equivalent to the fact that either z is an initial subword of y (in particular, for z = 1), or y z. Remark 1.1. For arbitrary associative words x and y, x ≺ y implies x < y. If x < y then either x ≺ y, or y is a proper initial subword of x. In particular, if x and y are equal in length, then x < y and x ≺ y are equivalent.If an associative word u is strictly larger than any one of its cyclic permutations, that is, wv < u for u = vw with v, w = 1, then u is called a regular word (see [2]).Let Reg(X) be a set of regular associative words, Words * (X) be a set of non-associative words, and b − : Words * (X) → Words(X) be the parentheses-removing map. According to [2], a non-associative word u is regular if it is a letter, or the following conditions are satisfied:(1) an associative word obtained by removing the parentheses in u is regular;(2) if u = xy, for non-associative words x and y of non-zero length, then x and y are regular nonassociative words; *
