Basis-conjugating automorphisms of a free group and associated Lie algebras

Abstract

Let F n = x 1 , . . . , x n denote the free group with generators {x 1 , . . . , x n }. Nielsen and Magnus described generators for the kernel of the canonical epimorphism from the automorphism group of F n to the general linear group over the integers. In particular among them are the automorphisms χ k,i which conjugate the generator x k by the generator x i leaving the x j fixed for j = k. A computation of the cohomology ring as well as the Lie algebra obtained from the descending central series of the group generated by χ k,i for i < k is given here. Partial results are obtained for the group generated by all χ k,i . 20F28; 20F40, 20J06

On Certain Questions of the Free Group Automorphisms Theory

“…Comparing Alexander polynomials of two poly-free groups Cb + 4 and P 4 we prove that these groups are not isomorphic, despite the fact that they have a lot of common properties. This answers the question of Cohen-Pakianathan-Vershinin-Wu from [8]. The questions of linearity of subgroups of Aut(F n ) are considered.…”

“…However, in the case of the group Cb + n , the situation is different. Lie algebras and cohomology rings of groups Cb + n were described in [8]. It was also shown in [8] that the group Cb + n is isomorphic to the pure braid group P n for n = 2, 3.…”

“…For instance, the groups Cb + n and P n are stably isomorphic, namely, suspensions over their classifying spaces ΣK(Cb + n , 1) and ΣK(P n , 1) are homotopically equivalent for all n ≥ 1 (see [8]). The conjecture that Cb + n and P n are isomorphic for all n ≥ 1 comes naturally (question 1, [8]). One of the main results of this paper is the proof that the groups Cb + 4 and P 4 are non-isomorphic (Theorem 1).…”

A reduction of the target of the Johnson homomorphisms of the automorphism group of a free group

“…Recently, Cohen-Pakianathan [3,4], Farb [5] and Kawazumi [12] independently showed that the abelianization of IA n is a free abelian group, and the Magnus generators above induce a basis of it. More precisely, they showed…”

“…It was conjectured that A n (k) = A n (k) for each k ≥ 1 by Andreadakis, who showed that A 2 (k) = A 2 (k) for each k ≥ 1 and A 3 (3) = A 3 (3) in [1]. Now, we have A n (2) = A n (2) due to Cohen-Pakianathan [3,4], Farb [5] and Kawazumi [12]. (See (3) below.)…”