3
$\begingroup$

I am reading Combinatorial Group Theory In Homotopy Theory I by Fred Cohen (preprint available on web page). Here is an extract of the paper:

enter image description here

Cohen called $A^R_n$ "a standard tool used in combinatorial group theory ..." but I have not read about this non-commutative analogue of exterior algebras anywhere else. Besides Cohens paper, is there any other reference to this algebra? I will be interested in the special case when $R=\mathbb{Z}$; if there is any reference on this algebra and in particular the automorphism group $\mathrm{Aut}(A^\mathbb{Z}_n)$ it would be great.

$\endgroup$

1 Answer 1

3
$\begingroup$

Cohen has already discussed such noncommutative analogues of exterior algebras much earlier, e.g., in the book Handbook of Algebraic Topology from $1995$, see the section before Theorem $13.1$. Note that this reference is not cited in Cohen's preprint "Combinatorial Group Theory In Homotopy Theory I".

Another reference is the paper Natural transformations of tensor algebras and representations of combinatorial groups by Grbic, and Wu; but again this is based on Cohen's definition.

$\endgroup$
5
  • 1
    $\begingroup$ Thank you for the reference! Is it really (as Cohen said) a standard tool in combinatorial group theory? I seldom see it except in some papers by a few algebraic topologists such as F.R. Cohen and J. Wu. $\endgroup$
    – Zuriel
    Dec 11, 2014 at 15:06
  • $\begingroup$ This depends on the interpretation of the term "combinatorial group theory", which is rather general. $\endgroup$ Dec 11, 2014 at 15:41
  • $\begingroup$ So can I assume that this algebra is not found in most (all) text books on combinatorial group theory? $\endgroup$
    – Zuriel
    Dec 11, 2014 at 15:43
  • 1
    $\begingroup$ You can check this yourself in the literature. For example, look at the "Recommended Literature" on combinatorial group theory here. $\endgroup$ Dec 11, 2014 at 15:54
  • $\begingroup$ Many thanks again for your kind assistance!! $\endgroup$
    – Zuriel
    Dec 11, 2014 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.