19
$\begingroup$

The following algebraic structure came up when I was thinking about invariants of coloured knots. The elements are all elements of a noncommutative free group $F$, and the operations are:

  1. $a^b= b^{-1}ab$, taking the conjugate in $F$.
  2. $[a,b]= aba^{-1}b^{-1}$, taking the commutator of two elements in $F$.

And that's all. (If I were allowing only conjugation then this structure would be a conjugation quandle, but I'm also allowing to take commutators, but not to take products- the group product is not part of the structure). Are such structures at all studied or known?

Question: Is the full set of relations in this structure known (in the sense of universal algebra)? Is there a proof in the literature?
Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Have you looked up the "Lie ring method" in group theory? It seems related. You have an identity something like $[a,b^{-1},c]^{b}[b,c^{-1},a]^{c}[c,a^{-1},b]^{a}] = 1$, which is reminiscent of the Jacobi identity ( I may have got the formula wrong from memory, but it can be found in group theory texts, and may be due to P. Hall). Taking the direct sum of the quotients of successive terms of the lower central series gives a Lie ring like structure which has been looked at in the literature. $\endgroup$
    – Geoff Robinson
    Commented Dec 4, 2012 at 8:09
  • $\begingroup$ IIRC that identity is due to Witt (or both?) (You get from it to Jacobi's by setting $a=\exp(At)$ and so on and computing the third derivative of the left hand side of he Witt identity) $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 4, 2012 at 12:36
  • $\begingroup$ I don't think I have Witt-Hall because I can't multiply. I can only conjugate and take commutators. $\endgroup$
    – Daniel Moskovich
    Commented Dec 4, 2012 at 13:58
  • 1
    $\begingroup$ It's probably not quite what you are asking, but I'll mention it anyway: A lot of the relations among commutators (and with conjugation, and especially the interaction of commutators and powers) are considered in the study of "commutator collection" and "basic commutators". The big work on them is Ward's "Basic Commutators", Philos. Trans. Roy. Soc. London Series A, vol 264 (1969), 343-412, MR 0251148 $\endgroup$
    – Arturo Magidin
    Commented Dec 5, 2012 at 6:51
  • 1
    $\begingroup$ If I understand correctly, the question is not yet answered? $\endgroup$
    – YCor
    Commented Sep 3, 2019 at 8:10

1 Answer 1

Reset to default
12
$\begingroup$

There is a notion of multiplicative Lie algebras introduced here: Ellis, Graham J. On five well-known commutator identities. J. Austral. Math. Soc. Ser. A 54 (1993), no. 1, 1–19. The signature there does include multiplication, though. The problem of finding axioms was solved there (I think somebody finally proved that the five standard commutator identities suffice). If the product operation is removed from the signature, the correct first question would be if the class is first order axiomatizable. It is not clear. One can write a bunch of axioms which certainly hold, but this list is not complete:

  1. $[x,x]=[y,y]$ (call this element 1)

  2. $[x,1]=[1,x]=1$

  3. $[[x,y], [y,x]]=1$

  4. $[x,y]=1, [x,z]=1 \to [x,[y,z]]=1$

  5. $[x,y]^z=[x^z,y^z]$

  6. $x^z=x \leftrightarrow [x,z]=1$

  7. $\exists x,y, [x,y]\ne 1 \to \exists x \exists a, [x,a]=1, a\ne 1, a\ne x$

(The last axiom follows from the fact that if all elements of a group are of order 2, then the group is Abelian.)

I think it is clear that the class cannot be axiomatized only by universal formulas (because it is not closed under taking subalgebras). It is closed under taking ultraproducts, which is a good news.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .