What are the relations between conjugates and commutators? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T03:13:14Z http://mathoverflow.net/feeds/question/115366 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115366/what-are-the-relations-between-conjugates-and-commutators What are the relations between conjugates and commutators? Daniel Moskovich 2012-12-04T07:19:10Z 2012-12-18T11:03:25Z <p>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: </p> <ol> <li>$a^b= b^{-1}ab$, taking the conjugate in $F$.</li> <li>$[a,b]= aba^{-1}b^{-1}$, taking the commutator of two elements in $F$.</li> </ol> <p>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?</p> <blockquote> <b>Question</b>: Is the full set of relations in this structure known (in the sense of universal algebra)? Is there a proof in the literature? </blockquote> http://mathoverflow.net/questions/115366/what-are-the-relations-between-conjugates-and-commutators/116693#116693 Answer by Mark Sapir for What are the relations between conjugates and commutators? Mark Sapir 2012-12-18T11:03:25Z 2012-12-18T11:03:25Z <p>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:</p> <ol> <li><p>$[x,x]=[y,y]$ (call this element 1)</p></li> <li><p>$[x,1]=[1,x]=1$</p></li> <li><p>$[[x,y], [y,x]]=1$</p></li> <li><p>$[x,y]=1, [x,z]=1 \to [x,[y,z]]=1$</p></li> <li><p>$[x,y]^z=[x^z,y^z]$</p></li> <li><p>$x^z=x \leftrightarrow [x,z]=1$</p></li> <li><p>$\exists x,y, [x,y]\ne 1 \to \exists x \exists a, [x,a]=1, a\ne 1, a\ne x$</p></li> </ol> <p>(The last axiom follows from the fact that if all elements of a group are of order 2, then the group is Abelian.) </p> <p>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. </p>