Let $G$ be a group and set $[x,y]:= x^{-1}y^{-1}xy$ as usual and consider it as a binary operation.

Question: Is there a description of the identities that the operation $[.,.]$ satisfies for all groups?

Just to clarify, those identities should not involve ordinary group multiplication, conjugation or inversion (such as the Hall-Witt identity and various other identities) but only commutators and the neutral element.

Of course one has identities of the form $[x,x]=1$ and $[[x,y],[y,x]]=1$ (as pointed out in a comment), but there are also more complicated ones. As an example, one can check the following three-variable identity $$[ [[x,y], z],[z,[y,x]]] = [ [[x,y], z],[[x,y],[z,[y,x]]]]$$ and derive one other of similar type. Are all other identities derived from this?

  • 5
    $\begingroup$ Since $[y,x]=[x,y]^{-1}$, a simpler relation is $[[x,y],[y,x]]=1$. $\endgroup$ – Ilya Bogdanov Mar 11 '16 at 10:33
  • 4
    $\begingroup$ Relevant link: mathoverflow.net/a/81316/8430 $\endgroup$ – Suvrit Mar 11 '16 at 15:02
  • 2
    $\begingroup$ One can ask whether the set of such identities is finitely generated (under some suitable operations). Namely, let $F$ be the free magma on countably many generators. Let's say that a subset $R$ of $F\times F$ (where we think of an element $(u,v)$ of $F\times F$ as an identity) is coherent if $(u,v)\in R$ implies $(v,u)\in R$, $(wu,wv)\in R$, $(uw,vw)\in R$ for all $w\in F$, and $R$ is stable by substitution (i.e. $(u,v)\in R$ and $f$ implies $(f(u),f(v))\in R$ for every endomorphism $f$ of $R$). (...) $\endgroup$ – YCor Mar 11 '16 at 17:59
  • 2
    $\begingroup$ (...) Observe that if we define $R$ as the set of pairs $(u,v)$ such that $g(u)=g(v)$ for every group $G$ and every homomorphism $F\to (G,[.,.])$, then $R$ is exactly the set of identities you want to describe. It is a coherent subset of $F\times F$ and one question is to describe a nice generating subset of $R$ as a coherent subset and in particular if there is a generating subset. $\endgroup$ – YCor Mar 11 '16 at 18:03
  • 2
    $\begingroup$ @AndreasThom: I doubt it. In general, a fragment of a finitely axiomatized variety (equational theory) obtained by restricting the signature need not be finitely axiomatized. $\endgroup$ – Emil Jeřábek Mar 12 '16 at 9:36

The paper Commutators of flows and fields contains the following result:

  • Let $M$ be a manifold, let $\phi^i:\Bbb R\times M\supset U_{\phi^i}\to M$ be smooth mappings for $i=1,\dots,k$ where each $U_{\phi^i}$ is an open neighborhood of $\{0\}\times M$ in $\Bbb R\times M$, such that each $\phi^i_t$ is a diffeomorphism on its domain, $\phi^i_0=Id_M$, and $\frac{\partial}{\partial t}|_0\, \phi^i_t=X_i\in\mathfrak X(M)$. We put $[\phi^i_t,\phi^j_t] :=(\phi^j_t)^{-1}\circ(\phi^i_t)^{-1}\circ\phi^j_t\circ\phi^i_t.$ Then for each formal bracket expression $B$ of length $k$ we have \begin{align} 0&= \tfrac{\partial^\ell}{\partial t^\ell}|_0 B(\phi^1_t,\dots,\phi^k_t)\quad\text{ for }1\le\ell<k,\\ B(X_1,\dots,X_k)&=\tfrac1{k!} \tfrac{\partial^k}{\partial t^k}|_0 B(\phi^1_t,\dots,\phi^k_t)\in \mathfrak X(M) \end{align}

This suggests that any relation holding for Lie brackets holds also for the group commutator.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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