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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?

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    $\begingroup$ Since $[y,x]=[x,y]^{-1}$, a simpler relation is $[[x,y],[y,x]]=1$. $\endgroup$ Mar 11, 2016 at 10:33
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    $\begingroup$ Relevant link: mathoverflow.net/a/81316/8430 $\endgroup$
    – Suvrit
    Mar 11, 2016 at 15:02
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    $\begingroup$ 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, 2016 at 17:59
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    $\begingroup$ (...) 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, 2016 at 18:03
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    $\begingroup$ @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. $\endgroup$ Mar 12, 2016 at 9:36

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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.

#Edit: I am not sure about the suggestion above. Cor.12 of the paper points in this direction.

A more precise conclusion is: The algebraic structure for the group commutator (alone) has as quotient the algebraic structure of Lie brackets (involving brackets alone). What is the kernel? Is there a kernel?

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  • $\begingroup$ It seems like the converse should be true, isn't it? If you differentiate an equality of formal expression of group commutator you obtain the corresponding relation for the Lie algebra. This suggests that for lie groups the relations are exactly the ones coming from Lie Algebras. Am I wrong? $\endgroup$ May 26, 2021 at 13:13
  • $\begingroup$ The formula I know as a translation of the Jacobi identity into group commutators involves conjugates (and says that the product of three triple commutators is 1). I'd be curious if one can do better. $\endgroup$
    – YCor
    May 28, 2021 at 8:28
  • $\begingroup$ @YCor: The Jacobi identity involves also sums, which we should not use in answering this question. So my theorem does not apply. The OP could allow a group operation which leads to addition like Trotter's fomula or BCH. Is there a cancellation which leads from reapted BCH applied to group commutators to the formula you mentioned? $\endgroup$ May 28, 2021 at 9:56

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