# Identities of commutators

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?

• Since $[y,x]=[x,y]^{-1}$, a simpler relation is $[[x,y],[y,x]]=1$. – Ilya Bogdanov Mar 11 '16 at 10:33
• Relevant link: mathoverflow.net/a/81316/8430 – Suvrit Mar 11 '16 at 15:02
• 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 $R$). (...) – YCor Mar 11 '16 at 17:59
• (...) 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. – YCor Mar 11 '16 at 18:03
• @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. – Emil Jeřábek Mar 12 '16 at 9:36

## 1 Answer

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.