Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Although this question is mostly out of curiosity (as of now), I hope it is nevertheless suitable for MO.

This very recent (and still open) question about the Hall-Witt identity led me to wonder:

  1. Does some related generalization to 4 or more variables exist?

And also,

  1. Does that then imply a 4 variable (or more) Jacobi like identity?
share|improve this question

1 Answer 1

up vote 10 down vote accepted

There is a sense in which there are no further identities. Let me explain.

In Section 5.2.3 of my paper "An infinite presentation of the Torelli group" (available on my webpage), I construct a sort of presentation of the commutator subgroup of a free group where relations are relations like the Witt-Hall relations. I actually was interested in the commutator subgroup as a subgroup of the fundamental group of a surface, so my generators are all things of the form $[x,y$] where $x$ and $y$ are simple closed curves that only intersect at the basepoint. However, I'm fairly certain you could adapt the proof there to prove the following.

Fix a nonabelian free group $F$. Let $S$ be the set of all elements of the form $[x,y]$, where $x$ and $y$ form part of a basis (any basis) for $F$. Observe that $S$ is infinite. Let $R$ be the set of all relations of the following forms. To save notation, we will denote by $[x,y]^w$ the element $[w^{-1}xw,w^{-1}yw] \in S$, where $[x,y] \in S$ and $w \in F$ is arbitrary.

  1. $[x,y]=[y,x]^{-1}$ if $[x,y] \in S$.

  2. $[x,y]=[z,w]$ if $[x,y] \in S$ and $[z,w] \in S$ happen to already be equal in $F$ (for example, $[yx,y] = [x,y]$).

  3. $[z,w]^{-1} [x,y] [z,w] = [x,y]^{[z,w]}$ if $[x,y] \in S$ and $[z,w] \in S$.

  4. $[xz,y] = [x,y]^z [z,y]$ if $x$ and $y$ and $z$ form part of a basis (any basis) for $F$.

  5. $[x,y]^z = [z,x] [z,y]^x [x,y] [x,z]^y [y,z]$ if $x$ and $y$ and $z$ form part of a basis (any basis) for $F.

This 5th relation is a variant on the usual Jacobi identity (put into a form useful for this presentation). The conclusion is then that $[F,F]$ has the presentation with generators $S$ and relations $R$.

The point of all this is that there are no other "deeper" commutator identities that are not consequences of the ones you know.

share|improve this answer
Thanks Andy for your knowledgeable answer. –  Suvrit Nov 19 '11 at 23:00
The last statement does not answer the question, which asks for commutator identities of certain form. Let $G$ be the free group with the free generators $x_1$, $\ldots$, $x_n$, and let $\gamma$ be the automorphism of $G$ rotating the generators, $x_1\mapsto x_2\mapsto\cdots\mapsto x_n\mapsto x_1$. The desired identities are $W(\gamma W)\cdots(\gamma^{n-1}W)=1$, where $W\in[G,G]$, with perhaps additional constraints on $W$. The answer for $W\in[G,G]$ is that $W=w(\gamma w)^{-1}$, where $l_1(w)=\cdots=l_n(w)$ and $l_i(w)$ is the sum of the exponents of $x_i^{\pm1}$ appearing in $w$. –  chizhek Aug 4 at 9:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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