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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?
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1 Answer 1

up vote 9 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.

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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

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