Does a four-variable analog of the Hall-Witt identity exist?

Lately I have been thinking about commutator formulas, sparked by rereading the following paragraph in Isaacs (p.125):

An amazing commutator formula is the Hall-Witt identity: [x,y1,z]y[y,z1,x]z[z,x1,y]x=1, which holds for any three elements of every group. One can think of the Hall-Witt formula as a kind of three-variable version of the much more elementary two-variable identity, [x,y][y,x]=1. This observation hints at the possibility that a corresponding four-variable formula might exist, but if there is such a four-variable identity, it has yet to be discovered.

To my knowledge a four-variable formula hasn’t been discovered since this was written. I was thinking about this and found myself unable to even decide whether or not I thought one could exist (i.e. whether one would, hypothetically, try to find an identity or disprove its existence). Thus, I am attempting to “write down the problem.”

How can one rigorously formulate the question, “Does a four-variable analog of the Hall-Witt identity exist?”

Let’s start by explicitly making the free group on 4 letters. Suppose we have a free generating set A={a,b,c,d}, so that S=AA1 and A1={a1,b1,c1,d1}. Let FA the group of all reduced words in S. Words are reduced when they have been can no longer be simplified by cancelling adjacent x and x1s (for xA).

Cyclically reduced words are words where the first and last letters are not inverse to each other. Every word is conjugate to a cyclically reduced word, so we can consider ^FA, the quotient of FA by the equivalence relation of being cyclically reduced. (Note that this is not the equivalence relation of being conjugate.)

Let Φ be the set of functions φ:A4^FA defining words in ^FA that contain at least one instance of each free generator or its inverse. Now let Ψ:Φ^FA be formally defined by Ψ(φ)(u,x,y,z)=φ(u,x,y,z)φ(x,y,z,u)φ(y,z,u,x)φ(z,u,x,y).
So, if a 4-variable Hall-Witt identity exists, it will be among the functions in the preimage of 0 (the function which just maps everything to the empty word) under Ψ.

Question: Assuming the above formulation is sound, does there exist a (nontrivial) four-variable analog to the Hall-Witt identity? Can this approach be used to further refine the question?

My use of “nontrivial” above is somewhat ambiguous: what I mean is that the identity should be made with commutators and conjugations in a free set of four letters, and done so in a way that it does not reduce to the two- or three-variable commutator identities by a substitution.


Progress.

Let W(a,b,c) Let’s chop W(a,b,c) in half and name the two parts. Define w_1(a,b,c)\triangleq a^{-1}b^{-1}ac^{-1}a^{-1} \hspace{30pt} \text{and} \hspace{30pt} w_2(a,b,c)\triangleq bab^{-1}cb, so that W(a,b,c)=w_1(a,b,c)w_2(a,b,c). Now, the Hall-Witt Identity can be written as W(x,y,z)W(y,z,x)W(z,x,y)=1, that is, \overbrace{\underbrace{x^{-1}y^{-1}xz^{-1}x^{-1}}_{w_1(x,y,z)}\underbrace{yxy^{-1}zy}_{w_2(x,y,z)}}^{W(x,y,z)} \overbrace{\underbrace{y^{-1}z^{-1}yx^{-1}y^{-1}}_{w_1(y,z,x)}\underbrace{zyz^{-1}xz}_{w_2(y,z,x)}}^{W(y,z,x)} \overbrace{\underbrace{z^{-1}x^{-1}zy^{-1}z^{-1}}_{w_1(z,x,y)}\underbrace{xzx^{-1}yx}_{w_2(z,x,y)}}^{W(z,x,y)} =1.
It’s clear the cancellation works by w_1(b,c,a)=w_2(a,b,c)^{-1}. This makes sense: we should be able to cyclically permute the overall word and have it still work, since 1^a=1. So, what we should be looking for are four letter strings w_1 and w_2 such that w_1(a,b,c,d) is the inverse of w_2(b,c,d,a).

Update:
In fact, the above observation is not just sufficient, but necessary for the existence of a four-variable Hall-Witt identity. Consider for example the case where W(a,b,c,d) would be split into three subwords, rather than two: W(a,b,c,d)=w_1(a,b,c,d)w_2(a,b,c,d)w_3(a,b,c,d). We would in this case have to insist that w_3(a,b,c,d)=w_1(b,c,d,a)^{-1}\hspace{14pt}\text{ and }\hspace{14pt}w_2(a,b,c,d)=w_2(b,c,d,a)^{-1}. After the w_1‘s and w_3‘s cancelled out, we’d be left with w_2(x,y,z,u)w_2(y,z,u,x)w_2(z,u,x,y)w_2(u,x,y,z)=1. But of course asking whether that that can happen is the same as asking whether W can exist, so eventually we will see a separation into two subwords, or we have a contradiction by infinite descent. If we divided up W(a,b,c,d) into n>3 subwords, we would reduce in the case of odd n to the case of a self-inverse word analogously to the n=3 case, or we would have an even number of subwords, which is the same thing as the 2 subword case.

So, it suffices to either find a word W(a,b,c,d)=w_1(a,b,c,d)w_2(a,b,c,d), nontrivial in each variable, such that w_1(a,b,c,d)=w_2(b,c,d,a)^{-1}, or to prove that such a word does not exist.

Answer

\renewcommand{\mod}[1]{~(\text{mod $#1$})}
\newcommand{congr}{\equiv}
The last statement in your question (which is not a mere question)
also suggests a solution of the more general problem,
namely the n-variable analog of the Hall-Witt formula for any n\geq 2.

Let x_1, \ldots, x_n be the variables.
If w=w(x_1,x_2,\ldots,x_n) is any word in x_i‘s and their inverses,
define the word \gamma w by
(\gamma w)(x_1,x_2,\ldots,x_n):=w(x_2,\ldots,x_n,x_1),
and set W:=w(\gamma w)^{-1}. Then
W(\gamma W)\cdots(\gamma^{n-1}W)=1~.
The requirement that W is nontrivial in each variable is easily satisfied.
In this way you can produce n-variable analogs of the Hall-Witt formula by truckload.

To make things more interesting, you may require, say,
that W must be representable by an expression
built from the variables and their inverses (the ‘tokens’)
by making commutators [u,v],
where u and v are already built expressions,
and by conjugations u^y,
where u is an already built expression, but not a token, and y is a token;
then the requirement is that W=u
for an expression built in this way that is not a single token.
Now the things are no longer so simple.
Am I far wrong in supposing that in fact you had some such additional conditions in mind
when you formulated your question?

I have no idea how to go about finding an example for n=4
or proving that it does not exist.
The only (rather weak) restriction on the word w I have found so far
is l_1=l_2=\cdots=l_n,
where l_i is the sum of the exponents
of the appearances of x_i^{\pm1} in w.

If we relax the condition imposed on W
and require only that W\in[G,G],
where G is the free group generated by the variables x_1, \ldots, x_n,
then we can give the full solution of the problem,
since for every w\in G we have w(\gamma w)^{-1}\in[G,G]
if and only if l_1(w)=l_2(w)=\cdots=l_n(w).
Here \gamma is the automorphism of G
that sends x_i to x_{i+1} for 1\leq i<n, and sends x_n to x_1.
We must also tell how the functions l_i: G\to\mathbb{Z} are defined.
Let A be the free abelian (additive) group generated by x_1, \ldots, x_n,
and let h\colon G\to A be the homomorphism
sending x_i\in G to x_i\in A for 1\leq i\leq n;
note that \ker h=[G,G].
For every w\in G we have h(w)=\sum_{i=1}^n l_i(w)x_i: this defines the l_i's.

For example, if w=x_1^{-1}x_2^{-1}\cdots x_{n-1}^{-1}x_n^{-1},
then W=w(\gamma w)^{-1}=[x_1,x_nx_{n-1}\cdots x_2].
When n=3 we obtain the identity

[x,zy][y,xz][z,yx]=1~,

which is a humble cousin of the Hall-Witt formula,
and is probably quite useless (nice as it is).
If you wish you can rewrite W=[x,zy]=x^{-1}y^{-1}z^{-1}xzy
as a product of iterated commutators, W=[x,y][x,z][[x,z],y],
or perhaps in the well-known form W=[x,y][x,z]^y.
(Mark that here [x,y]=x^{-1}y^{-1}xy;
some authors define the commutator as [x,y]:=xyx^{-1}y^{-1}.)
Note that the example w=w_1(a,b,c,d)=a^{-1}bc^{-1}b^{-1}dad^{-1}
provided by weux082690
has l_a(w)=l_b(w)=l_d(w)=0 but l_c(w)=-1,
thus w(\gamma w)^{-1}\notin[G,G].

By a sort of a retrograde progress, let us return to the beginning.
We are going to prove the following:

Let G be the free group with the free generators x_1, \ldots, x_n, n\geq 2,
and let \gamma be the automorphism
of G that rotates the generators one place to the right,
that is, \gamma x_i=x_{i+1} for 1\leq i<n and \gamma x_n=x_1.
Then g\in G has the property that g(\gamma g)\cdots(\gamma^{n-1}g)=1
iff there exists h\in G such that g=h(\gamma h)^{-1}.

Proof. ~The sufficiency is clear.

Necessity.
Let T be the set of 'tokens' \{x_1,x_1^{-1},\ldots,x_n,x_n^{-1}\},
and let T^* denote the free monoid (of 'words') generated by T;
the neutral element of T^* is the empty word \varepsilon.
Then G=T^*/{\sim}, where \sim is the congruence on the monoid T^*
generated by x_i^\alpha x_i^{-\alpha}\sim\varepsilon, 1\leq i\leq n, \alpha=\pm1.
The equivalence class of the empty word
is the multiplicative identity of the free group: \varepsilon/{\sim}=1_G=1.
Every equivalence class g\in G contains a unique reduced word \varrho(g)
which does not contain any pair of consecutive tokens that are inverses of each other.
Given a word w\in T^*, we obtain the reduced word \varrho(w/{\sim})
by repeatedly applying the reductions ux_i^{\alpha}x_i^{-\alpha}v\to uv,
1\leq i\leq n, \alpha=\pm1, u,v\in T^*;
it is easy to verify that this system of reductions is locally confluent,
thus it has the Church-Rosser property,
and so there is in fact a unique reduced word in each equivalence class.
The rotation \gamma of generators induces the double rotation of tokens
(the rotation of the tokens with the exponent 1
and also the rotation of the tokens with the exponent -1);
the automorphism of the free monoid T^* determined by this double rotation
we still denote by \gamma.
\quadLet w=x_{i_1}^{\alpha_1}x_{i_2}^{\alpha_2}\cdots x_{i_m}^{\alpha_m}\in T^*.
We denote by |w| the length m of the word w.
For each i, 1\leq i\leq n, we denote by l_i(w) the sum of the exponents
of all tokens x_i^{\pm1} appearing in w,
and by l(w) we denote the sum of all exponents
\alpha_1+\alpha_2+\cdots+\alpha_m=l_1(w)+\cdots+l_n(w).
Always |w|\congr l(w)\mod{2},
because every term in |w|-l(w)=(1-\alpha_1)+(1-\alpha_2)+\cdots+(1-\alpha_m)
is ether 0 or 2.
For each i, 1\leq i\leq n, we have l_i(uv)=l_i(u)+l_i(v) for all u,v\in T^*,
and l_i(w) is constant on every equivalence class g\in G.
\quadNow suppose that g\in G
has the property g(\gamma g)\cdots(\gamma^{n-1} g)=1,
and let w:=\varrho(g).
Then l_n(w)+l_n(\gamma w)+\cdots+l_n(\gamma^{n-1} w)=l_n(\varepsilon)=0;
since l_n(\gamma w)=l_{n-1}(w), \ldots, l_n(\gamma^{n-1} w)=l_1(w)
we have l(w)=l_1(w)+\cdots+l_n(w)=0,
therefore the word w is of even length, |w|=2k.
We split the word w as w=w_1w_2, where |w_1|=|w_2|=k.
We are assuming that k>0, since the case k=0 is trivial.
The reduction process must reduce the word

W := w_1w_2(\gamma w_1)(\gamma w_2)(\gamma^2w_1)(\gamma^2w_2)\cdots
(\gamma^{n-1}w_1)(\gamma^{n-1}w_2)

to the empty word.
Since the words w_1w_2, (\gamma w_1)(\gamma w_2), \ldots,
(\gamma^{n-1}w_1)(\gamma^{n-1}w_2) are reduced,
the only places in the word W where the reductions can be applied
are at the points of contact between subwords w_2 and \gamma w_1,
\gamma w_2 and \gamma^2 w_1, \ldots\,
Consider the effect of the reduction process on the subword w_2(\gamma w_1)
(mark that we can carry out the reductions in any order we choose,
the result will be always the same).
The first reduction eliminates some product t\,t^{-1}
from the center of w_2(\gamma w_1),
where t is the last token in w_2 and t^{-1} is the first token in \gamma w_1.
We are left with w_2'(\gamma w_1'), where |w_1'|=|w_2'|=k-1.
If k>1, there may be another reduction applicable
at the center of w_2'(\gamma w_1') (and nowhere else in this word), so we apply it, and so on.
In fact the reductions must proceed to the bitter end,
reducing the initial word w_2(\gamma w_1) to the final empty word.
For suppose that the reduction process stops after r<k reductions;
then the reduction process, applied to each of the subwords (\gamma w_2)(\gamma^2 w_1),
\ldots, (\gamma^{n-2}w_2)(\gamma^{n-1}w_1), will likewise stop after r reductions,
and we will have a nonempty reduced word on our hands, which cannot be,
because the word W must reduce to the empty word.
Let h:=w_1/{\sim}.
Since w_2(\gamma w_1)\sim\varepsilon,
it follows that w_2/{\sim}=(\gamma h)^{-1},
whence g=(w_1/{\sim})(w_2/{\sim})=h(\gamma h)^{-1}.~ Done.

Attribution
Source : Link , Question Author : Alexander Gruber , Answer Author : chizhek

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