I have a question about some quotient groups of the lower central series of a free group.

When there's a free group $F = \langle x_1,\cdots, x_n, y_1, \cdots, y_m\rangle $,

let $A$ be the subgroup generated by elements of the forms $[x_i, a]$ in $F_{k+1}/F_{k+2}$, with $1\leq i\leq n$ and $a\in F_{k}/F_{k+1}$; and $B$ be the subgroup generated by those $[y_j, b]$ of $F_{k+1}/F_{k+2}$, with $1\leq j\leq m$ and $b\in F_k/F_{k+1}$.

(May consider when $k\geq 2$.)

In this case, $A \cap B = 1$?

add. What is the rank of $A$ (and $B$) in the free abelian group $F_{k+1}/F_{k+2}$?

Thank you for your reading, considering and answering.

  • $\begingroup$ I added one more question~ $\endgroup$ – qkqh May 29 '13 at 20:06

I don't believe so.

I will associate my commutators on the left, so that $[a,b,c]=[[a,b],c]$; and use $[x,y]=x^{-1}y^{-1}xy$; I'm not sure it matters, but that's what I'm used to.

Say $k=2$; in $F/F_{4}$, the Hall-Witt identity becomes $$ [r,s,t][s,t,r][t,r,s]\equiv 1\pmod{F_{4}}.$$

Now consider the case where $a_1=[y_2,y_1]$, $b_1=[y_2,x_1]$, $b_2=[y_1,x_1]$. Then $$ [a_1,x_1] \equiv [y_2,y_1,x_1] \equiv [y_2,x_1,y_1][y_1,x_1,y_2]^{-1} \equiv [b_1,y_1][b_2,y_2]^{-1}\pmod{F_4}.$$ Now note that $[a_1,x_1]\in A$, but $[b_1,y_1][b_2,y_2]\in B$.

Similar problems are likely to arise with other values of $k$.

Edited: It seems your question was different from what I understood; rather, your $A$ is generated by all commutators of the form $[x_i,a]$ with $a\in F_k$ (modulo $F_{k+2}$), and your $B$ by all commutators of the form $[y_j,b]$ with $b\in F_k$ (modulo $F_{k+2}$).

$F_{k}/F_{k+1}$ is free abelian, with basis given by the basic commutators of weight $k$ on $x_1,\ldots,x_n,y_1,\ldots,y_m$; by the Witt Formula, the rank is $$M_{n+m}(k)=\frac{1}{k}\sum_{d|k}\mu(d)(n+m)^{k/d}$$ where $\mu$ is the Möbius function. The rank of $A$ is at most $nM_{n+m}k$, and the rank of $B$ is at most $mM_{n+m}(k)$. But in general they will be much smaller, since commutators of the form $[c,x_j]$ and $[c,y_j]$, where $c$ is a basic commutator of weight $k$, are seldom basic themselves; so you will have a fair amount of "collision". It would take some rather careful analysis with basic commutator collection to figure it out exactly. I would suggest looking at

  • Ward, James. Basic commutators. Philosoph. Trans. Roy. Soc. London Ser. A 264 (1969), 343-412, MR 0251148

to see if by any chance Ward has already computed these (or for information on techniques for computing it).


Let's look at the case of $k=2$. $F_2/F_3$ is freely generated by commutators of the forms $[x_j,x_i]$, $1\leq i\lt j\leq n$, $[y_s,y_r]$, $1\leq r\lt s\leq m$, and $[y_t,x_k]$, $1\leq k\leq n$ and $1\leq t\leq m$.

On the other hand, $F_3/F_4$ is freely generated by commutators of the forms

  1. $[x_j,x_i,x_k]$ with $1\leq i\lt j\leq n$, $i\leq k\leq n$;
  2. $[x_j,x_i,y_t]$ with $1\leq i\lt j \leq n$, $1\leq t\leq m$;
  3. $[y_t,x_k,x_i]$ with $1\leq k\leq i\leq n$, $1\leq t\leq m$;
  4. $[y_t,x_k,y_v]$ with $1\leq k\leq n$, $1\leq t\leq m$, $1\leq v\leq m$; and
  5. $[y_s,y_r,y_t]$ with $1\leq r\lt s\leq m$, $r\leq t\leq m$.

For $A$, we will obtain all commutators of type 1 and all commutators of type 3; no commutators of type 5; when we consider a commutator of the form $[y_s,y_r]$ and take the commutator with $x_i$, we obtain $[y_s,y_r,x_i] \equiv [y_s,x_i,y_r][y_r,x_i,y_s]^{-1}$. So commutators of type 4 are paired off, except we do not get the ones in which $t=v$. Finally, we obtain all commutators of type 2 because when we take $[y_t,x_j]$ for $a$, and take the commutator with $x_i$, we will obtain $[y_t,x_j,x_i]\equiv [y_t,x_i,x_j][x_j,x_i,y_t]^{-1}$; and since $[y_t,x_i,x_j]\in A$, we obtain the commutator $[x_j,x_i,y_t]$.

So, what is the rank of $A$? We have $\binom{n}{3}+2\binom{n}{2}$ commutators of type 1; we have $m\binom{n}{2}$ commutators of type 2 and $m\binom{n}{2}+mn$ of type 3; and we have $n\binom{m}{2}$ generators corresponding to commutators of type 4. This gives $$\mathrm{rank}(A) = \binom{n+1}{3} + (2m+1)\binom{n}{2}+n\binom{m+1}{2}.$$

As for $B$, the argument is similar: we get no commutators of type 1, all commutators of type 5 (of which there are $\binom{m}{3}+2\binom{m}{2}$); all commutators of type 2 (of which there are $m\binom{n}{2}$); we also get all of type 4, which gives $nm^2$ commutators; as for type 3, these cannot occur from expressions of the form $[b,y_j]$ with $b\in F_2/F_3$. So the rank of $B$ will be $$\mathrm{rank}(B) = \binom{m+1}{3}+\binom{m}{2} + m\binom{n}{2} + nm^2.$$

Finally, for the overlap: according to the Witt formula, the rank of $F_3/F_4$ is $$\frac{1}{3}(n+m)\left( (n+m)^2 - 1\right).$$ So this allows you to compute the rank of the overlap.

For larger values of $k$ the computations become more difficult, because not all basic commutators are of the form $[c,x_i]$ or $[c,y_j]$. But this should give you an idea of the kind of thing that is going on (and why I don't want to try doing this for $k=3$).

  • $\begingroup$ Oh, thank you for your kind answer. My $A$ means not a subgroup generated by $n$ element with some fixed $a_i$'s, but the one generated by all elements of that form $[x_i,a_i]$ where $a_i$'s range over $F_k/ F_{k+1}$. so, $A$ doesn't depend on the choice of $a_i$. sorry to give confusion. $\endgroup$ – qkqh May 30 '13 at 9:00
  • $\begingroup$ @qkqh: That's lousy notation, then. You are really looking at $\langle [F_k,x_1],\ldots,[F_k,x_n]\rangle\bmod F_{k+2}$, and similarly for $B$. I'll have to think about it. $\endgroup$ – Arturo Magidin May 30 '13 at 14:55
  • $\begingroup$ Thank you for your deep considering and editing my deficient questions. It seems to have small mistakes. In my thought, for $A$, there are $2n\binom{m}{2}$ commutators of type 4, and $B$ has $m\binom{n}{2}$ commutators of type 3, so \begin{eqnarray*} \textrm{rank}(B) &=& \binom{m+1}{3} + \binom{m}{2} + 2m\binom{n}{2} + nm^2 \\ &=& \binom{m+1}{3} + (2n+1)\binom{m}{2} + mn^2 \end{eqnarray*} Am I right? $\endgroup$ – qkqh Jun 9 '13 at 15:01
  • $\begingroup$ @qkqh: I'm wrong in type $4$ (there's no factor of $\frac{1}{2}$, but how do you get $2n\binom{m}{2}$ commutators of type 4? You choose two indices between $1$ and $m$ for the $y$s, which gives $\binom{m}{2}$; then you choose the index for $x$, which has $n$ possibilities. For this choice, we get the generator $[y_t,x_k,y_s][y_s,x_k,y_t]^{-1}$. So it's only $n\binom{m}{2}$. And I do not see how you get any commutators of type 3 in $B$ at all. $\endgroup$ – Arturo Magidin Jun 9 '13 at 19:57
  • $\begingroup$ Yes, you are right. I have wanted that rank$(A)$ and rank$(B)$ are symmetric (i.e. by changing $n$ and $m$, we get the other), and now it's right. Thank you $\endgroup$ – qkqh Jun 11 '13 at 3:17

Not necessarily. By the Jacobi Identity, in $F/F_4$ we have for example

$[x_1,[x_2,y_1]][x_2,[y_1,x_1] = [y_1,[x_1,x_2]]^{-1}$.

For your second question, the rank of $A$ is clearly at most $n$. For $k=2$, you could choose $a_i=x_i$, so it could be anything between 0 and $n$. For $k>2$, the relation above shows that it can be strictly less than $n$. It is possible that there is some lower bound in this case.

  • $\begingroup$ Thank you, I understand the first answer. And sorry, too. $\endgroup$ – qkqh May 30 '13 at 9:03
  • $\begingroup$ I have edit my definition of $A$ and $B$. $\endgroup$ – qkqh May 30 '13 at 9:10

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