This is a reference request, since the answer is probably well known, but I could not find it.

Given a finitely generated group $\Gamma$ with a generating set $S$, define the word norm $l = l_S : \Gamma \rightarrow \mathbb{N}$ to be $$l (g) = \min \lbrace k : \exists s_1,...,s_k \in S, g = s_1 ... s_k \rbrace ,$$ i.e., $l(g)$ is the distance between $e$ and $g$ in the Cayley graph of $\Gamma$ (w.r.t the generating set $S$).

My question: Let $\Gamma = \rm SL_3 (\mathbb{Z})$ (with some finite generating set). For $1 \leq i,j, \leq 3, i \neq j$ and $m \in \mathbb{Z}$, denote $e_{i,j} (m)$ to be the elementary matrix with $1$'s along the main diagonal, $m$ in the $(i,j)$-entry and $0$ in all other entries. What can one say about the growth rate of $l (e_{i,j} (m))$?

My naive attempt for an answer gives me $l (e_{i,j} (m)) = O (\log^3 (m))$:

  1. For convenience, we fix the generating set $S = \lbrace e_{i,j} (\pm 1), e_{i,j} (\pm 2) : 1 \leq i,j, \leq 3, i \neq j \rbrace$.

  2. Using commutator (Steinberg) relations $$ e_{i,j} (2^{2^{r+1}}) = [e_{i,k} (2^{2^{r}}), e_{k,j} ( 2^{2^{r}})]$$ it is not hard to show by induction on $r$ that $l (e_{i,j} (2^{2^r})) \leq 4^{r} $.

  3. Again by the commutator relations, it follows that for every $2^r \leq d < 2^{r+1}$ it holds that $l (e_{i,j} (2^{d})) \leq 4^{r+1} \leq 4 d^2$.

  4. Thus it follows that for every $2^d \leq m < 2^{d+1}$,
    $$l (e_{i,j} (m)) \leq \sum_{t =0}^d l (e_{i,j} (2^t)) \leq 4 \sum_{t =0}^d t^2 = 4 \frac{d (d+1) (2d +1)}{6} = O (\log^3 (m)).$$

As noted above - this computation is quite naive. Are there better known results?


1 Answer 1


The answer is that this is $\simeq\log(m)$. Where $f\sim g$ means that $f\preceq g\preceq f$ and $f\preceq g$ means that eventually $f\le cg$ for some $c>0$.

This is a particular case of a result of Lubotzky, Mozes and Raghunathan. But it's easy to prove directly.

First since the matrix norm of $e_{ij}(1)^m$ grows linearly and the matrix norm is submultiplicative, one immediately sees that $|e_{ij}(1)^m|\succeq \log(m)$.

Now we use that $m\ge 3$, so we can suppose $(i,j)=(1,3)$. Then consider the subgroup $\Gamma$ of matrices that are identity, except the entries 13,23 that are arbitrary, and the block 11 12 21 22 which is an arbitrary integral power of $\begin{pmatrix}2&1\\1& 1\end{pmatrix}$. Then $\Gamma$ is a cocompact lattice in the 3-dimensional group SOL. It is immediate that in the group SOL, if $v$ is in the normal abelian subgroup (written additively), then the word length of $v$ (with respect to any compact generating subset) is $\simeq \log(\|v\|)$. In particular, for fixed nonzero $v$ the word length of $mv$ (=$v^m$, switching back to multiplicative notation) is $\simeq\log(m)$. Since cocompact lattices are undistorted, we deduce that the same holds in $\Gamma$. Hence $|e_{13}(1)^m|\preceq \log(m)$ in $\Gamma$, and hence in the larger group $\mathrm{SL}_n(\mathbf{Z})$.

  • $\begingroup$ However if one fixes the generating subset, one can ask about the limit of $|u_{ij}(m)|/\log(m)$. I don't even know whether it converges (what's above says that the liminf is $>0$, the limsup is $<\infty$, and probably the proofs can provide, after some efforts, explicit distinct bounds). $\endgroup$
    – YCor
    Dec 16, 2021 at 9:24
  • $\begingroup$ Prefect answer - thanks! $\endgroup$ Dec 16, 2021 at 10:00

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