# Growth of the word norm for elementary matrices in $\rm SL_3 (\mathbb{Z})$

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?

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 $$c>0$$.
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 $$\mathrm{SL}_n(\mathbf{Z})$$.
• 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).