Logarithmic girth expander graphs of $SL_n({\mathbb {F}}_p)$

Abstract

We provide an explicit construction of finite 4-regular graphs $(\Gamma _k)_{k\in {\mathbb {N}}}$ with $\text {girth}\, \Gamma _k\rightarrow \infty$ as $k\rightarrow \infty$ and $\frac{\text {diam}\,\Gamma _k}{\text {girth}\,\Gamma _k}\leqslant D$ for some $D>0$ and all $k\in {\mathbb {N}}$. For each fixed dimension $n\geqslant 2,$ we find a pair of matrices in $SL_{n}({\mathbb {Z}})$ such that (i) they generate a free subgroup, (ii) their reductions $\bmod \, p$ generate $SL_{n}({\mathbb {F}}_{p})$ for all sufficiently large primes p, (iii) the corresponding Cayley graphs of $SL_{n}({\mathbb {F}}_{p})$ have girth at least $c_n\log p$ for some $c_n>0$. Relying on growth results (with no use of expansion properties of the involved graphs), we observe that the diameter of those Cayley graphs is at most $O(\log p)$. This gives infinite sequences of finite 4-regular Cayley graphs of $SL_n({\mathbb {F}}_p)$ as $p\rightarrow \infty$ with large girth and bounded diameter-by-girth ratio. These are the first explicit examples in all dimensions $n\geqslant 2$ (all prior examples were in $n=2$). Moreover, they happen to be expanders. Together with Margulis’ and Lubotzky–Phillips–Sarnak’s classical constructions, these new graphs are the only known explicit logarithmic girth Cayley graph expanders.

Appendix

For an interested reader, we give a detailed proof of Theorem 5.3.

Proof of Theorem 5.3

Fix $(n-1) = k\in {\mathbb {N}}$ and consider the matrix $A^{lr_{i}}B^{ls_{i}}$ with $l \geqslant 3k$ and $r_{i},s_{i}\in {\mathbb {Z}}\backslash \lbrace 0\rbrace$. Let $a,b\in {\mathbb {N}}$ with $a,b\geqslant 2$. Suppose

$$\begin{aligned} {\mathcal {P}}_{i} = A^{lr_{i}}B^{ls_{i}} = \begin{pmatrix} P_{11}(a,b) &{}\quad P_{12}(a,b) &{}\quad \ldots &{}\quad P_{1n}(a,b)\\ P_{21}(a,b) &{}\quad P_{22}(a,b) &{}\quad \ldots &{}\quad P_{2n}(a,b)\\ P_{31}(a,b) &{}\quad P_{32}(a,b) &{}\quad \ldots &{}\quad P_{3n}(a,b)\\ \vdots &{}\quad &{}\quad &{}\quad \vdots \\ &{}\quad &{}\quad \ldots &{}\quad P_{(n-1)n}(a,b)\\ P_{n1}(a,b) &{}\quad P_{n2}(a,b) &{}\quad \ldots &{}\quad P_{nn}(a,b)\\ \end{pmatrix} \in SL_{n}({\mathbb {Z}}), \end{aligned}$$

where the polynomials $P_{uv}(a,b), 1\leqslant u,v\leqslant n$ satisfy

• $P_{11}(a,b) = {lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k} + {lr_{i} \atopwithdelims ()k-1}{ls_{i} \atopwithdelims ()k-1}a^{k-1}b^{k-1}+ \cdots + {lr_{i} \atopwithdelims ()3}{ls_{i} \atopwithdelims ()3}a^{3}b^{3} + {lr_{i} \atopwithdelims ()2}{ls_{i} \atopwithdelims ()2}a^{2}b^{2} + {lr_{i} \atopwithdelims ()1}{ls_{i} \atopwithdelims ()1}ab~+~1$

• $P_{12}(a,b) = {lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k-1}a^{k}b^{k-1} + {lr_{i} \atopwithdelims ()k-1}{ls_{i} \atopwithdelims ()k-2}a^{k-1}b^{k-2}+ \cdots + {lr_{i} \atopwithdelims ()3}{ls_{i} \atopwithdelims ()2}a^{3}b^{2} + {lr_{i} \atopwithdelims ()2}{ls_{i} \atopwithdelims ()1}a^{2}b^{1} + {lr_{i} \atopwithdelims ()1}a$

  ...

  and in general

• $P_{uv}(a,b) = \Sigma _{u'=u,v'=v}^{u'=k+2-v,v'=k+1}{lr_{i} \atopwithdelims ()k-u'+1}{ls_{i} \atopwithdelims ()k-v'+1}a^{k-u'+1}b^{k-v'+1}$ with $u\leqslant v$

• $P_{uv}(a,b) = \Sigma _{u'=u,v'=v}^{u'=k+1, v' = k+2-u}{lr_{i} \atopwithdelims ()k-u+1}{ls_{i} \atopwithdelims ()k-v+1}a^{k-u'+1}b^{k-v'+1}$ with $u > v$

  ...

• $P_{nn}(a,b) = 1$

Then, we have the following inequalities for $l\geqslant 3k$.

• (1) $1 - \frac{1}{15}\leqslant \frac{P_{11}(a,b)}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}} \leqslant 1 + \frac{1}{15}$.

• (2) $|\frac{P_{12}(a,b)}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}}| \leqslant \frac{1}{4}(1+\frac{1}{4^{2}}+ \frac{1}{4^{4}}+\cdots +\frac{1}{4^{2k-2}}) < \frac{1}{4}.\frac{16}{15} .$

  ...

  and in general

• (3) $|\frac{P_{uv}(a,b)}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}}| \leqslant \frac{1}{4^{u+v-2}} \cdot \frac{16}{15}\,\, \forall uv > 1.$

We proceed as in the $n=4$ case and consider

$$\begin{aligned} Z = \prod _{i=1}^{t}{\mathcal {P}}_{i} = \prod _{i=1}^{t} A^{lr_{i}}B^{ls_{i}}, \hbox { for some } t\in {\mathbb {N}}, r_{i},s_{i}\in {\mathbb {Z}}\backslash \lbrace 0 \rbrace . \end{aligned}$$

We shall show, by induction, that $|z_{11}| > |z_{12}| + \cdots + |z_{1n}| + 1$, where the $z_{ij}, 1\leqslant i,j\leqslant n$ denote the elements of the matrix Z.

From the above inequalities, it is clear that $\Sigma _{v=2}^{n} |\frac{P_{1v}(a,b)}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}}|<\frac{1}{2}$ which gives

$$\begin{aligned} \frac{1}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}} + \Sigma _{v=2}^{n} \left| \frac{P_{1v}(a,b)}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}}\right| < \frac{P_{11}(a,b)}{{lr_{i} \atopwithdelims ()k}{ls_{i} \atopwithdelims ()k}a^{k}b^{k}} \end{aligned}$$

and in turn implies that the inductive assumption (basis of induction) holds.

For the main step of the induction, suppose we are already given

$$\begin{aligned} Z =\begin{pmatrix} z_{11} &{}\quad z_{12} &{}\quad \ldots &{}\quad z_{1n}\\ z_{21} &{}\quad z_{22} &{}\quad \ldots &{}\quad z_{2n}\\ \vdots &{}\quad \vdots &{}\quad &{}\quad \vdots \\ z_{n1} &{}\quad z_{n2} &{}\quad \ldots &{}\quad z_{nn}\\ \end{pmatrix} \end{aligned}$$

with $|z_{11}| > |z_{12}| + \cdots + |z_{1n}| + 1$.

Let

$$\begin{aligned} Z' = \begin{pmatrix} z'_{11} &{}\quad z'_{12} &{}\quad \ldots &{}\quad z'_{1n}\\ z'_{21} &{}\quad z'_{22} &{}\quad \ldots &{}\quad z'_{2n}\\ \vdots &{}\quad \vdots &{}\quad &{}\quad \vdots \\ z'_{n1} &{}\quad z'_{n2} &{}\quad \ldots &{}\quad z'_{nn}\\ \end{pmatrix} = Z\times A^{lr_{t+1}}B^{ls_{t+1}}. \end{aligned}$$

Expand the first row of $Z'$, i.e., $z'_{1j}, 1\leqslant j \leqslant n$ in terms of the first row of Z, $z_{1j}, 1\leqslant j\leqslant n$ and the elements of the matrix $A^{lr_{t+1}}B^{ls_{t+1}}$. Then, we can conclude by considering the inductive assumption $|z_{11}| > |z_{12}| + \cdots + |z_{1n}| + 1$ and the above inequalities for the matrix $A^{lr_{t+1}}B^{ls_{t+1}}$ that

$$\begin{aligned} |z'_{11}| > |z'_{12}|+ \cdots +|z'_{1n}| + 1. \end{aligned}$$

$\square$