Appendix: Spectral gap for the Apollonian group (by Péter P. Varjú)
In recent years some spectacular advances were made on estimating spectral gaps (to be defined below) of infinite co-volume subgroups of \(\operatorname {SL}(d,\mathbb {Z})\). Bourgain and Gamburd [6] proved uniform spectral gap estimates for Zariski-dense subgroups of \(\operatorname {SL}(2,\mathbb {Z})\) under the additional assumption that the modulus q is prime. One of the crucial ideas in their paper is the application of Helfgott’s triple-product theorem [28]. The result in [6] was generalized in a series of papers [5, 7, 10, 11, 48] and [40]. Some of these require the generalization of [28] obtained independently by Breuillard, Green and Tao [14] and Pyber and Szabó [39].
In particular, Bourgain and Varjú [10, Theorem 1] proved the spectral gap for Zariski-dense subgroups of \(\operatorname {SL}(d, \mathbb {Z})\) without any restriction for the modulus q. Salehi Golsefidy and Varjú [40, Theorem 1] obtained the result for Zariski-dense subgroups of perfect arithmetic groups, but only for square-free q. Unfortunately, these results do not cover Theorem 4.3; the first one is not applicable to the Apollonian group, the second one is restricted for the moduli.
In this appendix, we present an approach which differs from those discussed above. This is much simpler and probably would give better numerical results, but we do not pursue explicit bounds. However, our method depends on special properties of the Apollonian group and does not apply to general Zariski-dense subgroups.
Recall from Sect. 2 that the preimage of the Apollonian group under the homomorphism
$$\iota: \operatorname {SL}(2,\mathbb {C})\to \operatorname {SO}_F(\mathbb {R}) $$
is generated by the matrices
$$ \pm \left ( \begin{array}{c@{\quad}c}{1}&{4i}\\ {0}&{1} \end{array} \right ),\qquad\pm \left ( \begin{array}{c@{\quad}c} {2}&{-i}\\ {-i}&{0} \end{array} \right ), \qquad\pm \left ( \begin{array}{c@{\quad}c} {2+2i}&{4+3i}\\ {-i}&{-2i} \end{array} \right ). $$
(A.1)
We describe an automorphism of \(\operatorname {SL}(2,\mathbb {Z}[i])\) which transforms the above generators to matrices that will be more convenient to work with. Set 
. A simple calculation shows that the image of the matrices (A.1) under the map g↦A
−1
gA are
$$\pm \left ( \begin{array}{c@{\quad}c} {1}&{4i}\\ {0}&{1} \end{array} \right ),\qquad\pm \left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {-i}&{1} \end{array} \right ), \qquad\pm \left ( \begin{array}{c@{\quad}c} {1+2i}&{4i}\\ {-i}&{1-2i} \end{array} \right ). $$
We put
$$ \gamma _1=\left ( \begin{array}{c@{\quad}c} {1}&{4}\\ {0}&{1} \end{array} \right ),\qquad \gamma _2=\left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {1}&{1} \end{array} \right ), \qquad \gamma _3=\left ( \begin{array}{c@{\quad}c} {1+2i}&{4}\\ {1}&{1-2i} \end{array} \right ). $$
(A.2)
These are the image of (A.1) under the product of two isomorphism: first conjugation by A and then multiplication of the off-diagonal elements by −i and i. We denote by \(\bar{\varGamma }\) the group generated by \(\bar{S}=\{\pm \gamma _{1}^{\pm1},\pm \gamma _{2}^{\pm1},\pm \gamma _{3}^{\pm1}\}\). This is isomorphic to the group denoted by the same symbol in the paper.
First we recall two different notions of spectral gap. The notion, “geometric” spectral gap, has already been explained in Sect. 4.2. Recall that for an integer q, \(\bar{\varGamma}(q)\) denotes the kernel of the projection map \(\bar{\varGamma}\to \operatorname {SL}(2,\mathbb {Z}[i]/(q))\). We consider the Laplace Beltrami operator Δ on the hyperbolic orbifolds \(\bar{\varGamma}(q)\backslash \mathbb {H}^{3}\). We denote by λ
0(q)≤λ
1(q) the two smallest eigenvalues of Δ on \(\bar{\varGamma}(q)\backslash \mathbb {H}^{3}\). The geometric spectral gap is an inequality of the form λ
1(q)>λ
0(q)+ε for some ε>0 independent of q.
The other notion, “combinatorial” spectral gap is defined as follows. Let G be a finite group, and S a symmetric set of generators. Let T
G,S
be the Markov operator on the space L
2(G) defined by
$$T_{G,S}f(g)=\frac{1}{|S|}\sum_{\gamma \in S} f(\gamma g) $$
for f∈L
2(G) and g∈G. We denote by
$$\lambda _n'(G,S)\le\cdots\le \lambda _1'(G,S) \le \lambda _0'(G,S)=1 $$
the eigenvalues of T
G,S
in increasing order.
The operator \(Id-T_{\bar{\varGamma}/\bar{\varGamma}(q)}\) is a discrete analogue of the Laplacian Δ on \(\bar{\varGamma}(q)\backslash \mathbb {H}^{3}\). So by combinatorial spectral gap we mean the inequality
$$\lambda _1'\bigl(\bar{\varGamma}/\bar{\varGamma}(q),\bar{S} \bigr)<1-\varepsilon $$
for some ε>0 independent of q. To simplify notation, we will write \(\lambda _{1}'(q)=\lambda _{1}'(\bar{\varGamma}/\bar{\varGamma}(q),\bar{S})\).
The relation between the two notions is not just an analogy. It was shown by Brooks [15, Theorem 1] and Burger [17–19] that they are equivalent for the fundamental groups of a family of covers of a compact manifold. The orbifolds \(\bar{\varGamma}(q)\backslash \mathbb {H}^{3}\) are not compact, they even have infinite volume, however the equivalence can be extended to cover our example, see [13, Theorems 1.2 and 2.1].
We show that the congruence subgroups \(\bar{\varGamma}(q)\) of the Apollonian group have combinatorial spectral gap which implies Theorem 4.3 in light of [13, Theorems 1.2 and 2.1].
Theorem A.1
Let
\(\bar{\varGamma}\)
be the Apollonian group and
\(\lambda '_{1}(q)\)
be as above. There is an absolute constant
c>0 such that
\(\lambda _{1}'(q)<1-c\)
for all
q. I.e. the Apollonian group has combinatorial spectral gap.
Denote by Γ
1 and Γ
2 respectively, the groups generated by {γ
1,γ
2} and {γ
1,γ
3} respectively. Denote by \({\bf G}_{1}\) and \({\bf G}_{2}\) the Zariski-closures of Γ
1 and Γ
2 in \(\operatorname {Res}_{\mathbb {R}|\mathbb {C}} \operatorname {SL}(2,\mathbb {C})\), i.e. in \(\operatorname {SL}(2,\mathbb {C})\) considered an algebraic group over \(\mathbb {R}\).
As we will see later, \({\bf G}_{1}\) and \({\bf G}_{2}\) are isomorphic to \(\operatorname {SL}(2,\mathbb {R})\). Moreover Γ
1 and Γ
2 are lattices inside them. This feature of the Apollonian group was pointed out by Sarnak [42]. We exploit it heavily in our approach.
Due to a result going back to Selberg [44], Γ
1 and Γ
2 have geometric spectral gaps with respect to the congruence subgroups. From here we can deduce the combinatorial spectral gap using Brooks [15, Theorem 1] (see also [16, Theorem 1], where the non-compact case is considered.)
We transfer the combinatorial spectral gap property of Γ
1 and Γ
2 to the Apollonian group \(\bar{\varGamma}\) and conclude Theorem A.1. This is done in following two Lemmata:
Lemma A.2
Let
G
be a finite group and
S⊂G
a finite symmetric generating set. Let
G
1,G
2,…,G
k
be subgroups of
G
such that for every
g∈G
there are
g
1∈G
1,…,g
k
∈G
k
such that
g=g
1⋯g
k
. Then
$$1-\lambda _1'(G,S)\ge\min_{1\le i\le k} \biggl \{\frac{|S\cap G_{i}|}{|S|}\cdot \frac{1-\lambda _1'(G_{i},S\cap G_{i})}{2k^2} \biggr\}. $$
The above Lemma and its proof below is closely related to the well-known fact that if G is generated by S in k steps then one has \(\lambda '_{1}(G,S)\le1-1/|S|k^{2}\). This can be found for example in [21, Corollary 1 on page 2138]. After circulating an earlier version of this appendix, it was pointed out to me that an idea similar to Lemma A.2 has been used by Sarnak [41, Sect. 2.4], by Shalom [45], and also by Kassabov, Lubotzky and Nikolov [31].
Lemma A.3
Let
q≥2 be an integer. Then for every
\(g\in\bar{\varGamma}/\bar{\varGamma}(q)\), there are
\(g_{1},\ldots, g_{10^{13}}\in \varGamma _{1}/\varGamma _{1}(q)\)
and
\(h_{1},\ldots, h_{10^{13}}\in \varGamma _{2}/\varGamma _{2}(q)\)
such that
\(g=g_{1}h_{1}\cdots g_{10^{13}}h_{10^{13}}\).
Lemma A.3 enables us to apply Lemma A.2 with k=2⋅1013 and G
i
=Γ
1/Γ
1(q) for odd i and G
i
=Γ
2/Γ
2(q) for even i. Now [44] and [16, Theorem 1] provides us with lower bounds on
$$\begin{aligned} &1-\lambda _1'\bigl(\varGamma _1/ \varGamma _1(q),\bigl\{\pm \gamma _1^{\pm1},\pm \gamma _2^{\pm1}\bigr\}\bigr) \quad\mbox{and}\\ & 1- \lambda _1'\bigl(\varGamma _2/\varGamma _2(q), \bigl\{\pm \gamma _1^{\pm1},\pm \gamma _3^{\pm1} \bigr\}\bigr). \end{aligned}$$
Therefore Theorem A.1 is proved once the two Lemmata are proved.
Before we proceed with the proofs, we make two remarks. First, we note that instead of [44] we could just as well use [10, Theorem 1]. Second, we suggest that the constant 1013 in Lemma A.3 is not optimal. In particular, the argument we present would give 72 if the statement is checked for q=27⋅3, e.g. by a computer program. Certainly there is further room for improvement but we make no efforts to optimize the constants.
Proof of Lemma A.2
Denote by π the regular representation of G, i.e. we write
$$\pi(g_0)f(g)=f\bigl(g_0^{-1}g\bigr) $$
for f∈L
2(G) and g,g
0∈G. Let T
G,S
be the Markov operator defined above. Let f
0∈L
2(G) be an eigenfunction with ∥f
0∥2=1 corresponding to \(\lambda _{1}'(G,S)\). It is orthogonal to the constant and
$$\langle T_{G,S}f_0,f_0\rangle= \lambda _1'(G,S). $$
Since f
0 is orthogonal to the constant, we have
$$\sum_{g\in G}\bigl\langle\pi(g)f_0,f_0 \bigr\rangle=\big|\langle f_0,1\rangle\big|^2=0. $$
Thus there is g
0∈G such that 〈π(g
0)f
0,f
0〉≤0 and hence \(\|\pi(g_{0})f_{0}-f_{0}\|_{2}\ge\sqrt{2}\).
By the hypothesis of the lemma, there are g
i
∈G
i
for 1≤i≤k such that g
0=g
1⋯g
k
. By the triangle inequality, there is some 1≤i
0≤k such that
$$\big\|\pi(g_1\cdots g_{i_0-1})f_0- \pi(g_1\cdots g_{i_0})f_0\big\|_2\ge \sqrt{2}/k. $$
Since π is unitary, we have \(\|f_{0}-\pi(g_{i_{0}})f_{0}\|_{2}\ge\sqrt{2}/k\).
We write f
0=f
1+f
2 such that f
1 is invariant under the elements of \(G_{i_{0}}\) in the regular representation π and f
2 is orthogonal to the space of functions invariant under \(G_{i_{0}}\). Then
$$\sqrt{2}/k\le\big\|f_0-\pi(g_{i_0})f_0 \big\|_2 =\big\|f_2-\pi(g_{i_0})f_2 \big\|_2\le2\|f_2\|_2. $$
Thus \(\|f_{2}\|_{2}\ge1/\sqrt{2}k\).
Now we can write
$$\begin{aligned} \langle T_{G,S\cap G_{i_0}}f_0,f_0\rangle =& \|f_1\|_2^2+\langle T_{G,S\cap G_{i_0}}f_2,f_2 \rangle \\ \le&\|f_1\|_2^2+\lambda _1'(G_{i_0},S \cap G_{i_0})\|f_2\|_2^2 \\ =& 1-\bigl(1-\lambda _1'(G_{i_0},S\cap G_{i_0})\bigr)\|f_2\|_2^2. \end{aligned}$$
(A.3)
Since
$$T_{G,S}=\frac{|S\cap G_{i_0}|}{|S|}T_{G,S\cap G_{i_0}} +\frac{|S\backslash G_{i_0}|}{|S|}T_{G,S\backslash G_{i_0}}, $$
we have
$$ \langle T_{G,S}f_0,f_0 \rangle\le1- \frac{|S\cap G_{i_0}|}{|S|}\bigl(1-\langle T_{G,S\cap G_{i_0}}f_0,f_0 \rangle\bigr). $$
(A.4)
We combine (A.3), (A.4) and the estimate on ∥f
2∥2 and get
$$\langle T_{G,S}f_0,f_0\rangle\le1- \frac{|S\cap G_{i_0}|}{|S|}\cdot \frac{1-\lambda _1'(G_{i_0},S\cap G_{i_0})}{2k^2} $$
which was to be proved. □
Now we turn to the proof of Lemma A.3. It will be convenient to write
$$A_k(q)=\bigl\{g_1h_1\cdots g_kh_k:g_1,\ldots g_k\in \varGamma _1/\varGamma _1(q), h_1,\ldots h_k\in \varGamma _2/\varGamma _2(q)\bigr\}. $$
First we consider the case when q is the power of a prime; the general case will be easy to deduce from this.
Lemma A.4
Let
p
be a prime and
m
a positive integer. Then
\(A_{10^{13}}(p^{m})=\bar{\varGamma}/\bar{\varGamma}(p^{m})\).
We use different methods when p is 2 or 3 compared to when it is larger. First we consider the latter situation.
Proof of Lemma A.4 for p≥5
It is well-known and easy to check that the group generated by γ
1 and γ
2 is
$$ \varGamma _1=\left \{\left ( \begin{array}{c@{\quad}c} {a}&{b}\\ {c}&{d} \end{array} \right )\in \operatorname {SL}(2,\mathbb {Z}): b\equiv0\quad \operatorname {mod}4 \right \}. $$
(A.5)
Thus \(\varGamma _{1}/\varGamma _{1}(p^{m})= \operatorname {SL}(2,\mathbb {Z}/p^{m}\mathbb {Z})\) for p≠2.
By simple calculation:
$$\left ( \begin{array}{c@{\quad}c} {a^{-1}}&{0}\\ {0}&{a} \end{array} \right ) \left ( \begin{array}{c@{\quad}c} {\frac{1}{2}}&{0}\\ {\frac{-1}{8}}&{2} \end{array} \right ) \gamma _3^2 \left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {\frac{1}{8}}&{1} \end{array} \right ) \gamma _3^{-1} \left ( \begin{array}{c@{\quad}c} {a}&{0}\\ {0}&{a^{-1}} \end{array} \right ) =\left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {\frac{-3ia^2}{2}}&{1} \end{array} \right ). $$
Since p≠2 we can divide by 2 in the ring \(\mathbb {Z}/p^{m}\mathbb {Z}\), hence for (a,p)=1, the matrices in the above calculation are in Γ
1/Γ
1(p
m) except for γ
3. Therefore
$$\left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {\frac{-3ia^2}{2}}&{1} \end{array} \right )\in A_3 \bigl(p^m\bigr). $$
Using this, we want to show that
$$ \left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {a i}&{1} \end{array} \right )\in A_{12}\bigl(p^m\bigr) $$
(A.6)
for all \(a\in \mathbb {Z}/p^{m}\mathbb {Z}\). To do this, we need to show that for every element \(x\in \mathbb {Z}/p^{m}\mathbb {Z}\), we can find elements \(a_{1},\ldots, a_{k}\in \mathbb {Z}/p^{m}\mathbb {Z}\) for some 0≤k≤4, such that a
1,…,a
k
are not divisible by p and \(x=a_{1}^{2}+\cdots+a_{k}^{2}\). If m=1, this simply follows from the fact that any positive integer is a sum of at most 4 squares, and the a
i
can not be divisible by p since 0<a
i
≤x≤p and at least one of the inequalities are strict.
Suppose that m>1, \(x\in \mathbb {Z}/p^{m}\mathbb {Z}\) and \(a_{1}^{2}+\cdots+a_{k}^{2}\equiv x\operatorname {mod}p\) with none of a
1…a
k
divisible by p. Then by Hensel’s lemma (recall that p≠2), there is an \(a_{1}'\in \mathbb {Z}/p^{m}\mathbb {Z}\) such that
$$\bigl(a_1'\bigr)^2=a_1^2+ \bigl(x-a_1^2-\cdots-a_k^2 \bigr). $$
This proves the claim for arbitrary m≥1.
Multiplying (A.6) by a suitable unipotent element of Γ
1/Γ
1(p
m), we can get
$$\left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {a}&{1} \end{array} \right )\in A_{12} \bigl(p^m\bigr) $$
for \(a\in \mathbb {Z}[i]/(p^{m})\). We can prove the same for the upper triangular unipotents by a very similar argument.
Again, by simple calculation:
$$\left ( \begin{array}{c@{\quad}c} {1}&{a}\\ {0}&{1} \end{array} \right ) \left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {b}&{1} \end{array} \right ) \left ( \begin{array}{c@{\quad}c} {1}&{c}\\ {0}&{1} \end{array} \right ) =\left ( \begin{array}{c@{\quad}c} {1+ab}&{a+c+abc}\\ {b}&{1+bc} \end{array} \right ). $$
This shows that
$$\left ( \begin{array}{c@{\quad}c} {a'}&{b'}\\ {c'}&{d'} \end{array} \right )\in A_{36} \bigl(p^m\bigr) $$
for all \(a',b',c',d'\in \mathbb {Z}[i]/(p^{m})\), a′d′−b′c′=1, provided c′ is not divisible by a prime above p.
Thus, A
36(p
m) contains more than half of the group \(\bar{\varGamma}/\bar{\varGamma}(p^{m})\), hence
$$A_{72}\bigl(p^m\bigr)=\bar{\varGamma}/\bar{\varGamma} \bigl(p^m\bigr). $$
□
Proof of Lemma A.4 for p=2 and 3
We give the proof for p=2 and then explain the differences for p=3.
We prove by induction the following statement. For every m≥7 and \(g\in\bar{\varGamma}(2^{7})/\bar{\varGamma}(2^{m})\), there are g
1,g
2,g
3∈Γ
1(22)/Γ
1(2m) such that
$$g=g_1\gamma _3g_2\gamma _3^{-1} \gamma _3^2 g_3\gamma _3^{-2}. $$
For m=7 this is clear since we can take g
1=g
2=g
3=1. Now assume that m>7 and the statement holds for m−1. In this proof, we denote by 1 the multiplicative unit (identity matrix) and by 0 the matrix with all entries 0. Let \(g\in\bar{\varGamma}(2^{7})/\bar{\varGamma}(2^{m})\) be arbitrary. By the induction hypothesis, there is h
1,h
2,h
3∈Γ
1(22)/Γ
1(2m) such that
$$g-h_1\gamma _3h_2\gamma _3^{-1} \gamma _3^2 h_3\gamma _3^{-2}=2^{m-1}x, $$
where x can be considered as an element of \(\operatorname {Mat}(2,\mathbb {Z}[i]/(2))\), i.e. a 2×2 matrix with elements in \(\mathbb {Z}[i]/(2)\). Since g,h
1,h
2,h
3 has determinant 1 and congruent to the unit element mod 2, x has trace 0.
Now we look for suitable \(x_{1},x_{2},x_{3}\in \operatorname {Mat}(2,\mathbb {Z})\) such that
$$x_1+\gamma _3x_2\gamma _3^{-1}+ \gamma _3^2 x_3\gamma _3^{-2} \equiv2^{m-1}x\quad \operatorname {mod}2^m. $$
Moreover, we ensure that \(x_{i}\equiv0\quad \operatorname {mod}2^{m-4}\) and that \(\operatorname {Tr}(x_{i})\equiv0\quad \operatorname {mod}2^{m}\) for all i=1,2,3. Since m≥8, this implies that \(h_{i}+x_{i}\equiv1\quad \operatorname {mod}4\) and \(\det(h_{i}+x_{i})\equiv1 \quad \operatorname {mod}2^{m}\), hence h
i
+x
i
∈Γ
1(22)/Γ
1(2m). Recall (A.5) from the previous proof. If the matrices x
i
satisfy the claimed properties then
$$\begin{aligned} &(h_1+x_1)\gamma _3(h_2+x_2) \gamma _3^{-1}\gamma _3^2 (h_3+x_3) \gamma _3^{-2} \\ &\quad\equiv h_1\gamma _3h_2 \gamma _3^{-1}\gamma _3^2 h_3 \gamma _3^{-2}+ x_1+\gamma _3x_2 \gamma _3^{-1}+\gamma _3^2 x_3 \gamma _3^{-2}\equiv g\quad \operatorname {mod}2^m. \end{aligned}$$
The matrices x
1,x
2,x
3 can be chosen to be a suitable linear combination of the matrices in the following calculations, and this finishes the induction:
$$2^{m-1}\left ( \begin{array}{c@{\quad}c} {0}&{1}\\ {0}&{0} \end{array} \right )+ \gamma _3 0\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2} \equiv2^{m-1}\left ( \begin{array}{c@{\quad}c} {0}&{1}\\ {0}&{0} \end{array} \right )\quad \operatorname {mod}2^m, $$
$$2^{m-1}\left ( \begin{array}{c@{\quad}c} {0}&{0}\\ {1}&{0} \end{array} \right )+ \gamma _3 0\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2} 2^{m-1}\equiv \left ( \begin{array}{c@{\quad}c} {0}&{0}\\ {1}&{0} \end{array} \right )\quad \operatorname {mod}2^m, $$
$$2^{m-1}\left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {0}&{-1} \end{array} \right )+ \gamma _3 0\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2} \equiv2^{m-1}\left ( \begin{array}{c@{\quad}c} {1}&{0}\\ {0}&{-1} \end{array} \right )\quad \operatorname {mod}2^m, $$
$$\begin{aligned} &2^{m-2}\left ( \begin{array}{c@{\quad}c} {1}&{3}\\ {1}&{-1} \end{array} \right )+ \gamma _3 2^{m-2}\left ( \begin{array}{c@{\quad}c} {0}&{1}\\ {0}&{0} \end{array} \right )\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2} \\ &\quad\equiv2^{m-1}\left ( \begin{array}{c@{\quad}c} {-i}&{0}\\ {0}&{i} \end{array} \right )\quad \operatorname {mod}2^m, \\ &2^{m-3}\left ( \begin{array}{c@{\quad}c} {-4}&{0}\\ {3}&{4} \end{array} \right )+ \gamma _3 2^{m-3}\left ( \begin{array}{c@{\quad}c} {0}&{0}\\ {1}&{0} \end{array} \right )\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2}\\ &\quad \equiv2^{m-1}\left ( \begin{array}{c@{\quad}c} {0}&{0}\\ {i}&{0} \end{array} \right )\quad \operatorname {mod}2^m, \\ &2^{m-4}\left ( \begin{array}{c@{\quad}c} {2}&{15}\\ {4}&{-2} \end{array} \right )+ \gamma _3 0\gamma _3^{-1}+\gamma _3^2 2^{m-4}\left ( \begin{array}{c@{\quad}c} {0}&{1}\\ {0}&{0} \end{array} \right ) \gamma _3^{-2}\\ &\quad \equiv2^{m-1} \left ( \begin{array}{c@{\quad}c} {-i}&{i}\\ {0}&{i} \end{array} \right )\quad \operatorname {mod}2^m. \end{aligned}$$
Now we showed that
$$A_3\bigl(2^{m}\bigr)\supseteq\bar{\varGamma} \bigl(2^7\bigr)/\bar{\varGamma}\bigl(2^m\bigr). $$
The index of \(\bar{\varGamma}(2^{7})/\bar{\varGamma}(2^{m})\) in \(\bar{\varGamma}/\bar{\varGamma}(2^{m})\) is at most
$$\big| \operatorname {SL}\bigl(2,\mathbb {Z}[i]/\bigl(2^7\bigr)\bigr)\big|=46\cdot64^6. $$
This shows that
$$A_{10^{13}}\bigl(2^{m}\bigr)=\bar{\varGamma}/\bar{\varGamma} \bigl(2^m\bigr). $$
Now we turn to the case p=3. By the same argument, one can show that for every m≥1 and \(g\in\bar{\varGamma}(3)/\bar{\varGamma}(3^{m})\), there are g
1,g
2,g
3∈Γ
1/Γ
1(3m) such that
$$g=g_1\gamma _3g_2\gamma _3^{-1} \gamma _3^2 g_3\gamma _3^{-2}. $$
The only significant difference is that one needs to use the following identities:
$$\begin{aligned} &3^{m-1}\left ( \begin{array}{c@{\quad}c} {1}&{3}\\ {1}&{-1} \end{array} \right )+ \gamma _33^{m-1}\left ( \begin{array}{c@{\quad}c} {0}&{1}\\ {0}&{0} \end{array} \right )\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2}\\ &\quad \equiv3^{m-1}\left ( \begin{array}{c@{\quad}c} {i}&{i}\\ {0}&{-i} \end{array} \right )\quad \operatorname {mod}3^m, \\ &3^{m-1}\left ( \begin{array}{c@{\quad}c} {-4}&{16}\\ {3}&{4} \end{array} \right )+ \gamma _33^{m-1} \left ( \begin{array}{c@{\quad}c} {0}&{0}\\ {1}&{0} \end{array} \right )\gamma _3^{-1}+\gamma _3^2 0 \gamma _3^{-2}\\ &\quad \equiv3^{m-1}\left ( \begin{array}{c@{\quad}c} {i}&{0}\\ {-i}&{-i} \end{array} \right )\quad \operatorname {mod}3^m, \\ &3^{m-1}\left ( \begin{array}{c@{\quad}c} {2}&{15}\\ {4}&{-2} \end{array} \right )+ \gamma _3 0\gamma _3^{-1}+\gamma _3^23^{m-1} \left ( \begin{array}{c@{\quad}c} {0}&{1}\\ {0}&{0} \end{array} \right )\gamma _3^{-2}\\ &\quad\equiv3^{m-1}\left ( \begin{array}{c@{\quad}c} {i}&{-i}\\ {0}&{-i} \end{array} \right ) \quad \operatorname {mod}3^m. \end{aligned}$$
Using this claim, one can finish the proof as above. □
Proof of Lemma A.3
Let q be an integer and \(q=p_{1}^{m_{1}}\cdots p_{n}^{m_{n}}\) where p
i
are primes. We prove that
$$A_{10^{13}}(q)=A_{10^{13}}\bigl(p_1^{m_1} \bigr)\times\cdots\times A_{10^{13}}\bigl(p_n^{m_n} \bigr). $$
Let \(x\in A_{10^{13}}(p_{1}^{m_{1}})\times\cdots\times A_{10^{13}}(p_{n}^{m_{n}})\) be arbitrary. By definition, for each k, we can find elements \(g_{1}^{(k)},\ldots, g_{10^{13}}^{(k)}\in \varGamma _{1}/\varGamma _{1}(q)\) and \(h_{1}^{(k)},\ldots, h_{10^{13}}^{(k)}\in \varGamma _{2}/\varGamma _{2}(q)\) such that
$$x\equiv g_1^{(k)}h_1^{(k)}\cdots g_{10^{13}}^{(k)}h_{10^{13}}^{(k)}\quad \operatorname {mod}p_k^{m_k}. $$
Since Γ
1/Γ
1(p
m) and Γ
2/Γ
2(p
m) are the direct product of local factors, we can find elements \(g_{1},\ldots, g_{10^{13}}\in \varGamma _{1}/\varGamma _{1}(p^{m})\) and \(h_{1},\ldots, h_{10^{13}}\in \varGamma _{2}/\varGamma _{2}(p^{m})\) such that
$$g_i\equiv g_i^{(k)}\quad \operatorname {mod}p_k^{m_k}\quad\mbox{and}\quad h_i\equiv h_i^{(k)}\quad \operatorname {mod}p_k^{m_k} $$
for each i and k. Thus
$$x= g_1h_1\cdots g_{10^{13}}h_{10^{13}} \in A_{10^{13}}(q). $$
Using Lemma A.4 we get
$$\begin{aligned} \bar{\varGamma}/\bar{\varGamma}(q) \supset& A_{10^{13}}(q) \supset A_{10^{13}}\bigl(p_1^{m_1}\bigr)\times\cdots\times A_{10^{13}}\bigl(p_n^{m_n}\bigr) \\ =&\bar{\varGamma}/ \bar{\varGamma}\bigl(p_1^{m_1}\bigr)\times\cdots\times \bar{\varGamma }/\bar{\varGamma}\bigl(p_{n}^{m_n}\bigr). \end{aligned}$$
Obviously
$$\bar{\varGamma}/\bar{\varGamma}(q)\subset\bar{\varGamma}/\bar{\varGamma } \bigl(p_1^{m_1}\bigr)\times\cdots\times \bar{\varGamma}/ \bar{\varGamma}\bigl(p_{n}^{m_n}\bigr) $$
hence all these containments must be equality. □