Subsection 3.1 Linnik's theorem
¶Problem
Distribution of
\begin{equation*}
\mathscr H_d = \{ (x,y,z) \in \ZZ^3 : x^2 + y^2 + z^2 = d\}
\end{equation*}
as \(d \to \infty\text{.}\)
Theorem 3.1 Linnik
As \(d \to \infty\) squarefree such that \(d\not\equiv 7 \pmod 8\text{,}\) \(d \equiv \pm 1 \pmod 5\)
\begin{equation*}
\left\{ \frac {x}{\sqrt d} : x \in \mathscr H_d \right\}
\end{equation*}
is equidistributed w.r.t. the Lebesgue measure
\begin{equation*}
\operatorname{red}_\infty \colon \mathscr H_d \to S^2
\end{equation*}
\begin{equation*}
\frac{|\operatorname{red}_\infty\inv (\Omega)|}{|\mathscr H_d|} \to \operatorname{Vol} (\Omega)
\end{equation*}
in the sense that
\begin{equation*}
\frac{|\operatorname{red}_\infty\inv (\Omega)|}{|\mathscr H_d|} = \operatorname{Vol} (\Omega) ( 1 + o(1))
\end{equation*}
for any \(\Omega\) a reasonable subset of \(S^2\text{.}\)
The \(\pm 1\pmod 5\) condition comes from a splitting condition in \(\QQ(\sqrt{d})\) that we need to find an appropriate group action.
We will rather focus on the following variant
Theorem 3.2 Linnik variant
\begin{equation*}
\overline{\mathscr H}_d = \{ (x,y,x) \pmod q : (x,y,z)\in \mathscr H_d\}
\end{equation*}
with \(q\) fixed \(\gt 0\) and \((q,30) =1\) (non-essential). As \(d \to \infty\text{,}\) \(d\) squarefree such that \(d \not \equiv 7 \pmod 8\) and \(d \equiv \pm 1 \pmod 5\text{.}\) \(\overline {\mathscr H}_d(q)\) becomes equidistributed in \(\mathscr H_d(q) = \{ (x,y,z ) \in (\ZZ/q)^3 : x^2 + y^2 + z^2 = d\}\) w.r.t. counting measure, explicitly
\begin{equation*}
\frac{| \operatorname{red}\inv_q(x)|}{| \mathscr H_d|} = \frac{1}{|\mathscr H_d(q)|} ( 1+ o(1))
\end{equation*}
as \(d \to \infty\text{.}\)
Actually we will prove something a little strange.
Definition 3.3 Deviation
\(\infty\text{,}\) let \(\rho \gt 0\text{,}\) \(x\in S^2\) and consider the cap \(\Omega(x,\rho)\text{.}\)
\begin{equation*}
\operatorname{dev}_d(\Omega) = \frac {1}{\operatorname{Vol}(\Omega)} \frac{| \operatorname{red}\inv_\infty(\Omega)|}{|\mathscr H_d|} - 1
\end{equation*}
(we want to show this is \(o(1)\)).
\(q\text{,}\) let \(\bar x \in \mathscr H_d(q)\) then
\begin{equation*}
\operatorname{dev}_d(\bar x) = \frac {| \mathscr H_d(q)| | \operatorname{red}\inv_q(\bar x)|}{|\mathscr H_d|} - 1\text{.}
\end{equation*}
Theorem 3.4
Fix \(\delta, \eta \gt 0\) then for every
\begin{equation*}
\rho \ge d^{-1/4 + \eta}
\end{equation*}
\begin{equation*}
\frac 1q \ge d^{-1/4 + \eta}
\end{equation*}
\((q,30) = 1\text{,}\) then as \(d\to \infty\)
\begin{equation*}
\mu_{S^2}(x : \operatorname{dev}_d(\Omega(x, \rho)) \ge \delta) \to 0
\end{equation*}
\begin{equation*}
\frac{| \{ \bar x \in \mathscr H_d(q) : \operatorname{dev}_d(\bar x) \ge \delta \} |}{ |\mathscr H_d(q)|} \to 0\text{.}
\end{equation*}
How? Sketch of the argument:
Fix \(\delta ,\eta \gt 0\text{,}\)
\begin{equation*}
B_\delta = \{ \bar x \in \mathscr H_q(d) : \operatorname{dev}_d(\bar x) \gt \delta \}
\end{equation*}
and assume that
\begin{equation*}
|B_\delta| \ge \eta |\mathscr H_d(q)|\text{.}
\end{equation*}
Inputs
- There is an action of \(H(d) = \) class group of \(\QQ(\sqrt{-d})\) on \(\mathscr H_d(q)/ \specialorthogonal_3(\ZZ)\text{.}\) As \(d \equiv \pm 1 \pmod 5\) so \((5) = \ideal p \bar{\ideal p}\) in \(\QQ(\sqrt{-d})\text{:}\) This gives us a well defined dynamical system by considering the action
\begin{equation*}
\mathscr H_d(q) \ni x \xrightarrow {\ideal p} x_1 \to x_2 \to x_2 \to \cdots
\end{equation*}
and in reverse
\begin{equation*}
x_{-2} \to x_{-1} \to x \xrightarrow {\ideal p} x_1 \to x_2 \to x_2 \to \cdots
\end{equation*}
call this chain \(\gamma_X\) and for any \(l \gt 0 \in \ZZ\)
\begin{equation*}
\gamma_X^{(l)} = [x_{-l}, x_{-l + 1}, \ldots, x_{-1} , x,x_1, \ldots, x_l]
\end{equation*}
- Consider the average on
\begin{equation*}
B_\delta = \gamma_X^{(l)} \cap B_\delta\text{.}
\end{equation*}
On one hand the expected size of \(\gamma_X^{(l)} \cap B_\delta\) is
\begin{equation*}
\frac{1}{|\mathscr H_d|} \sum_{x \in \mathscr H_d} | \gamma_X^{(l)} \cap B_\delta|
\end{equation*}
on the other hand the expectation is
\begin{equation*}
\frac{(2l+ 1)}{|\mathscr H_d|} \operatorname{red}_q\inv (B_\delta)\text{.}
\end{equation*}
i.e.
\begin{equation*}
\frac{1}{|\mathscr H_d|} \sum_{x\in \mathscr H_d} \frac{|\gamma_X^{(l)} \cap B_\delta|}{2l + 1} = \frac{| \operatorname{red}\inv_q(B_\delta)|}{|\mathscr H_d|}
\end{equation*}
- Since we assume that the deviation set is large (\(\operatorname{dev}_d(x) \gt \delta\)) this implies on average
\begin{equation*}
\frac{| \gamma_X^{(l)} \cap B_\delta|}{2l+1}
\end{equation*}
is large, gives a lower bound for this.
- Then we count \(\gamma_X^{(l)}\) in another way, action of \(H(d)\) on \(\mathscr H_d\) and on \(\mathscr H_d(q)\text{.}\) action of \(\ideal p\) induces a graph structure on \(\mathscr H_d(q)\text{.}\) Then \(\gamma_X^{(l)}\) is a path on \(\mathscr H_d(q)\) which is non-backtracking. Count these: Inputs: for \(l \sim \log(d)\) the number of non-backtracking paths for which \(|\gamma_X^{(l)} \cap B_\delta|/(2l+1)\) is large is \(\gg d^{1/2 + \epsilon}\text{.}\) Count the total number of non-backtracking paths of length \(2l+1\) for which
\begin{equation*}
\left| \frac{|\gamma_X^{(l)} \cap B_\delta|}{2l+1 } - \frac{|B_\delta|}{|\mathscr H_d(q)|} \right| \gt \delta\eta/2
\end{equation*}
There can not be too many of these. This will give a contradiction. Will take some work and depend on many things.
Class group actions
\begin{equation*}
\mathscr H_d = \{ (x,y,z ) \in \ZZ^2: x^2 + y^2+ z^2 = d \}
\end{equation*}
Aim:
- \(\widetilde {\mathscr H}_d^*\) is a principal homogeneous space for \(H(d) = \Cl(\QQ(\sqrt{-d}))\text{.}\)
- Describe the action explicitly.
Definition 3.5 Principal homogeneous spaces
Let \(G\) be a group, \(X \ne \emptyset\) a set. \(X\) is a principal homogeneous space for \(G\) if
\begin{equation*}
G\acts X
\end{equation*}
in a transitive, free manner. \(X\) is also called a \(G\)-torsor.
Example 3.7
- \(G\) is a \(G\)-torsor
- \(G= V\) a \(n\)-dimensional vector space then \(\aff^n\) is a \(G\)-torsor
- \(G = \mu_n\) then \(X(2) = \{ x \in \CC: x^n = 2\}\)
Exercise 3.9
\begin{equation*}
\# \specialorthogonal_3(\ZZ) = 24
\end{equation*}
- \(\specialorthogonal_3(\ZZ)\) has a subgroup \(\specialorthogonal_3^+(\ZZ)\) of index 2 acting via even permutations.
Definition 3.10
\begin{equation*}
\widetilde{\mathscr H}_d = \specialorthogonal_3(\ZZ) \backslash \mathscr H_d
\end{equation*}
\begin{equation*}
\widetilde{\mathscr H}_d (q) = \specialorthogonal_3(\ZZ) \backslash \mathscr H_d(q)
\end{equation*}
\begin{equation*}
\widetilde S^2 = \specialorthogonal_3 (\ZZ) \backslash S^2
\end{equation*}
\begin{equation*}
\widetilde{\mathscr H}_d^* =
\begin{cases}
\specialorthogonal_3^+(\ZZ) \backslash \mathscr H_d ,\amp d \equiv 1,2 \pmod 4 \\
\specialorthogonal_3(\ZZ) \backslash \mathscr H_d ,\amp d \equiv 3 \pmod 4
\end{cases}
\end{equation*}
For all practical purposes we can basically assume that \(d \equiv 3 \pmod 4\text{.}\)
Proposition 3.11
\(\widetilde{\mathscr H}_d^* \) is a principal homogeneous space for \(H(d)\text{.}\) Idea: Venkov 1922: Use quaternion algebras.
Digression: Hamilton Quaternions
\begin{equation*}
B = \{ u + ia + jb + kc : u,a,b,c \in \RR\}
\end{equation*}
\begin{equation*}
\bar x = u - ia - j b - kc
\end{equation*}
\begin{equation*}
N(x) = x\bar x = u^2 + a^2 + b^2 + c^2
\end{equation*}
\begin{equation*}
\trace (x) = x + \bar x = 2u
\end{equation*}
\begin{equation*}
B^{(0)} = \{ x\in B : \trace (x) = 0\}
\end{equation*}
Note
\begin{equation*}
N(x) =a^2 + b^2 + c^2 \text{ if } x \in B^{(0)}\text{.}
\end{equation*}
\begin{equation*}
B^\times = \text{units} ,\, B^1 = \{ x \in B : N(x)= 1\}
\end{equation*}
\begin{equation*}
PB^\times = B^\times/ Z(B^\times)
\end{equation*}
Exercise 3.13
\begin{equation*}
1 \to Z(B^\times) \to B^\times \to PB^\times \simeq \specialorthogonal_3 (\QQ) \to 1
\end{equation*}
Definition 3.14 Hurwitz quaternions
\begin{equation*}
B(\ZZ) = \ZZ[ i ,j, \frac{i+j+ k}{2} ]
\end{equation*}
The Lipschitz quaternions are
\begin{equation*}
\ZZ[ i ,j, k]\text{.}
\end{equation*}
- \(B(\ZZ)\) is a maximal order
- \(B(\ZZ)\) is a Euclidean domain
- Lipschitz quaternions are not a Euclidean domain
Exercise 3.15
- Show these properties.
- Show \(\Cl(\ZZ\lb i, j, k \rb) =2\text{.}\)
2 implies \(B(\ZZ)\) is a PID
\begin{equation*}
(\QQ^3, a^2+ b^2+ c^2) \simeq (B^{(0)}, N)
\end{equation*}
as metric spaces
The image of
\begin{equation*}
B^\times(\ZZ) \to \specialorthogonal_3(\ZZ)
\end{equation*}
is \(\specialorthogonal_3^+(\ZZ)\text{.}\)
The action of \(H(d)\)
Start with \(x = (a,b,c) \in \mathscr H_d\text{.}\) This gives rise to an embedding
\begin{equation*}
\iota_x \colon \QQ(\sqrt{-d}) \hookrightarrow B
\end{equation*}
\begin{equation*}
\sqrt{-d} \mapsto x + ai + bj + ck
\end{equation*}
note \(x^2 = -d\text{.}\)
Let \(K= \QQ(\sqrt{-d})\) \(\iota_x\) is integral ion the sense that if
\begin{equation*}
\ints_x = B(\ZZ) \cap \QQ \lb x \rb
\end{equation*}
then
\begin{equation*}
\iota\inv (\ints_x) = \ints_K\text{.}
\end{equation*}
Exercise 3.16
\(I \in H(d) \leadsto y= y(x,I)\)
Consider \(B(\ZZ) \iota_x(I) = B(\ZZ) q\) for some \(q\in B(\QQ)\text{.}\) Since \(B(\ZZ)\) is a PID.
Then \(y = q xq\inv\in \mathscr H_d\)
To sum up:
For any \(x \in \mathscr H_d\) we have a well defined map
\begin{equation*}
\Cl(d) \to \widetilde{\mathscr H}_d^\times
\end{equation*}
\begin{equation*}
I \mapsto y(x,I)\text{.}
\end{equation*}
Claim 3.18
This map makes \(\widetilde{\mathscr H}_d^\times\) a principal homogeneous space over \(\Cl(d)\text{.}\)
Let \(x,y \in \mathscr H_d\text{.}\)
Construct an ideal \(\Lambda_{x\mapsto y}\) s.t. \(y(x,\Lambda_{x\mapsto y}) = y\text{.}\)
Definition 3.19
\begin{equation*}
\Lambda_{x\mapsto y} = \{ \lambda \in B(\ZZ) : x \lambda = \lambda y \}\text{.}
\end{equation*}
\(x, y \in \mathscr H_d\) Witt's theorem implies there exists \(q \in PB^\times(\QQ) = \specialorthogonal_3(\QQ)\) s.t.
Summary \(\widetilde {\mathscr H_d}^*\) is a principal homogeneous space for \(\Cl(d)\text{,}\) the class group of \(\QQ(\sqrt{-d})\text{.}\)
How? \(x\in\mathscr H_d\) defined
\begin{equation*}
\QQ(\sqrt{-d}) \xhookrightarrow i B(\QQ)
\end{equation*}
\begin{equation*}
\sqrt{-d} \mapsto x
\end{equation*}
any \(I \in \Cl(d) \leadsto B(\ZZ) i(I) = B(\ZZ) q\) for some \(q\text{.}\)
\begin{equation*}
B(\ZZ) = \{ x + ai +bj + ck : a,b,c,d \in \ZZ, \text{ or all }\in \frac 12\ZZ\smallsetminus \ZZ\}
\end{equation*}
\begin{equation*}
y= q\inv x q
\end{equation*}
\begin{equation*}
I x= y
\end{equation*}
\begin{equation*}
[ \Lambda_{x\mapsto y}]
\end{equation*}
inverse of this construction
\begin{equation*}
\specialorthogonal(3, \QQ) \simeq PB^\times (\QQ)
\end{equation*}
Proposition 3.21
\(\widetilde { \mathscr H_d}^*\) is a principal homogeneous space for \(\Cl(d)\text{.}\)
\begin{equation*}
\widetilde { \mathscr H_d} = \widetilde { \mathscr H_d}^*
\end{equation*}
if \(d \equiv 3 \pmod 4\text{.}\) The stabiliser of any point in \(\widetilde { \mathscr H_d}\) is the order 2 subgroup generated by \(\ideal p | (2)\text{.}\)
\begin{equation*}
\implies |\mathscr H_d | = 24 |h(d)| \text{ if } d \equiv 3 \pmod4
\end{equation*}
\begin{equation*}
|\mathscr H_d | = 12 |h(d)| \text{ if } d \equiv 1,2 \pmod4
\end{equation*}
Proof
Requires checking everything works as intended, it is local.
Explicit realization
\begin{equation*}
P \in \Cl(d)
\end{equation*}
\begin{equation*}
\leadsto M_P \in \specialorthogonal(3)
\end{equation*}
\begin{equation*}
x\in \mathscr H_d \implies Px = M_P x
\end{equation*}
\begin{equation*}
x = \begin{pmatrix} a\\b\\c \end{pmatrix}
\end{equation*}
Let
\begin{equation*}
P \bar P = (5)
\end{equation*}
because of the assumption on \(d\text{.}\)
\(P\) action, let \(x\in\mathscr H_d\text{.}\)
\begin{equation*}
N(P) = 5\leadsto q \in B(\ZZ)
\end{equation*}
\begin{equation*}
N(q) = 5 \implies 5 \in \{1\pm2i, 1\pm2j,1\pm2k\}B^\times(\ZZ)
\end{equation*}
two of these will give you the action of \(P, \bar P\text{.}\)
Example 3.22
\(q\) will act by \(q\inv x q\) acting on \(B^{(0)}\text{,}\) enough to check the action on \(i,j,k\text{.}\) Say
\begin{equation*}
q = 1+2i,\bar q = (1-2i)/5
\end{equation*}
\begin{equation*}
\bar q j q = \frac{(1-2i)}{5} j (1+2i)
\end{equation*}
\begin{equation*}
= \frac{j- 2k}{5} 1+2i
\end{equation*}
\begin{equation*}
j-2k - 2k - 2j????????
\end{equation*}
Basic construction
Let \(\mathscr A_5 = \{M_1^\pm, M_2^\pm, M_3^\pm\}\) starting with a solution \(x\in \mathscr H_d\) one gets two matrices \(\omega_1, \omega_{-1} \in\mathscr A_5\)
\begin{equation*}
x_{-3} \xleftarrow{\omega_{-3}} x_{-2} \xleftarrow{\omega_{-2}} x_{-1} \xleftarrow{\omega_{-1}} x \xrightarrow{\omega_{1}}x_1 \xrightarrow{\omega_{2}}x_2 \xrightarrow{\omega_{3}} x_3
\end{equation*}
where \(w_1\) comes from \(P\) and \(w_{-1}\) from \(\bar P\text{.}\)
This construction gives a well-defined (up to flip) path starting with a solution \(x \in \mathscr H_d\text{.}\)
Back to proof
Idea: if \(x\in\mathscr H_d\) then \(x\) and these paths are well defined mod \(p\text{.}\)
We are aiming for
Proposition 3.24
Let \(\Sigma(d,l,q)\) denote
\begin{equation*}
\#\{(x,x') \in \mathscr H_d\times \mathscr H_d : \gamma_x^{(l)} = \gamma_{x'}^{(l)} \text{ in } \mathscr H_d\}
\end{equation*}
where
\begin{equation*}
\gamma_x^{(l)}\in \mathscr H_d(q)
\end{equation*}
is the chain cut at \(\lb -l,l\rb\) and then reduced mod \(q\text{.}\) Then
\begin{equation*}
\sigma (l,d,q) \ll_\epsilon |\mathscr H_d| + d^\epsilon (1+ \frac{d}{q^2 5^{2l}})
\end{equation*}
Proof
Next time, Linnik's lemma \(\#\{xy = d\} \le d^\epsilon\text{.}\) One more input.
In \(\QQ(\sqrt d)\) we have \((5) = \ideal p \bar{ \ideal p}\)
\begin{equation*}
x \in \mathscr H_d = \{(x_1,x_2,x_3) : x_1^2 + x_2^2 + x_3^2 = d\}
\end{equation*}
there are 6 matrices \(\mathcal A = \{A^\pm, B^\pm, C^\pm\}\text{,}\) \(A,B,C \in \GL_2(\ZZ)\text{,}\) reduction of these \(\pmod q \acts \mathscr H_d(q)\text{.}\)
This induces a graph structure on \(\mathscr H_d(q)\text{,}\) (multiple edges between vertices are allowed).
\begin{equation*}
\mathscr H_d \to \mathscr H_d(q)
\end{equation*}
\begin{equation*}
[\ldots, x_{-1}, x_0, x_1, \ldots] \to \text{path in graph}
\end{equation*}
for \(l \in \NN\text{,}\)
\begin{equation*}
\gamma_x^{(l)} \subseteq \mathscr H_d(q)\text{.}
\end{equation*}
Linnik's basic lemma
Generalisation of the following
\begin{equation*}
r(d) = \sum_{a| d} 1 \le d^\epsilon\,\forall \epsilon\text{.}
\end{equation*}
Proposition 3.25 Linnik
Fix \(d\text{,}\) let \(c\in \ZZ\text{,}\) s.t. \(|c| \lt d\text{.}\) Then \(\forall \epsilon \gt 0\text{,}\)
\begin{equation*}
\#\{ (x_1,x_2) \in\mathscr H_d^2 : x_1\cdot x_2 = e\} \ll_\epsilon d^\epsilon\text{.}
\end{equation*}
Proof
Short detour on representations of quadratic forms by other quadratic forms.
Definition 3.26
Let \((Q, \ZZ^m)\text{,}\) \((R, \ZZ^n)\text{,}\) \(m\ge n\) be nondegenerate quadratic forms. Then \(Q\) is said to represent \(R\) if there exists a \(\ZZ\)-linear map
\begin{equation*}
\iota\colon \ZZ^n \to \ZZ^m
\end{equation*}
s.t.
\begin{equation*}
Q(\iota(x)) = R(x)\,\forall x\in \ZZ^n\text{.}
\end{equation*}
Example 3.27
\begin{equation*}
Q(x,y) = x \cdot y
\end{equation*}
\begin{equation*}
R(x) = d x^2
\end{equation*}
then let
\begin{equation*}
\iota\colon \ZZ \to \ZZ^2
\end{equation*}
\begin{equation*}
x \mapsto (ax,bx) \text{ where } ab =d
\end{equation*}
then
\begin{equation*}
Q(\iota(x)) = R(x)\text{.}
\end{equation*}
Let \(x_1, x_2 \in \mathscr H_d\) s.t. \(x_1\cdot x_2 =e\text{.}\)
\begin{equation*}
x_1=(a_1,b_1,c_1)\, x_2=(a_2,b_2,c_2)
\end{equation*}
consider
\begin{equation*}
\iota \colon \ZZ^2 \to \ZZ^3
\end{equation*}
\begin{equation*}
(u,v) \mapsto (u a_1 + v a_2, u b_1 + u b_2, u c_1 + v c_2) = u x_1 + v x_2
\end{equation*}
let \(Q(a,b,c) = a^2 + b^2 + c^2\text{.}\)
\begin{equation*}
Q(\iota(u,v)) = (u a_1 + v a_1)^2 + (u b_1 + v b_2)^2 + (u c_1 + v c_2)^2
\end{equation*}
\begin{equation*}
= du^2 + 2euv + dv^2
\end{equation*}
\begin{equation*}
= R_{d,e} (u,v)\text{.}
\end{equation*}
So the number of representations of \(R\) by \(Q/ \specialorthogonal_3(\ZZ)\) bounds number of solutions \(x_1 x_2 = e\text{.}\)
Theorem 3.28 Gordon Pell 1949
Let
\begin{equation*}
r(a,b,c) = \{ \text{reps. of } ax^2 + bxy + cy^2 \text{ by } x^2 + y^2 + z^2\}
\end{equation*}
then
\begin{equation*}
r(a,b,c) = 24 \cdot 2^\nu \prod_{p| 2(b^2 - 4ac)} r_p(a,b,c)
\end{equation*}
where
\begin{equation*}
\nu = \#\text{primes dividing } b^2 - 4ac
\end{equation*}
\begin{equation*}
r_p(a,b,c) = O(1) \text{ unless } p^2 | \gcd(a,b,c)
\end{equation*}
for us \(r(d,2e,d)\text{,}\) \(b^2 - 4ac = 4(e^2 - d^2)\) and
\begin{equation*}
r(d,2e,d) \ll_\epsilon d^\epsilon\text{.}
\end{equation*}
Exercise 3.29
Show this (recall \(d\) is a fundamental discriminant).
Lemma 3.30 Shadowing lemma
Let
\begin{equation*}
[x_{-l},\ldots, x_0, x_1 ,\ldots, x_l], [x'_{-l},\ldots, x'_0, x'_1 ,\ldots, x'_l]
\end{equation*}
be two marked paths in \(\mathscr H_d\text{.}\) Then these paths have the same reduction mod \(q\) iff \(x_0 = \pm x_0' \pmod{q5^l}\text{.}\)
Theorem 3.31
\begin{equation*}
\Sigma (d,l,q) = \#\{(x,x') \in \mathscr H_d^2 : \gamma_x^{(l)} = \gamma_{x'}^{(l)} \in \mathscr H_d(q)\}
\end{equation*}
\begin{equation*}
\implies \Sigma (d,l,q) \ll_\epsilon \# \mathscr H_d + d^\epsilon ( 1+ \frac{d}{q^2 5^{2l}}) \,\forall \epsilon \gt 0\text{.}
\end{equation*}
Proof
\begin{equation*}
\gamma_x^{(l)} = \gamma_{x'}^{(l)} \implies x = \pm x' \pmod{q5^l} \text{(by the Shadowing lemma)}
\end{equation*}
\begin{equation*}
\implies (x+x') ( x+ x') \equiv 0 \pmod{q^2 5^{2l}}
\end{equation*}
or
\begin{equation*}
\implies (x-x') ( x- x') \equiv 0 \pmod{q^2 5^{2l}}\text{.}
\end{equation*}
\begin{equation*}
\implies \Sigma(d,l,q) \ll 2\#\{(x,x') : x\ne x',\, x\cdot x' = d \pmod{q^2 5^{2l}}\}
\end{equation*}
recall
\begin{equation*}
\sum_{|e| \lt d,\,e = d \pmod{q^2 5^{2l}} } \{(x,x') : x\cdot x' = e\} \ll_\epsilon d^\epsilon (1+ \frac{d}{q^2 5^{2l}})\text{.}
\end{equation*}
Proof of the equidistribution statement
Recall:
\begin{equation*}
\operatorname{dev}_d(\bar x) = \frac { | \operatorname{red}\inv_q(\bar x)|}{|\mathscr H_d|/| \mathscr H_d(q)|} - 1
\end{equation*}
\begin{equation*}
\operatorname{dev}_d(\bar x) \to 0\text{ as } d \to \infty,\,d \equiv \pm 1 \pmod 5\,\, d\text{ fund. disc.}\text{.}
\end{equation*}
Sketch of proof
Assume \(B_\delta = \{ \bar x : \operatorname{dev}_d)\bar x) \\gt \delta\}\text{.}\) Lower bound by counting the number of paths that lie in \(\operatorname{red}\inv_q(B_\delta)\text{.}\) Upper bound by the expansion property of \(\mathscr H_d(q)\text{.}\)
Theorem 3.32 Linnik
Fix \(\nu, \delta\gt 0\) and let \(q \le d^{\frac 14 - \nu}\text{,}\) \((q,30) = 1\text{.}\) Then the fraction of \(\bar x \in \mathscr H_d(q)\) for which
\begin{equation*}
|\operatorname{dev}_d(\bar x)| \gt \delta
\end{equation*}
tends to 0 as \(d \to \infty\) through \(d \equiv \mp 1 \pmod 5\text{.}\)
Recall 3.3.
Last time
Proposition 3.33
Let
\begin{equation*}
\Sigma(d,l,q) = \# \{ (x,x') \in \mathscr H_d^2 : \gamma_x^{(l)} =\gamma_{x'}^{(l)}\}
\end{equation*}
then
\begin{equation*}
\Sigma(d,l,q) \ll_\epsilon| \mathscr H_d | + d^\epsilon \left(1+ \frac{d}{q^2 5^{2l}}\right)\text{.}
\end{equation*}
Proof of theorem
Let
\begin{equation*}
B_\delta = \{ \bar x\in \mathscr H_d (q) : \operatorname{dev}_d(\bar x) \gt \delta\}
\end{equation*}
and assume that
\begin{equation*}
B_\delta \ge \eta | \mathscr H_d(q)|
\end{equation*}
for some \(\eta \gt 0\text{.}\)
Observe that
\begin{equation*}
\frac{1}{|\mathscr H_d(q) | } \sum_{x\in \mathscr H_d } |\gamma_x^{(l)} \cap B_\delta | = (2l+ 1) \frac { | \operatorname{red}_q\inv (B_\delta)|}{|\mathscr H_d|}
\end{equation*}
or
\begin{equation}
\frac{1}{|\mathscr H_d(q) | } \sum_{x\in \mathscr H_d } \frac{ |\gamma_x^{(l)} \cap B_\delta | }{ (2l+ 1)} = \frac { | \operatorname{red}_q\inv (B_\delta)|}{|\mathscr H_d|}\tag{3.1}
\end{equation}
Choose an \(l\) s.t.
\begin{equation*}
\frac 15 |\mathscr H_d| \le q^2 5^{2l} \le 5 |\mathscr H_d|
\end{equation*}
note that we can indeed choose such an \(l\text{.}\)
By definition every \(\bar x \in B_\delta\) satisfies
\begin{equation*}
\operatorname{dev}_d(\bar x) \gt \delta
\end{equation*}
implies
\begin{equation*}
\frac{ |\operatorname{red}_q\inv(\bar x)|}{ |\mathscr H_d|/|\mathscr H_d(q)|} \gt 1 + \delta\text{.}
\end{equation*}
\begin{equation*}
| \operatorname{red}_q \inv(B_\delta)| \ge \frac{ |B_\delta | |\mathscr H_d| }{ |\mathscr H_d(q)|} (1+\delta)
\end{equation*}
so
\begin{equation*}
\frac{1}{|\mathscr H_d|} \sum_{x\in \mathscr H_d} \frac{|\gamma_x^{(l)} \cap B_\delta|}{(2l+1)} \gt \frac{|B_\delta|}{|\mathscr H_d(q)|}(1+\delta)
\end{equation*}
by the assumption that
\begin{equation*}
B_\delta \ge \eta | \mathscr H_d(q)|
\end{equation*}
we have this is
\begin{equation*}
\ge \frac{|B_\delta|}{|\mathscr H_d(q)|}+\eta\delta\text{.}
\end{equation*}
Let
\begin{equation*}
\Phi = \left\{ x\in \mathscr H_d : \frac{ | \gamma_x^{(l)} \cap B_\delta | }{ (2l + 1)} \gt \frac{|B_\delta|}{|\mathscr H_d(q)|} + \frac{\eta \delta }{2} \right\}
\end{equation*}
Note that
\begin{equation*}
\mathscr H_d = \Phi \sqcup \Phi^c
\end{equation*}
\begin{equation*}
\frac{1}{|\mathscr H_d|} \sum_{x\in \mathscr H_d} \frac{ | \gamma_x^{(l)} \cap B_\delta | }{ (2l + 1)} = \frac{1}{|\mathscr H_d|} \left( \sum_{x\in \Phi } \underbrace{\frac{ | \gamma_x^{(l)} \cap B_\delta | }{ (2l + 1)}}_{\le 1} + \sum_{x\in \Phi^c} \underbrace{\frac{ | \gamma_x^{(l)} \cap B_\delta | }{ (2l + 1)}}_{ \le \frac{|B_\delta|}{\mathscr H_d(q)|} + \frac{\eta \delta }{2} } \right)
\end{equation*}
\begin{equation*}
\le\frac{1}{|\mathscr H_d|} \left( |\Phi | + \overbrace{| \Phi^c|}^{=|\mathscr H_d| - |\Phi|} \left( \frac{|B_\delta|}{\mathscr H_d(q)|} + \frac{\eta \delta }{2} \right)\right)
\end{equation*}
\begin{equation}
\implies |\Phi| \ge \frac{\eta \delta }{2} | \mathscr H_d|\text{.}\label{eqn-phi-bd-linnik-pf}\tag{3.2}
\end{equation}
Now we count the number of marked non-backtracking paths on \(\mathscr H_d(q)\) of length \(2l+1\) (write \(MNBP(\mathscr H_d(q))\) for this quantity), which in addition satisfy
\begin{equation*}
\frac{|\gamma_x^{(l)} \cap B_\delta|}{(2l+1)} \gt \frac{|B_\delta|}{|\mathscr H_d(q) | } + \frac{\delta \eta}{2}\text{.}
\end{equation*}
By (3.2) we have \(\eta \delta |\mathscr H_d|/2\) of these paths “upstairs”.
Recall Siegel implies
\begin{equation*}
d^{\frac 12 - \epsilon} \lt | \mathscr H_d| \lt d^{\frac 12 + \epsilon}
\end{equation*}
so upstairs we have \(\gg_\epsilon \frac{\eta \delta }{2} d^{\frac 12 - \epsilon }\) paths.
Claim 1: The previous proposition implies not mant of these paths give the same path mod \(q\)
\begin{equation*}
MNBP(\mathscr H_d(q)) \gg_{\epsilon,\delta \eta} d^{\frac 12 - \epsilon}\text{.}
\end{equation*}
On the other hand the total number of marked non-backtracking paths on
\begin{equation*}
\mathscr H_d(q)
\end{equation*}
is
\begin{equation*}
| \mathscr H_d(q)| 6 \cdot 5^{2l -1}
\end{equation*}
also observe that \(|\mathscr H_d(q)| \ll q^2\) so the total number \(\sim q^2 5^{2l} \ll d^{1/2 + \epsilon}\)
\begin{equation*}
\frac 15 |\mathscr H_d| \lt q^2 5^{2l} \lt 5 |\mathscr H_d|
\end{equation*}
Final input is a Chernoff type bound.
Proposition 3.34
Fix \(\eta, \epsilon \gt 0\) for any subset
\begin{equation*}
B \subseteq \mathscr H_d(q)
\end{equation*}
s.t.
\begin{equation*}
| B | \ge \eta | \mathscr H_d(q)|
\end{equation*}
the fraction of non-backtracking paths of length \(2l + 1\) satisfying
\begin{equation*}
\left| \frac{| \gamma \cap B|}{2l + 1 } - \frac{|B | }{|\mathscr H_d(q)|} \right| \ge \epsilon
\end{equation*}
is
\begin{equation*}
\ll_{\epsilon,\eta} e^{-cl}
\end{equation*}
for some \(c =c(\epsilon, \eta) \gt 0\text{.}\)
Claim 2: This proposition implies that the fraction
\begin{equation*}
MNBP(\mathscr H_d(q)) \ll d^{\frac 12 - \tau}
\end{equation*}
for some \(\tau \gt 0\text{.}\) Then
\begin{equation*}
\gg_\epsilon d^{\frac 12 - \epsilon}
\end{equation*}
\begin{equation*}
\ll_\epsilon d^{\frac 12 - \tau}
\end{equation*}
give a contradiction.
We now check some claims made above.
Claim 3.35 Claim 1
Bound on \(\#\{(x,x') : \gamma_x ^{(l) } = \gamma_{x'}^{(l)}\}\) gives a bound on the number of distinct paths.
Proof
\begin{equation*}
\#\{ (a,a') \in A^2 : a \equiv a' \pmod q\} = \sum n_{a_i}
\end{equation*}
gives a lower bound on \(A \pmod q\text{.}\) To show this note that the
\begin{equation*}
| A(q)| = | A | - \sum_{i=1}^{|A|} \frac{n_{a_i} - 1}{n_{a_i}} - \sum_{i=1}^{|A|} \frac 1{n_{a_i}}
\end{equation*}
where \(n_{a_i} = \# \{ a \in A : a \equiv a_i \pmod q\}\text{.}\)
Claim 3.36 Claim 2
Follows from two points
\begin{equation*}
d^{\frac 12 - \epsilon} \ll q^2 5^{2l} \ll d^{\frac 12 + \epsilon}
\end{equation*}
\begin{equation*}
q \lt d^{\frac 14 - \nu}
\end{equation*}
Subsection 3.2 The adelic picture
Notation: \(\adeles\) is the ring of adeles of \(\QQ\)
\begin{equation*}
\prod_p' \QQ_p
\end{equation*}
\begin{equation*}
\adeles_f = \prod_{p \text{ finite}}' \QQ_p
\end{equation*}
so that
\begin{equation*}
\adeles = \RR \times \adeles_f\text{.}
\end{equation*}
\begin{equation*}
\widetilde \ZZ \colon \varprojlim_n \ZZ/n = \prod \ZZ_p\text{.}
\end{equation*}
\begin{equation*}
\widehat \ZZ \subseteq \adeles_f
\end{equation*}
maximal compact.
The genus of a quadratic space
For us,
\begin{equation*}
L\subseteq \QQ^3
\end{equation*}
a lattice, then
\begin{equation*}
\GL_3(\adeles_f) \ni g = \prod g_p
\end{equation*}
\begin{equation*}
g\cdot L = \{ \alpha \subseteq \QQ^3 : \alpha_p = g_p L_p \,\forall p\}
\end{equation*}
\begin{equation*}
\operatorname{genus}_{\specialorthogonal_3} (\mathcal L_p) = \specialorthogonal_3(\adeles_f) \mathcal L_0
\end{equation*}
\begin{equation*}
\mathcal L_0 = \ZZ^3,\,a^2+b^2 + c^2\text{.}
\end{equation*}
Proposition 3.37
There is one \(\specialorthogonal_3(\QQ)\) orbit of \(\mathcal L_0\text{.}\) i.e.
\begin{equation*}
\specialorthogonal_3(\adeles_f) = \specialorthogonal_3(\QQ) \specialorthogonal_3(\widehat\ZZ)
\end{equation*}
Proof
Based on quaternions having class number 1.
More definitions:
\begin{equation*}
P = \{ (\mathcal L, x) : \mathcal L \in \operatorname{genus}_{\specialorthogonal_3} (\mathcal L_0), x \in \mathcal L,\, x \cdot x = d\}
\end{equation*}
\begin{equation*}
P_{(q)} = \{ (\mathcal L, \bar x) : \mathcal L \in \operatorname{genus}_{\specialorthogonal_3} (\mathcal L_0), \bar x \in \mathcal L/q\mathcal L,\, \bar x \cdot \bar x = d \pmod q\}
\end{equation*}
\begin{equation*}
\widetilde{\mathscr H_d} \simeq \specialorthogonal_3 (\QQ) \backslash P
\end{equation*}
\begin{equation*}
K_f [q] = \ker(\specialorthogonal_3(\widehat \ZZ) \to \specialorthogonal_3(\ZZ/q\ZZ))
\end{equation*}
\begin{equation*}
K'_f [q] = \{ g \in \specialorthogonal_3(\widehat \ZZ) : g \bar x_q = \bar x_q\}
\end{equation*}
we fix a base point \(\bar x_q \in \mathscr H_d(q)\text{.}\) Then
\begin{equation*}
\widetilde{\mathscr H_d}(q) \simeq \specialorthogonal_3 (\QQ) \backslash P_{(q)} \simeq \specialorthogonal_3 (\QQ) \backslash \specialorthogonal_3(\adeles_f) / K_f'[q]
\end{equation*}
here we are heavily using the exceptional isomorphism
\begin{equation*}
\specialorthogonal_3 \simeq \PGL_2\text{.}
\end{equation*}
Claim 3.38
\begin{equation*}
\specialorthogonal_3(\ZZ) \backslash \mathscr H_d(q) \simeq \specialorthogonal_3(\QQ) \backslash P_{(q)}
\end{equation*}
Proof sketch
\begin{equation*}
\specialorthogonal_3(\QQ) \acts P_{(q)} \text{ diagonally}
\end{equation*}
\begin{equation*}
\specialorthogonal_3(\adeles_f) = \specialorthogonal_3(\QQ) \specialorthogonal_3(\widehat \ZZ)
\end{equation*}
every orbit has a representation \(\simeq (\mathcal L_0, \bar x),\,\bar x \in \mathscr H_d(q)\text{.}\)
\begin{equation*}
(\mathcal L_0, \bar x) \simeq (\mathcal L_0, \bar x')
\end{equation*}
\begin{equation*}
\iff
\end{equation*}
\begin{equation*}
\exists \gamma \in \underbrace{\specialorthogonal_3(\widehat \ZZ) \cap \specialorthogonal_3(\QQ)}_{\specialorthogonal_3(\ZZ)}
\end{equation*}
Claim 3.39
\begin{equation*}
\specialorthogonal_3(\QQ) \backslash P_{(q)} \simeq \specialorthogonal_3(\ZZ) \backslash \mathscr H_d(q) / K'_f[q]
\end{equation*}
Proof sketch
Observe
\begin{equation*}
\specialorthogonal_3(\adeles_f) \acts P_{(q)}
\end{equation*}
via
\begin{equation*}
\mathcal L \otimes \widehat \ZZ/ q\mathcal L \otimes \widehat \ZZ
\end{equation*}
This action is transitive, i.e.
\begin{equation*}
P_{(q)} = \specialorthogonal_3(\adeles_f) [ \mathcal L_0, \bar x_q]
\end{equation*}
What about
\begin{equation*}
\specialorthogonal_3(\ZZ) \backslash \mathscr H_d\text ?
\end{equation*}
Fix another base point \(x_0\in \mathscr H_d\)
\begin{equation*}
\specialorthogonal_3(\ZZ) \backslash \mathscr H_d \simeq \specialorthogonal_{x_0} (\QQ) \backslash \specialorthogonal_{x_0} (\adeles_f)/ \specialorthogonal_{x_0}(\widehat \ZZ)
\end{equation*}
where \(\specialorthogonal_{x_0}\) is the stabiliser of \(x_0\text{.}\)
\begin{equation*}
\xymatrix{
\mathscr H_d \acts \text{ class gp.} \ar[d] \\
\mathscr H_d(a) \acts \specialorthogonal_3(\adeles_f)
}
\end{equation*}
More concretely
\begin{equation*}
\specialorthogonal_{x_0} : \QQ^3 \simeq B^{(0)} (\QQ)
\end{equation*}
\begin{equation*}
\specialorthogonal_3 \simeq PB^\times
\end{equation*}
\begin{equation*}
\specialorthogonal_{x_0} \simeq PB^\times_{x_0} \simeq \langle a + b x_0 \rangle / Z
\end{equation*}
\begin{equation*}
\simeq \Res_{K/\QQ} \mathbf G_m / \mathbf G_m
\end{equation*}
where \(K = \QQ(\sqrt{-d})\text{.}\)
Recall
\begin{equation*}
K^\times \backslash \adeles_{K,f}^\times / \widehat \ints_K^\times \simeq \Pic(\ints_K)\text{.}
\end{equation*}
The graph structure on \(\mathscr H_d(q)\)
\(\specialorthogonal_3(\adeles_f)\) is a large group, but so is \(K_f' \lb q\rb \text{.}\) What is indeed true is
\begin{equation*}
\mathscr H_d(q) \cong \Gamma_{(3,q)} \specialorthogonal_3(\QQ_5) / K_5
\end{equation*}
analogous to
\begin{equation*}
\mathbf H = \specialorthogonal_2 \backslash \SL_2(\RR)/ \SL_2(\ZZ)\text{.}
\end{equation*}
Where
\begin{equation*}
\Gamma_{(3,q)} PB^\times(\QQ) \cap PB^\times(\QQ_5) K_f[3,q]
\end{equation*}
\begin{equation*}
K_{f} [3,q] = K_f[3] \cap K_f'[q] \subseteq K_f'[q]
\end{equation*}
Now
\begin{equation*}
\PGL_2(\QQ_5) / \PGL_2(\ZZ_5)
\end{equation*}
is a Bruhat-Tits tree, a 6-regular tree.
Automorphic spectrum of \(\PGL_2\) of \(\mathbf F_q(t)\)
Analogous to \(\SL_2(\ZZ) \backslash \HH\text{.}\)
Reference I. Efrat: Automorphic spectrum on the tree of \(\PGL_2\)
Setup: \(k = \FF_q\text{,}\) \(v_\infty = - \deg\) so \(|f|_\infty = q^{\deg (f)}\text{.}\)
\begin{equation*}
K = (k(t))_\infty = k((\frac 1t ))
\end{equation*}
\begin{equation*}
\ints_\infty = \ints = k [[ t \inv ]]
\end{equation*}
\begin{equation*}
\Gamma = \PGL_2(\FF_q[[t]])
\end{equation*}
Analogies:
\begin{equation*}
\HH \leftrightsquigarrow \PGL_2(K) / \PGL_2(\ints) = X
\end{equation*}
\begin{equation*}
\SL_2(\ZZ) \leftrightsquigarrow \Gamma
\end{equation*}
Problem: decompose \(L^2(\Gamma \backslash X)\text{.}\) In the case of
\begin{equation*}
\SL_2(\ZZ) \backslash \HH \leadsto L^2(\SL_2(\ZZ)\backslash \HH) = \underbrace{R}_{\text{Residual}} \oplus \underbrace{C}_{\text{cusp forms}} \oplus \underbrace{E}_{\text{Eisenstein}}\text{.}
\end{equation*}
Theorem 3.41
\begin{equation*}
L^2( \Gamma \backslash X) = R\oplus E
\end{equation*}
\(\dim(R) = 2,\,\dim(E) = \infty\) i.e. no cusp forms.
Little bit on \(X\) (a \((q+1)\)-regular tree)
Ref: Serre “Trees”.
Description
-
In terms of lattices
\begin{equation*}
L = \{ av_1 + bv_2 : a,b\in \ints\}
\end{equation*}
s.t.
\begin{equation*}
L\otimes_\ints K \simeq K^2\text{.}
\end{equation*}
Matrix
\begin{equation*}
\begin{pmatrix} v_1 \amp v_2 \end{pmatrix} \GL_2(K)
\end{equation*}
changing basis implies \(L\) a class in \(\GL_2(K) / \GL_2(\ints)\text{.}\)
\begin{equation*}
L_1 \sim L_2
\end{equation*}
if \(\exists a \in K^\times\) s.t. \(L_1 = a L_2\text{.}\)
\begin{equation*}
X = \{ \text{equivalence classes of lattices} \}
\end{equation*}
A class \(L\) is adjacent to \(L'\) \(\iff\) \(\exists\) representations \(L_0, L_0'\) s.t. \(L_0' \subseteq L_0\) and \(L_0/L_0' \simeq \FF_q\text{.}\)
-
Group theoretic
\begin{equation*}
B = \left\{ \begin{pmatrix} * \amp * \\ 0 \amp 1 \end{pmatrix} \right\}/K
\end{equation*}
Fact: \(\PGL_2(K) = B \cdot \PGL_2(\ints)\text{.}\) Each coset \(b \PGL_2(\ints)\) has a representative
\begin{equation*}
\begin{pmatrix} t^n \amp x \\ 0 \amp 1\end{pmatrix}
\end{equation*}
\(n\) unique and \(x \in K\text{.}\)
Vertices of \(X\)
\begin{equation*}
\begin{pmatrix} t^n \amp x \\ 0 \amp 1\end{pmatrix} \PGL_2(\ints)
\end{equation*}
\begin{equation*}
\begin{pmatrix} t^n \amp x \\ 0 \amp 1\end{pmatrix} \sim_{\text{adjacent}}
\begin{cases}
\begin{pmatrix} t^{n-1} \amp \xi t^n + x \\ 0 \amp 1\end{pmatrix},\, \xi \in \FF_q
\begin{pmatrix} t^{n+1} \amp x \\ 0 \amp 1\end{pmatrix}
\end{cases}
\end{equation*}
Fact:
\begin{equation*}
\Gamma\backslash X = \text{chain } \bullet -\bullet - \bullet - \cdots
\end{equation*}
Measures: Normalise Haar on \(\PGL_2\) s.t. \(\mu(\PGL_2(\ints)) = q(q-1)\text{.}\) \(\leadsto\) On \(\Gamma\backslash X\text{,}\) \(\mu\) gives vertices weights
\begin{equation*}
n=0: \frac{1}{q+1}
\end{equation*}
\begin{equation*}
n=1: \frac{1}{q}
\end{equation*}
\begin{equation*}
n=2: \frac{1}{q^2}
\end{equation*}
\begin{equation*}
\vdots
\end{equation*}
\begin{equation*}
n: \frac{1}{q^n}
\end{equation*}
Fact 3.43
The algebra of operators on functions on \(X\) that commute with the automorphisms of \(X\) is generated by
\begin{equation*}
(T(f))(s) = \sum_{s' \sim s} f(s')
\end{equation*}
Given \(f\colon \Gamma\backslash X \simeq \NN \to \CC\)
\begin{equation*}
(T(f))(n) = \begin{cases} (q+1)f(1),\amp n=0 \\ qf(n-1) + f(n+1),\amp n \gt 0 \end{cases}
\end{equation*}
\(T\) is self-adjoint with respect to \(\pair{}{}_\mu\text{.}\)
Eigenfunctions of \(T\)
\begin{equation*}
(T f)(n) = \lambda \cdot f(n),\,\forall n \in \NN\text{.}
\end{equation*}
\begin{equation*}
\implies \lambda f(0) = (q+1) f(1)
\end{equation*}
\begin{equation*}
\lambda f(n) = q f(n-1) + f(n+1),\,\forall n \gt 0
\end{equation*}
i.e.
\begin{equation*}
\begin{pmatrix} f(n+1) \\ f(n) \end{pmatrix} = \begin{pmatrix} \lambda \amp -q \\ 1 \amp 0 \end{pmatrix}^n\begin{pmatrix} f(1) \\ f(0) \end{pmatrix}\text{.}
\end{equation*}
Now fix a normalization
\begin{equation*}
f(0) = q+1
\end{equation*}
\begin{equation*}
f(1) = \lambda\text{.}
\end{equation*}
Characteristic values:
\begin{equation*}
x_1, x_2 = \frac{-\lambda \pm \sqrt{\lambda^2 - 4q}}{2}
\end{equation*}
note \(x_1 = x_2 \iff \lambda = \pm 2\sqrt q\text{.}\)
Eigenfunctions
\begin{equation*}
f_\lambda (n) = \begin{cases} \frac{1}{\sqrt{\lambda^2 - 4q}} \left(\lambda( x_1^n - x_2^n) - q(q+1)(x_1^{n-1} - x_2^{n-1})\right),\amp n \gt0 \\ q+1,\amp n=0 \end{cases}
\end{equation*}
for \(\lambda \ne \pm 2\sqrt q\) note \(f_{q+1} = q+1\text{.}\)
\begin{equation*}
f_{2\sqrt q} (n) = (q+1 - (q-1) n)q^{n/2}
\end{equation*}
\begin{equation*}
f_{-2\sqrt q} (n) = (-1)^n f_{2\sqrt q}
\end{equation*}
We now consider the space \(L^2( \Gamma\backslash X)\) with the operator \(T\text{.}\)
Question: which the \(f_\lambda\) are in \(L^2 (\Gamma\backslash X)\text{?}\)
Proposition 3.45
The only \(f_\lambda\in L^2\) with \(|\lambda| \gt 2 \sqrt q\) are \(\lambda = \pm(q+1)\text{.}\)
Proof
Normalise \(f_\lambda\) by
\begin{equation*}
(x_2 - x_1) f = x_1^{n-1} (\lambda x_1 - q(q+1)) - x_2^{n-1} (\lambda x_2 - q(q+1))
\end{equation*}
assume WLOG \(|x_1| \gt \sqrt q\) \(\implies\) \(|x_2| \lt \sqrt q\) (since \(|\lambda| \gt 2 \sqrt q\)) so \(x_2^n = o(q^{n/2})\text{.}\)
So \(\lambda x_1 - q(q+1) = 0\) (to be in \(L^2\)). So \(\lambda = \pm q(q+1)\text{.}\)
What about the rest?
Note: \(|\lambda| \lt 2 \sqrt q \implies x_1 = \bar x_2\text{.}\)
Renormalise: Let \(x_1 = \sqrt q e^{i\theta},x_2 = \sqrt q e^{-i\theta}\text{.}\)
\begin{equation*}
\lambda = 2\sqrt q \cos \theta,\,0 \lt \theta \lt \pi
\end{equation*}
\begin{equation*}
\tilde f_\theta(n) = \frac{x_1 - x_2}{2\sqrt q} f_{2\sqrt q \cos \theta} (n)
\end{equation*}
so
\begin{equation*}
\tilde f_\theta(n) = \begin{cases} q^{n/2} i (\sin(n+1)) \theta - q\sin(n-1) \theta\\ (q+1) i \sin \theta\end{cases}
\end{equation*}
Proposition 3.46
\begin{equation*}
\tilde f_\theta \not \in L^2
\end{equation*}
for \(0 \lt \theta \lt \pi\text{.}\)
Spectral decomposition of \(L^2(\Gamma \backslash X)\)
Continuous spectrum: Extend \(\tilde f_\theta\) to an odd function \(- \pi \lt \theta \lt \pi\text{.}\) Given \(\psi \in L^2(\Lambda\backslash X)\) consider
\begin{equation*}
F_\psi(n) = \frac{1}{2\pi} \int_{-\pi}^\pi \psi(\theta) \tilde f_\theta (n) \diff \theta\text{.}
\end{equation*}
Theorem 3.47
\(F_\psi \in L^2(\Gamma\backslash X)\text{.}\)
\begin{equation*}
\pair{F_\psi}{F_\phi}_\mu = \pair{\psi}{\phi} \frac{1}{2\pi} \int_0^\pi \psi(x) \overline{\phi(x)} ((q-1)^2 - 4 q \sin^2(x)) \diff x
\end{equation*}
Theorem 3.48
Let \(g \in L^2(\Gamma \backslash X)\text{.}\) Then
\begin{equation*}
g(n) = \pair{g}{u_1} u_1(n) + \pair{g}{u_2} u_2(n) + 2 \pi \int_0^\pi \pair{g }{\tilde f _\theta} \tilde f _\theta \frac{\diff \theta}{(q-1)^2 - 4q \sin^2 \theta}
\end{equation*}
where
\begin{equation*}
u_1 = \sqrt{\frac{q^2-1}{2q}},\, u_2 = \sqrt{\frac{q^2-1}{2q}} (-1)^n\text{.}
\end{equation*}