\begin{equation*}
I(s) = \int_{\Gamma\backslash \HH} h_m(z,-\bar z') E(z; s) y^k \diff \mu(z)
\end{equation*}
for \(\Re(s) \gt 1\text{.}\) (This “goes around the convergence issues”). Implies
\begin{equation*}
\Res_{s=1 } I(s) = \trace( T_m) \frac{c_k}{m^{k-1}} E(z; s) = \frac12 \zeta(2s) \pi^s \Gamma(s) \sum_{\gamma \in \Gamma_\infty \backslash \Gamma} \Im(\gamma z)^s\text{.}
\end{equation*}
Calculation of \(I(s)\)
First step:
Recall that there exists a one to one correspondence between (for fixed \(m,t\))
\begin{equation*}
\{A \in \Mat_{2\times 2 }(\ZZ) : \trace(A) = t,\,\det(A) =m\}
\end{equation*}
\begin{equation*}
\updownarrow
\end{equation*}
\begin{equation*}
\{\phi(u,v) = au^2 + buv + cv^2 : \Delta_\phi = | \phi| = b^2- 4ac= t^2 - 4m\}
\end{equation*}
\begin{equation*}
\begin{pmatrix} a\amp b \\ c\amp d \end{pmatrix} \mapsto \phi(u,v) = cu^2 + (d-a) uv - b^2v^2
\end{equation*}
\begin{equation*}
\phi(u,v) = au^2 + buv + cv^2 \mapsto \begin{pmatrix} \frac12 (t-b)\amp -c \\ a\amp \frac12 (t+b)\end{pmatrix}\text{.}
\end{equation*}
Using this
\begin{equation*}
y^k h_m(z,-\bar z) = \sum_{a,b,c,d\in \ZZ,\,ad-bc = m} \frac{y^k}{(c|z|^2 + d\bar z - az - b)^k}
\end{equation*}
\begin{equation*}
= \sum_{t\in \ZZ} \sum_{ad-bc = m,\, a+d = t} \frac{y^k}{(c|z|^2 + d\bar z - az - b)^k}
\end{equation*}
\begin{equation*}
= \sum_{t\in \ZZ} \sum_{\phi,\, |\phi | = t^2 - 4m} R_\phi(z,t)
\end{equation*}
where
\begin{equation*}
R_\phi(z,t) = \frac{y^k}{(a|z|^2 + bx + c - ity)^k}\text{.}
\end{equation*}
Exercise: substitute \(a= \frac 12 (t-b), b= -c , c = a, d= \frac12 (t+b)\text{.}\)
Proposition 2.9
For \(s\in \CC\) s.t. \(2-k \lt \Re(s) \lt k- 1\)
\begin{equation*}
\sum_{t\in \ZZ}\int_{\Gamma\backslash \HH} |E(z; s)|\left| \sum_{|\phi| = t^2 - 4m } R_\phi(z,t) \right| \diff \mu(z) \lt \infty
\end{equation*}
Recall that \(\SL_2(\ZZ)\) acts on binary quadratic forms \(\Phi\) by
\begin{equation*}
\gamma \phi(u,v) = \phi(au + cv, bu + dv)= \phi(\gamma^\transpose \begin{pmatrix}u\\v\end{pmatrix})\text{.}
\end{equation*}
Theorem 2.11 Geometric side of the Eichler-Selberg trace formula
For \(t,m\) fixed
\begin{equation*}
\int_{\Gamma\backslash\HH} \left( \sum_{|\phi| = t^2 - 4m} R_\phi (z,t)\right) E(z,s) \diff \mu(z) = \zeta(s, \Delta) \left\{I(\Delta, t, s) + I(\Delta, -t, s)\right\}
\end{equation*}
\begin{equation*}
+ \begin{cases}(-1)^{k/2} \frac{\Gamma(s+ k - 1) \zeta(s) \zeta(2s)}{(2\pi)^{s -1} \Gamma(k)}\amp \Delta = 0 \\ 0 \amp \Delta \ne 0\end{cases}
\end{equation*}
for \(\Delta = t^2 - 4m\text{.}\)
Where
\begin{equation*}
\zeta(s,\Delta) = \sum_{\phi \pmod \Gamma,\, |\phi| = \Delta} \sum_{(m,n) \in \ZZ^2 \smallsetminus\{(0,0\}/\Aut(\phi),\,\phi(m,n) \gt0} \frac{1}{\phi(m,n)^s}
\end{equation*}
\begin{equation*}
\Aut(\phi) = \{ \gamma\in \SL_2(\ZZ) : \gamma\phi = \phi\}
\end{equation*}
\begin{equation*}
I(\Delta, t, s) = \int_0^\infty \int_{-\infty}^\infty \frac{y^{k+s - 2}}{(x^2 + y^2 +ity -\frac14 \Delta)^k} \diff x \diff y
\end{equation*}
\begin{equation*}
= \frac{\Gamma(k- \frac 12)\Gamma(\frac 12)}{\Gamma(k)} \int_0^\infty \frac{y^{k+s -2}}{(y^2 + ity - \frac14 \Delta)^{k-\frac 12}} \diff y\text{.}
\end{equation*}
Proposition 2.13
Let \(\zeta(s, \Delta)\) be as above and \(\Re(s) \gt 1\text{.}\) Then
-
\begin{equation*}
\zeta(s, \Delta) = \zeta(2s)\sum_{a = 1}^\infty \frac{n(a)}{a^s}
\end{equation*}
wn(ae
\begin{equation*}
a(n) = \#\{b \pmod {2a} : b^2 \equiv \Delta \pmod{4a}\}\text{.}
\end{equation*}
- \(\zeta(s, \Delta)\) has meromorphic continuation and functional equation.
\begin{equation*}
\gamma(s, \Delta)\zeta(s, \Delta) = \gamma(1-s, \Delta) \zeta(1-s, \Delta)
\end{equation*}
where
\begin{equation*}
\gamma(s, \Delta) = \begin {cases} (2\pi)^{-s} |\Delta|^{s/2} \Gamma(s) \amp \Delta\lt 0 \\ \pi^{-s} \Delta^{s/2} \Gamma(s/2) \amp \Delta \gt 0\end{cases}\text{.}
\end{equation*}
-
\begin{equation*}
\zeta(s, \Delta) = \begin{cases} 0 \amp \Delta\equiv 2,3\pmod4 \\ \zeta(s)\zeta(2s-1) \amp \Delta = 0\\ \zeta(s) L_D(s) \sum_{d|f } \mu(d)\frac Dd d^{-s} \sigma_{1-2s}(\frac fd)\end{cases}
\end{equation*}
for \(D\) the fundamental discriminant, so that \(\Delta = D f^2\) and \(\sigma_\nu (n) = \sum_{d|n} d^\nu\text{,}\) \(\mu\) is the Möbius function, and \(L_D(s) = L(s, \legendre{D}{\cdot})\text{.}\)
\begin{equation*}
\Res_{s=1} \zeta(s, \Delta) = \frac{\pi}{\sqrt{|\Delta|}} H(|\Delta|)\text{.}
\end{equation*}
Proof
-
A slick proof involves noting that
\begin{equation*}
n(a) = \#\{\phi \in \Phi ,\,(m,n) \in \ZZ^2 : \phi(m,n) = 0, |\phi| = \Delta,\, \gcd(m,n) = 1\}
\end{equation*}
comes from the theory of quadratic forms.
Instead note that \(\Gamma \acts \Phi\) via \(\gamma \cdot \phi =(u,v) = \phi(\gamma^\transpose \begin{pmatrix} u\\v \end{pmatrix})\text{.}\) We also have \(\Gamma\acts \ZZ^2\) via \(\gamma \cdot (m,n) = \gamma \begin{pmatrix} m\\n \end{pmatrix}\text{.}\)
Note that
\begin{equation*}
(\gamma ^\transpose)\inv = \begin{pmatrix} 0 \amp 1 \\ -1 \amp 0 \end{pmatrix} \gamma \begin{pmatrix} 0 \amp -1 \\ 1 \amp 0\end{pmatrix}
\end{equation*}
(check!).
Define
\begin{equation*}
\pair{-}{-}\colon \Phi \times X
\end{equation*}
\begin{equation*}
\pair{\phi}{(m,n)} = \phi(n,-m)
\end{equation*}
\begin{equation*}
= \phi \left( \begin{pmatrix} 0 \amp 1 \\ -1 \amp 0 \end{pmatrix} \begin{pmatrix} m \\n \end{pmatrix} \right)\text{.}
\end{equation*}
Observe \(\pair{-}{-}\) is \(\Gamma\)-invariant! i.e.
\begin{equation*}
\pair{\phi}{(m,n)} = \pair{\gamma \phi }{ \gamma(m,n)}
\end{equation*}
for all \(\gamma \in \Gamma\text{.}\) Then
\begin{equation*}
\zeta(s, \Delta) = \sum_{\phi \pmod \Gamma,\, |\phi| = \Delta} \sum_{(m,n) \in X/\Aut(\phi),\,\phi(m,n) \gt 0} \frac{1}{\phi(m,n)^s},\,\Re(s) \gt 1
\end{equation*}
\begin{equation*}
= \sum_{\phi \in \Phi_\Delta/\Gamma} \sum_{x \in X/\Aut(\phi)} \frac{1}{\pair{\phi}{x}^s}
\end{equation*}
\begin{equation*}
= \sum_{(\phi,x) \in (\Phi_\Delta \times X)/\Gamma} \frac{1}{\pair{\phi}{x}^s}
\end{equation*}
\begin{equation}
= \sum_{x\in X/\Gamma} \sum_{\phi\in \Phi_\Delta/\Aut(x)} \frac{1}{\pair{\phi}{x}^s}\label{eqn-zeta-delta-calc}\tag{2.4}
\end{equation}
we are using Fubini to swap the order of summation as the pairing is invariant. Finally note that
\begin{equation*}
x= \pm(m,n)\sim_{\Gamma} \pm(0,\gcd(m,n)),\, r = \gcd(m,n) \gt 0
\end{equation*}
so (2.4)
\begin{equation*}
= \sum_{r= 1}^\infty \sum_{\phi \in \Phi_\Delta/\Aut((r,0))} \frac{1}{\phi(r,0)^s}
\end{equation*}
what is \(\Aut((r,0))\text{?}\) It is
\begin{equation*}
\Gamma_\infty^\transpose = \left\{\begin{pmatrix} 1 \amp 0 \\ \ast \amp 1 \end{pmatrix}\right\} \text{ as }\begin{pmatrix} 1 \amp 0 \\ \ast \amp 1 \end{pmatrix}\begin{pmatrix} 0\\ r \end{pmatrix} = \begin{pmatrix} 0\\ r \end{pmatrix}
\end{equation*}
continuing we have
\begin{equation*}
\zeta(2s) \sum_{a= 1}^\infty \frac{1}{a^s} \left\{ \sum_{b \pmod{2a},\, b^2 \equiv \Delta \pmod{4a}} 1\right\}
\end{equation*}
Exercise, show this is the sum from before. Hint, consider, explicitly, the action of \(\gamma\in \Gamma_\infty^\transpose\) on \(\phi(u,v)\text{.}\)
Back to the proof of the trace formula.
Lemma 2.14
Let \(k \in \ZZ_{\gt 2},\,k\equiv 0 \pmod 2,\,\Delta \in \ZZ,\,\Delta \equiv 0,1 \pmod4,\,t\in \RR\) s.t. \(t^2 \gt \Delta\text{.}\) For each \(\phi \in \Phi_\Delta\) let
\begin{equation*}
R_\phi (t,z) = \frac{y^k}{(a|z|^2 + bx + c - ity )^k}
\end{equation*}
Then for \(s \in \CC\text{,}\) \(s \ne 1 \text{,}\) \(1-k \lt \Re(s) \lt k\text{.}\)
\begin{equation*}
\int_{\Gamma \backslash \HH} \left(\sum_{|\phi| = \Delta} R_\phi(t,z)\right) E(z, s) \diff \mu (z)
\end{equation*}
\begin{equation*}
= \zeta(s,\Delta) \{I_k(\Delta, t, s) + I_k(\Delta, -t, s)\}
\end{equation*}
\begin{equation*}
+ \begin{cases}(-1)^{k/2} \frac{\Gamma(s+ k - 1) \zeta(s) \zeta(2s)}{(2\pi)^{s -1} \Gamma(k)}\amp \Delta = 0 \\ 0 \amp \Delta \ne 0\end{cases}
\end{equation*}
Proof
- Note \(R_{\gamma\phi}(t,z) = R_\phi (t, \gamma^\transpose z)\) which means that the integral is well defined (check this).
-
\begin{equation*}
\int_{\Gamma\backslash \HH} \left( \sum_{|\phi| = \Delta} R_\phi (t,z)\right) E(z; s) \diff \mu (z)
\end{equation*}
\begin{equation*}
= \int_{\Gamma\backslash \HH} \sum_{|\phi| = \Delta} \sum_{(m,n) \in \ZZ^2 \smallsetminus \{(0,0)\}} R_\phi(t, z) \frac{y^s}{|mz+n|^{2s}} \diff \mu(z)
\end{equation*}
unfold whenever you can!
- Recall \(\Gamma \acts \Phi_\Delta \times X\) fixing \(\pair{\phi}{(m,n)} = \phi(m,n)\text{.}\) We can break the sum into \(\phi(m,n) \gt 0,\, \phi(m,n) \lt 0\text{.}\)
- Fact: The action of \(\Gamma\) on the above is free, (exercise: prove this fact, hint: first take \((m,n) \to (0,r)\text{,}\) and analyse the action of the stabiliser on \((0,r)\)).
-
\begin{equation*}
\int_{\Gamma \backslash \HH} \sum_{|\phi| = \Delta} \sum_{(m,n) \in X,\, \phi(m,n) \gt 0} R_\phi(t,z) \frac{y^s}{|mz+n|^s} \diff \mu(z)
\end{equation*}
\begin{equation*}
\sum_{(\Phi_\Delta \times X)/\Gamma} \int_\HH R_\phi (t,z) \frac{y^s}{|mz+n|^{2s}} \diff \mu(z)\text{.}
\end{equation*}
- Check
\begin{equation*}
z\mapsto \frac{nx - \frac 12 bn +cm}{-mz + an - \frac 12 bm}
\end{equation*}
finishes the proof.
- To check other cases use
\begin{equation*}
R_{-\phi} (z,t) = R_\phi(z,-t)
\end{equation*}
and
\begin{equation*}
\pair{\phi}{(m,n)} = 0
\end{equation*}
needs to be considered separately.
We really need to check some convergence here, but lets instead use the trace formula for something useful, to determine its value.
Another application: Equidistribution of Hecke eigenvalues.
Following Serre '97, very readable article.
Equidistribution:
Definition 2.17
Let \((\Omega, \mu)\) be a measure space
\begin{equation*}
\Omega\text{ compact}
\end{equation*}
\begin{equation*}
\mu\text{ positive Radon measure}
\end{equation*}
\begin{equation*}
\int_\Omega \diff \mu = 1
\end{equation*}
Example 2.18
\begin{equation*}
[-1,1],\,\frac{\diff x}{2} = \diff \mu
\end{equation*}
Definition 2.20 \(\delta\) measures
Let \(L\) be a sequence of indices, going to \(\infty\) (this will be the indexing set for the weight \(k\)). For \(\lambda \in L\) let \(I_\lambda\) be a non-empty finite set of cardinality
\begin{equation*}
\#I_\lambda = d_\lambda
\end{equation*}
(this will be the set of eigenvalues for each \(k\)). Let
\begin{equation*}
\mathbf X _\lambda = (x_{i,\lambda}),\,i\in I_\lambda\text{.}
\end{equation*}
We define
\begin{equation*}
\delta _{\mathbf X_\lambda} = \frac{1}{d_\lambda} \sum_{i\in I_\lambda} \delta_{x_{i,\lambda}}
\end{equation*}
note
\begin{equation*}
\pair{f}{\delta _{\mathbf X_\lambda}} = \frac{1}{d_\lambda} \sum_{i\in I_\lambda} f(x_{i,\lambda})\text{.}
\end{equation*}
The family \(\mathbf X_\lambda\) for \(\lambda \in L\) is called \(\mu\)-equidistributed (or equidistributed with respect to the measure \(\mu\)) if
\begin{equation*}
\lim_{\lambda \to \infty} \delta_{\mathbf X_\lambda} = \mu
\end{equation*}
the limit is in the weak-\(\ast\) sense i.e.
\begin{equation*}
\lim_{\lambda \to \infty} \pair{f}{\delta_{\mathbf X_\lambda}} = \pair{f}{\mu},\,\forall f \in C(\Omega, \RR)\text{.}
\end{equation*}
So far these are all general statements. Let us put ourselves in the following situation:
For each \(\lambda \in L\) we have a linear operator
\begin{equation*}
H_\lambda,\,\rank (H_\lambda) = d_\lambda \lt \infty
\end{equation*}
whose eigenvalues are in \(\Omega\text{.}\) And \(I_\lambda = \{ 1,2,\ldots, d_\lambda\}\text{,}\) \(\mathbf X_\lambda = \{x_{1,\lambda}, \ldots, x_{d_\lambda,\lambda}\}\) the eigenvalues of \(H_\lambda\) suitably normalised counted with multiplicity.
Proposition 2.21
TFAE
\begin{equation*}
\{x_\lambda\}_{\lambda \in L} \text{ is }\mu \text{ equidistributed on } \Omega
\end{equation*}
- For all polynomials \(P\) with \(\RR\) coefficients
\begin{equation}
\frac{\trace(P(H_\lambda))}{d_\lambda} \to_{\lambda \to \infty} \pair{P}{\mu}\label{eqn-equidist-poly}\tag{2.5}
\end{equation}
- For all \(m \in \ZZ_{\ge 0}\) there exists a polynomial \(P\) of degree \(m\) s.t. (2.5) is satisfied.
Proof
Now we will choose a special set of polynomials to work with.
Symmetric polynomials:
Let \(\Omega= \lb -2 , 2 \rb \leftrightarrow 2 \cos \lb -\pi, \pi\rb = \trace(\specialunitary(2,\CC))\text{,}\) this is really Satake parameters. For \(x\in \lb -2, 2\rb\) we have a corresponding \(2\cos (\phi)\text{.}\) Let
\begin{equation*}
X_n (x) = e^{in\phi} + e^{i(n-2)\phi} + \cdots + e^{-in\phi} = \trace (\Sym^n(U)),\,x = \trace(u)\text{.}
\end{equation*}
Clebesh-Gordon: (Decomposition of tensor powers of irreducible representations of \(\specialunitary(2, \CC)\)).
\begin{equation*}
X_nX_m = \sum_{0\le r \le \min\{n,m\}} X_{n+m - 2r} = x_{n+m} + x_{n+m-2} +\cdots + x_{|n-m|}\text{.}
\end{equation*}
Exercise 2.22
Several measures:
- Sato-Tate:
\begin{equation*}
\mu_\infty = \frac 1 \pi \sqrt{1-\frac{x^2}{4}} \diff x
\end{equation*}
\begin{equation*}
= \frac 2 \pi \sin^2(\phi) \diff \phi
\end{equation*}
properties
\begin{equation*}
\int_{\Omega} \diff \mu_\infty = 1
\end{equation*}
\begin{equation*}
\pair{X_n}{ \mu_\infty} = \begin{cases} 1\amp n = 0,\\0\amp\text{otw}\end{cases}
\end{equation*}
\begin{equation*}
\pair{X_nX_m}{ \mu_\infty} = \delta_{n,m}\text{.}
\end{equation*}
Exercise: prove this, with orthogonality of characters.
- \(p\)-adic Plancherel measure: Let \(q \in \RR_{\gt 1}\) and define
\begin{equation*}
f_q(x) = \sum_{n=0}^\infty q^{-n} X_{2n}(x) = \frac{q+1}{(q^{1/2} + q^{-1/2})^2 - x^2}
\end{equation*}
\begin{equation*}
\mu_q = f_q \mu_\infty
\end{equation*}
note:
\begin{equation*}
\lim_{q\to 1^+} \mu_q = \frac{\diff \phi}{\pi}
\end{equation*}
\begin{equation*}
\lim_{q \to \infty} \mu_q = \mu_\infty
\end{equation*}
\(\mu_q\) is positive and has total mass 1. note:
\begin{equation*}
\pair{X_n}{\mu_q} = \pair{X_n}{\sum_{m=0}^\infty q^{-m} X_{2m} \mu_\infty}
\end{equation*}
\begin{equation*}
= \sum_{m=0}^\infty q^{-m} \pair{X_nX_{2m} }{\mu_\infty}
\end{equation*}
\begin{equation*}
= \begin{cases} q^{-n/2} \amp n\equiv 0 \pmod 2 \\ 0 \amp \text{otw}\end{cases}\text{.}
\end{equation*}
Theorem 2.23 Serre '97
\(k \equiv 0 \pmod 2\) let \(s(k) = \dim(S_k(1))\)
\begin{equation*}
T_k'(n) = \frac{T_k(n)}{n^{k-1/2}}
\end{equation*}
so that the eigenvalues are in \(\lb -2, 2\rb\) by Deligne. Letting \(\mathbf X(k,p)\) be the eigenvalues of \(T_k'(p)\text{.}\) Then \(\mathbf X(k,p)\) are equidistributed with respect to \(\mu_p\text{.}\)
\begin{equation*}
\mu_p = \frac{p+1}{(p^{1/2} + p^{-1/2})^2 - x^2} \frac 1 \pi \sqrt{1-\frac{x^2}{4}} \diff x\text{.}
\end{equation*}
Proof
Claim:
\begin{equation*}
\lim_{k\to \infty,k\equiv 0 \pmod 2} \frac{\tr(T_k'(n))}{s(k)}
\end{equation*}
\begin{equation*}
= \lim_{k\to \infty,k\equiv 0 \pmod 2} \frac{\tr(T_k'(n))}{(k-1)/12} = \begin{cases} n^{-1/2} \amp \text{ if } n \text{ is a square}\\ 0 \amp\text{otw} \end{cases}
\end{equation*}
Assume the claim for now: Note that
\begin{equation*}
T_k'(p^m) = X_m(T_k'(p))
\end{equation*}
as
\begin{equation*}
\sum_{m=0}^\infty T_k'(p^m)t^m = \frac{1}{1- T_k'(p)t + t^2}
\end{equation*}
\begin{equation*}
\sum_{m=0}^\infty X_m(T_k'(p))t^m
\end{equation*}
\begin{equation*}
= \lim_{k\to \infty,k\equiv 0 \pmod 2} \frac{\tr(T_k'(n))}{s(k)} = \begin{cases} p^{-m/2} \amp m \equiv 0 \pmod 2\\ 0 \amp\text{otw} \end{cases}
\end{equation*}
\begin{equation*}
= \pair{X_m}{\mu_p}
\end{equation*}
by the claim and previous calculation.
Proof of the claim: Recall that the Eichler-Selberg trace formula says that
\begin{gather}
\tr(T_k(n)) = \frac{k-1}{12} n^{k/2 -1} \delta_{n= \square}\label{eqn-serre-equi-1}\tag{2.6}\\
- \frac12 \sum_{t^2 \lt 4n} P_k(t,n) H(4n - t^2)\label{eqn-serre-equi-2}\tag{2.7}\\
- \frac 12 \sum_{dd' = n,\,d,d' \gt 0} \min(d,d')^{k-1}\label{eqn-serre-equi-3}\tag{2.8}
\end{gather}
Sublemma: (2.7) \(= O(n^{k/2})\)
Proof: (2.7) \(= O_n(\sum_{t^2 \lt 4n} P_k(t,n) )\)
\begin{equation*}
P_k(t,n) = \frac{\rho^{k-1} - \bar \rho^{k-1}}{\rho - \bar \rho},\, \rho = \frac{t + i \sqrt{4n - t^2}}{2}
\end{equation*}
\begin{equation*}
| \rho^{k-1} - \bar \rho^{k-1}| \le 2|\rho^{k-1}| = O((4n- t^2)^{(k-1)/2})
\end{equation*}
\begin{equation*}
\rho - \bar \rho = \sqrt{n - t^2}
\end{equation*}
Sublemma: (2.8) \(= O(n^{(k-1)/2})\) Proof: exercise.
The equations above combined with the fact \(n^{k-1/2} \to n^{-1/2} \delta_{n=\square}\text{.}\)