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Section 6.5 Archimedean Local Heights II (Stevan)

Last time: \(\pair \cdot\cdot \colon \Div^0(S) \times \Div^0(S) \to \RR\text{.}\) It is bi-additive and symmetric

\begin{equation*} b = \divisor(f),\,a = \sum n_i x_i \end{equation*}
\begin{equation*} \pair ab = \sum_j n_j \log|f(x_i)|^2 \end{equation*}

if \(x_0\ne y_0 \in S\)

\begin{equation*} G(x,y) = \pair{x-x_0}{y-y_0} \end{equation*}

if

\begin{equation*} a =\sum n_ix_i,\,b = \sum m_j y_j \end{equation*}

then

\begin{equation*} \pair a b = \sum n_im_j G(x_i,y_j)\text{.} \end{equation*}

\(G \) is continuous, harmonic, has residue \(+1\) at \(\text{,}\) \(-1\) at \(x_0\text{.}\) And has logarithmic singularities.

Now we pass to modular curves, \(S = X_0(N) = \HH^*/\Gamma _0(N)\text{.}\)

\begin{equation*} G(\gamma z,\gamma z') = G(z,z') \,\forall \gamma ,\gamma ' \in \Gamma _0(N)\text{.} \end{equation*}

Continuous and harmonic when \(z \not \in \gamma z',\,\gamma \in \Gamma _0(N)\text{.}\)

\begin{equation*} G(z,z')= e_z\log|z-z'|^2 + O(1) \end{equation*}
\begin{equation*} z'\to z,\,z\text{fixed } \end{equation*}
\begin{equation*} e_z =\#\Stab_z \Gamma _0(N) \end{equation*}
\begin{equation*} G(z,z') = 4 \pi y +O(1) \text{ as } z\to \infty \end{equation*}
\begin{equation*} G(z,z') = 4 \pi y/N|z|^2 +O(1) \text{ as } z\to 0 \end{equation*}

Want \(G\text{.}\)

\begin{equation*} g(\gamma z,\gamma z') = g(z,z'),\,\gamma \in \SL_2(\ZZ) \end{equation*}

with \(g(z,z')\) continuous and harmonic in \(z,z'\text{.}\)

\begin{equation*} g(z,z' ) = \log|z-z'|^2 + O(1),\,z'\to z\text{.} \end{equation*}

A natural guess is

\begin{equation*} g(z,z') = \log \left | \frac{z-z'}{\bar z- z'}\right |^2 \end{equation*}
\begin{equation*} G(z,z') = \sum_{\gamma \in \Gamma _0(N)} g(z,\gamma z') \end{equation*}

however this does not converge.

Instead of asking for harmonic \(\Delta ^2 g = 0\) we instead ask that \(\Delta ^2 g = \epsilon g\) and let \(\epsilon \to 0\text{.}\) This gives us a differential equation to solve.

\(g\) is a function only of the hyperbolic distance between the points,

\begin{equation*} t = 1 + \frac{|z-z'|^2}{2yy'} \end{equation*}
\begin{equation*} ((1-t^2) \frac{\diff^2}{\diff t^2} - 2t \frac{\diff}{\diff t} + \epsilon )Q(t) = 0 \end{equation*}
\begin{equation*} Q_{s-1}(t) = \frac{\Gamma (s)^2}{2\Gamma (2s)} \left(\frac{2}{1+t}\right)^2 F(s,s; 2s, \frac{2}{1+t}),\,t\gt 1,\,\epsilon = s(s-1),\,s\gt1 \end{equation*}

where

\begin{equation*} F(a,b; c, z)= \sum_{n\ge 0} \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!} \end{equation*}

where

\begin{equation*} (w)_n = \begin{cases} w(w+1)\cdots(w+n-1),\amp n\gt 0,\\1\\amp n=0\end{cases}\text{.} \end{equation*}

So

\begin{equation*} Q_{s-1}(t) = -\frac12 \log(t-1) + O(1),\,t \to 1 \end{equation*}
\begin{equation*} Q_{s-1}(t) = O(t^{-s}),\,t \to \infty \end{equation*}

and so we may now set

\begin{equation*} g_s(z,z') = -2Q_{s-1}(t) = -2 Q_{s-1}\left(1+ \frac{|z-z'|^2}{2yy'}\right) \end{equation*}
\begin{equation*} G_{N,s}(z,z') = \sum_{\gamma \in \Gamma _0(N)} g_s(z,\gamma z') \end{equation*}

And our final Green's function is

\begin{equation*} G(z,z') = \lim_{s\to 1} (G_{N,s}(z,z') + 4\pi E_N(w_N z,s) + 2\pi E_N(z', s)+\frac{K_N}{s-1}) +C \end{equation*}
\begin{equation*} E_N(z,s) = \sum_{\gamma \in \begin{pmatrix}\ast\amp\ast\\0\amp\ast\end{pmatrix}\backslash \Gamma _0(N)} \Im(\gamma z)^s,\,z\in \HH,\,\Re(s) \gt 1 \end{equation*}
\begin{equation*} K_N = -\frac{12}{\lb \SL_2(\ZZ) : \Gamma _0(N)\rb } = \text{a residue of }G_{N,s} \text{ at } s=1,\,G_N(z) = -\frac{1}{N} \end{equation*}
\begin{equation*} =\kappa_{N}\left[\log N+2 \log 2-2 \gamma+2 \frac{\zeta^{\prime}}{\zeta}(2)-2 \sum_{p | N} \frac{p \log p}{p^{2}-1}\right] \end{equation*}
\begin{equation*} E_{N}(z, s)=N^{-s} \prod_{p | N}\left(1-p^{-2 s}\right)^{-1} \cdot \sum_{d | N} \frac{\mu(d)}{d^{s}} E\left(\frac{N}{d} z, s\right) \end{equation*}

Asymptotics

\begin{equation*} E(z, s)=y^{s}+\phi(s) y^{1-s}+O\left(e^{-y}\right) \quad(y=\operatorname{Im}(z) \rightarrow \infty) \end{equation*}

where

\begin{equation*} \phi(s)=\frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(s-\frac{1}{2}\right)}{\Gamma(s)} \frac{\zeta(2 s-1)}{\zeta(2 s)} \end{equation*}

and importantly

\begin{equation*} G\left(z, z^{\prime}\right)=G\left(w_{N} z^{\prime}, w_{N} z\right)\text{.} \end{equation*}

Now let \(x\) be a Heegner point and take

\begin{equation*} c= (x) - (\infty ) \end{equation*}
\begin{equation*} d = (x)-(0) \end{equation*}

then \(\sigma \in \Gal HK \leftrightarrow \mathscr A\in \Cl_K\text{.}\) If \(\gcd(M,N) = 1\) then

\begin{equation*} r_{\mathscr A} (m ) = 0 \implies |c| \cap |T_m (d)| = \emptyset\text{.} \end{equation*}

What we want:

\begin{equation*} S = \pair c {T_md}_\infty = \sum_{v|\infty } \pair c {T_md^\sigma }_v \end{equation*}

sum goes over \(h_K\) archimidean places of \(K\text{.}\)

\begin{equation*} S = \sum_{\mathscr{A}_{1}, \mathscr{A}_{2} \in \mathrm{Cl}_{K} , \mathscr{A}_{1} \mathscr{A}_{2}^{-1}=\mathscr{A}, \mathscr{A}_{1} \mathscr{A}_{2}[\mathfrak{n}]^{-1}=\mathscr{B}} G_{N . s}^{m}\left(\tau_{\mathscr{A}_{1}, \mathbf{n}}, \tau_{\mathscr{A}_{2}, \mathrm{n}}\right) \end{equation*}

we will compute \(S\) using the above

\begin{equation*} \begin{aligned}\left\langle c, T_{m} d^{\sigma}\right\rangle_{\infty}=\amp \lim _{s \rightarrow 1}\left[\gamma_{N, s}^{m}(\mathscr{A})+4 \pi \sigma_{1}(m) \sum_{\mathscr{A}_{1} \in \mathrm{Cl}_{K}} E_{N}\left(w_{N} \tau_{\mathscr{A}_{1}, \mathbf{n}}, S\right)\right.\\ \amp\left.+4 \pi m^{s} \sigma_{1-2 s}(m) \sum_{\mathscr{A}_{2} \in \mathrm{Cl}_{K}} E_{N}\left(\tau_{\mathscr{A}_{2}, \mathbf{n}}, S\right)+\frac{h_{K} \sigma_{1}(m) \kappa_{N}}{S-1}\right] \\ \amp-h_{K} \sigma_{1}(m) \hat{\lambda}_{N}+2 h_{K} \sigma_{1}(m) \kappa_{N} \end{aligned} \end{equation*}

and apply

\begin{equation*} \begin{aligned} \sum_{\mathscr{A} \in \mathrm{Cl}_{K}} E_{N}\left(w_{N} \tau_{\mathscr{A}, \mathfrak{w}}, s\right) \amp=\sum_{\mathscr{A} \in \mathrm{Cl}_{K_ { K }}} E_{N}\left(\tau_{\mathscr{A}, \mathfrak{n}}, s\right) \\ \amp=N^{-s} \prod_{p | N}\left(1-p^{-2 s}\right)^{-1} \sum_{d \backslash N} \frac{\mu(d)}{d^{s}} \sum_{\mathscr{A} \in \mathbb{Cl}_{K}} E\left(\frac{N}{d} \tau_{\mathscr{A}, \mathfrak{n}}, s\right) \end{aligned} \end{equation*}

so we must only compute

\begin{equation*} \sum_{\mathscr{A} \in \mathrm{Cl}_{K}} E\left(\frac{N}{d} \tau_{\mathscr{A}, \mathrm{n}}, s\right)\text{.} \end{equation*}

\(\tau _{\mathscr A, \mathfrak n}\) is a solution of

\begin{equation*} a\tau ^2 + b\tau + c = 0 \end{equation*}

of discriminant \(D = b^2 -4ac\text{.}\) So all other \(\frac Nd \tau _{\mathscr A, \mathfrak n}\) is a solution of some quadratic equation of discriminant \(D = b^2 -4ac\text{.}\) From analytic number theory we have the formula

\begin{equation*} E\left(\tau_{\mathscr{A}}, s\right)=2^{-s}|D|^{s / 2} u \zeta(2 s)^{-1} \zeta_{K}(\mathscr{A}, s) \end{equation*}

with \(u\) half the number of units in \(K\text{.}\)

\begin{equation*} \zeta_{K}(\mathscr{A}, s)=\sum_{\mathfrak{a} \text { integral } ,\,[\mathfrak{a}]=\mathscr{A}} \frac{1}{N(\mathfrak{a})^{s}} \end{equation*}

as we also have

\begin{equation*} \sum_{\mathscr{A}} \zeta_{K}(\mathscr{A}, s)=\zeta_{K}(s)\text{.} \end{equation*}

we then get

\begin{equation*} \begin{aligned}\left\langle c, T_{m} d^{\sigma}\right\rangle_{\infty}=\amp \lim _{s \rightarrow 1}\left[\gamma_{N, s}^{m}(\mathscr{A})+\frac{2^{2-s}|D|^{s / 2} \pi u}{N^{s} \prod_{p | N}\left(1+p^{-s}\right)}\left(\sigma_{1}(m)+m^{s} \sigma_{1-2 s}(m)\right) \frac{\zeta_{K}(s)}{\zeta(2 s)}\right.\\ \amp\left.+\frac{h_{K} \sigma_{1}(m) \kappa_{N}}{s-1}\right]-h_{K} \sigma_{1}(m) \lambda_{N}+2 h_{K} \sigma_{1}(m) \kappa_{N} \end{aligned} \end{equation*}

where we may substitute

\begin{equation*} \begin{aligned} \zeta_{K}(s) \amp=\zeta(s) L(s, \varepsilon) \\ \amp=\left(\frac{1}{s-1}+\gamma+O(s-1)\right)\left(L(1, \varepsilon)+L^{\prime}(1, \varepsilon)(s-1)+O(s-1)^{2}\right) \end{aligned} \end{equation*}

to finally obtain

\begin{equation*} \begin{aligned}\left\langle c, T_{m} d^{\sigma}\right\rangle_{\infty}=\amp \lim _{s \rightarrow 1}\left[\gamma_{N, s}^{m}(\mathscr{A})-\frac{h_{K} \sigma_{1}(m) \kappa_{N}}{s-1}\right] \\ \amp+h_{K} \kappa_{N}\left[\sigma_{1}(m)\left(\log \frac{N}{|D|}+2 \sum_{p | N} \frac{\log p}{p^{2}-1}+2+2 \frac{\zeta^{\prime}}{\zeta}(2)-2 \frac{L^{\prime}}{L}(1, \varepsilon)\right)\right.\\ \amp\left.+\sum_{d | m} d \log \frac{m}{d^{2}}\right] \end{aligned}\text{.} \end{equation*}

To compute the archimidean local height when the supports are not disjoint (i.e. \(r_{\mathscr A}(m) \ne 0\)) We consider the simplified case of

\begin{equation*} \{x\} = |a| \cap |b| \end{equation*}

then

\begin{equation*} \pair a b_{v,g} = \lim_{y\to x} (\pair {a_y}{b} - \ord_x(a) \ord_x(b) \log |g(y)|_v) \end{equation*}

where \(g\) is a uniformizer at \(x\) i.e. \(\ord_x(g) =1\text{.}\) and \(a_y\) is the divisor \(a\) with \(y\) in place of \(x\text{.}\)

\begin{equation*} a = n_x x + \cdots \end{equation*}
\begin{equation*} a_y = n_x y + \cdots \end{equation*}

so \(|a_y | \cap |b| = \emptyset\text{.}\) If \(g'\) is another uniformizer at \(x\) then

\begin{equation*} \sum_{v} \pair a b_{v,g} -\pair ab_{v,g'} = \ord_x(a) \ord_x(b) \log|g'/g(x)|_v \end{equation*}

so this gives a well defined global height by the product formula.

In our setting we have

\begin{equation*} c = (x) -(\infty ),\,d = (x) - (0) \end{equation*}

and \(\ord_x(c) = 1,\,\ord_x(T_m(d^\sigma )) = r_{\mathscr A}(m)\) so under this definition

\begin{equation*} \pair c {T_md^\sigma} = \lim_{y\to x} \pair {c_y}{T_md^\sigma } - r_{\mathscr A} (m)\log(|g(y)|_v) \end{equation*}
\begin{equation*} \omega=\eta^{4}(z) \frac{d q}{q}=2 \pi i \eta^{4}(z) d z \end{equation*}

with

\begin{equation*} \eta(z)=q^{1/{24}} \prod_{n}\left(1-q^{n}\right) \end{equation*}

the Dedekind eta-function.

So for \(v\) a complex places

\begin{equation*} \log |g(y)|_{v}-u \log \left|2 \pi i \eta^{4}(z)(w-z)\right|_{v} \rightarrow 0 \end{equation*}

as \(y \to x\text{.}\)