\( \newcommand{\contra}{\Rightarrow\!\Leftarrow} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Zeq}{\mathbb{Z}_{\geq 0}} \newcommand{\Zg}{\mathbb{Z}_{>0}} \newcommand{\Req}{\mathbb{R}_{\geq 0}} \newcommand{\Rg}{\mathbb{R}_{>0}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\O}{\mathcal{O}} \newcommand{\C}{\mathbb{C}} \newcommand{\A}{\mathbb{A}} \newcommand{\P}{\mathbb{P}} \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Proj}{Proj} \DeclareMathOperator{\Ob}{Ob} \DeclareMathOperator{\Mor}{Mor} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\colim}{colim} \)

Analytic Number Theory Spring 2022

Table of Contents

\( \newcommand{\contra}{\Rightarrow\!\Leftarrow} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Zeq}{\mathbb{Z}_{\geq 0}} \newcommand{\Zg}{\mathbb{Z}_{>0}} \newcommand{\Req}{\mathbb{R}_{\geq 0}} \newcommand{\Rg}{\mathbb{R}_{>0}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\O}{\mathcal{O}} \newcommand{\C}{\mathbb{C}} \newcommand{\A}{\mathbb{A}} \newcommand{\H}{\mathcal{H}} \renewcommand{\P}{\mathbb{P}} \renewcommand{\mod}{\text{ mod }} \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Proj}{Proj} \DeclareMathOperator{\Ob}{Ob} \DeclareMathOperator{\Mor}{Mor} \DeclareMathOperator{\Hom}{Hom} \)

1. Dirichlet Characters

We have shown: if \(\chi \text{ mod } q\) is primitive, then \[ \sum_{a=1}^q \chi(a)e^{\frac{2\pi i a \ell}{q}} = \overline{\chi(\ell)}\sum_{a=1}^q \chi(a)e^{\frac{2\pi i a}{q}} \] since we if the character is primitive then we can just send \(\ell \mapsto \ell^{-1}\), which only works if \(\ell \neq 0 \text{ mod } q\). But if \(\ell = 0 \text{ mod } q\) and \(\chi\) is primitive, then both sides vanish, so everything is ok.

Lemma: If \(\chi\) is primitive, then \[ G(\chi) = \sum_{a= 1}^q \chi(a) e^{\frac{2\pi i a}{q}} \] satisfies that \(|\ G(\chi)| = q^{1/2}\).

Proof: Consider \[ \sum_{\ell = 1}^q \left(\sum_{a=1}^q \chi(a) e^{\frac{2\pi i a \ell}{q}} \cdot \sum_{b=1}^q e^{-\frac{2\pi i b \ell}{q}}\right) \] We evaluate in two different ways. The first way is reducing the sum to \[ \sum_{\ell = 1}^q (\overline{\chi(\ell)} G(\chi)) (\chi(\ell) \overline{G(\chi)}) = \varphi(q) \vert \ G(\chi) | \ ^2 \] and the second is \[ \sum_{a = 1}^q\chi(a) \sum_{b=1}^q\overline{\chi(b)}\sum_{\ell=1}^q e^{\frac{2\pi i (a - b) \ell}{q}} = \sum_{a,b, a = b \text{ mod } q} \vert \ \chi(a) \vert \ ^2 \varphi(q) = q \varphi(q) \] So we get what we want.

1.1. Functional Equation and Analytic Continuations of \(L(s, \chi)\)

Recall that \(L(s, \chi) = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}\). We need to handle the case when \(\chi\) is even and odd seperately.

Let us define \(\theta_\chi(y) = \sum_{n \in \Z} \chi(n) e^{-\pi n^2 y}\), and compute

\begin{align*} \int_0^\infty \theta_\chi (y) y^{\frac{s}{2}} \frac{dy}{y} &= 2 \int_0^\infty \sum_{n=1}^\infty \chi(n)e^{-\pi n^2 y}y^{\frac{s}{2}}\frac{dy}{y} \\ &= 2 \pi^{-\frac{s}{2}}\sum_{n=1}^\infty\frac{\chi(n)}{n^s}\int_0^\infty e^{-y}y^{\frac{s}{2}}\frac{ds}{y} \\ &= 2\pi^{-\frac{s}{2}}\Gamma \left(\frac{s}{2}\right)L(s, \chi) \end{align*}

Now note that

\begin{align*} \theta_\chi (y)) &= \sum_{a=1}^q \chi(a) \sum_{\ell }e^{-\pi (a + \ell q)^2 y} \end{align*}

and when we take the Fourier transform, we see that \[ \widehat{F}(\ell) = \int_{-\infty}^\infty e^{-\pi(a + u q)^2 y}e^{-2 \pi i u \ell}du = \frac{e^{\frac{2\pi i a \ell}{q}}}{qy^{1/2}} e^{- \pi \left(\frac{\ell}{q y^{1/2}}\right)^2} \]

We can now get a functional equation for \(\theta_\chi\). In particular, we have that

\begin{align*} \theta_\chi(y) &= \sum_{a=1}^q \chi(a) \sum_{\ell = -\infty}^\infty e^{-\pi(a + \ell q)^2 q} = \sum_{a = 1}^q\chi(a) \sum_{\ell} F(a) \\ &= \frac{1}{q y^{1/2}} \sum_{a=1}^q \chi(a) \sum_{\ell} e^{\frac{2 \pi i a \ell}{q}}e^{-\pi \frac{\ell ^2}{q^2 y}} \\ &= \frac{1}{q y^{1/2}} \sum_{\ell} \sum_{a=1}^q \chi(a) e^{\frac{2 \pi i a \ell}{q}}e^{-\pi \frac{\ell ^2}{q^2 y}} \\ &= \frac{G(\chi)}{q y^{1/2}} \sum_{\ell} \overline{\chi(\ell)} e^{-\pi \frac{\ell ^2}{q^2 y}} \\ \end{align*}

so we can see that \[ \theta_\chi (y) = \frac{G(\chi)}{q y^{1/2}} \theta_{\overline{\chi}}\left(\frac{1}{q^2y}\right) \]

After we take a Mellin transform and do Riemann’s trick, splitting the domain of integration on \(1/q\), we see that if \(\Lambda(s, \chi) = \pi^{-\frac{s}{2}}\Gamma(\frac{s}{2})L(s, \chi)\)qs/2, \(\Lambda\) is entire! Furthermore, we have that \[ \Lambda(s, \chi) = \frac{q^{1/2}}{G(\overline{\chi})} \Lambda(1 - s, \overline{\chi}) \]

To handle the odd case, we need to do a little bit more work. We have to chance the test function \(e^{\pi u^2 y}\) to an odd Schwarz function, and we will win. In particular, \(ue^{-\pi u^2 y}\) is a good idea. Doing everything as above, we have that in this case, \[ \Lambda(s, \chi) = \pi^{-s/2} \Gamma \left(\frac{s+1}{2}\right)L(s, \chi)q^{s/2} = \frac{-i q^{1/2}}{G(\chi)}\Lambda(1-s, \overline{\chi}) \]

2. The Infinitude of Primes in Arithmetic Progression

Fix \(\chi\) a character mod \(q\). Then, \[ \frac{L'}{L}(s, \chi) = \sum_{n=1}^\infty -\frac{\Lambda(n)}{n^s}\chi(n) \] and the orthogonality relations yield that \[ \frac{1}{\varphi(q)}\sum_{\chi \mod q}\frac{-L'}{L}(s, \chi)\overline{\chi(a)} = \sum_{n = a \mod q} \frac{\Lambda(n)}{n^s} \] As long as we can prove that \[ \lim_{s \rightarrow 1} \frac{1}{\varphi(q)}\sum_{\chi \mod q} -\frac{L'}{L}(s, \chi) = \infty \] Certainly this is true for the trivial character, which is the zeta function multiplied by a constant. So all we need to show is that \(L(1, \chi) \neq 0\) for any character \(\chi \) mod \(q\). But this is actually hard, so we’re going to prove it a different way.

Pick \(\epsilon > 0\) fixed, and the interval \([1 - \epsilon, 1]\). If there is a real character \(\chi\) mod \(d\) which vanishes at \(B \in [ 1- \epsilon, 1]\), i.e. \(L(\beta, \chi\) = 0, then we call \(\beta\) a Siegel zero. Either there are no Siegel zeros, which implies that \(L(1, \chi) > 0\) for all \(\chi\). Otherwise, there must be a Siegel zero, with some associated character \(\chi'\).

Now, we consider a simple lemma: consider \[ \xi(s) = \zeta(s)L(s, \chi)L(s, \chi')L(s, \chi \cdot \chi') = \sum_{n=1}^\infty\frac{a(n)}{m^s} \] Then, \(a(1) = 1\) and \(a(n) \geq 0\) for all \(n \geq 2\). By the Euler product, \[ \xi(s) = \prod_{p} \left(1 - \frac{1}{p^{s}}\right)^{-1} \left(1 - \frac{\chi(p)}{p^{s}}\right)^{-1} \left(1 - \frac{\chi'}{p^{s}}\right)^{-1} \left(1 - \frac{\chi' \cdot \chi}{p^{s}}\right)^{-1} \] And since these are real valued characters, you can just try all of the combinations to see that the lemma holds. But then, it must be true that \(\xi(\beta) =0 \), since \(L(\beta, \chi') = 0\). Then, we have that \[ 1 \ll \frac{1}{2\pi i}\int_{2 - i \infty}^{2 + i \infty} \xi(s + \beta) d^{10s}\frac{ds}{s(s+1)(s+2)(s+3)(s+4)} \] and shifting the line of integration we see that this becomes \[ \frac{L(1, \chi)L(1, \chi')L(1, \chi \cdot \chi')d^{10(1 - \beta)}}{(1 - \beta)(2 - \beta)(3 - \beta)(4 - \beta)(5 - \beta)} + \frac{1}{2\pi i}\int_{-\beta - i \infty}^{\beta + i \infty} \frac{\xi(s + \beta) d^{10s}ds}{s(s+1)(s+2)(s+3)(s+4)} \] and in particular this means that if \(\chi'\) is a character mod \(q'\), \[ 1 \ll \frac{L(1, \chi)\log(q')\log(dq')\vert \ d \vert \ ^{10\epsilon}}{\epsilon} + O(\vert \ d \vert \ ^{-5}) \] and so \[ L(1, \chi) \gg_\epsilon \vert \ d \vert \ ^{-\epsilon} \]

3. Modular Forms

Def: Let \(G\) act on \(X\). An automorphic function \(f: X \rightarrow \C\) satisfies that \(f(g x) = f(x)\) for all \(g \in G, x \in X\).

In particular, we want to consider the action of \(SL(2, \Z)\) on the upper half plane \(\H = \{z \mid \Im(z) > 0 \ \}\), defined by \[ \begin{bmatrix}a & b \\ c & d\end{bmatrix} g = \frac{az + b}{cz + d} \] Note that this is well-defined, since we can compute that \[ \Im \left(\frac{az + b}{cz + d}\right) = \frac{y}{| \ c z + d| \ ^2} > 0 \]

Def: We say that \(G\) acts discretely on \(X\) if for every pair of compact sets \(A, B\), there are only finitely many \(g \in G\) such that \((g A) \cap B \neq \emptyset\).

That \(SL(2, \Z)\) acts discretely on \(\H\) follows immedaitely from the previous lemma.

Def: A modular form of weight \(k\) is a holomorphic function \(f: \H \rightarrow \C\); which satisfies for any \(M \in SL(2, \Z)\), \[ f \left(\begin{bmatrix}a & b \\ c & d\end{bmatrix} z\right) = (cz + d)^kf(z) \]

We also want to consider \(PSL(2, \Z)\), which is \(SL(2, \Z) / \pm 1\), which also acts on \(\H\).

Def: Let \(G\) act on \(X\). We say that \(x, x' \in X\) are congruent mod \(G\) if there is some \(g \in G\) such that \(x' = gx\).

Def: Let \(G\) act on \(X\). Then a fundamental domain \(D \subset X\) satisfies the following conditions:

  1. For every \(x \in X\) there exists \(g \in G\) such that \(gx \in D\)
  2. Every \(x \neq x'\) with \(x, x' \in D\) are inequivalent mod \(G\).

Let \(D\) be a connected fundamental domain for \(PSL(2, \Z)\) acting on \(\H\). Then we can choose it to be the following shaded region:

Now consider the most general case of automorphic forms: \[ f(gx) = j(g,x) \cdot f(x), \ \forall g\in G, x \in X \] It must be true that \(j(e, x) = 1\), and \(j(g, g'x) = j(g, g'x) \cdot j(g', x)\) (whenever \(f\) is not identically zero). The latter condition is called the cocycle condition. It is easy to check that \(j(g, z) = cz + d\) and its powers satisfies both conditions.

3.1. Eisenstein Series

Consider \[ E_k(z) = \sum_{\substack{c,d \in \Z \\ (c,d) = 1}} \frac{1}{(cz + d)^k} \] which converges for \(k > 2\), and is nontrivial for \(k \neq 0 \mod 2\). This should be a modular form of weight \(k\). Let \(\Gamma_\infty = \left\{ \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix} \mid m \in \Z \right\}\). Then, I claim that \[ E_k(z) = \sum_{\gamma \in \Gamma_\infty \backslash SL(2, \Z)}j(\gamma, z)^{-k} \] which is not hard to see; then it becomes also clear that \(E_k\) is an automorphic form via the cocycle condition.

3.2. Ramanujan Tau

If we let \[ G_k(z) = 2 \zeta(k) + 2 \frac{(2\pi i)^k}{(k-1)!} \sum_{n=1}^\infty \sigma_{k-1}(n)e^{2\pi i nz} = 2 \zeta(k)E_k(z) \] where \(\sigma_k(n) = \sum_{d \mid n}d^k\).

We call \[ \Delta(z) = \frac{2^6}{3^3}(E_4^3(z) - E_6^2(z)) = \sum_{n=1}^\infty \tau(n)e^{2\pi i nz} \] the Ramanujan \(\Delta\).

Now, he made some conjectures about the \(\tau\) function:

  1. \(\tau(mn) = \tau(m) \tau(n)\), if \((m, n) = 1\)
  2. \(\tau(p^{k+1}) = \tau(p) \tau(p^k) - p^{11}\tau(p^{k-1})\)
  3. \(| \ \tau(p) | \ \leq 2p^{11/2}\)

Now, assuming the above, then \(\sum_{n=1}^\infty \frac{\tau(n)}{n^s}\) converges for \(\Re(s) > \frac{13}{2}\), and this has a Euler product \[ \prod_{p} (1 - \tau(p)p^{-s} + p^{11-2s})^{-1} = \prod_{p} \left(1 - \frac{\alpha(p)}{p}\right)^{-1}\left(1 - \frac{\beta(p)}{p}\right)^{-1} \] where \(\tau(p) = \alpha(p) + \beta(p)\) and \(\alpha(p)\beta(p) = p^{11}\). The above conjectures are all true (but (3) is really hard, involving the Weil conjectures).

4. \(L\)-functions and Modular Forms

Let \(f(z)\) be a modular form of weight \(k\) for \(SL(2, \Z)\). Then it is periodic, and has a Fourier expansion \(f(z) = \sum_{n=0}^\infty a_ne^{2\pi i n z}\). Then we may define the \(L\)-function associated to \(f\) as \[ L(s, f) = \sum_{n=1}^\infty \frac{a_n}{n^s} \] Now, Deligne proved with the Weil conjectures that \(| \ a_p | \ < 2p^{\frac{k-1}{2}}\), but it is a much easier result that \(| \ a_p | \ \ll p^{k/2}\), which is good enough for convergence.

Thm: \(L(s, f)\) has a meromorphic continuation to all \(s \in \C\), and for cusp forms \(L(s, f)\) is entire.

To see this, consider \[ f(iy) = \sum_{n = 0}^\infty a_ne^{2\pi n y} \] and choose \(\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \in SL(2, \Z)\) such that \[ f(iy^{-1}) = (iy)^kf(iy) \implies f(iy^{-1}) = i^k y^k f(iy) \]

Then, we just follow Riemann, and consider the Mellin transform \[ \int_0^\infty (f(iy) - a_0)y^{s+\frac{k}{2}-1}dy = \sum_{n=1}^\infty a_n \int_0^\infty e^{2\pi n y}y^{s + \frac{k}{2} - 1}dy = (2\pi)^{-s}\Gamma(s)L(s, f) = \Lambda(s,f) \] Then you split the integral as usual can do some more work and arrive at \[ \Lambda(s,f) = i^k \Lambda(k-s, f) \] which has poles with residue \(a_0\) at \(s = \pm \frac{k}{2}\).

Author: David Chen

Created: 2022-04-19 Tue 11:42