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

# Analytic Number Theory Spring 2022

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

## 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 $$s = \pm \frac{k}{2}$$.

Created: 2022-04-19 Tue 11:42