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\title{Modular Forms, PS 6}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
Suppose that $l\in\N$.  Recall that
$$J_l(x)=\sum_{n=0}^{\infty}(-1)^n\frac{(x/2)^{2n+l}}{n!(n+l)!}.$$
We have
\begin{align*}
        \int_{-1/2}^{1/2}e^{2\pi i l t}e^{-ix\sin(2\pi t)}\,\mathrm{d}t&=\sum_{n=0}^{\infty}\frac{(-ix)^n}{n!}\int_{-1/2}^{1/2}e^{2\pi i l t}\sin^n(2\pi t)\,\mathrm{d}t\\
        &=\sum_{n=0}^{\infty}\sum_{j=0}^n(-1)^j\frac{(x/2)^n}{n!}\binom{n}{j}\int_{-1/2}^{1/2}e^{2\pi i l t}e^{2\pi i (2j-n)}\,\mathrm{d}t\\
        &=\sum_{n=0}^{\infty}\sum_{j=0}^n(-1)^j\frac{(x/2)^n}{n!}\binom{n}{j}\int_{-1/2}^{1/2}e^{2\pi i (l+2j-n)}\,\mathrm{d}t.
\end{align*}
Now observe that the integral is zero unless $n=l+2j$ (in which case it is 1), and that for any $j$ there is exactly one value of $n$ for which this holds.
Thus, we may reindex the sum over $j$ to obtain
\begin{align*}
        \int_{-1/2}^{1/2}e^{2\pi i l t}e^{-ix\sin(2\pi t)}\,\mathrm{d}t&=\sum_{j=0}^{\infty}(-1)^j\frac{(x/2)^{2j+l}}{(2j+l)!}\binom{2j+l}{j}\\
        &=\sum_{j=0}^{\infty}(-1)^j\frac{(x/2)^{2j+l}}{j!(j+l)!}\\
        &=J_l(x).
\end{align*}
We thus have
\begin{align*}
    |J_l(x)|&\leq \int_{-1/2}^{1/2}\left|e^{2\pi i l t}e^{-ix\sin(2\pi t)}\right|\,\mathrm{d}t\\
    &\leq \sup_{t\in [-1/2,1/2]}\left|e^{2\pi i l t}e^{-ix\sin(2\pi t)}\right|\\
    &\leq 1
\end{align*}
since both exponentials have purely imaginary argument.
\end{exer}

\begin{exer}
    Let $c=qr$ with $(q,r)=1$.  Then there exist integers $u,v$ such that $qu+rv=1.$ Observe that
    \begin{align*}
        qu&\equiv 1\mod r\\
        rv&\equiv 1\mod q,
    \end{align*}
    so that $u=\bar{q},v=\bar{r}$.  We have
    \begin{align*}
        a&=qua+rva\\
        \bar{a}&=qu\bar{a}+rv\bar{a}.
    \end{align*}
    Notice also that since $a\bar{a}\equiv 1\mod qr$ we have $a\bar{a}\equiv 1\mod r$ and $a\bar{a}\equiv 1\mod q.$
    Finally, since $(r,q)=1$ we see by the chinese remainder theorem that as $a$ ranges over all values mod $c$, $(a \mod r,a\mod q)$ ranges over all pairs
    $(s\mod r,t\mod q)$.  Thus, we have
    \begin{align*}
        S(m,n,c)&=\sum_{a\bar{a}\equiv 1\mod c}e\left(\frac{am+\bar{a}n}{c}\right)\\
        &=\sum_{a\bar{a}\equiv 1\mod c}e\left(\frac{quam+rvam+qu\bar{a}n+rv\bar{a}n}{rq}\right)\\
        &=\sum_{a\bar{a}\equiv 1\mod c}e\left(\frac{uam+u\bar{a}n}{r}+\frac{vam+v\bar{a}n}{q}\right)\\
        &=\sum_{a\bar{a}\equiv 1\mod c}e\left(\frac{am\bar{q}+\bar{a}n\bar{q}}{r}\right)e\left(\frac{am\bar{r}+v\bar{a}n\bar{r}}{q}\right).\\
    \end{align*}
    Now since the value of
    $$\frac{am\bar{q}+\bar{a}n\bar{q}}{r}\mod 1$$
    depends only on $a\mod r$ and similarly the value of
    $$\frac{am\bar{r}+v\bar{a}n\bar{r}}{q}\mod 1$$
    depends only on $a\mod q$, by our observations above we have
    \begin{align*}
        S(m,n,c)&=\sum_{s\bar{s}\equiv 1\mod r}e\left(\frac{sm\bar{q}+\bar{s}n\bar{q}}{r}\right)\sum_{t\bar{t}\equiv 1\mod q}
        e\left(\frac{tm\bar{r}+v\bar{t}n\bar{r}}{q}\right)\\
        &=S(m\bar{q},n\bar{q},r)S(m\bar{r},n\bar{r},q).
    \end{align*}
\end{exer}

\begin{exer}
    We first prove some preliminary formulae.  Suppose that $(a,c)=1$.  Then
    \begin{align}
        S(am,n,c)=S(m,an,c).\label{lem1}
    \end{align}
    For we have
    \begin{align*}
        S(am,n,c)&=\sum_{x\bar{x}\equiv 1\mod c}e\left(\frac{amx+n\bar{x}}{c}\right)\\
        &=\sum_{x\bar{x}\equiv 1\mod c}e\left(\frac{mx+n\overline{x\bar{a}}}{c}\right)\\
                &=\sum_{x\bar{x}\equiv 1\mod c}e\left(\frac{mx+na\bar{x}}{c}\right)\\
                &=S(m,an,c),
    \end{align*}
    where we have used the fact that as $x$ runs over all units modulo $c$ so does $ax$ since $(a,c)=1$.
    Now if $p\not\|(m,n)$ it follows from (\ref{lem1}) that
    $$S(m,n,p^{\a})=S(mn,1,p^{\a})$$ for any $\a \geq 1.$  If, on the other hand, $p|(m,n)$ then we have
    \begin{align*}
        S(m,n,p^{\a})&=\sum_{x\bar{x}\equiv 1\mod p^{\a} }e\left(\frac{mx+n\bar{x}}{p^{\a}}\right)\\
        &=\sum_{\substack{x\bar{x}\equiv 1\mod p^{\a-1}\\k_1,k_2 \mod p} }e\left(\frac{m(x+k_1p^{\a-1})+n(\bar{x}+k_2p^{\a-1})}{p^{\a}}\right),
    \end{align*}
    where $k_1$ runs over all integers mod $p$ and $k_2$ is uniquely determined by $k_1$ and the condition
    $$(x+p^{\a-1}k_1)(\bar{x}+p^{\a-1}k_2)\equiv 1\mod p^{\a}$$ where $x\bar{x}\equiv 1\mod p$.
    We are implicitly using Hensel's lemma to rewrite the sum in this way.  We then have
    \begin{align}
        S(m,n,p^{\a})&=\sum_{x\bar{x}\equiv 1\mod p^{\a-1}}e\left(\frac{mx+n\bar{x}}{p^{\a}}\right)
        \sum_{k_1,k_2\mod p}e\left(\frac{mk_1+nk_2}{p}\right)\nonumber \\
        &=\sum_{x\bar{x}\equiv 1\mod p^{\a-1}}e\left(\frac{(m/p)x+(n/p)\bar{x}}{p^{\a-1}}\right)
        \sum_{k_1,k_2\mod p}1\nonumber\\
        \intertext{since $p|m,n$, so that the above is equal to}
        &pS(m/p,n/p,p^{\a-1})\label{lem3}
    \end{align}
    since $k_2$ is uniquely determined by $k_1$ and $k_1$ ranges over all values modulo $p$.
    Finally, observe that if $p|(m,n)$ then
    \begin{align*}
        S(mn,1,p^{\a})&=\sum_{x\bar{x}\equiv 1\mod p^{\a}}e\left(
        \frac{mnx+\bar{x}}{p^{\a}}\right)\\
        &=\sum_{\substack{x\bar{x}\equiv 1\mod p^{\a-1}\\k_1,k_2\mod p}}e\left(
        \frac{mn(x+k_1p^{\a-1})+(\bar{x}+k_2p^{\a-1})}{p^{\a}}\right)\\
        &=\sum_{x\bar{x}\equiv 1\mod p^{\a-1}}e\left(\frac{mnx+\bar{x}}{p^{\a}}\right)\sum_{k_1,k_2\mod p}e\left(\frac{mnk_1+k_2}{p}\right).
    \end{align*}
    But since $p|m,n$ we have
    \begin{align*}
        \sum_{k_1,k_2\mod p}e\left(\frac{mnk_1+k_2}{p}\right)&=\sum_{k_1,k_2\mod p}e\left(\frac{k_2}{p}\right)\\
        &=0
    \end{align*}
    since $k_2$ ranges over all values modulo $p$.  Hence if $p|(m,n)$ then
    \begin{align}
    S(mn,1,p^{\a})=0.\label{lem2}
    \end{align}
    We are now prepared to prove the main statement.  We first show by induction that
    $$S(m,n,p^{\a})=\sum_{d|(m,n,p^{\a})}dS(mn/d^2,1,p^{\a}/d).$$  For $\a=0$ the result follows by (\ref{lem1}).
    Further, if $p\not|(m,n)$ the result follows from (\ref{lem1}), so we may assume that $p|(m,n)$.
    If $\a=1$, we have
    \begin{align*}
        S(mn,1,p)+pS(mn/p^2,1,1)&=p\sum_{x\bar{x}\equiv 1\mod 1}e\left((mn/p^2)x+\bar{x}\right)+
        \sum_{x\bar{x}\equiv 1\mod p}e\left(\frac{mnx+\bar{x}}{p}\right)\\
        &=p+\sum_{x\bar{x}\equiv 1\mod p}e\left(\bar{x}/p\right),\\
        &=p-1\\
        \intertext{while}
        S(m,n,p)&=\sum_{x\bar{x}\equiv 1\mod p} e\left(\frac{mx+n\bar{x}}{p}\right)\\
        &=p-1
    \end{align*}
    since $p|m,n$.

    If $\a>1$ then we have
    \begin{align*}
        \sum_{d|(m,n,p^{\a})}dS(mn/d^2,1,p^a/d)&=S(mn,1,p^{\a})+\sum_{d|(m,n,p^{\a-1})}pdS(mn/p^2d^2,1,p^{\a}/pd)\\
        &=p\sum_{d|(m,n,p^{\a-1})}dS((m/p)(n/p)/d^2,1,p^{\a-1}/d)\\
        \intertext{by (\ref{lem2}) and our assumption that $p|(m,n)$}
        &=pS(m/p,n/p,p^{\a-1})\\
        \intertext{by our induction hypothesis}
        &=S(m,n,p^{\a})
    \end{align*}
    by(\ref{lem3}).  Therefore, for any $p,\a$ we have
    \begin{align}
        S(m,n,p^{\a})&=\sum_{d|(m,n,p^{\a})}dS(mn/d^2,1,p^{\a}/d).\label{fin1}
    \end{align}

      Now let $c_1,c_2$ be arbitrary with $(c_1,c_2)=1$.  Then by Exercise 2 we have
    \begin{align*}
        S(m,n,c_1c_2)=S(m\bar{c_1},n\bar{c_1},c_2)S(m\bar{c_2},n\bar{c_2},c_1).
    \end{align*}
    Suppose that we have
    \begin{align*}
        S(m\bar{c_1},n\bar{c_1},c_2)&=\sum_{d_1|(m\bar{c_1},n\bar{c_1},c_2)}d_1S(mn\bar{c_1}^2/d_1^2,1,c_2/d_1)\\
        S(m\bar{c_2},n\bar{c_2},c_1)&=\sum_{d_2|(m\bar{c_2},n\bar{c_2},c_1)}d_2S(mn\bar{c_1}^2/d_2^2,1,c_2/d_2).
    \end{align*}
    Observe that since $(c_1,c_2)=1$ we may rewrite both sums as
        \begin{align*}
        &\sum_{d_1|(m,n,c_2)}d_1S(mn\bar{c_1}^2/d_1^2,1,c_2/d_1)\\
        &\sum_{d_2|(m,n,c_1)}d_2S(mn\bar{c_2}^2/d_2^2,1,c_1/d_2).
    \end{align*}
    It follows by Exercise 2 that
    \begin{align*}
        S(m,n,c_1c_2)&=\sum_{\substack{d_1|(m,n,c_2)\\d_2|(m,n,c_1)}}d_1d_2S(mn\bar{c_1}^2/d_1^2,1,c_2/d_1)S(mn\bar{c_2}^2/d_2^2,1,c_1/d_2)\\
        &=\sum_{\substack{d_1|(m,n,c_2)\\d_2|(m,n,c_1)}}d_1d_2S(mn\bar{c_1}^2/d_1^2,1,c_2/d_1)S(mn\bar{c_2}^2/d_2^2,1,c_1/d_2)\\
        &=\sum_{\substack{d_1|(m,n,c_2)\\d_2|(m,n,c_1)}}d_1d_2S(mn/(d_1d_2)^2(\bar{c_1}^2/\bar{d_2}^2),1,c_2/d_1)
        S(mn/(d_1d_2)^2(\bar{c_2}^2/\bar{d_1}^2),1,c_1/d_2)\\
        &=\sum_{\substack{d_1|(m,n,c_2)\\d_2|(m,n,c_1)}}d_1d_2S(mn/(d_1d_2)^2(\bar{c_1}/\bar{d_2}),(\bar{c_1}/\bar{d_2}),c_2/d_1)
        S(mn/(d_1d_2)^2(\bar{c_2}/\bar{d_1}),(\bar{c_2}/\bar{d_1}),c_1/d_2)\\
        \intertext{by Exercise 1}
        &=\sum_{\substack{d_1|(m,n,c_2)\\d_2|(m,n,c_1)}}d_1d_2S(mn/(d_1d_2)^2,1,c_1c_2/d_1d_2).\\
    \end{align*}
    Again, since $(c_1,c_2)=1,$ as $d_1,d_2$ run over all divisors of $(m,n,c_1),(m,n,c_2)$ respectively, $d_1d_2$
    runs over all divisors of $(m,n,c_1c_2)$.  It follows that the above sum is
    \begin{align}
        S(m,n,c_1c_2)&=\sum_{d|(m,n,c_1c_2)}dS(mn/d^2,1,c_1c_2/d)\label{fin2}
    \end{align}
    Therefore, by (\ref{fin1}) and (\ref{fin2}) we have
    $$ S(m,n,c)=\sum_{d|(m,n,c)}dS(mn/d^2,1,c)$$
    for all positive integers $c$ and any $m,n$.
\end{exer}

\begin{exer}
    Let us first show that for any prime $p$ and any $\e>0$ there exists a constant $C_p(\e)$ independent of $n$
    such that
    $$d(p^n)\leq C_p(\e)p^{n\e}.$$  We know that $d(p^n)=n+1$.  Since $p\geq 2,$ we obviously have
    $$\lim_{n\rightarrow \infty} \frac{n+1}{p^{n\e}}=0$$ (by, say, l'Hopital's rule).  Since
    $$\frac{n+1}{p^{n\e}}$$ is a continuous function of $n$, it follows that
    $$\sup_{n\geq 0}\frac{n+1}{p^{n\e}}=y <\infty.$$  Clearly $y$ depends only on $p,\e$, so setting
    $C_p(\e)=y+1$ gives he desired constant.  Now observe that for $p>2^{1/\e}$ we have
    $$\frac{n+1}{p^{n\e}}<\frac{n+1}{2^n}\leq 1$$ for all $n\geq 0.$  Therefore, for any $p>2^{1/\e}$
    we may take $C_p(\e)=1$.  In any case, we always have $C_p(\e)\geq 1$.  Therefore, we have
    \begin{align*}
    d\left(\prod_{i=1}^r p_i^{a_i}\right)&=\prod_{i=1}^r d(p_i^){a_i}\\
    &<\prod_{i=1}^r C_{p_i}(\e)p_i^{a_i\e}\\
    &<\prod_{p} C_p(\e)\left(\prod_{i=1}^rp_i^{a_i}\right)^{\e}\\
    &=\prod_{p< 2^{1/\e}} C_p(\e) \left(\prod_{i=1}^rp_i^{a_i}\right)^{\e}.
    \end{align*}
    Setting
    $$C(\e)=\prod_{p< 2^{1/\e}} C_p(\e)$$
    (and noting that $C(\e)$ depends only on $\e$) we see that for any $\e>0$
    $$d(m)<C(\e)m^{\e}.$$
\end{exer}

\begin{exer}
    Let $f\in S_k(\G)$ and write $f$ in terms of an orthogonal basis for $S_k(\G)$, that is,
    $$f=\sum_{j=1}^r c_j f_j.$$  Recall that we have
    $$\frac{\G(k-1)}{(4\pi n)^{k-1}}\sum_{j=1}^{r} \frac{|a_{f_j}(n)|^2}{\|f_j\|^2}=1+2\pi i^k\sum_{c=1}^{\infty}\frac{S(n,n,c)}
    {c}J_{k-1}\left(\frac{4\pi n}{c}\right).$$
    Also recall the bounds
    $$J_{k-1}\left(\frac{4\pi n}{c}\right)\ll_k \begin{cases}\left(\frac{n}{c}\right)^{k-1} & \text{if $c>n$}\\\sqrt{\frac{c}{n}} &\text{if $c<n$}\end{cases}$$
    and
    $$|S(n,n,c)|\leq d(c)(c,n)^{1/2}c^{1/2}.$$
    It follows that
    $$1+2\pi i^k\sum_{c=1}^{\infty}\frac{S(n,n,c)}{c}J_{k-1}\left(\frac{4\pi n}{c}\right)\ll
    1+\sum_{c<n}\frac{d(c)(c,n)^{1/2}}{\sqrt{n}}+\sum_{c>n}\frac{d(c)(c,n)^{1/2}}{\sqrt{c}}\left(\frac{n}{c}\right)^{k-1}.$$
    This suggests that we estimate the sum
    $$S_j(n):=\sum_{jn<c<(j+1)n}d(c)(c,n)^{1/2}.$$
    By Exercise $4$ we have, for any $\e>0$
    \begin{align*}
        S_j(n)&\leq C(\e)\sum _{jn<c<(j+1)n}c^{\e}(c,n)^{1/2}\\
        &\ll((j+1)n)^{\e}\sum _{jn<c<(j+1)n}(c,n)^{1/2}.
    \end{align*}
    Now there are about $x/a$ integers less than $x$ that are divisible by $a$.  Hence, there are about
    $$\frac{(j+1)n}{a}-\frac{jn}{a}=\frac{n}{a}$$ integers $c$ between $jn$ and $(j+1)n$ that are divisible by $a$.
    Therefore
    \begin{align*}
        S_j(n)&\ll ((j+1)n)^{\e} \sum_{a|n} \frac{n}{a} a^{1/2}\\
        &\ll (j+1)^{\e} n^{1/2+\e}\sum_{d|n} d^{1/2}\\
        &\ll (j+1)^{\e} n^{1+\e}d(n)\\
        &\ll (j+1)^{\e} n^{1+2\e}.
    \end{align*}
    Replacing $2\e$ by $\e$ gives
    $$\sum_{c<n}\frac{d(c)(c,n)^{1/2}}{\sqrt{n}}=\frac{S_0(n)}{\sqrt{n}}\ll \frac{n^{1+\e}}{\sqrt{n}}\ll n^{1/2+\e}.$$
    Similarly, we have
    \begin{align*}
        \sum_{c>n}\frac{d(c)(c,n)^{1/2}}{\sqrt{c}}\left(\frac{n}{c}\right)^{k-1}&=\sum_{j=1}^{\infty}
        \sum_{jn\leq c< (j+1)n}\frac{d(c)(c,n)^{1/2}}{\sqrt{c}}\left(\frac{n}{c}\right)^{k-1}\\
        &\ll \sum_{j=1}^{\infty} \sum_{jn\leq c< (j+1)n}\frac{d(c)(c,n)^{1/2}}{\sqrt{jn}}\left(\frac{1}{j}\right)^{k-1}\\
        &\ll n^{1/2+\e} \sum_{j=1}^{\infty} \frac{(j+1)^{\e}}{\sqrt{j}}\left(\frac{1}{j}\right)^{k-1}\\
        &\ll n^{1/2+\e}
    \end{align*}
    since the sum in $j$ is clearly absolutely convergent and independent of $n$.  Therefore, we have
    $$\frac{\G(k-1)}{(4\pi n)^{k-1}}\sum_{j=1}^{r} \frac{|a_{f_j}(n)|^2}{\|f_j\|^2}\ll n^{1/2+\e},$$
    so that
    $$\sum_{j=1}^{r} \frac{|a_{f_j}(n)|^2}{\|f_j\|^2}\ll n^{k-1+1/2+\e}.$$
    It follows that
    $$\sup_{1\leq j\leq r} |a_{f_j}(n)|^2\ll n^{k-1+1/2+\e}$$ so that
    $$\sup_{1\leq j\leq r} |a_{f_j}(n)| \ll n^{\frac{k-1}{2}+1/4+\e}$$ for any $\e>0$.
    Since $f=c_1f_1+\cdots +c_r f_r$ we have
    $$|a_f(n)|\leq |c_1||a_{f_1}(n)|+\cdots +|c_r||a_{f_r}(n)|\ll n^{\frac{k-1}{2}+1/4+\e}.$$

\end{exer}


\end{document}