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\newtheorem{defn}{Definition}
\newtheorem{exer}{Exercise}
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\newtheorem{lem}{Lemma}
\newtheorem{cor}{Corollary}
\newtheorem{prop}{Proposition}

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\title{Modular Forms, PS 3}
\author{Bryden Cais}
\begin{document}
\maketitle


\begin{exer}
Recall that
$$E_{2k}(z)=1-\frac{2k}{B_k}\sum_{n=1}^{\infty}\sigma_{k-1}(n)q^n,$$
where $B_k$ is defined by the generating function
$$\frac{t}{e^t-1}=\sum_{k=0}^{\infty}B_k\frac{t^k}{k!}.$$
It is not difficult to determine the first few $B_k$.  We have
\begin{align*}
1&=\sum_{m=0}^{\infty}\frac{t^{m}}{(m+1)!}\sum_{k=0}^{\infty}B_k\frac{t^k}{k!}\\
&=\sum_{n=0}^{\infty}\sum_{m+k=n}\frac{B_k}{k!(m+1)!}t^n,\\
\end{align*}
which gives that $B_0=1$ and
$$\sum_{k=0}^n \binom{n+1}{k}B_k=0.$$
From this, we can readily calculate the $B_k$ and find, in particular, that
\begin{align*}
    B_2&=\frac{1}{6} & B_4&=\frac{1}{30} & B_6 &=\frac{1}{42}\\
    B_8&=\frac{1}{30} & B_{10} &=\frac{5}{66} & B_{12} &=\frac{691}{2730}.
\end{align*}
Now consider the function
$$f(z):=E_{12}(z)-E_6^2(z)$$
Clearly $f(z)$ is a modular form of weight 12.  We readily determine the first few terms of the $q$ expansion
of $f$:
\begin{align*}
f(z)&=1+\frac{65520}{691}\sum_{n=1}^{\infty}\sigma_{11}(n)q^n-\left(1-504\sum_{n=1}^{\infty}\sigma_{5}(n)q^n\right)^2\\
&=\frac{762048}{691}q+\cdots.
\end{align*}
Observe that $f$ is thus a cusp form of weight 12.  Since $S_{12}(\G)$ is one dimensional, with $\Delta(z)=q+\cdots$ as a basis,
we must have
$$E_{12}(z)-E_6^2(z)=\frac{2^6 3^5 7^2}{691}\Delta(z).$$
Now observe that the coefficients of $E_6$ are integers.  Therefore, the coefficients of $691E_6^2(z)$ are all divisible by $691$.
By comparing coefficients modulo 691, we therefore have
$$65520\sigma_{11}(n)\equiv 762048\tau(n)\mod 691.$$  Since $65520\equiv 762048\equiv 566\mod (691)$, we see that
$\tau(n)\equiv \sigma_{11}(n)\mod 691$.
\end{exer}
\begin{exer}\label{ex3}
    Let $f\in \mathcal{M}_k(\G)$.  Then $f(z+1)=f(z)$ and $f(-1/z)=z^k f(z)$.  Differentiating these identities
    gives $f^{'}(z+1)=f^{'}(z)$ and $f^{'}(-1/z)=z^{k+2} f^{'}(z)+kz^{k+1}f(z)$.  Recall that $E_2(z)$ satisfies $E_2(z+1)=E_2(z)$
    and
    $$E_2(-1/z)=z^2 E_2(z)+\frac{12z}{2\pi i}.$$
    Now define
    $$g(z)=\frac{f^{'}(z)}{2\pi i}-\frac{k}{12}f(z)E_2(z).$$  We clearly have $g(z+1)=g(z)$.  Using our formulae above, we find that
    \begin{align*}
    g(-1/z)&=\frac{z^{k+2} f^{'}(z)+kz^{k+1}f(z)}{2\pi i}-\frac{k}{12}z^k f(z)\left(z^2 E_2(z)+\frac{12z}{2\pi i}\right)\\
    &=z^{k+2}\frac{f^{'}(z)}{2\pi i}+\frac{k}{2\pi i}z^{k+1}f(z)-\frac{k}{12}\frac{12z}{2\pi i} z^k f(z)-\frac{k}{12}z^{k+2}f(z)E_2(z)\\
    &=z^{k+2}\frac{f^{'}(z)}{2\pi i}-\frac{k}{12}z^{k+2}f(z)E_2(z)\\
    &=z^{k+2}g(z).
    \end{align*}
    Therefore, since $z\mapsto z+1$ and $z\mapsto -1/z$ generate $\G$, we see that $g(z)\in\mathcal{M}_{k+2}(\G)$.
\end{exer}

\begin{exer}
    Let us now use Exercise \ref{ex3} to derive ``Ramanujan type'' relations between $\s_5(n),\s_3(n)$ and $\s_1(n)$, and between
    $\s_7(n),\s_5(n),$ and $\s_1(n)$.
    We know that for $k<6$ the space $\mathcal{M}_{2k}(\G)$ is one dimensional, a basis being given by $E_{2k}.$
    Therefore, since, by Exercise \ref{ex3}, the functions
    \begin{align*}
    f_6(z)&:=\frac{E_4^{'}(z)}{2\pi i}-\frac{1}{3}E_4(z)E_2(z),\\
    \intertext{and}
    f_8(z)&:=\frac{E_6^{'}(z)}{2\pi i}-\frac{1}{2}E_6(z)E_2(z)\\
    \end{align*}
    are modular forms of weights $6$ and $8$ respectively, we have
    \begin{align}
        f_6(z)&=c_6 E_6(z),\label{rel6}\\
        \intertext{and}
        f_8(z)&=c_8 E_8(z)\label{rel8}
    \end{align}
    for some constants $c_6,c_8$.  To determine these constants, we write down the $q$ series expansions of both sides
    of equations (\ref{rel6}) and (\ref{rel8}) and equate constant terms.  We therefore obtain
    \begin{align*}
        1-504\sum_{k=1}^{\infty}\s_5(k)q^k&=\left(1+240\sum_{n=1}^{\infty}\s_3(n)q^n\right)
        \left(1-24\sum_{m=1}^{\infty}\s_1(m)q^m\right)-720\sum_{n=1}^{\infty} k\s_3(k)q^k\\\
        \intertext{and}
        1+480\sum_{k=1}^{\infty} \s_7(k)q^k &=\left(1-504\sum_{n=1}^{\infty}\s_5(n)q^n\right)
        \left(1-24\sum_{m=1}^{\infty}\s_1(m)q^m\right)+1008\sum_{k=1}^{\infty}k\s_5(k)q^k.
    \end{align*}
    Now equate coefficients of $q^k$ on both sides of the above two formulae to obtain
    \begin{align*}
        7\s_5(k)&=80\sum_{\substack{m+n=k\\m,n\geq 0}}\s_3(m)\s_1(n)+10k\s_3(k)\\
        10\s_7(k)&=252\sum_{\substack{m+n=k\\m,n\geq 0}}\s_5(m)\s_1(n)+21k\s_5(k),
    \end{align*}
    where we define
    \begin{align*}
    \s_1(0)&=-\frac{1}{24},\\
    \s_3(0)&=\frac{1}{240},\\
    \intertext{and}
    \s_5(0)&=-\frac{1}{504}.
    \end{align*}
\end{exer}

\begin{exer}
    Given that there exist $\alpha_p,\beta_p$ such that
    $$\tau(p)=p^{11/2}(\alpha_p+\beta_p),$$ we show that
    \begin{eqnarray}
    \sum_{n=0}^{\infty}\frac{\tau(p^n)}{p^{n(s+11/2)}}=\left(1-\frac{\alpha_p}{p^s}\right)^{-1}\left(1-\frac{\beta_p}{p^s}\right)^{-1}.\label{tausum}
    \end{eqnarray}
    First let us prove by induction that
    $$\tau(p^n)=p^{11n/2}\sum_{\substack{m+k=n\\m,k\geq 0}}\alpha_p^m\beta_p^k.$$
    We are already given the identity for $n=0,1$ so suppose that the identity holds for $n=r$.
    Using the multiplicativity of the $\tau$ function, we have
    \begin{align*}
        \tau(p^{r+1})&=\tau(p)\tau(p^r)-p^{11}\tau(p^{r-1})\\
        &=p^{11(r+1)/2}(\alpha_p+\beta_p)\sum_{k=0}^r\alpha_p^k\beta_p^{r-k}-p^{11(r+1)/2}\sum_{k=0}^{r-1}\alpha_p^k\beta_p^{r-k-1}\\
        &=p^{11(r+1)/2}\left(\sum_{k=0}^r\alpha_p^{k+1}\beta_p^{r-k}+\sum_{k=0}^r\alpha_p^{k}\beta_p^{r+1-k}-\sum_{k=0}^{r-1}
        \alpha_p^k\beta_p^{r-1-k}\right)\\
        &=p^{11(r+1)/2}\sum_{k=0}^{r+1}\alpha_p^{k}\beta_p^{r+1-k},
    \end{align*}
    which is the required result.
    Now, expanding the right hand side of (\ref{tausum})as a geometric series in $p^{-s},$ we have
    \begin{align*}
    \left(1-\frac{\alpha_p}{p^s}\right)^{-1}\left(1-\frac{\beta_p}{p^s}\right)^{-1}&=
    \left(1+\frac{\alpha_p}{p^s}+\frac{\alpha_p^2}{p^{2s}}+\cdots\right)\left(1+\frac{\beta_p}{p^s}+\frac{\beta_p^2}{p^{2s}}+\cdots\right)\\
    &=1+\sum_{n=1}^{\infty}\left(p^{11n/2}\sum_{\substack{m+k=n\\m,k\geq 0}}\alpha_p^m\beta_p^k\right)p^{-n(s+11/2)}\\
    &=\sum_{n=0}^{\infty}\frac{\tau(p^n)}{p^{n(s+11/2)}},
    \end{align*}
    and we are done.
\end{exer}

\begin{exer}
    Suppose that for all integers $n$ and primes $p$ we have
    $$\tau(p)\tau(n)=\sum_{d|(n,p)}d^{11}\tau(np/d^2).$$
    We show by induction that
    $$\tau(p^r)\tau(n)=\sum_{d|(n,p^r)}d^{11}\tau(np^r/d^2)$$
    for all $r>0$.  We are given the result for $r=0,1$.  We have
    \begin{align}
        \tau(p^r)\tau(n)&=\tau(p)\tau(p^{r-1})\tau(n)-p^{11}\tau(p^{r-2})\tau(n)\\
        &=\sum_{d|(n,p^{r-1})}d^{11}\tau(p)\tau(np^{r-1}/d^2)-\sum_{d|(n,p^{r-2})}(dp)^{11}\tau(np^r/(dp)^2)\\
        &=\sum_{d|(n,p^{r-1})}\sum_{e|(np^{r-1}/d^2,p)}(ed)^{11}\tau(np^r/(ed)^2)-\sum_{dp|(np,p^{r-1})}(dp)^{11}\tau(np^r/(dp)^2)\\
        &=\sum_{d|(n,p^{r-1})}\sum_{e|(np^{r-1}/d^2,p)}(ed)^{11}\tau(np^r/(ed)^2)-\sum_{\substack{d|(np,p^{r-1})\\d> 1}}d^{11}\tau(np^r/d^2)\label{mult}.
    \end{align}
    Now observe that every divisor of $(n,p^r)$, with the exception of $1$ and $p^r$ if it divides $n$ can be represented as $ed$
    where $d|(n,p^{r-1})$ and $e|(np^{r-1}/d^2,p)$
    in exactly two ways.  To see this, let $s$ be the highest power of $p$ dividing $n$.
    \begin{enumerate}
        \item If $s< r-1$, then $(n,p^{r-1})=p^s$.  Then for $t\leq s$, we can write $p^t=ed$ where $d=p^t$ and $e=1$ or where $d=p^{t-1}$
        and $e=p$ (This is possible since $p|(p^{r+s-1-2(t-1)},p)$ since $r-2>s>t$).  Since $e$ must divide $p$, these are
        the only two such representations.  Observe,however, that we obtain one more term that does not correspond to a divisor
        of $(n,p^r)=p^s.$  This is the term $d=p^s=(n,p^{r-1})$ and $e=p|(np^{r-1}/p^{2s},p)$ since $r-s-1\geq 1$.

        \item If $s> r-1$, then $(n,p^{r-1})=p^{r-1}$.  For $t< r$ we have $p^t=ed$ with $d=p^t|(n,p^{r-1})$ and $e=1$
        or with $d=p^{t-1}$ and $e=p|(np^{r-1}/d^2,p)$ as before.  This gives precisely two representations for $t<r$.
        If $t=r$, there is a unique such representation: $d=p^{r-1},e=p$.

        \item If $s=r-1$, then $(n,p^{r-1})=(n,p^r)=p^{r-1}$.  For $t<r$ there are two representations, as before, while if
        $d=p^{r-1}$ then $(np^{r-1}/d^2,p)=1$ so we have a unique representation of the divisor $p^{r-1}$ of $(n,p^r)$.
    \end{enumerate}
    We therefore can write equality (\ref{mult}) as
    \begin{enumerate}
    \item If $s\geq r-1$:
    \begin{align}
        \tau(p^r)\tau(n)&=
        \tau(np^r)+(n,p^r)^{11}\tau(n/(n,p^r))+2\sum_{\substack{d|(n,p^r)\\1<d <(n,p^r)}}d^{11}\tau(np^r/d^2)
        -\sum_{\substack{d|(np,p^{r-1})\\d> 1}}d^{11}\tau(np^r/d^2)
    \end{align}

    \item If $s< r-1$:
    \begin{align}
        \tau(p^r)\tau(n)&=
        \tau(np^r)+2\sum_{\substack{d|(n,p^r)\\d> 1}}d^{11}\tau(np^r/d^2)+p^{11(s+1)}\tau(np^{r-2s-2})-\sum_{\substack{d|(np,p^{r-1})\\d> 1}}d^{11}\tau(np^r/d^2)
    \end{align}

    \end{enumerate}
    We now claim that the divisors of $(np,p^{r-1})$ are essentially in one to one correspondence with the nontrivial divisors of $(n,p^r)$.
    As before, let $s$ be the highest power of $p$ dividing $n$.
    \begin{enumerate}
        \item If $s< r-1$ then $(np,p^{r-1})=p^{s+1}$ while $(n,p^r)=p^{s}$.  Therefore,
        $$\sum_{\substack{d|(np,p^{r-1})\\d> 1}}d^{11}\tau(np^r/d^2)=\sum_{\substack{d|(n,p^r)\\d> 1}}d^{11}
        \tau(np^r/d^2)+p^{11(s+1)}\tau(np^r/p^{2(s+1)}).$$

        \item If $s> r-1$ then $(np,p^{r-1})=p^{r-1}$ and $(n,p^r)=p^r$.  In this case,
        $$\sum_{\substack{d|(np,p^{r-1})\\d> 1}}d^{11}\tau(np^r/d^2)=\sum_{\substack{d|(n,p^r)\\1<d<(n,p^r)}}d^{11}
        \tau(np^r/d^2).$$

        \item If $s= r-1$ then $(np,p^{r-1})=p^{r-1}$ and $(n,p^r)=p^{r-1}$.  In this case we have an actual bijection.
    \end{enumerate}

    Putting all three cases together and combining with the two cases above, we find that in any case,
    \begin{align*}
        \tau(p^r)\tau(n)&=
        \tau(np^r)+(n,p^r)^{11}\tau(n/(n,p^r))+\sum_{\substack{d|(n,p^r)\\1<d <(n,p^r)}}d^{11}\tau(np^r/d^2)\\
        &=\sum_{d|(n,p^r)}d^{11}\tau(np^r/d^2),
    \end{align*}
    which is what we wanted to prove.
    Now observe that for $p,q$ relatively prime we have
    \begin{align*}
        \tau(p^r)\tau(q^s)\tau(n)&=\sum_{d|(n,q^s)}d^{11}\tau(p^r)\tau(nq^s/d^2)\\
        &=\sum_{d|(n,q^s)}\sum_{e|(nq^s/d^2,p^r)}(ed)^{11}\tau(nq^sp^r/(ed)^2)\\
        &=\sum_{d|(n,q^sp^r)}d^{11}\tau(np^rq^s/d^2),
    \end{align*}
    since the divisors of $(n,p^rq^s)$ are in one to one correspondence with pairs of divisors $d,e$ of
    $(n,p^r),(n,q^s)$ respectively.
    It therefore follows that for any $m,n$ one has
    $$\tau(m)\tau(n)=\sum_{d|(m,n)}d^{11}\tau(mn/d^2).$$



\end{exer}

\end{document}