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\newtheorem{exer}{Exercise}
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\newtheorem{cor}{Corollary}
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\title{Modular Forms, PS 2}
\author{Bryden Cais}
\begin{document}
\maketitle


\begin{exer}
    Show that
    $$\z(k)=-\frac{(2\pi i)^k B_k}{2k!},$$
    where $B_k$ is defined by
    $$\frac{x}{e^x-1}=\sum_{k=0}^{\infty}B_k\frac{x^k}{k!}.$$
\end{exer}

Recall that we proved the formula
$$f(w)=\sum_{n\in\Z}\frac{1}{(n+w)^k}=\sum_{n=1}^{\infty}\frac{(-2\pi i)^k n^{k-1}}{(k-1)!}e(nw)$$
in class.  We write this as
$$f(w)=\frac{1}{w^k}+\sum_{n=1}^{\infty}\left\{\frac{1}{(n+w)^k}+\frac{1}{(-n+w)^k}\right\}=\sum_{n=1}^{\infty}\frac{(-2\pi i)^k n^{k-1}}{(k-1)!}e(nw).$$
Observe that equality holds wherever the two sides above converge.  Moreover, for $k=1$ the left hand summand is $O(1/n^2)$ and therefore
converges (provided we do not rearrange the sum).  The right hand side obviously converges to
\begin{align*}
-2\pi i\sum_{n=1}^{\infty}e(nw)&=-2\pi i\frac{1}{e(-w)-1}\\
&=\frac{1}{w}\sum_{k=0}^{\infty}B_k\frac{(-2\pi i w)^k}{k!},
\end{align*}
while we can write the left hand side as
\begin{align*}
    \frac{1}{w}-2w\sum_{n=1}^{\infty}\left(\frac{1}{n^2}\frac{1}{(1-w^2/n^2)}\right)&=
    \frac{1}{w}-2w\sum_{n=1}^{\infty}\left(\frac{1}{n^2}+\frac{w^2}{n^4}+\frac{w^4}{n^6}+\cdots\right)\\
    &=\frac{1}{w}-2w\sum_{k=0}^{\infty}\z(2k+2)w^{2k},
\end{align*}
where the change in order of summation is justified by absolute convergence.  Now compare coefficients of $w^{2k}$ to find that
$$\zeta(2k)=-\frac{(-2\pi i)^{2k}B_{2k}}{2(2k)!},$$ as required.

\begin{exer}
    Show that $\sigma_{k}$ is multiplicative; more generally, show that
    $$\sigma_{k}(m)\sigma_{k}(n)=\sum_{d|(m,n)}\sigma_{k}(\frac{mn}{d^2}).$$
\end{exer}

Recall that if $\varphi$ is a multiplicative function then so is
$$\psi(n):=\sum_{d_n} \varphi(n).$$  To see this, suppose that $(m,n)=1$.  We then have
\begin{align*}
    \psi(mn)&=\sum_{d|mn}\varphi(d)\\
    &=\sum_{d_1|m,d_2|n}\varphi(d_1d_2)\\
    &=\sum_{d_1|m}\sum_{d_2|n}\varphi(m)\varphi(n)\\
    &=\psi(m)\psi(n),
\end{align*}
where we have used the fact that if $(m,n)=1$ then the divisors of $mn$ are in bijective correspondence
with pairs of divisors $d_1,d_2$ of $m,n$ respectively.  This follows from the euclidian algorithm.  It is therefore
immediate that
$$\sigma_{k}(n)=\sum_{d|n}d^k$$ is multiplicative, since $d^k$ is.
Let $p$ be prime and suppose that $a\leq b$ are nonnegative integers.  Then we have
\begin{align*}
    \sigma_k(p^a)\sigma_k(p^b)&=\left(\sum_{r=0}^a p^{kr}\right)\left(\sum_{s=0}^b p^{ks}\right)\\
    &=\frac{1-p^{k(a+1)}}{1-p^k}\frac{1-p^{k(b+1)}}{1-p^k}\\
    &=\frac{1}{1-p^k}\left(\frac{1-p^{k(a+1)}}{1-p^k}-\frac{p^{k(b+1)}-p^{k(a+b+2)}}{1-p^k}\right)\\
    &=\frac{1}{1-p^k}\left(\frac{1-p^{k(a+1)}}{1-p^k}-p^{k(a+b+1)}\frac{1-p^{-k(a+1)}}{1-p^{-k}}\right)\\
    &=\frac{1}{1-p^k}\left(\sum_{j=0}^a p^{jk}-p^{(a+b+1)k}\sum_{j=0}^a p^{-jk}\right)\\
    &=\sum_{j=0}^a p^{jk}\frac{1-p^{(a+b+1-2j)k}}{1-p^k}\\
    &=\sum_{d|(p^a,p^b)} d^k \sigma_k\left(\frac{p^{a+b}}{d^2}\right).
\end{align*}
Now let
$$a=\prod_{i}p_i^{a_i},\quad\text{and}\quad b=\prod_{i}p_i^{b_i},$$
where $i$ ranges over a finite set and the $a_i,b_i$ are nonnegative integers.  Then we have
\begin{align*}
    \sigma_k(a)\sigma_k(b)&=\prod_{i}\sigma_k(p_i^{a_i})\sigma_k(p_i^{b_i})\\
    \intertext{since $\sigma_k$ is multiplicative,}
    &=\prod_i\sum_{d_i|(p_i^{a_i},p_i^{b_i})}d_i^k \sigma_k\left(\frac{p_i^{a_i+b_i}}{d_i^2}\right)\\
    &=\sum_{\prod d_i|\prod (p_i^{a_i},p_i^{b_i})}(\prod d_i)^k\prod \sigma_k\left(\frac{p_i^{a_i+b_i}}{d_i^2}\right)\\
    &=\sum_{\prod d_i|\prod (p_i^{a_i},p_i^{b_i})}(\prod d_i)^k\sigma_k\left(\prod \frac{p_i^{a_i+b_i}}{d_i^2}\right),
\end{align*}
where we have used the fact that $p_i\not=p_j$ for $i\neq j$ and therefore that $(p_i^{a_i+b_i}/d_i^2,p_j^{a_j+b_j}/d_j^2)=1$.
Finally, observe that the divisors of $(\prod (p_i^{a_i},\prod p_i^{b_i})$ are in bijective correspondence with lists $d_i$
of divisors of $(p_i^{a_i},p_i^{b_i})$.  Therefore, we see that the above expression is equal to
$$\sum_{d|(a,b)}d^k\sigma_k\left(\frac{ab}{d^2}\right)$$
and we are done.

\begin{exer}
    Show that
    $$n^{k}<\sigma_{k}(n)<\zeta(k)n^{k}$$ and that these are the sharpest asymptotics one can hope for in the sense that
    \begin{align*}
        \liminf \frac{\sigma_{k}(n)}{n^{k}}&=1\\
        \intertext{and}
        \limsup \frac{\sigma_{k}(n)}{n^{k-1}}&=\z(k).
    \end{align*}
\end{exer}

For any $n>1$ one has
$$n^k<\sum_{d|n}d^k\leq n^k+\left(\frac{n}{2}\right)^k+\left(\frac{n}{3}\right)^k+\ldots+1$$
and therefore that
\begin{eqnarray}
1<\frac{\sigma_k(n)}{n^k}\leq 1+\frac{1}{2^k}+\frac{1}{3^k}+\ldots+\frac{1}{n^k}<\z(k).\label{ineq}
\end{eqnarray}
Observe that if $p$ is prime then
$$\frac{\sigma_k(p)}{p^k}=1+\frac{1}{p^k}\longrightarrow 1$$ as $p\longrightarrow \infty$, and
$$1+\frac{1}{2^k}+\ldots+\frac{1}{m^k}<\frac{\sigma_k(m!)}{(m!)^k}<\z(k),$$
and therefore that $\sigma_k(m!)/(m!)^k\longrightarrow \z(k)$ as $m\longrightarrow\infty.$
Combining these results with (\ref{ineq}) shows that
    \begin{align*}
        \liminf \frac{\sigma_{k}(n)}{n^{k}}&=1\\
        \intertext{and}
        \limsup \frac{\sigma_{k}(n)}{n^{k}}&=\z(k),
    \end{align*}
as desired.

\begin{exer}
    Show that $\G:=\SL{\Z}$ acts discontinuously on $\H$.
\end{exer}

Suppose that $\G$ does not act discontinuously on $\H$.  Then there exists some $z\in\H$ such that the
orbit $\G z$ has a limit point in $\H$.  This implies that there exists a cauchy sequence
$\left\{g_i z\right\}_{i=1}^{\infty}$ with $g_i\in\G$ and $g_i\neq g_j$ if $i\neq j$.
Write $$g_i=\begin{pmatrix}a_i & b_i\\c_i & d_i\end{pmatrix}$$
and recall that we have
$$\Im(g_iz)=\Im(z)\frac{1}{|c_iz+d_i|^2}.$$
Now observe that since $a_i,b_i,c_i,d_i\in\Z$ there are only finitely many $g_i\in\G$
with bounded entries.  Put differently, for any $M>0,$ there exists some $i$ such that
all the entries of $g_i$ are larger than $M$ (since there are infinitely many $g_i$).
This implies that
$$\lim_{i\rightarrow\infty}\Im(g_iz)=0,$$
and therefore that $g_iz$ does not tend to a limit in $\H$, contradicting our hypothesis.
Therefore, $\G$ acts discretely on $\H$.

\begin{exer}
    Let $Q$ be a quadratic form $Q(x,y)=ax^2+bxy+cy^2$ with $a,b,c\in\Z$.  We have an action of $\G$ on
    quadratic forms given by
    $$g x=\alpha x+\beta y,\quad\text{and}\quad g y=\gamma x+\delta y$$ for
    $$g=\begin{pmatrix}\alpha & \beta \\ \gamma & \delta\end{pmatrix}\in\G$$
    and we say that two forms are equivalent if they are equivalent under this action.  What is the connection
    between this and the fundamental domain we constructed for $\SL{\Z}$?
\end{exer}

Of course, there are many connections.  Observe that since any $\SL{\Z}$ transformation is invertible,
equivalent quadratic forms represent the same integers.  Observe that we have
$$g\frac{x}{y}=\frac{\alpha x+\beta y}{\gamma x+\delta y}=\frac{\alpha(x/y)+\beta}{\gamma(x/y)+\delta}$$
if $y\neq 0$.  Setting $x/y=z$ we see that our quadratic form $Q$ gives rise to a quadratic polynomial
$Q(z)=az^2+bz+c$ and that the action of $\G$ on $Q$ is now just the usual action of $\G$ on $\H\cup \P(\Q).$
Furthermore, the discriminant of $Q$, defined to be $D=b^2-4ac$ is just the discriminant of $Q(z)$.  Since $\SL{\Z}$
is generated by $z\mapsto -1/z$ and $z\mapsto z+1$, we see that these transformations give us all equivalent
quadratic forms.

Now suppose that $D<0$.  Then the roots of $az^2+cz+d$ are complex, so that one root, say $\tau$ lies in $\H$.
We have just seen that two quadratic forms are equivalent if and only if their respective roots (in $\H$)
are $\G$ equivalent.  In particular, two quadratic forms are equivalent if and only if the $j$ invariants of their
roots (in $\H$) are equal.  Now we have a map from the set of equivalence classes $S$ of quadratic forms
to the set class group of the ring of integers of $\Q(\tau)$ given by
$$ax^2+bxy+cx\mapsto a(1,\tau)$$.
We have just shown that this map is injective.  It is also a surjective map since if $(u,v)$ is any fractional ideal
then we can assume that $u/v\in\H$ has minimal polynomial $az^2+bz+c$.  Then $ax^2+bxy+cy^2$ has discriminant $D$ and
maps to $a(1,\tau)$.  In particular, this shows that the number of equivalence classes of quadratic forms of discriminant $D$
is just the class number of the quadratic field $\Q(\tau)$ where $\tau$ is a root in $\H$ of $az^2+bz+c$ for any form
$Q=ax^2+bxy+cy^2$ in the class.

\end{document}