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\newtheorem{defn}{Definition}
\newtheorem{exer}{Exercise}
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\newtheorem{cor}{Corollary}
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\title{Modular Forms, PS 4}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
    Let $e_1=\Delta E_4^3,e_2=\Delta^2$.  Then clearly $e_1,e_2\in S_{24}(\G)$.  Moreover, we know that $S_{24}(\G)$ is two-dimensional.
    Moreover, $e_1,e_2$ are linearly independent since $E_4^3,\Delta$ are not multiples of one another (recall the definition of
    $\Delta$).  Therefore, $e_1,e_2$ is a basis for $S_{24}(\G).$  Recall that if
    $$f=\sum_{n=1}^{\infty}a_n q^n\in S_{24}(\G)$$ then
    $$T(m)f=\sum_{n=1}^{\infty} b_n(m)q^n=\sum_{n=1}^{\infty} \left(\sum_{d|(m,n)}d^{23} a_{mn/d^2}\right)q^n$$
    is also a cusp form of weight $24$.  When $m=2$, we have at once
    \begin{align}
    b_n(2)&=
        \begin{cases}
            a_{2n} & \text{if $n$ is odd}\\
            a_{2n}+2^{23} a_{n/2} &\text{otherwise}
        \end{cases}.
        \label{cases}
    \end{align}

    Using \texttt{PARI}, we easily compute
    \begin{multline}
        e_1=q + 696q^2 + 162252q^3 + 12831808q^4 + 34188270q^5 - 882400608q^6 - 833380744q^7\\
         \shoveright{+ 25487347200q^8 +\cdots}\\
    \end{multline}
    \begin{align*}
    e_2=q^2 - 48q^3 + 1080q^4 - 15040q^5 + 143820q^6 - 985824q^7 + 4857920q^8+\cdots,
    \end{align*}
    and using (\ref{cases}), we therefore have
    \begin{align*}
        T(2)e_1&=696q+ 21220416q^2 - 882400608q^3+31325818368q^4+\cdots\\
        T(2)e_2&=q+ 1080q^2+ 143820q^3+13246528q^4+\cdots.
    \end{align*}
    A quick computation now shows
    \begin{align*}
    &T(2)e_1-696e_1-20736000e_2=O(q^5)\\
    &T(2)e_2-e_1-384e_2=O(q^5).
    \end{align*}
    Observe that the left hand sides above are both cusp forms with a zero of order at least $5$ at infinity.  But we have
    $$\ord_{\infty}(f)\leq 24/6=4,$$ for any modular form of weight $24$ that is not identically 0.  It follows
    that
    \begin{align*}
    T(2)e_1&=696e_1+20736000e_2\\
    T(2)e_2&=e_1+384e_2,
    \end{align*}
    and hence that the matrix of $T(2)$ in our basis is
    $$T_2=\begin{pmatrix}696 & 1\\ 20736000 & 384\end{pmatrix}.$$
    The characteristic polynomial of $T_2$ is therefore
    $$p(x)=x^2-1080 x-20468736,$$
    so that the eigenvalues of $T_2$ are
    \begin{align*}
    \lambda_1&=540+12\sqrt{144169}\\
    \lambda_2&=540-12\sqrt{144169}.
    \end{align*}
    We then determine the corresponding eigenvectors:
    \begin{align*}
    v_1&=\begin{pmatrix}1\\-156+12\sqrt{144169}\end{pmatrix}\\
     v_2&=\begin{pmatrix}1\\-156-12\sqrt{144169}\end{pmatrix}.
    \end{align*}
    We therefore set
    \begin{align*}
        f_1&=\Delta E_4^3-(156-12\sqrt{144169})\Delta^2\\
        f_2&=\Delta E_4^3-(156+12\sqrt{144169})\Delta^2.
    \end{align*}
    By our calculations above, we have
    \begin{align*}
    T(2)f_1&=\lambda_1 f_1\\
    T(2)f_2&=\lambda_2 f_2.
    \end{align*}
    Observe that $\lambda_1,\lambda_2$ lie in a totally real quadratic field and are algebraic integers.  Moreover,
    the galois group of this field has one nontrivial automorphism, $\sigma$, and
    $$\lambda_1=\sigma\lambda_2,$$ from which it follows that
    $$f_1=\sigma f_2,$$ since $\Delta,E_4$ have integer coefficients.  Finally, we observe that since $\lambda_1\neq \lambda_2,$
    we have found a Hecke eigenbasis for $S_{24}(\G).$
\end{exer}

\begin{exer}
    We approach this naively.  Taking the usual region for $\G/\H$, we have that the hyperbolic area of $\G/\H$ is
    \begin{align*}
        \int_{x=-1/2}^{1/2}\int_{y=\sqrt{1-x^2}}^{\infty}\frac{\mathrm{d}y\,\mathrm{d}x}{y^2}&=
                \int_{x=-1/2}^{1/2}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\
                &=2\sin^{-1}(1/2)\\
                &=\frac{\pi}{3}.
    \end{align*}
\end{exer}

\begin{exer}
    We have
    \begin{align*}
        \int_{\R}e^{-\pi x^2 y} e^{-2\pi i n x}\,\mathrm{d}x&=\int_{\R}e^{-\pi y(x+in/y)^2- \pi n^2/y}\,\mathrm{d}x\\
        &=e^{- \pi n^2/y}\int_{\R}e^{-\pi yx^2}\,\mathrm{d}x\\
        &=\frac{e^{- \pi n^2/y}}{\sqrt{\pi y}}\int_{\R}e^{-x^2}\,\mathrm{d}x\\
        &=\frac{e^{- \pi n^2/y}}{\sqrt{y}}.
    \end{align*}
    The Poisson Summation formula then tells us (since $e^{-\pi n^2 y}$ is a rapidly decreasing function of $n$)
    that
    \begin{align*}
        \sum_{n\in\Z}e^{-\pi y n^2}=\frac{1}{\sqrt{y}}\sum_{n\in\Z} e^{-\pi n^2/y}.
    \end{align*}
    Analytic continuation then gives us the formula for all $y$ with $\Re(y)>0$.
\end{exer}

\begin{exer}
    We now give some intuition as to why the formula of Exercise 3 is likely.  Let us compute the integral
    of $e^{-\pi x^2}$ from $x=-a$ to $a$ as a Riemann sum.  This is:
    \begin{align*}
        \lim_{N\rightarrow \infty}\frac{a}{N}\sum_{j=-N}^{N}e^{-\pi (ja/N)^2 }.
    \end{align*}
    Now since $e^{-\pi x^2}$ is rapidly decreasing, we may truncate the sum at some large value of $N$; the limit will
    then approximate the integral very closely.  Therefore,
    \begin{align*}
        \lim_{y\rightarrow 0} \sqrt{y}\sum_{j=-N}^{N}e^{-\pi j^2 y}&\approx \int_{-N}^N e^{-\pi x^2}\,\mathrm{d}x\\
        &\approx \int_{-\infty}^{\infty} e^{-\pi x^2}\,\mathrm{d}x\\
        &=1\\
        &=\lim_{y\rightarrow 0} \sum_{n\in\Z} e^{-\pi n^2/y}.
    \end{align*}
    We therefore see that the transformation formula proved in Exercise 3 is likely, and may be interpreted as
    a statement about approximating integrals.
\end{exer}


\end{document}
