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Let $\Gamma$ be the $SL(2, \mathbb Z)$ group, $M_k(\Gamma)$ the space of modular forms of weight $k$ and $S_k(\Gamma)$ the subspace of cusp forms.

It is known (for ex. Neal Koblitz's book "Introduction to Elliptic Curves", chapter III section 2) that we have $$ S_k(\Gamma) = \Delta M_{k-12}(\Gamma) $$ and $M_k(\Gamma) = \mathbb C E_k$ for $k = 4,6,8,10,14$. Therefore, we have for $k = 16$ that $$ S_{16} = \mathbb C \Delta E_4. $$ where $$ \Delta = q \prod_{n=1}^{\infty}(1-q^n)^24 = \sum_{n=1}^{\infty} \tau(n) q^n $$ is the discriminant and $$ E_4 = 1 - \frac{2k}{B_k} \sum_{n=1}^{\infty} \sigma_{k-1}(n)q^n $$ is the 4-Eisenstein series.

An exercise from my Elliptic Curves course asks for a normalized simultaneous eigenform (for all Hecke operators) that generates $S_{16}(\Gamma)$.

What I've tried so far: showing that $\Delta E_4$ is a simultaneous eigenform explicitly so I can find the coefficient for $q$ and normalize it. I've been trying to use the fact that simultaneous eigenforms of weight $k$ given by $ \sum_{n=1}^{\infty} c(n) q^n$ satisfy $$ c(m)c(n) = \sum_{d | (n,m)} d^{k-1} C\left(\frac{nm}{d}\right) $$ but I'm honestly lost in calculations.

Is there any other way to do this? Any help is appreciated!

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    $\begingroup$ Each $T_m$ sends $S_{16}(SL_2(\Bbb{Z}))=\Bbb{C}\Delta E_4$ to itself so it is clear that $\Delta E_4$ is an eigenfunction of all the $T_m$. $\endgroup$
    – reuns
    Feb 27, 2022 at 15:06
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    $\begingroup$ I think it would be better to first show that $S_{16}$ is one-dimensional. Then, it immediately follows that any element of $S_{16}$ is an eigenform. This is the method used to prove that $\Delta$ is an eigenform! $\endgroup$
    – Mathmo123
    Feb 27, 2022 at 15:07
  • $\begingroup$ Oooh yes i'm so dumb. Thank you very much. I knew I was missing something simple. $\endgroup$
    – fbianchini
    Feb 27, 2022 at 15:10

1 Answer 1

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Each $T_m$ sends $S_{16}(SL_2(\Bbb{Z}))=\Bbb{C}\Delta E_4$ to itself so it is clear that $\Delta E_4$ is an eigenfunction of all the $T_m$.

When saying so it is implicit that we are using a few results:

  • The construction of $E_4=r \sum_{a,b\ne 0,0} \frac1{(az+b)^4}$ which is in $M_4(SL_2(\Bbb{Z}))$

  • We need the construction of $\Delta\in S_{12}(SL_2(\Bbb{Z}))$ as a $q$-product (with only one simple zero at $i\infty$) to get that any $f\in S_{16}(SL_2(\Bbb{Z}))$ is such that $f/\Delta\in M_4(SL_2(\Bbb{Z}))$.

  • $M_4(SL_2(\Bbb{Z}))$ is automatically one-dimensional because otherwise there is some non-zero $g\in S_4(SL_2(\Bbb{Z}))$ so $g/\Delta \in M_{-8}(SL_2(\Bbb{Z}))$ and $E_4^2 g/\Delta $ is a non-constant everywhere holomorphic modular function which is impossible by the maximum modulus principle.

    Whence $ S_{16}(SL_2(\Bbb{Z}))=\Bbb{C}\Delta E_4$.

  • That $T_m$ is a linear map $M_{16}(SL_2(\Bbb{Z}))\to M_{16}(SL_2(\Bbb{Z}))$ (which evidently maps the cusp forms to themselves)

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