# Proving $\Delta E_4$ is a simultaneous eigenfunction of weight 16 for $SL(2, \mathbb Z)$

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!

• 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 $T_m$. Commented Feb 27, 2022 at 15:06
• 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! Commented Feb 27, 2022 at 15:07
• Oooh yes i'm so dumb. Thank you very much. I knew I was missing something simple. Commented Feb 27, 2022 at 15:10

## 1 Answer

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)