First of all some notations :
$M_k(SL_2(\mathbb{Z}))$ denotes the set of all modular forms of weight k on $SL_2(\mathbb{Z}) $ .
For every even $k\geq 4$
$$E_k(z)=1-\frac{2k}{B_k}\sum_{n=1}^{\infty}\sigma_{k-1 }(n)q^n$$
is a modular form of weight k for the full modular group $SL_2(\mathbb{Z}) $ satisfying $E_k(i\infty)=1$ .
Now the exercise :
Let $r_k=\dim M_k(SL_2(\mathbb{Z}))$ and $l=k-12r_k+12$ . For any modular form $f \in M_k(SL_2(\mathbb{Z})) $ , show that
$$fE_l^{-1} \in M_{12r_k-12}(SL_2(\mathbb{Z})).$$
Proof : Since $f \in M_k(SL_2(\mathbb{Z}))$ , it is a linear combination of $$ E_4^aE_6^b$$ such that $4a + 6b = k$. Now $E_l^{-1}=\frac{E_{14-l}}{E_{14}}$ , so that $ fE_l^{-1}$ is a linear combination of terms like $$ E_4^aE_6^b \frac{E_{14-l}}{E_{14}}. \tag1 \label1$$
We have $E_{14}=E_4^2E_6$ and $E_{14-l}$ is one of $E_0, E_4,E_6,E_8,E_{10}$ or $E_{14}$. Recalling that
$E_8^2=E_4^2$ and $E_{10}=E_4E_6$, we see that $\frac{E_{14-l}}{E_{14}}$ is one of $\frac{1}{E_4^2E_6} ,\frac{1}{E_4E_6},\frac{1}{E_4^2},\frac{1}{E_6}$ , or $1$.
If $a\geq 2,b\geq1$ , then the term in(1) is holomorphic .
Thus, we need only consider the five cases $$(a,b) \in \lbrace(0,1),(1,0),(1,1),(2,0),(2,1) \rbrace .$$
But in all these cases , $r_k=1 $ so that $l=k\leq14$ and the result is evident.
- Why can I not say that f is a modular form of weight k and $E_l$ is one of weight l so that the quotient is of weight $k-l=12r_k-12$?
- In the proof why is $E_{14-l}$ one of $E_0,E_4...$
Is it because $4a+6b=k$?
I think $\frac{E_{14-l}}{E_{14}}$ can also be $\frac{1}{E_4}$. - Why do we have to consider the case (2,1)? I thought for this case (1) is already holomorphic.
- Why is the result evident? Because $fE_l^{-1}$ is of weight 0 ?