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Why are the basis of the space of modular form in $SL_2(\mathbb{Z})$ of weight 12 ($M_{12}(SL_2(\mathbb{Z})$)

$E_3^4$ and $\Delta$?

There is a theorem that says that the generators of modular form of weight $k$ are $E_4^{\alpha} \cdot E_6^{\beta}$ with $4\alpha +6\beta=k$. So shouldn't the basis be $E_4^3$ and $E_6^2$??

I don't know how the $\Delta$ appears. All I know is that $\Delta$ is of weight 12 and it has integral coefficients in the q-expansion just like $E_4$ and $E_6$

I can't seem to grasp this basic idea. Does it have to do with

$M_k = S_k + \mathbb{C}G_k$? ( the plus '+' is actually the tensor product.)

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    $\begingroup$ Vector spaces generally have more than one basis! The fact that $E_4^3, E_6^2$ is a basis doesn't mean $E_4^3, \Delta$ can't also be one. $\endgroup$ Oct 26, 2016 at 8:59
  • $\begingroup$ What is $E_3$ by the way (the defining series for $E_k$ is $0$ for odd $k$) ? Is it a typo for $E_4^3$ ? $\endgroup$
    – reuns
    Oct 26, 2016 at 9:00
  • $\begingroup$ Yes it's $E_4^3$ $\endgroup$
    – Tasmia
    Oct 26, 2016 at 13:13
  • $\begingroup$ Thank you it's true we can have different generators $\endgroup$
    – Tasmia
    Oct 26, 2016 at 17:08

1 Answer 1

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Yes, the Theorem is correct: a basis of $\mathcal{M}_k$ are all monomials $E_4^aE_6^b$ with $4a+6b=k$, and $a,b\ge 0$ integers. With $k=12=4a+6b$ we have either $(a,b)=(3,0)$ or $(a,b)=(0,2)$, because $a,b\ge0$ are integers. So the dimension is $2$. Of course, the Theorem gives, in general that the space $\mathcal{M}_k$ has dimension $$ \dim \mathcal{M}_k=\begin{cases} [k/12] & \text{ if } k\equiv 2\bmod 12\\ [k/12]+1 & \text{ if } k\not\equiv 2\bmod 12\end{cases} $$ Since $k=12$ is not congruent to $2$ modulo $12$ we obtain dimension $1+1=2$.

As to $\Delta$, there is the subspace $S_k\subseteq \mathcal{M}_k$ of cusp forms. For $k=12$, $\dim S_{12}=1$ with basis $\Delta$.

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    $\begingroup$ And the basis of Hecke eigenforms is $E_{12},\Delta$, where $691 E_{12} = 441 E_4^3+ 250 E_6^2$ and $\Delta=\frac{E_4^3-E_6^2}{1728}$ $\endgroup$
    – reuns
    Oct 26, 2016 at 9:01
  • $\begingroup$ Yes that makes sense, thank you so much! $\endgroup$
    – Tasmia
    Oct 26, 2016 at 17:08

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