2
$\begingroup$

I'm aware that a basis of the modular forms of weight $k$ over $SL_2(\mathbb{Z})$ is $\{E_{4}^iE_{6}^j: 4i + 6j = k\}$ where $E_4$ and $E_6$ are the 4th and 6th Eisenstein series respectively. I'm wondering if a similar basis exists for principal congruence subgroups. In particular, does there exist a similar basis for the set of modular forms of weight k over $\Gamma(2)$?

$\endgroup$
1
  • $\begingroup$ It is more difficult. We know $\Delta \in S_{12}(\Gamma(1))$ has a product formula (thus no zeros on $\mathcal{H}$) and a simple zero at $i\infty$, thus for any $f \in M_{k+12}(\Gamma(1))$ then $\frac{f-f(i\infty) E_{12}}{\Delta} \in M_k(\Gamma(1))$. When $f\in M_k(\Gamma(n))$ for lowering the weight, what we'd need is finding some $g \in M_k(\Gamma(n))$ such that $f-g$ has a zero at every cusp. $\endgroup$
    – reuns
    Oct 11, 2017 at 15:50

1 Answer 1

2
$\begingroup$

The group $\Gamma(2)$ is conjugate to $\Gamma_0(4)$; if $f(\tau)$ is a modular form for $\Gamma(2)$, then $f(2\tau)$ is modular of level $\Gamma_0(4)$ and vice versa. So it suffices to compute the ring of modular forms of level $\Gamma_0(4)$. Sage has functionality to do this in a single step:

sage: ModularFormsRing(Gamma0(4)).generators()

[(2, 1 + 24*q^2 + 24*q^4 + 96*q^6 + 24*q^8 + O(q^10)),
(2, q + 4*q^3 + 6*q^5 + 8*q^7 + 13*q^9 + O(q^10))]

What this means is that the ring of modular forms for $\Gamma_0(4)$ is generated by the two forms $F, G$ with the given $q$-expansions, both of which live in weight 2. (In fact, both of them are Eisenstein series: they are linear combinations of $H(\tau)$ and $H(2\tau)$, where $H$ is the unique Eisenstein series of weight 2 and level 2.) So the space of forms of weight $2k$ is spanned by the sums $\{F^a G^b : a+b=k\}$, which are linearly independent.

What is making this work is that the modular curve $X(2) \cong X_0(4)$ has genus 0; whenever you have a genus 0 modular curve, you will get a similar isomorphism of the ring of modular forms onto a polynomial ring in two generators. (Edit: Sorry, that bit was wrong; it fails for $\Gamma(5)$, even though $\Gamma(5)$ does have genus 0.)

$\endgroup$
5
  • $\begingroup$ Thank you for the help! $\endgroup$
    – User93
    Oct 16, 2017 at 15:56
  • $\begingroup$ One last question, I can't seem to find what the unique Eisenstein series you mention looks like. Diamond and Shurman only give representation formulas for $k \geq 3$. Would you happen to have one or a reference? $\endgroup$
    – User93
    Oct 16, 2017 at 16:17
  • $\begingroup$ wstein.org/books/modform/modform/… $\endgroup$ Oct 16, 2017 at 19:05
  • $\begingroup$ Wait, if two subgroups are conjugate to one another then what's the relationship between their modular forms? It seems you're saying if g is the conjugating element then $$f( g \tau)$$ is a modular form of the conjugate subgroup but I don't see why. $\endgroup$
    – Camilo
    Oct 11, 2019 at 6:02
  • $\begingroup$ It is $f \mid_k g$, i.e. $j(g, \tau)^{-k} f(g \tau)$, which is a modular form for the conjugate subgroup. In this particular example $j(g, \tau)$ is a constant so we can ignore it. $\endgroup$ Oct 12, 2019 at 8:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .