9
$\begingroup$

Every reference I've seen on modular forms seems to jump from the general definition of a modular form for congruence subgroups to studying modular forms just for $\Gamma_0(N)$.

Once upon a time, I saw something like

$$M_k(\Gamma(N)) \cong\bigoplus_{\chi\mod N} M_k(N^2,\chi)$$

where $M_k(\Gamma(N))$ is the vector space of all weight $k$ modular forms for $\Gamma(N)$, and on the right side $\chi$ ranges over all dirichlet characters mod $N$, and $M_k(N^2,\chi)$ is the vector space of weight $k$ modular forms $f$ satisfying $$f(\gamma z) = \chi(d)(cz+d)^kf(z)$$ for all $\gamma = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\in \Gamma_0(N^2)$.

However, I don't see how this makes sense, since $\Gamma(N)$ is not a subgroup of $\Gamma_0(N^2)$.

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ If you conjugate $\Gamma(N)$ by $\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$ you get a subgroup of $\Gamma_0(N^2)$. $\endgroup$
    – Tyler Lawson
    Oct 26, 2016 at 16:54
  • 1
    $\begingroup$ See Section 3.5 of my paper: arxiv.org/pdf/1609.06740.pdf (this is for Maass forms instead of holomorphic modular forms, but the same argument applies in both cases). $\endgroup$
    – Peter Humphries
    Oct 26, 2016 at 17:07
  • $\begingroup$ @TylerLawson Is it obvious that conjugate of $\Gamma(N)$ by $\begin{bmatrix} 1 & 0 \\ 0 & N\end{bmatrix}$ should be (a) normal inside $\Gamma_0(N)$, and (b) have quotient isomorphic to $(\mathbb{Z}/N\mathbb{Z})^\times$? $\endgroup$
    – stupid_question_bot
    Oct 26, 2016 at 17:26
  • 1
    $\begingroup$ This question is a bit silly. It might "suffice" to do Gamma_0(N) if you're interested in, say, elliptic curves, but it certainly does not suffice if you're interested in, say, weight 1 modular forms. Whether it not it suffices will depend on the context. A slightly more interesting question is why Gamma_1(N) suffices, but this has been answered in the comments above; any modular form for a congruence subgroup gives rise to an automorphic representation which will contain an element invariant under Gamma_1(N). $\endgroup$
    – znt
    Oct 26, 2016 at 22:11
  • 2
    $\begingroup$ @znt, I think you're reading "modular forms just for $\Gamma_0(N)$" to mean modular forms of level $q$ with trivial nebentypus (often denoted by $\mathcal{M}_k\left(\Gamma_0(N)\right)$), whereas the question clearly means modular forms of level $q$ with any nebentypus $\chi$ (often written as $\mathcal{M}_k\left(\Gamma_0(N),\chi\right)$ instead of $\mathcal{M}_k(N,\chi)$ to emphasise this point). Of course, each of these spaces is a subspace of $\mathcal{M}_k\left(\Gamma_1(N)\right)$. $\endgroup$
    – Peter Humphries
    Oct 26, 2016 at 23:06

1 Answer 1

Reset to default
7
$\begingroup$

Conjugating $\Gamma(N)$ by $\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix}$ $$\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & N \end{bmatrix} \displaystyle\Gamma(N)\scriptstyle\begin{bmatrix} 1 & 0 \\ 0 & 1/N \end{bmatrix} \scriptstyle \ \ = \ \ \underbrace{\left\{\ \scriptstyle\begin{bmatrix} aN+1 & b \\ cN^2 & dN+1\end{bmatrix} \in SL_2(\mathbb{Z})\right\}}_{\displaystyle\tilde{\Gamma}_1(N^2)}\displaystyle \ \supset \Gamma_1(N^2)$$

Define the linear operator $T : M_k(\Gamma(N)) \to M_k(\tilde{\Gamma}_1(N^2)), \ \ T f(\tau) = f (N\tau)$,

and the usual linear operators for showing $M_k(\Gamma_1(N^2)) =\displaystyle \bigoplus_{\chi \bmod N^2}M_k(\Gamma_0(N^2),\chi)$

  • For $gcd(d,N^2)=1$, let $\langle d \rangle : M_k(\Gamma_1(N^2)) \to M_k(\Gamma_1(N^2)) $, $\ \ \langle d \rangle g = g|_k\gamma, \quad \gamma \in \Gamma_0(N^2), \quad\gamma_d \equiv d \bmod N^2$ (which is well-defined, not depending on the chosen $\gamma$). Note that $\langle d d' \rangle = \langle d \rangle\langle d' \rangle$

  • And for a $\chi \bmod N^2$ : $$ \pi_\chi g= \frac{1}{\varphi(N^2)}\sum_{\begin{array}{l}d \bmod N^2\\gcd(d,N^2) =1\end{array}} \overline{\chi(d) }\langle d \rangle g$$ $\pi_\chi$ is an orthogonal projection $M_k(\Gamma_1(N^2)) \to M_k(\Gamma_0(N^2),\chi)$, and $\displaystyle\sum_{\chi \bmod N^2} \pi_\chi g=\langle 1\rangle g= g$ and for any $\chi \ne \chi'$ : $\pi_\chi \pi_{\chi'} = 0$

  • Finally, $Tf = \langle dN+1\rangle T f$, so that $\langle dN+d'\rangle T f= \langle d'\rangle T f$ and hence $\pi_\chi T f = 0$ whenever $\chi$ isn't a character $\bmod N$.

Thus $\sum_{\chi \bmod N} \pi_\chi Tf = Tf$, and together with $M_k(\Gamma_0(N^2),\chi) \subset M_k(\tilde{\Gamma}_1(N^2))$ for any $\chi \bmod N$,

it means that $$M_k(\Gamma(N)) \simeq M_k(\tilde{\Gamma}_1(N^2)) =\bigoplus_{\chi \bmod N}M_k(\Gamma_0(N^2),\chi)$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ (I'd like to know if you had a different idea for showing it) $\endgroup$
    – reuns
    Oct 26, 2016 at 23:21
  • $\begingroup$ There's an error here. The map $T$ does not map modular forms for $\Gamma(N)$ to modular forms for $\tilde{\Gamma}_{1}(N^{2})$, rather it maps in the other direction. You can see this because if $f$ is a modular form for $\Gamma(N)$, then since $\Gamma(N)$ contains $\begin{bmatrix} 1 & N \\ 0 & 1 \end{bmatrix}$ we will have $f(\tau+N) = f(\tau)$ and so $f$ will have an expansion in terms of $q^{1/N}$. With $T$ as it is defined, $Tf$ will have an expansion in powers of $q^{1/N^{2}}$. $\endgroup$
    – Jeremy Rouse
    Oct 26, 2016 at 23:40
  • $\begingroup$ The answer has now been edited to fix the error. $\endgroup$
    – Jeremy Rouse
    Oct 27, 2016 at 21:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.