I am looking for references that discuss Hecke operators $T_n$ acting on modular forms for the principal congruence subgroup $\Gamma(N)$ of the modular group $SL(2,Z)$ and am happy to restrict to the case that $(n,N)=1$. Most textbooks (Diamond and Shurman, Koblitz etc.) that discuss Hecke operators for congruence subgroups specialize to $\Gamma_0(N)$ or $\Gamma_1(N)$ which contain $T: \tau \rightarrow \tau+1$, but I am specifically interested in the action of Hecke operators on modular forms which are invariant under $T^N$ but not under $T$ and are thus forms for $\Gamma(N)$ but not for $\Gamma_0(N)$ or $\Gamma_1(N)$. It seems reasonably clear how to obtain the answer using the double coset formulation of Hecke operators, but I have not found a reference which writes this out explicitly in terms of the relation between the coefficients of the Fourier expansion of the form at the cusp at infinity and the coefficients of its image under $T_n$.

  • $\begingroup$ Miyake also works out the case $\Gamma_0(N)$, so I do not know. Have you looked at Shimura, introduction to the theory of automorphic form? It may contain what you want, but I don't have it here and can't check. $\endgroup$
    – Joël
    Commented Oct 11, 2017 at 1:26
  • $\begingroup$ There's some information/references, as well as applications (with the trace formula) in the recent paper of Kaplan and Petrow "Elliptic curves over a finite field and the trace formula", arxiv.org/abs/1510.03980 $\endgroup$ Commented Oct 11, 2017 at 5:15
  • $\begingroup$ Thanks. I did find something useful in Chapter 3 of Shimura. $\endgroup$ Commented Oct 11, 2017 at 15:44

2 Answers 2


The reason why Hecke theory for $\Gamma(N)$ doesn't get much treatment in the literature is because you can easily reduce it to the $\Gamma_1(N)$ case. More precisely, you can conjugate $\Gamma(N)$ by $\begin{pmatrix} N & 0 \\ 0 & 1\end{pmatrix}$ to get a group intermediate between $\Gamma_0(N^2)$ and $\Gamma_1(N^2)$.

This has come up before (in the context of explicit calcuations): see this question.


The Hecke operators $T(n)$ and the dual Hecke operators $T'(n)$ acting as correspondences on the modular curve $Y(N)$ are defined by Kato in $p$-adic Hodge theory and values of zeta functions of modular forms, section 2.9 (in Kato's notation $Y(N)=Y(N,N)$). The action of $T(p)$ on Fourier expansions is given in section 4.9, there he also describes the relation between his definition and other definitions in the litterature.

Actually, when you conjugate the double coset $\Gamma(N) \begin{pmatrix} n & 0 \\ 0 & 1 \end{pmatrix} \Gamma(N)$ in $\mathrm{GL}_2(\mathbf{Q})$ by the matrix $\begin{pmatrix} N & 0 \\ 0 & 1 \end{pmatrix}$, you get the double coset $\Gamma \begin{pmatrix} n & 0 \\ 0 & 1 \end{pmatrix} \Gamma$, where $\Gamma$ is this subgroup intermediate between $\Gamma_1(N^2)$ and $\Gamma_0(N^2)$. So it should be a simple exercise to check that Kato's definition agrees with David's answer.

  • $\begingroup$ In section 4.9 of Kato he does give the action of $T(p)$ on Fourier coefficients, but in the case $p|n$ for the coefficient $b_n$ he has an assumption about the action of $\begin{pmatrix} 1/p & 0 \\ 0 & p \end{pmatrix}^*$ on the modular function $f$. If $p^2=1 ~mod~N$ then I believe the assumption is trivially satisfied, but if this is not the case I don't see that the assumption is true in general. In this case is it true that the Hecke operator still exists but does not have a simple expression in terms of coefficients of the $q$ expansion? $\endgroup$ Commented Nov 11, 2017 at 17:07
  • $\begingroup$ @Jeff Harvey The assumption $(\begin{smallmatrix} 1/p & 0 \\ 0 & p \end{smallmatrix})^* f = \varepsilon(p) f$ is trivially satisfied when $p=1 \pmod{N}$ but not otherwise (this matrix is a diamond operator). You're right that in general there is no simple expression for Hecke operators in terms of $q$-expansions. $\endgroup$ Commented Nov 13, 2017 at 8:35

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.