A result due to B. Conrad (http://math.stanford.edu/~conrad/papers/prasanna-inv.pdf, Theorem A.1) states that the Atkin-Lehner operator $w_{Q,k}$ is $\mathbb{Z}[1/Q]$-integral on $M_k(\Gamma_0(N))$. In other words, if $f\in M_k(\Gamma_0(N))$ has coefficients in $\mathbb{Z}[1/Q]$ then $w_{Q,k}(f)$ has coefficients in $\mathbb{Z}[1/Q]$.

Does anyone know of a corresponding result over $\Gamma_1(N)$? My guess is that $w_{k,Q}$ is $\mathbb{Z}[1/Q][\zeta_Q]$-integral on $M_k(\Gamma_1(N))$ but I can't find a reference for this.

I am however, aware of a weaker result (Theorem 5.4 in https://arxiv.org/abs/1807.00391), which implies that $w_{k,Q}$ is $\mathbb{Q}(\zeta_Q)$-integral on $M_k(\Gamma_1(N))$.

  • $\begingroup$ In the preprint The Manin constant and the modular degree math.u-psud.fr/~cesnavicius/Manin-degree.pdf the authors have proved bounds on the denominators of modular forms at the cusps, using adelic techniques (see Section 4). Maybe you could email them and ask whether their results are sufficient to prove what you want. $\endgroup$ – François Brunault Feb 28 '20 at 8:49
  • $\begingroup$ It is quite straightforward to prove that $w_Q$ is $\mathbf{Z}[1/N, \zeta_Q]$-integral on forms of level $\Gamma_1(N)$; the hard work in Conrad's note is to deal with integrality at primes dividing $N$ but not $Q$. Would this weaker result be sufficient for your purposes? $\endgroup$ – David Loeffler Mar 8 '20 at 9:16
  • $\begingroup$ @DavidLoeffler actually yes that result would be sufficient. Would you be able to provide an outline of why it's true? $\endgroup$ – Daniel Johnston Mar 8 '20 at 22:12

Theorem. Let $\ell$ be prime, and $Q, R \ge 1$ such that $(\ell, Q, R)$ are pairwise coprime. Let $N = QR$ and for simplicity assume $N \ge 4$. Then $W_Q$ preserves $M_k(\Gamma_1(N), \mathbf{Z}[1/N, \zeta_Q])$.

Proof. Let $M_k^{\mathrm{wk}}(\Gamma_1(N), A)$ denote the space of weakly modular forms (possibly meromorphic at the cusps) with $q$-expansions in the ring $A$. If $A$ is a subring of $\mathbf{C}$ then $$M_k(\Gamma_1(N), A) = M_k^{\mathrm{wk}}(\Gamma_1(N), A) \cap M_k(\Gamma_1(N), \mathbf{C})$$ so it suffces to show that $M_k^{\mathrm{wk}}(\Gamma_1(N), \mathbf{Z}[1/N, \zeta_Q])$ is stable under $W_Q$.

If $A$ is a $\mathbf{Z}[1/N]$-algebra, then elements of $M_k^{\mathrm{wk}}(\Gamma_1(N), A)$ can be interpreted as rules sending "test objects" $(E, P_Q, P_R, \omega) / B$ to elements of $B$. Here $B$ is a $A$-algebra, $E$ is an elliptic curve over $B$, $P_Q$ and $P_R$ are points of exact order $Q$, $R$ respectively, and $\omega$ is a global differential on $E$. These have to satisfy various conditions (the main ones are compatibility with base change and homogeneity in $\omega$ of weight $k$).

So it suffices to show that $W_Q$ makes sense on test objects if $A = \mathbf{Z}[1/N, \zeta_Q]$. The map will send $(E, P_Q, P_R, \omega) / B$ to $(E', P_Q', P_R', \omega')/B$, where all but one of these objects are simple to define:

  • $E' = E/\langle P_Q \rangle$
  • $P_R' = \pi(P_R)$, where $\pi : E \to E / \langle P_Q\rangle$ is the quotient map
  • $\omega' = \pi_*(\omega)$

The hard one is $P_Q'$: one checks that $E[Q] / \langle P_Q\rangle$ has a unique generator $P_Q'$ characterised by the Weil pairing $\langle P_Q, P_Q'\rangle = \zeta_Q$ (and since it is unique, its formation is compatible with base-change). $\square$

(Caveat: this construction gives an operator whose square is something like $\langle Q\rangle_R Q^k$ where $\langle -\rangle_R$ denotes the diamond operator for something that is 1 (mod Q) and Q (mod R). Some people prefer to normalise away the $Q^k$, but this may not be possible if $Q$ is odd without introducing an extraneous $\sqrt{Q}$ into your ring.)

  • $\begingroup$ Thank you very much, this is very useful! Do you know whether there is an even simpler integrality statement if we instead restrict to modular forms with character? That is, would the Atkin-Lehner operator $w_{Q,k}$ be $\mathbb{Z}[1/N,\zeta_d]$-integral for some $d$ related to the conductor of the character? If there isn't a clear answer to this I'm happy to make a new question. $\endgroup$ – Daniel Johnston Apr 8 '20 at 0:47
  • 1
    $\begingroup$ If you're working with $\Gamma_0(N) \cap \Gamma_1(M)$ (i.e. level $N$ and characters of conductor dividing $M$) then the exact same argument shows $w_Q$ is defined over $\mathbf{Z}[1/N, \zeta_d]$ where $d = GCD(M, Q)$. (But this isn't really an "integrality" result as such, because the $1/N$ is still there, it's just controlling the roots of unity a little better.) $\endgroup$ – David Loeffler Apr 8 '20 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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