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(Sorry for my poor english...)

Let $N$ be a positive integer and $f=\sum_{n=1}^{\infty} a(n)q^n\in S_{k}(\Gamma_0(N))\cap K[[q]]$ be a cusp form with number field $\mathbb{Q}(\xi_{N})\subset K$ where $\xi_N$ be a $N$-th root of unity. For $\sigma\in Gal(K/\mathbb{Q})$, $f^{\sigma}$ is defined by \begin{equation} f^{\sigma}=\sum_{n=1}^{\infty}\sigma(a(n))q^n. \end{equation} I already know that $f^{\sigma}\in S_{k}(\Gamma_0(N))$.

Q. Do $f$ and $f^{\sigma}$ have the same order at each cusp $s$ of $\Gamma_0(N)$?

EDIT : Let $K\subset \mathbb{Q}(\xi_N)$. Do $f$ and $f^{\sigma}$ have the same order at each cusp $s$ of $\Gamma_0(N)$?

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    $\begingroup$ There are newforms $f$ on $\Gamma_0(N)$ such that $f$ and $f^\sigma$ do not always have the same order of vanishing at the cusps. I have examples at level $567,625,891$. But here, do you assume that $f$ is a newform? Do you really assume that the field of coefficients of $f$ contains the $N$-th cyclotomic field? $\endgroup$
    – François Brunault
    Aug 5, 2019 at 17:03
  • $\begingroup$ How do you prove your claims ? $\endgroup$
    – reuns
    Aug 5, 2019 at 21:35
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    $\begingroup$ @reuns Here is a reference: arxiv.org/abs/1609.08939v4 (see page 5 for the examples I mention). These newforms $f$ have different orders of vanishing at conjugate cusps. It follows that $f$ and $f^\sigma$ have different order of vanishing at the same cusp. $\endgroup$
    – François Brunault
    Aug 5, 2019 at 21:51
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    $\begingroup$ @ililiil That's right. For any modular form $f$, any cusp $x$ and any $\sigma \in \mathrm{Aut}(\mathbb{C})$, the order of vanishing of $f^\sigma$ at $\sigma(x)$ is equal to that of $f$ at $x$. This follows from the fact that the modular curve $X_0(N)$ is an algebraic curve defined over $\mathbb{Q}$. Cusp forms of weight 2 are differential forms on $X_0(N)$ so are also algebraic. For general weight $k>2$, modular forms are sections of certain line bundles on $X_0(N)$, so are again algebraic. Once you know algebraicity, the proof of this result is formal. $\endgroup$
    – François Brunault
    Aug 6, 2019 at 8:31
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    $\begingroup$ @ililiil For an explicit formula for $\sigma(x)$ you can look at Glenn Stevens's book Arithmetic on modular curves, the first sections are available on Google Books. $\endgroup$
    – François Brunault
    Aug 6, 2019 at 9:39

2 Answers 2

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We denote by $e_f(x) \geq 1$ the order of vanishing of a cusp form $f$ at a cusp $x$.

There is a newform $f$ of weight 2 and level $\Gamma_0(625)$ with coefficients in $\mathbb{Q}(\sqrt{5})$ and Fourier expansion \begin{equation*} f = q + \left(\frac{-1-\sqrt{5}}{2}\right) q^2 + \left(\frac{-3-\sqrt{5}}{2}\right) q^3 + \left(\frac{-1+\sqrt{5}}{2}\right) q^4 \ldots \end{equation*} One can show that \begin{equation*} e_f(a/25) = \begin{cases} >1 & \textrm{if } a \equiv \pm 1 \textrm{ mod } 5 \\ 1 & \textrm{if } a \equiv \pm 2 \textrm{ mod } 5. \end{cases} \end{equation*} The cusps $a/25$ are Galois conjugate. If we denote by $\sigma$ the non-trivial automorphism of $\mathbb{Q}(\sqrt{5})$, we get $e_{f^\sigma}(1/25)=1$ while $e_f(1/25)>1$. Note here that $\mathbb{Q}(\sqrt{5})$ is contained in the cyclotomic field $\mathbb{Q}(\zeta_{625})$.

I should add that determining the order of vanishing of a modular form at an arbitrary cusp is a difficult question in general. You can certainly guess it numerically by just estimating the decay rate of the modular form at the given cusp. However, proving it rigorously is much more delicate. For the modular form above this involves computations with the local automorphic representation associated to $f$, which in this case is supercuspidal and is described by an irreducible representation of $\mathrm{GL}_2(\mathbb{Z}/25\mathbb{Z})$ of dimension 20 with coefficients in $\mathbb{Q}(\sqrt{5})$.

For other examples, one can show that if $f$ is a newform on $\Gamma_0(p^4)$ with $p \geq 5$ prime, such that the local automorphic representation $\pi_{f,p}$ is a (ramified) principal series, then $e_f(a/p^2)$ is equal to 1 for approximately half of the $a$ in $(\mathbb{Z}/p^2\mathbb{Z})^\times$, while it is $>1$ for the other half. This follows from the vanishing of certain character sums modulo $p^2$ proved by Paul Nelson (unpublished). So we get further examples by looking at newforms of conductor divisible by a high power of a prime. The examples of level 567 and 891 in my comment are of this kind.

There are algorithms to compute the Fourier expansion of a modular form at arbitrary cusps (see this MO question), but they are either numerical or may become slow when the conductor gets large.

For theoretical results, you can look at the article of Corbett and Saha, On the order of vanishing of newforms at cusps.

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  • $\begingroup$ Thanks. After I read the condition of the Theorem 1.3, I have one more question. Is there any result when $N$ is a square-free integer and $f$ is not a newform? $\endgroup$
    – ililiil
    Aug 6, 2019 at 12:30
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    $\begingroup$ @ililiil Yes: in the case $N$ is squarefree, all the cusps are rational, so the orders of vanishing coincide. $\endgroup$
    – François Brunault
    Aug 6, 2019 at 13:35
  • $\begingroup$ Nice...! Is it true that when $N=4M$ with $M$ is an odd-square free? $\endgroup$
    – ililiil
    Aug 7, 2019 at 1:36
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    $\begingroup$ @ililiil The denominator of a cusp $a/b$ with $(a,b,N)=1$ is defined as $(b,N)$. As an exercise, you can show that the cusps of given denominator $d$ are Galois conjugate, and their field of definition is $\mathbb{Q}(\zeta_{(d,N/d)})$ (see Prop 2.2.3 of Cremona's book homepages.warwick.ac.uk/~masgaj/book/fulltext for the cusps, and Stevens for the Galois action). So in your case $N=4M$, all the cusps are again rational. $\endgroup$
    – François Brunault
    Aug 7, 2019 at 8:03
  • $\begingroup$ Thanks for kindly reply...!! $\endgroup$
    – ililiil
    Aug 7, 2019 at 9:00
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Just to confirm Francois's answer: in a few seconds Pari/GP gives the following:

? mf=mfinit([625,2],0); /* initialize new space */

? mffields(mf)[1]

% = y^2-y-1 /* field $\mathbb Q(\sqrt{5})$ */

? F=mfeigenbasis(mf)[1]; /* corresponding eigenform */

? mfcuspval(mf,F,1/25)

% = [1,2] /* Valuation 1 for one embedding, 2 for the other */

? mfcuspval(mf,F,2/25)

% = [2,1] /* Valuation 2 for one embedding, 1 for the other */

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  • $\begingroup$ Thanks. There are actually 4 newforms with coefficients in $\mathbb{Q}(\sqrt{5})$, and if I have the same version of your modular forms package, then the newform you selected is the quadratic twist of the $f$ from my answer. In any case, the quadratic twist also works. Interestingly the same phenomenon appears for the two other newforms, although their local components are principal series, not supercuspidals. It deserves a more systematic study... $\endgroup$
    – François Brunault
    Aug 6, 2019 at 10:57

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