Fourier Expansions at different cusps
Consider a modular form $\renewcommand{\Im}{\operatorname{Im}} \renewcommand{\Re}{\operatorname{Re}} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator*{\Res}{Res} \newenvironment{psmallmatrix} {\left(\begin{smallmatrix}} {\end{smallmatrix}\right)} f$ on a Fuchsian group $\Gamma \subset \SL(2, \mathbb{R})$ with finite covolume, $\mu(\Gamma \backslash \mathcal{H}) < \infty$. The cusps of $\mathcal{H}$ with respect to $\Gamma$ are points $s$ of $\mathbb{P}^1(\mathbb{R}) = \mathbb{R} \cup \{ \infty \}$ such that there is at least one parabolic $\gamma \neq 1$ in $\Gamma$ that fixes $s$. It is common to refer to the orbit $\Gamma s$ as "the cusp" $s$, and (confusingly) it is also common to refer to any cusp $t$ in the orbit $\Gamma s$ as a representative of the cusp. Similarly, it is common to refer to "the Fourier expansion" of $f$ at $s$.
This conflation leads to confusion.
The full story when $\Gamma$ is a congruence subgroup in $\SL(2, \mathbb{Z})$ is covered in newform theory (cf. Atkin and Lehner's work). By choosing nice cusp representatives and nice scaling matrices, expansions for $\Gamma_0(N)$ for squarefree $N$ are well-behaved.
The problem is that choosing random cusp representatives and random scaling matrices give slightly different expressions for Fourier expansions at cusps. In this note, I describe what goes wrong.
I write this down because I forget the details all the time.
Choices of Scaling matrices
Suppose $f$ is a weight $k$ modular form on $\Gamma$. For $\gamma = \begin{psmallmatrix} a & b \\ c & d \end{psmallmatrix} \in \GL(2, \mathbb{R})^+$, we define the weight $k$ slash operator \begin{equation} (f | \gamma) := (\det \gamma)^{k/2} (cz + d)^{-k} f(\gamma z). \end{equation} This is a right group action, and one can directly verify the computation $(f | \gamma_1) | \gamma_2 = f | (\gamma_1 \gamma_2)$.
Fix a cusp $s \in \mathbb{P}^1(\mathbb{R})$ and $\sigma \in \SL(2, \mathbb{R})$ with $\sigma \infty = s$. Then the modularity of $f$ at the (orbit of cusps equivalent under $\Gamma$ to the) cusp $s$ is determined by the behavior of $f | \sigma$. Such a $\sigma$ is called a scaling matrix for the cusp $s$.
The literature is divided on whether to call $\sigma$ or $\sigma^{-1}$ the scaling matrix. As a result, any time I refer to a scaling matrix I do the same quick look to make sure that neighborhoods behave as expected. The point is that \begin{equation} (f | \sigma)(z) = (cz + d)^{-k} f(\sigma z) \end{equation} maps a neighborhood of $z$ near $\infty$ to a neighborhood of $\sigma z$ near $s$. This is what we want, as the behavior of $f | \sigma$ near $\infty$ can be described naturally with a Fourier series, and this series describes the behavior of $f$ near $s$. (I really do this verification every time a scaling matrix appears).
The point is that $f | \sigma$ is modular with respect to the group $\sigma^{-1} \Gamma \sigma$, and $\sigma^{-1} \Gamma \sigma$ contains an element acting like a translation $z \mapsto z + h$ for some $h$. Thus $f | \sigma$ has a Fourier expansion \begin{equation} (f | \sigma)(z) = \sum a_\sigma(n) e^{2 \pi i n z / h}. \end{equation} The modularity of $f | \sigma$ is quickly checked (and for similar reasons as above, it's very easy to put the inverse in the wrong spot): \begin{equation} (f | \sigma) | (\sigma^{-1} \gamma \sigma) = f | (\sigma \sigma^{-1}) \gamma \sigma = (f | \gamma) | \sigma = f | \sigma. \end{equation} The fact that $\sigma^{-1} \Gamma \sigma$ contains an element acting like a translation has various proofs depending on the level of generality we consider.
When $\Gamma$ is a congruence subgroup, one can show that $s$ is rational and we can take $\sigma \in \SL(2, \mathbb{Z})$. The stabilizer of $\infty$ in $\SL(2, \mathbb{Z})$ is generated by $\left\langle \begin{psmallmatrix} 1 & 1 \\ 0 & 1 \end{psmallmatrix} \right\rangle$. Morally, writing out the conditions for the conjugate of $\begin{psmallmatrix} 1 & 1 \\ 0 & 1 \end{psmallmatrix}^n$ by $\sigma^{-1}$ to be in $\Gamma$ clearly gives a set of congruence conditions, and hence the stabilizer of $\infty$ in $\sigma^{-1} \Gamma \sigma$ is congruence. It is both possible to make this more explicit and to actually use this idea as the basis of an algorithm to compute the congruence subgroup. This is described in Stein's book on computational modular forms, for example. More generally, showing that $\sigma^{-1} \Gamma \sigma$ is congruence when $\Gamma$ is congruence will appear in any book on holomorphic modular forms.
For a general Fuchsian group $\Gamma$ acting discontinuously on $\mathcal{H}$ with finite covolume, abstract complex dynamics shows that the stabilizer of any point is a cyclic group (or annoyingly $\pm 1$ times a cyclic group, depending on whether $-1 \in \Gamma$ or not). (It is more correct to say that the image of the stabilizer in $\mathrm{PSL}(2, \mathbb{R})$ is cyclic). Further, the group will consist exclusively of exactly one type of element: either elliptic, parabolic, hyperbolic. We call a point $x$ elliptic, parabolic, or hyperbolic (with respect to $\Gamma$) if the stabilizer in $\Gamma$ of $x$ is nontrivial and made up of the corresponding type of matrices. Elliptic points have finite cyclic groups; hyperbolic points and parabolic points have infinite cyclic groups. From this point of view, cusps are precisely the parabolic points in $\mathbb{P}^1(\mathbb{R})$, and the generator of the stabilizer gives the necessary translation.
I don't prove this result in that level of generality here. Instead, we'll just prove the parabolic statement.
Let $s \in \mathbb{P}^1(\mathbb{R})$ be a cusp in $\Gamma \backslash \mathcal{H}$. Fix $\sigma \in \SL(2, \mathbb{R})$ with $\sigma \infty = s$. Let $\Gamma_s$ denote the stabilizer of $s$ in $\Gamma$. Then $\Gamma_s$ consists only of parabolic matrices and \begin{equation} \{ \pm 1 \} \sigma^{-1} \Gamma_s \sigma = \left\{ \pm \begin{pmatrix} 1 & h \\ 0 & 1 \end{pmatrix}^n : n \in \mathbb{Z} \right\} \end{equation} for some $h > 0$.
WLOG, we can assume $s = \infty$ by taking $\sigma^{-1} \Gamma \sigma$ in place of $\Gamma$. As traces are preserved by conjugation, this preserves parabolic elements. Any element of the stabilizer of $\infty$ in $\SL(2, \mathbb{R})$ must take the form \begin{equation} \pm \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \end{equation} for some $a, b \in \mathbb{R}$. As $s$ is a cusp, at least one of these elements is parabolic (i.e. has trace with absolute value $2$). This necessarily implies that $\lvert a\rvert = 1$ and that $b \neq 0$. Without loss of generality (as we multiply everything by $\pm 1$ throughout), we write this as \begin{equation} \pm T := \pm \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \in \pm \Gamma_\infty. \end{equation} Thus at least one parabolic element is in the stabilizer.
Now suppose there is another matrix $M \in \Gamma_\infty$ that isn't parabolic. We can write this element as \begin{equation} \begin{pmatrix} A & B \\ 0 & A^{-1} \end{pmatrix} \end{equation} with $A \neq \pm 1$. After possibly considering $M^{-1}$ instead, we can assume that $\lvert A \rvert < 1$. Then a short computation shows \begin{equation} (M T M^{-1})^n = M T^n M^{-1} = \begin{pmatrix} 1 & A^{2n} b \\ 0 & 1 \end{pmatrix}. \end{equation} As $M$ and $T$ are in $\Gamma_\infty$, $(M T^n M^{-1}) \in \Gamma_\infty \subset \Gamma$ for all $n$. But taking a sequence as $n \to \infty$ shows that the identity matrix is an accumulation point in $\Gamma$, contradicting that $\Gamma$ is discrete. Hence all matrices in $\Gamma_\infty$ are parabolic.
This shows that \begin{equation} \pm \Gamma_\infty \subset \left\{ \pm \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} : t \in \mathbb{R} \right\} \cong \{\pm 1\} \times \mathbb{R}, \end{equation} where the last relation is an isomorphism of topological groups. As $\Gamma$ is discrete, so is $\Gamma_\infty$. The only nontrivial discrete subgroups of $\mathbb{R}$ are isomorphic to $\mathbb{Z}$.\footnote{Idea of proof: if there are $\alpha, \beta$ in the subgroup with $\alpha \neq \frac{p}{q} \beta$ for a rational $\frac{p}{q}$, then linear combinations of $\alpha$ and $\beta$ are dense in $\mathbb{R}$ and hence not discrete. Similarly, rational denominators for relations between elements must be bounded above, as otherwise there is an accumulation point.}
Calling the generator $\begin{psmallmatrix} 1 & h \\ 0 & 1 \end{psmallmatrix}$ completes the proof. $\diamondsuit$
We've now shown that $f | \sigma$ has a Fourier expansion because $f | \sigma$ is modular with respect to $\sigma^{-1} \Gamma \sigma$, which contains a translation $z \mapsto z + h$. We call the minimal such $h$ (the generator guaranteed by the proposition) the width of the cusp $s$.
But there are multiple choices of scaling matrix $\sigma$. There are infinitely many.
How do different choices of scaling matrices for a cusp $s$ affect the resulting Fourier expansions "at $s$"?
First, we consider the various scaling matrices. We continue to have a fixed cusp $s \in \mathbb{P}^1(\mathbb{R})$.
Suppose $\sigma_1$ and $\sigma_2$ are two matrices in $\SL(2, \mathbb{R})$ with $\sigma_1 \infty = \sigma_2 \infty = s$. Then $\sigma_1^{-1} \sigma_2 \infty = \infty$, and so $\sigma_1^{-1} \sigma_2$ is in the stabilizer of $\infty$ in $\SL(2, \mathbb{R})$, which we can denote by $\SL(2, \mathbb{R})_\infty$. In particular, if $\sigma_1$ is any matrix in $\SL(2, \mathbb{R})$ with $\sigma_1 \infty = s$, then any matrix in the coset $\sigma_1 \SL(2, \mathbb{R})_\infty$ will also send $\infty$ to $s$.
This is very broad. We can see that the "width" $h$ of the cusp $s$ is not an inherent property of the cusp. For perhaps the simplest possible example, take $s = \infty$, but don't take $\sigma = 1$. Instead, take $\sigma = \begin{psmallmatrix} n & 1 \\ 0 & 1/n \end{psmallmatrix}$. We can compute that \begin{equation} \sigma \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \sigma^{-1} = \begin{pmatrix} 1 & n^2 \\ 0 & 1 \end{pmatrix}. \end{equation} This is made more confusing by inconsistent definitions of scaling matrices in the literature. Some mathematicians define the scaling matrix to also make the corresponding width of the cusp $1$, always. (For example, Iwaniec's Topics in Classical Automorphic Forms does this and is one of my favorite references even though this plays poorly with arithmetic properties).
Thus choosing two different scaling matrices $\sigma_1$ and $\sigma_2$ could lead to two Fourier expansions at the same cusp $s \in \mathbb{P}^1(\mathbb{R})$ that look like \begin{align} f_{\sigma_1}(z) &= \sum a_{\sigma_1}(n) e^{2 \pi i n / h_1} \\ f_{\sigma_2}(z) &= \sum a_{\sigma_2}(n) e^{2 \pi i n / h_2} \end{align} with different widths $h_j$, and hence are very different.
We can say more. As $\sigma_1^{-1} \sigma_2 \in \SL(2, \mathbb{R})_\infty$, we can write \begin{equation} \sigma_1^{-1} \sigma_2 = \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \implies \sigma_1^{-1} = \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \sigma_2^{-1}. \end{equation} Proposition 2 guarantees that there is $\gamma \in \Gamma_s$ such that \begin{equation} \sigma_2^{-1} \gamma \sigma_2 \in \left\{ \begin{pmatrix} 1 & h_2 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & h_2 \\ 0 & -1 \end{pmatrix} \right\}. \end{equation} I will assume with small loss of generality that $\sigma_2^{-1} \gamma \sigma_2 = \begin{psmallmatrix} 1&h_2 \\ 0 & 1 \end{psmallmatrix}$: the corresponding work with $-1$s essentially has each $1$ replaced by a $-1$. This is one definition of a regular or irregular cusp, but we don't use this here.
Then we compute \begin{align} \sigma_1^{-1} \gamma \sigma_1 &= \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \sigma_2^{-1} \gamma \sigma_2 \begin{pmatrix} a^{-1} & b \\ 0 & a \end{pmatrix} = \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} \begin{pmatrix} 1 & h_2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a^{-1} & b \\ 0 & a \end{pmatrix} \\ &= \begin{pmatrix} 1 & a^2 h \\ 0 & 1 \end{pmatrix}. \end{align} We note also that this is easily reversible and shows that the widths of the two different Fourier expansions corresponding to $\sigma_1$ and $\sigma_2$ differ by a factor of $a^2$.
The analogous description when $\Gamma$ is a congruence subgroup is much easier here, as the $\sigma_j$ can be taken from $\SL(2, \mathbb{Z})$. Then $\sigma_1^{-1} \sigma_2$ must be in $\SL(2, \mathbb{Z})_\infty$, which is exactly $\left\langle\begin{psmallmatrix} 1&1\\0&1 \end{psmallmatrix}\right\rangle$. Thus there is a one-parameter, discrete, family of choices of scattering matrices.
Let's now examine the effect on the Fourier expansions. Fix $\sigma_1$ with $\sigma_1 \infty = s \in \mathbb{P}^1(\mathbb{R})$ and let \begin{equation} \sigma_2 = \sigma_1 \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} =: \sigma_1 \alpha, \end{equation} which also has $\sigma_2 \infty = s$. On the one hand, \begin{equation} (f | \sigma_1)(z) = \sum a_{\sigma_1}(n) e^{2 \pi i n z / h_1}. \end{equation} We now compute \begin{equation} (f | \sigma_2)(z) = \big((f | \sigma_1) | \alpha \big) (z) = a^{-k} (f | \sigma_1) (z + b) = a^{-k} \sum a_{\sigma_1}(n) e^{2 \pi i n b / h_1} e^{2 \pi i n z / h_1}. \end{equation} On the other hand, \begin{equation} (f | \sigma_2)(z) = \sum a_{\sigma_2}(n) e^{2 \pi i n z / h_2}, \end{equation} (where $h_2 = a^2 h_1$). When $h_2 \neq h_1$, the coefficients can't be directly compared. But the constant coefficients must agree (up to a factor of $a^k$) and whether the series are meromorphic or holomorphic must agree (i.e. whether there are any coefficients with $n < 0$). I note this last part because none of the discussion so far has used that $f$ is a modular form, and could actually have been a meromorphic function on $\Gamma \backslash \mathcal{H}$ — and this shows that the ambiguity in Fourier expansions doesn't affect the holomorphic properties of $f$.
If we restrict to expansions with the same width $h$, we can say more. This is equivalent to choosing $a = \pm 1$ above, or equivalently to choosing $\sigma_2 = \sigma_1 \begin{psmallmatrix} 1 & 1 \\ 0 & 1 \end{psmallmatrix}^n$ for some $n$. As alluded to in Remark 4, this is always the case when $\Gamma$ is a congruence subgroup. In this case, we have different Fourier expansions over the same basis, and we see that the $n$th coefficients differ by the $n$th power of the root of unity $e^{2 \pi i b / h}$.
This is annoying in the sense that the series aren't simply multiples of each other, but instead have a complicated twisted property. But on the other hand, the relationship is very predicatable and determined.
Summary
Different scaling matrices absolutely affect the Fourier expansions, sometimes in radically different ways. If scaling matrices can be chosen to be in $\SL(2, \mathbb{Z})$, then the $n$th Fourier coefficient in the Dirichlet series is defined up to multiplication by the $n$th power of an $h$th root of unity, where $h$ is the width of the cusp with respect to these matrices.
Choices of Cusp Representatives
Now suppose $s_1$ and $s_2$ are two representatives of the same cusp. Stated differently, fix a cusp $s_1$ and take $\gamma \in \Gamma$, and consider the other cusp representative $\gamma s_1 = s_2$. Fix a scaling matrix $\sigma_1 \in \SL(2, \mathbb{R})$ with $\sigma_1 \infty = s_1$, and take the corresponding scaling matrix $\sigma_2 = \gamma \sigma_1$ for $s_2$.
On the one hand, a short computation shows that \begin{equation} f | \sigma_2 = f | (\gamma \sigma_1) = (f | \gamma) | \sigma_1 = f |\sigma_1, \end{equation} so the analytic behavior in neighborhoods of $s_1$ and $s_2$ are identical. The Fourier expansions can be obtained via integration of $f | \sigma$ against appropriate weights, and hence the Fourier expansions are also identical.
This is different behavior than the case of two different scaling matrices for the same cusp. In that case, there was no reason to expect $f | \sigma_1 = f | \sigma_2$. Instead, each expansion was a reasonable candidate for the expansion at the cusp.
Less clean way of seeing the widths agree
I was initially confused at this identification and how nice it seemed. I worried about whether each of $\sigma_1$ and $\sigma_2$ had their own, different, natural widths.
This is wrong, as the fact that $f | \sigma_1 = f | \sigma_2$ makes clear. But I wanted to see it another way. This is a short exercise.
Fix a cusp $s_1 \in \mathbb{P}^1(\mathbb{R})$ of a Fuchsian group $\Gamma$ with cofinite volume. Fix a scaling matrix $\sigma_1 \in \SL(2, \mathbb{R})$ with $\sigma_1 \infty = s_1$. Fix $\gamma \in \Gamma$. Let $s_2 = \gamma s_1$ and $\sigma_2 = \gamma \sigma_1$ denote another cusp representative for the same cusp and a distinguished corresponding scaling matrix. Then the width of $s_1$ with respect to $\sigma_1$ is equal to the width of $s_2$ with respect to $\sigma_2$.
Use Proposition 2. To $s_1$ is associated a width $h$ such that \begin{equation}\label{eq:s1} \{ \pm 1 \} \sigma_1^{-1} \Gamma_{s_1} \sigma_1 = \left\{ \pm \begin{pmatrix} 1 & h \\ 0 & 1 \end{pmatrix}^n : n \in \mathbb{Z} \right\}. \end{equation} The Fourier expansion of a modular form $f$ at $s_1$ with the scaling matrix $\sigma_1$ is in powers of $e^{2 \pi i z/h}$.
We first ask: what is the width at the cusp $s_2$ with respect to the scaling matrix $\sigma_2 = \gamma \sigma_1$? Working directly from \eqref{eq:s1}, we find \begin{equation} \{ \pm 1 \} \sigma_1^{-1} \Gamma_{s_1} \sigma_1 = \{ \pm 1 \} \sigma_2^{-1} \big(\gamma \Gamma_{s_1} \gamma^{-1}\big) \sigma_2 = \left\{ \pm \begin{pmatrix} 1 & h \\ 0 & 1 \end{pmatrix}^n : n \in \mathbb{Z} \right\}. \end{equation} As $\Gamma_{s_2} = \gamma \Gamma_{s_1} \gamma^{-1}$, this shows that the width $h$ of $s_2$ with respect to $\sigma_2$ is the same as the width of $s_1$ with respect to $\sigma_1$.
Summary
Choosing different cusp representatives (with distinguished scaling matrices that are "compatibly chosen") yields the exact same Fourier expansions.
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