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I'm reading Diamond and Shurman's book about modular forms, in particular I'm reading the proof of the dimension formula of the vector spaces $\mathcal{M}_{k}(\Gamma)$ and $\mathcal{S}_{k}(\Gamma)$ where $k$ is odd and $\Gamma$ is a congruence subgroup. The problem is that it seems to me that this formula depends on the number of regular cusps and irregular cusps of a given automorphic form and not just on the vector spaces $\mathcal{M}_{k}(\Gamma)$ and $\mathcal{S}_{k}(\Gamma)$. In other words, choosing a different automorphic form would imply a different dimension formula. How should I understand this part?

I will try to summarize what I understand from the proof. Assume that $-I\notin{\Gamma}$ (otherwise we would have $\mathcal{A}_{k}(\Gamma)=\{0\}$ and consequently also $\mathcal{M}_{k}(\Gamma)$ and $\mathcal{S}_{k}(\Gamma)$ are trivial). Roughly speaking, the steps of the proof are the following:

1) we should use the isomorphism of vector spaces $\omega:\mathcal{A}_{n}(\Gamma)\rightarrow{\Omega^{\otimes{n/2}}(X(\Gamma))}$ (where we assume that $n$ is even) in order to consider the images $\omega(\mathcal{M}_{2k}(\Gamma))$ and $\omega(\mathcal{S}_{2k}(\Gamma))$.

2) We choose ANY nonzero element $f\in{\mathcal{A}_{k}(\Gamma)}$ and take $f^{2}$ in the previous isomorphism with $n=2k$ and its image $\omega'=\omega(f^{2})$. Denote the regular cusps by $x_{i}$, the irregular cusps by $x_{i}'$ and the elliptic points of period 3 by $x_{3,i}$, and denote by $\epsilon_{\infty}^{\mathrm{reg}},\epsilon_{\infty}^{\mathrm{irr}}$ and $\epsilon_{3,i}$ the number of regular cusps, irregular cusps and elliptic points of period 3, respectively. Recall that the definition of regular and irregular cusp depends on $f$. Namely, if $s\in{\mathbb{Q}\cup\{\infty\}}$ and $\pi:\mathcal{H}^{*}\rightarrow{X(\Gamma)}$ sending $x$ to $\Gamma{x}$, then the order $v_{\pi(s)}(f)$ is defined as either $v_{s}(f)/2$ or $v_{s}(f)$ depending on extra conditions, and when this value is not integral then it is an irregular cusp, and when it is integral then we call it regular cusp.

3) Define the divisor $\mathrm{div}(d\tau)=-\sum_{i}\frac{2}{3}x_{3,i}-\sum_{i}x_{i}-\sum_{i}x_{i}'$, then prove that $\mathrm{div}(\omega')=2\mathrm{div}(f)+k\mathrm{div}(d\tau)$ and then prove that this implies $\mathrm{div}(f)=\frac{1}{2}\mathrm{div}(\omega')+\sum_{i}\frac{k}{3}x_{3,i}+\sum_{i}\frac{k}{2}x_{i}+\sum_{i}\frac{k}{2}x_{i}'$ (*). Also, recall that the degree of $\frac{1}{2}\mathrm{div}(\omega')$ is $k(g-1)$ where $g$ is the genus of $X(\Gamma)$.

4) Since the coefficients of $\mathrm{div}(f)$ are not necessarily integral (by definition, an irregular cusp will arise when the order of $f$ at such cusp is not integral), we will consider the divisor $\lfloor{\mathrm{div}(f)}\rfloor$ obtained by applying the greatest integer function to each coefficient, and then by noticing that $\mathcal{M_{k}(\Gamma)}\cong{L(\lfloor{\mathrm{div}(f)}\rfloor)}$ and $\mathcal{S}_{k}(\Gamma)\cong{L(\lfloor{\mathrm{div}(f)-\sum_{i}x_{i}-\sum_{i}\frac{1}{2}x_{i}'}\rfloor)}$ and that the degree of the divisors $\lfloor{\mathrm{div}(f)}\rfloor$ and $\lfloor{\mathrm{div}(f)-\sum_{i}x_{i}-\sum_{i}\frac{1}{2}x_{i}'}\rfloor$ is greater than $2g-2$, then the simpler version of Riemann-Roch, namely $l(D)=\mathrm{deg}(D)-g+1$ combined with (*) will yield the dimension formulas $\mathrm{dim}(\mathcal{M}_{k}(\Gamma))=(k-1)(g-1)+\sum_{i}\lfloor{\frac{k}{3}}\rfloor\epsilon_{3,i}+\sum_{i}\frac{k}{2}\epsilon_{\infty}^{\mathrm{reg}}+\sum_{i}\frac{k-1}{2}\epsilon_{\infty}^{\mathrm{irr}}$ and $\mathrm{dim}(\mathcal{S}_{k}(\Gamma))=(k-1)(g-1)+\sum_{i}\lfloor{\frac{k}{3}}\rfloor\epsilon_{3,i}+\sum_{i}\frac{k-2}{2}\epsilon_{\infty}^{\mathrm{reg}}+\sum_{i}\frac{k-1}{2}\epsilon_{\infty}^{\mathrm{irr}}$

As you can see, the previous steps imply that the dimension formula depends on the chosen $f$. How do I fix this?

Any help is appreciated.

Edit: Please be kind and tell me what part I'm not understanding correctly regarding the definition of regular and irregular cusps. The following can be found in page 75 of the book A First Course in Modular Forms by Diamond and Shurman:

First we take any automorphic form $f\in{\mathcal{A}_{k}(\Gamma)}$ and take $s\in{\mathbb{Q}\cup\{\infty\}}$ together with any matrix $\alpha\in{SL_{2}(\mathbb{Z})}$ such that $\alpha(s)=\infty$. The order of $f$ at $s$ is defined as $v_{s}(f):=v_{\infty}(f[\alpha]_{k})$, i.e. the order of $f[\alpha]_{k}$ at infinity, and it can be shown that it does not depend on the chosen $\alpha$.

Next, the width $h$ of the cusp $\pi(\infty)$ is characterized by the condition $\{\pm{I}\}\Gamma_{\infty}=\{\pm{I}\}<\begin{pmatrix} 1 & h \\ 0 & 1 \\ \end{pmatrix}>$ where $h$ a positive integer. The definition of the order of $f$ at $\pi(\infty)$ is then defined as $v_{\pi(\infty)}(f):=v_{\infty}(f)/2$ when $k$ is odd and $\Gamma_{\infty}=<-\begin{pmatrix} 1 & h \\ 0 & 1 \\ \end{pmatrix}>$, otherwise it is defined as $v_{\pi(\infty)}(f):=v_{\infty}(f)$. Using this, we define $v_{\pi(s)}(f):=v_{s}(f)/2$ if $(\alpha\Gamma\alpha^{-1})_{\infty}=<-\begin{pmatrix} 1 & h \\ 0 & 1 \\ \end{pmatrix}>$ and $k$ is odd, otherwise it's defined as $v_{\pi(s)}(f):=v_{s}(f)$.

Finally, referring to the value $v_{\pi(s)}(f)$ (formula 3.3 in page 75) the book says "This can be half-integral in the exceptional case, when $\pi(s)$ or $s$ itself is called an irregular cusp of $\Gamma$". This notion is repeated at the beginning of section 3.6 'Dimension Formulas for Odd $k$' (page 89), where the book says "Since $k$ is odd and $-I\notin{\Gamma}$, the modular curve $X(\Gamma)$ has two possible types of cusps as discussed in section 3.2, the regular cusps where $v_{\pi(s)}(f)$ is integral and the irregular cusps where $v_{\pi(s)}(f)$ is half-integral".

So it seems to me that the notion of regular and irregular cusp makes use of the chosen automorphic form $f$.

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    $\begingroup$ There is evidence of considerable confusion in your question. The definition of "regular/irregular cusps" has nothing to do with a choice of a specific cusp form $f$. $\endgroup$ Mar 24, 2016 at 8:01
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    $\begingroup$ @ David Loeffler Could you please tell me your definition of regular and irregular cusp? I edited my post (see the end of my post) in order to explain why it seems to me that this notion makes use of the chosen automorphic form $f$. I'm just following pages 75 and 89 from the aforementioned book, so I will appreciate it if you explain me what is wrong in my understanding. $\endgroup$ Mar 24, 2016 at 17:30
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    $\begingroup$ A cusp is regular if its stabiliser is generated by an element conjugate to $\begin{pmatrix} 1 & h \\ 0 & 1 \end{pmatrix}$ for some $h$, and irregular if the generator is conjugate to $-\begin{pmatrix} 1 & h \\ 0 & 1 \end{pmatrix}$. $\endgroup$ Mar 25, 2016 at 8:43
  • $\begingroup$ PS: In the passage you quote from Diamond + Shurman, I suspect "half-integral" just means "a rational number of the form $k/2$ with $k$ an integer" -- rather than "a rational number of the form $k / 2$ with $k$ an odd integer", which is also a common interpretation of "half-integral". (I don't have my copy of Diamond + Shurman with me to check.) $\endgroup$ Mar 25, 2016 at 11:34
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    $\begingroup$ I realise this thread is a few years old, but I've recently encountered a similar confusion and wanted to add something I had to figure out for myself. In the case where $k$ is odd and we are at an irregular cusp, the order of any nonzero $f\in{\mathcal{A}_{k}(\Gamma)}$ is $m/2$ where $m$ is necessarily odd. Since $f(\tau + h) = -f(\tau)$, the coefficients of the even terms in the Fourier expansion vanish (even terms are fixed by the translation $\tau + h$ and odd terms are negated by it). $\endgroup$
    – Sam Rice
    Jan 17, 2021 at 13:20

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