# Vanishing order of a modular form at an irregular cusp: why is it treated like a valuation?

I am studying Diamond and Shurman’s book on modular forms and I do not understand the computation of dimension formulas for odd weights (page 90).

To make the setting more precise, let $$\Gamma$$ be a congruence subgroup with $$-I_2 \notin \Gamma$$, $$f$$ a nonzero automorphic form for $$\Gamma$$ of odd weight $$k$$.

To compute the dimension of $$\mathcal{M}_k(\Gamma)$$, Diamond and Shurman point out that a modular form $$g$$ of weight $$k$$ for $$\Gamma$$ can be written as $$f_0f$$ for some meromorphic function $$f_0$$ on $$X(\Gamma)$$.

That $$g$$ must be holomorphic on $$\mathbb{H}$$ and at cusps can then be rewritten as a condition on $$f_0$$, more precisely that $$f_0 =0$$ or that $$\mathrm{div}(f_0) +D \geq 0$$ for some divisor $$D$$ on $$X(\Gamma)$$.

So far, so good, that approach worked perfectly well for the spaces of modular forms of even weight.

My problem occurs at the irregular cusps.

The book reads basically as follows (I can’t see where the reasoning is spelled out for modular forms and not cusp forms, but it doesn’t really matter): suppose $$s \in X(\Gamma)$$ is an irregular cusp. Then $$g$$ vanishes at $$s$$ iff $$\nu_s(g) \geq 1/2$$ iff $$\nu_s(f_0f) \geq 1/2$$ iff $$\nu_s(f_0)+\nu_s(f)-1/2 \geq 0$$ (*).

Now I’ll explain what my problem is. Assume that $$s$$ is the image of infinity (to make matters simpler – one just needs to conjugate $$\Gamma$$). Let $$h>0$$ be the width of $$s$$, so that $$-\begin{bmatrix}1&h\\0&1\end{bmatrix} \in \Gamma$$ (so that $$g(z+h)=-g(z)$$ and $$2h$$ is the period of $$g$$).

Then by the Diamond-Shurman notation (defined pages 72 and 75), for an automorphic form $$\psi$$ of weight $$l$$ for $$\Gamma$$, $$\nu_s(\psi)$$ is $$m/2$$ if $$l$$ is odd or $$m$$ is $$l$$ is even, where $$|\psi(z)| \sim ce^{-2\pi m\mathrm{Im}(z)/(2h)}$$ (with $$c >0$$) as $$\mathrm{Im}(z) \rightarrow \infty$$.

In particular, since $$m(\psi_1\psi_2)=m(\psi_1)+m(\psi_2)$$, it means that, because $$k$$ is odd, $$\nu_s(f_0f)=\frac{\nu_s(f_0)}{2}+\nu_s(f)$$. But this contradicts “directly” the last equivalence of (*)!

(And, of course, it means that $$\nu_s$$ is not a valuation on the graded ring of automorphic forms, which is also a little disturbing given the notation).

So could you please tell me what I understand incorrectly?

Thank you.

Perhaps unsurprisingly, the point that was problematic is the definition of $$\nu_s$$ at an irregular cusp $$s$$. Indeed, if that cusp is irregular with width $$h$$, the coordinate around it is (if $$s=\infty$$, to simplify) $$e^{2i\pi z/h}$$. But the smallest $$h’$$ such that $$\begin{bmatrix}1 & h’\\0&1\end{bmatrix}$$ is in the congruence subgroup is always $$h’=2h$$.
In particular, at this cusp, the leading term of $$\psi(z)$$ at the cusp is $$(e^{2i\pi z/(2h)})^m$$, and thus is the coordinate to the $$m/2$$-th power, whether the level is odd or even.
If the definition of $$\nu_s$$ is chosen to be “$$m/2$$ whatever the level” (where $$m$$ is the same as in the question) rather than “$$m/2$$ at odd level and $$m$$ at even level” then we get a valuation back and the later manipulations work without any problem.