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I am reading Serre's a course in arithmetic and I am very confused about the invariance of the order of a modular function (as a meromorphic function) under $SL_2(\mathbb Z)$ action:

The order at $p$ of a meromorphic function $f$ is $v_p(f):=n$, where $n$ is the integer such that $f/(z-p)^n$ is analytic and nonzero at $p$. The following is a screenshot of the book:

enter image description here

Serre says the invariance follows from the identity in the definition of modular forms, but I don't see why. Specifically,

let $g=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ be a matrix in $SL_2(\mathbb Z)$, let $v_p(f)=n$, the goal is to show that

$$\frac{f(z)}{(z-g.p)^n}=\frac{f(\frac{az+b}{cz+d})}{(cz+d)^{2k}(z-\frac{ap+b}{cp+d})^n}$$

is nonvanishing and holomorphic at $p$. But it is very unclear to me why this should be true. Maybe I am looking at it in the wrong way. Thanks for help!


Update:

From Parthiv's answer below it seems that the interpretation should really be for $g \in SL_2(\mathbb Z)$

(1) $v_p(f)=v_p(f\circ g)$

instead of

(2) $v_p(f)=v_{g(p)}(f)$

But now I wish to see a counterexample for (2) (I don't have a handy example for modular functions).

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    $\begingroup$ I thought about it. I think it's obvious why the second interpretation shouldn't hold. Say $f(z)$ has a zero of order $1$ at $p$, then we have that $h(z)$ in $f(z)=h(z)(z-p)$ is holomorphic and nonzero at $p$. So for $g$ such that $gp \neq p$, we have that $\frac{(z-p)h(z)}{(z-gp)}$ is zero at $p$. $\endgroup$ – Parthiv Basu Aug 9 '19 at 18:16
  • $\begingroup$ @ParthivBasu Thanks! And is it easy to come up with a modular function $f$ with zero of order $1$ at $p$? $\endgroup$ – No One Aug 9 '19 at 18:24
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    $\begingroup$ Finding the zeros of modular functions in general is a catastrophically hard problem. But they do exist. I don't have any simple examples, you have to confront the literature. $\endgroup$ – Parthiv Basu Aug 9 '19 at 18:44
  • $\begingroup$ @ParthivBasu Thanks! $\endgroup$ – No One Aug 9 '19 at 18:45
  • $\begingroup$ I went through my notes again. The modular form $E_4$ has a simple zero at $\rho := e^{2\pi i /3}$. In my notes this is proven using the valence formula. But there is a much simpler to see that this is a zero. The fact that $\rho^2 + \rho + 1 =0$ implies for the lattice $L_{\rho} = \rho \mathbb{Z} + \mathbb{Z}$ that $\rho L_\rho = L_\rho$. But then $E_4(\rho) = \rho^4 E_4(\rho)$, which implies $E_4(\rho) =0 $. $\endgroup$ – Parthiv Basu Aug 9 '19 at 21:55
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A modular function of weight $k$ (odd weighted modular functions are identically zero but a priori we don't know that) is meromorphic on the upper half-plane and satisfies $f(gz) = (cz+d)^k f(z)$ for all $g \in SL_2(\mathbb{Z})$. Since $cz+d$ is holomorphic and not equal to zero on the upper half-plane, we have that $(cz+d)^k f(z)$ and $f(z)$ have the same order. By $n = v_{g(p)}(f)$, Serre means $n$ such that $\frac{f(gz)}{(z-p)^n} = (cz+d)^k \frac{f(z)}{(z-p)^n}$ is holomorphic and nonzero in $p$.

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  • $\begingroup$ Are you sure the concerned formula for $n = v_{g(p)}(f)$ is $\frac{f(gz)}{(z-p)^n}$ instead of $\frac{f(z)}{(z-g(p))^n}$? I think we are supposed to replace $p$ by $g(p)$ instead of $f$ by $f(g())$. Otherwise we should use $v_p(f\circ g)$ $\endgroup$ – No One Aug 9 '19 at 17:14
  • $\begingroup$ Yes I am sure about it. This property is mentioned at the beginning of every text on modular forms. So it'd be weird if Serre has something else in mind. $\endgroup$ – Parthiv Basu Aug 9 '19 at 17:18
  • $\begingroup$ Actually Serre also says $v_p(f)$ depends only on the image of $p$ in $\mathbb H /G$ where $G=PSL_2(\mathbb Z)$ (the last sentence of the picture. I forgot to include the other half). I don't know how to justify this. $\endgroup$ – No One Aug 9 '19 at 17:21
  • $\begingroup$ Are you familiar with fundamental domains? See in particular the section on fundamental domain for the full modular group (en.m.wikipedia.org/wiki/Fundamental_domain) $\endgroup$ – Parthiv Basu Aug 9 '19 at 17:28
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    $\begingroup$ Sure, thanks for your help! $\endgroup$ – No One Aug 9 '19 at 17:47

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