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


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!


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).

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $p$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Sure, thanks for your help! $\endgroup$ – No One Aug 9 '19 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.