# The order of a modular form is invariant under the action of $SL_2(\mathbb Z)$

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:

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

(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).

• 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 $p$. – Parthiv Basu Aug 9 '19 at 18:16
• @ParthivBasu Thanks! And is it easy to come up with a modular function $f$ with zero of order $1$ at $p$? – No One Aug 9 '19 at 18:24
• 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. – Parthiv Basu Aug 9 '19 at 18:44
• @ParthivBasu Thanks! – No One Aug 9 '19 at 18:45
• 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 $E_4(\rho) =0$. – 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 $$p$$.
• 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 $v_p(f\circ g)$ – No One Aug 9 '19 at 17:14
• 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. – No One Aug 9 '19 at 17:21