In Exercise 1.3.3 from Automorphic Forms and Representations, Daniel Bump gives the hint as follows:
Observe that $\gamma(\rho)=\rho$ where $\gamma= \begin{pmatrix} 1&1\\
-1& \\ \end{pmatrix}$, and apply $$f\left(\frac{az+b}{cz+d}\right)=(cz+d)^kf(z)\ \ \text{for}\ \
\begin{pmatrix} a&b\\ c&d\\
\end{pmatrix}\in\mathrm{SL}(2,\mathbb{Z}).$$
However, I feel confused about the hint and wonder how to prove it.
- For $\rho=e^{\frac{2\pi i}{3}}$ a complex number and $\gamma$ a
matrix, how to understand the equation $\gamma(\rho)=\rho$ ? - Having applied the formula, I get
$$f\left(\frac{z+1}{-z}\right)=(-z)^kf(z).$$ For
$z=\rho=e^{\frac{2\pi i}{3}}$, it follows that
$$f\left(-1-e^{-\frac{2\pi i}{3}}\right)=(-1)^ke^{\frac{2k\pi
i}{3}}f\left(e^{\frac{2\pi i}{3}}\right).$$ How to demonstrate that
$f(\rho)=0$?
Best Answer
Simple proof:
By definition, we have
$$ f(\rho)=f(\gamma(\rho))=(-\rho)^kf(\rho). $$
Because $k$ is automatically even, we have
$$ (1-\rho^k)f(\rho)=0 $$
Since $\rho^k\ne1$ if and only if $3\nmid k$, we conclude that $f(\rho)=0$ in this case.
My original proof:
Let $\nu_p(f)$ denote the unique integer such that $f(z)(z-p)^{-\nu_p(f)}$ is analytic and nonzero at $z=p$, so when $f$ is a modular form of weight $k$ on $SL(2,\mathbb Z)$, there is
$$ \frac12\nu_i(f)+\frac13\nu_\rho(f)+\nu_\infty(f)+\sum_{\substack{p\in\mathcal F\\p\ne i,\rho}}\nu_p(f)={k\over12}. $$
Consequently, multiplying both sides by $12$ gives ($k$ is assumed to be even)
$$ 4\nu_\rho(f)\equiv k-6\nu_i(f)\pmod{12} $$
Suppose $v_\rho(f)=0$. Then the right hand side will be $\equiv0\pmod3$, so
$$ k\equiv6\nu_i(f)\equiv0\pmod3. $$
Therefore, we proved the contrapositive: If $f(\rho)\ne0$, then $3|k$.