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Something is wrong here. I have an eta-quotient

$$g(z) := \eta^{2}(z)\eta(2z)\eta(4z)\eta^{2}(8z),$$

which belongs to $S_{3}(8, \chi)$ according to page 3 of http://sites.science.oregonstate.edu/~math_reu/proceedings/REU_Proceedings/Proceedings2018/NickAsiminaBen.pdf (this is also found in many other texts on the same subject) (the character can be found explicitly there too).

However, page 4 (again a standard result on eta-quotients) tells me that the order of $g$ at the cusps $1/2, 1/4$ of $\Gamma_{0}(8)$ are both $1/2$. What is going on here?

Freddie
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1 Answers1

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$\Gamma_0(8),\chi$ doesn't lead to a Riemann surface. You should look at the cusps of $\Gamma_1(8)\setminus \Bbb{H}^*$ and use the $\Gamma_0(8),\chi$-invariance to relate certain cusps together. Once confortable with it you can refer to rational order of vanishing at the $\Gamma_0(8)$ cusps. Concretely if $\chi$ is of order $o$ then $f^o\in M_{ko}(\Gamma_0(8))$ and the order of vanishing of $f$ at $\Gamma_0(8) .\frac{a}b$ is $1/o$ the order of vanishing of $f^o$.

For $g\in M_K(\Gamma)$ and $\gamma\in SL_2(\Bbb{Z})$ then $g|_k \gamma$ is (weight $K$) $\gamma^{-1}\Gamma\gamma$ invariant. Let $m$ be the least positive integer such that $\pmatrix{1&m\\0&1}\in \gamma^{-1}\Gamma\gamma$.

Then $\phi(s)= \gamma(m\frac{\log s}{2i\pi})$ is a chart from $|s|<1/m$ to the modular curve $\Gamma\setminus\Bbb{H}^*$ such that $\phi(0)=\gamma(i\infty)$

And the order of vanishing of $g$ at the cusp $\gamma(i\infty)$ is the $r$ such that $$g|_K \gamma(\phi(s)) \sim C s^r \text{ as } s\to 0$$

reuns
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  • Okay, so though $\Gamma_{0}(8)\setminus \mathbb{H}^{*}$ is a compact Riemann surface, the $g$ above doesn't necessarily (due to the looser automorphy factor condition with $\chi$) correspond to a weight-$3$ form with this surface? – Freddie Feb 21 '20 at 14:33
  • Sure. The point is that $m$ won't be the same for $\gamma^{-1}\Gamma_0(8)\gamma$ and $\gamma^{-1}\ker(\chi)\gamma$. – reuns Feb 21 '20 at 15:23