Square root of the automorphy factor for integral weight modular form

26 Views Asked by At

Let $j(\alpha,z):=cz+d$, where $z\in\mathbb{H}$ and $\alpha=\begin{pmatrix}a&b\\c&d \end{pmatrix}\in SL_2(\mathbb{R})$.

Then we know that $j(\alpha\beta,z)=j(\alpha,\beta z)j(\beta,z)$, $j(\alpha^{-1},z)=j(\alpha,\alpha^{-1}z)^{-1}$ for all $\alpha,\beta\in SL_2(\mathbb{R})$ and $z\in\mathbb{H}$

However, $\sqrt{j(\alpha\beta,z)}\neq\sqrt{j(\alpha,\beta z)}\sqrt{j(\beta,z)}$. They may differ by $-1$. Here we take the principal branch of the square root.

Recently I read some paper related to half integral weight modular forms.

If I understand correctly the arguments in paper, the following formula should hold.

$\sqrt{j(\alpha^{-1}\beta\alpha,z)}=\frac{\sqrt{j(\beta,\alpha z)}\sqrt{j(\alpha,z)}}{\sqrt{j(\alpha,\alpha^{-1}\beta\alpha z)}}$

Is this right? I think that they may differ by $-1$.

Any idea or hint will be thankful.