1
$\begingroup$
  1. The problem statement, all variables and given/known data

I need to show that

enter image description here

transforms with modular weight $2$ for $SL_2(Z)$

We have the theorem that it is sufficient to check the generators S and T enter image description here

We have that E_2 is (whilst holomorphic) fails to transform with modular weight $2$ as it has this extra term when checking for $S$:

where transforming with weight $2$ means (for $S$):

$f(S.\tau)=\tau^{2}f(\tau)$

Therefore we expect a cancellation from the $Im (\tau)$ term

We have

MY QUESTION

  • I follow enter image description here

this solution and this is fine - I guess I am doing something pretty stupid, but we have the following formula

enter image description here

for $G=SL_2(R)$ and since $Z \in R$ we have $SL_2(Z) \in SL_2(R)$, so the above also holds for $SL_2(Z) $ and so this gives:

$Im(S.\tau)=\frac{Im(\tau)}{\tau^2}$

  • so $1/Im(\tau)$ transforms with modular weight $2$ itself, and so is not cancelling
  • I perhaps thought there may have been an issue that $\infty$ is not taken into account, but it also says the action on $G$ extends to $\infty$ , as I've said I follow the above solution, but have no idea what is wrong with using this formula.

Many thanks in advance

$\endgroup$
  • $\begingroup$ What is your definition for $E_2(\tau)$ ? $\endgroup$ – reuns May 16 '17 at 15:44
  • $\begingroup$ If $a,b,c,d$ are real and $ad-bc=1$ then $$Im(\frac{a\tau+b}{c\tau+d}) = Im(\frac{(a\tau+b)(c\overline{\tau}+d)}{|c\tau+d|^2}) = \frac{1}{|c\tau+d|^2} Im(ac|\tau|^2 + ad \tau+bc\overline{\tau}+bd) = \frac{1}{|c\tau+d|^2} (ac |\tau|^2+bd+(ad-bc)Im(\tau))=\frac{1}{|c\tau+d|^2} (ac |\tau|^2+bd+Im(\tau))$$ $\endgroup$ – reuns May 16 '17 at 15:56

Your Answer

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

Browse other questions tagged or ask your own question.