Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

I've a question regarding the Petersson operator.

We have the following definition and the lemma.

Definition

Let $k, m \in \mathbb{Z}$ and $\phi: \mathbb{H} \times \mathbb{C} \rightarrow \mathbb{C}$.

For

$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \in \Gamma_1$ und $\begin{pmatrix} \lambda & \mu \end{pmatrix} \in \mathbb{Z}^{2}$

we set

i) $\left( \phi|_{k,m} \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \right) (\tau,z) := (c\tau+d)^{-k} e^{2\pi i m \frac{-cz^2}{c\tau+d}} \phi(\frac{a\tau+b}{c\tau+d},\frac{z}{c\tau+d})$ `and

ii) $( \phi|_m \begin{pmatrix} \lambda & \mu \\ \end{pmatrix}) (\tau,z) := e^{2\pi i m (\lambda^2\tau + 2 \lambda z)} \phi(\tau, z + \lambda\tau+\mu) $

Lemma

Let $k, m \in \mathbb{Z}$ and $\phi: \mathbb{H} \times \mathbb{C} \rightarrow \mathbb{C}$.

$M, M_1, M_2 \in \Gamma_1;\; X, X_1, X_2 \in \mathbb{Z}^2$ satisfy

i) $(\phi|_{k,m} M_1)|_{k,m} M_2 = \phi|_{k,m} (M_1 M_2)$

ii) $(\phi|_m X_1)|_m X_2 = \phi|_m (X_1 + X_2)$

iii) $( \phi|_{k,m} M)|_m XM = (\phi|_m X)|_{k,m}M $

I've to prove that

$ \phi|(M,X)|(M',X') = \phi|(MM',XM'+X')$

Unfortunately, after spending a lot of time, I am not clear how I can prove this. The main problem is that I don't know which calculation rule I can use. The lemma only deals with one variable ($M$ or $X$) but not with 2 variables $(M,X)$.

Do you have any hints how I could proceed.

Thanks and regards, Jan

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

I think the definition you want for $\phi|(M,X)$ is $(\phi |_{k,m} M) |_m X$.So your left hand side involving $\phi |(M,X) | (M',X')$ is the composition of four of these operators. Then the problem isn't too bad given your (i)-(iii).

share|improve this answer
    
Hi Matt, thanks a lot for your answer. I had a similar idea, but with $\phi|(M,X)$ is $(\phi|_m X)|_{k,m} M$ but I am not sure whether I can write it like this. I think, I have to put i) and ii) from the definition together. Therefore, my guess was $\phi|(M,X)$ is $(\phi|_m X)|_{k,m} M$. However, I haven't found a calculation rule which proves that my guess goes to the right direction. Do you have any hint? Thanks a lot, Jan –  Jan Apr 7 '11 at 21:03
    
Actually, I want to prove, that definition defines an action of the semi direct product $SL_2(\mathbb{Z}) \ltimes \mathbb{Z}^2$ I've already shown, that the semi direct product has the group law $(X,M)(M',X') = (MM', XM'+X')$. Now, I have to prove $ \phi|(M,X)|(M',X') = \phi|(MM',XM'+X')$ If this is proven, then according to the definition of an action ( $s \circ (g_1 \circ g2) = (g_1 g_2) \circ s $), i) and ii) of the definition above jointly define an action. –  Jan Apr 7 '11 at 21:14
    
Are you sure you are checking what you need to check? Keep in mind that this is a right action just like for usual modular forms. It looks like you are under the impression it is a left action. Does this help? –  Matt Young Apr 8 '11 at 0:21
    
Actually, my main problem is, that I don't know why the definition of $\phi|(M,X)$ should be $(\phi|_{k,m}M)|_m X$ (as you've written). Do you know the rules, why this should be true? Could you be so kind and explain a little bit how you did this? Thanks a lot –  Jan Apr 8 '11 at 7:26
    
The definition is chosen so that your desired identity is true; the point is that you then get a group action of the semi-direct product of $\SL_2(\mathbb{Z})$ with $\mathbb{Z}^2$. –  Matt Young Apr 8 '11 at 15:17
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.