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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

flag	asked Apr 7 2011 at 18:43 Jan 13●3

Matt Young · Accepted Answer · 2011-04-07 19:55:33Z UTC

2

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).

answered Apr 7 2011 at 19:55

Matt Young
1,917●1●6●12

	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 2011 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 2011 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 2011 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 2011 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 2011 at 15:17

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Modular form - Petersson operator

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