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Let $f$ be a modular form on $\Gamma_0(4)$ of weight $k\in\tfrac 12\mathbb{Z}$ with trivial Nebentypus.

Is it true that if you twist $f$ by $\tfrac 12$, i.e. look at the function $g$ with $g(\tau)=f(\tau+\tfrac 12)$, this is again a modular form of the same weight $k$ on $\Gamma_0(4)$, but this time with Nebentypus $\chi$, the non-trivial Dirichlet character modulo 4?

Edit: You can show that $g$ is a modular form with trivial character on $\Gamma_0(16)$:

Consider the operators $V_m$ and $U_m$. If $f(\tau)=\sum\limits_{n=0}^\infty \alpha_f(n)q^n$ is the Fourier expansion of $f$ then these operators have the following effect: $$ (V_mf)(\tau)=\sum \alpha_f(n)q^{mn}\quad\text{and}\quad (U_mf)(\tau)=\sum\alpha_f(mn)q^n.$$ They both send modular forms on $\Gamma_0(N)$ to modular forms on $\Gamma_0(mN)$ of the same weight. (cf. http://people.mpim-bonn.mpg.de/zagier/files/scanned/IntroductionToModularForms/fulltext.pdf, page 251 in the book, 14 in the pdf). If you apply $U_2$ and $V_2$ to $f$ you get the form $f_{ev}(\tau)=\sum\alpha_f(2n)q^{2n}$ on $\Gamma_0(16)$ and claearly we have $g(\tau)=2f_{ev}(\tau)-f(\tau)$.

I suppose there should either be a reference for this, which I have not been able to find, or a simple proof that I don't see right now.

Thank you very much for your help!

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I actually alreasy struggle to see that it is a modular form of $\Gamma_0(16)$. Could you at least include that computation? –  Marc Palm Apr 26 '13 at 8:22
    
I see. Thanks. On which page can I find these operators? Please always include precise reference on this forum, i.e., pages. –  Marc Palm Apr 26 '13 at 9:39
    
Sorry about that, I added them. Maybe this is not that perfect reference because there are no proofs there, but one can figure most of this out more or less easily. –  MHMertens Apr 26 '13 at 9:51
    
So you mean $\Gamma_0(8)$. Quote from that text (next sentence): If $m$ divides the $N$, then the mapping goes from $\Gamma_0(N)$ to $\Gamma_0(N)$, so you get something without twist. Haven't you missed that passage or simply messed up with your notation? –  Marc Palm Apr 26 '13 at 10:34
    
Dear MHMertens, Note that $\GL_2(\mathbb Q)^+$ ($2 \times 2$ matrices with positive determinant) acts on the space of modular forms of some fixed weight $k$ via the usual "slash" action. If $f$ is of level $\Gamma$, then $\gamma f$ (for $\gamma$ in $\GL_2(\mathbb Q)$ is $\gamma \Gamma \gamma^{-1}$ (since being of level $\Gamma$ is the same as being invariant by $\Gamma$ under the weight $k$ slash action). So in a question like yours, there is always some level for $f|_k \gamma$, and to figure out what it is, is just a matter of computing the conjugate $\gamma \Gamma \gamma^{-1}$. Regards, –  Emerton Apr 26 '13 at 20:18

2 Answers 2

up vote 4 down vote accepted

EDIT : In what follows the weight $k$ is assumed to be an integer.

The matrix $\begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}$ normalizes the group $\Gamma_0(4)$, so in your case $g$ is still a modular form on $\Gamma_0(4)$ with trivial Nebentypus.

More generally if you start with $f$ on $\Gamma_0(N)$ and twist it by $1/m$ then you get a form on $\Gamma_0(N') \cap \Gamma_1(m)$ with $N'=\operatorname{lcm}(N,m^2)$.

In general, when you consider $g(z)=f(z+1/m)$, you are twisting $f$ by the additive character $\alpha(n)=\exp(2\pi i n/m)$. You can always write $\alpha$ as a linear combination of (not necessarily primitive) Dirichlet characters $\chi$ of level dividing $m$. Thus you can write $g$ as a linear combination of twists $f \otimes \chi$ for such $\chi$ (up to bad Euler factors). If you want to do this completely explicitly then the formulas are quite complicated in general, see Merel, Symboles de Manin et valeurs de fonctions L, Section 2.5. At some point, it may be useful to switch to the adelic language and work with the automorphic representation associated to the newform $f$.

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Note that $f \otimes \chi$ has Nebentypus $\chi^2$ so this indicated that in general, one cannot expect $g$ to have some Nebentypus. Somehow the case $N=4$, $m=2$ is quite particular. –  François Brunault Apr 26 '13 at 10:58
    
OK, I should have been able to figure that out by myself. But there is still one little point of confusion: According to the dimension formula in H. Cohen's "Sums Involving the Values at Negative Integers of L-Functions of Quadratic Characters" (link.springer.com/article/10.1007%2FBF01436180#page-1), we have dimension 1 for weight 1/2 and trivial character. The space is spanned by the ordinary theta series $\theta(\tau)=\sum_{n\in\mathbb{Z}}q^{n^2}$. But your answer would imply that also the "twisted" theta series, which is obviously not a multiple of $\theta$, belongs to this space. –  MHMertens Apr 26 '13 at 11:40
    
What I wrote is correct if the weight is an integer, I'm not sure about the case of half-integral weight. –  François Brunault Apr 26 '13 at 13:17

First of all, I realized today that I don't need this fact for half-integral weight, integral weight is enough, thus: problem solved! Thanks to all who replied.

Second: In the case of half-integral weight, I now have hints, that in that case it doesn't work at all:

The group acting in this case is not $\Gamma_0(4)$ but the Fuchsian group often denoted by $\Delta_0(4)$ (see e.g. Shimura's paper "On modular forms of half integral weight" (http://www.jstor.org/stable/1970831, p. 447). This group consists of pairs of elements of $\Gamma_0(4)$ and the automorphic factor of the ordinary theta-series. Unfortunately, this group does not have a normalizing element of the form $(\begin{pmatrix} 1 & \tfrac 12 \newline 0 & 1\end{pmatrix},\varphi(\tau))$ for any possible automorphic factor $\varphi$ (this can be calculated directly from the group law of the Fuchsian group), thus I would now assume that my question has to be answered negatively in case of half-integral weight.

Best regards, MHMertens

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