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!

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? $\endgroup$ – Marc Palm Apr 26 '13 at 10:34somelevel for $f|_k \gamma$, and to figure out what it is, is just a matter of computing the conjugate $\gamma \Gamma \gamma^{-1}$. Regards, $\endgroup$ – Emerton Apr 26 '13 at 20:18