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Write $U_u=\left(\begin{array}{cc}1& u/D\\ 0& 1\end{array}\right)$ and $\tau (\chi )={\sum }_{u\mathrm{mod}D}\chi (u)e^{2\pi iu/D}$. The finite Fourier transform of a primitive Dirichlet character givesFor $(n,D)=1$ this follows by multiplying residues by $n$. For a nonunit $n$, primitivity supplies a unit $v\equiv 1(\mathrm{mod}D/p)$ with nontrivial character value, for a prime $p\mid (n,D)$; multiplication by $v$ leaves the exponential unchanged and forces the sum to vanish. Orthogonality of the finite exponentials gives ${\sum }_{n\mathrm{mod}D}|{\sum }_u\chi (u)e^{2\pi inu/D}{|}^2=D{\sum }_u|\chi (u){|}^2=D\phi (D)$. The primitive transform identity makes the left side $\phi (D)|\tau (\chi ){|}^2$. Thus the primitive Gauss sum of a Dirichlet character is nonzero, with $|\tau (\chi ){|}^2=D$. Consequently the twist is the finite rational-translation sumThis representation will prove both the level statement and the Fricke formula.

Take $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N{D}^2)$, and choose $v\equiv d^{2}u(\mathrm{mod}D)$. Since $ad\equiv 1(\mathrm{mod}D)$, this means $av\equiv ud(\mathrm{mod}D)$. Direct multiplication givesEvery entry is integral, its determinant is one, and its lower-left entry is divisible by $N$. Thus ${\gamma }_u\in {\Gamma }_0(N)$. Its lower-right entry is congruent to $d$ modulo $N$, so $f|{\gamma }_u=\psi (d)f$ by its nebentypus character. Using the right-action property of the slash operator,The map $u\mapsto v=d^{2}u$ permutes the residue classes. Reindexing gives $\overline{\chi }(u)=\chi (d{)}^2\overline{\chi }(v)$, soThis proves the required modular weight and character on ${\Gamma }_0(N{D}^2)$.

Cuspidality must also be checked. For any modular cusp matrix $\sigma \in S{L}_2(\mathbb{Z})$ and any rational translation $U_u$, choose $\rho \in S{L}_2(\mathbb{Z})$ with $\rho \infty =U_u\sigma \infty$. Then ${\rho }^{-1}{U}_u\sigma$ is upper triangular with positive dilation. Since $f|\rho$ decays exponentially at infinity, so does $f|U_u\sigma$. The finite sum defining $f_{\chi }$ therefore tends to zero at every modular cusp; its transformation law supplies a modular cusp period, so its holomorphic Fourier expansion has zero constant term there. Hence the primitive twist at coprime level is

Now put $W_M=\left(\begin{array}{cc}0& -1\\ M& 0\end{array}\right)$ and $g=f{|}_k{W}_N$, always using the determinant factor specified in the question. For each unit $u$ choose a unit $v$ with $Nuv\equiv -1(\mathrm{mod}D)$, possible because $(N,D)=1$. The crucial integral matrix factorization isThe right side multiplies out to $\left(\begin{array}{cc}uND& -1\\ N{D}^2& 0\end{array}\right)$, so the identity is exact. Its determinant-one matrix has lower-right entry $D$, and therefore $f|{\gamma }_u=\psi (D)f$. A positive scalar matrix $DI$ acts trivially under the determinant-normalized slash operator, since its determinant contributes $D^k$ and its denominator contributes $D^{-k}$. ThusThe inverse-residue substitution is $u\equiv -(Nv{)}^{-1}(\mathrm{mod}D)$, giving $\overline{\chi }(u)=\chi (-N)\chi (v)$. The sum over $v$ is $\tau (\chi )g_{\overline{\chi }}$. Hence the Fricke transform of a primitive coprime twist isThe second version uses $\tau (\chi )\tau (\overline{\chi })=\chi (-1)D$, which follows from $\overline{\tau (\chi )}=\chi (-1)\tau (\overline{\chi })$ and $|\tau (\chi ){|}^2=D$. Here $g_{\overline{\chi }}=\sum b(n)\overline{\chi }(n)q^n$ if $g=\sum b(n)q^n$; it is not the complex conjugate function $\overline{g}$. The coprimality of $N,D$ is used both in the matrix factorization and in evaluating $\psi (D)$. For $D=1$ the twist is trivial and the formula simply reads $f|W_N=g$.

The modular modular weight specifies the power $(cz+d{)}^k$ in a modular form's transformation law. Multiplication adds modular weights. At level one, the central matrix $-I$ excludes nonzero odd-weight forms; a nebentypus character at higher level changes this condition to $\psi (-1)=(-1{)}^k$.