I am trying to interpret the symbol $L( \frac{1}{2}, \pi \times \chi)$ where $\chi = \mathbb{A}^\times / \mathbb{Q}^\times$ and $\pi$ is a cuspidal representation of $GL_2( \mathbb{A})$ (where $\mathbb{A}$ are the adeles over $\mathbb{Q}$), something like $$ \mathbb{A} = \prod' \mathbb{Q}_p $$

where we use a "restricted" product in order to avoid axiom of choice difficulties<\del>.

I was told cusp forms match cuspidal representations of $GL(2, \mathbb{A})$ Let me write down a cusp form:

$$ \theta(z; u) = \sum_{(a,b) \in \mathbb{Z}^2} (a^2 - b^2) \,e^{2\pi i (a^2 + b^2 ) \, z}$$

and I can even write down the L-function without too much thought

$$ L(\tfrac{1}{2}, \pi \times \chi) = \sum_{(a,b) \in \mathbb{Z}^2} \frac{a^2 - b^2}{a^2 + b^2} =? \, 0$$

It might not even be correct.

Since $u(a,b)=a^2-b^2$ is a harmonic polynomial in two variables, the theta function I have written should be a modular form on $\Gamma_0(4)$ of weight $2+1/2+1/2=3$. And it should be a cusp form. Not 100% sure about $z=\frac{1}{2}$.

  • $\begingroup$ Could you explain more how did u obtain your L value formula "without too much thought"? Where is chi in your formula? $\endgroup$
    – Q-Zhang
    Nov 19, 2017 at 17:49
  • $\begingroup$ @QingZhang $\chi \equiv 1$ identically. Part of my question is to write out $\pi$. $\endgroup$
    – cactus314
    Nov 19, 2017 at 17:50
  • $\begingroup$ Could u explain more on your last formula? $\endgroup$
    – Q-Zhang
    Nov 19, 2017 at 18:01
  • $\begingroup$ Why is the right side convergent? $\endgroup$
    – Q-Zhang
    Nov 19, 2017 at 18:02
  • $\begingroup$ @QingZhang Like I said I didn't put much thought, but then I really don't understand the definition of the automorphic L-function. $\endgroup$
    – cactus314
    Nov 19, 2017 at 18:04

1 Answer 1


For $f(z) = \sum_{n=0}^\infty a_n e^{2i \pi nz} \in M_k(\Gamma_0(N))$ then $$f \otimes \chi(z) = \sum_{n=0}^\infty a_n \chi(n) e^{2i \pi nz} =\frac{1}{q} \sum_{m=1}^q \hat{\chi(m)} f(z+\frac{m}{q}) \in M_k(\Gamma_0(N q^2),\chi^2)$$ for any primitive Dirichlet character modulo $q$, where $ \hat{\chi(m)} = \sum_{n=1}^q \chi(n) e^{-2i \pi nm/q} = \overline{\chi(m)} \hat{\chi(1)}=\overline{\chi(m)}G(\chi)$. For $\chi$ non-primitive $\hat{\chi(m)}$ is more complicated (not completely multiplicative) and hence $f \otimes \chi$ is only modular for $\Gamma_1(N q^2)$.

Proof : for $\gamma = {\scriptstyle\begin{pmatrix}a & b\\ cNq^2 & d \end{pmatrix}} \in \Gamma_0(Nq^2)$ and $T^{m/q} = {\scriptstyle\begin{pmatrix}1 & \frac{m}q\\ 0 & 1 \end{pmatrix}}$ then $T^{m/q} \gamma T^{-d^2 m/q} =\gamma_m \in \Gamma_0(Nq^2)$. Thus $f \otimes \chi|_k \gamma (z)$ $=\frac{1}{q} \sum_{m=1}^q \hat{\chi(m)} f |_k T^{m/q}\gamma (z)=\frac{1}{q} \sum_{m=1}^q \hat{\chi(m)} f|_k \gamma_m T^{d^2 m/q} (z)$ $=\frac{1}{q} \sum_{m=1}^q \hat{\chi(m)} f|_k T^{d^2 m/q} (z)= \ldots$

You can interpret $f \mapsto f \otimes \chi$ as a double coset operator, similar to the Hecke operators.

For automorphic representations, the meaning is the same, except we can interpret it as a tensor product of representations (looking first at twists of Artin L-functions is easier).

See also the Rankin-Selberg convolution, allowing to twist by any automorphic representation/L-function.

  • $\begingroup$ do you have a name or e-mail ? i could try to respond here. what Automorphic representations correspond to spherical harmonics? $\endgroup$
    – cactus314
    Dec 30, 2017 at 2:07
  • $\begingroup$ does the Dirichlet series in this question have a representation as an automorphic form? math.stackexchange.com/questions/2570742/… fortunately your twist works for any series $\endgroup$
    – cactus314
    Dec 30, 2017 at 2:10
  • $\begingroup$ i still have no idea what an automorphic form is so $f \otimes \chi $ is a better object for me, since i manipulate the series. $\endgroup$
    – cactus314
    Dec 30, 2017 at 2:21

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