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$\DeclareMathOperator\GL{GL} \DeclareMathOperator\Sp{Sp} \DeclareMathOperator\Ind{Ind}\DeclareMathOperator\B{B}$Let $F$ be a number field and $G_n$ the symplectic group over a $2n$-dimensional symplectic space over $F$. Let $\mathbb{A}$ be the adele ring of $F$.

Cconsider an automorphic cuspidal representation $\pi$ of $G_n(\mathbb{A})$. Let $B$ be the Borel subgroup of $G_n$ and $N$ be its unipotent radical.

Consider an additive character $\psi$ of $\mathbb{A}$ and let $\{\psi_c\}_{c \in \mathbb{A}}$ be the family of generic characters of $N(\mathbb{A})$ associated to $\psi$.

Then I am wondering whether there is a generic character $\psi_c$ for some $c \in \mathbb{A}$ such that the Whittacker fourier coefficient of $\pi$ with respect to $\psi_c$ is non-zero?

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    $\begingroup$ Unless you impose additional hypotheses on $\pi$, the answer is "no". There are many automorphic representations which are not generic at some place. For example, if $n \geq 2$, if $F = \mathbf{Q}$ is the rational numbers, and if $\pi$ is generated by a classical cuspidal holomorphic Siegel modular form, the answer is no. The reason is that every generic linear functional on $\pi_\infty$ vanishes. $\endgroup$
    – Joseph
    Commented Jul 17 at 16:54
  • $\begingroup$ @Joseph, thanks for the comment. Then is it possible that all the fourier coefficients of $f$ are zero for every $f$ in $\pi$ though $\pi$ is non-zero? $\endgroup$
    – Andrew
    Commented Jul 18 at 2:32
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    $\begingroup$ Right, every Fourier coefficient that comes from taking a character of the unipotent radical $N$ of the Borel can vanish for every $f$ in $\pi$. The "problem" is that $N$ is very large. Let $N_1 = [N,N]$ be the commutator subgroup. Then every character of $N$ is trivial on $N_1$. So, in terms of of Fourier coefficients, you are asking for an $f$ in $\pi$ to have a nontrivial constant term along $N_1$. But $N_1$ can be quite big, so you are asking for too much. If you replace $N$ with the unipotent radical of the Siegel parabolic, you will find nonvanishing Fourier coefficients. $\endgroup$
    – Joseph
    Commented Jul 18 at 3:18
  • $\begingroup$ @Joseph, thanks again for the comment. If one replace $N$ with the unipotent radical of the Siegel parabolic, how do we know the nonvanishing of the Fourier-coefficients? In such case, the character of $N$ is $\psi(a_{n,n+1})$ for $a\in N$. Right? $\endgroup$
    – Andrew
    Commented Jul 18 at 8:55
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    $\begingroup$ Let $U$ denote the unipotent radical of the Siegel parabolic. I only claim that there is some character $\chi$ of $U$ for which the $\chi$ Fourier coefficient of your cusp form $\varphi$ is nonzero. This follows immediately from the fact that $U$ is abelian. Specifically, you can Fourier expand $\varphi$ along $U$. If $\varphi$ is nonzero, then because $\varphi$ is a sum of its Fourier coefficients, at least one Fourier coefficient is nonzero. $\endgroup$
    – Joseph
    Commented Jul 18 at 13:39

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