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Let $G=GL_{n}$ and $N$ the maximal unipotent subgroup, $\mathbb{A}$ the ring of adeles on a number field $F$.

We fix a non trivial character $\psi:F\backslash\mathbb{A}\rightarrow \mathbb{C}^{*}$. We can extend it to a character of $N(F)\backslash N(\mathbb{A})$ by: $\psi(u)=\psi(u_{1,2}+\dots +u_{n-1,n})$.

We consider a character $\chi:N(F)\backslash N(\mathbb{A})\rightarrow \mathbb{C}^{*}$.

Then we know that there exists $(a_{1},\dots, a_{n-1})\in F^{n-1}$ such that:

$\chi(u)=\psi(a_{1}u_{1,2}+\dots +a_{n-1}u_{n-1,n})$.

How can we show that there exists $\gamma\in Q_{n-1}(F)\backslash GL_{n-1}(F)$ such that $\chi=\psi(\gamma u\gamma^{-1})$?

where $Q_{n-1}$ is the mirabolic of $GL_{n-1}$ and we embed $GL_{n-1}$ to $GL_{n}$ by the upper left corner.

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  • $\begingroup$ I fixed the typo that $u_{12}$ mis-printed as $a_{12}$. But, still, there is something amiss: depending what is meant by "character on" that quotient, there might be others than ones given by that formula. The formula does give all "generic" characters, for non-zero $a_i$. And the whole mirabolic is not needed to act transitively on generic characters, only the maximal split torus therein. If the question is really intended to be about getting Fourier expansions of cuspforms, then it is missing some key features... clarify? $\endgroup$
    – paul garrett
    Commented Apr 3, 2014 at 15:23
  • $\begingroup$ my question is about all characters, not only generic ones. $\endgroup$
    – prochet
    Commented Apr 5, 2014 at 8:45
  • $\begingroup$ I am still confused by the ulterior goals. For example, $\psi(u_{13}+u_{23})$ is a reasonable character on the unipotent radical of the minimal parabolic in $GL_3$. Also, the mirabolic does not stabilize the collection of all characters, etc. $\endgroup$
    – paul garrett
    Commented Apr 5, 2014 at 15:41

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