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Let $F$ be a $p$-adic field. Let $E$ be a quadratic extension of $F$ and $G$ be a quasi-split unitary group $U(2n)$ or $U(2n+1)$ over with respect to $E/F$. Let $N_{E/F}$ be a norm map.

Let $B=TU$ be a Borel subgroup of $G$, where $T$ is a maximal torus of $B$.

Let $\psi$ be a non-trivial additive character of $F$ and let $\psi_E(x)=\psi(\frac{x+ \bar{x}}{2})$ on $E$. For $a\in F$, let $\psi_U^a$ be a character of $U$ defined by

$\psi_U^a(x)=\psi_E(x_{12}+\cdots + ax_{n,n+1})$.

I am wondering that every $T$-orbits of generic characters of $U$ is bijection with $F^{\times}/N_{E/F}(E^{\times})$ given by $a \mapsto \psi_U^a$.

A similar things is true for symplectic group but I don’t know whether it is true for unitary groups.

I guess it differs from when $G_n=U(2n)$ or $G_n=U(2n+1)$.

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    $\begingroup$ It's more or less that, at least when $p\not=2$ (I am a bit nervous about the theory when $\mathrm{char}(F)=2$. Maybe everything is just fine as long as $E/F$ is separable ...). $T(F)$-orbits of generic characters on U are in bijection with regular unipotent orbits which are also in bijection with regular nilpotent orbits. For the latter orbits, they are classified by $\operatorname{ker}(H^1(F,Z(G))\rightarrow H^1(F,G))$. The map is an isomorphism for $G=U(2n+1)$ but is zero for $G=U(2n)$. In the $U(2n)$ case we get $H^1(F,Z(G))=H^1(F,U(1))=F^{\times}/N_{E/F}(E^{\times})$. $\endgroup$
    – Cheng-Chiang Tsai
    Commented Sep 17, 2023 at 16:51
  • $\begingroup$ @Cheng-ChiangTsai, Oh, thank you for such great answer! Then for $U(2n+1)$, $T$-orbit is unique and for $U(2n)$, $T$-orbits are in correspondence with $F^{\times}/N_{E/F}(E^{\times})$. Again, I thank you very much! $\endgroup$
    – Andrew
    Commented Sep 20, 2023 at 1:59

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