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Let $E/F$ be a quadratic extension of number fields and $V,\langle,\rangle$ is a hermition vector space over $E$. Let $\mathbb{A}$ be the adele ring of $F$.

Assume that there is a hermitian line $e \in V$ such that $\langle e,e\rangle=1$ and let $V'$ be the orthogonal complement of $\langle e\rangle$ in $V$.

Let $P_0'$ be a minimal parabolic subgroup of $U(V')$ and $K= \prod_v K_v$ a ‘good’ maximal compact subgroup of $U(V')$ such that $U(V')=P_0'K$. (Here, the definition of ‘good’ is from Section I.1.4 of the book "Spectral decomposition and Eisenstein series" by Waldspurger and Mœglin. That is, for any standard parabolic subgroup $P'$ of $U(V')$, $P'(\mathbb{A}) \cap K=(M'(\mathbb{A}) \cap K)(U'(\mathbb{A}) \cap K))$ and $M'(\mathbb{A}) \cap K$ is still a maximal compact subgroup of $M'(\mathbb{A})$.)

Then I have two questions.

  1. What does $K(\mathbb{A})$ look like? I guessed $K_v=U(V')(\mathfrak{o}_v)$ for non-archimean $v$ and $K_v=U(p,q)$ for some $(p,q)$ such that $p+q=n$, when $v$ is archimedean. But if so, it is independent of the choice of $P_0'$. Is this right?

  2. If we choose a good maximal compact subgroup $K'(\mathbb{A})$ of $U(V')$, then can we choose a good maximal compact subgroup $K$ of $U(V)$ such that $K' \subset K$? (Here, $P_0$ is a minimal parabolic subgroup of $U(V)$ contained in the parabolic subgroup of $U(V)$ which stabilizes the flag of $V'$ which defines $P_0'$.)

Any comments are highly appreciated!

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    $\begingroup$ TeX note: please use \langle\rangle for inner products, not <>. Compare $\langle e, e\rangle = 1$ \langle e, e\rangle = 1 to $<e, e> = 1$ <e, e> = 1. I have edited accordingly. $\endgroup$
    – LSpice
    Apr 21, 2021 at 22:11
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    $\begingroup$ @LSpice, thank you very much for correction! I will keep your advice! $\endgroup$
    – Andrew
    Apr 23, 2021 at 9:16

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