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Let $E/F$ be a quadratic extension of number fields. If $V$ is a hermitian space over $E$, let $V=X+V_0+Y$ be its Witt decomposition, where $X,Y$ are maximal totally isotropic subspaces and $V_0$ is anisotropic kernel of $V$ respectively.

Let $P(X)$ be the parabolic subgroup of $U(V)$ stabilizing $X$. Then we can decompose $U(V)=P(X)K$ where $K$ is a maximal compact subgroup of $U(V)$.

I am wondering what is the Iwasawa decomposition of $U(V_0)$. How can we decompose it? I also want to know what is the parabolic subgroup of $U(V_0)$.

Any comments will be highly appreciated!

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  • $\begingroup$ Unless I misunderstand something here... $U(V_o)$ is compact, no? And has no proper ($F$-rational) parabolics. $\endgroup$
    – paul garrett
    Commented Mar 14, 2020 at 21:06
  • $\begingroup$ @Paul, You are right! I think there is no proper parabolic and so no proper Iwasawa decomposition. Thank you very much for comment. $\endgroup$
    – Monty
    Commented Mar 15, 2020 at 5:05
  • $\begingroup$ @Paul, from your comment, I think that $U(1)$ has no proper parabolic subgroup. Right? $\endgroup$
    – Monty
    Commented Mar 15, 2020 at 5:53
  • $\begingroup$ Right: with a non-degenerate $E$-one-dimensional space, the corresponding unitary group has no proper parabolics... because it has no non-trivial isotropic subspaces. $\endgroup$
    – paul garrett
    Commented Mar 15, 2020 at 15:38
  • $\begingroup$ @Paul, thank you for clear explanation. I always learn much from your comment and answers!:) $\endgroup$
    – Monty
    Commented Mar 16, 2020 at 11:45

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