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Let $G$ be a connected reductive group over a number field $F$ and $P_0$ its minimal parabolic subgroup. Then we call a parabolic subgroup $P$ of $G$ is standard if $P_0 \subset P$.

Let $K$ be a fixed good maximal compact subgroup of $G(\mathbb{A}_F)$ such that $G=PK$ for all standard parabolic subgroup $P=UM$. (here $U,M$ are the unipotent radical and Levi of $P$.

If $Q=U_Q M_Q$ is another standard parabolic group of $G$ such that $Q \subset P$, I am wondering whether $M_P=(U_Q \cap M_P)(M_Q )(K \cap M_P)$.

Any comments are highly appreciated!

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    $\begingroup$ What do you mean by $G=PK$? Your $P$ is an algebraic subgroup of $G$, hence it is an algebraic group over the number field $F$, while $K$ is a maximal compact subgroup of some topological group - what group? In order to define $PK$, you groups $P$ and $K$ must be subgroups of the same group - what group? Please edit your question. $\endgroup$ – Mikhail Borovoi Aug 27 '19 at 19:45
  • $\begingroup$ OP probably means that ``$K$ is a maximal compact subgroup of $G$''. Then $G=PK$ is an Iwasawa decomposition of $G$. If OP, by $M$, meant Levis in the corresponding groups, then the last assertion is true. That assertion may probably be derived from the Iwasawa decomposition of $M_P$. $\endgroup$ – Subhajit Jana Aug 28 '19 at 8:01
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    $\begingroup$ @SubhajitJana The issue is you do not have an Iwasawa decomposition over arbitrary fields. E.g., one often puts the discrete topology on $G(\mathbb Q)$, in which case compact subgroups are just the finite subgroups. $\endgroup$ – Kimball Aug 28 '19 at 10:57
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    $\begingroup$ In what group is $K$ contained? The number field $F$ has infinitely many completions $F_v$. Do you mean that $K$ is a maximal compact subgroup in $G(F_v)$ for some $v$? Is your $F_v$ isomorphic to $\mathbb R$, or to $\mathbb C$, or is it a nonarchmedean local field? Please edit your question again. $\endgroup$ – Mikhail Borovoi Aug 28 '19 at 17:14
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    $\begingroup$ Is your field $F$ indeed a number field, or you mean a local field? $\endgroup$ – Mikhail Borovoi Aug 28 '19 at 17:17
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I'll take a stab at interpreting this.

Firstly on the level of algebraic groups, note that since $Q\subset P$ are standard, we can fix a Levi decomposition $T_0U_0=B_0$ of a minimal parabolic of $G$ and require that our Levi decompositions $Q= M_QU_Q$ and $P=M_PU_P$ satisfy $T_0\subset M_Q\subset M_P$. Then $M\cap Q= M_Q\cdot (U_Q\cap M)$ is a parabolic subgroup of $M_P$.

On the level of adelic points, we have the maximal compact subgroup $K\subset G(\mathbb{A})$ satisfying $G(\mathbb{A})=P(\mathbb{A})K$ for any standard parabolic subgroup $P$. We are free to choose $K$ so that $K_M=M(\mathbb{A})\cap K$ is a good maximal compact subgroup of $M(\mathbb{A})$ (this is, for example, stated in Moeglin-Waldsurger I.1.4). We then have an Iwasawa decomposition $$M(\mathbb{A})=(M\cap Q)(\mathbb{A})K_M.$$ This is as close to what you ask as I can imagine since $(M\cap Q)(\mathbb{A}) = M_Q(\mathbb{A})\cdot (U_Q\cap M)(\mathbb{A})$.

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