2
$\begingroup$

Let $G$ be a connected reductive group over a number field $F$ and fix a minimal parabolic subgroup $P_0$ of $G$. 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 $P$.

Then Harish-chandra defined a height function $H_G:G(\mathbb{A}_F) \to \mathcal{a}_G$. (here $\mathcal{a}_G$ is the real vector space spanned by coroots of $G$. For the precise definition of Height function, please see 16 page of the Arthur's book 'An introduction to the trace formula'.)

For a standard parabolic subgroup $P \subset G$, Harish-Chandra height function $H_P : G(\mathbb{A_F}) \to \mathbb{a}_P$ is defined by $$H_P(g)=H_{M_P}(m)$$ where $g=umk$ where $u \in N_P, m\in M_P, k\in K$ and $G=N_PM_PK$ is the Iwasawa decomposition of $P$.

Let $G^1(\mathbb{A}_F)=\{g \in G(\mathbb{A}_F) \ | \ H_G(g)=0\}$.

Then I have two questions;

  1. Is this true that $U_P(\mathbb{A}_F) \subset G^1(\mathbb{A}_F)$ for all standard parabolic subgroups $P \subset G$?

  2. For an element $g \in G(\mathbb{A}_F)^1$, we decompose $g=umk$ using Iwasawa decomposition with respect to $P$ as above. Then $u$ should be $1$ and $m \in M_P(\mathbb{A}_F)^1$?

I think that both are true but I can't prove it.

Any comments are greatly welcome!

$\endgroup$
5
  • $\begingroup$ What is $K$ here? Do you mean $A$? $\endgroup$
    – Asaf
    Commented Aug 29, 2019 at 17:31
  • $\begingroup$ @Asaf, Oh, sorry. I added the definition of $K$. $\endgroup$
    – Monty
    Commented Aug 29, 2019 at 18:52
  • 1
    $\begingroup$ Every element of $U_{P}$ is unipotent, hence in its inclusion to $G$ is still unipotent, hence under every morphism $GL_{1}$ it will be trivial. P.S. Iwasawa decomposition a-priori defined only for semisimple groups, hence it is not correct to consider $K$ in the decomposition of $P$ you mentioned. $\endgroup$
    – Asaf
    Commented Aug 29, 2019 at 22:08
  • $\begingroup$ This has nothing to do with automorphic forms or the trace formula. Removing unnecessary tags. $\endgroup$ Commented Aug 30, 2019 at 14:22
  • $\begingroup$ @Asaf, Thank you for ur answer. I edited the question as ur comment. $\endgroup$
    – Monty
    Commented Aug 30, 2019 at 15:36

1 Answer 1

1
$\begingroup$
  1. Is it true that $U_P(\mathbb A) \subseteq G(\mathbb A_F)^1$ for standard parabolic subgroups $P \subseteq G$?

Yes. Let $x \in U_P(\mathbb A)$ be written as $x = umk$ where $u \in U_P(\mathbb A), m \in M_P(\mathbb A)$ and $k \in K$. Clearly $u = x, m = 1$ and $k = 1$. Then $H_P(x) = H_{M_P}(m) = 0$ so $x \in G(\mathbb A)^1$.

Edit: If $H_P(g) = 0$ then $H_G(g) = 0$.

Observe that if $P_1 \subseteq P_2$ then $M_{P_1} \subseteq M_{P_2}$ and the restriction homomorphism $X(M_{P_2}) \to X(M_{P_1})$ is injective [Clay notes $\S 5$]. Thus $H_{P_2}(g) = 0$, which is equivalent to $\chi(g) = 0 \ \forall \chi \in X(M_{P_2})$ is true whenever $\chi(g) = 0 \ \forall \chi \in X(M_{P_1})$, which is equivalent to $H_{P_1}(g) = 0$.

  1. On the other hand, the second question is not true. For a quick counterexample, take $G = SL(2), P$ to be the minimal parabolic subgroup of upper triangular matrices and $g = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Note that $G(\mathbb A) = G(\mathbb A)^1$ and $g \in U_P(\mathbb A)$ but $g \neq 1$.
$\endgroup$
3
  • $\begingroup$ Thank you for the answer on the second question. But for the first question, why $H_P(x)=0$ implies $H_G(x)=0$? $\endgroup$
    – Monty
    Commented Aug 29, 2019 at 21:34
  • $\begingroup$ Edited answer with details. $\endgroup$ Commented Aug 30, 2019 at 14:17
  • $\begingroup$ Wow, thank you very much!😊 $\endgroup$
    – Monty
    Commented Aug 30, 2019 at 15:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .