This might be stupid question to some experts who works in the realm of automorphic form.

Let $K$ be a number field and $\mathbb{A}$ is a adele ring of $K$. Let $G$ be a connected reductive group defined over $K$. Moeglin and Waldspurger defined the norm function on $G(\mathbb{A})$ as follows:

For $GL_{2n}(\mathbb{A})$, its norm function is defined by $||g||=\prod_v \text{sup} \{|g_{rs}|_v : r,x=1,...,2n\}$, where we write $g=(g_{rs})$. (See Section 2.2 of the book 'Spectral decomposition and Eisenstein series' by Moeglin and Waldspurger.)

Fix an embedding $i' :G \to GL_n$ and let $i':G \to SL_{2n}$ be the embedding defined by

$i(g)=\begin{pmatrix} i(g) & 0 \\ 0 & i(g)^{-1} \end{pmatrix}$.

Define the norm function $\| \cdot \|$ on $G(\mathbb{A})$ as $||g||=|i(g)|$.

Then I am wondering whether the followings are true.

  1. Let $N$ be a unipotent subgroup of $G$. There exists a constant $c>0$ such that $\|g\|<c$ for all $g \in N(\mathbb{A})$?

  2. There exists a constant $c>0$ such that $\|g\|<c$ for all $g \in G(F)$.

I think that they are both true but I can't find any reference which proves these.

Does this hold for some simple reason? Or if there is a reference dealing with this, I would appreciate if you let me know it.

Thank you in advance.


1 Answer 1


For the first question,(and hence for the second,) it is false.

For example, you can take elements that are upper triangular unipotent and have trivial non-archimedean components as its counterexamples.

However, it is true that the norm is lower bounded. It is easily proven if you use the fact that the norm is inverse-invariant, and the triangle inequality.

  • $\begingroup$ Sorry, I saw this reply late. Your answer really helped me. Thank you very much! $\endgroup$
    – Andrew
    May 27, 2021 at 2:12

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