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I am reading Arthur's book "Introductionto the trace formula".

In reading the book, two small question has arised and so I would like to ask it.

  1. Let $G$ be a connected reductive group over $\mathbb{Q}$ and $P=M_PN_P$ a standard parabolic subgroup. (here $M_P$ is Levi subgroup and $N_P$ is the unipotent subgroup of $P$)

I am wondering whether $G(\mathbb{Q})$ acts on $N_P(\mathbb{A})$. Because it looks that Arthur used such fact in some argument.

Is it really true?

  1. Let $N$ be an arbitrary unipotent group defined over $\mathbb{Q}$ which has $G$-action over $\mathbb{Q}$. Denote $G$-action by $\rho \colon G \to Aut(N)$.

The for arbitrary $g\in G(\mathbb{A})$, let $n'=\rho(g^{-1})(n)$. Then I heard that two measures $dn$ and $dn'$ has the relation $dn'=\delta_{\rho}(g)dn$ for some character $\delta_{\rho} \colon G(\mathbb{A}) \to \mathbb{C}^{\times}$.

I am wondering the explicit formula for the character $\delta_{\rho}$.

For these two questions, any comments will be greatly appreciated.

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    $\begingroup$ Could you be more precise about how Arthur seems to use this? At first glance, I would say that the action should be of $M_P(\mathbb{Q})$ on $N_P(\mathbb{A})$, not $G$. For example, if $G=\mathrm{GL}(2)$ and $P=B$ is the Borel, there is no natural action of $G$ on $N\cong \mathbb{A}$. $\endgroup$ Commented Aug 20, 2019 at 20:05
  • $\begingroup$ @WSL, In page 76 of Arthur's book, Arthur wrote "The kernel $K_{P,\chi}(x,y)$ defined in Sec.12 is invariant under left translation of either variable by $N_P(\mathbb{A})$". I think this is the part where Arthur used the fact $G(\mathbb{Q})$ acts on $N_P(\mathbb{A})$. Without this, can we verify this sentence? $\endgroup$
    – Monty
    Commented Aug 21, 2019 at 10:48
  • $\begingroup$ The kernel under consideration is given as a sum of (integrals of) Eisenstein series induced from $P$, and is formally invariant under $N_P$ as these Eisenstein series are. He is not claiming that the entire automorphic kernel $K_f$ is invariant, which is not true. $\endgroup$ Commented Aug 21, 2019 at 11:40
  • $\begingroup$ @WSL, Why Eisenstein series are left $N_P(\mathbb{A})$ invariant? It is not obvious from the definition of Eisensterin series. $\endgroup$
    – Monty
    Commented Aug 21, 2019 at 12:39
  • $\begingroup$ Ah you’re right: I was being sloppy and thinking about the representation. Still, I expect that the definition for the kernel he gives will solve your question. Can you post the formula from Section 12? $\endgroup$ Commented Aug 21, 2019 at 12:45

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I will try to answer both your questions in the context of Arthur's notes.

There does not exist a canonical action of a general connected reductive group $G$ on $N_P$ where $P = N_P M_P$ is a parabolic subgroup. One however looks at the exact sequence $$ 1 \to N_P \to P \to M_P \to 1 $$ which gives an action of $M_P$ on $N_P$ by conjugation.

Suppose $\text d n$ is a Haar measure on the unimodular group $N_P(\mathbb A)$. For every $m \in M_P(\mathbb A)$, we have another measure on $N_P(\mathbb A)$, namely $\text d(mnm^{-1})$. By the uniqueness of Haar measures, there is a positive number $\delta_m$ such that $\text d(mnm^{-1}) = \delta_m \text dn$. It is not difficult to show that the map $m \mapsto \delta_m$ is a homomorphism of groups. There must thus be a character $\delta : M_P(\mathbb A) \to \mathbb R^*_{> 0}$ which we call the modulus character.

This modulus character plays an important role in the theory of automorphic forms. For instance, Arthur uses this when making a change of variables involving the truncated kernel. (If I recall correctly, Cartier's notes in the Corvallis proceedings talk about the modulus character).

Given $P = M_P N_P$, there is an explicit formula for this modulus character $\delta = \delta_{M_P}$. It is a good exercise to show that this equals half of the sum of positive roots of $(P, A_P)$ where $A_P$ is the split part of the center of $M_P$. Try this out when $G = SL(2)$. Perhaps look at the Lie algebra picture.

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  • $\begingroup$ Thank you Abhi. I am studying Arthur’s book these days. Your comments are really helpful to me. Thank you very much!😊 $\endgroup$
    – Monty
    Commented Sep 1, 2019 at 17:38

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