All Questions
7 questions
3
votes
0
answers
242
views
Explicit expression of automorphic representations as automorphic forms
Let‘s take $G=GL_n$ over a number field $F$ for example.
It's already known that every irreducible automorphic representation $\pi$ is a irreducible component of a induced representation $I(G,P;\...
- 111
1
vote
0
answers
242
views
Constant coefficient of Eisenstein series
Let $\chi$ be a character of $\mathbb{Q}^{\times}\setminus\mathbb{A}^{\times}$. We define an induced representation of $Mp_2(\mathbb{A})\simeq SL_2(\mathbb{A})\times \mathbb{C}^1$,
$$I(s,\chi) := \{\...
- 103
2
votes
1
answer
300
views
Question on the residual representation
Let $G=SO_n$ and fix a borel subgroup $P_0$ of $G$. Let $P=MN$ be a standard maximal parabolic subgroup $G$ and $\sigma$ a cuspidal representation of $M$
Consider the normalized parabolic induced ...
- 1,759
1
vote
1
answer
173
views
Factorizability of Subquotients of Principal Series Representations
Fix number field $F$, its ring of adeles $\mathbb{A}$, a "nice" algebraic group defined over $F$ (at least reductive but for my purposes I can assume simple and simply connected) and a parabolic ...
- 141
2
votes
1
answer
429
views
A computation about Whittaker functions and Eisenstein series
I have some questions about the computation of Eisenstein series and Whittaker functions in the book. The question is on page 29, Theorem 4.3.
My questions are in the following.
(1) I think that $B(...
- 6,201
4
votes
1
answer
203
views
Intertwining Operators Associated to Simple Reflections
Let $G$ be a quasi-split reductive group, over a local field, with a Borel subgroup $B=T\cdot N$ and the associated Weyl group $W$. Given a family of induced representations $\pi_s = Ind_B^G \chi\cdot ...
- 141
17
votes
4
answers
2k
views
Where do the real analytic Eisenstein series live?
In obtaining the spectral decomposition of $L^2(\Gamma \backslash G)$ where $G=SL_2(\mathbb{R})$, and $\Gamma$ is an arithmetic subgroup (I am satisfied with $\Gamma = SL (2,\mathbb{Z})$) we have a ...
