bio | website | plusepsilon.de |
---|---|---|
location | Uni Hamburg | |
age | 29 | |
visits | member for | 3 years, 11 months |
seen | Sep 25 at 13:24 | |
stats | profile views | 8,279 |
Sep 30 |
awarded | Explainer |
Sep 19 |
comment |
Are there any simple, interesting consequences to motivate the local Langlands correspondence?
@PaulSiegel: I have edited the question. I was probably thinking "global understanding requires local understanding". That is not necessarily so, I guess. |
Sep 19 |
revised |
Are there any simple, interesting consequences to motivate the local Langlands correspondence?
added 24 characters in body |
Aug 17 |
awarded | Nice Question |
Aug 7 |
awarded | Popular Question |
Jul 2 |
awarded | Inquisitive |
Jul 2 |
awarded | Curious |
Jun 30 |
comment |
Eisenstein series over a definite division algebra
Ah okay, I see my mistake. I was thinking $GL(1)$ over a division algebra, you are considering $GL(2)$. Moeglin-Waldspurger do not consider this setting in their book? |
Jun 28 |
awarded | Popular Question |
Jun 27 |
comment |
Eisenstein series over a definite division algebra
Please help me to understand what the cusp is. I recall that quaternion division algebras give rise to compact quotients, hence no cusps and Eisenstein series. The trace formula becomes simple and one gets the Jacquet-Langlands correspondence with forms with at least two square-integrable components. |
Jun 19 |
comment |
Complex zeros of $\zeta'(s)/\zeta(s) + \zeta'(1-s)/\zeta(1-s) $ = simpler expression (except at zeta zeros)
How do cancel the Euler products? Seems like you only got the $\Gamma$ factor. |
Jun 19 |
comment |
Complex zeros of $\zeta'(s)/\zeta(s) + \zeta'(1-s)/\zeta(1-s) $ = simpler expression (except at zeta zeros)
$\xi(s) = \xi(1-s)$, isn't it? What is your definition of $\xi$? |
Jun 14 |
answered | Are there any simple, interesting consequences to motivate the local Langlands correspondence? |
Jun 14 |
comment |
Are there any simple, interesting consequences to motivate the local Langlands correspondence?
Is there a local version of the Taniyama-Shimura conjecture? Also it proves the Ramanujan conjecture in the global setting, but this does not apply to the local steting at all, where actually non-tempered things play a role. |
Jun 12 |
revised |
Regularity assumption in the simple trace formula
added 25 characters in body |
Jun 12 |
answered | Regularity assumption in the simple trace formula |
Jun 11 |
comment |
Analytic criteria for the support of the Plancherel measure for SL(2,Qp), spherical functions
Characteristic functions are not smooth. Growth considerations are pretty analytic to me. For any thing else, there are not that many analytic differences between the characters. It's like $x \mapsto e^{sx}$ for $s$ varying and $s$ being imaginary means temperedness. |
Jun 10 |
answered | Analytic criteria for the support of the Plancherel measure for SL(2,Qp), spherical functions |
Jun 10 |
comment |
Spherical functions for sl(2,Q_p)
I gave the formula for the distribution. It is general. If you want to find a function on $G$ corresponding to this, I can't tell where to look for explicit computations. My feeling is that finding some useful computational analogy can only done to some extent. $P$-adic Gamma factors do not look like real or complex Gamma-functions. |
Jun 10 |
comment |
Spherical functions for sl(2,Q_p)
Being in the left regular correspond to being tempered, so you will not get the trivial representation nor the non-tempered principal series. But it does not matter, my computation assume $\mu$ general, not necessarily unitary. You probably means character seen as a function, which is locally integrable - not as a distribution. Yes you can do a similiar things whether as long as you assume admissible. |