Timeline for How badly can strong multiplicity one fail in the theory of automorphic representations?
Current License: CC BY-SA 3.0
12 events
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| Jun 25 at 10:32 | comment | added | David Corwin | This more recent paper msp.org/pjm/2016/285-2/pjm-v285-n2-p06-s.pdf says a bit more about what happens in the $\mathrm{SL}_2$ case: It says that "if we require further that a pair of cuspidal representations $\pi$ and $\pi'$ of $\mathrm{SL}_2$ have the same local components at the archimedean places and the places above $2$, and they are generic with respect to the same additive character, then they also satisfy the strong multiplicity one property. | |
| Sep 9, 2011 at 1:32 | answer | added | Joël | timeline score: 5 | |
| Sep 8, 2011 at 16:34 | answer | added | Kevin Buzzard | timeline score: 1 | |
| Sep 6, 2011 at 21:08 | answer | added | Moshe Adrian | timeline score: 4 | |
| Sep 6, 2011 at 18:14 | answer | added | anonymous | timeline score: 1 | |
| Sep 6, 2011 at 18:10 | comment | added | Kevin Buzzard | (by which I mean "not in general") | |
| Sep 6, 2011 at 18:10 | comment | added | Kevin Buzzard | Re: Q0. It is just an extremely obfuscated way of asking the following question: "if $T$ is a torus over a global field $k$ and if $S$ is a finite set of places of $k$, is $T(k)$ dense in $\prod_{v\in S}T(k_v)$?". The answer to this is surely "no". | |
| Sep 6, 2011 at 18:05 | comment | added | Kevin Buzzard | David -- I'm sure that unramified Frobenii will generate a Langlands group. That's not the issue. The issue is that if I have two homomorphisms $\rho_1$ and $\rho_2$ of $L$ into an arbitrary group $X$, and if we know $\rho_1(g)$ and $\rho_2(g)$ are conjugate for a dense set of $g$, we cannot conclude that $\rho_1$ and $\rho_2$ are conjugate, or even (in situations where this makes sense) that their semisimplifications are conjugate. This fails already for $X=PGL(n)$. Hence now if $D$ is a random subgroup of $L$ (a local decomp group) we can't deduce that $\rho_1|D$ and $\rho_2|D$ are conj. | |
| Sep 6, 2011 at 16:35 | comment | added | David Hansen | Thanks! I wonder if some "purely automorphic" analogue of Chebotarev can be expected in the setting of the automorphic Langlands group $L_F$, namely whether the image of $L_{F_v}$ for a density one set of $v$'s topologically generates $L_F$. (Of course this is predicated on defining $L_F$, so I suppose I am working with Arthur's extremely conditional definition...) | |
| Sep 6, 2011 at 16:29 | comment | added | Kevin Buzzard | (1) I am in a far more general situation than one where we get Galois representations. (2) Even if we get Galois representations, one cannot use Cebotarev! It only applies when the Galois representation is into GL_n, i.e. when $G=GL(n)$. For general $G$ all sorts of funny things can happen, because local conjugacy does not imply global gonjugacy. (3) For non-cuspidal $\pi$, local-global is in some sense not even true -- it's only true up to some sort of semisimplification. For example strong mult 1 fails for non-cuspidal reps of GL(2). | |
| Sep 6, 2011 at 16:25 | comment | added | David Hansen | Re. 4: If $\pi$ and $\pi'$ both give Galois representations, and $\pi_v \simeq \pi'_v$ at a density one set of places, then the Galois reps. agree by Chebotarev, so doesn't this force $\pi_v, \pi'_v$ into the same L-packet at all finite places? I am sure you know this style of argument, so am I missing something here? | |
| Sep 6, 2011 at 16:00 | history | asked | Kevin Buzzard | CC BY-SA 3.0 |