Let $G$ be a connected reductive group over a global field $k$, and let $\pi=\otimes_w\pi_w$ and $\pi'=\otimes_w\pi'_w$ be two automorphic representations for $G$, where here of course $w$ is ranging over all the places of $k$.

Assume now that $\pi_w\cong\pi'_w$ for all but finitely many places $w$ of $k$. I think people say "$\pi$ and $\pi'$ are nearly equivalent".

If ($G=GL(n)$ and $\pi$ is cuspidal), or if ($G=GL(n)$ and $\pi$ and $\pi'$ occur discretely in $L^2$), then this would force $\pi_v\cong\pi'_v$ for all places $v$. But in general this "strong multiplicity one" phenomenon does not occur. Indeed even if $G=GL(2)$ we can have $\pi_v\not\cong\pi'_v$ for a non-zero finite set of $v$: if $\pi$ is 1-dimensional then $\pi'$ can be Steinberg at $v$, for example. For groups other than $GL(2)$ we can even have $\pi$ and $\pi'$ cuspidal, with $\pi_v\not\cong\pi'_v$ -- this even happens if $G=SL(2)$: "strong multiplicity one" can fail here.

So here's a vague question. We've established that $\pi$ and $\pi'$ nearly equivalent does not imply $\pi_v\cong\pi'_v$ for all $v$. But can we say anything about the relationship between $\pi_v$ and $\pi'_v$?

But I am not a fan of vague questions so here are some more precise ones, together with some guesses for answers. Say $\pi_w\cong\pi'_w$ for almost all $w$, but $\pi_v\not\cong\pi'_v$.

0) Do $\pi_v$ and $\pi'_v$ necesarily have the same central character? [this should be an easy warm-up. It's just the question of whether tori satisfy some sort of strong mult 1. I feel a bit lame not being able to figure this out :-/]

1) Are $\pi_v$ and $\pi'_v$ necessarily in the same Bernstein component? [my guess is "no"; I half-suspect that for $G=GSp(4)$ one can have $\pi_v$ supercuspidal and $\pi'_v$ not, but my source is "I think someone once told me this" and it would be nice to have a more concrete one].

2) If $v$ is infinite, do $\pi_v$ and $\pi'_v$ have the same infinitesimal character? [My guess is "this is known for $GL(n)$, and might follow from a super-optimistic version of Langlands functoriality for general $G$ but perhaps I am being a bit too optimistic."]

3) If $v$ is infinite, are $\pi_v$ and $\pi'_v$ in the same local $L$-packet as defined by Langlands? [I have very little understanding of local $L$-packets at infinity and daren't hazard a guess.]

4) Back to general $v$. Should one expect that $\pi_v$ and $\pi'_v$ are in the same "packet" in some way? I write this in quotes because I don't know that I can give a definition of $L$-packet or $A$-packet in this generality. So here I daren't even have an opinion.

I'd be interested to know in anything that is proved or conjectured.

  • $\begingroup$ 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? $\endgroup$ Sep 6, 2011 at 16:25
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    $\begingroup$ (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). $\endgroup$ Sep 6, 2011 at 16:29
  • $\begingroup$ 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...) $\endgroup$ Sep 6, 2011 at 16:35
  • $\begingroup$ 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. $\endgroup$ Sep 6, 2011 at 18:05
  • $\begingroup$ 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". $\endgroup$ Sep 6, 2011 at 18:10

4 Answers 4


I confirm what "someone once told you" about question 1 (so now "two people once told you" or perhaps "someone twice told you"). This phenomenon ($\pi_\nu$ supercuspidal, and $\pi'_\nu$ principal series, even unramified) occurs for example when $\pi$ and $\pi'$ are the non-tempered endoscopic representation in the discrete (or even cuspidal) spectrum that some people like to deform, for $U(3)$ and $GSP(4)$ and their inner forms. There is an article by Rogawski, "The multiplicity formula for A-packets" in the book "the Zeta Function of Picard Modular Surfaces", where he describes in details such an example for each of the two inner forms of $U(3)$ attached to a quadratic imaginary field $E$. You have analog examples for $GSP_4$, where $\pi$ is a Saito-Kurokawa lift of a modular form.

  • $\begingroup$ Thanks Joel. I think it was Kai-Wen who told me before, so it really is two people now, so it must be true. Automorphic representations are far more crazy than I had realised. Do you know where I can read about the Sp_4 examples? I can construct the Saito-Kurokawa $\pi$'s -- this is mentioned in a Langlands Corvallis article. Maybe I should look there next. $\endgroup$ Sep 9, 2011 at 7:09

Regarding relationships between 0), 2), 3), and 4) (but not really an answer to any of them without additional insights) :

a) It's predicted (see Borel's Corvallis article page 44) that if two representations belong to the same local "$L$-packet", then they have the same central character.

b) Let $G(\mathbb{R})$ be a real group that has relative discrete series representations. Fix a central character $\chi$ and an infinitesimal character $\lambda$. Then the set of relative discrete series representations of $G(\mathbb{R})$ with central character $\chi$ and infinitesimal character $\lambda$ is an $L$-packet of $G(\mathbb{R})$, and every relative discrete series $L$-packet of $G(\mathbb{R})$ is of this form.

c) Icing on whatever cake exists : For explicit constructions of central characters associated to Langlands parameters : Given a discrete Langlands parameter for a real group $G(\mathbb{R})$, Langlands has constructed the central character of the associated $L$-packet. If $G(F)$ is $p$-adic such that the maximal torus in its center is anisotropic, Gross/Reeder (Section 8 of their recent paper for this, which also contains Langlands' above construction for real groups) have constructed a central character attached to a discrete Langlands parameter for $G(F)$. fwiw : Borel (in Corvallis) gave a construction of a central character attached to a Langlands parameter of an arbitrary connected reductive group over a real or $p$-adic field $k$, but it is noted in the first paragraph of Section 8 Gross/Reeder that Borel's construction "omits an essential point, namely the vanishing of the Schur multiplier of $Gal(\overline{k} / k)$, due to Tate".

  • $\begingroup$ Thanks for these comments. There is more than one Gross/Reader paper. I am guessing you mean "Arithmetic invariants of discrete Langlands parameters". $\endgroup$ Sep 7, 2011 at 11:18
  • $\begingroup$ Yes, thank you, the "Arithmetic invariants..." paper. $\endgroup$ Sep 7, 2011 at 13:46

They are in the same Arthur packet. The A-packets are designed exactly to answer this question. I think the answers to questions 1,2,3 are "no", "yes conjecturally", and "no": see book of Adams Barbasch Vogan or Vogan's survey article on the local Langlands conjectures. Note that, although the A-packet is not a union of L-packets in general, it at least contains a canonical L-packet.

[Oops: answer edited. This only applies to "automorphic" in the sense of "appearing discretely in the L^2-spectrum". Misunderstood the focus of question. Explicit reference for 1 is e.g. 1.11 of Gan-Gurevich "Non-tempered A-packets of G_2: liftings from SL_2-tilde."]

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    $\begingroup$ I thought that Arthur packets somehow didn't cover all representations and one had to put some sort of restriction on the automorphic representations one considered. Let me say that I know zilch about A-packets. Do I have to restrict to some subset of $\pi$s before A-packet stuff applies? $\endgroup$ Sep 6, 2011 at 18:21
  • $\begingroup$ PS I own A-B-V. Can you be more precise about a reference? Or give an explicit $G$ for which (1) and (3) should fail? Is $PGSp(4)$ a good candidate? $\endgroup$ Sep 6, 2011 at 18:23
  • $\begingroup$ Just to reiterate Kevin's question from his first comment: Are you restricting to some particular $\pi$s here? For example, already in the case of $GL_2$, in the non-cuspidal spectrum we have reps. which are nearly equivalent but not isomorphic (as Kevin notes), and the A-packets are singletons in this case. E.g. are you restricting to automorphic reps. in the L^2-spectrum? Regards, Matthew $\endgroup$
    – Emerton
    Sep 6, 2011 at 18:31
  • $\begingroup$ Aah yes of course Matt -- (3) is obviously false in general because it fails already for $GL(2)$. $\endgroup$ Sep 6, 2011 at 18:52

For what it's worth I can now answer Q0. I believe it is not true in general that $\pi_v$ and $\pi'_v$ will have to have the same central character. We can let $G$ be a torus $T$. If $S$ is a finite set of places of $k$ then we can set $\pi_w=\pi'_w=1$ for $w\not\in S$. In fact let's set $\pi'=1$. Can we find $\pi$ with $\pi\not=1$? This just boils down to the question of whether $T(k)$ is dense in $\prod_{v\in S}T(k_v)$. This is "bien connu" not to be true in general, according to the first paragraph of this paper of Colliot-Thelene and Suresh available here:

Skorobogatov tells me that Milne's Arithmetic Duality Theorems is the place to look for this well-known stuff. Thanks to Ambrus Pal for pointing out the C-T--S paper.

But even better -- the paper linked to above even answers, in the function field case, the stronger question of whether the induced map $T(k)\to\prod_{v\in S}T(k_v)/M$ is surjective, where $M$ is the maximal compact subgroup -- even this may fail. This means, I believe, that even tori give examples where $\pi_v$ and $\pi'_v$ are not in the same Bernstein component, so we get another counterexample to (1). Note however that it only applies in the function field setting (this is where the C-T--S counterexample takes place).

  • $\begingroup$ In Example 3 on pg. 122 of Algebraic Groups and their Birational Invariants by Voskresenskii, there is an explicit description of a torus which does not satisfy "weak approximation." The example is $T=R^{(1)}_{L/\mathbb{Q}}(\mathbb{G}_m)$, where $R^{(1)}$ denotes the norm 1 elements of $\mathrm{Res}_{L/\mathbb{Q}}$, and $L$ is any biquadratic extension in which some prime has no splitting. $\endgroup$ Sep 9, 2011 at 21:02
  • $\begingroup$ Thanks -- this sounds like another reference for the "well-known" part. What does "some prime has no splitting" mean? $\endgroup$ Sep 10, 2011 at 8:02
  • $\begingroup$ Sorry, I'm not sure what to call it. I mean that the field can be written as $L>K>\mathbb{Q}$, where $p$ is totally ramified in $K/\mathbb{Q}$ and inert in $L/K$. $\endgroup$ Sep 10, 2011 at 8:38
  • $\begingroup$ Oh -- you mean there's just one prime above $p$. Thanks for the clarification! $\endgroup$ Sep 10, 2011 at 15:12
  • $\begingroup$ Heh, yes I suppose that is an easier way of saying it. $\endgroup$ Sep 12, 2011 at 20:11

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