In short: what does Labesse-Langlands say?

Slightly more precise: what are the cuspidal automorphic representations of $SL_2(\mathbf{A}_{\mathbf{Q}})$, together with multiplicities? Let's say that I have a complete list of the cuspidal automorphic representations of $GL_2/\mathbf{Q}$ and I want to try and deduce what is happening for $SL_2$. I am looking for "concrete examples of the phenomena that occur".

Now let me show my ignorance more fully. My understanding from trying to read Labesse-Langlands is the following. The local story looks something like this: if $\pi$ is a smooth irreducible admissible representation of $GL(2,\mathbf{Q}{}_p)$ then its restriction to $SL(2,\mathbf{Q}{}_p)$ is either irreducible, or splits as a direct sum of 2 non-isomorphic representations, or, occasionally, as a direct sum of 4. One interesting case is the unramified principal series with Satake parameter $X^2+c$ for any $c$; this splits into two pieces (if I've understood correctly) (and furthermore these are the only unramified principal series which do not remain irreducible under restriction). Does precisely one of these pieces have an $SL(2,\mathbf{Z}{}_p)$-fixed vector? And does the other one have a fixed vector for the other hyperspecial max compact (more precisely, for a hyperspecial in the other conj class)? Have I got this right?

Packets: Local $L$-packets are precisely the J-H factors for $SL(2,\mathbf{Q}{}_p)$ showing up in an irreducible $GL(2,\mathbf{Q}{}_p)$-representation. So they have size 1, 2, or 4.

Now globally. a global $L$-packet is a restricted product of local $L$-packets (all but finitely many of the components had better have an invariant vector under our fixed hyperspecial max compact coming from a global integral model). Note that global automorphic $L$-packets might be infinite (because $a_p$ can be 0 for infinitely many $p$ in the modular form case). My understanding is that it is generally the case that one element of a global $L$-packet is automorphic if and only if all of them are, and in this case, again, generally, each one shows up in the automorphic forms with the same multiplicity. What is this multiplicity? Does it depend?

Finally, my understanding is that the above principle (multiplicities all being equal) fails precisely when $\pi$ is induced from a grossencharacter on a quadratic extension of $\mathbf{Q}$. In this case there seems to be error terms in [LL]. Can someone explain an explicit example where they can say precisely which elements of the packet are automorphic, and what the multiplicities which which the automorphic representations occur in the space of cusp forms?

I find it very tough reading papers of Langlands. My instinct usually would be to press on and try and work out some examples myself (which is no doubt what I'll do anyway), but I thought I'd ask here first to see what happens (I know from experience that there's a non-zero chance that someone will point me to a website containing 10 lectures on Labesse-Langlands...)

Edit: I guess that there's no reason why I shouldn't replace "cuspidal" by "lies in the discrete series" with the above (in the sense that the questions then still seem to make sense, I still understand everything (in some sense) for $GL_2$ and I still don't know the answers for $SL_2$)

  • $\begingroup$ So I am genuinely surprised and pleased that I could ask for an overview of such a technical area and get such helpful responses within a day or two. Thanks everyone. $\endgroup$ – Kevin Buzzard Nov 29 '09 at 23:11

Warning: not an expert, so could be major mistakes in this.

The multiplicity is one for every element in the global packet, in the non-CM case. In the CM case, half of the packet has multiplicity one and the other half has multiplicity zero.

For the general story, I'd suggest looking at Arthur's conjectures, which at least conjecturally give a very nice picture. In general one should talk of Arthur packets, not Langlands packets. They coincide in your case.

For CM representations on $SL_2$, the associated global parameter has dihedral image inside $PGL_2(\mathbb{C})$. Its centralizer $S$ has size $2$. According to Arthur, the obstruction to an element of the global packet being automorphic is valued in the dual of $S$; thus, "one-half" is automorphic. More precisely, the set of irreducible representations in a global $L$-packet is a principal homogeneous space for a certain product of $Z/2Z$s (in the obvious way: one $Z/2Z$ for each $p$ where $a_p = 0$; for simplicity, suppose that local multiplicity $4$ doesn't occur). The automorphic ones correspond exactly to one of the fibers of the summation map to Z/2Z. Thus, if you take an automorphic representation in this packet, and switch it at one place to the other element of the local packet, it won't be automorphic any more.

Concretely: Start with $\Pi$ an automorphic representation for $GL_2$, it maps by restriction to the space of automorphic forms for $SL_2$. The above remarks suggest that this restriction map must have a huge kernel when $\Pi$ is CM. But one can almost see this by hand: $\Pi$ is isomorphic to its twist by a certain quadratic character $\omega$. This, and multiplicity one for $GL_2$, means that, for $f \in \Pi$, the function $g \mapsto f(g) \omega(\det g)$ also belongs to $\Pi$. This forces extra vanishing, although I didn't work out the details: For instance, if $\omega$ is everywhere unramified and $f$ the spherical vector, then $f(g) \omega (\det g)$ must be proportional to $f$, so $f$ vanishes whenever $\omega(\det g)$ is not $1$, i.e. various translates of $f$ do vanish when restricted to $SL_2$.

Remark/warning: If you are concerned with multiplicity one, there is a further totally distinct phenomenon that causes it to fail for $\mathrm{SL}_n, n \geq 3$: Two non-conjugate homomorphisms of a finite group into $\mathrm{PGL}_n$ can be conjugate element by element. There's a paper of Blasius about this. It has nothing to do with packets.

  • $\begingroup$ Thanks. Yes I know about the "other" $SL_n$ phenomenon: there is a paper by Erez Lapid that takes Blasius' work even further. What is irking me a little about $SL_2$ is that my gut feeling is that this "not every element of the packet is automorphic" phenomenon should perhaps manifest itself in a completely classical way when considering cuspidal modular forms with only the Hecke operators coming from SL_2: the CM forms should perhaps behave differently, But somehow I don't see it. Maybe I am still not quite seeing what's truly happening. Thanks for the very informative answer though. $\endgroup$ – Kevin Buzzard Nov 29 '09 at 22:54
  • $\begingroup$ I agree with you, but couldn't see it either; the "vanishing" was the best I could do. Here's an interesting case which might be helpful to think about: Consider $SL_2$ over a real quadratic field $F$. It gives us some complex two-dimensional (Hilbert modular) variety. In this case, an elliptic curve over $F$ would contribute four dimensions to the $H^2$, one dimension in Hodge numbers $(0,2),(2,0)$ and two dimensions in Hodge number $(1,1)$. A CM elliptic curve only contributes half of this. What's going on in direct terms? I couldn't see this. $\endgroup$ – moonface Nov 30 '09 at 0:53

The multiplicity of any cusp form in the spectrum is one; this follows from the analysis in LL and the results proven in Ramakrishnan's paper "Modularity of the Rankin-Selberg L-series and multiplicity one for SL2," in vol. 152 of Annals. However this certainly doesn't answer your more subtle question about L-packets.

  • $\begingroup$ Thanks. This doesn't surprise me, but in some sense my underlying confusion remains. I only stayed away from non-cuspidal forms just to try and simplify things (I don't even know if the notion of multiplicity makes sense in the continuous part of the spectrum anyway, and suspect it doesn't). So is the issue highlighted in LL that there are packets such that some, but not all, of the elements are automorphic? Do any of these gadgets correspond to holomorphic cuspidal modular forms in the more classical sense? And if so, can I see this phenomenon from a more classical point of view? $\endgroup$ – Kevin Buzzard Nov 28 '09 at 18:11

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