Let $\mathbb{G}$ be a reductive group defined over a number field $K$, let $Z$ be its center, and let $\mathbb{A}:=\mathbb{A}_K$ be the ring of adeles of $K$. Reasonably, we care about the $\mathbb{G}(\mathbb{A})$-representation: $L^2(\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A}))$. It naturally contains the sub-representations $$L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash\mathbb{G}(\mathbb{A}),\omega):=\{f\in L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash\mathbb{G}(\mathbb{A}))|\,\,\,|f|\in L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A})), \forall z\in Z(\mathbb{A}), g \in \mathbb{G}(\mathbb{A})\,\,\, f(zg)=\omega(z)f(g)\} $$

for every $\omega$ a unitary character of $Z(\mathbb{A})$. In fact $L^2(\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A}))$ is the direct integral of these subrepresentations.

I understand that it is generally desirable to deal with $L^2(\mathbb{G}(K)\backslash \mathbb{G}(\mathbb{A}))$ by decomposing it into the cuspidal part, which is going to be discrete, and the Eisenstein part, which is (I think!) continuous. In order to define this cuspidal part, people define $L^2_0(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G} (\mathbb{A}),\omega)$ to be the subrepresentation of $L^2(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G} (\mathbb{A}),\omega)$ of all of the functions $f$ such that for every $K$-parabolic subgroup $\mathbb{P}$ of $\mathbb{G}$, whose unipotent radical we will call $N$, satisfies that for almost all $g\in\mathbb{G}(\mathbb{A})$ the integral $\int_{N(K)\backslash N(\mathbb{A})} f(gn)dn$ is $0$.

The definition of a cuspidal representation is then an irreducible unitary subrepresentation of $L^2_0(Z(\mathbb{A})\mathbb{G}(K)\backslash \mathbb{G} (\mathbb{A}), \omega)$ for some central character $\omega$.

I feel that I really do not understand the intuition behind the condition with the parabolic subgroups. Parabolic subgroups and their unipotent radicals seem like very formal constructions to me, but I bet there is some geometric intuition that I'm missing. Is there some geometry that should be in the back of my mind that explains the condition $\int_{N(K)\backslash N(\mathbb{A})} f(gn)dn=0$? How does this condition relate to being zeros at the cusps via the classic definition of cusp-forms?


In addition to Paul Garrett's answer, I address your last paragraph in a special example:

Strong approximation gives a homeomorphism $SL_2(Z) \backslash H \cong Z(A) GL_2(Q) \backslash GL_2(A) / \prod_p GL_2(Z_p) \times O(2)$.

Lets $f$ corresponds to $\tilde{f}$. This translates

$$ \int_{0}^1 f( y + t)\; d t = \int\limits_{N(F)\backslash N(A)} \tilde{f}(ng_y)\; dn.$$

This implies that the zero-th Fourier coefficient vanishes, but that does not imply that the functions vanish at the cusps for Maass forms. It only does for modular forms.

  • $\begingroup$ Hi. This is exactly the type of thing I'm looking for. I'm having a hard time parsing your statement, though. What is $N$ in your notation. Is it the unipotent radical of some Borel? Of every? What is a reference for this kind of intuition? $\endgroup$ – Jonas Aherne Apr 10 '14 at 20:00
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    $\begingroup$ Yes the unipotent radiacal of any Borel subgroup (defined over $F$). Note that your are allowed to conjugate by elements of $GL_2(F)$. There are may expositions on how to move between classical and adelic language, see e.g. Bump's book. $\endgroup$ – Marc Palm Apr 10 '14 at 20:30
  • $\begingroup$ Sorry, this is silly, but when you say "any" do you mean "some" or "every"? (or "some and therefore every"?) I always find that word ambiguous in mathematical contexts... $\endgroup$ – Jonas Aherne Apr 10 '14 at 23:04
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    $\begingroup$ "some and therefore every" $\endgroup$ – Marc Palm Apr 11 '14 at 8:52

First, one should be a little careful about saying that $L^2(G_k\backslash G_\mathbb A)$ has $L^2(Z_\mathbb A G_k\backslash G_\mathbb A,\omega)$ inside it... since appearing as direct integral "integrands" is not a very strong commitment. If $G$ has non-compact center, $L^2(G_k\backslash G_\mathbb A)$ will have no discrete spectrum at all... which gives the wrong impression (by Selberg et al's proof of various forms of "Weyl's Law" for arithmetic subgroups, namely, that the bulk of the spectrum is cuspidal, hence, discrete).

That Gelfand condition about integrals over all unipotent radicals being $0$ is far from obviously "the right thing". It is less a leap to understand that instead of "cusps" we should think of "parabolics" (etc).

The constant terms along various parabolics correspond to "going to infinity" in the variety of fashions possible in general, in higher-rank groups.

Still, yes, it is mildly amazing that vanishing of all constant terms guarantees discreteness. In general, it is easy to fail to prove this... :)

I think that Y. Colin de Verdiere's argument, cast into general form by Jacquet, that appears in Moeglin-Waldspurger's book, is potentially the clearest in terms of describing the causality, as it shows a somewhat more general thing, that square-integrable automorphic forms ($K$-finite, $\mathfrak z$-finite) all of whose various constant terms vanish above some fixed height(s), is already discrete. This follows by proving that the resolvent for Casimir on such a space is compact, which follows by proving a sort of Rellich compactness lemma for an inclusion of a Sobolev $H^1$ into $H^o=L^2$, as appears in Lax-Phillips' book on Automorphic Scattering, about page 204 and following. (The earlier parts are not essential to understanding what's happening just there.)

The rough explanation I've heard, and sometimes repeat, although it doesn't truly explain so much, is that (eventual) vanishing of all constant terms says that the given afm has $0$ "average mass" "at infinity", so that it behaves as though it lived on a compact manifold, where a simpler Rellich lemma would apply (by a smooth partition of unity, and reducing to the essentially elementary case of a product of circles, and Fourier series).

The historical version of "holomorphic cuspform" (also in the Siegel and Hilbert modular cases) played on some good fortune, in some regards. If it seems lucky, you're probably right.

EDIT: in response to comment/query... No, there is no general rubric that says that parabolic subgroups determine spectral features of automorphic forms. Plausibly, in a different universe, the stratification of automorphic $L^2$ could be different. Thus, although the rational-rank-one case was relatively easy to (optimistically) extrapolate from the $SL_2(\mathbb R)$ case, where the cross-sections going out to the point-cusps were elementary objects, all the pseudo-down-to-earth ideas about "going to infinity" and "cusp" that seemed to be decisive for elliptic modular forms, and for Maass' waveforms, rather abruptly not only "fail", but fail qualitatively.

Thus (to my mind) Langlands' earning his spot at IAS in the 1960s, for, among other things, carrying out Selberg's highly-optimistic sketch of automorphic spectral decompositions. A number of important, critical surprises: there're not just two sorts of things, cuspforms and "continuous spectrum", and then a little leftover, constants, but a whole range of things. Yes, as has only been proven in recent years, the discrete spectrum dominates, and the discrete spectrum is dominated by cuspforms. But, first, there are cuspidal-data Eisenstein series, apparently not anticipated by Selberg. But, as is the subtlest part of Langlands' SLN 544, and addressed completely only for $GL_n$, in Moeglin-Waldspurger's 1989 paper, there are many non-constant $L^2$ residues of Eisenstein series, for $GL_n$ at least called Speh forms, because Birgit Speh discovered the corresponding repns of real Lie groups $GL_n(\mathbb R)$.

The "constant term" along a parabolic $P$ is the trivial-character Fourier component along the unipotent radical of $P$. It is not obvious that this shadow of the thing should be important, but, yes, one proves... "the theory of the constant term"... that the aggregate of the constant terms of a $K$-finite, $\mathfrak z$-finite automorphic form determines its asymptotic behavior "at infinity".

Indeed, decompositions along other subgroups are very interesting, especially for number theoretic applications.

However, as it happens, it seems that no other decompositions-along-subgroups adequately distinguish non-compact quotients from compact... and compact quotients have discrete spectrum with respect to their invariant Laplacian, or Laplacian on suitable vector bundles.

Perhaps in a different universe the non-compactness of interesting arithmetic quotients would have been mediated by different sorts of subgroups, but in this universe the collection of all parabolics seems to do the job. Yes, this was not obvious, and Gelfand deserves substantial credit for formulating things this way...

  • $\begingroup$ Hi Paul. Thanks for your answer! I actually asked my question right after reading your short note "Automorphic Representations and L-functions", which I found to be very clear! Since I am not immersed in the culture yet, can you explain what you mean by "constant terms along various parabolics"? I still don't understand the geometric picture corresponding to parabolics (as opposed to, say, other subgroups of $\mathbb{G}$), and I also don't understand where the phrase "constant terms" comes from... $\endgroup$ – Jonas Aherne Apr 7 '14 at 1:06
  • $\begingroup$ Hi @Jonas, I'll respond at length later in the day tomorrow. $\endgroup$ – paul garrett Apr 7 '14 at 1:18
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    $\begingroup$ Okay, thanks! I really appreciate it! $\endgroup$ – Jonas Aherne Apr 7 '14 at 1:39

I think the simplest answer is this. Every reductive group $G$ comes together with a bunch of smaller reductive groups - the Levi subgroups (they are smaller in the sense that their semi-simple rank is smaller). Now, you given such a subgroup $M$ you have a way to construct representations of $G$ out of representations of $M$ (in the context of automorphic forms this procedure is called Eisenstein series). However, not all representations of $G$ will appear in this way (for strictly smaller $M$) -- cuspidal representations are exactly those "which have nothing to do" with smaller Levi subgroups (so you can't construct them out of something simpler).


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