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Two questions concerning the decomposition of $L_0^2(GL_2({\mathbb{Q}}) \backslash GL_2(A), \psi)$, where $\psi$ is a Hecke character on the adelic ring $A$:

  1. It is known that when $\psi$ is trivial on $\mathbb{R}^\times_{+}$, there is an explicit correspondence between classical modular forms and irreducible constituents of the regular representation of $L_0^2(GL_2({\mathbb{Q}}) \backslash, GL_2(A), \psi)$. Does similar correspondence (with other generalized automorphic forms) exist when $\psi$ is non-trivial on $\mathbb{R}^\times_{+}$?

  2. What is the role play by Maass forms in such a decomposition? Can we also construct a correspondence just as we did for modular forms?

Thank you!

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For 1. I am a bit confused. Classical cuspidal eigenforms (not modular forms) give rise to irreducible subrepresentations of the space you're interested in, but there are other irreducible subrepresentations not coming from classical eigenforms -- namely those coming from Maass forms. Does this answer 2. for you? For 1. why can't you just twist to reduce to the situation where $\psi$ is trivial on the positive reals? –  Kevin Buzzard Jun 6 '11 at 18:45
    
Thank you for your response, I didn't think of twisting the character. –  Hsueh-Yung Lin Jun 6 '11 at 20:20
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2 Answers

up vote 11 down vote accepted

Originally, I wrote: Assuming that I understand the intent of the question properly, it is intended that psi be a Hecke character on the center of GL(2, adeles). The question seems to grant that we understand the situation with non-trivial Hecke characters subject only to the requirement that they be trivial on the "ray" ("the non-compact part" of the idele class group) denoted R-times-sub-plus. Then all other cases can be reduced to this by tensoring with characters of the form |det|^{it} for suitable real t. (Note that for L^2 to make sense the central character should be unitary.)

That is, nothing really new comes up.

Edit: Upon further consideration, I think that perhaps the question had some implicit questions in it, which can be answered:

The whole space of L^2 automorphic forms, even with trivial central character, includes many things that are not immediately classical holomorphic automorphic/modular forms, nor Maass/wave-forms. The (relatively easy) structure theory of representations of GL(2,R) or SL(2,R) shows that a Gamma-invariant K-finite Casimir-Laplacian eigenfunctions is either a holomorphic/anti-holomorphic automorphic form, a Maass waveform, or... significantly... a Lie algebra derivative of one of those.

The central character issue is somewhat secondary to this classification, which itself is not so hard.

In particular, perhaps we have been lucky that historical events presented us with "vectors" which generate all the relevant representations.

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Thank you for your answer. –  Hsueh-Yung Lin Jun 6 '11 at 20:17
    
Dear Paul Garrett, I do not understand what you mean by the last sentence. Could you please elaborate how would you generate all the Maass forms by concrete "vectors". What do you mean by relevant representations? –  plusepsilon.de Jun 7 '11 at 7:11
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@pm: If you apply the Maass raising and lowering operators on a Maass cusp form, and you lift these to SL(2,R) appropriately, then you obtain an orthonormal basis of an automorphic representation: these are exactly those vectors (up to scalar multiple) in the representation which transform by a character under SO(2,R). By "relevant representations" Paul meant that the automorphic representations so obtained from classical modular forms (holomorphic, Maass, Eisenstein) exhaust all infinite-dimensional automorphic representations of SL(2,R), we don't miss anything. –  GH from MO Jun 8 '11 at 23:03
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Perhaps it should be noted that this is a more genral phenomena:

The Schur lemma tells you that the restriction of an irreducible representation of a locally compact group $G$ to the centrum $Z$ is always isomorphic to a character.

Assummin fixed central character, is a technical requirement, but important since $GL_2(F) \backslash GL_2(A)$ has not finite quotient measure, but $GL_2(F) Z(A) \backslash GL_2(A)$ does. Alternatively, you can look $GL_2(A)^1$, i.e. the kernel $g \mapsto | \det g|_A$, since $GL_2(F) \backslash GL_2(A)^1$ has finite volume as well.

This is a necessary point, if you want to analyse the representations via the Arthur trace formula, which does make sense for finite volume quotients only.

Maass (cusp) forms and modular (cusp) forms give both (cuspidal) automorphic representations and the cuspidal ones give vectors of $L_0$. The main point is the observation: $$ GL_2(\mathbb{Q}) z(A) \backslash GL_2(A) /\prod_p GL_2(\mathbb{Z}_p) O(2)$$ $$= PSL_2( \mathbb{Z}) \backslash \mathbb{H}.$$ An explanation, how to see this, is in Gelbarts "Adeles Groups ... " book, if I recall it correctly,and there is an article of Kudla in Cogdell et al. "Introduction to Langlands program". I think you should consider these only as different objects, if you want argue with algebraic geometry, which applies to holomorphic stuff only, but not if you want to do representation theory. Edit due to the comments: Maass forms and modular forms seen as representations look different at the archimedean primes, i.e. have a "different" representation of $GL_2(\mathbb{R})$ there, i.e. prinicpal series vs. discrete series.

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Dear pm, There are significant differences between Masss forms and modular forms from a rep'n theoretic perspective as well, namely the very different behavious of their local factors at $\infty$. Regards, Matthew –  Emerton Jun 7 '11 at 7:52
    
Dear Matthew, you mean that the gamma factors look different and hence the analytic conductor is different? –  plusepsilon.de Jun 7 '11 at 7:57
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I believe Matthew is referring to the fact that the $(\mathfrak{g},K)$-module (i.e. the infinite component) of the automorphic representation attached to a cusp form is a discrete series representation (if $k\geq2$, limit of discrete series for $k=1$), whereas, that for a Maass form is a principal series. These are two quite different objects from a representation theory point of view. –  Rob Harron Jun 7 '11 at 13:36
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