Automorphic Forms on product of groups $G\times H$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T08:16:18Z http://mathoverflow.net/feeds/question/86636 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86636/automorphic-forms-on-product-of-groups-g-times-h Automorphic Forms on product of groups $G\times H$ unknown (google) 2012-01-25T15:29:42Z 2012-04-26T20:24:35Z <p>Dear all, I have some difficulty in understanding the notion of automorphic forms on product of groups. </p> <p>Let $G$, $H$ be two reductive groups defined over a number field $F$. Let $\mathcal{A}(G)$ be the space of automorphic forms on $G(F)\backslash G(\mathbb{A})$ and $L^{2}(G):=L^{2}(G(F)\backslash G(\mathbb{A})^{1})$ as in e.g. J. Arthur's papers . When are the following isomorphisms true (as representations)? $$\mathcal{A}(G\times H)\cong\mathcal{A}(G)\otimes\mathcal{A}(H)$$ and $$L^{2}(G\times H)\simeq L^{2}(G)\hat{\otimes}L^{2}(H)$$</p> <p>How to obtain the spectral decomposition of $L^{2}(G\times H)$ (as direct integral of irreducibles) from the spectral decompositon of $L^{2}(G)$ and $L^{2}(H)$ ?</p> <p>The examples I have in mind is $G=GL_{n}$ and $H=GL_{m}$ or $G=SO_{n}$ and $H=SO_{m}$. Are they true in these case?</p> <p>Note that the isomorphisms here are both for representations! So the question may be not so trivial.</p> <p>Thank you so much! Any comments, example or non-example will be welcomed!</p> http://mathoverflow.net/questions/86636/automorphic-forms-on-product-of-groups-g-times-h/86645#86645 Answer by Vita Kala for Automorphic Forms on product of groups $G\times H$ Vita Kala 2012-01-25T17:53:27Z 2012-01-25T17:53:27Z <p>I think that it is true.</p> <p>When you look just at irreducible admissible representations over local fields, it's true by theorem of Flath (Corvallis, part 1, pp 179 - 183) - this is the theorem used in decomposing a global representation into a product of local ones.</p> <p>This of course doesn't answer the global question, but (at least to me) it suggests it should be true. Moreover, if we assume the (highly conjectural) global Langlands correspondence, the corresponding statement about representations of (again conjectural) global Langlands group is obvious, suggesting that the answer should be yes.</p> http://mathoverflow.net/questions/86636/automorphic-forms-on-product-of-groups-g-times-h/86693#86693 Answer by Marc Palm for Automorphic Forms on product of groups $G\times H$ Marc Palm 2012-01-26T06:49:21Z 2012-04-26T20:24:35Z <p>See Theorem 3, page 491 of Barut and Raczka "Theory of Group representations ..."</p> <p>I am not sure, why people downvoted but I refer to a theorem of Mackey: </p> <p>Let $G_1,G_2$ be two separable locally compact groups, and $H_1, H_2$ two closed subgroup, $\pi_1, \pi_2$ two unitary reps of $H_1$ resp. $H_2$, then a result of Mackey yields that</p> <p>$$Ind_{H_1 \times H_2}^{G_1 \times G_2} \left( \pi_1 \otimes \pi_2 \right) \cong Ind_{H_1}^{G_1} \pi_1 \otimes Ind_{H_2}^{G_2} \pi_2.$$</p> <p>Taking $\pi_j$ to be the trivial representation answer the OP's question affirmative in the most general situation.</p> <p>Note that $Ind_H^G 1 = \mathcal{L}^2(H \backslash G)$ by definition, so one gets everything what one needs. The proof goes over two pages, probably it is easier in the presence of type 1 representation, since the direct integral decomposition into irreducible is not unique otherwise, but it does not seem that the reference assumes type 1. Schur's lemma is probably the key in the proof, at least it will be in the analogous statements in the compact group case.</p> http://mathoverflow.net/questions/86636/automorphic-forms-on-product-of-groups-g-times-h/91504#91504 Answer by paul garrett for Automorphic Forms on product of groups $G\times H$ paul garrett 2012-03-18T02:08:38Z 2012-03-20T15:40:40Z <p>Even though it is perhaps not surprising for the applications to repns of reductive Lie groups or reductive adele groups, and of reductive p-adic groups, yes, irreducible unitaries of products $G\times H$ are (completed) tensor products of irreducible unitaries of the factors. However, I think this is not "trivially" true, because it depends on showing that these groups are "type I", meaning that "factor repns" are actually isotypic. Many naturally-occurring groups fail to have this property. This property for p-adic reducitve groups was completely proven (in the supercuspidal case) only as late as 1974, by J. Bernstein. Yes, Harish-Chandra proved type-I-ness for reductive Lie several years earlier.</p> <p>But, then, granting that, the $L^2$ version of your question is affirmative.</p> <p>Probably the more general assertion (about spaces of moderate-growth functions) has an affirmative answer, but I would not know how to prove it quickly.</p> <p>The abstract question about decomposition of irreducible repns of products on larger classes of TVSs is essentially open, I think, even for reductive groups, unless something has happened in the last decade or two.</p> <p>Edit: for examples like theta kernels, it depends partly on how one chooses to define the things. Even in simpler situations, outside of $L^2$ there is a loss of "semi-simplicity", for example, residues of Eisenstein series are quotients with typically uncomplemented kernels. If the Segal-Shale-Weil/oscillator repn is taken to be the <em>unitary</em> one, then since the groups are type I there is a direct integral decomposition into isotypic components (factor repns), on general principles. There is no a-priori guarantee about multiplicites... although many results are known (Howe-conjecture things, first-occurrence stuff due to Kudla-Rallis and others) about the structure.</p> <p>Edit-edit: ... which reminds me of a hazard: for example, with real-anisotropic orthogonal groups larger than the symplectic groups in pairings, the trivial repn of the orthogonal group maps (Siegel-Weil) to a copy of a holomorphic discrete series containing Siegel-type Eisenstein series. The point is that the repn is unitary, but the Eisenstein series is not in automorphic $L^2$. Of course there is no paradox, but there is some risk of saying irrelevant/silly things. A similar minor hazard is already present for non-compact arithmetic quotients, since Eisenstein series enter the spectral decomposition "continuously", are not in $L^2$ individually, but of necessity generate unitary repns "abstractly", as would any "generalized eigenfunction" entering a decomposition of a Hilbert space.</p> <p>Another edit: About Type I... Certainly one direction, that a tensor product of irreducible unitaries of $G,H$ is irreducible unitary, is not difficult. It is the other direction, that an irreducible unitary of $G\times H$ necessarily factors as (completion of) $\pi_1\otimes \pi_2$, that requires something (I'm pretty sure!) In looking at the natural argument to prove such a factorization, at some point one finds that the big repn restricted to $G$ has endomorphism algebra with center consisting only of scalars. <em>If</em> we can conclude that the repn is isotypic (=sum or integral of a single irreducible), then the factorization will succeed. Otherwise, the argument stops, in any case. Alain Robert's book on repns of locally compact groups mentions some not-type-I groups. I do not know enough about them to make a counter-example to a claim that this hypothesis is not necessary. Indeed, conceivably the failure of the proof mechanism does not deny the conclusion...?</p>