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Hi,

Let $A$ denote the adeles of $Q$. I know how to decompose $L^2(SL(2,A)/SL(2,Q))$ into irreducible $SL(2,A)$-representations. What is the difference between this decomposition and the corresponding decomposition for $GL(2)$? Can I deduce the $GL(2)$-case from the $SL(2)$-case?

Thanks for answering this basic question.

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Have a look at mathoverflow.net/questions/7059/… –  GH from MO Sep 19 '12 at 19:19
    
See Knapp's article "REPRESENTATIONS OF GL2(R) AND GL2(C)" in the Corvallis proceedings. –  Asaf Sep 20 '12 at 21:41
    
Nobody knows truly how to decompose $L^2(SL(2,A)/SL(2,Q))$ into irreducibles, one only knows how the orthogonal complement of the subspace of cupsidal automorphic forms decomposes into one-dimensional representations and Eisenstein series. –  Marc Palm Oct 4 '12 at 12:32

1 Answer 1

I think GH's link and K.Buzzard summary of Labesse-Langlands explains that there is no easy comparison possible.

But I think at the heart of your question is something else. Are you asking about a decomposition into cuspidal, continuous and residual part? If yes, the decomposition is completly analogous (of course for technical convenience you should rather fix a central character - say trivial - in GL(2)).

You get

1) a direct sum of cuspidal representation (all single multiplicity)

2) a sum over the one-dimensional representations $\chi \circ \det$ with $\chi$ Hecke character and $\chi^2 =1$ resp. for SL(2) only the trivial rep.

3) a direct integral over parabolic induced representation $\chi_1,\chi_2$ with $\chi_j$ Hecke quasi character and $\chi_1 \chi_2 =1$

The proof in Gelbart-Jacquet "Analytic ..." translates easily to this situation.

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