Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


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.

share|improve this question
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. –  plusepsilon.de Oct 4 '12 at 12:32
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.