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Let $G$ be an locally compact group $G$, then every irreucible representations $\pi$ is isomorphic to $\omega_{\pi} \otimes \pi'$, where $\omega_{\pi}$ is the central character of $\pi$ and $\pi'$ an irreducible representation with trivial central character, hence it is sufficient to classify, construct, analyse ... only the irreducible representation of $G/Z$.

Let $G$ now be a reductive group over a global field, why is it sufficient to study the representation theory of $G(\mathbb{A})^1$, where $G(\mathbb{A})^1$ is the intersection of all kernels of $x \mapsto |\chi(x) |_{\mathbb{A}}$ for $\chi$ a rational character $\chi$, in order to understand all automorphic representations?

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I don't understand the first sentence of this question. The central character of $\pi$ is not a representation of $G$, it is a 1-dimensional representation of $Z$ that in some cases may not be extendable to a representation of $G$ at all (e.g. if $\pi$ is the 2-dimensional irred representation of the dihedral group $D_8$). The question seems to be saying "we can obviously reduce the representation theory of $G$ to the representation theory of $G/Z$ so why do these automorphic people do it another way?". But I think the argument reducing to $G/Z$ doesn't work. –  Kevin Buzzard Mar 21 '12 at 10:05
    
This is a good argument, I was actually thinking to much in the context say of $GL(n)$, where the above argument works, but I see now that this is special to matrix groups;) Thanks for pointing out. –  plusepsilon.de Mar 21 '12 at 10:27
    
I don't think it's true for $GL(n)$. Can you explain more? I think that what I said above applies just as well to $GL(n)$. You can't split off the centre as a quotient. –  Kevin Buzzard Mar 21 '12 at 11:23
    
I have assumed the direct integral decomposition $L^2(G) = \int^\oplus_{\widehat{Z}} Ind_Z^G \omega d \omega,$ where $\widehat{Z}$ is the Pontryagin dual of $Z$ and $d \omega$ an appropiate chosen Haar measure on $\widehat{Z}$. Now, I thought that I can assume that $ Ind_Z^G 1$ as $C_c(G)$ module is equivalent to $\omega^{-1} \otimes_Z Ind_Z^G \omega$ as a $C_c(Z) \otimes_Z C_c(G/Z)$ module, where I identify $C_c(G)$ and $C_c(Z) \otimes_Z C_c(G/Z)$? Is there still a mistake in the argument? But, I see your point that $ \cdot SL(2,F) \rtimes \\{ \pm 1 \\} \cong GL(2,F)$... –  plusepsilon.de Mar 21 '12 at 11:48
    
I'm afraid I don't know enough analysis to understand what you are doing. But the representation theory of $GL(2,k)$ for $k$ a field of characteristic not two has two "parts" -- those where $-I$ (minus the $2\times 2$ identity) acts as $+1$ and those where it acts as $-1$. You can't twist to move from one to the other, because you twist by $\chi\circ det$ and $det$ is trivial on $-I$. –  Kevin Buzzard Mar 21 '12 at 11:54
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up vote 6 down vote accepted

At least over a number field, there is a section of the map to those norms of characters, so the adele group is a product of $G^1$ with a product of some "rays" $(0,\infty)$ whose representation theory we know.

Indeed, "removing" those rays does remove some "spurious" spectral decomposition mess from the automorphic spectrum.

It is also convenient that the intersection of the adele points of the center with $G^1$, modulo the rational points of the center, is compact, so the corresponding decomposition is discrete.

Edit: what I intended by "spurious" is that the repn theory of $(0,\infty)$ is elementary, and its entanglement with automorphic repn theory is usually inessential, although potentially creating notational trouble, as opposed to serious mathematical issues. A definite disadvantage to $G^1$ is that it is no longer a (colimit of) product(s) of local groups. (Issues about non-unique maximal compacts in p-adic reductive groups are inescapable, but the Iwahori is unique.)

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Thanks for the answer. So $G$ is a semi direct production of $G^1$ by the rays $G^1 \rtimes (0, \infty)^R$, I guess. Relying on the Mackey machine, this would it make necessary to study projective reps in general. Why not here? What do you mean by "spurious"? Do you mean one removes some junk, which is easy to understand, but requires more book keeping? To be honest, I have some doubts that $G^1$ is as good as the choice of $G/Z$, since there is only maximal compact at non archimedean places in $G^1$, whereas there are two in $G/Z$. Both are are necessary for constructing the supercuspidals. –  plusepsilon.de Mar 20 '12 at 9:09
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