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Let us fix a representation $\pi_\infty$ of GL(n,$\mathbb R$).

Let us fix a character $\chi$ of K, where K is a compact subgroup of $GL(n,\mathbb A_{finite})$.

$$K=\Pi_{v<\infty}K_v$$

$K_v$ is $GL(n,\mathbb Z_v)$ for almost all v.

For automorphic forms of ($\chi$, K), we require that

for $\phi: GL(n,\mathbb {A_Q})\to \mathbb C$

$\phi(x\gamma)=\chi(\gamma)\phi(x)$ for any $x\in GL(n,\mathbb{A_Q})$ and any $\gamma \in K$.

How many cuspidal automorphic representations of GL(n,$\mathbb{A_Q}$) with character $\chi$ and with $\pi_\infty$ at infinity place are there?

I am expecting answer to be "finitely many". Who and where is this proved?

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By "character $\chi$" do you mean "central character $\chi$". If so, then imagine that $n = 2$. By fixing $\pi_{\infty}$, you have (in classical terms) fixed the weight. By fixing $\chi$ you have (in classical terms) fixed the nebentypus. Does this modify your expectation? –  Emerton Oct 10 '11 at 3:15
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Something that is true could be extracted from Harish-Chandra's theorem (see "Automorphic forms on Semisimple Lie Groups", LNM 68) that the space of automorphic forms $A(\Gamma,\xi,J,K)$ is finite dimensional (see, e.g., Borel and Jacquet's article in Corvallis for the definition). With certain choices of $\xi$, $K$, and $J$, this is the space of automorphic forms on $\Gamma\backslash G$ that generate automorphic representations with fixed $\pi_\infty$. If you allow $\Gamma$ to vary, I think you should end up with an infinite-dimensional space ... –  B R Oct 10 '11 at 5:16
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... But I don't think this has been proven in general, and I think it requires a certain amount of effort even for modular forms (calculate then use dimension formulas, courtesy of Reimann-Roch), much less Maass forms (I have no clue what to do), but I could be wrong. –  B R Oct 10 '11 at 5:16
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7-adic, I'm not sure what you mean by an "automorphic representation with character $\chi$ coming from a compact subgroup". Can you elaborate? –  B R Oct 10 '11 at 5:27
    
modifying the original post and elaborating. K is a compact subgroup of $GL(n,\mathbb A_{finite})$. $$K=\Pi_{v<\infty}K_v$$ $K_v$ is $GL(n,\mathbb Z_v)$ for almost all v. For automorphic forms of ($\chi$, K), we require that for $\phi: GL(n,\mathbb {A_Q})\to \mathbb C$ $\phi(x\gamma)=\chi(\gamma)\phi(x)$ for any $x\in GL(n,\mathbb{A_Q})$ and any $\gamma \in K$. –  7-adic Oct 10 '11 at 6:18
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1 Answer

up vote 6 down vote accepted

This is precisely the content of Harish-Chandra's theorem ("Automorphic forms on Semisimple Lie Groups", LNM 68, 1968), proven for general reductive groups:

Fix a finite-dimensional representation $\delta$ of $K_\infty$, an ideal $J$ of finite co-dimension in ${\mathcal Z}({\mathfrak g})$, a compact open subgroup $L$ of $K_{\rm fin}$ (the maximal compact subgroup of $G_{\mathbb A_{\rm fin}})$, and a central character $\omega$, then the space of $K$-finite automorphic forms $f$ with central character $\omega$ and $K_\infty$-type $\delta$ that are right $L$-invariant and annihilated by $J$ is finite dimensional.

To see how this corresponds to your case, fix a representation $\pi_\infty$. For simplicity, assume $\pi_\infty$ is spherical. Thus $\delta$ is the trivial representation. Any vector $\phi$ generating $\pi_\infty$ will be an eigenvector for the Casimir operator, $C\phi=\lambda\phi$, so we take $J$ to be the ideal generated by $(C-\lambda)$. And since $\chi(K)$ is a compact totally disconnected subgroup of ${\mathbb C}^\times$, the kernel of $\chi$ must be a compact-open subgroup $L$, so any $\phi$ will be right $L$-invariant. The only thing left is the central character. The archimedean part is fixed as well as the restriction to a compact open subgroup, so there should only be finitely many options left.

A proof for $SL_2({\mathbb R})$ can be found in Chapter 8 of Borel's "Automorphic forms on $SL_2({\mathbb R})$". The general argument is sketched in section 8 of Borel's contribution to Sarnak and Shahidi's "Automorphic Forms and Applications".

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If memory serves, then another reference (with no proof, of course, but with a discussion of Harish-Chandra's result and how it can be translated into the language of automorphic representations) is in the Borel--Jacquet article "Automorphic forms and automorphic representations" in the first of the Corvalis volumes. –  Emerton Oct 10 '11 at 15:51
    
Emerton, indeed! That's where I first went to find the statement of the theorem (which I then quoted in my original comments). –  B R Oct 10 '11 at 16:36
    
to BR and Emerton, thank you so much! –  7-adic Oct 10 '11 at 19:50
    
7-adic, you are welcome. –  B R Oct 10 '11 at 19:59
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