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Fix a global field $F$. Does every automorphic representation of $GL(n)$ appear as an arbitrary twists in the continuous spectrum of $GL(m)$, $m>n$?

What happens for the automorphic representations of $G$ reductive, do they appear in $GL(n)$ for $n$ big?

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2 Answers 2

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In general, I don't think that a representation of $GL(n)$ can be viewed directly as a representation of $GL(m)$ (for $m > n$), so it doesn't quite make sense to say that it appears in the continuous spectrum of $GL(m)$. What happens is that automorphic representations on $GL(n_i)$ (where $\sum_i n_i = m$) can be combined into a representations of $GL(m)$ via parabolic induction, and these will appear in appear in the automorphic spectrum of $GL(m)$. (See Langlands's article On the notion of an automorphic representation.)

Some, but not all, of these parabolically induced representations will be in the continuous part of $L^2$ for $GL(m)$. Roughly, the idea is that one can twist the representations of the various $GL(n_i)$-representations independently, and so the parabolically induced representations form a family parameterized by these twists. A certain subfamily will lie in the continuous spectrum for L^2$.

E.g. if we were inducing a pair of characters on $GL(1)$ to get reps. of $GL(2)$, the (suitably normalized) inductions of the unitary characters would lie in the continuous spectrum of $L^2$.

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In brief, the answer to your first question is "yes", and a reformulated version of your second question has answer "conjecturally, yes".

For the first, more precisely, Eisenstein series formed with automorphic cuspidal repns as inducing data appear in the spectral decomposition of larger GL(n)s.

For the second, it is part of Langlands' conjectures that all the L-functions for all reductive groups appear as standard L-functions for suitable GL(n). Not the literal automorphic functions or repns, but the "data" in a suitable sense.

There is a lot more that can be said...

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