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Nov 29, 2021 at 21:42 comment added Andrew @Loeffler, I think $S_{\pi}$, the set of places $v$ such that $\pi_v$ is square integrable, is finite because unramified representation can never be square-integrable. Since the number of elements of the global Vogan A-packet of $\pi$ for $Mp(2)$ is $2^{|S_{\pi}|}$, the number of global cuspidal representations in the packet seems also finite.
Nov 29, 2021 at 12:54 comment added Andrew @Loeffler, Thank you for the comment. May I ask some stupid quesiton? If $\pi$ is a unitary irreducible cuspidal automorphic representation of $GL(2)$, I am wondering whether there are infinitely many places $v$ such that $\pi_v$ is discrete series representation. Because the local A-packet for $Mp(2)$ associated to $\pi_v$ is singleton iff $\pi$ is not discrete series, if such places $v$, which makes $\pi_v$ discrete series, are infinitely many, then there are infinitely many cuspidal representations in the global Vogan A-packet for $Mp(2)$ associated to $\pi$.
Nov 29, 2021 at 7:50 comment added David Loeffler I'm pretty sure this is not for a general linear group $GL_n$, but for some other reductive group -- apparently $Sp_{2(m-n)}$. (For $GL_n$ I believe all $A$-packets are finite, because they're singletons!) In general a global packet is a product of local packets, and all the local guys are finite, but since the set of places is infinite, the global packet can be infinite -- you can construct explicit examples using restriction to $SL_2$ of automorphic reps of $GL_2$ (the packet has more than one element at any unramified place where the Hecke eigenvalue is 0).
Nov 27, 2021 at 14:50 comment added Andrew @Kimball, Page 40 line 2~3 of the following paper. math.purdue.edu/~liu2053/S6.pdf
Nov 27, 2021 at 13:39 comment added Kimball What is the precise context in which it is asserted to be infinite?
Nov 27, 2021 at 9:26 history edited YCor
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Nov 27, 2021 at 7:55 history asked Andrew CC BY-SA 4.0