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In this question, we will follow the notations and definitions from the book "Automorphic Representations and L-Functions for the General Linear Group" by Goldfeld and Hundley. For simplicity, let's focus of $GL_2$. The definition of an (cuspidal) automorphic representation with central character $\omega$ is given in Definition 5.1.12, (5.1.14), where $\omega$ is a unitary Hecke character $$\omega: \mathbb{Q}^{\times} \backslash \mathbb{A}_{\mathbb{Q}}^{\times} \rightarrow \mathbb{C}^{\times}$$

i.e. it is a smooth $(\mathfrak{g},K_{\infty}) \times GL(2,\mathbb{A}_{\text{finite}})$-module which is also isomorphic to a subquotient of the complex vector space of adelic (cusp) automorphic forms. As I literally know nothing about the theory of automorphic representations, I have two questions to bother the community.

  1. Suppose we have two cuspidal automorphic representations with the same central character $\omega$, and if they are isomorphic as $(\mathfrak{g},K_{\infty}) $ modules, are they isomorphic as automorphic representations? If not, do the two automorphic representations have any relations?

  2. Given a $(\mathfrak{g},K_{\infty}) $ module, how to check whether there exists a cuspidal automorphic representation whose local representation at $\infty$ is just this module? E.g. in order for this to be true, are there restrictions on this module? On the other hand, if there exists such a cuspidal automorphic representation, is it unique?

Any references are welcomed!

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  • $\begingroup$ Did you look at the $GL_1$ case, if $\chi$ is a Dirichlet character modulo $q^k$ with $q$ prime then $\omega(x) = sign(x_\infty)^{(1-\chi(-1))/2} \chi(\frac{x_q}{q^{v_q(x_q)}} \bmod q^{k}) \prod_{p \ne q} \chi(p)^{-v_p(x_p)}$ is an automorphic form and representation. $\endgroup$
    – reuns
    Commented Jan 6, 2019 at 17:13
  • $\begingroup$ @reuns Yes I did, and this also shows the $\infty$ part only contains very limited information. I thought the $GL_2$ case might be very different. $\endgroup$
    – Wenzhe
    Commented Jan 6, 2019 at 19:07

1 Answer 1

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1) No, two distinct cuspidal automorphic representations can have the same underlying $(\mathfrak{g},K)$-module. In particular, if the $(\mathfrak{g},K)$-module is a weight $k$ discrete series, then the cuspidal automorphic representations corresponds to a holomorphic newform of weight $k$, and there are plenty of these. On the other hand, if the $(\mathfrak{g},K)$-module is a principal series representation, then there conjecturally are very few cuspidal automorphic representations that share the same $(\mathfrak{g},K)$-module, though they do exist: the known ones are quadratic twists. See my recent paper for further discussion on this problem.

2) For weight $k$ discrete series, there are formulae for the dimensions of spaces of holomorphic cusp forms of weight $k$, so this is easily done. For principal series, this is a very hard question. These correspond to Maass cusp forms, so you are asking whether given a positive real number $\lambda$, there exists a Maass cusp form with Laplacian eigenvalue $\lambda$. We only explicitly know a very thin sequence of such Laplacian eigenvalues, namely those corresponding to Hecke Größencharaktere, whose Laplacian eigenvalues are related to logarithms of fundamental units of real quadratic fields.

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  • $\begingroup$ Hi Peter, I have a question on your answer of 1). If I assume that the two cuspidal representations also share one of (resp almost all, all) their principal series component? Can we then say that they are the same? Also, the weight $k$ you're talking about is contained in the $\omega$ character right? $\endgroup$ Commented Jan 9, 2023 at 21:17
  • $\begingroup$ No, you can have two automorphic representations that are quadratic twists of one another and have the same local components at 50% of their places. And no, the weight of a discrete series is not contained in the central character, except that there is a compatibility $\omega(-1) = (-1)^k$. $\endgroup$ Commented Jan 9, 2023 at 22:11
  • $\begingroup$ Ok thank you very much! $\endgroup$ Commented Jan 9, 2023 at 23:24

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