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Short version: is there a canonical way to adelize a classical Hecke eigenform automorphic form when the adelic quotient has many components? If not, what are the different "choices", how many, etc.?

Some sketched details: Let $A$ be a central simple algebra over a number field $F$, e.g. $A$ is the matrix algebra over $F$. Let $\mathcal{O} \subset A$ be an order, e.g. the matrix ring over the ring of integers $\mathfrak{o}$ of $F$. The adelic quotient $A^\times \backslash \hat{A}^\times / \hat{\mathcal{O}}^\times$ is (assuming some Eichler condition) a disjoint union parametrised by the class group $F^\times_{>0} \backslash \hat{F}^\times / \operatorname{nr}(\hat{\mathcal{O}}^\times)$ (see Thm. 28.5.5 in Voight's Quaternion Algebras). This class group can be non-trivial if $F$ has non-trivial narrow class group or if the order $\mathcal{O}$ is small, more precisely if the image of the local norm $\operatorname{nr}(\mathcal{O}^\times_\mathfrak{p})$ fails to be the whole unit group $\mathfrak{o}_\mathfrak{p}^\times$, where $\mathfrak{p}$ is a prime of $F$.

Each such component corresponds to a locally symmetric space (e.g. for quaternion algebras this would be an arithmetic quotient of the upper half plane). In this way, each adelic automorphic form is given by a tuple of classical automorphic forms on these components (see e.g. Shimura's 1978 Hilbert modular forms paper).

If one has a classical automorphic form on one of these components and we assume that it is also a Hecke eigenform, then is there a canonical way to choose forms on the other components so that the corresponding adelic form is a Hecke eigenform. Of course, Hecke eigenform here has two meanings, referring to the classical and to the global Hecke algebra accordingly. Are there any references for this?

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    $\begingroup$ Every character of the class group gives a function on the disjoint union that is constant on each component, and I think multiplying such a character preserves the Hecke eigen-property. So there is no canonical way to do this. But I think there is a canonical way up to multiplication by characters... $\endgroup$
    – Will Sawin
    Commented Apr 6, 2022 at 19:08
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    $\begingroup$ The classical picture is explained in papers of Shimura (e.g., his special values paper I think). The Hecke operators do not preserve components, so I do not know what you mean by having an eigenform on one component. $\endgroup$
    – Kimball
    Commented Apr 6, 2022 at 21:31
  • $\begingroup$ @Kimball Yes, Shimura describes a bit what's happening with the Hecke operators. Classically, he shows that there is a kind of system of Hecke algebras, where some operators move between groups/quotients as you say (Shimura calls them $R_{\lambda \mu}$, $\mu, \lambda$ being indices of the components). My question was: if an automorphic form is a Hecke eigenform for just one $R_{\lambda \lambda}$, then how could we best lift this to an adelic eigenform for the entire Hecke algebra? $\endgroup$
    – Radu T
    Commented Apr 7, 2022 at 7:05

1 Answer 1

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$\newcommand{\p}{\mathfrak{p}}$Let $C$ be the class group parametrising the components, say $X = \bigcup_{c\in C}X_c$. Then the Hecke operator $T_\p$ sends component $X_c$ to $X_{c\p}$. In particular, the Hecke operators preserving the components are the $T_\p$ where the class of $\p$ in $C$ is trivial. If $f$ is an eigenform on one component $X_c$, then you can extend it by $0$ on the other components, but that is usually not going to be an eigenform for the whole Hecke algebra. If you look at the automorphic representation generated by this extension, it will in general not be irreducible, but it is going to be a finite sum $\bigoplus_{\chi}\pi\otimes\chi$ where $\pi$ is an automorphic representation and $\chi$ ranges over some subset of the characters of $C$.

Proof of the last statement: Let $\pi$ and $\pi'$ be two irreducible representations occurring in the decomposition of the representation generated by $f$. Then the $T_\p$-eigenvalues of $\bigoplus_{\chi \in C}\pi \otimes \chi$ and $\bigoplus_{\chi \in C}\pi' \otimes \chi$ agree for almost all $\p$ (determined by $f$ if the class of $\p$ is trivial in $C$, and $0$ otherwise), so these representations are isomorphic, and $\pi'$ is one of the $\pi\otimes\chi$. Together with the multiplicity one theorem, this proves the claim.

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  • $\begingroup$ This is also what I had in mind. Thank you! What I would also really like to know is if there is a Hecke eigenform in one of the $\pi \otimes \chi$ which "deadelizes" to my original $f$. So writing one of these Hecke eigenforms as a tuple, is there one which is $f$ on the original component? $\endgroup$
    – Radu T
    Commented Apr 7, 2022 at 7:13
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    $\begingroup$ There is one in each of them. Let's say that $f$ is a newform. If you pick $g$ a new vector in one of the $\pi\otimes \chi$, and restrict it to $X_c$, then this restriction will be $f$. But the values on the other components will differ depending on $\chi$. Said differently, there are several (but finitely many) extensions of $f$ to the whole $X$ that are eigenforms for the whole Hecke algebra, and their Hecke eigenvalues differ by twists by the caracters $\chi$ of $C$. However, if you want the extension by $0$, you have to take a linear combination of these various extended eigenforms. $\endgroup$
    – Aurel
    Commented Apr 7, 2022 at 9:54
  • $\begingroup$ Correction: the restriction will be a multiple of $f$, but of course you may have to multiply $g$ by the appropriate scalar. $\endgroup$
    – Aurel
    Commented Apr 7, 2022 at 9:56
  • $\begingroup$ Question: let's say there are two primes $\mathfrak{p}$ and $\mathfrak{q}$ that are both in the same class. Then the image of $(f, 0, \ldots, 0)$ would be of the form $(0, g_{\mathfrak{p}}, \ldots, 0)$ and $(0, g_\mathfrak{q}, \ldots, 0)$, respectively, and I have the impression that your answer would imply that $g_\mathfrak{p}$ is a multiple of $g_\mathfrak{q}$. I can see how this is true if there would be a way of inverting Hecke operators so that we can use that $f$ is an eigenform. But I can't see how this works. Any thoughts on this? $\endgroup$
    – Radu T
    Commented Apr 7, 2022 at 13:36
  • $\begingroup$ I have a paper in preparation with Alex Bartel where we go over this kind of considerations in some details; if you are interested in seeing a (rough) draft you may contact me by email. Regarding your specific question: let $T_0$ be the algebra generated by the $T_\mathfrak{p}$ for $\mathfrak{p}$ trivial in $C$ and $T_1$ the vector space generated by the Hecke operators sending one component to the other one you are considering, both inside the full Hecke algebra acting on the Hecke-span of $(f,0,...,0)$ (which is finite-dimensional). Let $m$ be the order of the corresponding class in $C$. $\endgroup$
    – Aurel
    Commented Apr 9, 2022 at 22:24

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