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In the theory of automorphic representations one says that G satisfies a "multiplicity one" property if every cuspidal representation occurs with multiplicity one in $L^2(G(F)\backslash G(A))$.

One also says that G satisfies "strong multiplicity one" if a cuspidal representation is uniquely determined up to isomorphism by its behaviour at cofinitely many places of F.

My question is: does the latter always imply the former (as the name would suggest), and is there a decent reason?

This question was prompted while thinking about Jacquet-Langlands, whose proof (via the trace formula) seems to give multiplicity one as a by-product and whose statement also directly implies strong multiplicity one (for inner forms of GL2 using the result for GL2 itself). However, I didn't see any way to deduce multiplicity one directly from the usual statement.

I also can't see how strong multiplicity one for classical modular forms should directly imply the q-expansion principle (which I guess is roughly the same as multiplicity one). Apologies if I'm missing something extremely obvious on this dozy Sunday evening.

Thanks, Tom.

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  • $\begingroup$ The $q$-expansion principle has nothing to do with multiplicity one (at least if you mean "$q$-expansion principle" in the sense that arises in arithmetic geometry, controlling the field or ring generated by the $q$-expansion coefficients in terms of the first few of them, which is a rather different thing from saying exactly what specific such coefficients are equal to). Put another way, multiplicity one is an analytic or representation-theoretic fact, whereas the $q$-expansion principle is an algebraic (or algebro-geometric) fact. Quite different beasts. $\endgroup$
    – user28172
    Feb 25, 2013 at 6:21
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    $\begingroup$ @Tom: I thought "strong multiplicity one" means that not only are the two representations isomorphic, but also equal (now clearly implying your "multiplicity one"). At least this is Theorem 3.3.6 in Bump's Automorphic Forms and Representations (for the case $G=GL_n$). In any case (like $GL_n$), if you have uniqueness of Whittaker models, then automorphic functions are determined by their Whittaker coefficients, so if two cuspidal representations are isomorphic - their functions will have the same Whittaker coefficients up to a constant, and hence the representations are in fact equal. I think. $\endgroup$
    – Dror Speiser
    Feb 25, 2013 at 11:44
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    $\begingroup$ @Nosr When I say 'q-expansion principle' here I just mean the fact that a q-expansion uniquely determines its modular form, sorry. @Dror Ah, that would define my question nicely out of existence (and possibly give anecdotal evidence that with my definition it's not true...). $\endgroup$
    – Tom Lovering
    Feb 25, 2013 at 14:11

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I'm not sufficiently privileged to leave comments yet, but in support of the semantics I'll mentioned the following paper: http://www2.math.ou.edu/~rschmidt/papers/sqfree.pdf

In this paper, Schmidt refers to his Corollaries 3.16 and 3.17 as "strong multiplicity one theorems". This would agree with Tom's definition of "strong multiplicity one". In the case of $GSp(4)$ "multiplicity one" is not known (representations do not necessarily have Whittaker models, which complicates things), so even though Schmidt can say isomorphic at cofintely many place implies isomorphic, he is not able to say that the global representations are literally equal because the latter may (but are not expected to) occur with multiplicity greater than one in $L^2(GSp(4, \mathbb{Q}) \backslash GSp(4, \mathbb{A})$.

So, from an appropriate point of view, Bump's statement of Theorem 3.3.6 is a "multiplicity one" + "strong multiplicity one" (the former of which is known to be true for $GL_2$ of course).

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    $\begingroup$ Nice answer. So if we interpret at Tom does, strong multiplicity one is not always known to imply multiplicity one. Interesting. Now one might ask: even if it is known, is it true conjecturally that strong multiplicity one implies multiplicity one? I guess some specialist of Arthur's conjectures could answer this one easily... $\endgroup$
    – Joël
    Feb 25, 2013 at 14:51

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