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Are automorphic quotient for quaternion algebras always compact (safe the totally split case)?

Is there any good reference for proof of this fact, or easy arguments to say do?

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2 Answers 2

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Weil's "Adeles and algebraic groups" proves this for general division algebras. It is labelled "Fujisaki's Lemma", and is a natural division-algebra generalization of the analogue for number fields themselves, namely, that $\mathbb J^1/k^\times$ is compact, where $\mathbb J^1$ is the collection of ideles of idele-norm $1$. For number fields, this compactness implies finiteness of (generalized) class number, and implies the (generalized) units theorem.

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    $\begingroup$ For those with a physical book rather than a computer-searchable .pdf, this is Lemma 3.1.1. $\endgroup$
    – nfdc23
    Commented Jan 15, 2018 at 23:28
  • $\begingroup$ Paul, I thought Fujisaki's lemma is true for any central simple algebra (not just division algebras), including in this case $B=\mathrm{M}_2(F)$. So the OP need not exclude the ``totally split'' case. Right? But the orbifolds obtained from quotients by discrete subgroups are only compact when $B$ is a division algebra (by the theorem of Hey). So there are two layers here, I think. $\endgroup$
    – John Voight
    Commented Jan 17, 2018 at 0:21
  • $\begingroup$ @JohnVoight, perhaps there is ambiguity in what "Fujisaki's lemma" refers to. For me, it only refers to the division algebra case. I don't off-hand know what more sophisticated analogous assertion would be true in the non-compact quotient case. In particular, although I may just be out of the loop, the impression I have is that there is a measure-theoretic argument that gives compactness in the division algebra case, and would not have an obvious extension otherwise. Perhaps this is just a naming convention... I'm curious: what further "F's lemma" exists in the wild? $\endgroup$
    – paul garrett
    Commented Jan 17, 2018 at 0:28
  • $\begingroup$ @paulgarrett, you're right, in Fujisaki's article "On the Zeta-Function of the Simple Algebra over the Field of Rational Numbers", the statement is Theorem 8.3, pg. 599, and at the beginning of section 8, he supposes that $B$ is a division algebra! I must be wrong about the matrix algebra case. $\endgroup$
    – John Voight
    Commented Jan 18, 2018 at 3:17
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What does ``automorphic quotient'' mean?

Does my book (http://quatalg.org), Main Theorem 38.4.3 (a theorem of Hey) answer your question? A quaternion algebra $B$ over a field $F$ is either isomorphic to $\mathrm{M}_2(F)$ (is this your ``totally split case''?) or is a division algebra over $F$, and in the latter case, Hey's theorem applies.

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    $\begingroup$ Usually, "automorphic quotient" means adelic points mod rational points, though in this case I suppose the OP means also mod center. $\endgroup$
    – Kimball
    Commented Jan 16, 2018 at 4:55

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