This is a reference request.

Let $A$ be an anisotropic quaternion algebra over $\mathbf Q$. Let $\mathcal O_A$ be a maximal order in $A$. Then $\mathrm{SL}_2(A)$ acts transitively on the right on the set $\mathbf P^1(A)$ of one dimensional left subspaces of $A^2$.

Let $h_A$ be the cardinal of the set (1) of left ideal classes of $\mathcal O_A$.

It is not hard to show that when $h_A=1$, one also has $\sharp[~\mathbf P^1(A)~/~\mathrm{SL}_2(\mathcal O_A)~]=1$.

More generally, computer assisted experiments for $A$ of small discriminant seem to indicate that the equality $$\sharp[~\mathbf P^1(A)~/~\mathrm{SL}_2(\mathcal O_A)~]=h_A^2$$ should hold, but after some unfruitful draft padding, I have been unable to prove it. Would someone here know where to find a proof of this undoubtedly classical fact (2), and even better, name its discoverer ?



Notes :

(0) The inclusion $\mathrm{SL}_2(\mathcal O_A)\to \mathrm{GL}_2(\mathcal O_A)$ is an equality.

(1) In the simpler case when $A$ is a quadratic field, the ideal classes form a group and this point is essential in the proof of the classical equality $\sharp[~\mathbf P^1(A)~/~\mathrm{SL}_2(\mathcal O_A)~]=h_A$.

(2) Some googling indicates that many authors studied automorphic cuspidal things in this quaternionic setting, so that this quotient has certainly been studied, but I could not find the statement.


1 Answer 1


The formula holds, and seems to be due to Krafft and Osenberg : Eisensteinreihen für einige arithmetisch definierte Untergruppen von SL2(H). (German), Math. Z. 204 (1990), no. 3, 425–449.

(See here : EuDML (free), or here : Springer.)

Just after Satz 2.2 :

"Die Zahl der Spitzzenbahnen ist also das Quadrat der (einseitigen) Klassen-zahl von $\mathcal O$".

  • $\begingroup$ This paper is freely accessible online, at EuDML, see eudml.org/doc/183781 $\endgroup$
    – ACL
    Nov 2, 2014 at 18:38
  • $\begingroup$ @ACL Sorry, I did not notice that the access via Springer was not free ... I'll edit the link. $\endgroup$
    – few_reps
    Nov 2, 2014 at 18:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .