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Consider a rational Quaternion algebra $M$ over $\mathbb{Q}$ that does not split at $\infty$. For example take the rational Hamilton quaternions $M=\mathbb{Q}(-1,-1)$.
For the adele ring $\mathbb{A}$ we define $M(\mathbb{A}):= M\otimes_{\mathbb{Z}} \mathbb{A}$ and $G=G(\mathbb{A}):= M(\mathbb{A})^\times / \mathbb{A}^\times$. Analogous we can define $G(\mathbb{Q})$. This is then a cocompact lattice in $G$. For a $\gamma \in G(\mathbb{Q})$ we can consider the centralizer $G_\gamma$ of $\gamma $ in $G$. If $\gamma$ is a generic semisimple element and I am not mistaken, $G_\gamma$ is an algebraic torus defined over $\mathbb{Q}$.

Is there a way to compute the Tamagawa number of these tori? Thanks a lot in advance.

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    $\begingroup$ You should look at Vignéras, Arithmétique des Algèbres de Quaternions. A lot of computations are done there. I believe there is a translated version available online. A good part of this book can be found in Maclachlan and Reid, The Arithmetic of Hyperbolic 3-manifolds $\endgroup$
    – Desiderius Severus
    Commented Jan 31, 2018 at 14:14
  • $\begingroup$ What does "generic" mean? Your element $\gamma$ is an element of the quaternion algebra (up to rescaling), and so if $\gamma \not \in \mathbb{Q}$, then $\mathbb{Q}[\gamma] = K$ is an imaginary quadratic field. What is the role of the quaternion algebra here? Every imaginary quadratic field with the property that $2$ is not split in $K$ arises this way, so your question is about $K^\times/\mathbb{Q}^\times$ for these fields? My basic understanding is that the Tamagawa number is given by a class number, is that what you're after? $\endgroup$
    – John Voight
    Commented Jan 31, 2018 at 20:44
  • $\begingroup$ Tam = Pic / Sha, src = Ono. :) $\endgroup$
    – Marty
    Commented Feb 1, 2018 at 4:53
  • $\begingroup$ @DesideriusSeverus Thanks, I will have a close look at Vignéras (or rather the translation). $\endgroup$
    – user113771
    Commented Feb 1, 2018 at 8:45
  • $\begingroup$ @JohnVoight I need the quaternion algebra as I am interested in the centralizer $G_\gamma$. This can be larger than $\mathbb{Q}[\gamma]$. By generic I mean elements unlike $\gamma=1$ for which the centralizer is the whole Quaternion algebra. $\endgroup$
    – user113771
    Commented Feb 1, 2018 at 8:45

1 Answer 1

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Here are some more details. As John Voight said, the quaternion algebra is kind of irrelevant here. If $\gamma$ is a regular semisimple element, then its centralizer is a torus ${\mathbf T}$ over ${\mathbb Q}$ satisfying $${\mathbf T}({\mathbb Q}) = K^\times / {\mathbb Q}^\times,$$ where $K = {\mathbb Q}(\gamma)$ is the centralizer of $\gamma$ in the quaternion algebra. In terms of algebraic groups, ${\mathbf T}$ fits into a short exact sequence of tori, $$1 \rightarrow {\mathbf G}_m \rightarrow {\mathbf R}_{K/{\mathbb Q}} {\mathbf G}_m \rightarrow {\mathbf T} \rightarrow 1.$$

There aren't too many rank-one tori. Since ${\mathbf T}$ is nonsplit, the character lattice of ${\mathbf T}$ is ${\mathbb Z}$, with the unique nontrivial action of $Gal(K/{\mathbb Q})$. Thus ${\mathbf T}$ is also isomorphic to the norm-one torus ${\mathbf R}_{K/{\mathbb Q}}^1 {\mathbf G}_m$.

It happens that norm-one tori arising from cyclic extensions have trivial Sha. This is in Platonov and Rapinchuk, I think, and probably goes back to Tate or something. The basic idea is that Sha captures the failure of the Hasse principle. In the context above, $$Sha({\mathbf T}) \cong \frac{\left( {\mathbb Q}^\times \cap N_{K/{\mathbb Q}} {\mathbb A}_K^\times \right) }{ N_{K / {\mathbb Q}} K^\times}.$$ This is trivial, by the Hasse Norm Theorem (since $K / {\mathbb Q}$ is cyclic).

So we have $$Tam({\mathbf T}) = \frac{\# Pic({\mathbf T})}{\# Sha({\mathbf T})} = \# Pic({\mathbf T}).$$

Explicitly, the Picard group $Pic({\mathbf T})$ in this context is the group $H^1(Gal(K/{\mathbb Q}), {\mathbb Z})$, where the order-two Galois group acts by the nontrivial automorphism. This is a group of order two. Hence $$Tam({\mathbf T}) = 2.$$

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