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.



Computing Tamagawa number of torus in Quaternion algebra
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
$$Tam({\mathbf T}) = 2.$$


