Estimating volumes in the trace formula

Question

Let $G$ be a reductive group. In many instances of the trace formula, elliptic terms corresponding to the $G(k)$-conjugacy class of $\gamma \in G(k)$ are weighted by the following volumes $$v_\gamma = \mathrm{vol}(G_\gamma(k) \backslash G_\gamma(\mathbb{A}))$$

I am interested in knowing how to handle $v_\gamma$ in practice. In very few cases I am able to explicitly compute the quotient (namely, $GL(2)$ or $SL(2)$) for diagonal element, what is possible up to conjugation by semisimplicity).

Is there any general method to compute these volumes, or results concerning their values or bounds (for instance depending on the eigenvalues of $\gamma$, on its determinant, and on quantities attached to $k$)?

I am interested in inner forms of general linear groups, unitary groups and symplectic $GSp(4)$.

Answer 1

Up to a normalisation factor for the measure you fixed on $G_\gamma(\mathbb A)$, this is the Tamagawa number of $G_\gamma$.

There is indeed a formula for this. It is $$ \frac{ \left| \pi_0\left( Z(\hat{G})^\Gamma\right) \right|}{ \left|\operatorname{ker}^1\left(F, Z(\hat{G}) \right)\right|}$$ where $\hat{G}$ is the Langlands dual group, $Z$ is its centre, $\Gamma$ is the Galois group acting on it, and $\operatorname{ker}^1$ refers to the kernel of the map from Galois cohomology to local Galois cohomology at each place.

For a proof see Stable trace formula, cuspidate tempered terms, section 5 where Kottwitz calculates the ratio of the Tamagawa number to the Tamagawa number of the universal cover of the derived subgroup, and Tamagawa numbers, where he shows that the Tamagawa number of the universal cover of the derived subgroup is $1$ (with a conditionality that has since been removed).