Finiteness of the volume $G(k) \backslash G(\mathbb A)^1$ from that of $G(k) Z(\mathbb A) \backslash G(\mathbb A)$

Question

It follows from the construction of a Siegel domain that $G(k) Z(\mathbb A) \backslash G(\mathbb A)$ has finite volume, where $G$ is a connected, reductive group over a number field $k$, $Z$ is the center of $G$, and $\mathbb A$ is the ring of adeles of $k$. I have also heard that $G(k) \backslash G(\mathbb A)^1$ has finite volume, where $G(\mathbb A)^1$ is the group of $g \in G(\mathbb A)$ for which $|\chi(g)| = 1$ for all $k$-rational characters $\chi$ of $G$. Does the finiteness of one of these volumes imply that of the other?

Answer 1

Yes. This follows from the fact that $G(\mathbb A)$ is the direct product of $G(\mathbb A)^1$ and $A_G(\mathbb R)^0$, where $A_G$ is the split component of $G$ (the unique maximal split subtorus of the center $Z$ of $G$), and $A_G(\mathbb R)^0$ denotes the connected component of the identity.

Since $G(\mathbb A) = Z(\mathbb A) G(\mathbb A)^1 = G(k) Z(\mathbb A) G(\mathbb A)^1$, we can write

$$G(k)Z(\mathbb A) \backslash G(\mathbb A) = G(k)Z(\mathbb A) \backslash \Big( G(k)Z(\mathbb A) G(\mathbb A)^1\Big) = \Big(G(k)Z(\mathbb A) \cap G(\mathbb A)^1 \Big) \backslash G(\mathbb A)^1 = G(k) \Big(Z(\mathbb A) \cap G(\mathbb A)^1\Big) \backslash G(\mathbb A)^1$$

where $G(k)Z(\mathbb A) \cap G(\mathbb A)^1 = G(k).\Big(Z(\mathbb A) \cap G(\mathbb A)^1\Big)$ follows from the fact that $G(k) \subset G(\mathbb A)^1$.

Now, let $H = Z(\mathbb A) \cap G(\mathbb A)^1$, so that $$G(k) Z(\mathbb A) \backslash G(\mathbb A) = [G(k)H ]\backslash G(\mathbb A)^1$$

We have

$$\operatorname{vol}(G(k) H \backslash G(\mathbb A)^1) = \operatorname{vol}(G(k) \backslash G(\mathbb A)^1) \operatorname{vol}(G(k) \backslash G(k)H) = $$

and writing $G(k) \backslash G(k)H = [G(k) \cap H] \backslash H$, we see that we can rewrite the above as

$$\operatorname{vol}(G(k) Z(\mathbb A) )\backslash G(\mathbb A) = \operatorname{vol}(G(k) \cap H \backslash H) \operatorname{vol}(G(k) \backslash G(\mathbb A)^1)$$ Thus the finiteness of the volume of one of the quotients is equivalent to that of the other, provided we can show that $[G(k) \cap H] \backslash H$ has finite volume.

In fact, $[G(k) \cap H] \backslash H$ is actually compact. Indeed, $H$ is the almost direct product of the adelic points of an anisotropic torus (which is compact) and the piece $A_G(\mathbb A)^1$ of the split component $A_G(\mathbb A)$ which has norm $1$ on all $k$-rational characters. The compactness of $A_G(k) \backslash A_G(\mathbb A)^1$, as a finite product of copies of $\mathbb A^{\ast 1}/k^{\ast}$, implies the compactness of the given quotient.