# A finite-volume problem in the adelic group of a global field

I am reading Zeta functions of simple algebras by Roger Godement and Hervé Jacquet. In p139-140 they introduce a version of the theory of reduction. To get the finiteness of the Siegel domain module center, I wish to prove this assertion:

Let $$F$$ be a global filed and $$Y$$ be a closed subset of $$\mathbb{A}^\times_F$$, such that $$|\cdot |_\mathbb{A}|_Y: Y\to \mathbb{R}_+^\times$$ is proper (the preimege of a compact subset is compact). Fixed some positive real number, and let $$X:=\{x\in Y: |x|_\mathbb{A}\geq t\}$$ then $$\int_{\mathbb{A}^\times_F}\mathbb{1}_X(x)|x|_\mathbb{A}^{-1}\mathrm{d}\mu_{\mathbb{A}^\times_F}(x)$$is finite.

It seems that we need some "growth condition". More explicitly:

Let $$\mathbb{A}_F^1$$ be the kernel of $$|\cdot |_\mathbb{A}|_{\mathbb{A}_F^\times}:\mathbb{A}_F^\times \to \mathbb{R}_+^\times$$, and notice that the image of this map in $$\mathbb{R}_+^\times$$ may not be the whole space (function filed case). Then by the theory of Haar measure on homogeneous space we can get: $$\int_{\mathbb{A}^\times_F}\mathbb{1}_X(x)|x|_\mathbb{A}^{-1}\mathrm{d}\mu_{\mathbb{A}^\times_F}(x)=\int_{\mathbb{A}^\times_F/\mathbb{A}_F^1}(\int_{\mathbb{A}_F^1}\mathbb{1}_X(xa)|x|_\mathbb{A}^{-1}\mathrm{d}\mu_{\mathbb{A}^1_F}(a))\mathrm{d}\mu_{\mathbb{A}^\times_F/\mathbb{A}_F^1}(|x|_\mathbb{A})$$

To simplify we fixed some notation: $$x\in\mathbb{A}^\times_F, S(x):=\{a\in\mathbb{A}_F^1: xa\in X\}, V(|x|_\mathbb{A}):=V(x):=\mu_{\mathbb{A}^1_F}(S(x))$$

(One can check for $$r\in\mathbb{R}_+^\times$$, $$V(r)$$ is well-defined.)

Then we have: $$\int_{\mathbb{A}^\times_F}\mathbb{1}_X(x)|x|_\mathbb{A}^{-1}\mathrm{d}\mu_{\mathbb{A}^\times_F}(x)=\int_{\mathbb{A}^\times_F/\mathbb{A}_F^1}V(r)r^{-1}\mathrm{d}\mu_{\mathbb{A}^\times_F/\mathbb{A}_F^1}(r)$$.

Now it becomes clear that we need growth condition on $$V(r)$$.

I have some remarks or peoblems:

1. I don't know how to use "closed subset" condition. (This might be the key) (Notice that the topology of $$\mathbb{A}^\times$$ is weaker than the restriction topology into $$\mathbb{A}$$. )

I also tried to made a counterexample before:

$$F=\mathbb{Q}, X_n:=\prod_{p|n}p^{-mv_p(n)-1}\mathbb{Z}_p^\times\cdot\prod_{p\nmid n}\mathbb{Z}_p^\times\cdot\{\text{all possible}\ r\in\mathbb{R}_+ , \text {s.t} |\cdot|_\mathbb{A}\in[n,n+1] \}$$

Then $$\mu(X_n)=\frac{n^{m+1}}{\phi(n)}(\ln(n+1)-\ln(n))$$, $$X=\cup X_n$$

I don't know whether $$X$$ is closed

1. In number field case, We have decomposition $$\mathbb{A}^\times=\mathbb{A}^\times_{\geq 0}\times\mathbb{A}^1$$, where $$\mathbb{A}^\times_{\geq 0}$$ consists of elements in $$\mathbb{A}^\times$$ which has the same positive real number in infinite places and $$1$$ in finite places. This means we have a good representatives of $$\mathbb{A}^\times_F/\mathbb{A}_F^1=\mathbb{R}_+^\times$$ in $$\mathbb{A}^\times$$. And for $$x\in \mathbb{A}^\times_{\geq 0}$$ with $$|x|_\mathbb{A}\geq t$$ we can choose $$S(x)$$ to be a fundamental domain of $$\mathbb{A}_F^1/F^\times$$ which is compact (this is the theory of reduction of $$GL(1)$$). (the reference book for this may be Chap.13.5 (p235) of London Mathematical Society Lecture Note Series 467- The Genesis of the Langlands Program )

2. I don't know how to deal with the function field case. Since although $$V(x)$$ only depends on $$|x|_\mathbb{A}$$, $$S(x)$$ not. Any ideas about suitable condition for the finiteness? (This is where I actually want help. )

I think this is not enough (although it’s only a sketch of a counterexample). Consider the number field case, and let, for every $$t>0$$, $$x_t\in \mathbb{A}_F^{\times}$$ be equal to $$1$$ everywhere except a fixed infinite place such that $$|x_t|_{\mathbb{A}}=t$$.

Let $$E$$ be the closure of a (precompact measurable) fundamental domain for $$\mathbb{A}^1_F/F^{\times}$$. Let $$f_1,\ldots,f_n$$ be an enumeration of the elements of $$F^{\times}$$.

For each $$n \geq 1$$, let $$Y_n=\{x_t,\,n \leq t \leq n+1\}\{f_1,\ldots,f_{2^{n^2}}\}E$$, and $$Y=\cup_{n \geq 1}{Y_n}$$. Then $$Y$$ should satisfy the hypotheses but I expect your integral (for eg $$t=1$$) to be infinite.

• I don't know whether the $X$ in the counterexample is closed (which I think is very important condition) Commented Nov 8, 2021 at 6:21
• I add a conterexample I made before in the question. Still I don't know if the $X$ is closed. Commented Nov 8, 2021 at 6:30
• Yes, your $X$ is closed for the following general reason: let $f:X \rightarrow [1,\infty)$ be a continuous map of Hausdorff spaces, and let $A \subset X$ be such that $A \cap f^{-1}([n,n+1])$ is compact for all $n \geq 1$. Then $A$ is closed. (because $A \cap f^{-1}((n-1/2,n+3/2))$ is closed in $f^{-1}((n-1/2,n+3/2))$ because it’s the trace of a compact, so you can show that $X \backslash A$ is locally open hence open so $A$ is closed). Commented Nov 8, 2021 at 8:52

Firstly, the counterexample is available. See Mindlack's comment in his answer.

Here I will give a slightly general version of his statement.

Let $$X$$ be a Hausdorff space and $$Y$$ be a locally compact space, and $$\phi: X\to Y$$ be a continuous map. Let $$A\subset X$$ satisfy $$\phi|_A$$ is proper, then $$A$$ is closed.

Pf. $$\forall y\in Y$$, choose any $$K_y$$ to be a compact neighborhood of $$y$$, then $$A\cap \phi^{-1}(K_y)$$ is compact (since $$\phi|_A$$ is proper), then closed (since $$X$$ is Hausdorff).

Then $$\forall x\in X\setminus A$$, let $$K$$ be a compact neighborhood of $$\phi(x)$$, and $$U$$ an open neighborhood of $$\phi(x)$$ contained in $$K$$, we have that $$\phi^{-1}(U)\setminus(A\cap \phi^{-1}(K))$$ is open and contained in $$X\setminus A$$ and containing $$x$$. So $$A$$ is closed. Q.E.D.

Secondly, I will give a reasonable additional condition on $$Y$$ for function field case.

The key point is that $$\mathbb{A}^\times\simeq \mathbb{A}^1\times\mathbb{Z}$$. (See Weil's Basic Number Theory, p75 Cor.1 of The.5 .) More precisely, we can choose $$z_1\in\mathbb{A}^\times$$ such that $$|z_1|_\mathbb{A}=Q$$, a power of $$p$$ (the char. of the function field), the generator of "$$\mathbb{Z}$$", i.e. $$\mathbb{A}^\times= \mathbb{A}^1\times$$. This is similar to the number filed case $$\mathbb{A}^\times= \mathbb{A}^1\times\mathbb{A}^\times_{\geq0}$$.

Then we can set $$Y=\mathbb{A}^\times_{\geq0}\times D$$ or $$Y=\times D$$, resp. for the number field case or the function field case, resp. , where $$D$$ is a fixed compact fundamental domain for $$\mathbb{A}^1/F^{\times}$$, with volume $$d:=\mu_{\mathbb{A}^1}(D)$$.

then the integral is equal to $$d\cdot \int_{t}^{\infty}r^{-1}\mathrm{d}^\times r$$ or $$d\cdot\sum_{m\in\mathbb{Z}, m\geq\log_Q t}Q^{-m}$$ resp., hence finite.

Finally, the reason why such $$Y$$ can be used to construct the Siegel domain is beyond the question. The ref. can again be Chap.13.5 (p235) of London Mathematical Society Lecture Note Series 467- The Genesis of the Langlands Program.