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. )


2 Answers 2


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.

  • $\begingroup$ I don't know whether the $X$ in the counterexample is closed (which I think is very important condition) $\endgroup$ Commented Nov 8, 2021 at 6:21
  • $\begingroup$ I add a conterexample I made before in the question. Still I don't know if the $X$ is closed. $\endgroup$ Commented Nov 8, 2021 at 6:30
  • 1
    $\begingroup$ 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). $\endgroup$
    – Aphelli
    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. $\mathbb{A}^\times= \mathbb{A}^1\times<Z_1>$. 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 $Y=<z_1>\times D$, 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.


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