Let $G$ be a semisimple algebraic group over a number field $F$ with trivial center. Let $\mathfrak S \subset G(\mathbb A)$ be a Siegel domain (defined in terms of a given maximal split torus and minimal parabolic in $G(\mathbb A)$, for a reference see for example pg. 37 of Arthur's introduction to the trace formula).

My question is, from the following two facts, is it possible to deduce the finiteness of the volume of $G(F) \backslash G(\mathbb A)$?

  • $\operatorname{meas}(\mathfrak S) < \infty$ (follows from a standard computation from the definition of $\mathfrak S$)

  • For sufficiently large $\mathfrak S$, we have $G(\mathbb A) = G(F) \mathfrak S$.

I have seen computations (which I think are) to the effect of

$$\int\limits_{G(F) \backslash G(\mathbb A)} dg = \int\limits_{G(F) \backslash G(F) \mathfrak S} dg = \int\limits_{(G(F) \cap \mathfrak S) \backslash \mathfrak S} dg \leq \int\limits_{\mathfrak S} dg < \infty$$

but was uncertain of the legality of this reasoning. For one thing, $\mathfrak S$ and $G(F) \cap \mathfrak S$ are not groups, and although one can transfer the right invariant measure on $G(F) \backslash G(\mathbb A)$ to one on $(G(F) \cap \mathfrak S) \backslash \mathfrak S$ via the natural bijection

$$(G(F) \cap \mathfrak S) \backslash \mathfrak S \rightarrow G(F) \backslash G(\mathbb A)$$

$$(G(F) \cap \mathfrak S)x \mapsto G(F)x$$

it is not obvious how this measure on $(G(F) \cap \mathfrak S) \backslash \mathfrak S$ compares to the on $G(\mathbb A)$ with which $\int\limits_{\mathfrak S} dg$ is calculated.

  • $\begingroup$ I'm not sure what $(G(F) \cap \mathfrak S)\backslash\mathfrak S$ should mean; one cannot generally take a quotient by a subset. Perhaps simply identify points in $\mathfrak S$ that are $G(F)$-translates of one another; or perhaps just $G(F)\backslash G(F)\mathfrak S$. Anyway, if we take $\omega$ as in the definition of $\mathfrak S$ in your reference to be open (in $P_0(\mathbb A)$), then $\mathfrak S$ is open (in $G(\mathbb A)$), and that makes measures behave much more nicely. (EDIT: Hmm, we can't take it to be compact and open. This requires a bit more care, but should still work.) $\endgroup$
    – LSpice
    Jan 30, 2021 at 22:47
  • $\begingroup$ I may not have represented the arguments I have seen correctly. Gelbart's book on the trace formula just says $$\int\limits_{G(F) \backslash G(\mathbb A)} dg \leq \int\limits_{\mathfrak S} dg < \infty$$ for which I tried to fill in the details with a "second isomorphism theorem" type argument. $\endgroup$
    – D_S
    Jan 30, 2021 at 22:52
  • 2
    $\begingroup$ Please don't crosspost: math.stackexchange.com/q/4006223/11323 $\endgroup$
    – Kimball
    Jan 31, 2021 at 2:09
  • 1
    $\begingroup$ More precisely, wait between posting on one site and posting on another (so that the site with the best chance of answering gets a look at it first before the other ones have to) and link between the two posts (so that effort is not wasted on one site duplicating the work already done on another). $\endgroup$
    – Will Sawin
    Jan 31, 2021 at 2:12

2 Answers 2


$\DeclareMathOperator\intr{int}\DeclareMathOperator\meas{meas}$This is one—probably not the best—way of thinking about it. (I am used to the $p$-adic world, where one can think about (real-valued) measures much less subtly than in the real world.) I hope someone will come along and write something more elegant.

Since $G(F)$ is discrete, so that its Haar measure may be taken to be the counting measure, the way we know we have the correct measure on $G(F)\backslash G(\mathbb A)$ is that $$ \int_{G(F)\backslash G(\mathbb A)} \sum_{h \in G(F)} f(h g)\,\mathrm d\dot g = \int_{G(\mathbb A)} f(g)\mathrm dg $$ for all $f \in \operatorname C_c^\infty(G(\mathbb A))$. Now take $\mathfrak S$ and $\mathfrak S'$ such that $G(\mathbb A) = G(F)\mathfrak S'$ and $\mathfrak S' \subseteq \intr(\mathfrak S)$. Write $\mathfrak S'$ and $\mathfrak S$ as countable, increasing unions of compacts $\mathfrak S'_k$ and $\mathfrak S_k$ such that $\mathfrak S'_k \subseteq \intr(\mathfrak S_k)$ for all $k$; and, for each $k$, let $f_k \in \operatorname C_c^\infty(G(\mathbb A))$ be a non-negative function that is $1$ on $\mathfrak S'_k$ and $0$ outside $\mathfrak S_k$. Notice that, for each $k$ and each $g \in \mathfrak S'_k$, we have that $$ \sum_{h \in G(F)} f_k(h g) \ge f_k(g) = 1, $$ so that \begin{multline*} \meas_{\mathrm d\dot g}(G(F)\backslash G(\mathbb A)) = \lim_{k \to \infty} \meas_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'_k) \le \lim_{k \to \infty} \int_{G(F)\backslash G(\mathbb A)} \sum_{h \in G(F)} f_k(h g)\,\mathrm d\dot g \\ = \lim_{k \to \infty} \int_{G(\mathbb A)} f_k(g)\mathrm dg \le \lim_{k \to \infty} \meas_{\mathrm dg}(\mathfrak S_k) = \meas_{\mathrm dg}(\mathfrak S) < \infty. \end{multline*}

  • 1
    $\begingroup$ Thanks for your answer. Could you explain why $meas_{\mathrm d\dot g}(G(F)\backslash G(F)\mathfrak S'_k) \le \int_{G(F)\backslash G(\mathbb A)} \int_{G(F)} f_k(h g)\mathrm dh\,\mathrm d\dot g$? $\endgroup$
    – D_S
    Jan 30, 2021 at 23:17
  • $\begingroup$ I have edited in an explanation. $\endgroup$
    – LSpice
    Jan 31, 2021 at 3:10

I'm not good at English, and my idea may be not true but I would like to share. I think it works in more general (but number-theoretical?) setting.


$G$: 2nd countable locally compact Hausdorff group

$\Gamma$: its discrete subgroup (automatically countable)

$\pi$: canonical projection $G$ $\to$ $G$/$\Gamma$

$B$: A Borel subset of $G$/$\Gamma$

We take the normalized Haar measure of $\Gamma$ i.e. all the points have measure 1.

Lemma 1 (Fundamental set lemma)

If $A$ is a Borel subset of $G$ which is mapped onto $B$ by $\pi$ then there exists a Borel subset of $A$ such that it maps $B$ bijectively.


Take an open subset $U$ of $G$ such that $U^{-1}U \cap \Gamma ={1}$ and {$U_n$} be its countable left translations which cover G. Then the restrictions of $\pi$ to each $U_n$ are all bijective onto their images.

Define {$X_n$} inductively by

$X_1$=$A$ $\cap$ $U_1$

$X_{n+1}$ = $A$ $\cap$ $U_{n+1}$ $\cap$ ($\cup_{k \leq n} X_k \Gamma )^c$

Let $X$ = $\cup$ $X_n$

By definition, $X_n$'s are mutually disjoint and X is mapped bijectively onto its image by $\pi$.

We claim that $X$ maps onto $B$. Then the Lemma 1 follows. Take $b\in B$, then $\pi^{-1}(b) \cap A$ intersects with some $U_n$. Take such $n$ minimal. Then, if $n=1$ then OK, otherwise if $m < n$, then $\pi^{-1}(b) \cap A$ does not intersect with $U_m$, so $\pi^{-1}(b) \cap X_n$ is not empty. Hence the claim. □

Now, finiteness result will be achieved by the next lemma.

Lemma 2

If $X$ is a Borel subset which maps to $B$ bijectively, then

$meas_{G/\Gamma}(B)$ = $meas_{G}(X)$

sketch of the prf.

By the LSpice's first formula ("coset Fubini") it follows by direct computation.

I will explain more precisely.

LSpice's formula is true for compactly supported continuous functions by the construction of the quotient measure. Also, by regularity of measures (e.g. Rudin "Real and Complex Analysis 2.18"), the equality extends firstly to the case that $f$ is simple functions by using countable sum of measurable positive functions are measurable, then arbitrary positive measurable functions (e.g. Rudin loc. cit. 1.33).

Then just substitute the characteristic function of X to the formula to get the result.□

Proof of the disired result

If S is Siegel set, then by Lemma 1 we can take Borel subset X of S which bijects onto quotient space. $meas_{G/\Gamma}(G/\Gamma)$ = $meas_{G}(X)$ $\leq$ $meas_{G}(S)$


Using the "$U$" of Lemma 1, it can be shown that canonical projection is locally homeomorphism, and hence by $\sigma$-compactness, it maps Borel measurable set to Borel measurable set. So, in fact, we don't need the hypothesis that B is measurable in Lemma 2 and it can be used to fill the gap in the proof of Lemma 2.

  • $\begingroup$ Thank you for the excellent explanation $\endgroup$
    – D_S
    Mar 6, 2021 at 6:40
  • $\begingroup$ You're welcome! $\endgroup$
    – Aut
    Mar 6, 2021 at 14:50

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