4
$\begingroup$

Now let $\textbf{G}$ be some connected semisimple linear algebraic group over a number field $F$. Let $G_{\infty}$ be $\textbf{G}(\mathbb{R}\otimes_{\mathbb{Q}} F)$. Let $K_{\infty}$ be a maximal compact subgroup of $G_{\infty}$. We fix an embedding $G\hookrightarrow GL_N$ for some $N$. Let $\mathfrak{P}$ denote a prime of $\mathcal{O}_F$ lying over $p$. Let $G(\mathfrak{P})$ be the intersection of $G_{\infty}$ with the congruence subgroup of $GL_N(\mathcal{O}_F)$ at level $\mathfrak{P}$. Let $\Gamma$ be an arithmetic lattice of $G_{\infty}$.
Let us denote
$\Gamma (\mathfrak{P}):=\Gamma \cap G(\mathfrak{P})$. Let us denote by $e$ and $f$ the ramification and inertia degrees of $\mathfrak{P}$ in $F$. Hence we get that $[F_{\mathfrak{P}}:\mathbb{Q}_p]=ef$ where $F_{\mathfrak{P}}$ is the completion of $F$ at $\mathfrak{P}$. We define $G_k:=G \cap (1+p^kM_{efN}(\mathbb{Z}_p))$ where $G:=\varprojlim_k\Gamma /\Gamma({\mathfrak{P}}^k)$ For $k\geqslant 0$, we define $Y_k:=\Gamma (\mathfrak{P}^{ek})\backslash G_{\infty} /K_{\infty}$.

Then let $vol(\mathfrak{P}^{ek})$ be the volume of $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$.

Let $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$ be compact.

Then I want to show that $vol(\mathfrak{P}^{ek})\sim [G:G_k]$. Here $\sim$ denotes that both of them are similar order as $k \rightarrow \infty$. It will be helpful if I can get some hints or some reference on this fact. Thank you in advance.

$\endgroup$
3
  • 3
    $\begingroup$ I am a bit puzzled by the amount of notation, but look at the canonical map from $\Gamma(\mathfrak P^{ek})\backslash G_\infty$ to $\Gamma\backslash G_\infty$. It should be a (possibly ramified) covering of degree $[G:G_k]$, and this should imply your claim. $\endgroup$
    – ACL
    Jul 5, 2015 at 11:43
  • $\begingroup$ @ACL: I understand that the covering map $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$ to $\Gamma \backslash G_{\infty}$ has deck transformation group $G/G_k$. But how does it imply the statement on volume? I see the following: We can cover $\Gamma\backslash G_{\infty}$ by finitely many open balls whose pullback by the covering map covers $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$ and $vol$( $\Gamma(\mathfrak{P}^{ek})\backslash G_{\infty}$) is then bounded by $[G:G_k]\cdot vol(\Gamma\backslash G_{\infty})$. But how to prove that they are $\sim$, i.e 'similar ' not just bounded. $\endgroup$ Jul 5, 2015 at 14:55
  • $\begingroup$ The deck transformation group preserves the measure. See Paul Garrett's answer. $\endgroup$
    – ACL
    Jul 5, 2015 at 16:43

1 Answer 1

2
+50
$\begingroup$

For a unimodular topological group $G$, and for discrete subgroups $\Theta\subset\Gamma\subset G$, for $f\in C^o_c(\Theta\backslash G)$, it is true that $$ \int_{\Theta\backslash G} f \;=\; \int_{\Gamma\backslash G}\sum_{\gamma \in \Theta\backslash\Gamma} f\circ \gamma $$ in the sense that choice of right $G$-invariant measure on one quotient uniquely determines the right $G$-invariant measure on the other. This follows from the surjectivity of the obvious averaging map from compactly-supported continuous functions on the one quotient to the other.

$\endgroup$
1
  • $\begingroup$ @garrett: Thanks for the explanation $\endgroup$ Jul 5, 2015 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.