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This is a followup to my previous question here.

Let $G = \operatorname{GL}_n(F)$ for $F$ a $p$-adic field. Let $Z$ be the center of $G$, and let $f: G \rightarrow \mathbb{C}$ be a function which is locally constant and compact modulo $Z$.

This means that there exists a compact set $\Omega \subseteq G$ such that $\{ g \in G : f(g) \neq 0 \}$ is contained in the product set $Z.\Omega$. Equivalently, the image of the closure of $\{g \in G : f(g) \neq 0 \}$ in $G/Z$ is compact.

Assume also that there exists a continuous homomorphism $\omega: Z \rightarrow \mathbb{C}^{\ast}$ such that $f(zg) = \omega(z)f(g)$ for all $g \in G, z \in Z$.

Let $U$ be the group of upper triangular unipotent matrices in $G$, and define

$$F(g) = \int\limits_U f(ug)du$$

I want to understand why the integral $F$ converges absolutely. Already I have a proof, found in my previous question, based on the fact that a certain homomorphism $C_c^{\infty}(G) \rightarrow \textrm{c-Ind}_Z^G(\omega)$ is surjective. But I don't want to use this fact.

The paper I'm reading claims that the integral converges "since the orbit $UZg$ is closed in $G$." I would like to understand why we know the integral converges on account of this explanation. I would appreciate any hint on how to approach this.

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  • $\begingroup$ The point is that the map product map $Z\times U\longrightarrow ZU$ is a homeomorphism. The same holds true if you replace $Z$ by $T$, the group of diagonal matrices. More generally if $P=LU$ is a parabolic subgroup of $p$-adic reductive group ($U$ is the unipotent radical and $L$ a Levi component), the map $L\times U\longrightarrow P$ is a homeomorphism. $\endgroup$ Feb 20, 2018 at 10:08
  • $\begingroup$ I know that $ZU$ is a direct product, but I still don't understand how this implies convergence. Can you explain further? $\endgroup$
    – D_S
    Feb 20, 2018 at 14:31

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I use your notation. Take $v\in U$ in the support of $u\mapsto f(ug)$, then $vg\in Z\Omega$. So there exists $z\in Z$ such that $zv\in \Omega g^{-1}\cap ZU$, a compact subset of $ZU$. Since $Z\times U\longrightarrow ZU$ is a homeomorphism, the projection ${\rm pr}_U$: $ZU\longrightarrow U$ is continuous. Hence $v\in {\rm pr}_U (\Omega g^{-1})$, a compact subset of $U$.

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