Making sense out of intertwining operators defined by a vector valued integral

Question

Let $G$ be the rational points of a connected, reductive group over a $p$-adic field $F$. Let $S$ be a maximal split torus of $G$ with $\Delta$ a set of simple roots corresponding to a minimal parabolic. Let $(\pi,V)$ be an irreducible, admissible representation of a Levi subgroup $M$ of $G$. For $P = MN$ corresponding to some set of simple roots $\theta \subseteq \Delta$, we have the induced representation

$$I(\nu,\pi) = \operatorname{Ind}_{MN}^G \pi \otimes q^{\langle \nu + \rho, H_M(-) \rangle}$$

where $\nu$ is in the complexified real Lie algebra of $M$, and $\rho$ is half the sum of the roots of $S$ in $N$ counting multiplicity. For $w$ in the Weyl group of $S$, sending $\theta$ to $\theta' \subseteq \Delta$, we get an intertwining operator $A: I(\nu,\pi) \rightarrow I(w(\nu),w(\pi))$ defined by an integral

$$Af(g) = \int\limits_{N_w} f(\dot w^{-1}ng)dn$$

where $N_w$ is generated by the root subgroups of those roots which are made negative by $w^{-1}$, and $\dot w$ is a nice choice of representative for $w$.

So, I understand formally why this integral intertwines the action of $G$ on each space. What I don't understand and haven't yet found a reference for is how to make sense of the integral itself. It is a vector valued integral, over a function with values in the underlying space of $\pi$. Since we are dealing with smooth representations of $p$-adic groups, this space need not have any topology associated with it.

Usually, vector valued integrals for $p$-adic representations are finite sums, taken over compact sets. I don't believe this should be the case here. If one is stating everything precisely, how should we make sense out of this intertwining operator?

Answer 1

It's not really that the space of locally constant, compactly-supported (complex-valued) functions on a p-adic (or other totally disconnected) group has no topology. Rather, it has a canonical topology with some convenient features. That is, that space is a "strict" (filtered) colimit of finite-dimensional spaces. That is, it is a countable ascending union of finite-dimensional spaces. Finite-dimensional spaces have unique topological vector space topologies. This colimit has a unique (topological vector space) topology. The reason that for many purposes we can ignore the topology is that every linear map from that space to any other topological vector space is continuous: this follows from the corresponding fact for finite-dimensional TVS's and the definition of "colimit".

As in the question, there are times when the topology matters to some degree. Happily, it is quasi-complete (despite not being complete-metric), which is sufficient for various version of vector-valued integrals to make sense, whether Bochner-style integrals or Gelfand-Pettis-style.

As an easier case, it is standard that a compactly-supported, continuous, $V$-valued function (for $V$ locally convex, quasi-complete) $F$ has an integral with expected properties. As with L. Schwartz' treatment of "Schwartz functions" as extending to a suitable one-point compactification, if/when we can compactify the physical set on which we integrate, then we can immediately apply this simplest case of vector-valued integration.

(Of course, the integrals defining these intertwinings only converge for parameters in some cone, and must be meromorphically continued...)

EDIT: to be clear(er), as in comments, no, indeed, the functions in the induced repn are not compactly supported (although compactly supported mod $P$), but they are completely determined by their values on the maximal compact, and this gives the finite-dimensional feature... But/and one easily finds in the literature misleading remarks about spaces of test functions on t.d. groups "having no topology" or "having discrete topology", and I intended partly to address that.