Convergence of the intertwining operator as a vector valued integral

Question

Let $G$ be a connected, reductive group over a $p$-adic field with parabolic $P = MN$ defined by a set of simple roots $\theta \subset \Delta$. For $(\pi,V)$ a representation of $M$, and $\nu \in \mathfrak a_{M,\mathbb C}^{\ast}$, we have the induced representation

$$I(\nu,\pi) = \operatorname{Ind}_P^G \pi q^{\langle \nu+\rho,H_M(-)\rangle}$$ of $G$. For $w$ in the Weyl group sending $\theta$ to $\theta' \subset \Delta$, and $P' = M'N'$ corresponding to $\theta'$, we have the intertwining operator $A = A(\nu,\sigma,w): I(\nu,\pi) \rightarrow I(w(\nu),w(\pi))$ defined by

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

where $N_w$ is generated by the root subgroups of positive roots made negative by $w^{-1}$. The given integration takes place in the vector space $V$, and I am trying to understand:

• What is the meaning of this vector valued integral?

• Why does the integral converge (whatever that means, depending on the answer to my first question) for $\nu$ in a suitable cone?

I had asked a question about the meaning of the integral before, but I am sorry to say that after all this time I still do not understand what is going on. Paul Garrett provided an answer in which he suggested that we should not think of $V$ as having the discrete topology, but having a locally convex, quasi-complete topological vector space structure (coming as a colimit of its f.d. subspaces) in which one could make sense of the integral as a Pettis integral. That is, we should show that there exists a vector $v = A(f)(g)$ in $V$ with the property that for all $v^{\ast}$ in the algebraic dual of $V$,

$$\langle v^{\ast},v \rangle \rangle = \int\limits_N \langle v^{\ast}, f(w^{-1}ng)\rangle dn$$

He also suggested that taking a good maximal compact subgroup $K$ of $G$, so that we have $G = PK = P'K$, we could use the fact that elements of the induced representation are determined by their effect on $K$ to reduce to the case where the vector valued integrals are just finite sums. I still have not figured out how to do this, and wanted to ask math overflow again for help.

These intertwining operators are unfortunately still very much a mystery to me, and I have not seen any reference explain them rigorously.

Answer 1

A reference for this material is Waldspurger's article "La formule de Plancherel pour les groupes p-adiques, d’après Harish-Chandra," (pdf). See section IV.1.

Here I will make a few remarks only about the definition. All serious mathematical arguments (e.g. convergence) are contained in the above reference (in particular Theorem IV.1.1).

Assume V is admissible. The definition of $$ \int_N f(w^{-1}ng)\,dn=v $$ where v∈V is that $$ \int_N \langle f(w^{-1}ng),\check{v}\rangle\,dn=\langle v,\check{v}\rangle $$ for all $\check{v}$ in the contragredient of V. If we ask this for all $\check{v}$ in the algebraic dual of V, then I believe that condition is too strong.

To check the integral converges, it suffices to check the integral of $\langle f(w^{-1}ng),\check{v}\rangle$ converges for all $\check{v}$. For once you have this convergence, it defines v in the algebraic dual of the contragredient of V. It's not hard to see that v is a smooth vector, so lies in the double contragredient, which is the same as V since V is admissible.

Answer 2

EDIT: This doesn't work, because $n \mapsto f(k_n)$ is not well defined.

There is another way I was just thinking of this vector valued integral, and several people (in particular, Paul Garrett) have already explained things to me in this way, but I was not able to understand what they were saying at the time. I will write what I have, and if it's wrong, hopefully someone will point out my error.

So we choose a maximal compact open subgroup $K$ of $G$, in good position relative to $P$, so that we have $G = PK$. For each $n \in N_w$, we choose $p_n \in P$ and $k_n \in K$ such that $w^{-1}n = p_nk_n$. Define $f_{\nu}(n) = q^{\langle \nu + \rho, H_M(p_n)\rangle}$. This is a well defined locally constant function on $N_w$. Also, for a given $f \in I(\nu,\pi)$, the map $n \mapsto f(k_n)$ is a well defined locally constant function $N_w \rightarrow V$, and we have

$$f(w^{-1}n) = f_{\nu}(n)f(k_n) \in V$$

Since $f|_K:K \rightarrow V$ is locally constant, $\{f(k_n) : n \in N_w \}$ is a finite set.

Now to make sense out of $\int\limits_{N_w} f(w^{-1}ng)dn$, we may replace $f$ by $R_g(f)$ and assume $g = 1$. Thus we want to make sense out of $\int\limits_{N_w} f(w^{-1}n)dn$. For $v \in V$, define

$$ A_v = \{ n \in N_w : f(k_n) = v\}$$

which is an open set in $N_w$, and is empty for almost all $v$. Naively, I'm going to write

$$\int\limits_{N_w} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v} f(w^{-1}n)dn = \sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)f(k_n)dn$$

$$ = \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$$ where the sum over $V$ is really just a finite sum. So the vector valued integral over $N_w$ can be defined to be $ \sum\limits_{v \in V} \bigg(\int\limits_{A_v}f_{\nu}(n)dn \bigg)v$, provided each Lebesgue integral

$$\int\limits_{A_v}f_{\nu}(n)dn$$

converges. But each such Lebesgue integral converges if and only if

$$\sum\limits_{v \in V} \int\limits_{A_v}f_{\nu}(n)dn = \int\limits_{N_w} f_{\nu}(n)dn$$

converges. So the intertwining operator makes sense provided that $\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$.