# Convergence of the intertwining operator as a vector valued integral

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.

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.

• This was exactly what I was looking for, thank you! – D_S Oct 25 '18 at 4:52

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 $$\int\limits_{N_w} |f_{\nu}(n)|dn < \infty$$.

• Isn't $f$ locally constant, by the way induction is defined? – rj7k8 Dec 6 '18 at 3:34
• What I meant to write is $n \mapsto f(k_n)$ is not well defined, unless $\pi$ is the trivial representation. So what I wrote doesn't work. – D_S Dec 6 '18 at 4:47