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.


2 Answers 2


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.

  • 1
    $\begingroup$ This was exactly what I was looking for, thank you! $\endgroup$
    – D_S
    Oct 25, 2018 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


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

  • $\begingroup$ Isn't $f$ locally constant, by the way induction is defined? $\endgroup$
    – rj7k8
    Dec 6, 2018 at 3:34
  • $\begingroup$ 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. $\endgroup$
    – D_S
    Dec 6, 2018 at 4:47

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.