# The spherical Whittaker Inversion Theorem and the quantum non-periodic Toda
Lattice

Nolan R. Wallach

###### Abstract

In this paper the spherical case of the Whittaker Inversion Theorem is given a
relatively self-contained proof. This special case can be used as a help in
deciphering the handling of the continuous spectrum in the proof of the full
theorem. It also leads directly to the solution of the quantum non-periodic
Toda Lattice. This is also explained in detail in this paper.

##  1 Introduction

The main purpose of this paper is to give a complete, relatively self
contained, proof of the spherical Whittaker Inversion Theorem for real
reductive groups. This result is an important special case of the general
theorem but it is unencumbered by the complications caused by discrete
spectrum. Reading it can be used as a help in understanding the arguments used
in [W3] to handle the continuous spectrum. By relatively self contained I mean
that it will be based on two main results: The first is the Harish-Chandra
Plancherel Formula for $L^{2}(G/K)$ (actually the inversion formula) as
developed in the work of Helgason [He1],[He2].The second is the holomorphic
continuation of the Jacquet Integral for minimal parabolic subgroups and its
brilliant implication due to Raphaël Beauzart-Plessis [B]. The full theorem
was announced in the 1980’s (see [RRGII]) but a correct proof has only
recently appeared in [W3]. The applications of these results to the theory of
automorphic forms inundate the literature. We will include in the last section
of this paper this paper an application to the quantum non-periodic Toda
lattice (see [GW] for more background on the subject). If you are a physicist
or a mathematician who is not an expert in Representation Theory, I would
recommend that you read Section 7 of this paper first.

##  2 Some notation

Let $G$ be a real reductive group. In this paper there is no loss of
generality to take $G$ to be the identity component of a subgroup of
$GL(n,\mathbb{R})$ that is the locus of zeros of a set of polynomials on
$M_{n}(\mathbb{R})$ such that $G$ invariant under transpose ($g\mapsto
g^{T}$). We set $K=G\cap O(n)$ a maximal compact subgroup of $G$. We choose an
Iwasawa decomposition of $G$ given by $G=NAK$ with $N$ a maximal unipotent
subgroup (i.e. the elements of $N$ are of the form $I+X$ with $X$ nilpotent)
of $G$, $A$ a subgroup maximal among the subgroups of $G$ contained in the set
of symmetric positive definite matrices and $aNa^{-1}\subset N,a\in A$. Let
$\mathfrak{a}=Lie(A),\mathfrak{n}=Lie(N)$ and $\mathfrak{g}=Lie(G)$. Then
$ad(h)_{|\mathfrak{n}}$ for $h\in\mathfrak{a}$ simultaneously diagonalize
yielding the roots $\Phi^{+}$ of $\mathfrak{a}$ on $\mathfrak{n}$. The
exponential map $\exp:\mathfrak{a}\rightarrow A$ is a Lie group isomorphism
($\mathfrak{a}$ is a group under addition) set $\log:A\rightarrow\mathfrak{a}$
equal to the inverse map. We define for
$\lambda\in\mathfrak{a}_{\mathbb{C}}^{\ast},a\in A$.
$a^{\lambda}=e^{\lambda(\log(a))}.$ Define

$\rho(h)=\frac{1}{2}\mathrm{tr}(ad(h)_{|\mathfrak{n}}).$

If $x,y\in\mathfrak{g}$ then set $\left\langle
x,y\right\rangle=\mathrm{tr}(xy^{T})$ and $\left\|x\right\|=\left\langle
x,x\right\rangle^{\frac{1}{2}}$. Let $m=\dim\mathfrak{n}.$ On
$\wedge^{m}\mathfrak{g}$ we put the inner product induced by
$\left\langle...,...\right\rangle$. Let $Int(g)$ be the transformation of
$\,\mathfrak{g}$ given by $Int(g)x=gxg^{-1}$. Define for $g\in$ $G$,
$\left|g\right|$ to be the operator norm of $\wedge^{m}Int(g)$. Set $A_{G}$
equal to the subgroup of elements of the center of $G$ that are positive
definite. Then $KA_{G}[G,G]=G$. Define

$\left\|kag\right\|=e^{\left\|\log a\right\|}\left|g\right|^{\frac{1}{2}},k\in
K.a\in A_{G},g\in[G,G].$

If $g\in G$ then $g$ can be written in the form $g=k_{1}ak_{2}$ with $a\in A$
such such that $a^{\alpha}\geq 1$ for $\alpha\in\Phi^{+}$ and $k_{1},k_{2}\in
K$. One can check that if $g$ is of this form and if $g\in[G,G]$ then

$\left|g\right|=a^{2\rho}\text{. }$

It is easily seen that

$\left\|x\right\|\geq 1,\left\|xy\right\|\leq\left\|x\right\|\left\|y\right\|$

and that the sets

$\left\|x\right\|\leq r$

are compact for $r<\infty$. It is a bit harder to prove that there exists $d$
such that

$\int_{G}\left|x\right|^{-1}(1+\log\left\|x\right\|)^{-d}dx<\infty.$

Let $M=\\{k\in K|ka=ak,a\in A\\}$.

We now recall the Harish-Chandra Schwartz space. If $f\in C^{\infty}(G)$,
$x,y\in U(\mathfrak{g})$ $d\in\mathbb{R}$ then set

$p_{x,y,d}(f)=\sup\\{\left|R_{x}L_{y}f(g)\right|\left|g\right|^{\frac{1}{2}}(1+\log\left\|g\right\|)^{d}|g\in
G\\}.$

Here if $X\in U(\mathfrak{g})$ then $R_{X}f(g)=\frac{d}{dt}f(g\exp tX)_{|t=0}$
and $L_{Y}f(g)=\frac{d}{dt}f(\exp(-tX)g)_{|t=0}$ and since
$[L_{X},L_{Y}]=L_{[X,Y]}$ and $[R_{X},R_{Y}]=R_{[X,Y]}$ the universal mapping
property of the universal enveloping algebra allows us to define $L_{x}$ and
$L_{y}$ for all $x.y\in U(\mathfrak{g}).$

Define $\mathcal{C}(G)=\\{f\in C^{\infty}(G)|p_{x,y,d}(f)<\infty,x\in
U(\mathfrak{g}),d\in\mathbb{R\\}}$ endowed with the topology defined by the
semi-norms $p_{x,y,d}$.

##  3 The Plancherel Theorem for $\mathcal{C}(G/K)$

Note that the map $\theta:G\rightarrow G$ given by $\theta(g)=(g^{T})^{-1}$ is
an automorphism of $G$. Set $\bar{N}=\theta(N)$. We have the corresponding
Iwasawa decomposition $G=\bar{N}AK$. Define for $g=\bar{n}ak$,
$\bar{n}\in\bar{N},a\in A,k\in K,$ $a(g)=a,k(g)=k$. Since the Iwasawa
decomposition is unique, $a:G\rightarrow A$ and $k:G\rightarrow K$ are
$C^{\infty}$. If $f\in C^{\infty}(M\backslash K)$ define for
$\nu\in\mathfrak{a}_{\mathbb{C}}^{\ast}$

$f_{\nu}(\bar{n}ak)=a^{\nu-\rho}f(k).$

This defines $f_{\nu}$ as a $C^{\infty}$ function on $G$. Define an action of
$G$ on $C^{\infty}(M\backslash K)$ by

$\left(\pi_{\nu}(g)f\right)(k)=f_{\nu}(kg)\text{.}$

If we put the $C^{\infty}$ topology on $C^{\infty}(M\backslash K)$ then
$(\pi_{\nu},C^{\infty}(M\backslash K))$ is a smooth Fréchet representation of
$G$. We also put the $L^{2}$–inner product on $C^{\infty}(M\backslash K)$

$\left\langle u,w\right\rangle=\int_{K}u(k)\overline{w(k)}dk$

and find that

$\left\langle\pi_{\nu}(g)u,w\right\rangle=\left\langle
w,\pi_{-\bar{\nu}}(g^{-1})u\right\rangle$

where $\bar{\nu}(h)=\overline{\nu(h)}$. In particular, if
$\nu\in\mathfrak{a}^{\ast}$ then $(\pi_{i\nu},L^{2}(M\backslash K))$ is a
unitary representation. Let $f\in C_{c}^{\infty}(G/K)$ then set

$\left(\pi_{\nu}(f)u\right)(k)=\int_{G}u_{\nu}(kg)f(g)dg.$

If $\lambda\in\mathfrak{a}^{\ast}$ define $H_{\lambda}\in\mathfrak{a}^{\ast}$
by $\left\langle H_{\lambda},h\right\rangle=\lambda(h)$ for
$h\in\mathfrak{a}$. Define $(\lambda,\mu)=\left\langle
H_{\lambda},H_{\mu}\right\rangle$ and extend $(...,...)$ to
$\mathfrak{a}_{\mathbb{C}}^{\ast}$ bilinearly. Then set

$c(\nu)=\int_{N}a(n)^{\nu-\rho}dn$

where we leave, for the moment, the bi-invariant measure on $N$ unnormalized.
This integral converges absolutely for
$\nu\in\mathfrak{a}_{\mathbb{C}}^{\ast}$ such that

$\operatorname{Re}(\nu,\alpha)<0$

for all $\alpha\in\Phi^{+}$ and uniformly in compacta in this set. Thus
$c(\nu)$ defines a holomorphic function on this subset. It has a meromorphic
continuation to all of $\mathfrak{a}_{\mathbb{C}}^{\ast}$ indeed Gindikin-
Karpelovic derived an explicit formula (c.f. [He2] or [W2]) 8.10.18) if we set
for
$\alpha\in\Phi_{0}^{+}=\\{\beta\in\Phi^{+}|\frac{\beta}{2}\notin\Phi^{+}\\}$,

$c_{\alpha}(\nu)=\left\\{\begin{array}[c]{c}B(\frac{\dim\mathfrak{n}_{\alpha}}{2},\frac{(\nu,\alpha)}{(\alpha,\alpha)})\text{
if }2\alpha\notin\Phi^{+}\\\
B(\frac{\dim\mathfrak{n}_{\alpha}}{2},\frac{(\nu,\alpha)}{(\alpha,\alpha)})B(\frac{\dim\mathfrak{n}_{2\alpha}}{2},\frac{(\nu,\alpha)}{2(\alpha,\alpha)}+\frac{\dim\mathfrak{n}_{\alpha}+\dim\mathfrak{n}_{2\alpha}}{2})\text{
if }2\alpha\in\Phi^{+}\end{array}\right..$

Then the the measure on $N$ can be normalized so that

$c(\nu)=\prod_{\alpha\in\Phi_{0}^{+}}c_{\alpha}(\nu).$

Set for $\nu\in\mathfrak{a}^{\ast}$

$\mu(\nu)=\frac{1}{c(i\nu)c(-i\nu)}=\frac{1}{\left|c(i\nu)\right|^{2}}$

one can show using the formula for $c(\nu)$ and basic properties of the
$\Gamma$–function that that $\mu(\nu)\leq C(1+\left\|\nu\right\|^{r})$ for
some $r$.

We are now ready develop the Plancherel theorem for $G/K.$ We first recall a
special case of the Harish-Chandra Plancherel Theorem:

###### Theorem 1

The measure on $G$ and $\mathfrak{a}^{\ast}$ can be normalized so that if
$\phi\in\mathcal{C}(K\backslash G/K)$ (that is
$\phi(k_{1}gk_{2})=f(g),k_{1},k_{2}\in K$) then

$\phi(g)=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(L_{g^{-1}}\phi)1,1\right\rangle\mu(\nu)d\nu.$

There is a relatively elementary proof of this Theorem due to Anker [A] that
specifically proves this result. Harish-Chandra proved much more. We recall an
argument in Helgason [He2] section III.1, in the proof of the following
implication.

###### Theorem 2

Let $f\in\mathcal{C}(G/K)$ then with the same normalizations as in the
previous result we have

$f(g)=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(L_{g^{-1}}f)1,1\right\rangle\mu(\nu)d\nu.$

Note:The point here is that $f$ is not necessarily left $K$–finite (since
under that condition the result will be a special case of Harish-Chandra’s
Theorem).

Proof. Define for $g,x\in G$

$\phi(g,x)=\int_{K}f(gkx)dk.$

Noting that the Casimir operator, $C$, corresponding to the choice of $B$ on
$\mathfrak{g}$ yields the Laplacian of the Riemannian structure on $G$ given
by the inner product on $\mathfrak{g}/Lie(K)$ induced by $B$. This and Sobolev
theory implies that if $g$ is fixed then $x\mapsto\phi(g,x)$ is in
$\mathcal{C}(K\backslash G/K)$. Thus keeping $g$ fixed we have

$\phi(g,I)=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(\phi(g,\cdot))1,1\right\rangle\mu(\nu)d\nu.$

Now

$\left\langle\pi_{i\nu}(\phi(g,\cdot))1,1\right\rangle=\int_{G}\phi(g,x)\left\langle\pi_{i\nu}(x)1,1\right\rangle
dx=\int_{G}\int_{K}f(gkx)\left\langle\pi_{i\nu}(x)1,1\right\rangle dkdx$
$=\int_{G}\int_{K}f(gx)\left\langle\pi_{i\nu}(k^{-1}x)1,1\right\rangle
dkdx=\int_{G}f(gx)\left\langle\pi_{i\nu}(x)1,1\right\rangle
dx=\left\langle\pi_{i\nu}(L_{g^{-1}}f)1,1\right\rangle.$

Noting that $\phi(g,I)=f(g)$ completes the proof.

##  4 The holomorphic continuation of Jacquet integrals and a Theorem of
Beuzart-Plessis

Retain the notation of the previous sections. If $\chi:N\rightarrow S^{1}$ is
a unitary (one dimensional) character of $N$ we consider for $u\in
C^{\infty}(M\backslash K)$

$J_{\chi,\nu}(u)=\int_{N}\chi(n)^{-1}u_{\nu}(n)dn.$

Note that if $\chi=1$ and $u=1$ then the integral is the one that defines the
Harish-Chandra c-function. This implies that the integral defining
$J_{\chi,\nu}$ converges absolutely if $\operatorname{Re}(\nu,\alpha)<0$ for
all $\alpha\in\Phi^{+}$. Let $\Delta$ be the set of elements of $\Phi^{+}$
that appear in $\mathfrak{n}/[\mathfrak{n},\mathfrak{n}].$ Thus
$\mathfrak{n}/[\mathfrak{n},\mathfrak{n}]=\oplus_{\alpha\in\Delta}\left(\mathfrak{n}/[\mathfrak{n},\mathfrak{n}]\right)_{\alpha}$.
We say that $\chi$ is generic if its differential is non-zero on each of the
spaces $\left(\mathfrak{n}/[\mathfrak{n},\mathfrak{n}]\right)_{\alpha}$ with
$\alpha\in\Delta$. The Jacquet integrals are the $J_{\chi,\nu}$ with $\chi$
generic. The holomorphic continuation of Jacquet integrals in this generality
(non-K-finite $u$) was first proved in [W1] (c.f. [RRGII] Theorem 15.6.7).

###### Theorem 3

If $\chi$ is generic and if $u\in C^{\infty}(M\backslash K)$ then
$J_{\chi,\nu}(u)$ has a holomorphic continuation to
$\mathfrak{a}_{\mathbb{C}}^{\ast}$, furthermore the map $\nu\mapsto
J_{\chi,\nu}$ is a weakly holomorphic map of
$\mathfrak{a}_{\mathbb{C}}^{\ast}$ to $C^{\infty}(M\backslash K)^{\prime}$
(the continuous dual space).

If $\chi$ is not generic then one can use this result combined with the
meromorphic continuation of conical vectors to prove a meromorphic
continuation result.

The Theorem of Beuzart-Plessis [B] which is based on the holomorphy of
$\nu\mapsto J_{\chi,\nu}(1)$ is

###### Theorem 4

If $\chi$ is a generic character of $N$ then there exists $\varepsilon>0$ such
that

$\int_{\mathrm{\ker}(\chi)}a(n)^{-(1-\varepsilon)\rho}dn<\infty.$

We will also need the following results (see [W3] Theorem 43 for the full
details of a proof of Proposition 6) whose proof is complicated and uses parts
of the proof of the holomorphic continuation of the Jacquet integrals ([W1]).
We will just give an idea of why they are true.

###### Lemma 5

Assume that $\chi$ is generic. There exists a continuous semi-norm, $q$, on
$C^{\infty}(M\backslash K)$ such that

$\left|J_{\chi,\nu}(u)\right|\leq q(u)c(\operatorname{Re}\nu)$

if $u\in C^{\infty}(M\backslash K)$ and $\operatorname{Re}(\nu,\alpha)<0$ for
$\alpha\in\Phi^{+}$.

Proof. We have if $\nu$ satisfies the condition then

$J_{\chi,\nu}(u)=\int_{N}\chi(n)^{-1}u_{\nu}(n)dn=\int_{N}\chi(n)^{-1}a(n)^{\nu-\rho}u(k(n))dn$

so defining $q(u)=\sup_{k\in K}\left|u(k)\right|$ we have

$\left|J_{\chi,\nu}(u)\right|\leq
q(u)\int_{N}a(n)^{\operatorname{Re}\nu-\rho}d\nu=q(u)c(\operatorname{Re}\nu)\text{.}$

###### Proposition 6

Assume that $\chi$ is generic. And let $0<r<\infty.$ There exists a continuous
semi-norm, $q_{r}$, on $C^{\infty}(M\backslash K)$ and $m_{r}$ such that

$\left|J_{\chi,\nu}(u)\right|\leq
q_{r}(u)(1+\left\|\operatorname{Im}\nu\right\|)^{m_{r}}$

for $\nu\in\mathfrak{a}_{\mathbb{C}}^{\ast}$ such that
$0>\operatorname{Re}(\nu,\alpha)>-r(\rho,\alpha),\alpha\in\Phi^{+}$.

Proof. This is proved using an argument involving tensoring with finite
dimensional representations and details from the shift argument used in the
proof of the holomorphic continuation of $J_{\chi,\nu}(u)$ in Section 15.5 of
[RRGII].

###### Proposition 7

Assume that $\chi$ is generic. If $0<r<\infty$ is fixed and if
$f\in\mathcal{C}(G/K)$ then for each $m$ there exists $C_{l,r}$ such that

$|J_{\chi,i\nu-z\rho}(\pi_{i\nu}(f)1)|\leq C_{m,r}(1+\left\|\nu\right\|)^{-m}$

for $\nu\in\mathfrak{a}^{\ast}$ and $0\geq\operatorname{Re}z>-r$

Proof. We note that if $p$ is a continuous semi-norm on
$C^{\infty}(M\backslash K)$ then there exists $k$ and a constant $L$ such that

$p(u)\leq L_{P}\left\|(1+C_{K})^{k_{P}}u\right\|.$

Thus Proposition 6 implies that

$|J_{\chi,i\nu-z\rho}(\pi_{i\nu}(f)1)|\leq
q_{r}(\pi_{i\nu}(f)1)(1+\left\|\nu\right\|)^{m_{r}}$ $\leq
L_{q_{r}}\left\|\pi_{i\nu}(L_{(1+C_{K})^{k_{q_{r}}}}f)1\right\|(1+\left\|\nu\right\|)^{m_{r}}.$

Now applying Lemma 15 in Appendix 1 to $L_{(1+C_{K})^{k_{q_{r}}}}f$ with
$m=m_{r}+l$ completes the proof.

By analogy with the Harish-Chandra Schwartz space we have the Whittaker
Schwartz space : If $g\in G$ and $g=nak$ $n\in N,a\in A,k\in K$ then set
$a_{o}(g)=a$. That is $a_{o}(g)=a(\theta(g))^{-1}$.

$C^{\infty}(N\backslash G;\chi)=\\{f\in C^{\infty}(G)|f(ng)=\chi(n)f(g),n\in
N,g\\}.$

If $f\in C^{\infty}(N\backslash G;\chi),x\in U(\mathfrak{g})$ then set

$q_{x,d}(f)=\sup_{g\in G}a_{o}(g)^{-\rho}(1+\left\|\log
a_{o}(g)\right\|)^{d}|R_{x}f(g)|.$

Then $\mathcal{C}(N\backslash G;\chi)$ is the space of $f\in
C^{\infty}(N\backslash G;\chi)$ such that $q_{x,d}(f)<\infty$ for all $x,d$.

##  5 The key formula

In this section $f\in\mathcal{C}(G/K)$. Then we have seen that

$f(g)=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(L_{g^{-1}}f)1,1\right\rangle\mu(\nu)d\nu=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(f)1,\pi_{i\nu}(g)1\right\rangle\mu(\nu)d\nu.$

Thus

$\overline{f(g)}=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(g)1,\pi_{i\nu}(f)1\right\rangle\mu(\nu)d\nu.$

###### Theorem 8

$\int_{N}\chi(n)\overline{f(ng)}dn=\int_{\mathfrak{a}^{\ast}}J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)1)\overline{J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)1)}\mu(\nu)d\nu.$

Proof. In this proof we will use the notation $u(\nu)=\pi_{i\nu}(f)1\in
C^{\infty}(M\backslash K)$). Using the non-compact model for the unitary
principal series (c.f. [W2] 8.4.7) one has

$\left\langle\pi_{i\nu}(g)1,\pi_{i\nu}(f)1\right\rangle=\int_{N}a(ng)^{i\nu-\rho}a(n)^{-i\nu-\rho}\overline{u(\nu)k(n))}dn.$

Thus

$\int_{N}\chi(n)\overline{f(ng)}dn=\int_{N}\chi(n_{1})\int_{\mathfrak{a}^{\ast}}\int_{N}a(nn_{1}g)^{i\nu-\rho}a(n)^{-i\nu-\rho}\overline{u(\nu)_{i\nu}(k(n))}dn\mu(\nu)d\nu
dn_{1}.$

We first deform the parameter and consider

$\int_{N}\chi(n_{1})\int_{\mathfrak{a}^{\ast}}\int_{N}a(nn_{1}g)^{i\nu-(1+z)\rho}a(n)^{-i\nu-(1+z)\rho}\overline{u(\nu)_{i\nu}(k(n))}dn\mu(\nu)d\nu
dn_{1}$

for $\operatorname{Re}z<0$. If we put absolute values on all of the terms we
have

$\int_{N}\int_{\mathfrak{a}^{\ast}}\int_{N}a(nn_{1}g)^{-(1+\operatorname{Re}z)\rho}a(n)^{-(1+\operatorname{Re}z)\rho}\left|\overline{u(\nu)_{i\nu}(k(n))}\right|dn\mu(\nu)d\nu
dn_{1}$

and using Lemma 15 in Appendix 1 (and the notation therein) we have
$\left|\overline{u(\nu)_{i\nu}(k(n))}\right|\leq
C_{1,l}(1+\left\|\nu\right\|)^{-l}$ with $C_{1,l}<\infty$ Thus the integrand
is dominated by

$\int_{\mathfrak{a}^{\ast}}\int_{N}\int_{N}a(nn_{1}g)^{-(1+\operatorname{Re}z)\rho}a(n)^{-(1+\operatorname{Re}z)\rho}dn(1+\left\|\nu\right\|)^{-m}d\nu
dn_{1}$

since $\mu(\nu)\leq B(1+\left\|\nu\right\|)^{r}$ for some $r$ and we take $m$
to be greater than $\dim A$. which converges for $\operatorname{Re}z<0$.
noting that $a(ng)^{-\rho}\leq C_{\omega}a(n)^{-\rho}$ if $g\in\omega$ a
compact set we see that the integral of the absolute values is dominated by a
multiple of

$C_{\omega}^{(1+\operatorname{Re}z)\rho}\int_{N}\int_{N}a(n_{1})^{-(1+\operatorname{Re}z)\rho}a(n)^{-(1+\operatorname{Re}z)\rho}dndn_{1}<\infty.$

We can therefore do the deformed integral in any order. We choose

$\int_{N}\chi(n_{1})\chi(n)^{-1}\int_{\mathfrak{a}^{\ast}}\int_{N}a(n_{1}g)^{i\nu-(1+z)\rho}a(n)^{-i\nu-(1+z)\rho}\overline{u(\nu)(k(n))}dn\mu(\nu)d\nu
dn_{1}=$
$\int_{\mathfrak{a}^{\ast}}J_{\chi^{-1},i\nu-z\rho}(\pi_{i\nu-z\rho}(g)1)\overline{J_{\chi^{-1},i\nu-\bar{z}\rho}(\pi_{i\nu}(f)1)}\mu(\nu)d\nu.$

We are left with taking the limit under the integral sign $z\rightarrow 0$.
This will be done indirectly.

Let $x_{o}$ be perpendicular to $\ker d\chi$ relative to
$\left\langle...,...\right\rangle$ and assume that $d\chi(x_{o})=-i1.$ We
define

$\tau_{z}(t)=\int_{\ker\chi}\int_{\mathfrak{a}^{\ast}}\int_{N}a(n\exp
tx_{o}n_{1}g)^{i\nu-(1+z)\rho}a(n)^{-i\nu-(1+z)\rho}\overline{u(\nu)(k(n))}dn\mu(\nu)d\nu
dn_{1}$

for Re$z\geq 0$. Then Proposition 17 in the Appendix 1 implies that if
$\omega$ is a compact subset of $G$ then

$\left|\tau_{z}(t)\right|\leq
C_{\omega}^{(1+\operatorname{Re}z)}B(1+\left|t\right|)^{d}$

with $C_{\omega},d$ and $B$ finite. Fix $g\in G$. The estimate implies that we
can define a family of tempered distribution on $\mathbb{R}$ by

$T_{z}(\phi)=\int_{-\infty}^{\infty}\tau_{z}(t)\phi(t)dt$

for $\operatorname{Re}z\geq 0$. Dominated convergence implies that the map
$z\mapsto T_{z}$ is a weakly continuous map of $\\{z|\operatorname{Re}z\geq
0\\}$ to $\mathcal{S}^{\prime}(\mathbb{R)}$ (the space of tempered
distributions). Thus if $\mathcal{F}$ is the usual Fourier transform

$\mathcal{F}(\phi)(s)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ist}\phi(t)dt$

which is a continuous linear isomorphism of $\mathcal{S}(\mathbb{R}).$ If $T$
is a tempered distribution we define (as usual)
$\mathcal{F}(T)=T\circ\mathcal{F}$. If $T$ is given by integration by an
element of $L^{1}(\mathbb{R})$, i.e.
$T(\phi)=\int_{-\infty}^{\infty}\tau(t)\phi(t)dt$ with $\tau\in
L^{1}(\mathbb{R})$, then

$\mathcal{F}(T)(\phi)=\int_{-\infty}^{\infty}\mathcal{F(}\tau(t))\phi(t)dt$

If $\operatorname{Re}z>0$ then as we have seen above

$n_{1}\mapsto\int_{\mathfrak{a}^{\ast}}\int_{N}a(nn_{1}g)^{i\nu-(1+z)\rho}a(n)^{-i\nu-(1+z)\rho}\overline{u(\nu)(k(n))}dn\mu(\nu)d\nu$

defines an element of $L^{1}(N_{1}).$ So Fubini’s theorem implies that
$\tau_{z}\in L^{1}(\mathbb{R})$ if $\operatorname{Re}z>0$. If
$\operatorname{Re}z=0$ then

$\tau_{0}(t)=\int_{\ker\chi}\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(\exp
tx_{o}ng)1,\pi_{i\nu}(f)1\right\rangle\mu(\nu)d\nu
dn=\int_{\ker\chi}\overline{f(\exp tx_{o}ng)}dn.$

Since $f$, in particular, is in $\mathcal{C}(G/K)$ the function
$n\mapsto\overline{f(ng)}$ on $N$ is in $L^{1}(N)$. Thus if $z=0$ then
$\tau_{z}\in L^{1}(\mathbb{R})$. Let for $s,t\in\mathbb{R},$ $n\in\ker\chi,$
$\chi_{s}(\exp tx_{o}n)=e^{its}$ Then $\chi_{1}=\chi$ and $\chi_{s}$ is
generic for $s\neq 0$. Note that if $\operatorname{Re}z>0$ and $s\neq 0$ then

$\mathcal{F}(\tau_{z})(s)=\int_{\mathfrak{a}^{\ast}}J_{\chi_{s}^{-1},i\nu-z\rho}(\pi_{i\nu-z\rho}(g)1)\overline{J_{\chi_{s}^{-1},i\nu-\bar{z}\rho}(\pi_{i\nu}(f)1)}\mu(\nu)d\nu.$

Also if $z=0$ and $s\neq 0$ then

$\mathcal{F}(\tau_{0})(s)=\overline{\int_{N}\chi_{s}(n)^{-1}f(ng)dn}.$

Define for $s\neq 0$

$\sigma_{z}(s)=\int_{\mathfrak{a}^{\ast}}J_{\chi_{s}^{-1},i\nu-z\rho}(\pi_{i\nu-z\rho}(g)1)\overline{J_{\chi_{s}^{-1},i\nu-\bar{z}\rho}(\pi_{i\nu}(f)1)}\mu(\nu)d\nu$

then Proposition 7 implies that $\sigma_{z}$ is continuous in $z$ for $s\neq
0$ and $\operatorname{Re}z\geq 0,$ Furthermore, if $\operatorname{Re}z>0$ and
$s\neq 0,$ then $\sigma_{z}(s)=\mathcal{F}(\tau_{z})(s)$. We therefore have if
$\phi$ has support in $\mathbb{R}-\\{0\\},$

$\lim_{\begin{array}[c]{c}\operatorname{Re}z>0\\\ z\rightarrow
0\end{array}}\mathcal{F(\tau}_{z})(\phi)=\mathcal{F(\tau}_{0})(\phi)=\int_{-\infty}^{\infty}\int_{N}\chi_{s}(n)f((ng)dn\phi(s)ds.$

Also

$\lim_{\begin{array}[c]{c}\operatorname{Re}z>0\\\ z\rightarrow
0\end{array}}\int_{-\infty}^{\infty}\int_{\mathfrak{a}^{\ast}}J_{\chi_{s}^{-1},i\nu-z\rho}(\pi_{i\nu-z\rho}(g)1)\overline{J_{\chi_{s}^{-1},i\nu-\bar{z}\rho}(\pi_{i\nu}(f)1)}\mu(\nu)d\nu\phi(s)ds=\int_{-\infty}^{\infty}\sigma_{0}(s)\phi(s)ds$
$\int_{-\infty}^{\infty}\int_{\mathfrak{a}^{\ast}}J_{\chi_{s}^{-1},i\nu}(\pi_{i\nu}(g)1)\overline{J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)1)}\mu(\nu)d\nu\phi(s)ds$

This implies the theorem.

##  6 The spherical Whittaker Inversion Theorem.

Let if $\chi$ is a generic character of $N$ set $J_{\chi,\nu}$ equal to the
corresponding Jacquet integral. We will need the following

###### Lemma 9

Assume $\chi$ is generic. Let $\psi\in\mathcal{C}(N_{o}\backslash G;\chi)$ and
let $\varphi\in C_{c}^{\infty}(N)$ is such that

$\int_{N_{o}}\chi(n)^{-1}\varphi(n)dn=1.$

Set $f(nak)=\varphi(n)\psi(ak)$ for $n\in N_{o},a\in A_{o},k\in K$. Then
$f\in\mathcal{C}(G)$ and if $u\in C^{\infty}(M\backslash K)$ then

$J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)u)=\int_{N\backslash
G}J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)u)\psi(g)dg$

Proof. Appendix 2, Corollary 20 proves that $f\in\mathcal{C}(G)$. We calculate

$\int_{N\backslash
G}J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)u)\psi(g)dg==\int_{A\times
K}a^{-2\rho}J_{\chi^{-1},i\nu}(\pi_{i\nu}(ak)u)\psi(ak)dadk$
$=\int_{N}\chi(n)^{-1}\int_{A\times
K}a^{-2\rho}J_{\chi^{-1},i\nu}(\pi_{i\nu}(ak)u)\varphi(n)\psi(ak)dadkdn$
$=\int_{N\times A\times
K}a^{-2\rho}J_{\chi^{-1},i\nu}(\pi_{i\nu}(nak)u)\varphi(n)\psi(ak)dadkda=\int_{G}J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)u)f(g)dg.$

For each $\nu\in\mathfrak{a}^{\ast}$ $J_{\chi^{-1},i\nu}$ is a continuous
functional on $C^{\infty}(M\backslash K)$. Also since $J_{\chi,i\nu}$ is tame
in the sense of Theorem 15.2.5 in [RRGII], for each $\nu$ in
$\mathfrak{a}^{\ast}$ there exist $C$ and $d$ such that if $u\in
C^{\infty}(M\backslash K)$ then

$\left|J_{\chi^{-1}.i\nu}(\pi_{i\nu}(g)1)\right|\leq
C\left|g\right|^{-\frac{1}{2}}(1+\log\left\|g\right\|)^{d}.$

This implies that

$\int_{G}J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)u)f(g)dg=J_{\chi^{-1},i\nu}(\int_{G}f(g)\pi_{i\nu}(g)dg)=J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)u).$

###### Theorem 10

Let $\psi\in\mathcal{C}(N\backslash G/K;\chi)$. Set

$W_{\chi}(\nu,\psi)=\int_{N\backslash
G}\overline{J_{i\nu}(\pi_{i\nu}(g)1)}\psi(g)dg$

then

$\psi(g)=\int W_{\chi}(\nu)J_{i\nu}(\pi_{i\nu}(g)1)\mu(\nu)d\nu.$

Proof. Let $f$ be as in the preceding lemma for $\psi$. We observe that

$\overline{J_{i\nu}(\pi_{i\nu}(g)1)}=J_{\chi^{-1},-i\nu}(\pi_{-i\nu}(g)1).$

Thus the preceding lemma implies that

$W_{\chi}(\nu,\psi)=J_{\chi^{-1},-i\nu}(\pi_{i\nu}(f)1).$

The spherical Plancherel Theorem implies that

$f(g)=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(L_{g^{-1}}f)1,1\right\rangle\mu(\nu)dv=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(f)1,\pi_{i\nu}(g)1\right\rangle\mu(\nu)dv$

so

$\bar{f}(g)=\overline{f(g)}=\int_{\mathfrak{a}^{\ast}}\left\langle\pi_{i\nu}(g)1,\pi_{i\nu}(f)1\right\rangle\mu(\nu)dv.$

Theorem 8 implies that

$\bar{f}_{\chi^{-1}}(g)=\int_{\mathfrak{a}^{\ast}}J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)1)\overline{J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)1)}\mu(\nu)dv,$

Thus

$\psi(g)=f_{\chi}(g)=\int_{\mathfrak{a}^{\ast}}\overline{J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)1)}J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)1)\mu(\nu)dv.$

The first part of this proof implies that

$J_{\chi^{-1},i\nu}(\pi_{i\nu}(f)1)=W_{\chi}(-i\nu,\psi).$

Also,

$\overline{J_{\chi^{-1},i\nu}(\pi_{i\nu}(g)1)}=J_{-i\nu}(\pi_{-i\nu}(g)1).$

Hence

$\psi(g)=\int_{\mathfrak{a}^{\ast}}W_{\chi}(-\nu,\psi)J_{-i\nu}(\pi_{-\nu}(g)1)\mu(\nu)dv.$

This proves the theorem since $\mu(\nu)=\mu(-\nu)$.

###### Corollary 11

With the notation as in the previous theorem, if
$f,h\in\mathcal{C}(N\backslash G/K;\chi)$ then

$\int_{N\backslash
G}f(g)\overline{h(g)}dg=\int_{\mathfrak{a}^{\ast}}W_{\chi}(\nu,f)\overline{W_{\chi}(\nu,h)}d\nu.$

Proof. We calculate. The previous theorem implies that

$\int_{N\backslash G}f(g)\overline{h(g)}dg=\int_{N\backslash
G}\int_{\mathfrak{a}^{\ast}}W_{\chi}(\nu,f)J_{\chi,i\nu}(\pi_{i\nu}(g)1)d\nu\overline{h(g)}dg$
$=\int_{\mathfrak{a}^{\ast}}W_{\chi}(\nu,f)\overline{\int_{N\backslash
G}h(g)\overline{J_{\chi,i\nu}(\pi_{i\nu}(h)1)}dg}d\nu=\int_{\mathfrak{a}^{\ast}}W_{\chi}(\nu,f)\overline{W_{\chi}(\nu,h)}d\nu.$

##  7 The non-periodic Toda Lattice

The original non-periodic Toda Lattice is the Hamiltonian system with
Hamiltonian

$H(p,q)=\frac{1}{2}\sum_{i=1}^{n}p_{i}^{2}+\sum_{i=1}^{n-1}c_{i}^{2}e^{2(q_{i}-q_{i+1})}$

$c_{i}\in\mathbb{R}-\\{0\\}$. Using the quantization rules (here Planck’s
constant is normalized) $p_{j}\rightarrow i\frac{\partial}{\partial q_{j}}$
and $f(q)\rightarrow m_{f(q)}$ with $m_{f(q)}$ the operator on,
$C^{\infty}(\mathbb{R}^{n})$, given by multiplication by $f(q)$. Thus the
quantum Hamiltonian is

$\mathcal{H=}-\frac{1}{2}\Delta_{q}+\sum_{i=1}^{n-1}c_{i}^{2}e^{2(q_{i}-q_{i+1})}$

where

$\Delta_{q}=\sum_{i=1}^{n}\frac{\partial^{2}}{\partial q_{j}^{2}}.$

Consider the algebra, $\mathcal{A}$, of linear differential operators on
$\mathbb{R}^{n}$ with coefficients in the algebra generated by
$e^{(q_{i}-q_{i+1})},i=1,...,n-1$. We take as a domain for this algebra the
space, $\mathcal{T}$, of $f\in C^{\infty}(\mathbb{R}^{n})$ such that

$t_{m,d,x}(f)=\sup_{q\in\mathbb{R}^{n}}e^{\sum_{j=1}^{n-1}m_{i}(q_{i}-q_{i+1})}(1+\left\|q\right\|)^{d}|xf(q)|<\infty$

with $m=(m_{1},...,m_{n}),m_{i}$ and $d\in\mathbb{Z}_{\geq 0}$ and $x$ is a
constant coefficient differential operator on $\mathbb{R}^{n}$ and endowed
with the topology induced by these semi-norms. This space is invariant under
$\mathcal{A}$ and $\mathcal{A}$ acts continuously on it. In [GW] Toda 1
Section 2 it was shown the centralizer of $\mathcal{H}$ in $\mathcal{W}$ is an
algebra generated over $\mathbb{C}$ by $n$ elements
$D_{1}=\sum_{i=1}^{n}\frac{\partial}{\partial
q_{j}},D_{2}=\mathcal{H},...,D_{n}$ with algebraically independent symbols,
$\sigma(D_{1}),...,\sigma(D_{n})$ generators for the $S_{n}$ invariant
constant coefficient differential operators. A solution to the quantum Toda
lattice is thus a family $K_{\nu}(q)$,
$\nu\in\left(\mathbb{R}^{n}\right)^{\ast}$ such that

$\int_{\left(\mathbb{R}^{n}\right)^{\ast}}|K_{\nu}(q)f(q)|dq<\infty,f\in\mathcal{T}$

and

$D_{j}K_{\nu}=\sigma_{i}(\nu)K_{\nu},j=1,...,n.$

One has the following inversion formula: There exists a non-negative function
$\gamma(\nu)$ on $\left(\mathbb{R}^{n}\right)^{\ast}$ such that
$\left|\gamma(\nu)\right|\leq C(1+\left\|\nu\right\|)^{r}$ for
$\nu\in\left(\mathbb{R}^{n}\right)^{\ast}$ and such that if $f\in\mathcal{T}$
and if

$\mathcal{K}(f)(\nu)=\int_{\mathbb{R}^{n}}f(q)\overline{K_{\nu}(q)}dq$

then

$f(q)=\int_{\left(\mathbb{R}^{n}\right)^{\ast}}\mathcal{K}(f)(\nu)K_{\nu}(q)\gamma(\nu)d\nu.$

and if $f_{1},f_{2}\in\mathcal{T}$ then

$\int_{\mathbb{R}^{n}}f_{1}(x)\overline{f_{2}(x)}dx=\int_{\left(\mathbb{R}^{n}\right)^{\ast}}\mathcal{K}(f_{1})(\nu)\mathcal{K}(f_{2})(\nu)\gamma(\nu)d\nu$

We now return to the situation of the preceding sections. Let $G$ and the
notation be as in Section 2 so we assume that, in particular, $G\subset
GL(n,\mathbb{R})$ for some $n$. In particular, $\mathfrak{g=}Lie(G)$. Let $C$
be the Casimir operator corresponding to the invariant form

$B(X,Y)=\mathrm{tr}XY$

for $X,Y$ in $\mathfrak{g}$. That is, if $X_{1},...,X_{m}$ is a basis of
$\mathfrak{g}$ and $Y_{1},...,Y_{n}$ are defined by
$B(X_{i},Y_{j})=\delta_{ij}$ then

$C=\sum X_{i}Y_{i}\in U(\mathfrak{g}).$

Let $\theta,N,A,K,\mathfrak{n},\mathfrak{a},\mathfrak{k}$,$\Phi^{+},\Delta$ be
as before. Then $\mathfrak{n=}\sum_{\alpha\in\Phi^{+}}\mathfrak{n}_{\alpha}$
and let $X_{\alpha,j}$,$j=1,...,m_{\alpha}$ be an orthonormal basis of
$\mathfrak{n}_{\alpha}$ relative to the inner product

$\left\langle X,Y\right\rangle=-B(X,\theta Y).$

We define the generalized quantum non-periodic Toda Lattices associated with
$G$ to be the operator on $C^{\infty}(A)$ given by

$L_{c}=-\frac{\sum
h_{i}^{2}}{2}+\sum_{\alpha\in\Delta}c_{\alpha}^{2}a^{2\alpha}$

with $c_{\alpha}\in\mathbb{R}-\\{0\\}$. For $G=GL(n,\mathbb{R})$ then take $A$
to be the group of diagonal $n\times n$ matrices with positive coefficients
then and $N$ the group of upper triangular $n\times n$ matrices with ones on
the main diagonal. Then identifying $\mathfrak{a}$ with $\mathbb{R}^{n}$ via
the map

$(x_{1},...,x_{n})\mapsto\left[\begin{array}[c]{cccc}x_{1}&0&\cdots&0\\\
0&x_{2}&\cdots&0\\\ 0&0&\ddots&0\\\ 0&0&\cdots&x_{n}\end{array}\right]$

we have $\Delta=\\{\alpha_{1},...,\alpha_{n-1}\\}$ with
$\alpha_{i}(x)=x_{i}-x_{i-1}$. Thus

$L_{c}=-\frac{1}{2}\sum_{i=1}^{n}\frac{\partial^{2}}{\partial
x_{j}^{2}}+\sum_{i=1}^{n-1}c_{\alpha_{i}}^{2}e^{2(x_{i}-x_{i+1})}$

Set $K_{\nu}(x)=a^{-\rho}J_{i\nu}(\pi_{\nu}(\exp x)1)$ for $x\in\mathfrak{a}.$
Let $(...,...)$ also denote the complex bilinear extension of the dual form,
$(...,...)$ of $B_{|\mathfrak{a}}$ to $\mathfrak{a}_{\mathbb{C}}^{\ast}$.

###### Proposition 12

Let $\chi$ be a generic character of $N$. Set for $\alpha\in\Delta$

$c_{\alpha}^{2}=-\sum_{j=1}^{m_{\alpha}}\left(d\chi(X_{\alpha,j})\right)^{2}>0$

Let If $\nu\in\mathfrak{a}^{\ast}$ then

$L_{c}a^{-\rho}K_{i\nu}(a)=\frac{\left\|\nu\right\|^{2}}{2}a^{-\rho}K_{i\nu}(a).$

Proof. We have for $\mu\in\mathfrak{a}_{\mathbb{C}}^{\ast}$

$CJ_{\nu}(\pi_{\nu}(g)1)=J_{\nu}(\pi_{\nu}(g)d\pi_{\nu}(C)1)=((\nu,\nu)-(\rho,\rho))J_{\nu}(\pi_{\nu}(g)1).\text{
}$

The proposition now follows directly from the calculations in Appendix 3.

We note that if $\mathcal{W}(\mathfrak{a})$ is as in Appendix 2 and if
$\phi\in\mathcal{W}(\mathfrak{a})$ then the function $a^{-\rho}\phi$ is in the
space $\mathcal{T}(\mathfrak{a})$ defined as follows: Define for $u\in
C^{\infty}(\mathfrak{a})$ the semi-norm

$t_{d,m,x}(u)=\sup_{x\in\mathfrak{a}}e^{\sum_{\alpha\in\Delta}c_{\alpha}\alpha(h)}(1+\left\|h\right\|)^{d}\left|x\phi(h)\right|$

with $m=\\{m_{\alpha}|\alpha\in\Delta\\},m_{\alpha},d\in\mathbb{Z}_{\geq 0}$
and $x$ is a constant coefficient differential operator on $\mathfrak{a}$.
$\mathcal{T}(\mathfrak{a})$ is the space of all $u$ in
$C^{\infty}(\mathfrak{a})$ such that all of the $t_{d,m,x}(u)<\infty$ endowed
with the topology induced by these semi-norms. Then the map

$\mu\mapsto(h\mapsto e^{-\rho(h)}\mu(\exp h)$

defines a topological isomorphism of $\mathcal{W}(\mathfrak{a})$ onto
$\mathcal{T}(\mathfrak{a})$ with inverse $u\mapsto(a\mapsto e^{\rho}u(\log
a))$. The main results of the previous section can be stated in the following
form.

###### Theorem 13

If $u\in\mathcal{T}(\mathfrak{a}),\nu\in\mathfrak{a}^{\ast}$ set

$\mathcal{K}(u)(\nu)=\int_{\mathfrak{a}}u(h)\overline{K_{\nu}(h)}dh$

then

$u(h)=\int_{\mathfrak{a}^{\ast}}K_{\nu}(h)\mathcal{K}(u)(\nu)\mu(\nu)d\nu.$

Furthermore, if $u,w\in\mathcal{T}(\mathfrak{a})$ then

$\int_{\mathfrak{a}}u(h)\overline{w(h)}dh=\int_{\mathfrak{a}^{\ast}}\mathcal{K}(u)(\nu)\overline{\mathcal{K}(w)(\nu)}\mu(\nu)d\nu.$

Proof. We note that if $u\in\mathcal{T}(\mathfrak{a})$ then
$u(h)=e^{-\rho(h)}\phi(\exp h)$ with
$\phi\in\mathcal{W}=\mathcal{C}(N\backslash G/K;\chi)_{|A\text{.}}$ Thus if
$\psi(nak)=\chi(n)\phi(a)$ then

$\mathcal{K}(u)(\nu)=\int_{A}a^{-\rho}\phi(a)a^{-\rho}\overline{J_{i\nu}(\pi_{i\nu}(a)1)}da=\int_{A}a^{-2\rho}\phi(a)\overline{J_{i\nu}(\pi_{i\nu}(a)1)}da$
$=\int_{N\backslash
G}\psi(g)\overline{J_{i\nu}(\pi_{i\nu}(g)1)}dg=\mathcal{W}_{\chi}(\nu,\psi).$

The theorem says that

$\psi(g)=\int_{\mathfrak{a}^{\ast}}\mathcal{W}_{\chi}(\nu,\psi)J_{i\nu}(\pi_{i\nu}(g)1)\mu(\nu)d\nu.$

So

$u(h)=e^{-\rho(h)}\psi(\exp(h))=e^{-\rho(h)}\int_{\mathfrak{a}^{\ast}}\mathcal{K}(u)(\nu)J_{i\nu}(\pi_{i\nu}(\exp
h)1)\mu(\nu)d\nu=\int_{\mathfrak{a}^{\ast}}\mathcal{K}(u)(\nu)K_{\nu}(h)\mu(\nu)d\nu.$

The above formulas lead to the second assertion of the theorem.

The rest of this section involves recalling several results from [GW].Since
$\mathfrak{g}=\mathfrak{n}\oplus\mathfrak{a}\oplus\mathfrak{k}$ the Poincaré-
Birkhof-Witt Theorem implies that

$U(\mathfrak{g})=U(\mathfrak{n}\oplus\mathfrak{a)\oplus}U(\mathfrak{g})\mathfrak{k}\text{.}$

Let $p$ denote the projection of $U(\mathfrak{g})$ onto
$U(\mathfrak{n}\oplus\mathfrak{a)}$ corresponding to this direct sum
decomposition. If $x\in U(\mathfrak{g})^{\mathfrak{k}}$ (the centralizer of
$\mathfrak{k}$ in $U(\mathfrak{g)}$) and if $y\in U(\mathfrak{g})$ then
$y=p(y)+\sum u_{i}Y_{i}$ and $x=p(x)+\sum w_{i}Y_{i},$with
$Y_{i}\in\mathfrak{k},u_{i},w_{i}\in U(\mathfrak{g}).$ Thus

$yx=p(y)x+\sum u_{i}Y_{i}x=p(y)x+\sum u_{i}xY_{i}$ $=p(y)p(x)+p(y)\sum
w_{i}Y_{i}+\sum u_{i}xY_{i}.$

This $p(yx)=p(y)p(x)$. Consider the two sided ideal in
$U(\mathfrak{n}\oplus\mathfrak{a}),\mathcal{I=}U(\mathfrak{n}\oplus\mathfrak{a})[\mathfrak{n},\mathfrak{n}]$.
Then

$U(\mathfrak{n}\oplus\mathfrak{a})/\mathcal{I\cong}U(\mathfrak{n/}[\mathfrak{n},\mathfrak{n}]\oplus\mathfrak{a}).$

Let $\mu:$ $U(\mathfrak{n}\oplus\mathfrak{a})\rightarrow
U(\mathfrak{n/}[\mathfrak{n},\mathfrak{n}]\oplus\mathfrak{a})$ be the
corresponding surjection and $q:U(\mathfrak{g})^{\mathfrak{k}}\rightarrow
U(\mathfrak{n/}[\mathfrak{n},\mathfrak{n}]\oplus\mathfrak{a})$ be the
corresponding homomorphism, that is $q=\mu\circ p$. If $h\in\mathfrak{a}$ then
define $\partial_{h}f(x)=\frac{d}{dt}f(x+th)_{\backslash t=0}$. If
$\lambda\in\mathfrak{a}^{\ast}$ define $m_{\lambda}f=e^{\lambda}f$. Then
$[\partial_{h},m_{\lambda}]=\lambda(h)m_{\lambda}$. Let $\mathcal{A}$ be the
algebra of operators on $C^{\infty}(\mathfrak{a})$ generated by $\partial_{h}$
and $m_{\alpha}$ for $h\in\mathfrak{a}$ and $\alpha\in\Delta$. We define
$\tau(h)=\partial_{h}$ and $\tau(x)=d\chi(x)m_{\alpha}$ if
$x\in\mathfrak{n}/[\mathfrak{n}.\mathfrak{n]}_{\alpha}$. Then $\tau\circ q$
defines a homomorphism of
$U(\mathfrak{n/}[\mathfrak{n},\mathfrak{n}]\oplus\mathfrak{a})$ onto
$\mathcal{A}$. In [GW] we proved

###### Theorem 14

The centralizer of $\tau\circ q(C)$ in $\mathcal{A}$ is $\tau\circ
q(U(\mathfrak{g})^{\mathfrak{k}})$.

This result implies that the centralizer of $L_{m}$ in $\mathcal{A}$ is an
algebra generated by $\dim A$ elements with algebraically independent symbols.
The results of [GW] imply that if $D\in\mathcal{A}$ and $[D,L_{c}]=0$ then
$DK_{\nu}=\sigma(D,i\nu$)$K_{\nu}$. Thus in particular all of the assertions
for the quantum non-periodic Toda lattice have been proved.

##  8 Appendix 1: Some inequalities

The purpose of this appendix is to prove some estimates that will be used in
the body of the paper. The notation is as in Section 2.

###### Lemma 15

Let $f\in\mathcal{C}(G/K)$ then for each $l\geq 0$ and $D$ a constant
coefficient differential operator on $\mathfrak{a}^{\ast}$ there exists
$B_{D,l}$ such that

$\left|D(\pi_{i\nu}(f)1)(k)\right|\leq
B_{D,l}(1+\left\|\nu\right\|)^{-d},\nu\in\mathfrak{a}^{\ast},k\in K.$

Proof. By definition

$\left(\pi_{i\nu}(f)1\right)(k)=\int_{G}f(g)a(kg)^{i\nu-\rho}dg=\int_{G}f(k^{-1}g)a(g)^{i\nu-\rho}dg.$

Up to normalization of measures one has the standard integration formula (c.f.
[W2] 7.7.4)

$\int_{G}\varphi(g)dg=\int_{\bar{N}\times A\times
K}a^{2\rho}\varphi(\bar{n}ak)d\bar{n}dadk.$

Thus

$\left(\pi_{i\nu}(f)1\right)(k)=\int_{\bar{N}\times
A}a^{2\rho}f(k^{-1}\bar{n}a)a^{i\nu-\rho}d\bar{n}da=\int_{\bar{N}\times
A}a^{\rho}f(k^{-1}\bar{n}a)a^{i\nu}d\bar{n}da.$

The map $\varphi\mapsto(h\mapsto e^{\rho(h)}\int_{\bar{N}}\varphi(\bar{n}\exp
h)d\bar{n}=\varphi^{\bar{P}})$ is a continuous map of $\mathcal{C}(G/K)$ to
$\mathcal{S}(\mathfrak{a})$ (c.f. Theorem 7.2.1 [RRGI]..The map $f\mapsto
L_{k}f$ is a continuous map of $K$ to $\mathcal{C}(G/K)$ and the Fourier
transform, $\mathcal{F}$ , is a continuous map from
$\mathcal{S}(\mathfrak{a})$ to $\mathcal{S}(\mathfrak{a}^{\ast})$. So if $q$
is a continuous semi-norm on $\mathcal{S}(\mathfrak{a}^{\ast})$ then
$q(\mathcal{F}((R_{k}f)^{\bar{P}})\leq B_{q}.$ Thus if
$q_{D,l}(u)=\sup_{\nu\in\mathfrak{a}^{\ast}}(1+\left\|\nu\right\|)^{l}\left|D\alpha(\nu)\right|$
for $l\geq 0$ and $D$ a constant coefficient differential operator on
$\mathfrak{a}^{\ast}$ then since $K$ is compact

$\max q_{D,l}(\mathcal{F}((R_{k}f)^{\bar{P}}))\leq B_{D_{l},q_{l}}.$

This implies that

$\left|D\mathcal{F}((R_{k}f)^{\bar{P}})(\nu)\right|\leq
B_{D,q_{l}}(1+\left\|\nu\right\|)^{-l}.$

Unraveling the above we have

$\left(\pi_{i\nu}(f)1\right)(k)=\mathcal{F}((R_{k}f)^{\bar{P}})(-\nu)$

so the lemma is proved.

###### Lemma 16

If $x,y\in G$ then
$a(xg)^{-\rho}\leq\left|g\right|^{\frac{1}{2}}a(x)^{-\rho}.$

Proof. Let (as in section 2) $E=\wedge^{m}\mathfrak{g}$ ($m=\dim\mathfrak{n}$)
and let $u_{o}\in\wedge^{m}\mathfrak{\bar{n}}$ be a unit vector. If $g\in G$
with $g=\bar{n}a(g)k$ with $\bar{n}\in\overline{N}$ and $k\in K$ then
$\left\|\wedge^{m}g^{-1}u_{o}\right\|=\left\|\wedge^{m}k^{-1}\wedge^{m}a(g)^{-1}\wedge^{m}\bar{n}^{-1}u_{o}\right\|=a(g)^{2\rho}$.
This implies that if $x,g\in G$ then

$a(x)^{2\rho}=\left\|\wedge^{m}g\wedge^{m}g^{-1}\wedge^{m}x^{-1}v_{o}\right\|\leq\left|g\right|a(xg)^{2\rho}$

hence

$a(xg)^{-\rho}\leq\left|g\right|^{\frac{1}{2}}a(x)^{-\rho}.$

Let $x_{o}\in\mathfrak{n}$ be orthogonal to $[\mathfrak{n},\mathfrak{n}]$ and
such that $d\chi(x_{o})=i$.

###### Proposition 17

Let $f\in\mathcal{C}(G/K)$ and if $\nu\in\mathfrak{a}^{\ast}$ then set
$u(\nu)=\pi_{i\nu}(f)1\in C^{\infty}(M\backslash K).$ If $\omega$ is a compact
subset of $G$ then there exist constants $M,D,C_{\omega},d<\infty$ such that
if $g\in\omega$ then

$\left|\int_{\ker\chi}\int_{\mathfrak{a}^{\ast}}\int_{N}1_{i\nu-(1+z)\rho}(n_{1}\exp
tx_{o})ng)\overline{u(\nu)_{i\nu-(1+\bar{z}\rho)}(n_{1})}dn_{1}\mu(\nu)d\nu
dn\right|\leq C_{\omega}^{1+\operatorname{Re}z}DM(1+\left|t\right|)^{d}.$

Proof. We put the absolute values inside the integration. We are estimating

$I=\int_{\ker\chi}\int_{\mathfrak{a}^{\ast}}\int_{N}a(n_{1}n(\exp
tx_{o})g)^{-(1+\operatorname{Re}z)\rho}a(n_{1})^{-(1+\operatorname{Re}z)\rho}\left|u(\nu)(k(n_{1}))\right|dn_{1}\mu(\nu)d\nu
dn.$

We note that

$\mu(\nu)\leq B(1+\left\|\nu\right\|)^{r}$

for some $r$ and all $\nu\in\mathfrak{a}^{\ast}$. In Lemma 15 we showed that
there exists a constant $L$ such that

$\left|u(\nu)\right|\leq L_{m}(1+\left\|\nu\right\|)^{-r-m}$

with $m$ arbitrary we take $m$ to be any $m>\dim\mathfrak{a}.$ Thus

$I\leq M\int_{\ker\chi}\int_{N}a(n_{1}n(\exp
tx_{o})g)^{-(1+\operatorname{Re}z)\rho}a(n_{1})^{-(1+\operatorname{Re}z)\rho}dn_{1}\mu
dn$

with

$M=BL_{m}\int_{\mathfrak{a}^{\ast}}(1+\left\|\nu\right\|)^{-m}d\nu<\infty.$

Also, Lemma 16 implies that

$a(n_{1}n(\exp
tx_{o})g)^{-(1+\operatorname{Re}z)\rho}\leq\left|g\right|^{\frac{{}^{1+\operatorname{Re}z}}{2}}a(n_{1}n(\exp
tx_{o}))^{-(1+\operatorname{Re}z)\rho}\leq$
$\left|g\right|^{\frac{{}^{1+\operatorname{Re}z}}{2}}a(n_{1}n\exp
tx_{o}))^{-\rho}\leq\exp
tx_{o}|^{\frac{1}{2}}\left|g\right|^{\frac{{}^{1+\operatorname{Re}z}}{2}}a(n_{1}n)^{-\rho}.$

Since $\wedge^{m}Ad(extx_{o})$ is a polynomial in $t$ there exists a constant
$Q<\infty$ such that $\leq Q(1+\left|t\right|)^{d}$. Setting
$C_{\omega}=\max_{g\in\omega}\left|g\right|^{\frac{1}{2}}$

$I\leq
MC_{\omega}^{\frac{1+\operatorname{Re}z}{2}}Q(1+\left|t\right|)^{d}\int_{\ker\chi}\int_{N}a(n_{1}n)^{-(1+\operatorname{Re}z)\rho}a(n_{1})^{-(1+\operatorname{Re}z)\rho}dn_{1}dn$
$\leq
MC_{\omega}^{\frac{1+\operatorname{Re}z}{2}}Q(1+\left|t\right|)^{d}\int_{\ker\chi}\int_{N}a(n_{1}n)^{-\rho}a(n_{1})^{-\rho}dn_{1}dn$
$=MC_{\omega}^{\frac{1+\operatorname{Re}z}{2}}\left(Q(1+\left|t\right|)^{d}\right)^{\frac{1+\operatorname{Re}z}{2}}\int_{\ker\chi}\Xi(n)dn.$

As in the proof of Lemma 16 we have
$a(g)^{\rho}=\left\|\wedge^{m}Ad(g)^{-1}v_{o}\right\|^{\frac{1}{2}}\leq\left|g^{-1}\right|^{\frac{1}{2}}.$
Also,

$\Xi(x)\leq N\left|x\right|^{-\frac{1}{2}}(1+\log\left|x\right|)^{s}$

for some $s,N<\infty$ (c.f. Theorem 5.5.3 [RRGI]). Hence

$\Xi(x)=\Xi(x^{-1})\leq
N\left|x^{-1}\right|^{-\frac{1}{2}}(1+\log\left|x^{-1}\right|)^{s}\leq
N_{\varepsilon}\left|x^{-1}\right|^{-\frac{1}{2}+\varepsilon}\leq
N_{\varepsilon}a(x)^{-(1-2\varepsilon)}$

for each $\varepsilon>0$. Theorem 4 says that if $\varepsilon$ is sufficiently
small

$\int_{\text{ker}\chi}a(n)^{-(1-2\varepsilon)\rho}dn<\infty.$

Completing the proof.

##  9 Appendix 2: The restriction of $\mathcal{C}(N\backslash G/K;\chi)$ to
$A$

The purpose of this appendix is to give a complete description of the
restriction in its title.

###### Lemma 18

If $m=(m_{1},...,m_{l}),m_{i},d\in\mathbb{Z}_{\geq 0}$ then there exists a
continuous semi-norm, $q_{m,d}$,on $\mathcal{C}(N\backslash G;\chi)$ such that
if $f\in\mathcal{C}(N\backslash G;\chi)$ then $|f(\exp hk)|\leq
q_{m,d}(f)e^{\rho(h)}e^{-\sum m_{i}\alpha_{i}(h)}(1+\left\|h\right\|)^{d}$ for
$h\in\mathfrak{a}$.

Proof. Let $F=\\{i|m_{i}>0\\}$ and let $x_{1},...,x_{n}$ be a basis of
$\mathfrak{g}$. If $X\in\mathfrak{g}$ and if $k\in K$ then we can write
$Ad(k)X=\sum a_{i}(k,X)x_{i}$. Note that there exists $C$ such that

$\left|a_{i}(k,X)\right|\leq C\left\|X\right\|$

for all $k\in K$. Now let $X_{i}$ be an element of the $\alpha_{i}$ root space
in $\mathfrak{n}_{o}$ such that $d\chi(X_{i})=z_{i}\neq 0$. Then

$f(\exp(h)k)=z_{i}^{-1}L_{X_{i}}f(\exp(h)k)=z_{i}^{-1}\frac{d}{dt}_{|t=0}f(\exp(tX_{i})\exp(h)k)=$
$z_{i}^{-1}\frac{d}{dt}_{|t=0}f(\exp(h)\exp(tAd(\exp(-h))X_{i})k)=$
$z_{i}^{-1}\frac{d}{dt}_{|t=0}f(\exp(h)\exp(te^{-\alpha_{i}(h)}X_{i})k)=$
$z_{i}^{-1}\frac{d}{dt}_{|t=0}f(\exp(h)\exp(te^{-\alpha_{i}(h)}Ad(k^{-1})X_{i})k)=$
$e^{-\alpha_{i}(h)}z_{i}^{-1}\sum a_{j}(k^{-1},X_{i})R_{x_{i}}f(\exp(h)k).$

Iterating this argument yields an expression

$f(\exp(h)k)=e^{-\sum_{i\in F}m_{i}\alpha_{i}(h)}Z(k)f(ak)$

with $Z$ a smooth function from $K$ to $L=U^{\sum_{i\in
F}m_{i}}(\mathfrak{g})$ with $U^{j}(\mathfrak{g})$ the standard filtration. If
we choose a basis of $L$, $y_{1},...,y_{r}$ then we have

$Z(k)=\sum b_{i}(k)y_{i}$

with $b_{i}$ continuous functions on $K.$ Let $C_{j}=\max_{k\in
K}\left|b_{j}(k)\right|$. We have

$\left|f(\exp(h)k)\right|\leq e^{-\sum_{i\in
F}m_{i}\alpha_{i}(h)}\sum_{j}C_{j}\left|y_{j}f(\exp(h)k)\right|\leq$
$e^{-\sum_{i\in
F}m_{i}\alpha_{i}(h)}(1+\left\|h\right\|)^{-d}e^{\rho_{o}(h)}\sum_{j}C_{j}q_{d,y_{j}}(f).$

###### Lemma 19

Let $\psi\in C^{\infty}(G)$ be expressed in the form

$\psi(nak)=\sum_{i=1}^{r}\sum_{j=1}^{s}\phi_{i}(n)f_{ij}(a)\gamma_{j}(k)$

for $n\in N,a\in A,k\in K$ with $r,s<\infty$, $\phi_{i}\in
C_{c}^{\infty}(N),\gamma_{j}\in C^{\infty}(K)$ and $f_{ij}\in C^{\infty}(A)$
such that if $m=(m_{1},...,m_{l})$ with $m_{i}\in\mathbb{Z}_{\geq
0},d\in\mathbb{Z}_{\geq 0}$ and $x\in U(\mathfrak{a})$ then there exists
$C_{ij,m,d,x}$ such that

$\left|xf_{ij}(a)\right|\leq
C_{ij,m,d,x}a^{\rho}a^{-c_{1}\alpha_{1}-...-c_{l}\alpha_{l}}(1+\left\|\log
a\right\|)^{-d}.$

Then for $d$ $\in\mathbb{Z}_{\geq 0}$ there exists $B_{d}$ such that

$\left|\psi(g)\right|\leq
B_{d}\left|g\right|^{-\frac{1}{2}}(1+\log\left\|g\right\|)^{-d}.$

Also, if $x,y\in U(\mathfrak{g)}$ then $L_{x}R_{y}\psi$ is of the same form.

Proof. To prove the inequality we may assume that $r,s=1$ so

$\psi(nak)=\phi(n)f(a)\gamma(k)$

Let $\omega$ be the the supports of $\phi$. Let $c_{1}\geq 1$ be such that

$\max\\{\left\|n\right\|,\left\|n^{-1}\right\|\\}\leq
c_{1},\min\\{\left\|n\right\|,\left\|n^{-1}\right\|\\}\geq
c_{1}^{-1},n\in\omega.$

We note that $\left|n\right|^{\frac{1}{2}}=\left\|n\right\|$ and
$|k|=\left\|k\right\|=1.$ We have for $n\in N,a\in A.k\in K$

$\left|nak\right|=\left|na\right|\leq c_{1}\left|a\right|$

and

$\left|a\right|=\left|n^{-1}nak\right|\leq c_{1}\left|nak\right|.$

By the same argument we have the same inequalities for $\left\|...\right\|$.
If $h\in\mathfrak{a}$ let $s\in W(A)$ be such that $\alpha(sh)\geq
0,\alpha\in\Phi^{+}$. Then
$\left|a\right|^{\frac{1}{2}}=\exp(sh)^{\rho}=a^{s^{-1}\rho}$ so if $s_{o}$ is
the element of $W(A)$ such that $s_{o}\Phi^{+}=-\Phi^{+}$ then

$\left|nak\right|^{-\frac{1}{2}}\leq
c_{1}\left|a\right|^{-\frac{1}{2}}=c_{1}a^{s^{-1}s_{o}\rho}.$

Also note that $s^{-1}s_{o}\rho=\rho-\sum_{i=1}^{l}u_{i}\alpha_{i}$ with
$u_{i}\in\mathbb{Z}_{\geq 0}$. We also leave it to the reader to check that
there exists $c_{2}>0$ such that $\left\|a\right\|\leq e^{c_{2}\left\|\log
a\right\|}$ for $a\in A$. Thus $(1+\log\left\|a\right\|)\geq
c_{3}(1+\left\|\log a\right\|).$ With these observations in place we have

$\left|\psi(nak)\right|\leq\left(\sup_{n\in\omega,k\in
K}\left|\phi(n)\gamma(k)\right|\right)\left|f(a)\right|=c_{4}f(a)\leq
c_{4}C_{1,1,u,d}a^{\rho-\sum u_{i}\alpha_{i}}(1+\left\|\log a\right\|)^{-d}$
$=c_{4}C_{1,1,u,d}\left|a\right|^{-\frac{1}{2}}c_{3}^{-d}(1+\log\left\|a\right\|)^{d}.$

Thus if we take the maxima of the $C_{1,1,u,d}$ for the $s\in W(A)$ and
incorporate the constants that appear in the estimates at the beginning of the
proof we have

$\psi(g)|\leq C_{d}\left|g\right|^{-\frac{1}{2}}(1+\log\left\|g\right\|)^{-d}$

as asserted.

To complete the proof of the lemma we now we consider the derivatives. It is
enough to show that $R_{X}\psi$ and $L_{X}\psi$ are of the same form for
$X\in\mathfrak{g}$ We start with $R_{X}$. Again it is enough to show that if
$s,t=1$ then $R_{X}\psi$ is of the form indicated in the statement of the
lemma. Let $X_{1},...,X_{n}$ be a basis of $\mathfrak{g}$ such that
$X_{1},...,X_{r}\in\mathfrak{n}$ with $[h,X_{i}]=\beta_{i}(h)X_{i}$
$h\in\mathfrak{a}$, $X_{r+1},...,X_{r+l}\in\mathfrak{a}$,
$X_{r+l+1},...,X_{n}\in Lie(K)$. Then

$Ad(k)X=\sum c_{i}(k,X)X_{i}.$

We have

$R_{X}\psi(nak)=\frac{d}{dt_{|t=0}}\psi(nak\exp
tX)=\frac{d}{dt_{|t=0}}\psi(na\exp tAd(k)Xk)$
$=\sum_{i=1}^{n}c_{i}(k,X)\frac{d}{dt_{|t=0}}\psi(na\exp
tX_{i}k)=\sum_{i=1}^{r}c_{i}(k,X)a^{\beta_{i}}\left(R_{X_{i}}\phi(n)\right)f(a)\gamma(k)$
$+\sum_{i=r+1}^{r+l}c_{i}(k,X)\phi(n)\left(R_{X_{i}}f(a)\right)\gamma(k)=+\sum_{i=r+1}^{r+l}c_{i}(k,X)\phi(n)f(a)\left(L_{X_{i}}\gamma(k)\right)$

which is easily seen to be of the right form.

To handle the left derivative we consider a different basis $Y_{i}=X_{i}$,
$i=1,...,r+l,Y_{r+l+1},...,Y_{r+l+m}$ a basis of $Lie(M)$ and
$Y_{r+l+m+i}=\theta X_{i},i=1,...,r$. Then

$Ad(n^{-1})X=\sum d_{i}(n,X)Y_{i}$

so

$L_{X}\psi(nak)=-\sum_{i=1}^{n}d_{i}(n,X)\frac{d}{dt_{|t=0}}\psi(n\exp
tY_{i}ak)=-\sum_{i=1}^{r}d_{i}(n,X)\left(R_{i_{i}}\phi(n)\right)f(a)\gamma(k)$
$+\sum_{i=r+1}^{r+l}d_{i}(n,X)\phi(n)L_{Y_{i}}f(a)\gamma(k)+\sum_{i=r+l=1}^{r+l+m}d_{i}(n,X)\phi(n)f(a)L_{Y_{i}}\gamma(k)+$
$-\sum_{i=r+l+m+1}^{n}d_{i}(n,X)\frac{d}{dt_{|t=0}}\psi(n\exp tY_{i}ak).$

All but the last term are of the right form so we will show that it is also.
Set $\mu=r+l+m$ then

$\exp tY_{\mu+i}a=a\exp(tAd(a)^{-1}Y_{\mu+i})=a\exp(ta^{\beta_{i}}Y_{\mu+i}).$

So we are looking at

$-\sum_{i=1}^{r}d_{\mu+i}(n,X)a^{\beta_{i}}\frac{d}{dt_{|t=0}}\psi(na\exp
tY_{\mu+i}k)$

Now $Y_{\mu+i}+X_{i}=Z_{i}\in Lie(K).$ Thus $Y_{m+i}=Z_{i}-X_{i}$. So

$-\sum_{i=1}^{r}d_{\mu+i}(n,X)a^{\beta_{i}}\frac{d}{dt_{|t=0}}\psi(na\exp
tY_{\mu+i}k)=\sum_{i=1}^{r}d_{\mu+i}(n,X)a^{\beta_{i}}\frac{d}{dt_{|t=0}}\psi(na\exp
tX_{i}k)+$
$-\sum_{i=1}^{r}d_{i}(n,X)a^{\beta_{i}}\frac{d}{dt_{|t=0}}\psi(na\exp
tZ_{i}k)=\sum_{i=1}^{r}d_{\mu+i}(n,X)a^{2\beta_{i}}R_{X_{i}}\phi(n)f(a)\gamma(k)$
$-\sum_{i=1}^{r}d_{\mu+i}(n,X)\phi(n)f(a)L_{Z_{i}}\gamma(k).$

The result is, finally, proved.

###### Corollary 20

If $f\in\mathcal{C}(N\backslash G;\chi)$ is right $K$ finite then and if
$\phi\in C_{c}^{\infty}(N)$ then the function on $G$ defined by

$\psi(nak)=\phi(n)f(a)$

is in $\mathcal{C}(G)$.

Proof. Let $V=\mathrm{span}_{\mathbb{C}}\\{R_{k}f|k\in K\\}$. Then $\dim
V<\infty.$ Let $v_{1},...,v_{m}$ be a basis if $V$ then
$R_{k}f=\sum\gamma_{i}(k)v_{i}$. Thus

$f(ak)=\sum v_{i}(a)\gamma_{i}(k).$

Since, $v_{i}\in\mathcal{C}(N\backslash G;\chi)$ we see that $R_{x}v_{i|A}$
satisfies the inequalities for all $x\in U(\mathfrak{a})$. The result is now a
direct consequence of the definition of $\mathcal{C}(G)$ and Lemma 19 .

###### Theorem 21

If $\psi\in\mathcal{C}(G)$ then
$\psi_{\chi}(g)=\int_{N}\chi(n)^{-1}\psi(ng)dn$ defines an element of
$\mathcal{C}(N\backslash G;\chi)$.

Proof. The Harish-Chandra spherical function
$\Xi(g)=\left\langle\pi_{0}(g)1,1\right\rangle$ satisfies

$\Xi(g)\geq\left|g\right|^{-\frac{1}{2}}.$

Thus since

$\left|\psi(na)\right|\leq
C_{d}\left|na\right|^{-1/2}(1+\log\left\|na\right\|)^{-d}$

for all $d\geq 0$ we have

$\left|\psi_{\chi}(ak)\right|\leq
C_{d}\int_{N}\left|na\right|^{-\frac{1}{2}}(1+\log\left\|an\right\|)^{-d}dn\leq
C_{d}\int_{N}\Xi(na)(1+\log\left\|na\right\|)^{-d}dn.$

Also $\left\|na\right\|\geq\left\|a\right\|$ and

$\left\|n\right\|=\left\|naa^{-1}\right\|\leq\left\|a^{-1}\right\|\left\|an\right\|=\left\|a\right\|\left\|an\right\|\leq\left\|an\right\|^{2}.$

Thus

$\left|\psi_{\chi}(ak)\right|\leq
C_{d+r}\int_{N}\Xi(na)(1+\log\left\|na\right\|)^{-d}dn(1+\log\left\|a\right\|)^{-r}.$

Now in the proof of Theorem 7.2.1 in [RRGI] we have seen that there exists $d$
such that

$a^{-\rho}\int_{N}\Xi(na)(1+\log\left\|na\right\|)^{-d}dn\leq B<\infty$

Since $\left(R_{x}\psi\right)_{\chi}=R_{x}(\psi_{\chi})$ the theorem now
follow from the definition of $\mathcal{C}(N\backslash G;\chi)$.

If $f\in C^{\infty}(\mathfrak{a})$ define for $m=(m_{1},...,m_{l}),d,$
$m_{i},d\in\mathbb{Z}_{\geq 0}$ and $x$ a constant coefficient differential
operator on $\mathfrak{a}$

$w_{m,d}(f)=\sup_{h\in\mathfrak{a}}e^{-\rho(h)}e^{\sum
m_{i}\alpha_{i}(h)}(1+\left\|h\right\|)^{d}\left|xf(h)\right|.$

Then we set $\mathcal{W(}\mathfrak{a})$ equal to the space of all $f\in
C^{\infty}(\mathfrak{a})$ such that $w_{c,d}(f)<\infty$ for all
$m=(m_{1},...,m_{l}),d,$ $m_{i},d\in\mathbb{Z}_{\geq 0}$ endowed with the
topology determined by these semi-norms

###### Theorem 22

Assume that $\chi$ is generic. If $\psi\in C^{\infty}(N\backslash G/K;\chi)$
set $T(\psi)(h)=\psi(\exp h)$ for $h\in\mathfrak{a}$. Then
$T(\mathcal{C}(N\backslash G/K;\chi))\subset\mathcal{W(}\mathfrak{a})$ and $T$
defines a continuous isomorphism of $\mathcal{C}(N\backslash G;\chi)$ onto
$\mathcal{W(}\mathfrak{a})$.

Proof. Lemma 18 implies that $T(\mathcal{C}(N\backslash
G/K;\chi))\subset\mathcal{W(}\mathfrak{a})$. Using the definitions of the
semi-norms defining the topologies of $\mathcal{C}(N\backslash G/K;\chi)$ and
$\mathcal{W(}\mathfrak{a})$ the continuity of $T$ follows. $T$ is injective
since $G=NAK$. Let $\phi\in C_{c}^{\infty}(N)$ be such that

$\int_{N}\chi(n)^{-1}\phi(n)dn=1.$

If $f\in\mathcal{W(}\mathfrak{a})$ set $\psi(nak)=\phi(n)f(a).$ Lemma 19
implies that $\psi\in\mathcal{C}(G/K)$. Then
$\psi_{\chi}\in\mathcal{C}(N\backslash G/K;\chi)$ and $T(\psi_{\chi})=f$. Thus
the map is surjective. The open mapping theorem now implies that $T^{-1}$ is
continuous.

##  10 Appendix 3. The Whittaker radial component of the Casimir operator

Let $f\in C^{\infty}(N\backslash G/K;\chi)$ and let $\chi$ be a unitary
character of $N$. Let $\mathfrak{m}$ be the centralizer of $\mathfrak{a}$ in
$\mathfrak{k}$. In this appendix we will calculate the differential operator
on $A$ corresponding to the Casimir operator of $G$ on $C^{\infty}(N\backslash
G/K;\chi).$

Let $X_{\alpha,i},i=1,...,m_{\alpha}$ be a basis of $\mathfrak{n}_{\alpha}$
such that $B(X_{\alpha,i},\theta X_{\alpha,j})=-\delta_{ij}.$ Note, before we
start calculating, that this implies that

$[X_{\alpha,i},\theta X_{\alpha,j}]=-\delta_{ij}h_{\alpha}+m_{ij}$

with $m_{ij}\in\mathfrak{m}$ and $m_{ij}=0$ if $i=j$ .Let $h_{1},...,h_{l}$ be
an orthonormal basis of $\mathfrak{a}$ relative to
$\left\langle...,...\right\rangle.$ Then the Casimir operator of $G$ relative
to $B$ is

$C=-\sum_{j=1}^{m_{\alpha}}(X_{\alpha,j}\theta X_{\alpha,j}+\theta
X_{\alpha,j}X_{\alpha,j})+C_{\mathfrak{m}}+\sum_{i=1}^{l}h_{i}^{2}$

Where $C_{\mathfrak{m}}$ is the Casimir operator corresponding to
$B_{|\mathfrak{m}}$. Let $f\in C^{\infty}(N\backslash G/K,\chi)$ we wish to
calculate $R_{C}f(a)$ for $a\in A$. We observe that $X_{\alpha,i}+\theta
X_{\alpha,j}\in Lie(K)$ and

$\left(X_{\alpha,i}+\theta X_{\alpha,j}\right)^{2}=X_{\alpha,i}^{2}+\theta
X_{\alpha,i}^{2}+X_{\alpha,j}\theta X_{\alpha,j}+\theta
X_{\alpha,j}X_{\alpha,j}.$

Thus

$R_{X_{\alpha,j}\theta X_{\alpha,j}+\theta
X_{\alpha,j}X_{\alpha,j}}f=-R_{X_{\alpha,i}^{2}}f-R_{\theta
X_{\alpha,i}^{2}}f.$

Also,

$R_{\theta X_{\alpha,j}}R_{\theta X_{\alpha,j}}f=-R_{\theta
X_{\alpha,j}}R_{X_{\alpha,j}}f=-R_{h_{\alpha}}f+R_{X_{\alpha,j}}R_{\theta
X_{\alpha,j}}f=-R_{h_{\alpha}}f-R_{X_{\alpha,j}}^{2}f,$
$\left(R_{X_{\alpha,j}^{2}}f\right)(a)=a^{2\alpha}\left(L_{X_{\alpha,j}^{2}}f\right)(a)=2d\chi(X_{\alpha,j})^{2}a^{2\alpha}f(a)$

and

$R_{C_{\mathfrak{m}}}f=0.$

The upshot is that

$Cf(a)=2\sum_{\alpha\in\Phi^{+}}\sum_{j=1}^{m_{\alpha}}d\chi(X_{\alpha,j})^{2}a^{2\alpha}f(a)-\sum_{\alpha\in\Phi^{+}}m_{\alpha}h_{\alpha}f(a)+\sum_{i=1}^{l}h_{i}^{2}f(a).$

If $\alpha\notin\Delta$ then $d\chi(X_{\alpha,i})=0$ and
$\sum_{\alpha\in\Phi^{+}}m_{\alpha}h_{\alpha}=2h_{\rho}$, so

$Cf(a)=2\sum_{\alpha\in\Delta}\sum_{j=1}^{m_{\alpha}}d\chi(X_{\alpha,j})^{2}a^{2\alpha}f(a)-2h_{\rho.}f(a)+\sum_{i=1}^{l}h_{i}^{2}f(a).$

Noting that $d\chi=i\xi_{\chi}$ with $\xi_{\chi}\in\mathfrak{n}^{\ast}$ thus
if we set $\xi_{\chi,\alpha}=\xi_{\chi|\mathfrak{n}_{\alpha}}$

$\sum_{j=1}^{m_{\alpha}}d\chi(X_{\alpha,j})^{2}=-\left\|\xi_{\chi,\alpha}\right\|^{2}.$

We also have

$a^{\rho}\sum_{i=1}^{l}h_{i}^{2}a^{-\rho}=(\rho,\rho)-2h_{\rho}+\sum_{i=1}^{l}h_{i}^{2}.$

We have derived

###### Lemma 23

if $f\in C^{\infty}(N\backslash G/K;\chi)$ and $a\in A$ then

$\left(C+(\rho,\rho)\right)f(a)=a^{\rho}(\sum_{i=1}^{l}h_{i}^{2}-2\sum_{\alpha\in\Delta}\left\|\xi_{\chi,\alpha}\right\|^{2}a^{2\alpha})a^{-\rho}f(a).$

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