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\documentclass[twoside]{article}
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\pagestyle{myheadings} \markboth{\hfil Differential operators
on equivariant vector bundles\hfil EJDE--2000/59}
{EJDE--2000/59\hfil Anton Deitmar \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations}, Vol.~{\bf
2000}(2000), No.~59, pp.~1--8. \newline ISSN: 1072-6691. URL:
http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
\vspace{\bigskipamount} \\
%
Differential operators on equivariant vector bundles over symmetric spaces
%
\thanks{ {\em Mathematics Subject Classifications:} 43A85, 22E30, 35J45.
\hfil\break\indent {\em Key words:} invariant operators.
\hfil\break\indent \copyright 2000 Southwest Texas State
University and University of North Texas. \hfil\break\indent
Submitted May 16, 2000. Published September 1, 2000.} }
\date{}
%
\author{ Anton Deitmar }
\maketitle
\begin{abstract}
Generalizing the algebra of motion-invariant differential
operators on a symmetric space we study invariant operators on
equivariant vector bundles. We show that the eigenequation is
equivalent to the corresponding eigenequation with respect to
the larger algebra of all invariant operators. We compute the
possible eigencharacters and show that for invariant integral
operators the eigencharacter is given by the Abel transform.
We show that sufficiently regular operators are surjective,
i.e. that equations of the form $Df=u$ are solvable for all $u$.
\end{abstract}
\newcommand{\rez}[1]{\frac{1}{#1}}
\newcommand{\norm}[1]{\parallel #1 \parallel}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\section{Equivariant vector bundles} \label{equiv}
Let $X$ denote a manifold with a smooth action of a Lie group
$G$. An {\it equivariant vector bundle} $E$ over $X$ is a
smooth vector bundle $\pi : E\rightarrow X$ together with a
smooth action of $G$ on $E$ such that $\pi(gv)=g\pi(v)$ for
$v\in E$ and such that all maps of the fibres $g: E_x\rightarrow
E_{gx}$, $g\in G$, $x\in X$, are linear. An example is given by
$E=X\times V$, where $(\sigma ,V)$ is a finite dimensional
representation of $G$ and $G$ acts on $E$ by
$g(x,v)=(gx,\sigma(g)v)$.
Now assume $X$ to be a homogeneous space, i.e. $X=G/H$ for a
closed subgroup $H$ of $G$. Given an equivariant bundle $E$
over $X$ we get a representation of $H$ on the fibre over $eH$,
where $e$ denotes the neutral element of $G$. Conversely given
a representation $\tau$ of $H$ on a finite dimensional vector
space $V$ we let $H$ act from the right on $G\times V$ by
$(g,v)h= (gh,\tau(h)^{-1}v)$ and define $E_\tau= (G\times
V)/H$. This is a vector bundle over $X$. This construction
gives an equivalence of categories between the category of
equivariant vector bundles over $X$ and the category of finite
dimensional representations of $H$. If $\tau$ splits as a
direct sum $\tau=\tau_1\oplus\tau_2$ then $E_\tau\cong
E_{\tau_1}\oplus E_{\tau_2}$ and vice versa.
Now let $G$ denote a connected semisimple Lie group with finite
center and let $K$ denote a maximal compact subgroup. The
quotient $X=G/K$ is diffeomorphic to ${\mathbb R}^n$ for some
$n$. Using the Killing form of $G$ one defines a Riemannian
metric on $X$ such that the group $G$ acts by isometries. This
is the most general symmetric space without compact factors. See
\cite{helg} for further details.
\section{Differential operators}
Let $\hat{K}$ denote the set of isomorphism classes of
irreducible unitary representations of the group $K$. Since $K$
is compact, every $\tau\in\hat{K}$ is finite dimensional. We do
not distinguish between a class in $\hat{K}$ and a
representative. For $(\tau ,V_\tau)\in\hat{K}$ let
$E_\tau\rightarrow G/K$ denote the vector bundle as in section
\ref{equiv}. The group $G$ acts on the space of smooth sections
$\Gamma^\infty(E_\tau)$ of the bundle $E_\tau$ by
$g.s(x)=gs(g^{-1}x)$, $x\in X$, $g\in G$, $s\in
\Gamma^\infty(E_\tau)$. Hence $G$ also acts on differential
operators by conjugation. Let ${\cal D}_\tau$ denote the
algebra of $G$-invariant differential operators on $E_\tau$,
i.e. those operators $D$ that satisfy $D(g.s)=g.D(s)$ for any
$s\in\Gamma^\infty(E_\tau)$.
Let $C_\tau^\infty(G)$ denote the space of all infinitely often
differentiable maps from $G$ to $V$ with $f(kx)=\tau(k)f(x)$
for $k\in K$ and $x\in G$. The group $G$ acts on the space
$C_\tau^\infty(G)$ by translations from the right. The map
$s\mapsto f_s$ with $f_s(x)=xf_s(x^{-1})$ gives a
$G$-isomorphism of $\Gamma^\infty(E_\tau)$ to
$C_\tau^\infty(G)$. We conclude that the algebra ${\cal
D}_\tau$ acts on $C_\tau^\infty(G)$.
Let the compact group $K$ act on the space $C^\infty(G)$ of
smooth functions on $G$ by
$$
L_k(x) = f(k^{-1}x),\quad k\in K, \ x\in G.
$$
Then there is a decomposition into $K$-isotypes:
$$
C^\infty(G) = \bigoplus_{\tau\in\hat{K}} C^\infty(G)(\tau).
$$
The sum here means that finite sums on the right hand side are
dense in the Fr\'echet space $C^\infty(G)$. Further
$C^\infty(G)(\tau)$ is the space of functions in $C^\infty(G)$
that transform under $L$ according to $\tau$.
For $(\tau, V)\in \Hat{K}$ fix any nonzero $v^*$ in the dual
space $V^*$ and for $f\in C_\tau^\infty(G)$ let
$Bf(x)=v^*(f(x))$. Then $Bf\in C^\infty(G)(\tau)$ and the map
$B$ is a $G$-isomorphism. We have
$$
\Gamma^\infty(E_\tau)\ \cong\ C_\tau^\infty(G)\ \cong \
C^\infty(G)(\tau).
$$
Let ${\mathfrak g}$ denote the Lie algebra of $G$. The universal
enveloping algebra $U({\mathfrak g})$ may be viewed as the
algebra of all right invariant differential operators on $G$,
i.e. the algebra of all differential operators $D$ on $G$ such
that $R_gD=DR_g$ for any $g\in G$. Here for a function
$\varphi\in C^\infty(G)$ and $g\in G$ we have
$R_g\varphi(x)=\varphi(xg)$ for all $x\in G$. Let $U({\mathfrak
g})^K$ denote the subalgebra of all differential operators
which are $K$-invariant on the left side, i.e. which satisfy
$L_kD=DL_k$ for any $k\in K$. The algebra $U({\mathfrak g})^K$
leaves stable the decomposition of $C^\infty(G)$ and thus acts
on $C^\infty(G)(\tau)$. We therefore get an algebra homomorphism
$$
\varphi_\tau : U({\mathfrak g})^K\rightarrow {\cal D}_\tau.
$$
\begin{proposition}
The homomorphism $\varphi_\tau$ is surjective and the
intersection of all kernels $\ker(\varphi_\tau)$ for varying
$\tau$ is zero.
\end{proposition}
\noindent{\bf Proof:} Let the algebra $U({\mathfrak g})\otimes
{\rm End}(V)$ acts on the space $C^\infty(G,V)$ of smooth
functions on $G$ with values in $V$. The group $K$ acts on
$U({\mathfrak g})$ via the adjoint representation and on ${\rm
End}(V)$ by means of conjugation via $\tau$. Then the algebra of
$K$-invariants, $(U({\mathfrak g})\otimes{\rm End}(V))^K$ acts
on $C_\tau^\infty$. The annihilator $I$ of $C_\tau^\infty$ in
$U({\mathfrak g})\otimes{\rm End}(V)$ is generated by elements
of the form $XY\otimes T+X\otimes T\tau(Y)$ with $X\in
U({\mathfrak g})$, $Y\in {\mathfrak k}$ and $T\in{\rm End}(V)$.
Since $K$ is reductive we have
$$
{\cal D}_\tau\ \cong\ (U({\mathfrak g})\otimes{\rm
End}(V))/I)^K\ \cong\ (U({\mathfrak g})\otimes{\rm
End}(V))^K/I^K
$$
Since $\tau$ is irreducible and $K$ is connected it follows
that the representation of $U({\mathfrak k})$ induced by $\tau$
also is irreducible. Its image $\tau(U({\mathfrak k}))$ then
may be considered as a von Neumann algebra (it is weakly closed
since it is finite dimensional), hence it coincides with its
bicommutator. Its commutator, however, is ${\mathbb C} {\rm
Id}$, hence $\tau(U(k))={\rm End}(V)$. This implies that any
element of ${\cal D}_\tau$ can be written in the form $Z\otimes
1$ for some $Z\in U({\mathfrak g})$. It follows that $Z$ must
be in $U({\mathfrak g})^K$, which implies the surjectivity of
$\varphi_\tau$.
For the second assertion assume $X$ is in the intersection of
all kernels. Then, since $C^\infty(G) = \bigoplus_\tau
C^\infty(G)(\tau)$, we get that $Xf=0$ for every $f\in
C^\infty(G)$, this gives $X=0$. \hfill$\square$
\begin{corollary}
${\cal D}_\tau$ is finitely generated as ${\mathbb C}$-algebra.
\end{corollary}
\noindent{\bf Proof:} $U({\mathfrak g})$ has a natural
filtration by order. The associated graded version equals the
symmetric algebra $S({\mathfrak g})$ over ${\mathfrak g}$.
Since the adjoint action of $K$ preserves the filtration, we
have a filtration on $U({\mathfrak g})^K$ with graded version
$S({\mathfrak g})^K$. The latter is finitely generated by
invariant theory, hence the former is, too. \hfill$\square$
\subsection*{Examples}
For $\tau =1$ the trivial representation, the algebra ${\cal
D}_\tau$ is the algebra of $G$-invariant differential operators
on $G/K$. In this case ${\cal D}_\tau$ is isomorphic to the
polynomial ring in $r$ generators, where $r$ is the rank of the
symmetric space $G/K$ (see \cite{helg2} II.5).
Let $G=SO(n,1)^+$ the group of motions on the hyperbolic space
$H_n$. For $\tau=\wedge^p({\rm Ad})$ the space
$\Gamma^\infty(E_\tau)$ is just the space of $p$-differential
forms on $H_n$. For $p=0$ or $p=n$ the algebra ${\cal D}_\tau$
is the polynomial ring in one variable, generated by $\delta d$
and $d\delta$ respectively, where $d$ is the exterior
differential and $\delta$ its formal adjoint. For $n$ odd and
$p=(n\pm 1)/2$ the algebra ${\cal D}_\tau$ is generated by
$d\delta$ and $*d$ resp. $\delta d$ and $d*$, where $*$ is the
Hodge operator, with the generating relations
$$
d\delta *d = 0 = *dd\delta,\quad \delta dd* = d*\delta d=0.
$$
In all other cases ${\cal D}_\tau$ is generated by $d\delta$ and
$\delta d$ with the relations $d\delta \delta d=0=\delta
dd\delta$. Summarizing we get the structure of ${\cal D}_\tau$
as:
$$
{\cal D}_\tau \cong
\begin{cases} {\mathbb C}[x,y]/xy & \text{ for } 1\le p\le n-1,\\
{\mathbb C}[x] & \text{ for } p=0,n,
\end{cases}
$$
These results are proven as follows: First the algebras
generated by $d\delta$ and $\delta d$ give subalgebras of the
algebras of invariant operators with the given structure. One
then has to show that these are all, which is done by
considering the corresponding graded algebras of principal
symbols which are describable by an invariant problem. Explicit
calculations show that the dimensions of the graded parts
already give the dimensions of the graded parts of the
subalgebras.
\section{Integral operators}
A smooth function $\Phi : G\rightarrow {\rm End}(V)$ which
satisfies $\Phi(kxl)=\tau(k)\Phi(x)\tau(l)$ for $x\in G$,
$k,l\in K$ is called {\it $\tau$-sherical}. The algebra ${\cal
D}_\tau$ acts on the set of $\tau$-spherical functions.
Compactly supported $\tau$-spherical functions form an algebra
under convolution:
$$
\Phi * \Psi (x) = \int_G \Phi(y)\Psi(y^{-1}x)\,dy\,,
$$
where $dy$ denotes a Haar measure on $G$. This algebra is
denoted by ${\cal A}_\tau$. The algebra ${\cal A}_\tau$ acts on
$C_\tau^\infty(G)$ by $L_\Phi f= \Phi * f$ for $\Phi\in{\cal
A}_\tau$ and $f\in C_\tau^\infty$. The algebra ${\cal A}_\tau$
contains an approximate identity.
Let $D_1,\dots ,D_n$ be a set of generators of the algebra
${\cal D}_\tau$. For $z\in {\mathbb C}^n$ let
$$
{\rm Eig}(z) = \{ f\in C^\infty(G, {\rm End}(V)) |\ f\ {\rm
is}\ \tau-{\rm spherical\ and}\ D_jf=z_jf\}.
$$
\begin{proposition}
For any $z\in{\mathbb C}^n$ we have $\dim{\rm Eig}(z)\le 1$.
\end{proposition}
\noindent{\bf Proof:} By Lemma 1 in \cite{har1} any $f\in{\rm
Eig}(z)$ is analytic. So let $f\in{\rm Eig}(z)$ and
$H\in{\mathfrak g}$. If $H$ is small enough then
$$
f(\exp(H)) = \sum_{n\ge 0} \rez{n!}\left(
\frac{\partial}{\partial t}\right)^n f(\exp(tH))|_{t=0} \ =\
\sum_{n\ge 0} \rez{n!}H^nf(0).
$$
We get
$$
\int_K \tau(k)f(\exp(H))\tau(k^{-1})\,dk = \sum_{n\ge 0}\int_K
({\rm Ad}(k)H)^nf(e)\, dk\,.
$$
Now $\int_K({\rm Ad}(k)H)^n \,dk$ is a right invariant
differential operator mapping the space $C_\tau^\infty(G)$ to
itself, hence defines an element of the algebra ${\cal
D}_\tau$. So the values $ \int_K({\rm Ad}(k)H)^nf(e) \,dk$ only
depend on $z$ and not on $f$. We conclude that the function
${\rm tr} f(x)$ is determined by $z$ up to scalar. But the map
${\cal A}_\tau\rightarrow C^\infty(G)$, $f\rightarrow{\rm
tr}(f)$ is injective \cite{wa2}. \hfill$\square$
\begin{theorem}
Suppose $f\in C^\infty_\tau(G)$ is an eigenform for any
$D\in{\cal D}_\tau$. Then $f$ is an eigenform for every
$T\in{\cal A}_\tau$ with an eigenvalue only depending on $T$
and the eigenvalues on ${\cal D}_\tau$.
\end{theorem}
\noindent{\bf Proof:} For $\varphi\in C^\infty_\tau(G)$ and
$w\in V$ define $\varphi_w\in C^\infty(G,{\rm End}(V))$ by
$$
\varphi_w(x)v = \langle v,w\rangle \varphi(x),
$$
where $\langle .,.\rangle$ denotes the scalar product on $V$.
Fot $psi : G\rightarrow {\rm End}(V)$ let
$$
M\psi(x) = \int_K\psi(xk)\tau(k^{-1})\,dk\,.
$$
Now let $f$ be as in the theorem and assume $f(e)\ne 0$.
(Otherwiese replace $f$ by $R_gf$ for a suitable $g\in G$.) Fix
some $w\in V$ such that ${\rm tr} f_w(e)=1$. Now $M(f_w)$ lies
in ${\rm Eig}(z)$ for some $z$. Let $\Phi$ be $\tau$-spherical
and compactly supported. Since $L_\Phi M(f_w)=M((L_\Phi f)_w)$
we see that there is a $\lambda\in{\mathbb C}$ such that
$$
M((\lambda f-L_\Phi f)_w) = 0,
$$
and $\lambda$ does not depend on $f$ or $w$. The claim now
follows by the proposition. \hfill$\square$\medskip
Let ${\mathfrak a}$ denote the Lie algebra of the maximal
${\mathbb R}$-split torus $A$ of $G$. Let $G=KNA$ be a
corresponding Iwasawa decomposition. For $\lambda\in{\mathfrak
a}^*_{\mathbb C}$ the complex dual space of ${\mathfrak a}$ and
any $v\in V$ let
$$
p_{\lambda,v}(kna) = \tau(k)e^{\lambda(\log(a))} v.
$$
This defines $p_{\lambda ,v}\in C_\tau^\infty(G)$. Let $P=MAN$
be the minimal parabolic given by $A$ and $N$. Then $M$ is a
closed subgroup of $K$.
\begin{lemma}\label{3.5}
For every $T\in {\cal A}_\tau\oplus{\cal D}_\tau$ and every
$\lambda\in{\mathfrak a}_{\mathbb C}^*$ there is a
$S_\lambda(T)$ in ${\rm End}(V)$ such that for all $v\in V$
$$
T(p_{\lambda ,v})=p_{\lambda ,S_\lambda(T)v}.
$$
We have $S_\lambda(T)\tau(m)=\tau(m)S_\lambda(T)$ for all $m\in
M$
\end{lemma}
\noindent{\bf Proof:} The lemma is clear by group invariance
and the fact that $A$ normalizes $N$. \hfill$\square$\medskip
For a simultaneous eigenform $f\in C_\tau^\infty(G)$ of ${\cal
D}_\tau$ let $\chi_f$ denote the eigencharacter $\chi_f : {\cal
A}_\tau\oplus{\cal D}_\tau\rightarrow {\mathbb C}$ defined by
$$
Tf = \chi_f(T)f.
$$
\begin{theorem}
Let $f$ denote a bounded simultaneous eigenform. Then there is
a $v\in V$ and a $\lambda\in{\mathfrak a}_{\mathbb C}^*$ such
that $p_{\lambda,v}$ is an eigenform and
$$
\chi_f = \chi_{p_{\lambda,v}}.
$$
\end{theorem}
\noindent{\bf Proof:} The character $\chi_f$ is determined by
its values on ${\cal A}_\tau$. Let $p$ denote the trivial
seminorm on $G$, i.e. $p(g)=1$ for all $g\in G$. Set
$$
\norm{\Phi}_p=\int_G\norm{\Phi(y)}p(y)\,dy,
$$
where $\norm{.}$ is the norm on ${\rm End}(V)$. Assume
$\norm{f(e)}=1$ and $\norm{f(x)}\le M$, $x\in G$. Then we get
for $\Phi\in{\mathbb A}_\tau$:
\begin{eqnarray*}
|\chi_f(L_\Phi)| &=& \norm{\int_G \Phi(y)f(y^{-1})\,dy}\\
&\le & M\int_G\norm{\Phi(y)}\,dy\\
&\le& M\norm{\Phi}_p.
\end{eqnarray*}
Thus $\chi_f$ is a $p$-continuous representation of ${\cal
A}_\tau$. The claim now follows from the theorem of Glover
\cite{wa2}, p.~40. \hfill$\square$\medskip
We now give the computation of the eigencharacters of ${\cal
A}_\tau$. Let \\
$\rho=\rez{2}\sum_{\alpha >0}m_\alpha \alpha\in{\mathfrak a}^*$
be the usual modular shift, i.e. the sum runs over all positive
roots and $m_\alpha$ is the dimension of the root space to the
root $\alpha$.
\begin{theorem}
Let $S_\lambda(L_\Phi)$ denote the endomorphism of Lemma
\ref{3.5}. Then with
$$
g_{\Phi}(a) = a^{-\rho}\int_N \Phi(na) \,dn,
$$
(Abel transform), we get
$$
S_\lambda(L_\Phi) = \int_A a^{\rho-\lambda} g_{\Phi}(a) \,da,
$$
(Fourier transform on $A$). Moreover, $g_\Phi(a)$ is in the
center of $\tau|_M$ and $g_{\Phi *\Psi}=g_{\Phi}*g_\Psi$ with
$A$-convolution on the right hand side. The map $\Phi\rightarrow
g_\Phi$ is injective.
\end{theorem}
\noindent{\bf Proof:} A calculation using the integral formula
of the Iwasawa decomposition gives the first claim. The
injectivity is proved in \cite{wa2}, p.~35. \hfill$\square$
\section{Surjectivity of differential operators}
Let $\theta$ denote the Cartan involution fixing $K$ pointwise.
Let $\Gamma_1,\dots,\Gamma_r$ denote a complete system of
nonconjugate $\theta$-stable Cartan subgroups of $G$ and let
$A_i=\Gamma_i\cap\exp({\mathfrak p})$, where ${\mathfrak g}
={\mathfrak k}\oplus{\mathfrak p}$ denotes the polar
decomposition of ${\mathfrak g}$. Let $A$ denote one of the
$A_i$. Let $L=MA$ denote the centralizer of $A$ in $G$. Let
$\tau_M$ denote the restriction f $\tau$ to $K_M=K\cap M$. Let
$C^\infty(MA,\tau)$ denote the set of $\tau_M$-spherical
functions on $MA$. For $g\in C^\infty(MA,\tau_M)$ let
$g^\#(kman)=\tau(k)g(ma)$. The set of these functions $g^\#$ is
stable under ${\cal D}_\tau$ and we get a homomorphism
$$
\gamma : {\cal D}_\tau\rightarrow{\cal D}_{\tau_{M}}(MA) =
{\cal D}_{\tau_M}\otimes U({\mathfrak a}).
$$
Now every $T\in{\cal D}_{\tau_M}$ operates on the finite
dimensional space of rapidly decreasing cusp forms $^0{\cal
C}(M,\tau_M)$ as defined in \cite{har2}. So $\gamma(D)$ defines
an element $^0\gamma(D)\in |End(^0{\cal C}(M,\tau_M))\otimes
U({\mathfrak a})$. Let $\det(^0\gamma(D))\in U({\mathfrak a})$
denote the determinant. Call $D$ {\it regular} if
$\det(^0\gamma(D))\ne 0$ for all $A=A_i$.
\begin{theorem}\label{4.2}
If $D\in{\cal D}_\tau$ is regular, then $D$ is surjective as a
map from $C_\tau^\infty$ to $C_\tau^\infty$.
\end{theorem}
\noindent{\bf Proof:} We are going to formulate a vector bundle
version of Holmgren's uniqueness theorem. Let $X$ denote a real
analytuc manifold and $E$ an analytic vector bundle over $X$.
Let $P$ be a differential operator on $E$ with analytic
coefficients. For a susbset $A$ of $X$ let $N(A)$ denote the
set of normal vectors to $A$ in $T^*X$, (see \cite{hoer}, chap.
8). Consider the determinant of the principal symbol $\sigma_P$
as a map:
$$
\det \sigma_p : T^*X\rightarrow {\mathbb C}.
$$
\begin{proposition}
(Holmgren's principle) Let $u\in \Gamma_c^\infty(E)'$ be a
distribution with $Pu=0$. Then we have
$$
\det\sigma_P(N({\rm supp}(u)))=0.
$$
\end{proposition}
\noindent{\bf Proof:} Since the assertion is local in nature
the proof for the trivial bundle (\cite{hoer}, Theorem 8.6.5)
carries over to the present case. \hfill$\square$\medskip
Let $P$ be as above. A point $x\in X$ is called a {\it singular
point} of $P$ if
$$
\det\sigma_P(T^*_xX) = 0.
$$
Now consider the case when $X$ is a symmetric space $G/K$ and
let $E=E_\tau$ be a homogeneous bundle. Assume that $P$ is
invariant, i.e. $P\in{\cal D}_\tau$. Let $\nabla$ denote the
canonical homogeneous connection on $E$. For
$\lambda\in{\mathfrak a}_{\mathbb C}^*$ let
$p_\lambda(ank)=e^{\lambda(\log(a))}$.
\begin{lemma}
There is a section $\gamma_P$ of the bundle $S({\mathfrak
a})\otimes {\rm End}(E)$ such that
$$
P(p_\lambda s) = \gamma_P(\lambda)p_\lambda s,
$$
for all parallel sections $s$ of $E$. For this section
$\gamma_P$ and any parallel $s$ we have
$$
\sigma_P(dp_\lambda)s = \gamma_{P,m} s,
$$
where $m=\deg(P)$ and $\gamma_{P,m}$ is the principal part of
$\gamma_P$ with respect to the gradation of $S({\mathfrak a})$.
\end{lemma}
\noindent{\bf Proof:} The first part is well known, the second
follows from a calculation in Iwasawa coordinates.
\hfill$\square$\medskip
A differential operator $P$ on $E$ is called {\it $D$-convex},
if for any Weyl group stable compact convex subset ${\mathfrak
s}$ of ${\mathfrak a}$ and anny section $s$ of $E$ with compact
support and ${\rm supp}(Ps)\subset K\exp({\mathfrak s})K$ we
already have ${\rm supp}(s)\subset K\exp({\mathfrak s})K$.
\begin{theorem}\label{last}
Any $P\in {\cal D}_\tau$ with $\det\sigma_P\ne 0$ is $D$-convex.
\end{theorem}
\noindent{\bf Proof:} Let $P$ be as in the theorem. Since
$\det\sigma_P$ is $G$-invariant, it follows from
$\det\sigma_P\ne 0$ that $P$ has no singular points. Let for
$x\in X$, $x=k\exp(H)K$, $H\in{\mathfrak a}$,
\begin{eqnarray*}
\delta(x) &=& \inf \{ t>0 | \frac{H}{t}\in{\mathfrak s} \},\\
\delta(A) &=& \sup \{ \delta(x) | x\in A\},\ \ \ {\rm for}\
A\subset X.
\end{eqnarray*}
Assume $\delta({\rm supp} Ps)=1$ and $\delta({\rm supp} s)>>1$.
Let $x_0\in{\rm supp} s$ with $\delta(x_0)=\alpha$ and
$x_0=\exp(H)$, $H$ in the positive Weyl chamber of ${\mathfrak
a}$. Let $\lambda\in{\mathfrak a}_+^*$. With Kostants convexity
theorem we get
$$
p_\lambda(x)\ \le\ p_\lambda(x_0),
$$
for all $x\in{\rm supp} s$. So $dp_\lambda(x_0)\in N({\rm supp}
s)$. Hence $\det\sigma_P(dp_\lambda(x_0))=0$ for all
$\lambda>>0$, hence for all $\lambda$. By group invariance this
gives that $x_0$ is a singular point for $P$, a contradiction.
This proves Theroem \ref{last}. \hfill$\square$\medskip
Now regular operators $P$ satisfy the condition of Theroem
\ref{last}. They further admit fundamental solutions \cite{D}.
Now Theorem \ref{4.2} follows as in \cite{helg4}.
\hfill$\square$\medskip
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{\it Harmonic Analysis on Semisimple Lie Groups II},
Springer, Berlin-Heidelberg-New York, 1972.
\end{thebibliography}
\noindent
{\sc Anton Deitmar }\\
School of Mathematical Sciences, University of Exeter \\
Laver Building, North Park Road \\
Exeter, EX4 4QE \\
Devon, UK \\
email: A.H.J.Deitmar@exeter.ac.uk
\section*{Addendum: February 26, 2001.}
It was brought to my knowledge by Werner Hoffmann that the proof
of Theorem 4.4 of the paper {\it Differential operators on
equivariant vector bundles over symmetric spaces}, Electron. J.
Diff. Eqns., Vol. 2000, No. 59, pp. 1-8 (2000) was lacking.
Indeed the principal idea was shortened beyond recognition, so
that I have set up this erratum. \medskip
\noindent{\bf Proof of Theorem 4.4: } Let ${\mathfrak s}\subset{\mathfrak
a}$ be convex compact and stable under the Weyl group $W$. Let
$s$ be a smooth section of $E$ with compact support and ${\rm
supp}(Ps)\subset K\exp({\mathfrak s})K=X_{\mathfrak s}$.
Expanding $s$ as a sum of $K$-finite vectors we have that $s$ is
supported in the $K$-stable set $X_{\mathfrak s}$ if and only if
every summand is. Moreover, the operator $P$ may be applied
termwise to the sum and hence we see that we may assume $s$ to be
$K$-finite. Then the support of $s$ is $K$-invariant and it
suffices to show that ${\rm supp}(s)\cap AK$ is a subset of
$\exp({\mathfrak s})K$.
{\it Assume} that ${\rm supp}(s)\cap AK$ is not a subset of
$\exp({\mathfrak s})K$. Let
$$
{\rm supp}_{\mathfrak a}(s)=\{ Y\in{\mathfrak a} |
\exp(Y)K\in{\rm supp}(s)\}.
$$
Then ${\rm supp}_{\mathfrak a}(s)$ is not a subset of ${\mathfrak
s}$. Let $L$ be the set of all linear maps $\lambda:{\mathfrak
a}\rightarrow{\mathbb R}$ such that
$$
0 <\max_{Y\in{\mathfrak s}}\lambda(Y) <\max_{Y\in{\rm
supp}_{\mathfrak a}(s)}\lambda(Y).
$$
By the assumption $L$ is a nonempty open subset of the dual space
${\mathfrak a}^*$.
Recall that by the invariance of $P$ the map $\det\sigma_P :
T^*X\rightarrow{\mathbb C}$ is $G$-invariant. We identify
\begin{eqnarray*}
G\backslash T^*X &\cong& K\backslash T_{eK}^*X\\
&\cong& K\backslash{\mathfrak p}^*\\
&\cong& W\backslash{\mathfrak a}^*
\end{eqnarray*}
and so $\det\sigma_P$ is given by a $W$-invariant polynomial map
on ${\mathfrak a}^*$. Note that the first identification step is
given by mapping a vector $v\in T_{xK}^*$ to $x^{-1}v$.
Let $\lambda\in L$. Since $L$ is invariant under the Weyl group we
may choose $\lambda$ to be antidominant. Let $Y_0\in{\rm
supp}_{\mathfrak a}(s)$ be a point at which the maximum
$\max_{Y\in{\rm supp}_{\mathfrak a}(s)}\lambda(Y)$ is taken. Then
$Y_0\notin {\mathfrak s}$ and
$$
\lambda(Y)\le\lambda(Y_0)
$$
for any $Y\in{\rm supp}_{\mathfrak a}(s)$. Let $a_0=\exp(Y_0)$
then $a_0K\notin{\rm supp}(Ps)$ but $a_0K\in{\rm supp}(s)$. Let
$g_\lambda:X\rightarrow {\mathbb R}$ be defined by
$$
g_\lambda(naK)=\lambda(\log(a)).
$$
\begin{lemma}
For any $x\in{\rm supp}(s)$ we have $g_\lambda(x)\le
g_\lambda(a_0K)$.
\end{lemma}
\noindent{\bf Proof:} Write $x=naK$. We have to show that
$$
\lambda(\log(a))\le\lambda(Y_0).
$$
For this write $na=k_1\exp(Z)k_2$ for some $Z\in {\rm
supp}_{\mathfrak a}(s)$ and $k_1,k_2\in K$. By the convexity
theorem of van den Ban, E.
[{\it A convextity theorem for semisimple symmetric spaces}.
Pacific J. Math. 124, 21-55 (1986)]
we infer that $\log(a)$ is in the convex
hull of $WZ$. Therefore
$$
\lambda(\log(a))\le \max_{w\in W}\lambda(wZ)\le
\max_{Y\in{\rm supp}(s)}\lambda(Y)\le\lambda(Y_0).
$$
\quad\hfill$\square$\smallskip
The lemma implies that $dg_\lambda(a_0K)$ is normal to ${\rm
supp}(s)$. Therefore, by the Holmgren principle we get
$$
\det\sigma_P(dg_\lambda(a_0))= 0.
$$
The above identification $G\backslash T^*X\cong
W\backslash{\mathfrak a}^*$ defines a projection map
$T^*X\rightarrow W\backslash {\mathfrak a}^*$. Since the group $A$
is abelian and $g_\lambda$ is linear on $A$, the vector
$$
a^{-1} dg_\lambda(a)\in T_{eK}^*X
$$
does not depend on the point $a\in A$. Therefore the image of
$dg_\lambda(a_0)$ in $W\backslash{\mathfrak a}$ only depends on
$\lambda$ and not on $a_0$. The image of the map
$L\ni\lambda\mapsto dg_\lambda(a_0)\in W\backslash {\mathfrak
a}^*$ is an open set. Therefore the assertion
$$
\det\sigma_P(dg_\lambda(a_0))=0
$$
for any $\lambda$ leads to $\det\sigma_P=0$, a contradiction.
\hfill$\square$
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