0$, it is easy to check that
$$\pi(P)\subset\int_{0}^{\infty}t^{-\alpha}\pi(\delta_{t}H){dt\over t }$$
and that the right hand side defines closed injective operator (
we compute the integral applying function under the integral to a vector ---
the domain is
the set of all vectors for which the integral is convergent).
For $\alpha\leq 0$ we get the same conclusion using \bb (condition
$\alpha >-Q$ is used to prove that $P$ is a regular kernel).
We are going to build $H$. Let us recall that by [\Br] the set $\Gl=\{\pi_{l}:l\in\Gamma\}$ is closed
in the Fell topology of the space of representations. Fix $p\notin\Gamma$.
$Ad^{*}(G)\Gamma \subset \Gamma$ implies $\pi_{p}\notin\Gl$. By the
definition of Fell topology and density of $C^{\infty}_{c}(G)$ in $C^{*}(G)$,
there exists a function $\FA\in C^{\infty}_{c}(G)$ such that
$$\|\pi_{p}(F)\|=1$$
and for all $l\in\Gamma$
$$\|\pi_{l}(F)\|<1/10.$$
Replacing $\FA$ by $\FA^{*}*\FA$ we may assume that $\FA$ is
positive definite.
Choose $\phi\in C^{\infty}_{c}(\Bbb R)$ such that $\phi(1)=1$, $\phi\geq 0$,
$\supp(\phi)\subset[1/10,2]$. By the spectral theorem the operator
$$\phi(\pi_{p}(\FA))\ne0$$
while for all $l\in\Gamma$
$$\phi(\pi_{l}(\FA))=0.$$
Using functional calculus, as for example in [\HJ], we show
that there
exists a Schwartz class function $\FB$ on $G$ such that (as a convolution
operator on $L^{2}(G)$) $\FB=\phi(\FA)$.
Approximating $\phi$ by polynomials we see that for all $l$ we have
$$\phi(\pi_{l}(\FA))=\pi_{l}(\FB).$$
We also note that the set of $\pi$ such that $\pi(\FB)\ne0$ is
open (by definition). To summarize, we constructed $\FB$ such that
for all $\pi\in\Gl$
$$\pi(\FB)=0,$$
the set $U_{\FB}=\{\pi(\FB)\ne0\}$ is open and $\pi_{p}\in U_{\FB}$.
Since Fell topology has a countable basis, there exists a sequence
$\{\FB_{i}\}_{i\in\Bbb N}$ such that the complement of $\Gl$ is
the union of $U_{\FB_{i}}$. Therefore, putting $\FC=\sum a_{i}\FB_{i}$ where
$a_{i}$ are positive and small enough for $S$ to be a Schwartz class function
we see that for all $\pi$ in $\Gl$
$$\pi(\FC)=0$$
and $\pi(\FC)\ne0$ on the complement of $\Gl$.
To finish the proof we need the following lemma:
\p {\rb} { If $\pi$ is an irreducible unitary representation of $G$, the
sequence $\{g_{j}\}_{j\in\Bbb N}$ is dense in $G$, $f\in L^{1}(G)$,
$\pi(f)\ne0$, $\pi(f)\geq 0$,
the
sequence $\{c_{j}\}_{j\in\Bbb N}$ is positive and summable, then
$$A=\pi(\sum c_{j} \delta_{g_{j}^{-1}}f\delta_{g_{j}})$$
(where $\delta_{g_{j}}$ means convolution operator with unit mass at
$g_j$) is injective.}
{\it Proof.} Suppose, on the contrary that $A$ is not injective. Then, there exists
a nonzero $v$ such that
$$(Av,v)=\sum c_{j} (\pi(\delta_{g_{j}^{-1}})\pi(f)\pi(\delta_{g_{j}})v,v)=0,$$
hence for each $j$
$$
(\pi(f)^{1/2}\pi(\delta_{g_{j}})v,\pi(f)^{1/2}\pi(\delta_{g_{j}})v)=0,$$
or simply
$$\pi(f)^{1/2}\pi(\delta_{g_{j}})v=0.$$
Since $\pi$ is irreducible, closed linear span of $\pi(\delta_{g_{j}})v$
gives the whole space, so $\pi(f)^{1/2}=0$. As $\pi(f)$ is nonzero
this gives a contradiction.
Choosing $c_{j}=\exp(-j -|g_{j}|)$ and applying $\rb$ we conclude
that
$$
\FD=\sum c_{j} \delta_{g_{j}^{-1}}\FC\delta_{g_{j}}$$
is a Schwartz class function such that for $\pi$ in the complement
of $\Gl$ the operator $\pi(\FD)$ is injective and for $\pi$ in $\Gl$
$$
\pi(\FD)=0$$
which ends the proof.
{\bf Remark} If $P$ is a regular kernel of order $\alpha$,
$\Gamma=\{l:\pi_l(P)=0\}$, then $Ad^{*}(G)\Gamma \subset \Gamma$,
$\forall_{t>0}\delta_{t}\Gamma \subset \Gamma$ and
$\Gamma$ is closed. More precisely, if $\Re\alpha<0$, then $\Gamma-\{0\}$
is closed in $G^{*}-\{0\}$. The first condition is clear. $\Gamma$ is invariant
under dilations because $P$ is homogeneous. To see that
$\Gamma$ is closed assume first that $\Re \alpha>0$. Let $\phi_n$ be
an approximate unit in $L^1$ consisting of $C^{\infty}_{c}$
functions. $\phi_n*P\in L^1$ so $\Gamma_n=\{l:\pi_l(\phi_n*P)=0\}$
is closed. As $\Gamma=\bigcap_n \Gamma_n$ we see that $\Gamma$ is
closed. If $\Re \alpha \leq 0$, then we compose $P$ with a kernel $R$
such that $\pi_l(R)$ is injective on smooth vectors for all $l\ne 0$
and $P*R$ have order with positive real part.
\vskip0.3cm
{\bf An application}\sno=3\relax%
\nobreak\vskip0.4cm\nobreak\leavevmode%
As an application of our construction we will give an extension to $L^p$ of
a theorem by J. Nourrigat ([\Noe] Th\'eor\`eme 1.3). We need some setup to
state the theorem.
Let $\Omega$ be measure space with measure $\mu$. Assume $G$ act on
$\Omega$ preserving the measure. Let $\phi:G\times\Omega\mapsto \Bbb
C$ be a (measurable) cocycle for this
action, that is $|\phi|=1$ and for all $g_1,g_2\in G$ and all
$x\in\Omega$
$$
\phi(g_1g_2,x)=\phi(g_1,x)\phi(g_2,g_1^{-1}x).$$
Then the formula
$$
\pi(g)f(x)=\phi(g,x)f(g^{-1}x)$$
gives continuous representation of $G$ which act trough isometries on
$L^{p}(\Omega)$, $1\leq p < \infty$ (on $L^{\infty}$ we get isometries, but
the action is only weak-$*$ continuous). We say that $\pi$ is a {\it
cocycle representation}. The set of smooth vectors $C^{\infty}(\pi)$ is
defined as usual (of course it may depend on $p$). Let us note that
$\pi(C_{c}^{\infty}(G))(L^1\cap L^{\infty})$ is dense in $C^{\infty}(\pi)$
so we may
do all the calculations on the common core. We also note that the usual construction of $\pi_l$ gives cocycle representation so we may consider $\pi_l$ as
representations on $L^p$. Let us also sketch the proof of the
following well-known lemma:
\p {\kol} {If $W$ is a regular kernel of order $\alpha$,
$\Re(\alpha)=0$ and $W$ gives bounded operator on $L^{2}(G)$ then $W$
gives bounded operator on $L^{p}(G)$, $10}$, and $\Gamma$ be a closed subset of
$G^{*}$ such that $Ad^{*}(G)\Gamma \subset \Gamma$,
$\forall_{t>0}\delta_{t}\Gamma \subset \Gamma$.
Let $\pi$ be a cocycle representation of $G$ such that all irreducible
components of $\pi$ are of the form $\pi_l$ with $l\in\Gamma$.
Let $R$ be a regular kernel of order
$\alpha$, $\Re(\alpha)>0$ such that for all $l\in\Gamma$, $l\ne 0$ the operator
$\pi_{l}(R)$ is injective on $C^{\infty}$ vectors of $\pi_{l}$. Then for
every $1

-Q$ operator $P^{s}$ is given by the regular kernel of order $s$,
for small $s>0$ $P^s$ generates a semigroup of symmetric probability
measures (see [\CGGP] Theorem 6.1 and [\GlInv]). Choose $s_01$ we simply replace $R$ by $R^k$.
\nic {
the semigroup $e^{tB}$
%a Rockland operator
generated by $B$ satisfy
$$|X^{\gamma}e^{tB}\delta_{0}|\leq ...$$
We define kernel $K$ on $G\times R$ by the formula:
$$
=\int_0^{\infty}e^{-t}dt.$$
From the spectral theorem one easily checks that
$\|B^{1/2}KB^{1/2}\|_{L^2,L^2}\leq 1$. As $\|Rf\|\leq \|B^{1/2}f\|$
so
$$\|RKR^{*}\|_{L^2,L^2}\leq 1.$$
Similarly $\|KR^{*}\|_{L^2,L^2}\leq 1$,
$\|RK\|_{L^2,L^2}\leq 1$ and $\|K\|_{L^2,L^2}\leq 1$. Using ... we have
$\|AKR^{*}\|_{L^2,L^2}\leq C$ and $\|AK\|_{L^2,L^2}\leq C$. One easily
checks using the estimates for $e^{tB}\delta_{0}$ that both $AKR^{*}$
and $AK$ are $L^1$ kernels giving Calderon-Zygmund operators. Hence
both are bounded on $L^p(G\times R)$. Note that one may extend
representation $\pi_l$ to the representation $\tilde \pi_l$ of
$G\times R$ by the formula $\tilde \pi_l((x,t))=\pi_l(x)$. By the
transference principle we have
$$
\|\pi_l(A)f\|_{L^p}=\|\tilde \pi_l(AKR^{*}R+AK)f\|_{L^p}\leq$$
$$
\|\tilde \pi_l(AKR^{*})\tilde \pi_l(R)f\|_{L^p}+
\|\tilde \pi_l(AK)f\|_{L^p}$$
$$\leq
\|AKR^{*}\|_{L^p(G\times R),L^p(G\times R)}\|\pi_l(R)f\|_{L^p}+
\|AK\|_{L^p(G\times R),L^p(G\times R)}\|f\|_{L^p}\leq $$
$$
C(\|\pi_l(R)f\|_{L^p}+\|f\|_{L^p}.$$
}
\vfil
\vskip1cm
{\bf References\par}\nobreak\vskip1cm\nobreak\unvbox2
{Institute of Mathematics, \par Wroc\l{}aw University, \par pl. Grunwaldzki 2/4,
\par 50-384 Wroc\l{}aw, %\par
\hskip1cm Poland.
{\it e-mail}: {\tt hebisch@math.uni.wroc.pl}
{\it www}: {\tt http://www.math.uni.wroc.pl/{\char126}hebisch/Title.html}
}
and
{Institute of Mathematics,
Polish Academy of Sciences,
ul. \'Sniadeckich 8,
00-950 Warszawa,
\hskip1cm Poland}
\end