text:
\section{Introduction}
If $G$ is a locally compact abelian group with dual group $\hat G$, then the Fourier transform on $\hat G$ maps the group algebra $L^1(\hat G)$ injectively onto a subset $A(G)$ of the continuous functions on $G$. Also, the Fourier-Stieltjes transform on $\hat G$ maps the measure algebra $M(\hat G)$ injectively onto a subset $B(G)$ of the continuous functions on $G$. Using the usual identification $L^1(\hat G)\subseteq M(\hat G)$ we see that $A(G) \subseteq B(G)$. Every function in $B(G)$ is bounded, and every function in $A(G)$ vanishes at infinity. In the very special case when $\hat G = \R^n$, the fact that functions in $A(G)$ vanish at infinity is the Riemann-Lebesgue lemma.

In the paper \cite{eymard}, Eymard introduced the algebras $A(G)$ and $B(G)$ in the setting where $G$ is no longer assumed to be abelian. Let $G$ be a locally compact group. The \emph{Fourier-Stieltjes algebra} $B(G)$ is defined as the linear span of the continuous positive definite functions on $G$. There is a natural identification of $B(G)$ with the Banach space dual of the full group \Cs-algebra $C^*(G)$, and under this identification $B(G)$ inherits a norm with which it is a Banach space. The \emph{Fourier algebra} $A(G)$ is the closed subspace in $B(G)$ generated by the compactly supported functions in $B(G)$. Other descriptions of $A(G)$ and $B(G)$ are available (see Section~\ref{sec:fourier}). The Fourier and Fourier-Stieltjes algebras play an important role in non-commutative harmonic analysis.

For general locally compact groups it is still true that $A(G) \subseteq C_0(G)$ just as in the abelian case, and it is natural to ask whether every function in $B(G)$ which vanishes at infinity belongs to $A(G)$.

\begin{que}\label{q1}
Let $G$ be a locally compact group. Does the equality
\begin{align}\label{eq:q1}
A(G) = B(G) \cap C_0(G)
\end{align}
hold?
\end{que}
Of course, if $G$ is compact then $B(G) = A(G)$, and \eqref{eq:q1} obviously holds. But for non-compact groups the question is more delicate.

In 1916, Menchoff \cite{menchoff} proved the existence of a singular probability measure $\mu$ on the circle such that its Fourier-Stieltjes transform $\hat\mu$ satisfies $\hat\mu(n) \to 0$ as $|n|\to\infty$. In other words, $\hat\mu \in B(\Z) \cap C_0(\Z)$, but $\hat\mu\notin A(\Z)$, and thus the answer to Question~\ref{q1} is negative when $G$ is the group $\Z$ of integers. In 1966, Hewitt and Zuckerman \cite{MR0193435} proved that for any abelian locally compact group $G$ the answer to Question~\ref{q1} is always negative, unless $G$ is compact. In 1983 it was shown that for any discrete group $G$ one has $A(G) \neq B(G) \cap C_0(G)$, unless $G$ is finite (see \cite[p.~190]{MR690194} and \cite{MR552704}).

The first non-compact example of a group satisfying \eqref{eq:q1} was given by Khalil in \cite{MR0350330} and is the (non-unimodular) $ax+b$ group consisting of affine transformations $x\mapsto ax + b$ of the real line, where $a> 0$ and $b\in\R$. We remark that the $ax+b$ group is isomorphic to the minimal parabolic subgroup in the simple Lie group $\PSL_2(\R)$ of real rank one.

It is proved in \cite{MR0493175},\cite{MR552704} that if \eqref{eq:q1} holds for some second countable, locally compact group $G$, then the regular representation of $G$ is completely reducible, i.e., a direct sum of irreducible representations. For a while, this was thought to be a characterization of groups satisfying \eqref{eq:q1}, but this was shown not to be the case (see \cite{MR509261} or \cite{MR0486289}). However, it follows from the fact that second countable, locally compact groups satisfying \eqref{eq:q1} have completely reducible regular representations combined with \cite[Theorem~3.1]{MR1637396} that \eqref{eq:q1} fails for second countable, locally compact IN-groups, unless they are compact. Recall that an IN-group is a group which has a compact neighborhood of the identity which is invariant under all inner automorphisms. In particular, abelian, discrete and compact groups are all IN-groups.

It follows from Baggett's work \cite{MR735532} that if $G$ is a locally compact, second countable group which is also connected, unimodular and has a completely reducible regular representation, then $G$ is compact (see \cite[Theorem~3]{MR2459312}). In particular, Question~\ref{q1} has a negative answer for locally compact second countable connected unimodular groups which are non-compact. This gives an abundance of examples of groups where Question~\ref{q1} has a negative answer. An example given in \cite{MR0420149} of a unimodular group satisfying \eqref{eq:q1} shows that the assumption about connectedness cannot be removed from the previous statement, and of course the assumption about unimodularity cannot be removed as the $ax +b$ group shows.

It should be apparent from the above that there are plenty of examples of groups for which Question~\ref{q1} has a negative answer. In this paper we provide new examples of groups answering Question~\ref{q1} in the affirmative. Our main source of examples is formed by the minimal parabolic subgroups in connected simple Lie groups of real rank one. But first we give a more straightforward example which is a subgroup of $\SL_3(\R)$. The method of proof in this example can be seen as an easy version of what follows after. We prove the following.

\begin{thm}\label{thm:fourier-PP}
For the group
\begin{align}\label{eq:PP}
P =
\left\{\begin{pmatrix}
\lambda & a & c \\
0 & \lambda^{-1} & b \\
0 & 0 & 1
\end{pmatrix}
\middle\vert a,b,c\in\R,\ \lambda>0
\right\}
\end{align}
we have $A(P) = B(P) \cap C_0(P)$.
\end{thm}

If we think of $\SL_2(\R)\ltimes\R^2$ as a subgroup of $\SL_3(\R)$ in the following way
\begin{align*}
\left(\begin{array}{cc|c}
\multicolumn{2}{c|}{\multirow{2}{*}{$\SL_2(\R)$}} & \multirow{2}{*}{$\R^2$} \\
 & & \\
 \hline
\multicolumn{2}{c|}{0} & 1
\end{array}\right),
\end{align*}
then we can think of $P$ as a subgroup of $\SL_2(\R)\ltimes\R^2$. This viewpoint will be relevant in a forthcoming paper \cite{HK-WH-examples} by the author and U.~Haagerup.

Apart from the group in \eqref{eq:PP}, our examples of groups satisfying \eqref{eq:q1} arise in the following way. Let $n\geq 2$, let $G$ be one of the classical simple Lie groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ or the exceptional group $\FF$, and let $G = KAN$ be the Iwasawa decomposition. If $M$ is the centralizer of $A$ in $K$, then $P = MAN$ is the minimal parabolic subgroup of $G$. We refer to Section~\ref{sec:simple-groups} for explicit descriptions of the groups $G$, $K$, $A$, $N$ and $M$. We prove the following theorem concerning the Fourier algebra of the minimal parabolic subgroup.

\begin{thm}\label{thm:fourier-P}
Let $P$ be the minimal parabolic subgroup in one of the simple Lie groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ or $\FF$. Then $A(P) = B(P) \cap C_0(P)$.
\end{thm}

In order to prove Theorem~\ref{thm:fourier-PP} and Theorem~\ref{thm:fourier-P} we develop a general strategy for providing examples of groups that answer Question~\ref{q1} affirmatively. The strategy is based on (1) determining all irreducible representations of the group, (2) determining the irreducible subrepresentations of the regular representation and (3) disintegration theory. An often useful tool for (1) is the Mackey Machine (see \cite[Chapter 6]{Folland}).

Our strategy for proving Theorem~\ref{thm:fourier-PP} and Theorem~\ref{thm:fourier-P} is contained in the following theorem.

\begin{thm}\label{thm:strategy}
Let $G$ be a second countable, locally compact group satisfying the following two conditions.
\begin{enumerate}
	\item $G$ is type I.
	\item There is a non-compact, closed subgroup $H$ of $G$ such that every irreducible unitary representation of $G$ is either trivial on $H$ or is a subrepresentation of the left regular representation $\lambda_G$.
\end{enumerate}
Then
$$
A(G) = B(G) \cap C_0(G).
$$
In particular, the left regular representation $\lambda_G$ is completely reducible.
\end{thm}

It was pointed out to the author by T.~de~Laat that with the assumptions of Theorem~\ref{thm:strategy} one can deduce that $(G,H)$ has the \emph{relative Howe-Moore property} (defined in \cite{MR2854688}). In fact, condition (1) can be dropped, and condition (2) still implies that $(G,H)$ has the relative Howe-Moore property as is immediately seen from \cite[Proposition~2.3]{MR2854688} and the well-known fact that the regular representation is a $C_0$-representation.

Since we prove Theorem~\ref{thm:fourier-PP} and Theorem~\ref{thm:fourier-P} by verifying the conditions in Theorem~\ref{thm:strategy} for the groups in question, we obtain the following corollary.

\begin{cor}\label{cor:HM}
Let $P$ be the group in \eqref{eq:PP} or the minimal parabolic subgroup in one of the simple Lie groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ or $\FF$. Then there is a normal, non-compact closed subgroup $H$ in $P$ such that $(P,H)$ has the relative Howe-Moore property.
\end{cor}

The non-compact subgroup $H$ can be described explicitly. For a more precise statement see Corollary~\ref{cor:HM2} below. We refer to \cite{MR2854688} for a treatment of the relative Howe-Moore property.

In order to verify the two conditions in Theorem~\ref{thm:strategy} for the minimal parabolic subgroups $P$, we rely primarily on earlier work of J.A.~Wolf. In \cite{MR0422519} the irreducible representations of some parabolic subgroups are determined by employing the Mackey Machine, and the approach of \cite{MR0422519} carries over to our situation almost without changes. Using \cite{MR0342641} we can easily determine the irreducible subrepresentations of the regular representation of $P$.

The paper is organized as follows. In Section~\ref{sec:fourier} we describe the basic properties of the Fourier and Fourier-Stieltjes algebra, and Section~\ref{sec:strategy} contains the proof of Theorem~\ref{thm:strategy}. Section~\ref{sec:inv-meas} contains a few results to be used later when we verify condition~(2) of Theorem~\ref{thm:strategy} for the groups in question. In Section~\ref{sec:PP} we prove Theorem~\ref{thm:fourier-PP}. This includes determining all irreducible unitary representations of the group \eqref{eq:PP}, determining the Plancherel measure for the group and finally verifying conditions~(1) and (2) of Theorem~\ref{thm:strategy} for the group.

In Section~\ref{sec:simple-groups} we turn to the minimal parabolic subgroups $P$ in the simple Lie groups of real rank one that we will be working with. We give an explicit description of the groups as matrix groups (at least in the classical cases). In Section~\ref{sec:irreducibles} we describe the irreducible representations of the minimal parabolic subgroups, and then, in Section~\ref{sec:fourier-P}, we verify the two conditions in Theorem~\ref{thm:strategy} for the minimal parabolic subgroups. Theorem~\ref{thm:fourier-P} then follows immediately.

Section~\ref{sec:HM} contains the proof of Corollary~\ref{cor:HM} concerning the Howe-Moore property, and Section~\ref{sec:concluding} contains some concluding remarks.

\section{The Fourier and Fourier-Stieltjes algebra}\label{sec:fourier}

This section contains a brief description of the Fourier and Fourier-Stieltjes algebra of a locally compact group introduced by Eymard in \cite{eymard}. We refer to the original paper \cite{eymard} for more details. Let $G$ be a locally compact group equipped with a left Haar measure. By a representation of $G$ we always mean a continuous unitary representation of $G$ on some Hilbert space (except for the vector and spin representations in Section~\ref{sec:FF}). If $\pi$ is a unitary representation of $G$ on a Hilbert space $\mc H$, and $x,y\in\mc H$, then the continuous complex function
$$
\phi(g) = \la \pi(g)x,y\ra, \qquad (g\in G)
$$
is a \emph{matrix coefficient} of $\pi$. The Fourier-Stieltjes algebra of $G$ is denoted $B(G)$ and consists of the complex linear span of continuous positive definite functions on $G$. It coincides with the set of all matrix coefficients of representations of $G$,
$$
B(G) = \{ \la \pi(\cdot)x,y \ra \mid (\pi,\mc H) \text{ is a representation of } G \text{ and } x,y\in\mc H \}.
$$
Since the pointwise product of two positive definite functions is again positive definite, $B(G)$ is an algebra under pointwise multiplication. Given $\phi\in B(G)$, the map
$$
f\mapsto \la f,\phi\ra = \int_G f(x)\phi(x) \ dx
$$
is a linear functional on $L^1(G)$ which is bounded, when $L^1(G)$ is equipped with the universal \Cs-norm. Hence $\phi$ defines a functional on $C^*(G)$, the full group \Cs-algebra of $G$, and this gives the identification of $B(G)$ with $C^*(G)^*$ as vector spaces. The Fourier-Stieltjes algebra inherits the norm
$$
\|\phi\| = \sup\{ |\la f,\phi\ra| \mid f\in L^1(G),\ \|f\|_{C^*(G)} \leq 1 \}
$$
of $C^*(G)^*$ from this identification. With this norm $B(G)$ is a unital Banach algebra.

Given $\phi\in B(G)$, a representation $(\pi,\mc H)$ and $x,y\in\mc H$ such that $\phi(g) = \la \pi(g)x,y\ra$ we have
$$
\|\phi\| \leq \|x\|\|y\|,
$$
and conversely, it is always possible to find $(\pi,\mc H)$ and $x,y\in \mc H$ such that $\phi(g) = \la \pi(g)x,y\ra$ and $\|\phi\| = \|x\|\|y\|$.

The Fourier algebra of $G$ is denoted $A(G)$ and is the closure of the set of compactly supported functions in $B(G)$, and $A(G)$ is in fact an ideal. The Fourier algebra coincides with the set of all matrix coefficients of the left regular representation of $G$,
$$
A(G) = \{ \la \lambda(\cdot)x,y \ra \mid x,y\in L^2(G) \},
$$
and given any $\phi\in A(G)$, there are $x,y\in L^2(G)$ such that $\phi(g) = \la \lambda(g)x,y\ra$ and $\|\phi\| = \|x\|\|y\|$. This can be rephrased as follows. Given $\phi\in A(G)$, there are $f,h\in L^2(G)$ such that $\phi = f * \check h$ and $\|\phi\| = \|f\|\|h\|$, where $\check h(g) = h(g^{-1})$. This is often written as
$$
A(G) = L^2(G) * L^2(G).
$$
It is known that $\|\phi\|_\infty \leq \|\phi\|$ for any $\phi\in B(G)$, and hence $A(G) \subseteq C_0(G)$.

Although we will not study von Neumann algebras in this paper, we note that $A(G)$ may be identified with the predual of the group von Neumann algebra $L(G)$ of $G$. When $G$ is abelian, the Fourier transform provides an isometric isomorphism between $L^1(\hat G)$ and $A(G)$, and in this way $A(G)$ is identified isometrically with the predual of group von Neumann algebra $L(G) \simeq L^\infty(\hat G)$. In the non-abelian case it is still true that $A(G)$ identifies isometrically with the predual of the group von Neumann algebra via the duality
$$
\la T,\phi\ra = \la Tf,h\ra,
$$
where $T\in L(G)$ and $\phi = \bar h*\check f$ for some $f,h\in L^2(G)$.

\section{Proof of Theorem~\ref{thm:strategy}}\label{sec:strategy}

In this section we prove Theorem~\ref{thm:strategy}, which is the basis of proving Theorems~\ref{thm:fourier-PP} and \ref{thm:fourier-P}. We first prove that the conditions in Theorem~\ref{thm:strategy} ensure that the regular representation is completely reducible.

\begin{lem}
Let $G$ be a locally compact group. Any unitary representation of $G$ on a separable Hilbert space has at most countably many inequivalent (with respect to unitary equivalence) irreducible subrepresentations.
\end{lem}
\begin{proof}
Let $\pi$ be a unitary representation of $G$. The subrepresentations of $\pi$ are in correspondence with the projections in the commutant $\pi(G)'$, equivalent subrepresentations correspond to projections that are equivalent in $\pi(G)'$ (in the sense of Murray-von Neumann), and the irreducible subrepresentations correspond to minimal projections in $\pi(G)'$. It is therefore enough to show that a von Neumann algebra on a separable Hilbert space has at most countably many inequivalent minimal projections. Let $M$ be such a von Neumann algebra.

Recall that two minimal projections are inequivalent if and only if their central supports are orthogonal (see \cite[Proposition 6.1.8]{MR1468230}). Let $(p_i)_{i\in I}$ be a family of inequivalent minimal projections, and let $c_i$ be the central support of $p_i$. Then $(c_i)_{i\in I}$ is a family of orthogonal projections. By separability of the Hilbert space, $I$ must be countable. Hence there are at most countably many inequivalent minimal projections in $M$.
\end{proof}

\begin{cor}
Let $G$ be a locally compact, second countable group. Then the left regular representation of $G$ has at most countably many inequivalent irreducible subrepresentations.
\end{cor}
\begin{proof}
The left regular representation represents $G$ on the Hilbert space $L^2(G)$, which is separable, since $G$ is second countable. The statement now follows.
\end{proof}

We recall that a unitary representation is of type I, if the image of the representation generates a type I von Neumann algebra. A locally compact group is said to be of type I, if all its unitary representations are of type I (see \cite[Chapter~13]{MR0458185}). Disintegration theory works especially well in the setting of type I groups. We refer to \cite[Chapter~7]{Folland} for more on type I groups and disintegration theory. Several equivalent characterizations of type I groups can also be found in \cite[Chapter~9]{MR0458185}, but let us just mention one characterization here. The unitary equivalence classes of irreducible representations form a set $\hat G$ called the unitary dual of $G$. The dual $\hat G$ is equipped with the Mackey Borel structure, and $G$ is of type I if and only if $\hat G$ is a standard Borel space. When $G$ is abelian, the unitary dual coincides with the usual dual group.

\begin{prop}\label{prop:completely-reducible}
Let $G$ be a second countable, locally compact group satisfying the following two conditions.
\begin{enumerate}
	\item $G$ is type I.
	\item There is a non-compact, closed subgroup $H$ of $G$ such that every irreducible unitary representation of $G$ is either trivial on $H$ or is a subrepresentation of the left regular representation $\lambda_G$.
\end{enumerate}
Then the left regular representation $\lambda_G$ is completely reducible.
\end{prop}
\begin{proof}
For each $p\in\hat G$, we let $\pi_p$ denote a representative of the class $p$, and we assume that the choice of representative is made in a measurable way (\cite[Lemma 7.39]{Folland}). We write the left regular representation as a direct integral of irreducibles,
$$
\lambda_G = \int_{\hat G}^\oplus n_p\pi_p \ d\mu(p),
$$
where $\mu$ is a Borel measure on $\hat G$ and $n_p\in\{0,1,2,\ldots,\infty\}$ (see \cite[Theorem 7.40]{Folland}). Let $A = \{[p] \in\hat G \mid p(h) = 1 \text{ for all } h\in H \}$ and let $B = \hat G \setminus A$. It is not hard to check that $A\subseteq \hat G$ is a Borel set.

We note that if $\pi_p\in B$, then $\pi_p$ is a subrepresentation of $\lambda_G$. By the previous corollary, $B$ is countable. Since $\lambda_G$ has no subrepresentation which is trivial on a non-compact subgroup, we must have $\mu(A) = 0$. Then
$$
\lambda_G = \int_{B}^\oplus n_p\pi_p \ d\mu(p),
$$
and since $B$ is countable, $\lambda_G$ is a direct sum of irreducibles.
\end{proof}

\begin{lem}\label{lem:split}
Let $G$ be a locally compact, second countable group with left regular representation $\lambda$ and a closed subgroup $H$ such that
\begin{enumerate}
	\item $G$ is type I;
	\item Every irreducible unitary representation of $G$ is either trivial on $H$ or is a subrepresentation of $\lambda$;
	\item $\lambda$ is completely reducible.
	\end{enumerate}
Then every unitary representation $\pi$ of $G$ is a sum $\sigma_1\oplus\sigma_2$, where $\sigma_1$ is trivial on $H$ and $\sigma_2\leq \lambda^\infty$. Here $\lambda^\infty$ denotes a countably infinite direct sum $\lambda\oplus\lambda\oplus\cdots$ of the representation $\lambda$.
\end{lem}
\begin{proof}

For each $p\in\hat G$, we let $\pi_p$ denote a representative of the class $p$, and we assume that the choice of representative is made in a measurable way (\cite[Lemma 7.39]{Folland}). There is a decomposition
$$
\lambda \simeq \bigoplus_{p\in C} m_p \pi_p
$$
for some countable $C\subseteq \hat G$ and suitable multiplicities $m_p\in\{1,2,\ldots,\infty\}$. We may write $\pi$ is a direct integral of irreducibles,
$$
\pi = \int_{\hat G}^\oplus n_p\pi_p \ d\mu(p),
$$
where $\mu$ is a Borel measure on $\hat G$ and $n_p\in\{0,1,2,\ldots,\infty\}$ (see \cite[Theorem 7.40 ]{Folland}). Let $A = \{[p] \in\hat G \mid p(h) = 1 \text{ for all } h\in H \}$ and let $B = \hat G \setminus A$. Then $A\subseteq \hat G$ is a Borel set. It follows from our assumptions that $B\subseteq C$. If
$$
\sigma_1 = \int_{A}^\oplus n_p\pi_p \ d\mu(p), \qquad \sigma_2 = \int_{B}^\oplus n_p\pi_p \ d\mu(p),
$$
then we see that
$$
\pi = \sigma_1 \oplus \sigma_2,
$$
where $\sigma_1(g) = 1$ for every $g\in H$. Also,
$$
\sigma_2 \leq \bigoplus_{p\in B} n_p \pi_p \leq \bigoplus_{p\in C}(n_pm_p)\lambda \leq \lambda^\infty.
$$
\end{proof}

\begin{lem}\label{lem:AB}
Let $G$ be a locally compact group with left regular representation $\lambda$ and a closed, non-compact subgroup $H$. Suppose every unitary representation $\pi$ of $G$ is a sum $\sigma_1\oplus\sigma_2$, where $\sigma_1$ is trivial on $H$ and $\sigma_2\leq \lambda^\infty$. Then $A(G) = B(G)\cap C_0(G)$.
\end{lem}
\begin{proof}
The inclusion $A(G) \subseteq B(G) \cap C_0(G)$ holds for any locally compact group $G$. Suppose $\phi\in B(G)\cap C_0(G)$. Then there is a continuous, unitary representation $\pi$ of $G$ on some Hilbert space $\mc H$ and vectors $x,y\in\mc H$ such that
$$
\phi(g) = \la \pi(g)x,y\ra \qquad\text{for all } g\in G.
$$
By assumption we may split $\pi = \sigma_1\oplus\sigma_2$. Accordingly, we split $\phi = \phi_1 + \phi_2$, where $\phi_1$ is a coefficient of $\sigma_1$ etc. We will show that $\phi_1 = 0$ and $\phi_2\in A(G)$, which will complete the proof.

Since $\sigma_2\leq \lambda^\infty$, we see that $\phi_2$ is of the form
$$
\phi_2(g) = \sum_{i=1}^\infty \la \lambda(g)x_i,y_i\ra
$$
for some $x_i,y_i\in L^2(G)$ with $\sum_i \| x_i\|^2 < \infty$ and $\sum_i \| y_i\|^2 < \infty$. Each of the maps
$$
g\mapsto \la \lambda(g)x_i,y_i\ra
$$
is in $A(G)$ with norm at most $\|x_i\|\|y_i\|$. Since $A(G)$ is a Banach space and $\sum_i \|x_i\|\|y_i\| < \infty$, we deduce that $\phi_2 \in A(G)$, and in particular $\phi_2\in C_0(G)$. It then follows that $\phi_1\in C_0(G)$. Since $\sigma_1$ is trivial on $H$, we see that $\phi_1$ is constant on $H$ cosets. Since $H$ is non-compact, we deduce that $\phi_1 = 0$. Then $\phi = \phi_2\in A(G)$. This proves $B(G)\cap C_0(G) = A(G)$.
\end{proof}

Theorem~\ref{thm:strategy} is an easy consequence of the previous statements.

\begin{proof}[Proof of Theorem~\ref{thm:strategy}]
We assume that $G$ is a locally compact, second countable group satisfying the two conditions in the statement of the theorem. It follows from Proposition~\ref{prop:completely-reducible} that $\lambda_G$ is completely reducible. So by Lemma~\ref{lem:split}, every unitary representation $\pi$ of $G$ is a sum $\sigma_1\oplus\sigma_2$, where $\sigma_1$ is trivial on $H$ and $\sigma_2\leq \lambda^\infty$. From Lemma~\ref{lem:AB} we conclude that $A(G) = B(G) \cap C_0(G)$.
\end{proof}

\section{Invariant measures on homogeneous spaces}\label{sec:inv-meas}

To describe the irreducible representations of the groups $P$ in question, we rely on a general method known to the common man as the Mackey Machine. Essential in the Mackey Machine is the notion of induced representations. For a general introduction to the theory of induced representations we refer to \cite[Chapter 6]{Folland} which also contains a description of (a simple version of) the Mackey Machine. The general results about the Mackey Machine can be found in the original paper \cite{MR0098328}.

The construction of an induced representation from a closed subgroup $H$ to a group $G$ is more easily described when the homogeneous space $G/H$ admits an invariant measure for the $G$-action given by left translation. Regarding homogeneous spaces and invariant measures we record the following easy (and well-known) facts.

\begin{prop}\label{prop:cosetspaces}
Consider topological groups $G$, $N$, $H$, $K$, $A$, $B$ and topological spaces $X$ and $Y$.

\begin{enumerate}
	\item
Suppose $G$ is the semi-direct product $G = N\rtimes H$, where $N$ is normal in $G$. If $K\leq H$ is a closed subgroup of $H$, then there is a canonical isomorphism
$$
NH/NK \simeq H/K
$$
as $G$-spaces. Here the $G$-action on $H/K$ is the $H$-action, and $N$ acts trivially on $H/K$.

\item
Suppose $G = N\times H$, and $A\leq N$, $B\leq H$ are closed subgroups. Then there is a canonical isomorphism
$$
(N\times H)/(A\times B) \simeq N/A \times H/B
$$
as $G$-spaces, where the $G$-action on $N/A \times H/B$ is the product action of $N\times H$.

\item
Suppose $G\acts X$ and $H\acts Y$ have invariant, $\sigma$-finite Borel measures. Then the product $G\times H \acts X\times Y$ has an invariant, $\sigma$-finite Borel measure.

\item
Suppose $G$ is compact (or just amenable) and $X$ is compact. Then any action $G\acts X$ has an invariant probability measure.
\end{enumerate}
\end{prop}
\begin{proof}
\mbox{}
\begin{enumerate}
	\item The map $[nh]_{NK} \mapsto [h]_K$ is a well-defined, equivariant homeomorphism.
	\item The map $[(n,h)]_{A\times B} \mapsto ([n]_A , [h]_B)$ is a well-defined, equivariant homeomorphism.
	\item Take the product measure on $X\times Y$ of the invariant measures on $X$ and $Y$.
	\item This is Proposition~5.4 in \cite{MR767264}.
\end{enumerate}
\end{proof}

The following lemma will be relevant in Section~\ref{sec:PP} and Section~\ref{sec:fourier-P} when we verify condition (2) of Theorem~\ref{thm:strategy} for the minimal parabolic groups $P$.

\begin{lem}\label{lem:trivial-induction}
Let $G$ be a locally compact group with closed subgroups $N\subseteq H\subseteq G$, and suppose $N\triangleleft G$. If $\sigma$ is a unitary representation of $H$ which is trivial on $N$, and if $G/H$ admits a $G$-invariant measure, then the induced representation $\Ind_H^G \sigma$ is also trivial on $N$.
\end{lem}
\begin{proof}
Let $\mc H$ denote the Hilbert space of $\sigma$, and let $q:G\to G/H$ be the quotient map. The induced representation $\pi = \Ind_H^G\sigma$ acts on a completion of the space
$$
\mc F_0 = \left\{ f \in C(G,\mc H) \middle\vert \begin{array}{l} f(gh) = \sigma(h^{-1})f(g) \text{ for } g\in G, h\in H \\ \text{and } q(\supp f) \text{ is compact} \end{array}\right\}.
$$
Since $G/H$ admits an invariant measure, the action on $\mc F_0$ is simply given by left translation, $(\pi(x) f)(g) = f(x^{-1}g)$. With $f\in\mc F_0$, $g\in G$ and $n\in N$, we compute $(\pi(n) f)(g)$. We get
$$
(\pi(n) f)(g) = f(n^{-1}g) = f(g(g^{-1}n^{-1}g)) = \sigma(g^{-1}ng)f(g) = f(g),
$$
since $g^{-1}ng\in N$. It follows that $\pi(n) = 1$.
\end{proof}

\section{The first example}\label{sec:PP}

In this section we prove Theorem~\ref{thm:fourier-PP}. Let $P$ be the group defined in \eqref{eq:PP}. In the following proposition we describe the unitary dual of $P$, i.e. the equivalence classes of the irreducible representations of $P$. To do so we apply the Mackey Machine, which works particularly well in our case, where $P$ decomposes as a semidirect product $N_0\rtimes P_0$ with $N_0$ abelian. For an account on the Mackey Machine we refer to Chapter 6 in \cite{Folland}.

Consider the following closed subgroups of $P$.
\begin{align}
\label{eq:Pzero}
P_0 &= \left\{\begin{pmatrix}
\lambda & a & 0 \\
0 & \lambda^{-1} & 0 \\
0 & 0 & 1
\end{pmatrix}
\middle\vert\ a\in\R,\ \lambda>0
\right\}
\displaybreak[0] \\[15pt]
\label{eq:P1}
P_1 &= \left\{\begin{pmatrix}
1 & a & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\middle\vert\ a\in\R
\right\}
\displaybreak[0] \\[15pt]
N_0 &= \left\{\begin{pmatrix}
1 & 0 & c \\
0 & 1 & b \\
0 & 0 & 1
\end{pmatrix}
\middle\vert\ b,c\in\R
\right\}
\displaybreak[0] \\[15pt]
\label{eq:N1}
N_1 &= \left\{\begin{pmatrix}
1 & 0 & c \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\middle\vert\ c\in\R
\right\}
\end{align}
Observe that $P = N_0\rtimes P_0$. We note that $P_0$ is isomorphic to the $ax+b$ group, i.e. the group of affine transformations $x\mapsto ax + b$ of the real line, where $a>0$ and $b\in\R$. The dual of the $ax+b$ group is well-known (see for instance \cite[Section~6.7]{Folland}). The dual of $N_0 \simeq \R^2$ is $\hat N_0 \simeq \hat \R^2$ which we as usual identify with $\R^2$.

\begin{prop}\label{prop:unitary-dual}
Let $\pi$ be an irreducible representation of $P$. Then $\pi$ is equivalent to one of the following representations (and these are all inequivalent).
\begin{enumerate}
	\item $\pi_1 = \Ind_{N_0}^P(\nu)$, where $\nu\in\hat N_0$ is $\nu = (1,0)$.
	\item $\pi_2 = \Ind_{N_0}^P(\nu)$, where $\nu\in\hat N_0$ is $\nu = (-1,0)$.
	\item $\pi_{3,\rho} = \Ind_{N_0P_1}^P(\nu\rho)$, where $\nu\in\hat N_0$ is $\nu = (0,1)$ and $\rho$ is a character in $\hat{P_1} \simeq \R$.
	\item $\pi_{4,\rho} = \Ind_{N_0P_1}^P(\nu\rho)$, where $\nu\in\hat N_0$ is $\nu = (0,-1)$ and $\rho$ is a character in $\hat{P_1} \simeq \R$.
	\item $\pi_{5,\sigma} = \sigma \circ q$, where $\sigma \in \hat P_0$ and $q: P\to P_0$ is the quotient map.
\end{enumerate}
\end{prop}

\begin{proof}
We follow the strategy of the Mackey Machine as described in Theorem 6.42 in \cite{Folland}, which gives a complete description of the unitary dual of $P$. We think of $P = N_0\rtimes P_0$ as a subgroup of $\R^2\rtimes\SL_2(\R)$, where $\SL_2(\R)$ acts on $\R^2$ by matrix multiplication. The action $P_0 \acts N_0$ is then simply matrix multiplication, and the dual action $P_0\acts\hat N_0$ is given by $(p.\nu)(n) = \nu(p^{-1}.n)$ for $p\in P_0$, $\nu\in\hat N_0$ and $n\in N_0$. Under the usual identification $\hat N_0 \simeq \R^2$ we see that $p\in P_0$ acts on $\R^2$ by matrix multiplication by the transpose of the inverse of $p$. Thus, if $p$ has the form in \eqref{eq:Pzero}, then the action of $p$ on $\R^2$ is
$$
\begin{pmatrix}
	s \\ t
\end{pmatrix}
\mapsto
\begin{pmatrix}
	\lambda^{-1} & 0 \\
	-a & \lambda
\end{pmatrix}
\begin{pmatrix}
	s \\ t
\end{pmatrix}.
$$
There are five orbits in $\hat N_0$ under this action, which give the five alternatives in the proposition. The orbits are
\begin{alignat*}{2}
&\mc O_1 &&= \{ (s,t) \mid s > 0 \},
\displaybreak[0] \\
&\mc O_2 &&= \{ (s,t) \mid s < 0 \},
\displaybreak[0] \\
&\mc O_3 &&= \{ (0,t) \mid t > 0 \},
\displaybreak[0] \\
&\mc O_4 &&= \{ (0,t) \mid t < 0 \},
\displaybreak[0] \\
&\mc O_5 &&= \{ (0,0) \}.
\end{alignat*}

Since there are only finitely many orbits, the action of $P_0$ on $\hat N_0$ is regular. As representatives of the orbits we choose the points
$$
(1,0)\in\mc O_1,\ (-1,0)\in\mc O_2,\ (0,1)\in\mc O_3,\ (0,-1)\in\mc O_4,\ (0,0)\in\mc O_5.
$$

Case 1: $\nu = (1,0)$. In this case the stabilizer subgroup of $\nu$ inside $P_0$ is trivial, and hence we obtain the representation $\pi = \Ind_{N_0}^P(\nu)$.

Case 2: $\nu = (-1,0)$. This is similar to case 1.

Case 3: $\nu = (0,1)$. The stabilizer subgroup of $\nu$ inside $P_0$ is $P_1$, and hence we obtain $\pi = \Ind_{N_0P_1}^P(\nu\rho)$, where $\rho \in \hat{P_1}$. Here the representation $\nu\rho$ on $N_0P_1$ is given by
$$
(\nu\rho)(nh) = \nu(n)\rho(h), \qquad \text{for all }n\in N_0,\ h\in P_1.
$$

Case 4: $\nu = (0,-1)$. This is similar to case 3.

Case 5: $\nu = (0,0)$. In this case the stabilizer subgroup of $\nu$ inside $P_0$ is everything. It follows that $\pi$ is a representation which satisfies $\pi(n) = \la n,\nu\ra$ for every $n\in N_0$. In other words, $\pi$ is trivial on $N_0$ and factors to an irreducible representation $\sigma$ of $P_0$. That is, $\pi = \sigma \circ q$.
\begin{comment}
\begin{figure}[hbt
\begin{tikzpicture}[scale=1.5]
\draw[pattern=north west lines,pattern color=lightgray] (0,-2.5) rectangle (4,2.5);
\draw[pattern=north east lines,pattern color=lightgray] (-4,-2.5) rectangle (0,2.5);

\draw [<->] (0,2.6) -- (0,0) -- (4.1,0);
\draw [<->] (0,-2.6) -- (0,0) -- (-4.1,0);

\draw [line width=2] (0,0) -- (0,2.5);
\draw [line width=2] (0,0) -- (0,-2.5);

\node [above right] at (0,0) {$\mc O_5$};
\node [right] at (0,1.5) {$\mc O_3$};
\node [right] at (0,-1.5) {$\mc O_4$};
\node at (-2,0.5) {$\mc O_2$};
\node at (2,0.5) {$\mc O_1$};

\draw [fill] (0,0) circle (0.05);
\draw [fill] (1,0) circle (0.05);
\draw [fill] (-1,0) circle (0.05);
\draw [fill] (0,1) circle (0.05);
\draw [fill] (0,-1) circle (0.05);

\node [below right] at (0,0) {$(0,0)$};
\node [right] at (0,1) {$(0,1)$};
\node [right] at (0,-1) {$(0,-1)$};
\node [below right] at (1,0) {$(1,0)$};
\node [below left] at (-1,0) {$(-1,0)$};

\end{tikzpicture}

\caption{The orbit space of the dual action $P_0 \acts \hat N_0$.
\label{fig:orbit-space
\end{figure}
\end{comment}
\end{proof}

The Plancherel measure of a group describes how the left regular representation decomposes as a direct integral of irreducible representations. For example, the Plancherel measure of a locally compact abelian group is simply the Haar measure on the dual group. This is seen using the Fourier transform. The following proposition determines the Plancherel measure of $P$ and shows in particular that the measure is purely atomic. Hence the left regular representation of $P$ is completely reducible.

\begin{prop}\label{prop:left-regular-P}
The left regular representation $\lambda_P$ of $P$ is (equivalent to) the countably infinite direct sum of $\pi_1\oplus\pi_2$, where $\pi_1$ and $\pi_2$ are as in Proposition \ref{prop:unitary-dual}.
\end{prop}

\begin{proof}
Again it is useful to view $P$ as the semidirect product $P = N_0\rtimes P_0$. We follow the approach described in \cite[Section 1]{MR509261}. Their results are stated for the right regular representation, but everything works mutatis mutandis for the left regular. As before, the dual group $\hat N_0$ is identified with $\R^2$, and the Plancherel measure on $\hat N_0$ is simply Lebesgue measure. The orbits under the dual action $P_0\acts \hat N_0$ which have positive Lebesgue measure are $\mc O_1$ and $\mc O_2$, and their complement in $\hat N_0$ is a null set (the $y$-axis). The stabilizer subgroups inside $P_0$ of the points $(1,0)$ and $(-1,0)$ are trivial, so in particular these stabilizer subgroups have completely reducible regular representations. Thus criteria (a) and (b) of \cite{MR509261} are satisfied, and it follows from their calculation on page 595 that
$$
\lambda_P = \bigoplus_{n=1}^\infty (\pi_1 \oplus \pi_2).
$$
\end{proof}

\begin{lem}
Consider the groups $P = N_0 \rtimes P_0$ and $N_1 \subseteq P$. If $\pi$ is an irreducible unitary representation of $P$, then one (and only one) of the following holds.
\begin{enumerate}
	\item $\pi(g) = 1$ for every $g$ in the subgroup $N_1$,
	\item $\pi$ is a subrepresentation of $\lambda_P$.
\end{enumerate}
\end{lem}
\begin{proof}
We divide the proof into the cases according to the description in Proposition \ref{prop:unitary-dual}.

If $\pi = \pi_1$ or $\pi = \pi_2$, then it follows from Proposition \ref{prop:left-regular-P} that $\pi \leq \lambda_P$.

Suppose now $\pi = \pi_{3,\rho}$, where $\rho \in \hat P_1$. If we let $\nu = (0,1) \in \hat N_0$, then we see that $N_1 = \ker\nu$. Hence the representation $\nu\rho$ of $N_0P_1$ is trivial on $N_1$ which is normal in $P$. Since $N_0P_1$ is a normal subgroup of $P$, the homogeneous space $P/(N_0P_1)$ has a $P$-invariant measure, Haar measure. It follows from Lemma~\ref{lem:trivial-induction} that $\pi = \Ind_{N_0P_1}^P(\nu\rho)$ is trivial on $N_1$.

The case $\pi = \pi_{4,\rho}$ is similar to the previous case. We simply note that $\ker\bar\nu = N_1$, where $\bar\nu = (0,-1)$.

In the case $\pi = \pi_{5,\sigma}$, it is clear that $\pi(g) = 1$ for every $g\in N_0$, and hence in particular for every $g\in N_1$.
\end{proof}

\begin{lem}
The group $P$ is of type I.
\end{lem}
\begin{proof}
The group $P$ is a connected, real algebraic group, and such groups are of type I according to \cite[Theorem 1]{MR0099380}.

\end{proof}

We collect the previous results of this section in the following proposition, which together with Theorem~\ref{thm:strategy} immediately implies Theorem~\ref{thm:fourier-PP}.

\begin{prop}\label{prop:PBA}
Let $P$ and $N_1$ be the groups in \eqref{eq:PP} and \eqref{eq:N1}. The following holds.
\begin{enumerate}
	\item $P$ is of type I.
	\item Every irreducible unitary representation of $P$ is either trivial on the non-compact closed subgroup $N_1$ or is a subrepresentation of $\lambda_P$.
\end{enumerate}
\end{prop}

\section{Simple Lie groups of real rank one}\label{sec:simple-groups}
Let $G$ be a connected simple Lie group with finite center and of real rank one. Let $G = KAN$ be an Iwasawa decomposition of $G$. Then $K$ is a maximal compact subgroup, $A$ is abelian of dimension $1$, and $N$ is nilpotent. Let $M$ be the centralizer of $A$ in $K$, and let $P = MAN$ be the minimal parabolic subgroup of $G$.

It is known that $G$ is locally isomorphic to one of the classical groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ or the exceptional group $\FF$ (see for instance the list on p.~426 in \cite{MR1920389}), and we now describe these groups in more detail, including explicit descriptions of the Iwasawa subgroups and the minimal parabolic subgroup $P$.

\subsection{The classical cases}

Let $\mb F$ be one of the three finite-dimensional division algebras over the reals, the real field $\R$, the complex field $\C$ or the quaternion division ring $\mb H$. In the exceptional case treated later, $\mb F$ will be the non-associative real algebra $\mb O$ of octonions, also known as the Cayley algebra. We let $\Re\;\mb F$ and $\ImF$ denote the real and imaginary part of $\mb F$, so that $\mb F = \Re\;\mb F+\ImF$, and in the standard notation
$$
\Im\;\R = 0, \quad \Im\;\C = \R i, \quad \Im\;\mb H = \R i+\R j+\R k.
$$
We use the notation $\mb F'$ to denote the unit sphere in $\mb F$,
$$
\mb F' = \{x \in\mb F\mid  \|x \| = 1\},
$$
and $\mb F^* = \mb F \setminus \{0\}$.

Let $\mb F^{p,q}$ denote the real vector space $\mb F^{p+q}$ equipped with the hermitian form
$$
\la x,y\ra = \sum_{i=1}^p  x_i\bar y_i - \sum_{i=p+1}^{p+q}  x_i\bar y_i.
$$
We also think of $\mb F^{p,q}$ as a right $\mb F$-module. Of course, $\mb F^n = \mb F^{n,0}$. We write $w^t$ for the row vector which is the transpose of a column vector $w\in\mb F^n$. Also, $w^* = \bar w^t$ and $|w|^2 = w^*w = \la w,w\ra$, when $w\in\mb F^n$.

Let $\mr U(p,q,\mb F)$ denote the unitary group of $\mb F^{p,q}$, i.e. the square matrices over $\mb F$ of size $p+q$ that preserve the hermitian form. If $\mb F$ is $\R$ or $\C$, we will be concerned with the unitaries of determinant $1$. We write $\SU(p,q,\mb F)$ for this group. It is customary to write
\begin{align*}
\mr U(p,q,\mb R) = \mr O(p,q), \quad \SU(p,q,\mb R) = \SO(p,q), \\ \SU(p,q,\mb C) = \SU(p,q), \quad \mr U(p,q,\mb H) = \Sp(p,q),
\end{align*}
The groups $\SU(p,q)$ and $\Sp(p,q)$ are connected, and $\Sp(p,q)$ is even simply connected (see \cite[Section~I.17]{MR1920389}).

We let $\mr U_0(p,q,\mb F)$ denote the connected component of $\mr U(p,q,\mb F)$. Note that
$$
\mr U_0(n,\R) = \SO(n), \quad \mr U_0(n,\C) = \mr U(n), \quad \mr U_0(n,\mb H) = \Sp(n),
$$
and in particular $\mr U_0(1,\R) = \{1\}$. We remark that $\mr U_0(n,\mb F)$ acts transitively on the unit sphere in $\mb F^n$ except for the case $n = 1$ and $\mb F = \R$.

The following is taken from \cite{MR0374341}. Let $G$ be one of $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$. Then the subgroups related to the Iwasawa decomposition of $G$ are the following.

\begin{align}
K &=
\begin{pmatrix}
	k & 0 \\
	0 & \beta
\end{pmatrix} ,
\qquad
\begin{array}{l}
k\in\mr U_0(n,\mb F), \ \
\beta\in\mr U_0(1,\mb F), \\[5pt]
\beta\det k = 1 \text{ if } \mb F \neq \mb H
\end{array}
\displaybreak[0] \\[15pt]
M &= \begin{pmatrix}
	\beta & 0 & 0 \\
	0 & u & 0 \\
	0 & 0 & \beta
\end{pmatrix} , \qquad
\begin{array}{l}
u\in\mr U_0(n-1,\mb F), \ \
\beta\in\mr U_0(1,\mb F), \\[5pt]
\beta^2\det u = 1 \text{ if } \mb F \neq \mb H
\end{array}
\displaybreak[0] \\[15pt]
\notag
A &= \begin{pmatrix}
	\cosh t & 0 & \sinh t \\
	0 & I & 0 \\
	\sinh t & 0 & \cosh t
\end{pmatrix} , \quad
t\in\R
\displaybreak[0] \\[15pt]
\label{eq:N-orig}
N &= \begin{pmatrix}
1 + z - \tfrac{1}{2}|w|^2 & w^* & -z + \tfrac{1}{2}|w|^2 \\
-w & I & w \\
z-\tfrac{1}{2}|w|^2 & w^* & 1-z+\tfrac{1}{2}|w|^2
\end{pmatrix},
\quad
w\in\mb F^{n-1},\ \
 z \in \ImF
\end{align}
The subgroups $M$ and $A$ of $P$ commute. The group $N$ is normal in $P$, and $P$ is the semi-direct product of $MA$ and $N$. To describe the action of $M$ and $A$ on $N$, it will be easier to work with a group isomorphic to $P$ (but no longer a subgroup of $G$) obtained by conjugating $P$ by the orthogonal matrix
$$
\begin{pmatrix}
\tfrac{1}{\sqrt2} & 0 & \tfrac{1}{\sqrt2} \\
0 & I & 0 \\
-\tfrac{1}{\sqrt2} & 0 & \tfrac{1}{\sqrt2}
\end{pmatrix}.
$$
Then $A$ and $N$ become, with $\alpha = e^t$,
\begin{align}
A &= \begin{pmatrix}
	\alpha & 0 & 0 \\
	0 & I & 0 \\
	0 & 0 & \alpha^{-1}
\end{pmatrix} , \quad
\alpha>0
\displaybreak[0] \\[15pt]
\label{eq:N}
N &= \begin{pmatrix}
1 & w^t & z + \tfrac{1}{2}|w|^2 \\
0 & I & \bar w \\
0 & 0 & 1
\end{pmatrix},
\quad
w\in\mb F^{n-1},\ \
 z \in \ImF
\end{align}
while $M$ remains the same. We have chosen to rescale the parameter $z$ in \eqref{eq:N} by a factor of two compared with \eqref{eq:N-orig} and replace $w$ by its conjugate $\bar w$, so that the group law in $N$ matches the one from \cite{MR0422519}. We think of the group $N$ as $\mb F^{n-1} \times\ImF$ with group structure
$$
(w_1,z_1)(w_2,z_2) = (w_1+w_2, z_1 + z_2 + \Im \la w_1,w_2\ra )
$$
and write $(w,z)$ for the matrix in \eqref{eq:N}. The action $MA\acts N$ is given by
\begin{align}\label{eq:actionA}
\alpha.(w,z) = (\alpha w, \alpha^2 z)
\end{align}
and
\begin{align}\label{eq:actionM}
\diag(\beta,u,\beta).(w,z) = (uw\beta^{-1},\beta z\beta^{-1}).
\end{align}
Note that the three subsets $\{0\}\times \ImF$, $\{0\}\times \ImF^*$ and $\mb F^{n-1}\times \{0\}$ are invariant under the action $MA$. If $\mb F = \R$, then $N$ is abelian, and otherwise the center of $N$ is
$$
Z(N) = \{(0,z) \mid z \in\ImF \}.
$$

\subsection{The exceptional case}\label{sec:FF}
The exceptional group $\FF$ has a realization as automorphisms of a Jordan algebra. A detailed treatment of the group $\FF$ can be found in \cite{MR560851} including a description of the Iwasawa decomposition $\FF = KAN$ (see \cite[\S5~Theoreme~1]{MR560851}). Here we only describe the components $M$, $A$ and $N$ of the minimal parabolic subgroup $P = MAN$ and not the group $\FF$ itself. The group $P$ is best described using the octonion non-associative division algebra $\mb O$. For a detailed description of the octonions we refer to \cite[\S1]{MR560851}. Another reference is \cite{MR1886087,MR2132837}.

We recall that $\mb O$ is an 8-dimensional real vector space, and thus we usually identify $\mb O$ with $\R^8$. We use the notation $\bar y$ for the conjugate of $y\in\mb O$, and we let $\la x,y\ra = x\bar y$. The real bilinear form $(x | y) = \Re\la x,y\ra$ corresponds to the usual inner product on $\R^8$. The imaginary octonions $\ImO$ form a subspace identified with $\R^7$.

The group $N$ is $\mb O \times \ImO$ with group product
$$
(w_1,z_1)(w_2,z_2) = (w_1+w_2, z_1 + z_2 + \Im \la w_1,w_2\ra ).
$$
The center of $N$ is $Z(N) = \{(0,z) \mid z \in\ImO \}$, and the quotient $N/Z(N)$ is then isomorphic to $(\mb O,+)$. The group $N$ is connected and nilpotent.

The group $A$ is $\R_+$, and the action $A \acts N$ is given by
$$
\alpha.(w,z) = (\alpha w, \alpha^2 z), \quad \alpha\in \R_+.
$$
The group $M$ is the spin group $\Spin(7)$, which is the (2-sheeted) universal cover of $\SO(7)$. In order to describe the action $M\acts N$, we need to consider two orthogonal representations of $\Spin(7)$, the spin representation $\sigma:M\to\SO(8)$ and the vector representation $\nu:M\to\SO(7)$. Then the action $M\acts N$ is then given as
$$
u.(w,z) = (\sigma(u)w, \nu(u)z), \quad u\in \Spin(7).
$$
If we identify $\ImO$ with $\R^7$ in the usual way, then $\SO(7)$ acts on $\ImO$ by matrix multiplication. The vector representation $\nu$ is simply the covering homomorphism $\nu : \Spin(7)\to \SO(7)$. Under the identification of $\ImO$ with $\R^7$, the purely imaginary unit octonions are identified with the unit sphere $S^6$. Since $\SO(7)$ acts transitively on $S^6$, it follows that $MA$ acts transitively on $\ImO^*$.

The spin representation $\sigma:\Spin(7) \to \SO(8)$ gives a transitive action of $\Spin(7)$ on $S^7$ (see \cite[\S4~Lemme~1]{MR560851}).

The actions of $M$ and $A$ on $N$ commute and thus give an action $M\times A\acts N$. The group $P$ is the semidirect product $P = MA \ltimes N$.

Note that the three subsets $\{0\}\times \ImO$, $\{0\}\times \ImO^*$ and $\mb O\times \{0\}$ of $N$ are invariant under the action $MA$.

\section{The irreducible representations of parabolic subgroups}\label{sec:irreducibles}

In this section we describe the unitary dual of the minimal parabolic subgroups $P$ from the previous section. The result is contained in Theorem~\ref{thm:unitary-dual-P} and Theorem~\ref{thm:dual-FF-P}. We also prove that $P$ and $N$ are of type I.

\subsection{The classical cases}

Let $G$ be one of the classical groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$, and let $P = MAN$ be the minimal parabolic subgroup of $G$. To describe the irreducible representations of $P$ we rely on the work of \cite{MR0422519}, in which groups very similar to our $P$ are considered as well as many other groups. In fact, they consider the group $\tilde MAN$, where $\tilde M$ is
\begin{align*}
\tilde M &= \begin{pmatrix}
	\beta & 0 & 0 \\
	0 & u & 0 \\
	0 & 0 & \beta
\end{pmatrix} , \quad u\in\mr U(n-1,\mb F),\ \beta\in\mr U(1,\mb F)
\end{align*}
Their conclusion about the irreducible representations is contained in \cite[Proposition~7.8]{MR0422519}. Actually, if $\mb F = \mb H$, which is the case we are most interested in because of future applications \cite{HK-WH-examples}, then $\tilde M = M$, and Theorem~\ref{thm:unitary-dual-P} is a special case of \cite[Proposition~7.8]{MR0422519}.

The discussion below is based on Section~4 and 7 from \cite{MR0422519} to which we refer for proofs and more details. The arguments carry over without any challenges to our situation. The representations of $P$ fall into three series.

(1)
The subgroup $N$ is normal in $P$, and $P/N \simeq MA$. We let $q:P\to P/N$ denote the quotient map. Of course, any irreducible representation $\sigma$ of $P/N$ gives rise to the irreducible representation $\sigma\circ q$ of $P$, and these are precisely the irreducibles of $P$ that annihilate $N$.

(2)
Next we describe the irreducibles of $P$ arising from characters on $N$. Let $v\in\mb F^{n-1}$ be non-zero, and define the character $\chi_v$ on $N$ by
$$
\chi_v(w,z) = \exp(i\Re \la w,v \ra).
$$
The group $MA$ acts on $N$ by conjugation, and this induces a dual action of $MA$ on $\hat N$.
Let $L_v$ be the stabilizer of $\chi_v$ in $MA$ under this action. Then $\chi_v$ extends to a character of $N\rtimes L_v$ by the formula
$$
\chi_v(w,z,g) = \chi_v(w,z) = \exp(i\Re \la w,v \ra), \qquad (w,z,g)\in\mb F^{n-1}\times\ImF\times L_v.
$$
Let $\gamma$ be an irreducible representation of $L_v$. Extend $\gamma$ to be the irreducible representation of $N\rtimes L_v$ defined by letting $\gamma$ be trivial on $N$. Form the tensor product representation $\chi_v \tensor \gamma$ and induce this representation from $N\rtimes L_v$ to $P$ to get a representation $\pi_{2,v,\gamma}$ of $P$,
$$
\pi_{2,v,\gamma} = \Ind_{NL_v}^P (\chi_v \tensor \gamma).
$$

Before we move on to the last series in $\hat P$, we describe the action of $MA$ on the non-trivial characters on $N$ in more detail. The action is given by \eqref{eq:actionA} and \eqref{eq:actionM},
\begin{align}\label{eq:dual-char-action-A}
(u,\beta,\alpha).\chi_v = \chi_{v'} \qquad \text{where } v' = u^{-1}\alpha^{-1} v \beta.
\end{align}
We see that unless $G = \SO_0(2,1)$, the action of $MA$ on $\mb F^{n-1}\setminus \{0\}$ is transitive, and if $G = \SO_0(2,1)$, the action of $MA$ on $\R^*$ has two orbits $\R_+$ and $\R_-$. A set of representatives for the orbits $MA\acts \mb F^{n-1} \setminus\{0\}$ is then
$$
S_2 = \{-1,1\} \quad\text{if}\quad G = \SO_0(2,1) \qquad\text{and}\qquad
S_2 = \{1\} \quad\text{if}\quad G \neq \SO_0(2,1)
$$
The stabilizer $L_v$ is $L_v = \{ (u,\beta)\in M \mid uv = v\beta \}$,
and we note that $L_v \subseteq M$.

(3)
Finally, we consider representations that do not come from characters on $N$. This happens only when $\mb F \neq \R$. Let $m\in\ImF^*$, and define $\lambda:\ImF\to\R$ by $\lambda(z) = -\Re(m\bar z)$. Then $\lambda$ is a non-trivial $\R$-linear map. It is known that there exists an infinite dimensional irreducible representation $\eta_m$ of $N$, uniquely determined be the property
$$
\eta_m(w,z) = e^{i\lambda(z)}\eta_m(w,0), \qquad (w,z)\in\mb F^{n-1}\times\ImF.
$$
Moreover, $\eta_m$ is uniquely determined within unitary equivalence by the central character $\lambda$ (see \cite[Lemma~4.4]{MR0422519}). The group $MA$ acts on the classes of representations $\eta_m$. Let $L_m$ denote the stabilizer in $MA$ of the class $[\eta_m]$,
$$
L_m = \{ g\in MA \mid g.\eta_m \simeq \eta_m \}.
$$
Then $\eta_m$ extends to a representation of $N\rtimes L_m$ as discussed in \cite[Section~7]{MR0422519}, and the extension is of course still irreducible.

Let $\gamma$ be an irreducible representation of $L_m$. Extend $\gamma$ to be the irreducible representation of $N\rtimes L_m$ defined by letting $\gamma$ be trivial on $N$. Form the tensor product representation $\eta_m \tensor \gamma$ and induce this representation to get a representation $\pi_{3,m,\gamma}$ of $P$,
$$
\pi_{3,m,\gamma} = \Ind_{NL_m}^P (\eta_m \tensor \gamma).
$$
We now describe the action of $MA$ on the infinite dimensional representations of $N$ in more detail. Since $\eta_m$ is uniquely determined within unitary equivalence by $\lambda$ (or equivalently by $m$), the action is best described by the action $MA\acts\ImF^*$ given by \eqref{eq:actionA} and \eqref{eq:actionM},
\begin{align}\label{eq:dual-inf-action-A}
(u,\beta,\alpha).\eta_m = \eta_{m'}  \qquad \text{where } m' = \beta\alpha^{-2} m \beta^{-1}.
\end{align}
If $\mb F = \C$, there are two orbits under this action, $i\R_+$ and $i\R_-$, and if $\mb F = \mb H$, there is only one orbit $\ImF^*$. As a set of representatives for the orbits we choose
$$
S_3 = \{-i,i\} \quad\text{if}\quad \mb F = \C \qquad\text{and}\qquad
S_3 = \{i\} \quad\text{if}\quad \mb F = \mb H.
$$
The stabilizer of $m \in \{-i,i\}$ is
\begin{align*}
L_m &= \{(u,\beta) \in M \mid \beta \in \R + \R i \}
\end{align*}
We note that the stabilizer $L_m \subseteq M$.

The three constructions given above exhaust the unitary dual of $P$.
\begin{thm}\label{thm:unitary-dual-P}
Let $G$ be one of the classical groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$, let $\mb F$ be the corresponding division algebra ($\R$, $\C$ or $\mb H$), and let $P = MAN$ be the minimal parabolic subgroup of $G$. The irreducible representations of $P$ fall into three series as follows.

\begin{enumerate}
	\item The series
	$$
	\pi_{1,\sigma} = \sigma\circ q,
	$$
	where $q:P\to P/N$ is the quotient map, and $\sigma\in\hat{P/N} = \hat{MA}$. The classes are
	parametrized by $\sigma\in \hat{P/N}$.
	\item The series
	$$
	\pi_{2,v,\gamma} = \Ind_{NL_v}^P (\chi_v \otimes \gamma),
	$$
	where $v\in\mb F^{n-1}$ is non-zero and $\gamma\in\hat L_v$. The classes are parametrized by $v\in S_2$ and $\gamma\in \hat L_v$.

	\item The series	(only when $\mb F \neq \R$)
	$$
	\pi_{3,m,\gamma} = \Ind_{NL_m}^P (\eta_m \otimes \gamma)
	$$
	where $m\in \ImF^*$ and	$\gamma\in\hat L_m$.
	The classes are parametrized by $m\in S_3$ and $\gamma\in\hat L_m$.

\end{enumerate}
\end{thm}

\begin{lem}\label{lem:type-I-PN}
Consider the minimal parabolic subgroup $P = MAN$ in one of the classical groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$. Then $P$ and $N$ are of type I.
\end{lem}
\begin{proof}
It is known that connected nilpotent Lie groups are of type I (see \cite[Corollaire~4]{MR0115097}), and it follows that $N$ is of type I.

Theorem~9.3 in \cite{MR0098328} provides a way of establishing that $P$ is type I. First of all, $\hat N$ is a standard Borel space, because $N$ is of type I. The action $MA\acts \hat N$ has only finitely many orbits (the exact number depends on $\mb F$ and $n$), so in particular there is a Borel set in $\hat N$ which meets each orbit exactly once. By \cite[Theorem~9.2]{MR0098328} the action $MA\acts \hat N$ is \emph{regular}, that is, $N$ is \emph{regularly embedded} in $P$.

We now verify that when $\pi\in \hat N$, the stabilizer $L_\pi = \{ g\in MA \mid g.\pi \simeq \pi\}$ is of type I. Indeed, if $\pi$ is the trivial character on $N$, then $L_\pi = MA$ which is a direct product of the compact group $M$ and the abelian group $A$. Hence the stabilizer $MA$ is of type I. If $\pi$ is not the trivial character, then $\pi = \eta_m$ or $\pi = \chi_v$, where $m\in\mb F^*$ or $v\in\ImF^*$, and we already saw that $L_m$ and $L_v$ are closed subgroups of $M$ and hence compact. In particular the stabilizers are of type I.

According to \cite[Theorem~9.3]{MR0098328} we may now conclude that $P$ is of type I.
\end{proof}

\subsection{The exceptional case}
Let $P = MAN$ be the minimal parabolic subgroup of $\FF$. We will now describe the irreducible representations of $P$. Again, this is based on \cite{MR0422519}. They consider the group $\tilde MAN$, where $\tilde M = \Spin(7) \times \{\pm 1\}$. The complete description of the unitary dual of $\tilde MAN$ can be found in (8.12) and (8.15) in \cite{MR0422519}. The discussion below is based on Section~8 in \cite{MR0422519} to which we refer for proofs and more details. The representations fall into two series.

(1) Irreducible representations of $N$ that annihilate the center $Z(N) = \ImO$ are characters of the form $\chi_v$ for some $v\in\mb O$, where $\chi_v$ is given by
$$
\chi_v(w,z) = e^{i\Re \la w,v \ra} = e^{ i (w|v) }
$$
The group $MA$ acts on $N$, and this induces a dual action of $MA$ on $\hat N$.

Let $L_v$ be the stabilizer of $\chi_v$ in $MA$. Then $\chi_v$ extends to a character of $N\rtimes L_v$ by the formula
$$
\chi_v(w,z,g) = \chi_v(w,z) = e^{ i (w|v) }, \qquad (w,z,g)\in\mb O\times\ImO\times L_v.
$$
Let $\gamma$ be an irreducible representation of $L_v$. Extend $\gamma$ to be the irreducible representation of $N\rtimes L_v$ defined by letting $\gamma$ be trivial on $N$. Form the tensor product representation $\chi_v \tensor \gamma$ and induce this representation from $N\rtimes L_v$ to $P$ to get a representation $\pi_{1,v,\gamma}$ of $P$,
$$
\pi_{1,v,\gamma} = \Ind_{NL_v}^P (\chi_v \tensor \gamma).
$$
This representation $\pi_{1,v,\gamma}$ is a representation in the first series.

From the definition of the action $MA\acts N$ we see that
$$
(u,\alpha).\chi_v(w,z) = e^{ i ((\alpha\sigma(u))^{-1}w|v) } = e^{ i (w|\alpha\sigma(u)v) } = \chi_{\alpha\sigma(u)v}(w,z)
$$
Since $M$ acts transitively on $S^7 \subseteq\mb O$, we see that $MA$ acts transitively $\mb O^*$ and thus on the characters $\{\chi_v\}_{v\in\mb O^*}$.

If $v = 0$, the stabilizer $L_v$ is of course all of $MA$. Otherwise the stabilizer $L_v$ is
$$
L_v = \{ (u,\alpha) \in MA \mid \sigma(u)\alpha v = v\}.
$$
Since $\sigma(u)$ preserves the norm of elements in $\mb O$, we see that if $(u,\alpha) \in L_v$, then $\alpha = 1$. Hence $L_v \subseteq M$.

(2) Let $m\in\ImO^*$ be non-zero, and define $\lambda:\ImO\to\R$ by $\lambda(z) = -\Re(m\bar z)$. Then $\lambda$ is a non-trivial $\R$-linear map which is uniquely determined by $m$. Irreducible representations of $N$ that do not annihilate the center are infinite dimensional and of the form $\eta_m$ for some $m\in \ImO^*$, where $\eta_m$ is uniquely determined by the property
$$
\eta_m(w,z) = e^{i\lambda(z)} \eta_m(w,0), \quad (w,z)\in N.
$$
Moreover, the equivalence class of $\eta_m$ is uniquely determined by the central character $\lambda$ and hence by $m$. Since the action of $MA$ on $\ImO^*$ is transitive, $MA$ acts transitively on the set $\{\eta_m\}_{m\in\ImO^*}$.

Let $L_m$ denote the stabilizer in $MA$ of the class of $\eta_m$. Then
$$
L_m = \{ u \in M\mid \nu(u)m=m \}.
$$
It follows from \cite[Lemma~8.14]{MR0422519} that $\eta_m$ extends to a representation of $N\rtimes L_m$. Let $\gamma$ be an irreducible representation of $L_m$, and extend $\gamma$ to $N\rtimes L_m$ by letting $\gamma$ be trivial on $N$. Form the tensor product representation $\eta_m \tensor \gamma$ and induce this representation to get a representation $\pi_{2,m,\gamma}$ of $P$,
$$
\pi_{2,m,\gamma} = \Ind_{NL_m}^P (\eta_m \tensor \gamma).
$$

\begin{thm}\label{thm:dual-FF-P}
Let $P = MAN$ be the minimal parabolic subgroup of $\FF$ and let $\pi$ be an irreducible representation of $P$. Then $\pi$ is unitarily equivalent to one of the following.
\begin{enumerate}
	\item $\pi_{1,v,\gamma} = \Ind_{NL_v}^P (\chi_v \otimes \gamma)$ for some $v\in\mb O$ and $\gamma\in\hat L_v$.
	\item $\pi_{2,m,\gamma} = \Ind_{NL_m}^P (\eta_m \otimes \gamma)$ for some $m\in \ImO^*$ and	$\gamma\in\hat L_m$.
\end{enumerate}
\end{thm}

\begin{lem}\label{lem:FF-P-type-I}
Consider the minimal parabolic subgroup $P = MAN$ of $\FF$. Then $P$ and $N$ are of type I.
\end{lem}
\begin{proof}
Since $N$ is a connected nilpotent Lie group, $N$ is of type I (see \cite[Corollaire~4]{MR0115097}).

Theorem~9.3 in \cite{MR0098328} provides a way of establishing that $P$ is type I. First of all, $\hat N$ is a standard Borel space, because $N$ is of type I. The action $MA\acts \hat N$ has only three orbits,
$$
\mc O_1 = \{\chi_0\}, \quad
\mc O_2 = \{\chi_v\}_{v\in \mb O^*}, \quad
\mc O_3 = \{\eta_m\}_{m\in\ImO^*},
$$
where $\chi_0$ is the trivial representation. Then, clearly, there is a Borel set in $\hat N$ which meets each orbit exactly once. By \cite[Theorem~9.2]{MR0098328} the action $MA\acts \hat N$ is regular.

We now verify that when $\pi\in \hat N$, the stabilizer $L_\pi = \{ g\in MA \mid g.\pi \simeq \pi\}$ is of type I. Indeed, if $\pi = \eta_m$ or $\pi = \chi_v$, where $m\in\mb O^*$ or $v\in\ImO^*$, then we already saw that $L_m$ and $L_v$ are compact and in particular of type I. If $\pi = \chi_0$, then the stabilizer is $MA$ which is a direct product of the compact group $M$ and the abelian group $A$. Hence the stabilizer $MA$ is of type I.

According to \cite[Theorem~9.3]{MR0098328} we may now conclude that $P$ is of type I.
\end{proof}

\section{The Fourier algebra of \texorpdfstring{$P$}{P}}\label{sec:fourier-P}
In this section we verify the last condition in Theorem~\ref{thm:strategy} for the minimal parabolic subgroups $P$. The result is contained in Proposition~\ref{prop:strategy-P}.

Recall that part of the Peter-Weyl Theorem asserts that the left regular representation of a compact group is completely reducible, and every irreducible representation of the compact group occurs as a direct summand (see \cite[Theorem~5.12]{Folland}).

We now set out to determine which irreducible representations of $P$ that occur as subrepresentations of the left regular representation. For this we will rely on Corollary~11.1 in \cite{MR0342641}. In order to apply the corollary we first need to verify the assumptions I-IV from \cite{MR0342641}. For this it will suffice to observe that $N$ and $P$ are of type I (see Lemma~\ref{lem:type-I-PN} and Lemma~\ref{lem:FF-P-type-I}), and all stabilizers $L_v$ and $L_m$ are closed and contained in $M$, so they are compact and in particular of type I.

In the case of the representation $\pi_{1,v,\gamma}$, Corollary~11.1 in \cite{MR0342641} then applies to show that $\pi_{1,v,\gamma}$ is a subrepresentation of the left regular representation of $P$ if and only if $\gamma$ is a subrepresentation of the left regular representation of the stabilizer group $L_v$ and the orbit of $\chi_v$ inside $\hat N$ has positive Plancherel measure. Similar conclusions hold in the other cases.

\subsection{The classical cases}
We first consider the case where $\mb F = \R$.
\begin{lem}\label{lem:split-SO}
Let $G$ be the group $\SO_0(n,1)$, and let $P = MAN$ be the minimal parabolic subgroup of $G$. Any irreducible unitary representation $\pi$ of $P$ is either trivial on the non-compact subgroup $N$ or is a subrepresentation of $\lambda_P$.
\end{lem}
\begin{proof}
We divide the proof into the cases according to the description in Theorem~\ref{thm:unitary-dual-P}.

In the case $\pi = \pi_{1,\sigma}$, it is clear that $\pi(g) = 1$ for every $g\in N$.

Consider now a representation $\pi = \pi_{2,v,\gamma}$ where $v$ is non-zero. Since $L_v$ is compact, $\gamma\in\hat L_v$ is a subrepresentation of the regular representation of $L_v$. If $n\neq 2$, then the action of $MA$ on the non-zero characters of $N$ is transitive. In particular, the orbit has positive Plancherel measure in $\hat N$. If $n = 2$, then the orbit of $\chi_v$ is either $\R_+$ or $\R_-$ inside $\hat N \simeq \R$, and both of these sets have positive measure. By Corollary~11.1 in \cite{MR0342641} we conclude that $\pi$ is a subrepresentation of $\lambda_P$.

The case $\pi = \pi_{3,m,\gamma}$ does not occur, when $\mb F = \R$.
\end{proof}

From Proposition~\ref{prop:completely-reducible} we can now conclude that the left regular representation of $P$ is completely reducible. From the proof of Lemma~\ref{lem:split-SO} we then obtain the following.

\begin{cor}
Let $G$ be the group $\SO_0(n,1)$, and let $P = MAN$ be the minimal parabolic subgroup of $G$. The left regular representation of $P$ is completely reducible with the representations $\pi_{2,v,\gamma}$ as its subrepresentations. Here $v\in S_2$ and $\gamma\in \hat L_v$.
\end{cor}

When $\mb F$ equals $\C$ or $\mb H$ we have the following.

\begin{lem}\label{lem:split-SU}
Let $G$ be one of the groups $\SU(n,1)$, $\Sp(n,1)$, and let $P = MAN$ be the minimal parabolic subgroup of $G$. Any irreducible unitary representation $\pi$ of $P$ is either trivial on the non-compact subgroup $Z(N)$ or is a subrepresentation of $\lambda_P$.
\end{lem}
\begin{proof}
We divide the proof into the cases according to the description in Theorem~\ref{thm:unitary-dual-P}.

In the case $\pi = \pi_{1,\sigma}$, it is clear that $\pi(g) = 1$ for every $g\in N$, and hence in particular for every $g\in Z(N)$.

Suppose now $\pi = \pi_{2,v,\gamma}$. Since $\chi_v$ is trivial on $Z(N)$ and $\gamma$ is trivial on $N$, it follows that $\chi_v\tensor\gamma$ is trivial on $Z(N)$. Since $Z(N) \triangleleft P$, it now follows from Lemma~\ref{lem:trivial-induction} that $\pi$ is trivial on $Z(N)$, once we show that the homogeneous space $P / NL_v$ admits a $P$-invariant measure. Using Proposition~\ref{prop:cosetspaces} we find
$$
P / NL_v \simeq MA / L_v \simeq M/L_v \times A.
$$
The left translation action $A\acts A$ has the Haar measure as an invariant measure. Since $M$ is compact, the action $M\acts M/L_v$ has an invariant measure. It follows that $P\acts NL_v$ has an invariant measure, and then by Lemma~\ref{lem:trivial-induction} the representation $\pi$ is trivial on $Z(N)$.

Consider now a representation $\pi = \pi_{3,m,\gamma}$. Since $L_m$ is compact, $\gamma\in\hat L_m$ is a subrepresentation of the regular representation of $L_m$. It remains to show that the orbit of $\eta_m$ in $\hat N$ has positive Plancherel measure.

If $\mb F = \mb H$, then the third series of Theorem~\ref{thm:unitary-dual-P} forms a single orbit, which must then have positive Plancherel measure, because all other irreducible representations of $N$ are trivial on $Z(N)$ and hence must form a null set for the Plancherel measure.

If $\mb F = \C$, then the action of $MA$ on the representations $\{\eta_m \in \hat N \mid m\in \ImF^*\}$ has two orbits, so the simple argument for $\mb H$ does not apply. Luckily, the Plancherel measure of $N$ is well-known. In fact, $N$ is the Heisenberg group of dimension $2n -1$, and the Plancherel measure for the Heisenberg group can be found on p.~241 in \cite{Folland}. We see that the measure of the orbit of $\eta_i$ is
$$
\mu_N(P.\eta_i) = \int_0^\infty |m|^{n-1} \; dm.
$$
Hence the orbit of $\eta_i$ has positive, in fact infinite, measure. Similarly, the orbit of $\eta_{-i}$ has positive measure. By Corollary~11.1 in \cite{MR0342641} we conclude that $\pi$ is a subrepresentation of $\lambda_P$.
\end{proof}

From Proposition~\ref{prop:completely-reducible} we can now conclude that the left regular representation of $P$ is completely reducible. From the proof of Lemma~\ref{lem:split-SU} we then obtain the following.

\begin{cor}
Let $G$ be one of the classical groups $\SU(n,1)$, $\Sp(n,1)$, and let $P = MAN$ be the minimal parabolic subgroup of $G$. The left regular representation of $P$ is completely reducible with the representations $\pi_{3,m,\gamma}$ as its subrepresentations. Here $m\in S_3$ and $\gamma\in \hat L_m$.
\end{cor}

\subsection{The exceptional case}

\begin{lem}\label{lem:split-FF}
Let $P = MAN$ be the minimal parabolic subgroup of $\FF$. Any irreducible unitary representation $\pi$ of $P$ is either trivial on the non-compact subgroup $Z(N) \simeq \Im\mb O$ or is a subrepresentation of $\lambda_P$.
\end{lem}
\begin{proof}
Recall that any irreducible representation of $P$ is given as in Theorem~\ref{thm:dual-FF-P}. We will show that representations $\pi_{1,v,\gamma}$ are trivial on $Z(N)$ and that representations $\pi_{2,m,\gamma}$ are subrepresentations of $\lambda_P$.

Suppose first $\pi = \pi_{1,v,\gamma}$. The proof from Lemma~\ref{lem:split-SU} carries over verbatim and shows that $\pi$ is trivial on $Z(N)$.

Consider now a representation $\pi = \pi_{2,m,\gamma}$. Since $L_m$ is compact, $\gamma\in\hat L_m$ is a subrepresentation of the regular representation of $L_m$. It remains to show that the orbit of $\eta_m$ in $\hat N$ has positive Plancherel measure.

Clearly, the characters $\{\chi_v\}_{v\in\mb O}$ form a null set for the Plancherel measure on $\hat N$, because they are all trivial on the non-compact subgroup $Z(N)$. As mentioned in the proof of Lemma~\ref{lem:FF-P-type-I}, the complement of $\{\chi_v\}_{v\in\mb O}$ forms a single orbit in $\hat N$ under the action of $MA$. Thus, the orbit of $\eta_m$ must have positive Plancherel measure. By Corollary~11.1 in \cite{MR0342641} we conclude that $\pi$ is a subrepresentation of $\lambda_P$.
\end{proof}

\subsection{Conclusion}
The following proposition sums up the necessary results from this and last section so that we may apply Theorem~\ref{thm:strategy}.

\begin{prop}\label{prop:strategy-P}
Let $n\geq 2$, and let $G$ be one of the simple Lie groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ or $\FF$. Let $P = MAN$ be the minimal parabolic subgroup in $G$. The following holds.
\begin{enumerate}
	\item $P$ is type I.
	\item There is a non-compact, closed subgroup $H$ of $P$ such that every irreducible unitary representation of $P$ is either trivial on $H$ or is a subrepresentation of the regular representation $\lambda_P$.
\end{enumerate}
In fact, if $G = \SO_0(n,1)$, then one can take $H = N$, and otherwise one can take $H = Z(N)$.
\end{prop}

From Proposition~\ref{prop:strategy-P} and Theorem~\ref{thm:strategy} we immediately obtain Theorem~\ref{thm:fourier-P}.

\section{The relative Howe-Moore property}\label{sec:HM}

In this section we prove Corollary~\ref{cor:HM} concerning the relative Howe-Moore property.
We recall from \cite{MR2854688} that if $H$ is a closed subgroup of a locally compact group $G$, then the pair $(G,H)$ has the relative Howe-Moore property, if every representation $\pi$ of $G$ either has $H$-invariant vectors, or the restriction $\pi|_H$ is a $C_0$-representation, i.e. all coefficients of $\pi|_H$ vanish at infinity. Using a direct integral argument, it is proved in \cite[Proposition~2.3]{MR2854688} that it is sufficient to consider only irreducible representations of $G$.

From the results in the previous sections we easily obtain the following, which obviously implies Corollary~\ref{cor:HM}.

\begin{cor}\label{cor:HM2}
If $P$ is the group in \eqref{eq:PP} and $N_1$ is the group in \eqref{eq:N1}, then $N_1$ is a normal, non-compact closed subgroup of $P$, and $(P,N_1)$ has the relative Howe-Moore property.

Let $n\geq 2$. If $P = MAN$ the minimal parabolic subgroup in the simple Lie group $\SO_0(n,1)$, then $N$ is a normal, non-compact closed subgroup of $P$, and $(P,N)$ has the relative Howe-Moore property.

If $P = MAN$ the minimal parabolic subgroup in one of the simple Lie groups $\SU(n,1)$, $\Sp(n,1)$ or $\FF$, then $Z(N)$ is a normal, non-compact closed subgroup of $P$, and $(P,Z(N))$ has the relative Howe-Moore property.
\end{cor}
\begin{proof}
Apply Proposition~\ref{prop:PBA} or Proposition~\ref{prop:strategy-P}, respectively. Since any subrepresentation of the left regular representation $\lambda_P$ is a $C_0$-representation, we immediately obtain the result.
\end{proof}

\section{Concluding remarks}\label{sec:concluding}
Theorem~\ref{thm:fourier-P} shows that Question~\ref{q1} has a positive answer for the minimal parabolic subgroups $P = MAN$ in the groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ and $\FF$. One could ask if the same is true for the smaller groups $MN$, $AN$ or $N$. We will now discuss these cases. Recall from the introduction that a non-compact second countable connected unimodular groups never satisfy \eqref{eq:q1}.

Let $G$ be one of the groups classical groups $\SO_0(n,1)$, $\SU(n,1)$, $\Sp(n,1)$ with $n\geq 2$ or the exceptional group $\FF$. Let $\mb F$ be the corresponding division algebra, $\R$, $\C$, $\mb H$ or $\mb O$. We start by discussing the groups $N$. Since $N$ is nilpotent, $N$ is unimodular. Indeed, a locally compact group $G$ is unimodular if and only if $G/Z$ is unimodular, where $Z$ is the center of $G$ (see \cite[p.~92]{MR0175995}). Induction on the length of an upper central series then shows that all locally compact nilpotent groups are unimodular. Since $N$ is also connected and second countable, it follows that
$$
A(N) \neq B(N) \cap C_0(N).
$$

Next we discuss the groups $MN$. Since $MN$ is a semi-direct product of the unimodular group $N$ by the compact group $M$, we will argue that $MN$ itself is unimodular. Indeed, this follows directly from \cite[Proposition~23]{MR0175995} but we also include another argument here. If we use $\Delta_G$ to denote the modular function of a locally compact group $G$, then since $N$ is normal in $MN$, we see that the quotient space $MN/N$ has an invariant measure, Haar measure on $M$, and using \cite[Theorem~2.49]{Folland} we see that $\Delta_{MN}|N = \Delta_N = 1$. Also, since $M$ is compact, $\Delta_{MN}|M = 1$ by \cite[Proposition~2.27]{Folland}. Since $M$ and $N$ generate $MN$, it follows that $\Delta_{MN} = 1$. So $MN$ is connected and unimodular, and hence
$$
A(MN) \neq B(MN) \cap C_0(MN).
$$

Alternatively, one could show that all orbits in $\hat N$ under the action of $M$ have zero Plancherel measure. This type of argument will be used below for the groups $AN$.

For the groups $\SO_0(n,1)$, $\Sp(n,1)$ and $\FF$ it will usually also be the case that Question~\ref{q1} has a negative answer for the groups $AN$ as well. However, there is one exception. If $G = \SO_0(2,1)$, then $M$ is trivial and $P$ coincides with $AN$. Hence it follows from Theorem~\ref{thm:fourier-P} that Question~\ref{q1} has an affirmative answer for the group $AN$. In this special case let us remark that in fact $AN$ is isomorphic to the $ax + b$ group, and the result that $A(AN) = B(AN)\cap C_0(AN)$ is actually the original result of Khalil from \cite{MR0350330}.

The unimodularity argument used for the groups $N$ and $MN$ cannot be replicated for $AN$, since these groups are not unimodular (see \cite[(1.14)]{MR0507242}). As mentioned in the introduction, a group satisfying \eqref{eq:q1} has a completely reducible left regular representation, and in particular the left regular representation has irreducible subrepresentations. Then by \cite[Corollary~11.1]{MR0342641} at least one of the orbits of the action $A\acts\hat N$ must have positive Plancherel measure. To show that $A(AN) \neq B(AN) \cap C_0(AN)$ it therefore suffices to show that any orbit of $A\acts\hat N$ has zero Plancherel measure.

It this point we split the argument in cases. Consider first the case when $\mb F = \R$ and $n\geq 3$. Then $N \simeq \R^{n-1}$, and the Plancherel measure on $\hat N \simeq \R^{n-1}$ is the Lebesgue measure. Since $A$ acts on $\hat N$ by dilation, every orbit except $\{0\}$ is a a half-line. Since $n\geq 3$ every half-line in $\R^{n-1}$ has vanishing Lebesgue measure, and hence every orbit in $\hat N$ has vanishing Plancherel measure.

In the other cases $\mb F$ is $\C$, $\mb H$ or $\mb O$. For convenience, when $\mb F = \mb O$, we set $n = 2$. As mentioned earlier, the dual $\hat N$ then consists of the characters $\{\chi_v\}_{v\in\mb F^{n-1}}$ and the infinite dimensional representations $\hat N_r = \{\eta_m\}_{m\in\ImF^*}$.

Fortunately, the Plancherel measure for $N$ is known. It is described in \cite[Section~3]{MR996553}. Since the characters are trivial on the center $\ImF$ of $N$ which is non-compact, the characters form a null set for the Plancherel measure. Let $k$ be the dimension of $\ImF$ as a real vector space so that $k$ is either 1, 3 or 7. If we identify $\hat N_r$ with $\ImF^*$ which in turn is identified with the punctured Euclidean space $\R^k\setminus\{0\}$, then it follows from \cite[p.~524]{MR996553} that the Plancherel measure on $\hat N_r$ is absolutely continuous (has density) with respect to the Lebesgue measure.

Since $A$ acts on $\hat N_r$ by dilation, every orbit in $\hat N_r$ is a a half-line. Every half-line has vanishing Lebesgue measure, unless $k=1$, and hence every orbit in $\hat N_r$ has vanishing Plancherel measure, except when $\mb F = \C$. Combined with the fact that the characters have vanishing Plancherel measure, we conclude that every orbit in $\hat N$ has vanishing Plancherel measure. As pointed out, the argument breaks down when $\mb F = \C$.

We collect the discussion above in the following proposition.

\begin{prop}
Let $G$ be one of the simple Lie groups $\SO_0(n,1)$ ($n\geq 3$), $\Sp(n,1)$ ($n\geq 2$) or $\FF$. Let $G = KAN$ be the Iwasawa decomposition of $G$. Then if $H$ is either $N$, $MN$ or $AN$, then
$$
A(H) \neq B(H)\cap C_0(H).
$$
\end{prop}

Finally, we consider the group $AN$ in $G = \SU(n,1)$.

\begin{prop}
Let $G$ be the simple Lie group $\SU(n,1)$ ($n\geq 2$) with Iwasawa decomposition $G = KAN$. Then
$$
A(AN) = B(AN)\cap C_0(AN).
$$
\end{prop}
\begin{proof}
We will verify the conditions of Theorem~\ref{thm:strategy} for the group $AN$.

First we verify that $AN$ is a group of type I. We mimic the proof of Lemma~\ref{lem:type-I-PN}. Recall that $N$ is of type I, and hence $\hat N$ is a standard Borel space. Using the notation from Section~\ref{sec:irreducibles}, we identify $\hat N$ with the union of the characters $\{\chi_v\}_{v\in\C^{n-1}}$ and the infinite dimensional representations $\{\eta_m\}_{m\in i\R^*}$. The action $A\acts\hat N$ is described by \eqref{eq:dual-char-action-A} and \eqref{eq:dual-inf-action-A}, and it is easy to read off the orbits of the action.

The characters in $\hat N$, which we think of simply as $\C^{n-1}$, form an invariant subset whose orbits consist of the origin $\{0\}$ and half-lines originating at the origin. The infinite dimensional representations in $\hat N$, which we think of simply as $i\R^*$ also form an invariant subset which has two orbits, $i\R_+$ and $i\R_-$.

If $S$ denotes the unit sphere in $\C^{n-1} \simeq \R^{2n-2}$, then $R = \{0\}\cup S \cup \{i,-i\}$ is a set of representatives for the orbits of $A\acts \hat N$. We claim that $R$ is a Borel subset of $\hat N$. To see this, it suffices to prove that $S$ is a Borel subset, since points are always Borel subsets in a standard Borel space.

The group $N$ is the Heisenberg group of dimension $2n-1$, and the Fell topology on $\hat N$ is well-known (see e.g. \cite[Chapter~7]{Folland}). The characters $\{\chi_v\}_{v\in\C^{n-1}}$ form a closed subset in $\hat N$, and on the set of characters the Fell topology coincides with the Euclidean topology (on $\C^{n-1}$). In particular $S$ is closed in the Fell topology. By \cite[Theorem~7.6]{Folland}, the Mackey Borel structure on $\hat N$ is induced by the Fell topology, since $N$ is of type I. It follows that $S$ is a Borel set.

We may now conclude from \cite[Theorem~9.2]{MR0098328} that the action $A\acts \hat N$ is \emph{regular}, that is, $N$ is \emph{regularly embedded} in $AN$.

Next we verify that if $\pi\in \hat N$, then the stabilizer $L_\pi = \{ \alpha\in A \mid \alpha.\pi \simeq \pi\}$ is of type I. Indeed, if $\pi$ is the trivial character on $N$, then $L_\pi = A$ which is abelian group. Hence the stabilizer $A$ is of type I. If $\pi$ is not the trivial character, then the stabilizer $L_\pi$ is trivial. So all stabilizers are of type I. According to \cite[Theorem~9.3]{MR0098328} we may now conclude that $AN$ is of type I.

The unitary dual of $AN$ is described in \cite[Proposition~7.6]{MR0422519}. The irreducible representations of $AN$ fall into three series as follows (retaining earlier notation).

\begin{enumerate}
	\item The series
	$$
	\pi_{1,\sigma} = \sigma\circ q,
	$$
	where $q:AN\to A$ is the quotient map, and $\sigma\in\hat{A}$.
	\item The series
	$$
	\pi_{2,v} = \Ind_N^{AN} (\chi_v),
	$$
	where $v\in\C^{n-1}$ is non-zero. The classes are parametrized by the orbits of $A\acts\C^{n-1} \setminus\{0\}$.

	\item The series
	$$
	\pi_{3,m} = \Ind_N^{AN} (\eta_m)
	$$
	where $m\in i\R^*$.
	The classes are parametrized by $m\in \{i,-i\}$.
\end{enumerate}

We claim that $\pi_{1,\sigma}$ and $\pi_{2,v}$ are trivial on the center $Z(N)$ of $N$, and that $\pi_{3,m}$ is a subrepresentation of the regular representation.

Clearly, $\pi_{1,\sigma}$ annihilates $N$ and in particular $Z(N)$. Consider now a representation $\pi_{2,v} = \Ind_N^{AN} (\chi_v)$, where $v\in\C^{n-1}$ is non-zero. The character $\chi_v \in \hat N$ is trivial on $Z(N)$. Both $A$ and $N$ normalize $Z(N)$, so $Z(N)$ is normal in $AN$. The representation $\pi_{2,v}$ is induced from $N$ to $AN$, and when the quotient space $AN / N$ is identified with $A$ in the natural way, it is obvious that $AN / N$ carries an invariant measure for the $AN$-action, namely the Haar measure on $A$. From Lemma~\ref{lem:trivial-induction} it now follows that $\pi_{2,v}$ is trivial on $Z(N)$.

Finally, consider a representation $\pi_{3,m} = \Ind_N^{AN} (\eta_m)$ where $m\in i\R^*$. We will show that the orbit of $\eta_m$ in $\hat N$ has positive Plancherel measure, and then it follows from \cite[Corollary~11.1]{MR0342641} that $\Ind_N^{AN} (\eta_m)$ is a subrepresentation of the left regular representation of $AN$.

As mentioned before, the action of $A$ on the representations $\{\eta_m \in \hat N \mid m\in i\R^*\}$ has two orbits, $i\R_+$ and $i\R_-$. The Plancherel measure of $N$ is known and can be found on p.~241 in \cite{Folland}. We see that the measure of the orbit of $\eta_i$ is
$$
\mu_N(A.\eta_i) = \int_0^\infty |m|^{n-1} \; dm.
$$
Hence the orbit of $\eta_i$ has positive, in fact infinite, measure. Similarly, the orbit of $\eta_{-i}$ has positive measure. By \cite[Corollary~11.1]{MR0342641} we conclude that $\pi_{3,m}$ is a subrepresentation of the left regular representation of $AN$.

The conditions of Theorem~\ref{thm:strategy} have now been verified for the group $AN$, and our proof is complete.
\end{proof}

redpajama_set_name: RedPajamaArXiv


The DSO annual picnic will be held March 18, 2017, at the Mud Hole Road campground. Everyone is invited to join in the fun. There will be FREE food, we will be grilling hotdogs & hamburgers and lunch will be served around 11:30 am (central). Bring your lawn chairs and join us.


text:
Victoria Azarenka (born in 1989 in Minsk, Belarus - zodiac sign Leo) began playing tennis at age 7, and played her first professional match in 2003.
She landed on the WTA top 100 in 2006 and nabbed her first Grand Slam title in 2012.
At the 2012 Olympics in London, Azarenka won the gold medal in mixed doubles with Max Mirnyi, and the bronze medal in women's singles.
In 2012, Azarenka made it to the final round of the US Open, but lost American Serena Williams.
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@interface ZJSliderTransitionDelegateObj : NSObject<UIViewControllerAnimatedTransitioning>

@end

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You worked hard to purchase your home. We'll work hard to make sure your home is fully protected.
Purchasing a home - especially for the first timers - can be a confusing time. It all starts with finding a home you love and can afford, then finding a lender. The last step before moving into your new home is to obtain a Pike County, PA home insurance policy, and not just any policy. There are many different types of home insurance. We'll help you make the right decision.
Home insurance is all about protecting your family financially in the event of a disaster. There are different types of disasters such as flooding, house fires, etc. It's essential to get a home insurance plan that protects against all types of disasters.
For most of us, our home is our most valuable asset. Therefore, you must do everything in your power to protect that asset. Don't take any risks. Get insured properly right away. It's that important.
How Long Since Your Policy Went Into Effect?
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Conrad Insurance doesn't just insure family homes. We also help our clients get affordable insurance for vacation rental properties, secondary homes, etc. Regardless of how many homes you own - even the ones you rent out - we will get you the right homeowners insurance.
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This beef and broccoli meal prep is an easy meal prep option that's made in one pan! It has tons of flavor and is a satiating lunch or dinner option to keep you nourished and happy. Seriously, you won't be jealous of any takeout that's brought into the office… you'll have something much tastier and healthier!
Don't like meal prepping? This recipe is still just as good as an easy one pan meal for a fast weeknight dinner! It's a recent favorite for my husband and me.
I love using flank steak for meals like this. It is nice and tender and easy to work with. Skirt steak will likely work as well, but I recommend flank. Alternatively, a lot of grocery stores sell pre-cut stir fry beef at the butchers counter which is a nice short cut that works well for this dish. And I seriously love a short cut!
You'll use avocado oil for cooking the whole recipe, and the avocado, coconut aminos, and lime together for marinating the steak. It seals in a lot of flavors and gives it a light hint of acid from the lime, and soy-sauce flavor from the aminos. You can also add in some coconut sugar if desired, but leave this out if you're on a whole30.
Grab a big head of broccoli (or some smaller ones) and get lots of extra greens in this dish.
Of course, you can use regular rice if you tolerate it, but cauliflower rice keeps this paleo and whole30. You can buy cauliflower pre-riced, or use a food processor to rice it.


text:
Tamil Nadu Opposes Indian Neutrino Observatory in Tiger Corridor
Apr 17, 2022 by Monika Thakur in Animals
Tamil Nadu says big "No" to the Neutrino Observatory project to be setup in the Indian subcontinent
Last week, the Tamil Nadu government strongly urged the Centre to hold the proposal for setting up the Indian Neutrino Observatory (INO) in the state's Theni Valley or Cumbum Valley. MK Stalin, Chief Minister of Tamil Nadu, wrote a letter to Prime Minister, Narendra Modi, requesting to stop this project in the tiger corridor.
Over a decade ago, India started its journey to form a world-class scientific project called INO in Tamil Nadu. It is a particle physics research project. But it has again landed in controversy with some serious allegations. Some of the grave concerns are that the project planning is very unscientific.
Image: The Hindu
Moreover, it doesn't adhere to the general environmental safeguards. So, the Tamil Nadu government is now against the project because it could threaten the ecologically sensitive Western Ghats. Most importantly, they are worried that it could even impact the survival and natural habitat of the Indian National Animal – the Tiger.
Why government permitted the Indian Neutrino Observatory?
According to the Indian government, this project involves the formation of a proper two-kilometre tunnel. This tunnel will cut through the solid rock formation at the depth of around one kilometre from the hill's peak.
The advantage of having an underground tunnel is that it won't disturb the ecosystem or wildlife. And, with the development of the INO project, there's a worldwide interest to explore various allied and diverse fields, like cosmology, physics, and the universe's origin.
However, the TN government worries about the process of tunnel formation. That is because it requires a huge amount of dangerous explosives, which is not favourable for plants and animals.
Possible Threat To Tigers!
Reportedly, the site selected for the Neutrino project lies in the Mathikettan. This site has been mapped as the Periyar Tiger Corridor according to the National Tiger Conservation Authority (NCTA). According to the TN chief minister, the tiger corridor has a critical purpose to maintain proper genetic flow. But this will be impacted because of the project activities.
In 2017, the Tamil Nadu SEIAA opposed this project because the tunneling work involves the use of composite and hard rock mass that needs several explosives to break it. Even if the tunnel will be 1 km below the ground, the noise and vibrations caused by huge explosives can disturb the global diversity of natural plant and animal species may suffer.
Furthermore, the fragile ecology of the tiger habitat is likely to feel an adverse impact. It is because tigers' biological behavior makes them avoid the human presence and various other anthropogenic disturbances. So, even a minute of vibration in the land can completely impact their movement pattern. Besides that, the area even forms a significant water catchment and watershed of River Kottakudi and River Sambal. There are some small streams on Bodi Hill that join Periyar before finally draining into Vaigai Dam. Therefore, the government and wildlife conservation authorities are opposing the project citing the irreversible damage that it may cause.
As per the INO scientists, this ambitious project is most likely to give India a major lead worldwide in the neutrino science field. But this project is on hold for over 11 years now. And, the reasons for the stay are quite genuine as well.
So, it is up to the central government and the state government to come up with an ethical plan. The final INO project plan must not disturb the ecological diversity in the region.
Via: Mongabay
Tagged with: Biodiversity, Tiger Corridor, wildlife, Wildlife conservation
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text:
Jatwieź Mała – wieś w Polsce położona w województwie podlaskim, w powiecie sokólskim, w gminie Suchowola.

W latach 1975–1998 miejscowość administracyjnie należała do województwa białostockiego.

Wierni kościoła rzymskokatolickiego należą do parafii Matki Bożej Pocieszenia w Jatwiezi Dużej.

Przypisy

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What is the Soft Drinks Industry Levy (sugar tax)?
Why is it so important, and what does Jamie hope will happen now it's been introduced? Here, Jenny Rosborough, Jamie's Head of Nutrition, explains the impact of this momentous new levy.
The Soft Drinks Industry Levy officially comes into effect today – 6 April 2018 – and we've already seen signs of its success. Since the levy was announced in 2016, it has driven reformulation of products at an unprecedented rate (when brands reduce the level of sugar, fat and salt in their products) and it will also channel new money into child health. Win win!
What changes are we seeing?
Soft drink companies who make products containing more than 5% sugar will have to pay between 18p and 24p for every litre they produce. This means that nearly 50% of the sugar-sweetened soft drinks market have already reduced the level of sugar in their products, in order to avoid paying the levy.
Importantly, the money raised from the levy will be spent funding breakfast clubs and sports in schools.
Why do we need to restrict high-sugar drinks?
Many cans of soft drink contain 10 teaspoons of sugar or more, which is above the maximum daily sugar recommendation for adults! (The maximum recommendation is 5 teaspoons for 4-6 year olds, 6 teaspoons for ages 6-10 and 7 teaspoons for 11 years and over.) Soft drinks are the biggest source of sugar in the diets of children and teenagers in the UK, yet contain no nutritional value. These drinks contribute to an increased risk of tooth decay, weight gain and subsequently other diet-related diseases, like type-2 diabetes. So it was vital that the government took action when they did.
So, what next?
The soft drinks levy, backed by 69% of the public, is a significant moment for public health. It's a vital step forward in the effort to reduce the amount of sugar we consume as a population – which currently exceeds the government recommendation by two to three times. Plus, it's a sign that the government accepts that improvements to the food environment need to be tackled head-on, at national scale, if they are really serious about protecting children's health.
This is just the start of a range of measures needed to reduce the rate of childhood obesity – including tighter restrictions on the advertising of foods and drinks that contain high amounts of sugar, salt and saturated fat.
"This levy is one of the most important and symbolic moments so far in the fight to protect our kids against diet-related disease. This is a progressive levy, for good, which – let's remember – polled at 69% popularity, and will bring much-needed new money into the education system. My hope is that some of this money will be funnelled into food education to teach our kids how to cook from scratch, setting them up for life. But the real beauty of this levy is that it's pushed the industry to accelerate reformulation of high-sugar fizzy drinks, which is profoundly powerful for child health. These drinks are the single largest source of sugar and empty calories in our children's diets – action is essential!
"This levy was the only progressive, steely action proposed in the government's previous – and frankly disgraceful – childhood obesity strategy. But I'm hopeful that they'll now put Britain's kids first and deliver a brave and powerful chapter 2 of the childhood obesity strategy before the summer recess. Here's hoping – a strong and stable future starts with child health."
For more detail, visit our Sugary Drinks Levy FAQs. For information on the Soft Drinks Industry Levy and type-1 diabetes, we recommend the info on Diabetes UK website. For more information on artificial sweeteners, visit the NHS website's page.
NB: Sugars in this feature refers to 'free sugars', which includes sugars that are added to food and drink, as well as sugars that are naturally present in honey, syrups, fruit juices and fruit concentrates. It does not refer to sugars in milk products and whole fruit and vegetables. Read our guide for more information on different types of sugar.
Watch the video: UK sugar tax gets sour response (January 2022).
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