# A note on $L^{p}$-factorizations of representations

Pritam Ganguly<EMAIL_ADDRESS>Universität Paderborn, Institut für
Mathematik
Warburger Straße 100, 33098 Paderborn  ,  Bernhard Krötz<EMAIL_ADDRESS>Universität Paderborn, Institut für Mathematik
Warburger Straße 100, 33098 Paderborn  and  Job J. Kuit<EMAIL_ADDRESS>Universität Paderborn, Institut für Mathematik
Warburger Straße 100, 33098 Paderborn

###### Abstract.

In this paper we give an overview on $L^{p}$-factorizations of Lie group
representations and introduce the notion of smooth $L^{p}$-factorization.

###### 2000 Mathematics Subject Classification:

22F30, 22E46, 53C35, 22E40

This article is dedicated to the fond memories of Gerrit van Dijk.

##  1\. Introduction

Gerrit van Dijk had a liking for the theory of Gel’fand pairs and Gel’fand
pairs were a recurring theme in his research. He was interested in pairs
$(G,H)$ where $G$ is a noncompact semisimple Lie group and $H$ a non-compact
symmetric subgroup. The definition of what should be a Gel’fand pair for non-
compact $H$ is debatable as their is no natural Hecke algebra of
$H$-biinvariant function to our disposal. The notion Gerrit worked with
readily translates into the criterion that the space of $H$-invariant
distribution vectors attached to a unitary irreducible representation is at
most one-dimensional. A space that occurs regularly in his work is the
hyperbolic space

$X_{n}=\operatorname{SO}_{0}(n,1)/\operatorname{SO}_{0}(n-1,1)$

The pair
$(G,H)=\big{(}\operatorname{SO}_{0}(n,1),\operatorname{SO}_{0}(n-1,1)\big{)}$
happens to be not Gel’fand. However, if $\sigma$ is the involution stabilizing
$H$, then $H$ is an open proper subgroup of the fixed point subgroup
$G^{\sigma}$, and the pair $(G,G^{\sigma})$ does have the Gel’fand property. A
beautiful result of Gerrit [9] asserts that all rank one symmetric spaces have
the Gel’fand property with the exception of the $X_{n}$ above.

The space $X_{n}=G/H$ was considered recently in [7] where the failure of the
Gel’fand property is shown directly in case $n\geq 4$ and it is shown that
there are quantitative differences between the various $G$-equivariant
embeddings $\pi\hookrightarrow C^{\infty}(G/H)$ for certain irreducible
subrepresentations $\pi$ of $L^{2}(G/H)$.

This article we address some questions related to the phenomenon, named after
Kunze and Stein, that for semisimple Lie groups $G$ the convolution product

$C_{c}(G)\times L^{2}(G)\to L^{2}(G),\quad(f,h)\mapsto f*h$

extends to a continuous map

$L^{p}(G)\times L^{2}(G)\to L^{2}(G)$

for $1\leq p<2$. See [3]. In 1960, Kunze and Stein [8] discovered this result
in the case $G=SL_{2}(\mathbb{R})$ as a byproduct of their study of analytic
continuation of the principal series as uniformly bounded representations.

We set out to investigate whether a Kunze-Stein phenomenon could hold for
symmetric spaces $G/H$, i.e., whether the natural convolution map

$C_{c}(G)\times L^{2}(G/H)\to L^{2}(G/H),\quad(f,h)\mapsto f*h$

extends to a continuous map

$L^{p}(G)\times L^{2}(G/H)\to L^{2}(G/H)$

for some or all $1\leq p<2$.

The Kunze-Stein phenomenon may be put in a slightly more general framework by
saying that a representation $(\pi,E)$ of $G$ is $L^{p}$-factorizable if the
natural map

$C_{c}(G)\times E\to E,\quad(f,v)\mapsto\pi(f)v$

extends to a continuous map

$L^{p}(G)\times E\to E.$

We introduce a new notion: $(\pi,E)$ is smoothly-$L^{p}$-factorizable if the
natural map

$C_{c}^{\infty}(G)\times E^{\infty}\to E^{\infty},\quad(f,v)\mapsto\pi(f)v$

extends to a continuous map

$L^{p}(G)^{\infty}\times E^{\infty}\to E^{\infty}.$

Here $L^{p}(G)^{\infty}$ denotes the space of smooth vector for the (left and
right) regular of $G\times G$ on $L^{p}(G)$. Clearly every
$L^{p}$-factorizable representation is also smoothly-$L^{p}$-factorizable but
the converse is not clear. A natural question would be:

Open question: Is a unitary representation $L^{p}$-factorizable if it is
smoothly $L^{p}$-factorizable.

We consider this question worth of further investigation, in particular
because smooth factorizability appears to be easier to establish. For instance
we show by fully elementary means that the smooth Fréchet representation on
the Harish-Chandra Schwartz space is smoothly-$L^{p}$-factorizable for all
$1\leq p<2$.

We return to the hyperboloids and the main result of this article. We show
that $L^{2}(X_{n})$ is not smoothly-$L^{p}$-factorizable for
$p>\frac{n-1}{n-2}$ which implies in particular that $L^{2}(X_{n})$ is not
$L^{p}$-factorizable. This means that there is no straightforward
generalization of the Kunze-Stein phenomenon to symmetric spaces.

For a unimodular homogeneous space $X=G/H$ it follows from the work of Cowling
[3] that the Kunze-Stein phenomenon holds if and only if the unitary
representations occurring in the Plancherel decomposition of $L^{2}(X)$ are
$G$-tempered. Investigations into these type of spaces begun in [1]. For real
spherical spaces $X=G/H$ the spectral gap theorem [5] hints towards the
existence of a constant $1<p_{X}\leq 2$ such that $L^{2}(X)$ is
smoothly-$L^{p}$-factorizable for all $1\leq p<p_{X}$.

##  2\. Setup and Notation

Let $G$ be a unimodular Lie group. By a representation of $G$ we understand a
pair $(\pi,E)$ consisting of a complete locally convex topological vector
space $E$ and a group homomorphism $\pi:G\rightarrow\operatorname{GL}(E)$ such
that the associated action map

$G\times E\rightarrow E,(g,v)\mapsto\pi(g)v$

is continuous. If $E$ is a Banach, Hilbert or Fréchet space, then we say that
$(\pi,E)$ is a Banach, Hilbert or Fréchet representation, respectively.

Given a representation $(\pi,E)$ of $G$, we call a vector $v\in E$ smooth
provided the orbit map

$G\rightarrow E,\quad g\mapsto\pi(g)v$

is smooth. We write $E^{\infty}$ for the space of all smooth vectors, which
forms a $G$-invariant subspace of $E$. We denote the restricted action of $G$
on $E^{\infty}$ by $\pi^{\infty}$ and equip $E^{\infty}$ with the usual
topology so that $(\pi^{\infty},E^{\infty})$ forms a representation of $G$. To
each $\lambda\in E^{\prime}$ and $v\in E$, we associate the matrix coefficient

$m_{\lambda,v}:G\rightarrow\mathbb{C},\quad g\mapsto\lambda(\pi(g)v).$

We note that for a given $\lambda$ the matrix coefficient $m_{\lambda,v}$ is
continuous for every $v\in E$ and smooth for every $v\in E^{\infty}$. By abuse
of notation, we denote the associated algebra representation

$C_{c}(G)\times E\to E,\quad(f,v)\mapsto\int_{G}f(g)\pi(g)v\,dg$

by $\pi$ as well. Note that depending on the kind of space $E$ we are working
with, $\pi$ extends to larger algebra. For example, it is not difficult to see
that $\pi$ extends to a representation of $L^{1}(G)$ in case $(\pi,E)$ is a
bounded Banach representation, i.e., $(\pi,E)$ is a Banach representation so
that there exists a uniform bound on the operators norms of the operators in
$\pi(G)$.

##  3\. $L^{p}$-factorization

###  3.1. Definition and first results

######  Definition 3.1.

We say that a representation $(\pi,E)$ of $G$ is ${L^{p}}$-factorizable
provided that the canonical map

$C_{c}(G)\times E\rightarrow E,\quad(f,v)\mapsto\pi(f)v$

extends to a continuous bi-linear map

$L^{p}(G)\times E\rightarrow E.$

######  Proposition 3.2.

Let $1\leq p<\infty$ and let $1<p^{\prime}\leq\infty$ satisfy
$\frac{1}{p}+\frac{1}{p^{\prime}}=1$. Let further $(\pi,E)$ be a
representation of $G$ on a reflexive Banach space $E$. The following are
equivalent.

  1. (i)

$(\pi,E)$ is $L^{p}$-factorizable.

  2. (ii)

$m_{\lambda,v}\in L^{p^{\prime}}(G)$ for every $\lambda\in E^{\prime}$ and
$v\in E$.

If either holds true, then there exists a constant $C>0$ so that for every

$\|m_{\lambda,v}\|_{p^{\prime}}\leq
C\|\lambda\|_{E^{\prime}}\|v\|_{E}\qquad\big{(}\lambda\in E^{\prime},v\in
E\big{)}.$

###### Proof.

The proof is an straightforward adaption of the unitary case treated in [8,
Lemma 27]. We first assume that $(\pi,E)$ is $L^{p}$-factorizable, which by
definition implies that

$\|\pi(f)v\|_{E}\leq C\|f\|_{p}\|v\|_{E},\qquad\big{(}f\in L^{p}(G),v\in
E\big{)}.$

Now fix $\lambda\in E^{\prime}$, and $v\in E$. For $f\in C_{c}(G)$, by the
continuity of $\lambda$, we see that

$\int_{G}m_{\lambda,v}(g)f(g)\,dg=\lambda\left(\int_{G}f(g)\pi(g)v\,dg\right)=\lambda(\pi(f)v),$

whence we infer that there exists a $C>0$ so that

$\left|\int_{G}m_{\lambda,v}(g)f(g)\,dg\right|\leq C\|f\|_{p}\|v\|_{E}.$

Taking supremum over $\|f\|_{p}\leq 1$, by duality we conclude that
$m_{\lambda,v}\in L^{p^{\prime}}(G)$.

To prove the converse, for $v\in E$ and $f\in L^{p}(G)$, we define $\pi(f)v$
weakly by

$\lambda(\pi(f)v)=\int_{G}m_{\lambda,v}(g)f(g)\,dg\qquad(\lambda\in
E^{\prime}).$

Since by the hypothesis, $m_{\lambda,v}\in L^{p^{\prime}}(G)$, Hölder
inequality tells us that the integrals on the right are absolutely convergent.
Now to justify the well-definedness of $\pi(f)v$, we use closed graph theorem.
For a fixed $v\in E$, we consider the map

$T_{v}:E^{\prime}\rightarrow L^{p^{\prime}}(G),\quad\lambda\mapsto
m_{\lambda,v}.$

To prove $T_{v}$ is closed, let $\lambda_{n}\rightarrow\lambda$ in
$E^{\prime}$ and $T_{v}\lambda_{n}\rightarrow m$ in $L^{p^{\prime}}(G)$ for
$n\to\infty$. Then for any $g\in G$, we see that

$m_{\lambda_{n},v}(g)\rightarrow m_{\lambda,v}(g)\qquad(n\to\infty),$

hence we must have $m(g)=m_{\lambda,v}(g)$. It follows that $T_{v}$ is closed,
and hence it is a bounded operator by the closed graph theorem. It follows
that for every $v\in E$ there exists a $C_{v}>0$ so that

$|\lambda(\pi(f)v)|\leq
C_{v}\|\lambda\|_{E^{\prime}}\|f\|_{p}\qquad(\lambda\in E^{\prime}).$

Because of the assumption that $E$ is reflexive, this shows that $\pi(f)v$ is
well-defined.

Similarly, for each $\lambda\in E^{\prime}$, we can obtain an estimate

$|\lambda(\pi(f)v)|\leq
C\|\lambda\|_{E^{\prime}}\|v\|_{E}\|f\|_{p}\qquad\big{(}v\in E,f\in
L^{p}(G)\big{)},$ (3.1)

for some $C>0$, which together with another application of the closed graph
theorem allows us to conclude that $\pi(f)$ is a bounded operator on $E$. Now
to show that the continuity of the map

$L^{p}(G)\times E\rightarrow E,\quad(f,v)\rightarrow\pi(f)v,$

in view of the closed graph theorem, it is now enough to show that if
$f_{n}\rightarrow f$ in $L^{p}(G)$ and $\pi(f_{n})\rightarrow T$ in operator
norm topology for $n\to\infty$, then $T=\pi(f)$. From $\pi(f_{n})\rightarrow
T$ we see that

$\lambda(\pi(f_{n})v)=\lambda(Tv).$

But from (3.1), we have

$\lambda(\pi(f_{n})v)\rightarrow\lambda(\pi(f)v)$

whence we get $T=\pi(f)$ proving continuity.

The final estimate in the proposition on the matrix coefficients follow from
(3.1). ∎

######  Corollary 3.3.

Let $1\leq p<\infty$ and $(\pi,E)$ be a bounded representation on a reflexive
Banach space. If $(\pi,E)$ is $L^{p}$-factorizable then $(\pi,E)$ is
$L^{q}$-factorizable for all $1\leq q\leq p$.

###### Proof.

This follows from Proposition 3.2 and the facts that matrix coefficients of a
bounded Banach representations are bounded and $L^{p}(G)\cap
L^{\infty}(G)\subset L^{q}(G)$ for all $q\geq p$. ∎

###  3.2. Kunze-Stein phenomenon

Assume that $G$ is a connected noncompact semisimple Lie group with finite
center. Let $L$ denote the left regular representation $G$ on $L^{2}(G)$
defined by the action

$L(g)f(x)=f(g^{-1}x),~{}g\in G,~{}~{}f\in L^{2}(G).$

Clearly, given $f,h\in L^{2}(G)$, the associated matrix coefficient is given
by

$m^{L}_{f,h}(g)=\langle L(g)f,h\rangle=f\ast h.$

Therefore, by Proposition 3.2, it follows that $L$ is $L^{p}$-factorizable if
and only if there exists a $C>0$ so that

$\|f\ast h\|_{p^{\prime}}\leq C\|f\|_{2}\|h\|_{2}\qquad\big{(}f,h\in
L^{2}(G)\big{)}.$

In view of the duality, this is equivalent to

$\|f\ast h\|_{2}\leq C\|f\|_{p}\|h\|_{2},~{}~{}f\in L^{p}(G),~{}h\in
L^{2}(G).$

In other words, $L^{p}$-factorizability of $L$ is equivalent to convolution
extending to a continuous map

$L^{p}(G)\times L^{2}(G)\to L^{2}(G),\quad(f,h)\mapsto f*h.$

If the latter holds true, it is called the Kunze-Stein phenomenon. The Kunze-
Stein phenomenon holds true only when $1\leq p<2$ (see [3]). Hence the left
regular representation $L$ is $L^{p}$-factorizable if and only if $1\leq p<2$.

###  3.3. $L^{p}$-factorizability of unitary representations

We first consider direct integrals of unitary representations. We begin with a
definition.

######  Definition 3.4.

Let $(S,\mu)$ be a $\sigma$-finite measure space. We say that the family
$(\pi_{s},E_{s})_{s\in S}$ of representations of $G$ is ${L^{p}}$-factorizable
provided that that there exists a constant $C>0$ so that for almost every
$s\in S$ the canonical map

$C_{c}(G)\times E_{s}\rightarrow E_{s},\quad(f,v)\mapsto\pi_{s}(f)v$

satisfies

$\|\pi_{s}(f)v\|\leq C\|f\|_{p}\|v\|_{s}\qquad\big{(}f\in C_{c}(G),v\in
E_{s}\big{)}.$

Note that if a family $(\pi_{s},E_{s})_{s\in S}$ is $L^{p}$-factorizable, then
the representation $(\pi_{s},E_{s})$ is $L^{p}$-factorizable for almost every
$s\in S$.

######  Proposition 3.5.

Let $1\leq p<\infty$, let $(S,\mu)$ be a $\sigma$-finite measure space and let

$\Big{(}\pi=\int^{\oplus}_{S}\pi_{s}\,d\mu(s),\mathcal{H}=\int^{\oplus}_{S}\mathcal{H}_{s}\,d\mu(s)\Big{)}$

be a direct integral of a family $(\pi_{s},\mathcal{H}_{s})_{s\in S}$ of
unitary representations on separable Hilbert spaces. Then $(\pi,\mathcal{H})$
is $L^{p}$-factorizable if and only if the family
$(\pi_{s},\mathcal{H}_{s})_{s\in S}$ is $L^{p}$-factorizable.

###### Proof.

First assume that the family $(\pi_{s},\mathcal{H}_{s})_{s\in S}$ is
$L^{p}$-factorizable. Let $C>0$ be so that

$\|\pi_{s}(f)v\|_{s}\leq C\|v\|_{s}\|f\|_{p}\qquad\big{(}f\in
L^{p}(G),v\in\mathcal{H}_{s}\big{)}$

for almost every $s\in S$. For any $v\in\mathcal{H}$, by definition we have

$\displaystyle\|\pi(f)v\|^{2}=\int_{S}\|\pi_{s}(f)v_{s}\|^{2}\,d\mu(s)\leq
C^{2}\|f\|_{p}^{2}\int_{S}\|v_{s}\|^{2}_{s}\,d\mu(s)=C^{2}\|f\|_{p}^{2}\|v\|^{2}$

proving that $(\pi,\mathcal{H})$ is $L^{p}$ factorizable.

For the other implication, assume that $(\pi,\mathcal{H})$ is $L^{p}$
factorizable. Now using the separability of $L^{p}(G)$, choose a countable
dense subset $\\{h_{j}:j\in\mathbb{N}\\}\subset L^{p}(G)$. So, by assumption
there exists a $C>0$ so that for all $j\in\mathbb{N}$,

$\|\pi(h_{j})v\|\leq C\|h_{j}\|_{p}\|v\|\qquad(v\in\mathcal{H}),$

i.e.,

$\mathrm{ess\,sup}_{s\in
S}\|\pi_{s}(h_{j})\|_{\mathcal{H}_{s}\rightarrow\mathcal{H}_{s}}\leq
C\|h_{j}\|_{p}\qquad(j\in\mathbb{N}).$

For $j\in\mathbb{N}$ we define

$\Omega_{j}:=\\{s\in
S:\|\pi_{s}(h_{j})\|_{\mathcal{H}_{s}\rightarrow\mathcal{H}_{s}}>C\|h_{j}\|_{p}\\}.$

We note that $\mu(\Omega_{j})=0$. Let
$\Omega:=\cup_{j\in\mathbb{N}}\Omega_{j}$. Then clearly $\mu(\Omega)=0$.
Moreover, for every $j\in\mathbb{N}$ and all $s\in S\setminus\Omega$

$\|\pi_{s}(h_{j})u\|_{s}\leq
C\|u\|_{s}\|h_{j}\|_{p}\qquad(u\in\mathcal{H}_{s}).$

The assertion now follows using a standard density argument. ∎

Let $\widehat{G}$ be the set of all equivalence classes of irreducible unitary
representations of $G$ equipped with the usual Fell topology. Now the
disintegration theorem for unitary representations $(\pi,\mathcal{H})$ of a
type I group $G$ states that there exists a Borel measure $\mu$ on
$\widehat{G}$ such that $(\pi,\mathcal{H})$ is unitarily equivalent to the
direct integral representation

$\left(\int_{\widehat{G}}\pi_{s}\,d\mu(s),~{}\int_{\widehat{G}}\mathcal{H}_{s}\,d\mu(s)\right).$

See for example [10, Theorem 14.10.5] for more details. This, as an immediate
consequence of the above proposition, yields the following result.

######  Corollary 3.6.

Let $G$ be a type I Lie group. An unitary representation of $G$ is
$L^{p}$-factorizable if and only if the family of representations appearing in
its disintegration is $L^{p}$-factorizable.

As a consequence of this result, in the following we witness that there is a
large sub-class of type I non-compact groups which does not admit
$L^{p}$-factorizability property.

######  Corollary 3.7.

Let $G$ be a type $I$ Lie group with non-compact center. Then unitary
representations of $G$ are not $L^{p}$-factorizable for $p>1.$ In particular
this holds when $G$ is non-compact and abelian, or a Carnot Lie group.

###### Proof.

In view of Corollary 3.6, it is sufficient to prove that irreducible unitary
representations of $G$ are not $L^{p}$ factorizable. Fix $p>1.$ Let $(\pi,H)$
be an irreducible unitary representation of $G.$ So, by the Proposition 3.2,
it is enough to show that there exits matrix coefficient of
$(\pi,\mathcal{H})$ which is not in $L^{p^{\prime}}(G).$ Now from the Schur
lemma it follows that

$\pi(z)=C_{z}I\qquad\big{(}z\in Z(G)\big{)}$

where $C_{z}$ is constant such that $|C_{z}|=1$ for all $z\in Z(G).$

The center $Z(G)$ is a closed subgroup of $G$. It is easy to see that there
exists a submanifold $M$ of $G$ of dimension equal to the codimension of
$Z(G)$, so that the map

$Z(G)\times M\to G,\quad(z,m)\mapsto zm$

is a diffeomorphism onto a nonempty $Z(G)$-invariant open subset of $G$. The
pull-back of the Haar measure on $G$ along this diffeomorphism yields a
product measure of a Haar measure $dz$ on $Z(G)$ and a positive Radon measure
$d\nu$ on $M$. Since $Z(G)$ is non-compact, the volume of $Z(G)$ is infinite.
Now for any $u,v\in H$, and $(z,m)\in Z(G)\times M$, it follows that

$|\langle\pi(zm)u,v\rangle|=|\langle\pi(m)u,v\rangle|$

which yields

$\int_{G}|\langle\pi(g)u,v\rangle|^{p^{\prime}}dg\geq\int_{Z(G)\times
M}|\langle\pi(zm)u,v\rangle|^{p^{\prime}}dzd\nu(m)=\infty.$

This shows that the matrix coefficient are not in $L^{p^{\prime}}(G)$,
completing the proof. ∎

###  3.4. Semisimple symmetric spaces

Assume that $G$ is a connected semisimple Lie group. Let $\sigma$ be an
involution i.e., $\sigma\in\operatorname{Aut}(G)$, and $\sigma^{2}=I$. Suppose
$H$ is an open subgroup of the group $G^{\sigma}$ of fixed points for
$\sigma$. Then the pair $(G,H)$ is called a semisimple symmetric pair and the
associated homogeneous space $X:=G/H$ is a semisimple symmetric space.

As a special case, when $\sigma=\theta$ is a Cartan involution and
$H=K=G^{\theta}$, $X=G/K$ is a Riemannian symmetric space. Semisimple Lie
groups are examples of semisimple symmetric spaces. Indeed, if $G^{\prime}$ is
a connected semisimple Lie group, then take $G=G^{\prime}\times G^{\prime}$
and consider the action of $G$ on $G^{\prime}$ given by $(g_{1},g_{2})\cdot
x:=g_{1}xg_{2}^{-1}$. This induces the isomorphism of $G$-spaces

$X:=G/H\simeq G^{\prime},$

where $H=\operatorname{diag}(G^{\prime})$ is the stabilizer of the identity of
$G^{\prime}$ in $G$. The group $H$ equals the fixed point subgroup
$G^{\sigma}$ where $\sigma$ is the involution given by
$\sigma(g_{1},g_{2})=(g_{2},g_{1})$. This is referred to as the group case.

The goal of this section is to demonstrate that the Kunze-Stein phenomon does
not generalize to the class of semisimple symmetric spaces. To provide a
counter example we consider the $n$-dimensional one-sheeted hyperboloid $X$.
The space $X$ is homogeneous for the connected Lorentz group
$G=\mathrm{SO}_{0}(n,1)$.

Let $\sigma$ be the involution of $G$ defined by conjugation with the diagonal
matrix $\operatorname{diag}(-1,1,\dots,1)$. The subgroup $G^{\sigma}$ of $G$
of $\sigma$-fixed elements is the stabilizer of $\mathbb{R}x_{0}$ where
$x_{0}:=(1,0,\dots,0)\in\mathbb{R}^{n+1}$. One of the connected components of
$G^{\sigma}$ is given by the subgroup

$H=\begin{pmatrix}1&0\\\
0&\operatorname{SO}_{0}(n\\!-\\!1,1)\end{pmatrix}\subset G.$

Then the homogeneous space

$G/H=\operatorname{SO}_{0}(n,1)/\operatorname{SO}_{0}(n-1,1)$

is a symmetric space. Since $G$ acts transitively on the hyperboloid

$X:=\\{x\in\mathbb{R}^{n+1}\mid x_{1}^{2}+\dots+x_{n}^{2}-x_{n+1}^{2}=1\\},$

and $H$ is nothing but the stabilizer of $x_{0}$, it follows that $X\simeq
G/H$. We also remark that as shown by Van Dijk [9], though $(G,G^{\sigma})$ is
a Gelfand pair, $(G,H)$ is not and $X$ is the only rank one symmetric space,
which is not obtained as the homogeneous space of a Gelfand pair.

Let $\Delta_{X}$ stand for the $G$-invariant Laplace-Beltrami operator, which
is a scalar multiple of the Casimir element associated to the Lie group $G$.
For each $\lambda\in\mathbb{C}$ let $\mathcal{E}_{\lambda}(X)$ be the
eigenspace

$\mathcal{E}_{\lambda}(X)=\\{f\in
C^{\infty}(X)\mid\Delta_{X}f=(\lambda^{2}-\rho^{2})f\\},$

where

$\rho:=\tfrac{1}{2}(n-1).$

For every irreducible $H$-spherical representation of $(\pi,E)$ of $G$ there
exists a $\lambda\in\mathbb{C}$ and a $G$-equivariant injection
$E\hookrightarrow\mathcal{E}_{\lambda}(X)$. In view of the $G$-equivariant
symmetry $x\mapsto-x$, the space $C^{\infty}(X)$ admits a $G$-equivariant
decomposition

$C^{\infty}(X)=C_{\operatorname{even}}^{\infty}(X)\oplus
C_{\operatorname{odd}}^{\infty}(X).$

Let us write

$\mathcal{E}^{\operatorname{even}}_{\lambda}(X)=\mathcal{E}_{\lambda}(X)\cap
C_{\operatorname{even}}^{\infty}(X),\quad\mathcal{E}^{\operatorname{odd}}_{\lambda}(X)=\mathcal{E}_{\lambda}(X)\cap
C_{\operatorname{odd}}^{\infty}(X).$

We now recall the following result proved in [7, Theorem 1.2].

######  Theorem 3.8.

For every $0<\lambda<\rho$ with $\lambda\in\rho-\mathbb{N}$ the
$G$-representations $\mathcal{E}^{\operatorname{even}}_{\lambda}(X)$ and
$\mathcal{E}^{\operatorname{odd}}_{\lambda}(X)$ are irreducible and
infinitesimally equivalent. Moreover in this case,

  1. (1)

if $\lambda-\rho$ is even then $\mathcal{E}^{\operatorname{odd}}_{\lambda}(X)$
is contained in $L^{2}(X)$ and
$\mathcal{E}^{\operatorname{even}}_{\lambda}(X)$ is non-tempered, and

  2. (2)

if $\lambda-\rho$ is odd then $\mathcal{E}^{\operatorname{even}}_{\lambda}(X)$
is contained in $L^{2}(X)$ and $\mathcal{E}^{\operatorname{odd}}_{\lambda}(X)$
is non-tempered.

Next, we describe the $K$-types. The stabilizer of
$e_{n+1}:=(0,0,\cdots,1)\in\mathbb{R}^{n+1}$ is a maximal compact subgroup $K$
of $G$. Note that $K$ is isomorphic to $SO(n)$. It is convenient for our
purpose to work with the coordinate system on $X$ induced by the
diffeomorphism

$S^{n-1}\times\mathbb{R}{\to}X,\quad(y,t)\mapsto(y_{1}\cosh t,\dots,y_{n}\cosh
t,\sinh t)\in X.$ (3.2)

Now as described in [7, Section 2], the space $C_{K}^{\infty}(X)$ of
$K$-finite functions in $C^{\infty}(X)$ given by

$C_{K}^{\infty}(X)\underset{K}{\simeq}\oplus_{j=0}^{\infty}\big{(}\mathcal{H}_{j}\otimes
C^{\infty}(\mathbb{R})\big{)}$

where for each $j\geq 0$, $\mathcal{H}_{j}\subset C^{\infty}(S^{n-1})$ denotes
the space of spherical harmonics of degree $j$. So, the Harish-Chandra module
of $\mathcal{E}_{\lambda}(X)$ is given by

$\mathcal{E}_{\lambda,K}(X)=\mathcal{E}_{\lambda}(X)\cap C^{\infty}_{K}(X).$

Consequently,

$\mathcal{E}_{\lambda,K}^{\operatorname{even}}(X)=\oplus_{j=0}^{\infty}\,\mathcal{E}_{\lambda,j}^{\operatorname{even}}(X),\qquad\mathcal{E}_{\lambda,K}^{\operatorname{odd}}(X)=\oplus_{j=0}^{\infty}\,\mathcal{E}_{\lambda,j}^{\operatorname{odd}}(X)$

where
$\mathcal{E}_{\lambda,j}^{\operatorname{even}}(X)(\text{or}~{}\mathcal{E}_{\lambda,j}^{\operatorname{odd}}(X))$
denotes the space of all even (or odd) functions in
$\big{(}\mathcal{H}_{j}\otimes C^{\infty}(\mathbb{R})\big{)}$ and
$\mathcal{E}_{\lambda,j}^{\operatorname{even}}(X)\underset{K}{\simeq}\mathcal{E}_{\lambda,j}^{\operatorname{odd}}(X)\underset{K}{\simeq}\mathcal{H}_{j}$
are equivalent irreducible $K$-types for each $j$.

The following Proposition will be strengthened in Theorem 5.1 below.

######  Proposition 3.9.

Let $n\geq 4.$ The regular representation of $G$ on $L^{2}(X)$ is not
$L^{p}$-factorizable for some $1<p<2.$

###### Proof.

Since $n\geq 4,$ clearly $\rho>1.$ Take $0<\lambda<\rho$ with
$\lambda\in\rho-\mathbb{N}.$ By Theorem 3.8 there exists a subrepresentation
$E$ of $L^{2}(X)$ where $E=\mathcal{E}^{\operatorname{odd}}_{\lambda}(X)$ or
$\mathcal{E}^{\operatorname{even}}_{\lambda}(X)$ depending on $\lambda-\rho$
is even or odd. From the above discussion, it is clear that
$E^{K}\neq\\{0\\}.$ Now if $E$ is $G$-tempered, then we must have that all
matrix coefficients $m_{v,\eta}$ with $\eta\in E^{K}\neq\\{0\\}$, lie weakly
in $L^{2}(G/K).$ But then $E$ is a unitary principle series representation.
The infinitesimal character of unitary principle series representations are
such that $\lambda\in i\mathbb{R}$. This contradicts with the assumption
$0<\lambda<\rho$. It follows that $E$ is not $G$-tempered and hence the matrix
coefficients are not contained in $L^{2+\varepsilon}(G)$ for suffiently small
$\varepsilon>0$. The assertion now follows from Proposition 3.2. ∎

##  4\. Smooth-$L^{p}$-factorization

###  4.1. Definition

The left and right actions of $G$ on itself naturally induce a Banach
representation of $G\times G$ on $L^{p}(G)$. The space of all smooth vectors
for this representation in $L^{p}(G)$ is given by

$L^{p}(G)^{\infty}=\\{f\in C^{\infty}(G):L(u)R(v)f\in L^{p}(G),~{}~{}\forall
u,v\in\mathcal{U}(\mathfrak{g})\\},$

where $L$ and $R$ denote the left and right action of $U(\mathfrak{g})$,
respectively. Now if a Banach representation $(\pi,E)$ of $G$ is
$L^{p}$-factorizable, then the $G$-equivariance of the map $L^{p}(G)\times
E\rightarrow E$ yields that the canonical map

$L^{p}(G)^{\infty}\times E^{\infty}\rightarrow E^{\infty}$

is continuous. This motivates the following definition.

######  Definition 4.1.

We say that a representation $(\pi,E)$ of $G$ is smoothly-$L^{p}$-factorizable
provided that the canonical map

$C_{c}^{\infty}(G)\times E^{\infty}\rightarrow
E^{\infty},\quad(f,v)\rightarrow\pi(f)v$

extends to a continuous bi-linear map

$L^{p}(G)^{\infty}\times E^{\infty}\rightarrow E^{\infty}.$

In the remainder of this section we will show that the Harish-Chandra Schwartz
space of $G$ is smoothly-$L^{p}$-factorizable for all $1\leq p<2$.

###  4.2. Harish-Chandra Schwartz space

We fix a Cartan involution $\theta$ and denote by $K$ the maximal compact
subgroup which is fixed by $\theta$. By abuse of notation we write $\theta$
for the derived automorphism of the Lie algebra $\mathfrak{g}$ of $G$, which
induces the Cartan decomposition
$\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{s}$. Here
$\operatorname{Lie}(K)=\mathfrak{k}$. Let $\mathfrak{a}\subset\mathfrak{s}$ be
a maximal abelian subspace. We denote the set of restricted roots of
$\mathfrak{a}$ in $\mathfrak{g}$ by
$\Sigma\subset\mathfrak{a}^{*}\setminus\\{0\\}$ and record the following root
space decomposition

$\mathfrak{g}=\mathfrak{a}\oplus\mathfrak{m}\oplus\left(\bigoplus_{\alpha\in\Sigma}\mathfrak{g}^{\alpha}\right).$

Let $A$ be the connected subgroup of $G$ with Lie algebra $\mathfrak{a}$. Note
that the usual exponential map $\exp:\mathfrak{a}\rightarrow A$ is a
diffeomorphism and we write $\log:A\rightarrow\mathfrak{a}$ for its inverse.
With respect to a fixed choice of Weyl chamber $\mathfrak{a}^{+}$, we denote
the system of positive roots by $\Sigma^{+}$. We then write
$\mathfrak{n}:=\oplus_{\alpha\in\Sigma^{+}}\mathfrak{g}^{\alpha}$ and
$N:=\exp\mathfrak{n}$. Now in view of the Iwasawa decomposition $G=NAK$, we
define the projection ${\bf{a}}:G\rightarrow A$ by $g\in N{\bf{a}(g)}K$ for
$g\in G$. We let
$\rho:=\frac{1}{2}\sum_{\alpha\in\Sigma^{+}}\dim(g^{\alpha})\alpha\in\mathfrak{a}^{*}$
and for any $\lambda\in\mathfrak{a}^{*}$ and $a\in A$, we define
$a^{\lambda}=e^{\lambda(\log a)}$. Let $dk$ denote the invariant probability
measure on $K$ and let $da$ be a Haar measure on $A$. We normalize the Haar
measure on $G$ so that with respect to the polar decomposition
$G=K\overline{A^{+}}K$, where $A^{+}:=\exp(\mathfrak{a}^{+})$, we have

$\int_{G}\varphi(g)\,dg=\int_{K}\int_{\overline{A^{+}}}\int_{K}\varphi(k_{1}ak_{2})J(a)\,dk_{1}\,da\,dk_{2}\qquad\big{(}\varphi\in
C_{c}(G)\big{)}$ (4.1)

where

$J(a):=\prod_{\alpha\in\Sigma^{+}}\left(e^{\alpha(\log a}-e^{-\alpha(\log
a)}\right)^{m_{\alpha}}.$

Recall the Harish-Chandra $\Xi$ function

$\Xi(g):=\int_{K}{\bf{a}}(kg)^{\rho}\,dk.$

This is a $K$ biinvariant smooth function on $G$, which, in view of the polar
decomposition can be realised as a function on $\overline{A^{+}}$. As $\Xi$ is
a spherical function, it features the mean value property

$\int_{K}\Xi(xky)\,dk=\Xi(x)\Xi(y)\qquad(x,y\in G).$ (4.2)

We now record the following estimate

$a^{-\rho}\leq\Xi(a)\leq Ca^{-\rho}(1+\|\log
a\|)^{d}\qquad\big{(}a\in\overline{A^{+}}\big{)}$

where $C$ and $d$ are constants.

$r(g)=\|\log a\|\qquad\big{(}g=k_{1}ak_{2}\in K\overline{A^{+}}K\big{)}.$

It is not difficult to see that

$r(xy)\leq r(x)+r(y)\qquad(x,y\in G).$ (4.3)

For each $n\in\mathbb{Z}$, we let $\mathcal{C}_{n}(G)$ stand for the set of
all continuous functions $f$ on $G$ such that

$p_{n}(f):=\sup_{g\in G}|f(g)|\Xi(g)^{-1}(1+r(g))^{n}<\infty.$ (4.4)

For each $u,v\in U(\mathfrak{g})$, we consider the seminorm

$p_{n,u,v}(f):=p_{n}\big{(}L(u)R(v)f\big{)}.$

The Harish-Chandra Schwartz space $\mathcal{C}(G)$ is the space of all smooth
functions $f$ on $G$ for which $p_{n,u,v}(f)<\infty$ for all $u,v\in
U(\mathfrak{g})$, and $n\in\mathbb{Z}$. It is equipped with the Fréchet
topology generated by the semi norms $p_{n,u,v}$. It is worth noting that an
equivalent characterization of $\mathcal{C}(G)$ can be obtained by replacing
$p_{n}$ by the norms $q_{n}$ given by

$q_{n}(f):=\left(\int_{G}|f(g)|^{2}(1+r(g))^{n}\,dg\right)^{\frac{1}{2}}.$

Harish-Chandra proved that the convolution product defines a continuous map

$\mathcal{C}(G)\times\mathcal{C}(G)\to\mathcal{C}(G),\quad(f,h)\mapsto f*h.$

###  4.3. Bernstein’s invariant Sobolev lemma

Let $H\subset G$ be a closed subgroup of $G$. We consider the homogeneous
space $Z:=G/H$. Assume that $Z$ is unimodular, that is, $Z$ carries a positive
$G$-invariant Radon measure. We call a locally bounded function
$w:Z\rightarrow\mathbb{R}_{>0}$ a weight on $Z$ provided for any compact
subset $\Omega\subset G$, there exists $C>0$ such that

$w(gz)\leq Cw(z)\qquad(z\in Z,g\in\Omega).$

We fix a ball $B$ in $G$, by which, we always mean a compact symmetric
neighborhood of $\bf{1}$ in $G$. Define

${\bf{v}}(z)=\text{vol}_{Z}(B\cdot z)\qquad(z\in Z).$

It is not difficult to see that $\bf{v}$ is a weight. We call $\bf{v}$ the
volume weight on $Z$ corresponding to the ball $B$. Different choices of balls
in $G$ give rise to equivalent weights in the sense that if $v_{1}$ and
$v_{2}$ are two weights corresponding to balls $B_{1}$ and $B_{2}$,
respectively, then there exists a constant $c>0$ such that

$\frac{1}{c}v_{1}(z)\leq v_{2}(z)\leq cv_{1}(z)\qquad(z\in Z).$

######  Lemma 4.2 (Bernstein’s invariant Sobolev lemma).

Let $1\leq p<\infty$ and $k\in\mathbb{N}$ be such that $pk>\dim G$. Fix a ball
$B$ in $G$ and let $\bf{v}$ be the corresponding volume weight. Let
$\|\cdot\|_{p,k,B\cdot z}$ be the $k$’th $L^{p}$ Sobolev norm on $B\cdot z$.
Then for all $f\in C^{\infty}(Z)$, there exists constant $C_{B}>0$ such that

$|f(z)|\leq C_{B}{\bf{v}}(z)^{-\frac{1}{p}}\|f\|_{p,k,B\cdot z}\qquad(z\in
Z).$

###### Proof.

See [2, Sect. 3.4, Key Lemma] or [6, Lemma 4.2]. ∎

We apply the invariant Sobolev theorem to $G$ viewed as the homogeneous space
$G\times G/\operatorname{diag}(G)$. We choose a ball $B\subset G$ and define
$\bf{v}:G\to\mathbb{R}_{>0}$ be the volume weight corresponding to the ball
$B\times B$ of $G\times G$, i.e.,

${\bf{v}}(x)=\text{Vol}(BxB)\qquad(x\in G).$

In this context we record that

${\bf{v}}(x)\asymp a^{2\rho}\qquad(x\in KaK,a\in\overline{A^{+}}).$

Hence, in view of the above lemma, we observe that for every
$k>\frac{\dim(G\times G)}{p}$ there exists a $C_{B,k}>0$ so that

$|f(x)|\leq C_{B,k}{\bf{v}}(x)^{-\frac{1}{p}}\|f\|_{p,k,BxB}\leq
C_{B,k}{\bf{v}}(x)^{-\frac{1}{p}}\|f\|_{p,k}$ (4.5)

for any $f\in L^{p}(G)^{\infty}$.

######  Lemma 4.3.

Let $1\leq p<2$ and $k\in\mathbb{N}$ with $k>\frac{\dim(G\times G)}{p}$. For
each $n\in\mathbb{Z}$ there exists a constant $C_{n}>0$ so that

$p_{n}(f\ast h)\leq
C_{n}\|f\|_{p,k}~{}~{}p_{n}(h)\qquad\big{(}h\in\mathcal{C}_{n}(G),f\in
L^{p}(G)^{\infty}\big{)}.$

###### Proof.

Fix $n\in\mathbb{Z}$. Using (4.5), we observe that there exists a constant
$C>0$ so that for every $f\in L^{p}(G)^{\infty}$ and $h\in\mathcal{C}_{n}(G)$

$\displaystyle|f\ast h(x)|$
$\displaystyle=\left|\int_{G}f(g)h(g^{-1}x)\,dg\right|$ $\displaystyle\leq
C\|f\|_{p,k}p_{n}(h)\int_{G}{\bf{v}}(g)^{-\frac{1}{p}}\Xi(g^{-1}x)\left(1+r(g^{-1}x)\right)^{-n}\,dg.$
(4.6)

In view of (4.1) the last integral takes the form

$\displaystyle\int_{K}\int_{A^{+}}{\bf{v}}(a)^{-\frac{1}{p}}\Xi(a^{-1}kx)\left(1+r(a^{-1}kx)\right)^{-n}J(a)\,dk\,da.$

In order to estimate this, we first notice that for every $k\in K$, $a\in A$
and $x\in G$

$\left(1+r(a^{-1}kx)\right)^{-n}\leq(1+r(x))^{-n}(1+r(a))^{|n|}.$

Indeed, using (4.3), we first see that

$1+r(a^{-1}kx)\leq 1+r(a)+r(x)\leq(1+r(x))(1+r(a)).$

So, if $n<0$, then we have

$(1+r(a^{-1}kx))^{-n}\leq(1+r(x))^{-n}(1+r(a))^{-n}.$

Also we observe that

$1+r(x)=1+r(aa^{-1}kx)\leq(1+r(a))(1+r(a^{-1}kx)).$

Now if $n>0$, the above yields

$(1+r(a^{-1}kx))^{-n}\leq{(1+r(x))^{-n}}{(1+r(a))^{n}}$

proving the claim. Using this estimate we have

$\displaystyle\int_{K}\int_{A^{+}}{\bf{v}}(a)^{-\frac{1}{p}}\Xi(a^{-1}kx)\left(1+r(a^{-1}kx)\right)^{-n}J(a)\,dk\,da$
$\displaystyle\leq(1+r(x))^{-n}\int_{A^{+}}{\bf{v}}(a)^{-\frac{1}{p}}\left(\int_{K}\Xi(a^{-1}kx)dk\right)\left(1+r(a)\right)^{|n|}J(a)\,da$
$\displaystyle=(1+r(x))^{-n}\Xi(x)\int_{A^{+}}{\bf{v}}(a)^{-\frac{1}{p}}\Xi(a)\left(1+r(a)\right)^{|n|}J(a)\,da$

where in the last equality we have used the mean value property (4.2) of the
$\Xi$-function. Finally, denoting the last integral in the above equation by
$C_{n}$, from (4.3), we obtain

$|f\ast h(x)|(1+r(x))^{n}\Xi(x)^{-1}\leq CC_{n}\|f\|_{p,k}p_{n}(h)\qquad(x\in
G),$

completing the proof. ∎

As an immediate consequence we obtain the following result.

######  Corollary 4.4.

For $1\leq p<2$, representations $(L,\mathcal{C}_{n}(G))$ and
$(L,\mathcal{C}(G))$ are smooth $L^{p}$-factorizable.

##  5\. $L^{2}(X)$ is not smooth $L^{p}$-factorizable

In this section we let $X=X_{n}$ be the real hyperbolic space for $n\geq 4$.

######  Theorem 5.1.

$L^{2}(X)$ is not smooth $L^{p}$-factorizable for $p>\frac{n-1}{n-2}$.

###### Proof.

We saw earlier that there exists a $K$-spherical discrete series $E\subset
L^{2}(X)$ with spectral parameter $0<\lambda<\rho$ subject to
$\rho-\lambda\in\mathbb{N}$. Recall that $\rho=\frac{1}{2}(n-1)$. Let
$E^{K}=\mathbb{C}v_{K}$. For any $1\leq p<\infty$ we can define
$L^{p}$-Schwartz spaces $\mathcal{C}^{p}(G)$ simply by replacing the growth
constraint $\Xi$ with $\Xi^{2\over p}$ in the definition (4.4) of the Harish-
Chandra Schwartz space $\mathcal{C}(G)$. Note that
$\mathcal{C}(G)=\mathcal{C}^{2}(G)$ in our generalized framework. We denote by
$\mathcal{C}^{p}(G\mathbin{/\mkern-6.0mu/}K)\subset\mathcal{C}^{p}(G)$ the
subspace of $K$-bi-invariant functions. Assume now that $p>\frac{n-1}{n-2}$
and that $L^{2}(X)$ is smooth $L^{p}$-factorizable. As $\mathcal{C}^{p}(G)$ is
dense in $L^{p}(G)^{\infty}$ we would arrive via Dirac approximation at a
contradiction if these assumptions would yield

$\mathcal{C}^{p}(G\mathbin{/\mkern-6.0mu/}K)*v_{K}={0}$ (5.1)

To proceed we invoke the action of the Casimir element $\Omega$ which descends
to $\Delta_{X}$. Then we have $\Omega v_{K}=(\lambda^{2}-\rho^{2})v_{K}$. We
reformulate that in $(\Omega-c_{\lambda})v_{K}=0$ with
$c_{\lambda}=\lambda^{2}-\rho^{2}$. Now for
$f\in\mathcal{C}^{p}(G\mathbin{/\mkern-6.0mu/}K)$ we have

$0=f*(\Omega-c_{\lambda})v_{K}=(\Omega-c_{\lambda})f*v_{K}$

We claim that $(\Omega-c_{\lambda})$ is an invertible operator on
$\mathcal{C}^{p}(G\mathbin{/\mkern-6.0mu/}K)$ provided that $1<p<2$ is
sufficiently close to $2$. In fact, the Paley-Wiener Theorem for the Fourier
transform identifies $\mathcal{C}^{p}(G\mathbin{/\mkern-6.0mu/}K)$ with
certain holomorphic functions on the strip $|\mathrm{Re}\lambda|<\epsilon\rho$
for $\epsilon=\frac{2}{p}-1$, see [4, Th. 7.8.6]. This implies in particular
$(\Omega-c_{\lambda})$ is invertible provided
$|\mathrm{Re}\lambda|>(\frac{2}{p}-1)\rho$. For the discrete series parameter
$\lambda=\rho-1$ we obtain this condition for $p>\frac{n-1}{n-2}$ and the
desired contradiction. ∎

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