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\begin{document}
\begin{center}
{\Large{\bf Rigidity of action of compact quantum groups II}}\\
{\large {\bf Biswarup Das\footnote{Acknowledges support from UKIERI}, \bf Debashish Goswami\footnote{Partially supported by Swarnajayanti Fellowship from D.S.T. (Govt. of India)}, \bf Soumalya Joardar \footnote{Acknowledges support from CSIR}}}\\
Indian Statistical Institute\\
203, B. T. Road, Kolkata 700108\\
Email: goswamid@isical.ac.in\\
\end{center}
\begin{abstract}

Suppose that a compact quantum group
${\cal Q}$ acts faithfully and isometrically (in the sense of \cite{Goswami}) on a smooth compact, oriented, connected Riemannian manifold
$M$. If the manifold is stably parallelizable then it is shown that the
compact quantum group is necessarily commutative as a $C^{*}$ algebra i.e.
${\cal Q}\cong C(G)$ for some compact group $G$. Using this, it is also proved that the quantum isometry group of Rieffel deformation of such manifold $M$ must be a Rieffel-Wang deformation of $C(ISO (M))$.
\end{abstract}

 \section{Introduction}
 It is  a very important and interesting problem in the theory of quantum groups and noncommutative geometry to study  `quantum symmetries' of various classical and quantum structures. Indeed, symmetries of physical systems (classical or quantum) were conventionally modelled by group actions, and after the advent of quantum groups, group symmetries were naturally generalized to  symmetries given by quantum group action. In this context, it is  natural to think of quantum automorphism or the full quantum symmetry groups of various mathematical and physical structures. The underlying basic
 principle of defining a quantum automorphism group of a given mathematical structure consists of two steps : first, to identify (if possible) the group of automorphisms of the structure as a universal object in a
   suitable category, and then, try to look for the universal object in a similar but bigger category by replacing groups by quantum groups of appropriate type. Initiated by S. Wang who defined and studied quantum permutation groups of finite sets and quantum automorphism groups of finite dimensional algebras, such questions were taken up by a number of mathematicians including Banica, Bichon (see, e.g. \cite{ban_1}, \cite{ban_2}, \cite{wang}), and more recently in the framework of Connes' noncommutative geometry (\cite{con}) by Goswami, Bhowmick,  Skalski, Banica and others who have extensively studied the quantum group of isometries  (or quantum isometry group) defined in \cite{Goswami}  (see also \cite{qorient}, \cite{qiso_trans}, \cite{qdisc} etc.). In this context, it is important to compute such quantum isometry groups for classical (compact)  Riemannian manifolds. This will also allow one to compute quantum isometry groups of the noncommutative manifolds obtained by  Rieffel-type deformation of classical manifolds by the techniques developed in \cite{qiso_comp}.

   However, it was rather amazing to see that for all the connected classical manifolds including the spheres and the tori (with the usual Riemannian metrics) for which the quantum isometry groups were computed so far, the quantum isometry groups turned out to be the same as the classical isometry groups. In other words, there is no genuine (i.e. noncommutative as a $C^*$ algebra) compact quantum group which can act isometrically on such manifolds. It may be mentioned here that it is easy to have faithful isometric action of genuine compact quantum group on disconnected Riemannian manifolds with at least four components.
However, no examples of even  faithful continuous action by genuine compact quantum groups on $C(X)$ with $X$ being connected compact space were  known until recently, when H. Huang (\cite {Huchi}) constructed  examples of such action on  topological spaces which are typically obtained by topological connected sums of copies of some given compact metric space.
But  none of the examples constructed by Huang are smooth manifolds. In fact, his construction would fail if topological connected sum is replaced by a smooth gluing of copies of a given Riemannian manifold. On the other hand, it follows from the work of  Banica et al (\cite{banica_jyotish})  that most of known compact quantum groups, including the quantum permutation groups of Wang, can never act faithfully and isometrically on a connected compact Riemannian manifold. All these led the second author of the present paper to make the following conjecture in \cite{Rigidity}, where he also gave some supporting evidence to this conjecture, by proving non-existence of `linear' (see \cite{Rigidity} for the precise definition) action of any genuine compact quantum group on a large class of classical connected  manifolds which are homogeneous spaces of semisimple compact connected Lie groups.\\
{\bf Conjecture I:} It is not possible to have smooth faithful actions of genuine compact quantum groups on $C(M)$ when $M$ is a compact connected smooth manifold.\\

It is perhaps wise to consider first a slightly weaker form of this conjecture, namely:\\
{\bf Conjecture II:} For any compact, oriented, connected Riemannian manifold $M$, the quantum isometry group is classical, i.e. same as $C(ISO(M)).$\\

 The truth of the above conjectures will have two physical implications: firstly, it implies that for a classical system with phase-space modelled on a compact connected manifold, the generalized notion of symmetries in terms of quantum groups coincides with the conventional notion, i.e. symmetries coming from group actions. This gives some kind of consistency of the philosophy of thinking quantum group actions as symmetries. Secondly, it also allows us to describe all the  (quantum) symmetries of a physical model obtained by suitable deformation (at least for the Rieffel-type deformations) of a classical model with connected compact phase space, showing that such quantum symmetries are indeed deformations of the classical (group) symmetries of the original classical model.

 The goal of the present article is to prove Conjecture II for a quite large and important class of manifolds, namely stably parallelizable ones. This class includes for example all compact Lie groups and most of the compact symmetric spaces and also all hypersurfaces.

 As a biproduct, we have developed some new tools for studying quantum isometry
groups of classical manifolds which may also be useful in the more general
framework of noncommutative manifolds. For example, we have given a new
geometric characterization of isometric action by a CQG on compact Riemannian
manfiolds in terms of the Riemmanian inner-product, proving that a CQG action is
isometric if and only if it preserves the inner product. We have also extended
the standard trick of averaging to the framework of quantum group actions,
showing how to make a non-isometric CQG action $\alpha$ on $C(M)$, which is
smooth and satisfies an appropriate commutativity condition ensuring the
existnece of `differential' $d\alpha$ as an equivariant representation  on the
bimodule of one-forms, to be isometric w.r.t. a new `averaged' Riemannian
metric. We believe that all these new techniques will be extremely important for
the general  theory of quantum isometry groups in the long run.

\section{Preliminaries and definitions}
\subsection{Frechet algebra of smooth functions}
Let ${\cal A}$, ${\cal B}$ be  locally convex topological spaces with ${\cal
S}_{\cal A}$ and ${\cal S}_{\cal B}$ being the corresponding families of
seminorms. A seminorm on a *algebra is said to be compatible with * algebraic
structure if $p(ab)\leq p(a)p(b)$ and $p(a^{*})=p(a)$. We denote the algebraic
tensor product of ${\cal A}$ and ${\cal B}$ by ${\cal A} {\otimes} {\cal B}$. Equip
${\cal A} {\otimes} {\cal B}$ with the locally convex topology given by the family
of semi norms  $\{\gamma_{pq}: p\in{\cal S}_{\cal A},q\in{\cal S}_{\cal B}\}$
where $\gamma_{pq}(\xi)=inf\sum{p(a_{i})q(b_{i})}$ and the infimum is taken over
all possible expressions of $\xi=\sum{a_{i}\otimes b_{i}}$. We denote the completion
of ${\cal A} \otimes {\cal B}$ by ${\cal A} \hat{\otimes} {\cal B}$. In case ${\cal A}$
and ${\cal B}$ are algebras(* algebras) and the families of semi norms are also
compatible with algebraic(* algebraic) structure, $ {\cal A}\hat{\otimes} {\cal B}$
is also a topological algebra(*algebra). In case both the algebras are $C^{*}$
algebras ${\cal A} \hat{\otimes} {\cal B}$ is also a $C^{*}$ algebra. For further details of
tensor products of topological vector spaces see \cite{TVS}. Throughout this
paper Sp and $\bar{Sp}$ stand for the linear span and closed linear span of a
subset of a topological vector space respectively.  \\
 \\Let $M$ be a smooth $n$- dimensional compact Riemannian manifold with
boundary (in our case a compact manifold will mean a manifold with boundary
unless otherwise stated ). We denote the algebra of real(complex) valued smooth
functions on $M$ by $C^{\infty}(M)_{\mathbb{R}}$ $(C^{\infty}(M))$. Clearly
$C^{\infty}(M)$ is the complexification of $C^{\infty}(M)_{\mathbb{R}}$ . We
also equip it with a locally convex topology : we say a sequence $f_{n}\in
C^{\infty}(M)$ converges to an $f\in C^{\infty}(M)$ if the following holds:
\\Cover $M$ with finitely many coordinate neighborhoods
$U_{1},U_{2},....,U_{l}$(say). Then for any compact set $K$ within a single
coordinate neighborhood and a multi index $\alpha$,  $\partial^{\alpha}f_{n}
\rightarrow  \partial^{\alpha}f$ uniformly over $K$. Equivalently its a locally convex
topology described by a countable family of seminorms given by: \\ \\
$$p_{i}^{K,\alpha}=\sup_{x\in K}|\partial^{\alpha}f(x)|,$$ where $K$ is a
compact set within $U_{i}$, $\alpha$ is any multi index, $i=1,2,....l$. We
denote this topology by $\tau$. Below we collect a few well known results about
this.
\begin{ppsn}
\label{completeness}
$C^{\infty}(M)$ is complete in $\tau$-topology.\end{ppsn}
{\it Proof:}\\
For the case $C^{\infty}(\Omega)$, where $\Omega$ is an open subset of $\mathbb{R}^{n}$, see example 1.46 of \cite{Rudin}. For a compact manifold going to coordinate charts and using similar arguments, we can easily deduce the result.\\
 \mbox{}\hfill $\sqare$\vspace{1ex} \\
\begin{ppsn}
\label{top alg_iso}
Let $M$ and $N$ be two smooth compact $n$ dimensional manifolds with boundary. Let $F:M \rightarrow N$ be a diffeomorphism. Then $\pi_{F}:C^{\infty}(N) \rightarrow C^{\infty}(M)$ defined by, $\pi_{F}(f)(m) = f(F(m))$ for $f\in C^{\infty}(N)$ and $m\in M$, is an isomorphism between locally convex algebras $C^{\infty}(M) $ and $ C^{\infty}(N) $.
\end{ppsn}
Let $M$ and $N$ be two smooth compact manifolds with boundary. We can view the
elements of $C^{\infty}(M)\otimes C^{\infty}(N)$ as elements of $C^{\infty}(M\times
N )$in the following way:\\
Let $$\sum {f_{i}\otimes g_{i}}\in C^{\infty}(M)\otimes C^{\infty}(N). $$ Then$$\sum ({f_{i}\otimes g_{i}})(m,n)=\sum {f_{i}(m)g_{i}(n)}.$$ With the above notations we have:\\
\begin{ppsn}
\label{Frechet density_cross manifold}
$C^{\infty}(M)\hat{\otimes} C^{\infty}(N)\cong C^{\infty}(M\times N )$ and contains $C^{\infty}(M){\otimes} C^{\infty}(N)$ as a $\tau$ dense subalgebra.
\end{ppsn}

\begin{ppsn}
 let ${\cal Q}$ be a $C^{*}$ algebra. Then we denote the algebra of ${\cal Q}$ valued
smooth functions on $M$ with the obvious topology by $C^{\infty}(M,{\cal Q})$. Then
$C^{\infty}(M,{\cal Q})$ with its obvious locally convex topology is complete. In
fact with the above notations we have $C^{\infty}(M)\hat
{\otimes}{\cal Q}=C^{\infty}(M,{\cal Q})$.
 \end{ppsn}

 Let ${\cal E}_{1}$, ${\cal E}_{2}$ be two locally convex spaces with locally convex topologies $\tau_{1}$, $\tau_{2}$ respectively. Suppose $T$ is a $\tau_{1}-\tau_{2}$ continuous map between them. Also let ${\cal E}$ be another locally convex space. Denote the topology of the projective tensor product between ${\cal E}_{i}$ and ${\cal E}$ by $\tau^{'}_{i}$ respectively for $i=1,2$. Define $(T\otimes id): {\cal E}_{1}\otimes {\cal E} \rightarrow {\cal E}_{2}\otimes {\cal E}$ by $(T\otimes id)(\sum_{i=1}^{k}a_{i}\otimes b_{i})=\sum_{i=1}^{k}(T(a_{i}))\otimes b_{i}.$ Then we have
 \begin{ppsn}
 \label{Ext}
 $(T\otimes id)$ is $\tau^{'}_{1}-\tau^{'}_{2}$ continuous and hence extends to ${\cal E}_{1}\hat{\otimes} {\cal E}$. We will denote the extension by
 $(T\hat{\otimes} id)$.
 \end{ppsn}
\subsection{Basics of compact quantum groups}
A compact quantum group (CQG for short) is a  unital $C^*$ algebra ${\cal S}$ with a coassociative coproduct
(see \cite{Van}) $\Delta$ from ${\cal S}$ to ${\cal S} \hat{\otimes} {\cal S}$ (it is the same as injective tensor product)
such that each of the linear spans of $\Delta({\cal S})({\cal S} \hat{\otimes} 1)$ and that of $\Delta({\cal S}) (1 \hat{\otimes} {\cal S})$ is norm-dense in ${\cal S} \hat{\otimes} {\cal S}$.
From this condition, one can obtain a canonical dense unital $\ast$-subalgebra ${\cal S}_0$ of ${\cal S}$ on which linear maps $\kappa$ and
$\epsilon$ (called the antipode and the counit respectively) are defined making the above subalgebra a Hopf $\ast$ algebra. In fact, we shall always choose this dense
 Hopf $\ast$-algebra to be the algebra generated by the `matrix coefficients' of the (finite dimensional) irreducible unitary representations (to be defined
 shortly) of the CQG. The antipode is an anti-homomorphism and also satisfies $\kappa(a^*)=(\kappa^{-1}(a))^*$ for $a \in {\cal S}_0$.

 It is known that there is a unique state $h$ on a CQG ${\cal S}$ (called the Haar state) which is bi invariant in the sense that $({\rm id} \otimes h)\circ \Delta(a)=(h \otimes {\rm id}) \circ \Delta(a)=h(a)1$ for all $a$. The Haar state need not be faithful in general, though it is always faithful on ${\cal S}_0$ at least.

We say that a CQG ${\cal S}$ (with a coproduct $\Delta$) (co)acts on a unital $C^*$ algebra ${\cal C}$ if there is a unital $C^*$-homomorphism
$\beta : {\cal C} \rightarrow {\cal C} \hat{\otimes} {\cal S}$ such that ${\rm Span}\{ \beta({\cal C})(1 \hat{\otimes} {\cal S})\}$ is norm-dense in ${\cal C} \hat{\otimes} {\cal S}$, and it satisfies the coassociativity
 condition, i.e. $(\beta \otimes {\rm id}) \circ \beta=({\rm id} \otimes \Delta) \circ \beta$. It has been shown in \cite{Podles} that
 there is a unital dense $\ast$-subalgebra ${\cal C}_0$ of ${\cal C}$ such that $\beta$ maps ${\cal C}_0$ into ${\cal C}_0 \otimes {\cal S}_0$ (where ${\cal S}_0$ is the dense
 Hopf $\ast$-algebra mentioned before) and we also have
   $({\rm id} \otimes \epsilon)\circ
 \beta={\rm id}$ on ${\cal C}_0$. In fact, this subalgebra ${\cal C}_0$ comes from the canonical decomposition of ${\cal C}$ into subspaces on each of
which the action
 $\beta$ is equivalent to an irreducible representation. More precisely, ${\cal C}_0$ is the algebraic direct sum of finite dimensional vector spaces ${\cal C}^\pi_i$, say, where $i$ runs over some index set $J_i$, and $\pi$ runs over some subset (say $T$) of the set of (inequivalent) irreducible unitary representations of ${\cal S}$, and the restriction of $\beta$ to ${\cal C}^\pi_i$ is equivalent to the representation $\pi$. Let $\{ a^{(\pi,i)}_j,j=1,...,d_\pi \}$ (where $d_\pi$ is the dimension of the representation $\pi$) be a basis of ${\cal C}^\pi_i$ such that $\beta(a^{(\pi,i)}_j)=\sum_k a^{(\pi,i)}_k \otimes t^\pi_{jk}$,
  for elements $t^\pi_{jk}$ of ${\cal S}_0$. The elements $\{ t^\pi_{jk}, \pi \in T;~j,k=1,...,d_\pi \}$ are called the `matrix coefficients' of the action $\beta$.

 We say that the action $\beta$ is faithful if the $\ast$-subalgebra of ${\cal S}$ generated by elements of the form $(\omega \otimes {\rm id})(\beta(a))$, where $a \in {\cal C}$, $\omega$ being a bounded linear functional on ${\cal C}$, is norm-dense in ${\cal S}$, or, equivalently, the $\ast$-algebra generated by the matrix coefficients is norm-dense in ${\cal S}$.\\
 \begin{ppsn}
 If a CQG ${\cal S}$  acts faithfully on $C(X)$, where $X$ is some compact space,
then the corresponding reduced CQG ${\cal S}_r$ (which has a faithful Haar state)
must be a Kac algebra. In particular, the Haar state of ${\cal S}_r$ and hence of
${\cal S}$ is tracial.
 \end{ppsn}
 For a proof of this fact,  we refer the reader to the Theorem 3.2 of \cite{Metric}.

Let ${\cal H}$ be a Hilbert space and denote by ${\cal B}_{0}({\cal H})$ the $C^{*}$ algebra of compact operators on ${\cal H}$. Consider the multiplier algebra ${\cal M}({\cal B}_{0}({\cal H})\otimes {\cal S})$. This algebra has two natural embeddings into the multiplier algebra ${\cal M}({\cal B}_{0}({\cal H})\otimes {\cal S}\otimes {\cal S})$. The first one is obtained by extending the map $x\mapsto x\otimes 1$. The second one is obtained by composing this map with the flip on the last two factors. We will write $v^{12}$ and $v^{13}$ for the images of an element $v\in {\cal M}({\cal B}_{0}({\cal H})\otimes {\cal S}) $ by these two maps respectively.\\
\begin{dfn}
Let $({\cal S},\Delta)$ be a CQG. A unitary representation of ${\cal S}$ on a Hilbert space ${\cal H}$ is an element $U\in {\cal M}({\cal B}_{0}({\cal H})\otimes {\cal S})$ such that\\
$(id\otimes \Delta)U= U^{12}U^{13}$.
\end{dfn}
We recall some basic facts about representation of CQG's over finite dimensional vector spaces. For details we refer to \cite{Pseudogroup}. By a representation of a CQG ${\cal S}$ over a finite dimensional vector space $\cal K$ we mean a linear map $v:{\cal K}\rightarrow {{\cal K}}\otimes {\cal S}$ such that $$(v\otimes id)v= (id\otimes \Delta)v.$$ We say $v$ is a smooth representation of ${\cal S}$ if actually $v:{\cal K}\rightarrow{\cal K}\otimes {\cal S}_{0}$. Moreover a smooth representation $v$ is said to be non degenerate if $v_{e}= I_{B({\cal K})}$, where $v_{e}:= (id\otimes \epsilon)v$.
\begin{lmma}
\label{f-d rep1}
Let $v$ be a smooth non degenerate representation of ${\cal S}$ acting on ${\cal K}$. Then $v$ is invertible and $v^{-1}=(id\otimes \kappa)v$.
\end{lmma}
{\it Proof:}
see proposition 3.2 of \cite{Pseudogroup}.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
\begin{crlre}
\label{f-d rep2}
If $v$ is smooth non degenerate, Sp $v({\cal K})(1\otimes {\cal S}_{0})=({\cal K} \otimes {\cal S}_{0})$.
\end{crlre}
{\it Proof:}
For $\xi\in {\cal K} \otimes {\cal S}$ write $\xi$ as $v(id\otimes \kappa)v(\xi)$ which belongs to Sp $v({\cal K})(1\otimes {\cal S}_{0})$.\\

 \mbox{}\hfill $\sqare$\vspace{1ex}\\\\
If $v$ and $w$ are two finite dimensional representations of ${\cal S}$ on ${\cal K}$ and ${\cal L}$ respectively, then we can define tensor product of them as defined in \cite{Pseudogroup} which is again a representation on ${\cal K}\otimes {\cal L}$. We call the representation as $v \otimes w$. Then
\begin{lmma}
\label{f-d rep3}
If $v$ and $w$ are smooth and non degenerate, so is $v \otimes w$.
\end{lmma}
{\it Proof:}
follows from the fact that $(v \otimes w)_{e}=v_{e}\otimes w_{e}$.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}  \\
Now denote the tensor product of $v$ with itself by $v_{(2)}$.
\begin{crlre}
$v$ is a smooth non degenerate representation on a finite dimensional linear space ${\cal K}$. Then Sp $v_{(2)}({\cal K}\otimes {\cal K})(1\otimes 1\otimes {\cal S}_{0})= ({\cal K}\otimes {\cal K}\otimes {\cal S}_{0}).$
\end{crlre}
{\bf Fact}: Every irreducible unitary representation is finite dimensional. (see \cite{Van})
\\
\subsection{ $C^{\infty}(M)$ bimodule structure on exterior bundle}
Let $M$ be a compact oriented smooth manifold. Also let
$\Lambda^{k}(C^{\infty}(M))$ be the space of smooth $k$ forms on the manifold
$M$. We equip $\Lambda^{1}(C^{\infty}(M))$ with the natural locally convex
topology induced by the locally convex topology of $C^{\infty}(M)$ given by a
family of seminorms $\{p_{(U,(x_{1},\ldots,x_{n}), K)}\}$, where $
(U,(x_{1},\ldots,x_{n})) $  is  a  local  coordinate chart and $K$ is  a
compact subset and
$p_{(U,(x_{1},\ldots,x_{n}),K)}(\omega):= \sup_{x\in K}|f_{i}(x)|,$ where $f_{i}\in C^{\infty}(M)$ such that $\omega|_{U}=\sum_{i=1}^{n}f_{i}dx_{i}|_{U}.$\\
We also denote this topology by $\tau$. Then note that for any $\tau$ dense subalgebra ${\cal A}$ of $C^{\infty}(M)$, $\Omega^{1}({\cal A})$ is dense in $\Lambda^{1}(C^{\infty}(M)).$\\
Let $\Omega^{k}(C^{\infty}(M))_{u}$ be the space of universal k-forms on the manifold $M$ and $\delta$ be the derivation for the universal algebra of forms for $C^{\infty}(M)$ i.e $\delta: \Omega^{k}(C^{\infty}(M))_{u}\rightarrow \Omega^{k+1}(C^{\infty}(M))_{u}$(see \cite{Landi} for further details).\\

 Then by the universal property $\exists$ a surjective bimodule morphism $\pi\equiv\pi_{(1)}:\Omega^{1}(C^{\infty}(M))_{u}\rightarrow \Lambda^{1}(C^{\infty}(M))$, such that $\pi(\delta g)=dg.$\\
 $\Omega^{1}(C^{\infty}(M))_{u}$ has a $C^{\infty}(M)$ bimodule structure:\\
 $$f(\sum_{i=1}^{n}g_{i}\delta h_{i}) = \sum_{i=1}^{n}fg_{i}\delta h_{i}$$
 $$(\sum_{i=1}^{n}g_{i}\delta h_{i})f = \sum_{i=1}^{n} (g_{i}\delta (h_{i}f)-g_{i}h_{i}\delta f)$$
 As $M$ is compact, there is a Riemannian structure. Using the Riemannian
structure on $M$ we can equip $\Lambda^{1}(C^{\infty}(M))$ with a
$C^{\infty}(M)$ valued inner product
$<<\sum_{i=1}^{n}f_{i}dg_{i},\sum_{i=1}^{n}f_{i}^{'}dg_{i}^{'}>> \in C^{\infty}
(M)$ by the following prescription:\\ for $x\in M$ choose a coordinate
neighborhood $(U,x_{1},x_{2},....,x_{n})$ around $x$ such that
$dx_{1},dx_{2},...,dx_{n}$ is an orthonormal basis for $T_{x}^*M$. Note that the
topology does not depend upon any particular choice of the Riemannian metric.
Then
 $$<<\sum_{i=1}^{n}f_{i}dg_{i},\sum_{i=1}^{n}f_{i}^{'}dg_{i}^{'}>>(x) = (\sum_{i,j,k,l}\bar {f_{i}}f_{j}^{'}(\frac{\bar {\partial f_{i}}}{\partial x_{k}}\frac{{\partial f_{j}^{'}}}{\partial x_{l}}))(x)$$ We see that a sequence $\omega_{n}\rightarrow \omega$ in $\Lambda^{1}(C^{\infty}(M))$ if $<<\omega_{n}- \omega,\omega_{n}- \omega>> \rightarrow 0$ in $\tau$ topology of $C^{\infty}(M). $ With this $\Lambda^{1}(C^{\infty}(M))$ becomes a Hilbert module.\\
 \begin{lmma}
 \label{Frechet_den}
 Let ${\cal B}$ be a $\tau$ dense subalgebra of $C^{\infty}(M)$. Then $\Lambda^{1}({\cal B})$ is dense in the Hilbert module $\Lambda^{1}(C^{\infty}(M))$.
 \end{lmma}
 {\it Proof:}\\
 It is enough to approximate $fdg$ where $f,g\in C^{\infty}(M)$ by elements of $\Lambda^{1}({\cal B})$. By $\tau$ density of ${\cal B}$ in $C^{\infty}(M)$ we can choose sequences $f_{m},g_{m}\in {\cal B}$ such that $f_{m}\rightarrow^{\tau}f$ and $g_{m}\rightarrow^{\tau}g$. We show $f_{m}dg_{m}\rightarrow fdg$ in the Hilbert module i.e. $<<f_{m}dg_{m}-fdg,f_{m}dg_{m}-fdg>>\rightarrow 0$ in $\tau$ topology of $C^{\infty}(M)$.\\
Fix a trivializing neighborhood $U$ of $M$ and a compact set $K$ within it, with the coordinates $x_{1},x_{2},...,x_{n}$. We shall estimate $sup_{x\in K}|\frac{\partial}{\partial x_{1}}<<fdg-f_{m}dg_{m},fdg-f_{m}dg_{m}>>(x)|$.\\
Observe that
 \begin{eqnarray*}
 &&\frac{\partial}{\partial x_{1}}<<fdg-f_{m}dg_{m},fdg-f_{m}dg_{m}>>(x)\\
 &=& \frac{\partial}{\partial x_{1}}<<(f-f_{m})dg_{m}+f_{m}(dg-dg_{m}),(f-f_{m})dg-f_{m}(dg-dg_{m})>>(x)\\
 &=& I+II+III+IV\\
 \end{eqnarray*}
 Where
 \begin{eqnarray*}
 I&=& \frac{\partial}{\partial x_{1}}[\sum_{i}(f-f_{m})(x)\overline{(f-f_{m})}(x)\frac{\partial g}{\partial x_{i}}\overline{\frac{\partial g}{\partial x_{i}}}(x)]\\
 &=&\sum_{i} \frac{\partial}{\partial x_{1}}(\overline{(f-f_{m})}(f-f_{m}))(x)(\frac{\partial g}{\partial x_{i}}\overline{\frac{\partial g}{\partial x_{i}}})(x)\\
 &+& \sum_{i}(f-f_{m})(x)\overline{(f-f_{m})}(x) \frac{\partial}{\partial x_{1}}(\frac{\partial g}{\partial x_{i}}\overline{\frac{\partial g}{\partial x_{i}}})(x)\\
 \end{eqnarray*}
 \begin{eqnarray*}
 II&=&\frac{\partial}{\partial x_{1}}[\sum_{i}\overline{(f-f_{m})}(x)\overline{\frac{\partial g}{\partial x_{i}}}(x)f_{m}(x)\frac{\partial}{\partial x_{i}}(g-g_{m})(x)]\\
 &=& \sum_{i}\overline{(f-f_{m})}(x)\frac{\partial}{\partial x_{1}}(\overline{\frac{\partial g}{\partial x_{i}}}(x)f_{m}(x))\frac{\partial}{\partial x_{i}}(g-g_{m})(x)\\
 &+& \sum_{i}\frac{\partial}{\partial x_{1}}\overline{(f-f_{m})}(x)(\overline{\frac{\partial g}{\partial x_{i}}}(x)f_{m}(x))\frac{\partial}{\partial x_{i}}(g-g_{m})(x)\\
 &+& \sum_{i}\overline{(f-f_{m})}(x)(\overline{\frac{\partial g}{\partial x_{i}}}(x)f_{m}(x))\frac{\partial}{\partial x_{1}}\frac{\partial}{\partial x_{i}}(g-g_{m})(x).
 \end{eqnarray*}
 and so on.

Using the above expressions and the definition of convergence in $\tau$ topology which implies in particular $\frac{\partial f_{m}}{\partial x_{i}}\rightarrow \frac{\partial f}{\partial x_{i}}$ and $\frac{\partial g_{m}}{\partial x_{i}}\rightarrow \frac{\partial g}{\partial x_{i}}$ uniformly over $K$ for all $i$, we conclude that $\Lambda^{1}({\cal B})$ is dense in $\Lambda^{1}(C^{\infty}(M))$.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
For the following discussion reader might consult \cite{Landi}.\\
 Let ${\cal E}_{1}$ and ${\cal E}_{2}$ be two Hilbert modules over a topological algebra ${\cal A}$. Then we can form the interior tensor product of ${\cal E}_{1},{\cal E}_{2}$ by the following prescription:\\\\
 Define an ${\cal A}$-valued inner product on ${\cal E}_{1}\otimes {\cal E}_{2}$ by $$<<\sum{\xi_{i}\otimes \eta_{i}},\sum{\xi_{j}^{'}\otimes \eta_{j}^{'}}>> = \sum_{i,j}<<\eta_{i},<<\xi_{i},\xi_{j}^{'}>>\eta_{j}^{'}>>$$ Let ${\cal I}= \{\xi\in {\cal E}_{1}\otimes {\cal E}_{2}$ such that $<<\xi,\xi>>=0\}$. Then ${\cal I}$ is a submodule of ${\cal E}_{1}\otimes {\cal E}_{2}$ (It has an obvious ${\cal A-{\cal A}}$ bimodule structure). We denote the completion of $\frac{{\cal E}_{1}\otimes {\cal E}_{2}}{\cal I}$ by ${\cal E}_{1}\otimes_{in} {\cal E}_{2}$.\\ \\
 Now with the notations introduced before, let $\Omega^{2}(C^{\infty}(M))_{u}=\Omega^{1}(C^{\infty}(M))_{u}\otimes_{C^{\infty}(M)} \Omega^{1}(C^{\infty}(M))_{u}$ and $\Omega^{k}(C^{\infty}(M))_{u}=\Omega^{k-1}(C^{\infty}(M))_{u}\otimes_{C^{\infty}(M)} \Omega^{1}(C^{\infty}(M))_{u}$.\\
 and $\Omega^{k}(C^{\infty}(M))=\Omega^{k-1}(C^{\infty}(M)) \otimes_{in} \Omega^{1}(C^{\infty}(M))$, for $k \geq 2$, $\Omega^{1}(C^{\infty}(M))\equiv \Lambda^{1}(C^{\infty}(M))$. $$\dot{\Omega}(C^{\infty}(M))=\oplus_{k\geq 0} \Omega^{k}(C^{\infty}(M)).$$

By the universality of $\Omega^{2}(C^{\infty}(M))_{u}$, we have a surjective bimodule morphism $\pi_{(2)}:\Omega^{2}(C^{\infty}(M))_{u}\rightarrow \Omega^{2}(C^{\infty}(M))$.\\
Let ${\cal J}_{2}$ be a submodule of $\Omega^{2}(C^{\infty}(M))$ given by ${\cal J}_{2}= \{\pi_{(2)}(\delta\omega)|\pi(\omega)=0 \ for \ \omega \in \Omega^{1}(C^{\infty}(M))_{u}\}$ . In fact it is closed. Denote $\frac{\Omega^{2}(C^{\infty}(M))}{{\cal J}_{2}}$ by $\Lambda^{2}(C^{\infty}(M)).$ Similarly $\Lambda^{k}(C^{\infty}(M))= \frac{\Omega^{k}(C^{\infty}(M))}{{\cal J}_{k}}$ where ${\cal J}_{k}= \{\pi_{(k)}(\delta\omega)| \pi_{(k-1)} (\omega)= 0 \ for \ \omega \in \Omega^{k-1}(C^{\infty}(M))_{u}\} $. If $\omega$ and $\eta$ belong to $\Omega^{1}(C^{\infty}(M))$, sometimes we denote the image of $\omega\otimes \eta$ in $\Omega^{2}(C^{\infty}(M))$ by $\omega \eta$ and in $\Lambda^{2}(C^{\infty}(M))$ by $\omega \wedge \eta$. Similar notations will be used for products in $\Omega^{k}(C^{\infty}(M))$ and $\Lambda^{k}(C^{\infty}(M))$. With this, the familiar de Rham differential is given by
 \begin{eqnarray*}
 &&d:\Lambda^{k}(C^{\infty}(M))\rightarrow \Lambda^{k+1}(C^{\infty}(M))\\
 &&[\pi_{(k)}(\omega)]\rightarrow [\pi_{(k+1)}(\delta \omega)] \ ([\xi]:=\xi+{\cal J}_{k} \ for \ \xi\in \Omega^{k}(C^{\infty}(M)))
 \end{eqnarray*}

\begin{lmma}
\label{decomposition_forms}
$\Omega^{k}(C^{\infty}(M))= \Lambda^{k}(C^{\infty}(M))\oplus {\cal J}_{k}.$\end{lmma}
{\it Proof:}\\
We do the proof for $k=2$. For $k\geq 3$, proofs will be similar.\\
Let $\pi_{(2)}(\delta f\otimes \delta g)\in  \Omega^{2}(C^{\infty}(M))$.
Observe that
\begin{eqnarray*}
&& \pi_{(2)} (\delta(\delta(gf)))=0\\
&\Rightarrow& \pi_{(2)}(\delta(\delta g f+g\delta f))=0\\
&\Rightarrow& \pi_{(2)}(\delta g \otimes \delta f + \delta(\delta gf))=0\\
&\Rightarrow& \pi_{(2)}(\delta(\delta (g)f))=-\pi_{(2)}(\delta g \otimes \delta f)\\
\end{eqnarray*}
Hence $\pi_{(2)}(\delta(f\delta g-\delta gf))=\pi_{(2)}(\delta f\otimes \delta g +
\delta g\otimes \delta f)$. But $\pi(f\delta g - \delta gf)=0$ in
$\Omega^{1}(C^{\infty}(M))$. So  $\frac{1}{2}\pi_{(2)}(\delta f\otimes \delta g +
\delta g\otimes \delta f)\in {\cal J}_{(2)}$. Similarly $\frac{1}{2}\pi_{(2)}(\delta
f\otimes \delta g - \delta g\otimes \delta f)\in \Lambda^{2}(C^{\infty}(M))$.\\
Hence $\pi_{(2)}(\eta \delta f\otimes \delta g)= \frac{1}{2}\pi_{(2)} (\eta (\delta f\otimes \delta g + \delta g\otimes \delta f)) + \frac{1}{2} \pi_{(2)}(\eta (\delta f\otimes \delta g - \delta g\otimes \delta f))$.\\ Also by definition $<<\pi_{(2)}(\delta f\otimes \delta g + \delta g\otimes \delta f),\pi_{(2)}(\delta f\otimes \delta g - \delta g\otimes \delta f)>>=0$ \\
\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
Let ${\cal A}$ be any $\tau$ dense subalgebra of $C^{\infty}(M)$. Then observe that as our arguments of the above proof is completely algebraic, we can make the following corollary:\\
\begin{crlre}
$\Omega^{k}({\cal A})= \Lambda^{k}({\cal A}) \oplus {{\cal J}}^{{\cal A}}_{k}$, where $\Omega^{1}({\cal A})=\{\sum f_{i}dg_{i}, f_{i},g_{i}\in {\cal A}\}$,  $$\Omega^{k}({\cal A})=\Omega^{k-1}({\cal A})\otimes_{{\cal A}} \Omega^{1}({\cal A}),$$
$${{\cal J}}^{{\cal A}}_{k}=\{\pi_{(k)}(\delta\omega)| \pi_{(k-1)}\omega= 0 \ for \ \omega \in \Omega^{k-1}({\cal A})_{u}\} \ and$$
$$\Lambda^{k}({\cal A})=\frac{\Omega^{k}({\cal A})}{{{\cal J}}^{{\cal A}}_{k}}$$
\end{crlre}
Also note that $\Omega^{k}({\cal A})$, $\Lambda^{k}({\cal A})$ and ${\cal J}^{{\cal A}}_{k}$ are dense in the Hilbert modules $\Omega^{k}(C^{\infty}(M))$, $\Lambda^{k}(C^{\infty}(M))$ and ${\cal J}_{k}$ respectively. For example, the density of ${\cal J}^{{\cal A}}_{2}$ follows from the algebraic description ${\cal J}^{{\cal A}}_{2}={\rm Sp}\{ dfdg+dgdf, f,g\in {\cal A}\}$, which is contained in the proof of the Lemma \ref{decomposition_forms}. \\
We define $p:\Omega^{2}(C^{\infty}(M))\rightarrow {\cal J}_{2}$ by $$p\pi_{(2)}(\sum_{i=1}^{k}h_{i}\delta f_{i}\otimes \delta g_{i})=\frac{1}{2}\pi_{(2)}(\sum_{i=1}^{k}h_{i}(\delta f_{i}\otimes \delta g_{i}+\delta g_{i}\otimes \delta f_{i})).$$
Then $p$ is a projection onto the closed complemented submodule ${\cal J}_{2}$ of $\Omega^{2}(C^{\infty}(M))$. \\ \\
Now if ${\cal Q}$ is a $C^{*}$ algebra then $\Lambda^{2}(C^{\infty}(M))\otimes {\cal Q}$ has a natural $C^{\infty}(M)\hat{\otimes}{\cal Q}$ bimodule structure. The left action is given by $$(\sum_{i} f_{i}\otimes q_{i})(\sum_{j}[\pi_{(2)}(\omega_{j})]\otimes q_{j}^{'})= (\sum_{i,j}[\pi_{(2)}(f_{i}\omega_{j})]\otimes q_{i}q_{j}^{'} )$$
The right action is similarly given. The inner product is given by
$$<<\sum_{i}\omega_{i}\otimes q_{i},\sum_{j}\omega_{j}^{'}\otimes
q_{j}^{'}>>=\sum_{i,j}<<\omega_{i},\omega_{j}^{'}>>\otimes q_{i}^{*}q_{j}^{'}.$$
Topology on $\Lambda^{2}(C^{\infty}(M))\otimes {\cal Q}$ is given by requiring
$\omega_{n}\rightarrow \omega$ if and only if $<<\omega_{n}- \omega,\omega_{n}-
\omega>>\rightarrow 0$ in $C^{\infty}(M)\hat{\otimes}{\cal Q}$ or $C^{\infty}(M,{\cal Q})$. We
denote the corresponding completion by $\Lambda^{2}(C^{\infty}(M))\hat{\otimes}
{\cal Q}$.\\
Now we introduce the Hodge star operator, which is a pointwise isometry $* =
*_{x}:\Lambda^{k}T^{*}_{x}M \rightarrow \Lambda^{n-k}T^{*}_{x}M $. Choose a
positively oriented orthonormal basis $\{\theta^{1}, \theta^{2},..., \theta^{n}
\}$ of $T^{*}_{x}M$. Since $*$ is a linear transformation it is enough to define
$*$ on a basis element $\theta^{i_{1}}\wedge \theta^{i_{2}}\wedge...\wedge
\theta^{i_{k}}(i_{1}<i_{2}<...<i_{k})$ of $\Lambda^{k}T^{*}_{x}M$. Note that \\
\begin{eqnarray*}
dvol(x)&=& \sqrt{det (<\theta^{i},\theta^{j}>)}\theta^{1}\wedge \theta^{2}\wedge ...\wedge \theta^{n}\\
&=& \theta^{1}\wedge \theta^{2}\wedge ...\wedge \theta^{n}
\end{eqnarray*}
\begin{dfn}
$*(\theta^{i_{1}}\wedge \theta^{i_{2}}\wedge...\wedge \theta^{i_{k}})=\theta^{j_{1}}\wedge \theta^{j_{2}}\wedge...\wedge \theta^{j_{n-k}}$ where $\theta^{i_{1}}\wedge \theta^{i_{2}}\wedge...\wedge \theta^{i_{k}}\wedge\theta^{j_{1}}..\wedge \theta^{j_{n-k}}= dvol(x)$.\\
\end{dfn}
Since we are using $\mathbb{C}$ as the scalar field, we would like to define $\bar{\omega}$ for a $k$ form $\omega$. In the set-up introduced just before the definition we have some scalars $c_{i_{1},...,i_{k}}$ such that $\omega(x)=\sum c_{i_{1},...,i_{k}}\theta^{i_{1}}\wedge \theta^{i_{2}}\wedge...\wedge \theta^{i_{k}} $. Then define $\bar{\omega}$ to be $\bar{\omega}(x)=\sum \bar{c}_{i_{1},...,i_{k}}\theta^{i_{1}}\wedge \theta^{i_{2}}\wedge...\wedge \theta^{i_{k}}.$
Then the equation $<<\omega,\eta>>=*(\bar{\omega} \wedge *\eta)$ defines an inner product on the Hilbert module $\Lambda^{k}(C^{\infty}(M))$ for all $k=1,...,n$ which is the same as the $C^{\infty}(M)$ valued inner product defined earlier. Then the Hodge star operator is a unitary between two Hilbert modules $\Lambda^{k}(C^{\infty}(M))$ and $\Lambda^{n-k}(C^{\infty}(M))$ i.e. $<<*\omega,*\eta>>= <<\omega,\eta>>$. For further details about the Hodge star operator we refer the reader to \cite{Rosen}.

 Hence we have $$(*\otimes id):\Lambda^{k}(C^{\infty}(M))\otimes {\cal Q}\rightarrow \Lambda^{n-k}(C^{\infty}(M))\otimes{\cal Q}.$$ Since Hodge$*$ operator is an isometry, $(*\otimes id)$ is continuous with respect to the Hilbert module structure of $ \dot{\Lambda}(C^{\infty}(M))\hat\otimes{\cal Q}$. So we have $$(*\otimes id):\Lambda^{k}(C^{\infty}(M))\hat{\otimes} {\cal Q}\rightarrow \Lambda^{n-k}(C^{\infty}(M))\hat{\otimes} {\cal Q}.$$
\subsection{\bf Representation of CQG on a Hilbert module over a topological algebra}
Let ${\cal E}$ be a Hilbert bimodule over a topological algebra ${\cal A}$ and ${\cal Q}$ be a compact quantum group. $ {\cal E}\otimes {\cal Q} $ has a natural ${\cal A}\otimes {\cal Q}$ bimodule structure. We denote the corresponding completion by ${\cal E}\hat\otimes {\cal Q}$. Let $\alpha$ be an action of ${\cal Q}$ on ${\cal A}$  i.e. $\alpha:{\cal A}\rightarrow {\cal A}\hat\otimes {\cal Q}$ such that $(\alpha\otimes id)\alpha=(id\otimes \Delta)\alpha$ and Sp $\{\alpha({\cal A} )(1 \otimes {\cal Q} )\}$ is dense in ${\cal A}  \hat{\otimes}{\cal Q} $. Then \\
\begin{dfn}
 $\Gamma$ is an $\alpha$-equivariant representation of ${\cal Q}$ on ${\cal E}$ if $\Gamma:{\cal E}\rightarrow {\cal E}\hat\otimes {\cal Q}$ be such that\\ 1. $\Gamma(a.\xi)=\alpha(a)\Gamma(\xi)$ and $\Gamma(\xi.a)=\Gamma(\xi)\alpha(a)$ where $a\in{\cal A}$. \\
2. $\tilde{\Gamma}$ defined by $\tilde{\Gamma}(\xi\otimes q)=\Gamma(\xi)(1\otimes q)$ satisfies $$(id\otimes \Delta)\tilde{\Gamma}=\tilde{\Gamma}^{12}\tilde{\Gamma}^{13}.$$
3. $<<\Gamma(\xi),\Gamma(\xi^{'})>>=\alpha<<\xi,\xi^{'}>>$.\\
4. $\bar{Sp} \Gamma({\cal E})(1\otimes {\cal Q})={\cal E} \hat{\otimes} {\cal Q}$.
\end{dfn}
Sometimes we will simply call $\Gamma$ to be just unitary if the action is understood from the situation.
\begin{lmma}
\label{unitary_rep}
 Now assume that $\Gamma$ is an $\alpha$ equivariant unitary representation of ${\cal Q}$ on ${\cal E}$ such that ${\cal E}$ has a dense submodule ${\cal B}$ over which $\Gamma$ is algebraic i.e. $\Gamma({\cal B})\subseteq {\cal B}\otimes {\cal Q}_{0}$ and  where ${\cal Q}_{0}$ is the canonical hopf algebra of ${\cal Q}$, such that ${\cal B}$ is a direct sum of finite dimensional linear spaces. Also assume that over each of those finite dimensional linear spaces $\Gamma$ is total. Then it is easy to see that Sp $\Gamma({\cal B})(1\otimes {\cal Q}_{0})= {\cal B}\otimes {\cal Q}_{0}$. Using Sweedler's notation we can define $\Gamma_{(2)}: {\cal B}\otimes {\cal B}\rightarrow {\cal B}\otimes {\cal B}\otimes {\cal Q}_{0} $ as $$\Gamma_{(2)}(\xi\otimes \eta)= \xi_{(1)}\otimes \eta_{(1)}\otimes \xi_{(2)} \eta_{(2)}.$$
With the above assumptions on $\Gamma$, $\Gamma_{(2)}$ is $\alpha$ equivariant over ${\cal B}\otimes {\cal B}$.
\end{lmma}
{\it Proof:}\\
claim: $<<\Gamma_{(2)}(\omega\otimes \eta),\Gamma_{(2)}(\omega^{'}\otimes \eta^{'})>>= <<\Gamma (\eta),<<\Gamma(\omega),\Gamma(\omega^{'})>>\Gamma(\eta^{'})>>$.\\
{\it proof of the claim}: $$ <<\Gamma_{(2)}(\omega\otimes \eta),\Gamma_{(2)}(\omega^{'}\otimes \eta^{'}))>>=<<\omega_{1}\otimes \eta_{1},\omega_{1}^{'}\otimes \eta_{1}^{'}>>\otimes \eta_{2}^{*} \omega_{2}^{*}\omega_{2}^{'}\eta_{2}^{'}.$$ On the other hand by  R.H.S.=$<<\eta_{1},<<\omega_{1},\omega_{1}^{'}>>\eta_{1}^{'}>>\otimes  \eta_{2}^{*} \omega_{2}^{*}\omega_{2}^{'}\eta_{2}^{'}$ which is same as L.H.S. by the definition of the inner product on ${\cal E}\otimes {\cal E}$.\\
Now as $\Gamma$ is $\alpha$ equivariant $<<\Gamma(\omega),\Gamma(\omega^{'})>>=\alpha<<\omega,\omega^{'}>>.$ Using this and the fact that $<<\Gamma(\eta),\alpha(a)\Gamma(\eta^{'})>>=\alpha(a)<<\Gamma(\eta),\Gamma(\eta^{'})>>$ for all $\eta,\eta^{'}\in {\cal E}$ and $a\in {\cal A}$, we complete the proof of the lemma.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
\begin{crlre}
\label{equi_rep}
$\Gamma_{(2)}:{\cal E}\otimes_{in}{\cal E}\rightarrow {\cal E}\otimes_{in}{\cal E}\hat{\otimes}{\cal Q}$ is $\alpha$ equivariant unitary representation on ${\cal E}\otimes_{in}{\cal E}$
\end{crlre}
{\it Proof:}\\
We note that $\Gamma_{(2)}$ is nothing but tensor product of the representation
$\Gamma$ with itself. Now by assumption ${\cal B}$ is a direct sum of finite
dimensional linear spaces over each of which $\Gamma$ is a representation and
total. Then $\Gamma_{(2)}$ is a representation over tensor products of those
finite dimensional spaces and hence total by Corollary 2.11 . So we conclude
that Sp $\Gamma_{(2)}({\cal B}\otimes {\cal B})(1\otimes {\cal Q}_{0})={\cal B}\otimes {\cal B}\otimes {\cal Q}_{0} $.
Now the claim follows from $\alpha$ equivariance of $\Gamma_{(2)}$ on ${\cal B}\otimes
{\cal B}$ and density of ${\cal B}\otimes {\cal B}$ in ${\cal E}\otimes_{in}{\cal E}$ and the density of
${\cal B}\otimes {\cal B}\otimes {\cal Q}_{0}$ in ${\cal E}\otimes_{in} {\cal E} \hat{\otimes} {\cal Q}.$\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
We can prove similar results for $\Gamma_{(k)}$(whose definition is obvious).\\
\begin{rmrk}
The converse of the lemma \ref{unitary_rep} holds in the following sense:\\
Given an $\alpha$ equivariant unitary representation $\Gamma:{\cal E}\rightarrow {\cal E}\hat {\otimes}{\cal Q}$ of a CQG ${\cal Q}$, $\exists$ a dense submodule ${\cal E}_{0}$ over which $\Gamma$ is algebraic. This follows by decomposition of ${\cal E}$ into spectral subspaces of the representation $\Gamma$.
\end{rmrk}

\subsection{\bf Basics of normal bundle}
We state some basic definitions and  facts about normal bundle of a  manifold
without boundary embedded in some euclidean space. For details of the topic we
refer to \cite{Shastri}.  Let $M\subseteq \mathbb{R}^{N}$ be a smooth embedded
submanifold of $\mathbb{R}^{N}$. For each point $x\in M$ define the space of
normals to $M$ at $x$ by $$ N_{x}(M)=\{v\in \mathbb{R}^{N}: v \perp
T_{x}(M)\}.$$
The total space ${\cal N}(M)$ of the normal bundle is defined by
$${\cal N}(M)=\{(x,v)\in M\times \mathbb{R}^{N}; v\perp T_{x}(M)\}$$ with the
projection $\pi$ on the first coordinate. Then define
${\cal N}_{\epsilon}(M)=\{(x,v)\in {\cal N}(M);||v||\leq \epsilon\}$. With the
introduced notations we have\\
Fact: ${\cal N}(M)$ is a manifold of dimension $N$. (see page no. 153 of \cite{Shastri}).\\
\begin{lmma}
\label{global_diff}
Let $B_{\epsilon}^{N-n}(0)$ be a closed euclidean $(N-n)$ ball of radius $\epsilon$. If $M$ is a compact n-manifold without boundary embedded in some euclidean space $\mathbb {R}^{N}$ such that it has trivial normal bundle, then there exists an $\epsilon\textgreater 0$ and a global diffeomorphism $F: M\times B_{\epsilon}^{N-n}(0)\rightarrow {\cal N}_{\epsilon}(M)\subseteq \mathbb{R}^{N}$  given by $$F(x,u_{1},u_{2},...,u_{N-n})=x+ \sum_{i=1}^{N-n}\langle\xi_{i}(x),u_{i}\rangle \xi_i(x)$$ where $(\xi_{1}(x),...,\xi_{N-n}(x))$ is an orthonormal basis of $N_{x}(M)$ for all $x$, and $x\mapsto \xi_{i}(x)$ is smooth.
\end{lmma}
This is a consequence of the tubular neighborhood lemma. For the proof see \cite{Shastri}.\\
\vspace{0.2in}\\
We now introduce the notion of stably parallelizable manifolds.
\begin{dfn}
A manifold $M$ is said to be stably parallelizable if its tangent bundle is stably trivial.
\end{dfn}
We recall the following from \cite{Singhof}:\\
\begin{ppsn}
A manifold $M$ is stably parallelizable if and only if it has trivial normal
bundle when embedded in a Euclidean space of dimension higher than twice the
dimension of $M$.
\end{ppsn}
{\bf Proof}: see discussion following the Theorem (7.2) of \cite{Kol}.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
We note that this class of manifolds is extremely rich. For example it includes ( see \cite{Singhof} and \cite{Kol} and references therein):\\
1. all Lie groups.\\
\vspace{0.05in}\\
2. all homotopy spheres.\\
\vspace{0.05in}\\
3. all hypersurfaces in Euclidean space.\\
\vspace{0.05in}\\
4. all homogeneous spaces of the form $G/H$, where $G$ is simple simply connected Lie group and $H$ is closed, connected, abelian subgroup (e.g. maximal torus).\\
\vspace{0.05in}\\
5. and also all $G/H$ as listed in Theorem 2 of \cite{Singhof}.

So in particular the result of this paper strengthens the result obtained in \cite{Rigidity}.
\section{Smooth action of CQG on a manifold}
Throughout Sections 3 and 4, $M$ denotes a smooth, compact, oriented manifold possibly with boundary.
\subsection{Decomposing a smooth action into its spectral subspaces}
\begin{dfn}
An action  $\alpha$ on $C(M)$ by a CQG ${\cal Q}$ is said to be smooth if $\alpha (C^{\infty}(M))\subseteq C^{\infty}(M,{\cal Q})$ and the span of $\{ \alpha(f)(1 \otimes q):~f \in C^\infty(M),~q \in {\cal Q}\}$ is $\tau$-dense in $C^\infty(M) \hat{\otimes} {\cal Q}\equiv C^\infty(M, {\cal Q})$.
\end{dfn}
\begin{lmma}
\label{frechet_cont}
If $\alpha$ is smooth, $\alpha$ is $\tau$ continuous.
\end{lmma}
{\it Proof:}\\
Let $f_{n}\rightarrow f$ in $\tau$ topology of $C^{\infty}(M)$ and $\alpha(f_{n})\rightarrow \xi$ in $\tau$ topology of $C^{\infty}(M,{\cal Q})$.\\
Then $f_{n}\rightarrow f$ in norm topology of $C(M)$. So by the $C^{*}$ continuity of $\alpha$, $\alpha(f_{n})\rightarrow \alpha(f)$.\\
Similarly, $\alpha(f_{n})\rightarrow \xi$ in the norm topology of $C(M,{\cal Q})$.\\
So $\alpha(f)= \xi$ and by the closed graph theorem $\alpha$ is $\tau$ continuous.\\
  \mbox{}\hfill $\sqare$\vspace{1ex} \\

For a smooth action of a CQG ${\cal Q}$ we have the following result whose proof goes along the same line as Theorem 1.5 of \cite{Podles}. However, for  completeness and the importance of the $\tau$ topology in the present paper we decided to sketch the proof briefly.
\begin{lmma}
\label{dense_alg}
For a smooth action  $\alpha$ of ${\cal Q}$ on $C(M)$ we have a $\tau$ dense subalgebra ${\cal A}$ of $C^{\infty}(M)$ over which the action is algebraic as in the previous sense.
\end{lmma}
{\it Proof:}\\
Step 1:  For any state $\phi\in {\cal Q}^{*}$,
$\alpha_{\phi}(C^{\infty}(M))\subseteq (C^{\infty}(M))$, where
$\alpha_{\phi}\equiv (id\otimes \phi)\alpha$.
We use the fact that $C^{\infty}(M)\otimes {\cal Q}$ is $\tau$ dense in $C^{\infty}(M,{\cal Q}).$\\
Let $C_{\phi}:=(id\otimes \phi):C^{\infty}(M)\otimes {\cal Q}\rightarrow C^{\infty}(M)$ be the contraction map with respect to $\phi$.
Then $C_{\phi}$ extends by $\tau$ continuity to the whole of $C^{\infty}(M,{\cal Q})$. Indeed for a multi index $\beta$,$x\in M$ choose a coordinate neighborhood $U$ of $x$ and a compact set $K$ within it.\\
Let $\sum{f_{i}\otimes q_{i}}\in C^{\infty}(M,{\cal Q})$. Using the fact that on $C^{\infty}(M)\otimes {\cal Q}, (id\otimes \phi)(\partial^{\beta}\otimes 1)=\partial^{\beta}(id \otimes \phi) $,we have
$$sup_{x\in K}|\partial^{\beta}C_{\phi}(\sum{f_{i}\otimes q_{i}})(x)| = sup_{x\in K}|C_{\phi}(\partial^{\beta}\otimes 1)(\sum{f_{i}\otimes q_{i}})(x)|.$$
Now since $C_{\phi}: C(M,{\cal Q})\rightarrow C(M)$ is a continuous map and $\partial^{\beta}f\in C(M)$,
$$|C_{\phi}(\sum({\partial^{\beta}f_{i}})\otimes q_{i})(x)|\leq C||\sum({\partial^{\beta}f_{i})(x)q_{i}}||$$ for some $C\textgreater 0$.\\
So if $\psi_{n}$ is $\tau$ Cauchy in $C^{\infty}(M,{\cal Q})$, $C_{\phi}(\psi_{n})$
is also so in $C^{\infty}(M)$. So for $\omega=\tau-lim \omega_{n}$ defining
$C_{\phi}(\omega)=\tau-lim C_{\phi}(\omega_{n})$ and using the completeness of
$C^{\infty}(M)$ in $\tau\textit{}$, we conclude that
$C_{\phi}(C^{\infty}(M,{\cal Q}))\subseteq C^{\infty}(M)$.\\
step2:   Now we can follow the lines of  proof of Theorem 1.5 of \cite{Podles}.
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
\subsection{Defining $d\alpha$ for a smooth action $\alpha$}
Define $d\alpha (df):= (d\otimes id)\alpha(f)$ for all $f\in C^{\infty}(M)$.
\begin{thm}
\label{def_dalpha}
$d\alpha$ extends to a well defined bimodule morphism from $\Omega^{1}(C^{\infty}(M))$ to $\Omega^{1}(C^{\infty}(M))\hat{\otimes} {{\cal Q}}$ satisfying $d\alpha(df)= (d\otimes id)\alpha(f)$ if and only if
\begin{eqnarray}
\label{dalpha_eq}
(\nu\otimes id)\alpha(f)\alpha(g)= \alpha(g) (\nu\otimes id)\alpha(f)
\end{eqnarray}
for all $f,g\in C^{\infty}(M)$ and all smooth vector fields $\nu$ on $M$.
\end{thm}
{\bf proof:}\\
Suppose that $d\alpha$ is a well defined bimodule morphism.\\
Then $d\alpha (df.g)= (d\otimes id)\alpha(f).\alpha(g)$.\\
Also $d\alpha (g.df)= \alpha(g).(d\otimes id)\alpha(f)$.\\
But $df.g= g.df$ in $\Omega^{1}(C^{\infty}(M))$.\\
Hence $(d\otimes id)\alpha(f).\alpha(g)= \alpha(g).(d\otimes id)\alpha(f), \ \forall f,g\in C^{\infty}(M)$\\
Observe that as $\nu$ is a smooth vector field, $\nu$ is a $\tau$ continuous map from $C^{\infty}(M)$ to $C^{\infty}(M)$. Thus it is enough to prove (1) for $f,g$ belong to the $\tau$ dense subalgebra ${\cal A}$ as in Lemma \ref{dense_alg}.\\
Let $\alpha(f)=f_{(1)}\otimes f_{(2)}$ and $\alpha(g)=g_{(1)}\otimes g_{(2)}$(Sweedler's
notation).\\
Let $x\in M$ and $(U,x_{1},..,x_{n})$ be a coordinate neighborhood around $x$.\\
Then $[(d\otimes id)\alpha(f)\alpha(g)](x)= \sum_{i=1}^{n}g_{(1)}(x)\frac{\partial f_{(1)}}{\partial x_{i}}(x)f_{(2)}g_{(2)} dx_{i}|_{x}$.\\
So
\begin{eqnarray}
&&[(d\otimes id)\alpha(f)\alpha(g)](x)=[\alpha(g)(d\otimes id)\alpha(f))](x) \nonumber \\
&\Rightarrow& g_{(1)}(x)\frac{\partial f_{(1)}}{\partial x_{i}}(x) f_{(2)}g_{(2)}= g_{(1)}(x)\frac{\partial f_{(1)}}{\partial x_{i}}(x) g_{(2)}f_{(2)}
\end{eqnarray}
for all $i=1(1)n$\\
Now let $a_{i}\in C^{\infty}(M)$ for $i=1(1)n$ such that $\nu(x)=\sum_{i=1}^{n}a_{i}(x)\frac{\partial}{\partial{x_{i}}}|_{x}$ for all $x\in U$.\\
So
\begin{eqnarray*}
&&[(\nu\otimes id)\alpha(f)\alpha(g)](x)\\
&=& \sum_{i=1}^{n}a_{i}(x)\frac{\partial f_{(1)}}{\partial x_{i}}(x)g_{(1)}(x) f_{(2)}g_{(2)}\\
\end{eqnarray*}
and
\begin{eqnarray*}
&&[\alpha(g)(\nu\otimes id)\alpha(f)](x)\\
&=& \sum_{i=1}^{n}a_{i}(x)\frac{\partial f_{(1)}}{\partial x_{i}}(x)g_{(1)}(x) g_{(2)}f_{(2)}\\
\end{eqnarray*}
Hence by (2) $[\alpha(g)(\nu\otimes id)\alpha(f)](x)=[(\nu\otimes id)\alpha(f)\alpha(g)](x)$ for all $x\in M$\\
i.e. $[\alpha(g)(\nu\otimes id)\alpha(f)]=[(\nu\otimes id)\alpha(f)\alpha(g)]$ for all $f,g\in {\cal A}$.

\vspace{0.025in}
Now let us prove the converse direction.\\
Let $x\in M$ and $(U,x_{1},...,x_{n})$ be a coordinate neighborhood around
it.\\
Choose smooth vector fields $\nu_{i}$'s on $M$ which are $\frac{\partial}{\partial x_{i}}$ on $U$.\\
So $[\alpha(g)(\nu_{i}\otimes id)\alpha(f)](x)=\frac{\partial f_{(1)}}{\partial x_{i}}(x)g_{(1)}(x)g_{(2)}f_{(2)}$ and $[(\nu_{i}\otimes id)\alpha(f)\alpha(g)](x)=\frac{\partial f_{(1)}}{\partial x_{i}}(x)g_{(1)}(x)f_{(2)}g_{(2)}$.\\
Hence by the assumption
\begin{eqnarray*}
&& \sum_{i}\frac{\partial f_{(1)}}{\partial x_{i}}(x)g_{(1)}(x)g_{(2)}f_{(2)}dx_{i}|_{x}=\sum_{i}\frac{\partial f_{(1)}}{\partial x_{i}}(x)g_{(1)}(x)f_{(2)}g_{(2)}dx_{i}|_{x} \\
&\Rightarrow & [(d\otimes id)\alpha(f)\alpha(g)](x)=[\alpha(g)(d\otimes id)\alpha(f)](x)\\
\end{eqnarray*}
Since $x$ is arbitrary, we conclude that $[\alpha(g)(d\otimes id)\alpha(f)]=[(d\otimes
id)\alpha(f)\alpha(g)]$ for all $f,g\in {\cal A}$.\\
So by $\tau$ continuity of $d$ and $\alpha$ we can prove the result for $f,g \in C^{\infty}(M)$.\\
We use the commutativity to deduce the following.
\begin{lmma}
\label{new}
For $F\in C^{\infty}(\mathbb{R}^{n})$ and $g_{1},g_{2},..,g_{n} \in C^{\infty}(M)$
\begin{eqnarray}
(d\otimes id)\alpha (F(g_{1},..., g_{n}))= \sum_{i=1}^{n}\alpha (\partial_{i}F(g_{1},..., g_{n}))(d\otimes id)\alpha(g_{i}),
\end{eqnarray}
where $\partial_{i}F$ denotes the partial derivative of $F$ with respect to the ith coordinate of $\mathbb{R}^{n}$.
\end{lmma}
{\it Proof}:\\
As $\{(g_{1}(x)\ldots g_{n}(x))|x\in M\}$ is a compact subset of $\mathbb{R}^{n}$, for $F\in C^{\infty}(\mathbb{R}^{n})$, we get a sequence of polynomials $P_{m}$ in $\mathbb{R}^{n}$ such that $P_{m}(g_{1},..., g_{n})$ converges to $F(g_{1},...,g_{n})$ in the $\tau$ topology of $C^{\infty}(M)$.\\
We see that for $P_{m}$,
\begin{eqnarray*}
&& (d\otimes id)\alpha(P_{m}(g_{1},..g_{n}))\\
&=& (d\otimes id)P_{m}(\alpha(g_{1}),...,\alpha(g_{n}))\\
&=&  \sum_{i=1}^{n}\alpha (\partial_{i}P_{m}(g_{1},..., g_{n}))(d\otimes id)\alpha(g_{i}),
\end{eqnarray*}
using $ (d\otimes id)\alpha(f)\alpha(g)=\alpha(g)(d\otimes id)\alpha(f)$ as well as the
Leibnitz rule  for $ (d\otimes id).$

As $\alpha$ is $\tau$ continuous, ${\tau}-lim \alpha(P_{m}(g_{1},...,g_{n}))=\alpha(F(g_{1},...g_{n}))$.\\
Also $(d\otimes id)$ is continuous from $C^{\infty}(M)\hat{\otimes} {\cal Q}$ to $\Omega^{1}(C^{\infty}(M))\hat{\otimes}{\cal Q}$.\\
So ${\tau}-lim (d\otimes id)\alpha (P_{m}(g_{1},...,g_{n}))=(d\otimes id)\alpha (F(g_{1},..,g_{n}))$.\\
Also $\partial_{i}P_{m}(g_{1},...,g_{n})\rightarrow^{\tau} \partial_{i}F(g_{1},...,g_{n})$.\\
So $lim_{\tau} (d\otimes id)\alpha (P_{m}(g_{1},...,g_{n}))=  \sum_{i=1}^{n}\alpha (\partial_{i}F(g_{1},..., g_{n}))(d\otimes id)\alpha(g_{i})$.\\
Hence $(d\otimes id)\alpha(F(g_{1},..,g_{n}))=  \sum_{i=1}^{n}\alpha (\partial_{i}F(g_{1},..., g_{n}))(d\otimes id)\alpha(g_{i})$.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
 \vspace{0.1in}\\

\begin{lmma}
\label{support}
Let $U$ be a coordinate neighborhood. Also let $g_{1}, g_{2},..., g_{n}\in C^{\infty}(M)$ be such that $(g_{1}|_{U},\ldots g_{n}|_{U})$ gives a local coordinate system on $U$.
Then
\begin{eqnarray*}
(d\otimes id)\alpha(f)=\sum_{j=1}^{n}\alpha(\partial_{g_{j}}f)(d\otimes id)\alpha(g_{j}),
\end{eqnarray*}
for all $f\in C^{\infty}(M)$ supported in $U$
\end{lmma}
{\bf Proof}:\\
Let $F\in C^{\infty}(\mathbb{R}^{n})\rightarrow \mathbb{R}$ be a smooth function such that $f(m)= F(g_{1}(m),....,g_{n}(m)) \ \forall m\in U.$\\
Choose $\chi\in C^{\infty}(M)$ with $\chi\equiv 1$ on $K=supp(f)$ and $supp(\chi)\subset U$. Then $\chi f=f$ as $\chi\equiv 1$ on $K$.\\

Hence $\chi F(g_{1},...,g_{n})= f ( \chi F=\chi f= f \ on \ U , \ \chi F=0 \ outside \ U)$.\\
Also $\chi^{2}F(g_{1},...,g_{n})= \chi F(g_{1},...,g_{n})$, since on $K$, $\chi^{2}=\chi=1$ and outside $K$, $\chi^{2}F(g_{1},...,g_{n})= \chi F(g_{1},...,g_{n})= 0$.\\
Let $T:=\alpha(\chi)$ and $S:=\alpha(F(g_{1},...,g_{n}))$. Also denote $(d\otimes id)\alpha(F(g_{1},...,g_{n}))$ by $S^{'}$ and $(d\otimes id)\alpha(\chi)$ by $T^{'}$.\\

\begin{eqnarray}
T^{2}S^{'}&=& \alpha(\chi^{2})(d\otimes id)\alpha(F(g_{1},...,g_{n})) \nonumber\\
&=& \alpha(\chi^{2})\sum_{i=1}^{n}\alpha (\partial_{i}F(g_{1},...,g_{n}))(d\otimes id)\alpha(g_{i})\nonumber \ (by \ (3))\\
&=& \alpha(\chi) \sum_{i=1}^{n} \alpha (\chi \partial_{i}F(g_{1},...,g_{n}))(d\otimes id)\alpha(g_{i}) \nonumber \\
&=& \alpha(\chi) \sum_{i=1}^{n}\alpha(\partial_{g_{i}}f)(d\otimes id)\alpha(g_{i}) \ (as \ supp(\partial_{g_{i}}f)\subset K).
\end{eqnarray}
\begin{eqnarray}
TS^{'}&=& \alpha(\chi)(d\otimes id)\alpha(F(g_{1},...,g_{n})) \nonumber\\
&=& \sum_{i=1}^{n} \alpha (\chi \partial_{i}F(g_{1},...,g_{n}))(d\otimes id)\alpha(g_{i}) \nonumber \\
&=& \sum_{i=1}^{n} \alpha (\chi^{2} \partial_{i}F(g_{1},...,g_{n}))(d\otimes id)\alpha(g_{i}) \nonumber \\
&=& \alpha(\chi)\sum_{i=1}^{n}\alpha(\partial_{g_{i}}f)(d\otimes id)\alpha(g_{i})
\end{eqnarray}
Combining (4) and (5) we get
\begin{eqnarray}
T^{2}S^{'}= TS^{'}
\end{eqnarray}
Now
\begin{eqnarray}
&&T^{2}S = TS \nonumber\\
& \Rightarrow & (d\otimes id)(T^{2}S)= (d\otimes id)TS\nonumber \\
& \Rightarrow & 2TT^{'}S + T^{2}S^{'} = TS^{'} + T^{'}S (by \ Leibnitz \ rule \
and \ (d\otimes id)\alpha(f).\alpha(g)=\alpha(g)(d\otimes id)\alpha(f) ) \nonumber \\
& \Rightarrow & 2TT^{'}S= T^{'}S \nonumber \ (by (6)) \\
& \Rightarrow & 2\alpha(\chi)(d\otimes id)\alpha(\chi)\alpha(F(g_{1},...,g_{n})) = (d\otimes id)\alpha(\chi)\alpha(F(g_{1},...,g_{n})) \nonumber \\
& \Rightarrow & 2\alpha(\chi^{2})(d\otimes id)\alpha(\chi)\alpha(F(g_{1},...,g_{n})) = \alpha(\chi) (d\otimes id)\alpha(\chi)\alpha(F(g_{1},...,g_{n})) \nonumber \\
& \Rightarrow & 2(d\otimes id)\alpha(\chi)\alpha(f) = (d\otimes id)\alpha(\chi)\alpha(f)( \ using \ the \ assumption \ and \ \chi^{2}F=f) \nonumber \\
&\Rightarrow & (d\otimes id)\alpha(\chi)\alpha(f) = 0
\end{eqnarray}
So
\begin{eqnarray*}
(d\otimes id)\alpha(f) &=& (d\otimes id)\alpha(\chi f)\\
&=& (d\otimes id)\alpha(\chi)\alpha(f) + \alpha(\chi)(d\otimes id)\alpha(f)\\
&=& \alpha(\chi)(d\otimes id)\alpha(f) ( \ by \ (7))\\
&=& \alpha(\chi)(d\otimes id)\alpha(\chi F(g_{1},..., g_{n}))\\
&=& \alpha(\chi)(d\otimes id)\alpha(\chi) \alpha(F(g_{1},...,g_{n}))+ \alpha(\chi^{2})(d\otimes id)\alpha(F(g_{1},...,g_{n}))\\
&=& (d\otimes id)\alpha(\chi)\alpha(f) + \alpha(\chi^{2})(d\otimes id)\alpha(F(g_{1},...,g_{n})) (Again \ by \ assumption)\\
&=& \alpha(\chi^{2})\sum_{i=1}^{n} \alpha(\partial_{i}F(g_{1},...,g_{n}))(d\otimes id)\alpha(g_{i}) (by \ (7) \ and \ (3))\\
&=& \sum_{i=1}^{n} \alpha(\chi^{2}\partial_{i}F(g_{1},...,g_{n}))(d\otimes id)\alpha(g_{i})\\
&=& \sum_{i=1}^{n} \alpha(\partial_{g_{i}}f)(d\otimes id)\alpha(g_{i})\\
\end{eqnarray*}
  \mbox{}\hfill $\sqare$\vspace{1ex}\\
 Now to complete the proof of the Theorem, we want to first define a bimodule
morphism $\beta$ which extends $d\alpha$ locally, i.e. we define
$\beta_{U}(\omega)$ for any coordinate neighborhood $U$ and any smooth 1-form
$\omega$ supported in $U$ as follows:\\
 Choose $C^{\infty}$ functions $g_{1},\ldots, g_{n}$ as before such that they
give a local coordinate system on $U$ and $\omega$ has the unique expression
$\omega= \sum_{j=1}^{n}\phi_{j}dg_{j}$. Then define $\beta_{U}(\omega):=
\sum_{j=1}^{n}\alpha(\phi_{j})(d\otimes id)\alpha(g_{j})$. We should verify the
following:
 {\bf Claim:} $\beta_{U}$ is independent of the choice of the coordinate functions $(g_{1},\ldots,g_{n})$, i.e. if $(h_{1},\ldots,h_{n})$ is another such set of coordinate functions on $U$ with $\omega=\sum_{j=1}^{n}\psi_{j}dh_{j}$ for some $\psi_{j}$'s in $C^{\infty}(M)$, then \\
 $\sum_{j=1}^{n}\alpha(\phi_{j})(d\otimes id)\alpha(g_{j})= \sum_{j=1}^{n}\alpha(\psi_{j})(d\otimes id)\alpha(h_{j})$.\\
 {\bf proof of the claim:}\\
 We work in the set up of the claim. Let $\chi$ be a smooth function which is
$1$ on the support of $\omega$ and $0$ outside $U$. We have
$F_{1},\ldots,F_{n}\in C^{\infty}(\mathbb{R}^{n})$ such that
$g_{j}=F_{j}(h_{1},\ldots, h_{n})$ for all $j=1,\ldots,n$ on $U$. Then $\chi
g_{j}=\chi F_{j}(h_{1},\ldots, h_{n})$ for all $j=1,\ldots,n$ .\\
 Hence $dg_{j}=\sum_{k=1}^{n}\partial_{h_{k}}(F_{j}(h_{1},\ldots,h_{n}))dh_{k}$ on $U$.\\
 That is $\omega=\sum_{j,k}\chi\phi_{j}\partial_{h_{k}}(F_{j}(h_{1},\ldots,h_{n}))dh_{k}$.\\
 So $\psi_{k}=\sum_{j}\chi\phi_{j}\partial_{h_{k}}(F_{j}(h_{1},\ldots,h_{n}))$.\\
 Also, note that, as $\chi\equiv 1$ on support of $\phi_{j}$ $\forall j$, we must have $\phi_{j}\partial_{h_{k}}(\chi)\equiv 0$, so $\chi \phi_{j}\partial_{h_{k}}(F_{j}(h_{1},\ldots,h_{n}))= \chi\phi_{j} \partial_{h_{k}}(\chi F_{j}(h_{1},\ldots,h_{n}))$. Thus

 \begin{eqnarray*}
 && \sum_{k}\alpha(\psi_{k})(d\otimes id)\alpha(h_{k})\\
 &=& \sum_{k,j}\alpha(\chi\phi_{j}\partial_{h_{k}}(F_{j}(h_{1},\ldots,h_{n})))(d\otimes id)\alpha(h_{k})\\
 &=& \sum_{k,j}\alpha(\phi_{j})\alpha (\partial_{h_{k}}(\chi F_{j}(h_{1},\ldots,h_{n})))(d\otimes id)\alpha(h_{k})\\
 &=& \sum_{j}\alpha(\phi_{j})(d\otimes id)\alpha(\chi F_{j}(h_{1},\ldots,h_{n})) \ (by \ Lemma \ 3.6)\\
 &=& \sum_{j}\alpha(\phi_{j})(d\otimes id)\alpha(\chi g_{j})\\
 &=& \sum_{j}\alpha(\phi_{j})(d\otimes id)\alpha(g_{j})\\
 \end{eqnarray*}
 Where the last step follows from Leibnitz rule and the fact that
 \begin{eqnarray*}
 &&\alpha(\phi_{j})(d\otimes id)(\alpha (\chi))\\
 &=&\sum_{k} \alpha(\phi_{j})\alpha(\partial_{h_{k}}(\chi))(d\otimes id)(\alpha
(h_{k}))\\
 &=& \sum_{k}\alpha(\phi_{j}\partial_{h_{k}}(\chi))(d\otimes id)(\alpha (h_{k}))\\
 &=&0 \ (using \ \phi_{j}\partial_{h_{k}}(\chi)\equiv0)\\
 \end{eqnarray*}
  Hence the definition is indeed independent of choice of coordinate system.\\
 Then for any two coordinate neighborhoods $U$ and $V$, $\beta_{U}(\omega)=\beta_{V}(\omega)$ for any $\omega$ supported in $U\cap V$. It also follows from the definition and Lemma \ref{support} that $\beta_{U}$ is a $C^{\infty}(M)$ bimodule morphism and $\beta_{U}(df)=(d\otimes id)\alpha(f)$ for all $f\in C^{\infty}(M)$ supported in $U$.\\
 Now we define $\beta$ globally by the following:\\
 Choose (and fix) a smooth partition of unity $\{\chi_{1},\ldots,\chi_{l}\}$ subordinate to a cover $\{U_{1},\ldots,U_{l}\}$ of the manifold $M$ such that each $U_{i}$ is a coordinate neighborhood. Define $\beta$ by:
 $$\beta(\omega):= \sum_{i=1}^{l}\beta_{U_{i}}(\chi_{i}\omega),$$
 for any smooth one form $\omega$.\\
 Then for any $f\in C^{\infty}(M)$,
 \begin{eqnarray*}
 \beta(df)&=& \sum_{i=1}^{l} \beta_{U_{i}}(\chi_{i}df)\\
 &=& \sum_{i=1}^{l} \beta_{U_{i}}(d(\chi_{i}f)-fd\chi_{i})\\
 &=& \sum_{i=1}^{l}[(d\otimes id)\alpha(\chi_{i}f)-\alpha(f)(d\otimes id)\alpha(\chi_{i})]\\
 &=& \sum_{i=1}^{l} \alpha(\chi_{i})(d\otimes id)\alpha(f) \ (by \ Leibnitz \ rule \
for \ (d\otimes id) \ and \\ && (d\otimes id)\alpha(f)\alpha(g)=\alpha(g)(d\otimes
id)\alpha(f))\\
 &=& (d\otimes id)\alpha(f)\\
 \end{eqnarray*}
 This completes the proof of the Theorem.\\
  \mbox{}\hfill $\sqare$\vspace{1ex}\\

\section{Isometric action of a CQG }
In this section, we consider a Riemannian structure on the compact manifold $M$, denoting the corresponding inner product on $T_xM$ ($x \in M$) by $\langle \cdot, \cdot \rangle_x$.
\subsection{Definition of isometric action}
We now discuss the notion of isometric action of CQG as introduced in \cite{Goswami} for compact manifolds without boundary and also consider its natural extension to compact manifolds with non trivial boundary. For a compact manifold (possibly with boundary) let ${\cal L}=-(*d)^{2}$ be the Hodge Laplacian restricted to $C^{\infty}(M)$.\\
Changing the   convention in \cite{Goswami}, in this paper we usually consider
${\cal L}$ as a $\tau$- continuous operator on $C^{\infty}(M)$ and call it the
"geometric Laplacian". We shall denote the corresponding extension to the
Hilbert space $L^{2}(M, \ dvol)$ by ${\cal L}_{2}$ (to be called the
$L_{2}$-Laplacian), also making the convention that we choose the Dirichlet
boundary condition for manifolds with boundary. That is, ${\cal L}_{2} = -d^{*}d$,
where for manifold with boundary we consider $d$ as the closure of the unbounded
operator with domain  ${\cal C}=\{f\in C^{\infty}(M):f|_{\partial M}=0 \}$.\\
\begin{dfn}
A smooth action $\alpha$ of a CQG ${\cal Q}$ on a compact manifold $M$ without boundary, is said to be isometric if it commutes with ${\cal L}_{2} = -d^{*}d$. For a compact manifold with boundary we call $\alpha$ to be isometric if it maps ${\cal C}$ into ${\cal C} \hat{\otimes} {\cal Q}$ and commutes with ${\cal L}_{2}$.
\end{dfn}
\begin{rmrk}
In case $M$ is a compact manifold without boundary, $C^{\infty}(M)$ is actually
a core for ${\cal L}_{2}$, hence a smooth action is isometric if and only if it
commutes with ${\cal L}$. For a manifold with boundary, commutation with ${\cal L}$ may
not be sufficient to imply that $\alpha$ is isometric. We also require the
condition that $\alpha({\cal C})\subset {\cal C} \hat{\otimes}{\cal Q}$. We can prove the
existence of QISO like in \cite{Goswami}. It is a consequence of the fact that
the Dirichlet Laplacian has discrete spectrum with finite dimensional eigen
spaces.
\end{rmrk}
\subsection{Geometric characterization of isometric action}
  Let $\alpha : C(M)\rightarrow C(M)\hat{\otimes}{\cal Q}$ be a smooth action (as introduced earlier) and let us fix a dense subalgebra ${\cal A}$ of $C^{\infty}(M)$ over which the action is algebraic i.e. $\alpha({\cal A})\subset ({\cal A}\otimes {\cal Q}_{0})$ and $Sp \ \alpha({\cal A})(1\otimes {\cal Q}_{0})={\cal A}\otimes {\cal Q}_{0}$. Recall for $f\in C^{\infty}(M)
\begin{eqnarray}
d\alpha(df):= (d\otimes id)\alpha(f).
\end{eqnarray}

\begin{thm}
\label{inner_prod_pres}
If $\alpha$ commutes with the geometric Laplacian ${\cal L}$ on ${\cal A}$, then $d\alpha$ extends to a bimodule morphism on $\Omega^{1}(C^{\infty}(M))$, which is also an $\alpha$ equivariant unitary representation on $\Omega^{1}(C^{\infty}(M))$, satisfying in particular
\begin{eqnarray}
<<d\alpha (\omega),d\alpha (\eta)>>=\alpha<<\omega,\eta>>
\end{eqnarray}
for all $\omega,\eta\in \Omega^{1}(C^{\infty}(M))$.

\end{thm}
{\it Proof:}\\
Let $a,a^{'},b,b^{'} \in {\cal A}$. Then\\
\begin{eqnarray*}
&&<<d\alpha(adb),d\alpha (a^{'}db^{'})>>\\
&=& <<a_{(1)}db_{(1)},a_{(1)}^{'}db_{(1)}^{'}>>\otimes a_{(2)}^{*}b_{(2)}^{*}b_{(2)}^{'}a_{(2)}^{'}\\
&=&\bar{a_{(1)}}[{\cal L}(\bar{b_{(1)}}b_{(1)}{'})-{\cal L}(\bar{b_{(1)}})b_{(1)}{'}-\bar{b_{(1)}}{\cal L}{b_{(1)}^{'}}]a_{(1)}^{'}\otimes a_{(2)}^{*}b_{(2)}^{*}b_{(2)}^{'}a_{(2)}^{'}
\end{eqnarray*}
On the other hand
\begin{eqnarray*}
&&\alpha<<adb,a^{'}db^{'}>>\\
&=&\alpha(\bar{a})\alpha<<db,db^{'}>>\alpha(a^{'})\\
&=&(\bar{a_{(1)}}\otimes a_{(2)}^{*})\alpha[{\cal L}(\bar{b}b^{'})-{\cal L}(\bar{b})b^{'}-\bar{b}{\cal L}(b^{'})](a_{(1)}^{'}\otimes a_{(2)}^{'})
\end{eqnarray*}
Now using the fact that $\alpha$ commutes with the Laplacian ${\cal L}$, we get\\
\begin{eqnarray*}
&& \alpha[{\cal L}(\bar{b}b^{'})-{\cal L}(\bar{b})b^{'}-\bar{b}{\cal L}(b^{'})]\\
&=&[{\cal L}\bar{(b_{(1)}}b_{(1)}^{'})-{\cal L}\bar{(b_{(1)})}b_{(1)}^{'}-\bar{b_{(1)}}{\cal L}(b_{(1)}^{'})]\otimes b_{(2)}^{*}b_{(2)}^{'}.
\end{eqnarray*}
Hence $$\alpha<<adb,a^{'}db^{'}>>= \bar{a_{(1)}}[{\cal L}(\bar{b_{(1)}}b_{(1)}^{'})-{\cal L}(\bar{b_{(1)}})b_{(1)}^{'}-\bar{b_{(1)}}{\cal L}(b_{(1)}^{'})]a_{(1)}^{'}\otimes a_{(2)}^{*}b_{(2)}^{*}b_{(2)}^{'}a_{(2)}^{'}.$$
Similarly we can prove the result for finite sums of the form
$\sum_{i=1}^{n}a_{i}db_{i}$. Now by \ref{Frechet_den}, $\Omega^{1}({\cal A})$
is dense in the Hilbert module $\Omega^{1}(C^{\infty}(M))$. So $d\alpha$ extends
to the whole of $\Omega^{1}(C^{\infty}(M))$ as an inner product preserving map.
It is clearly a bimodule morphism. Moreover, as Sp $\alpha({\cal A})(1\otimes
{\cal Q}_{0})={\cal A}\otimes {\cal Q}_{0}$, we have Sp $d\alpha(\Omega^{1}({\cal A}))(1\otimes
{\cal Q}_{0})=\Omega^{1}({\cal A})\otimes {\cal Q}_{0}$. it follows from density of
$\Omega^{1}({\cal A})$ in the Hilbert module $\Omega^{1}(C^{\infty}(M))$ that
$\overline{Sp} \ d\alpha(\Omega^{1}(C^{\infty}(M)))(1\otimes
{\cal Q})=\Omega^{1}(C^{\infty}(M))\hat{\otimes}{\cal Q}$.\\
  \mbox{}\hfill $\sqare$\vspace{1ex}\\

The main result of this section is converse to this, under the assumption of faithful Haar state of the CQG ${\cal Q}$. To this end we make the following definition.
\begin{dfn}
Call a smooth action $\alpha:C(M)\rightarrow C(M)\hat {\otimes}{\cal Q}$ inner product
preserving on ${\cal A}$ if ${\cal A}$ is a $\tau$-dense subalgabra of $C^{\infty}(M)$
and equation (9) holds with $f,g$ varying over  ${\cal A}$.
\end{dfn}
We need quite a few preparatory lemmas. Fix a smooth action $\alpha$ as in the
above definition which is inner product preserving on ${\cal A}$.\\
Moreover, assume for the rest of this subsection only that the CQG ${\cal Q}$ has {\it faithful Haar state}. It follows from the proof of the Theorem \ref{inner_prod_pres} that there exists an $\alpha$ equivariant unitary representation $d\alpha: \Omega^{1}(C^{\infty}(M))\rightarrow \Omega^{1}(C^{\infty}(M))\hat{\otimes}{\cal Q}$, given by $d\alpha (\sum_{i}f_{i}dg_{i})= \sum_{i}\alpha(f_{i})(d\otimes id)\alpha(g_{i})$. Similarly we can have representations over the Hilbert modules $\Omega^{k}(C^{\infty}(M))$ as discussed in section 1.4.

\begin{lmma}
\label{extension}
$d\alpha_{(k)}:\Omega^{k}(C^{\infty}(M))\rightarrow \Omega^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q} $ is an $\alpha$ equivariant unitary representation for all $k=1,\ldots,n$.
\end{lmma}
{\it Proof:}\\
By Lemma \ref{equi_rep}, setting ${\cal E}=\Omega^{1}(C^{\infty}(M))$ and ${\cal B}=\Omega^{1}({\cal A})$, we get $d\alpha_{(2)}:\Omega^{2}(C^{\infty}(M))\rightarrow \Omega^{2}(C^{\infty}(M))\hat{\otimes}{\cal Q}$ is an $\alpha$ equivariant unitary representation of ${\cal Q}$. Similar arguments will work for $k\geq 3$.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
Now define a scalar valued inner product on $\Omega^{2}(C^{\infty}(M))$ by the following:\\
let $\tau^{'}$ be the linear functional on $(C^{\infty}(M))$ coming from the volume form i.e.
$\tau^{'}(f)=\int_{M} fdvol$. Then $\tau^{'}$ is a faithful positive functional on $(C^{\infty}(M))$. Define $\tau=(\tau^{'}\otimes h) \alpha$ where $h$ is the haar state on ${\cal Q}$.\\
Then $\tau$ is a faithful $\alpha$ invariant positive functional on
$C^{\infty}(M)$. So make $\Omega^{2}(C^{\infty}(M))$ a pre Hilbert space by
setting $<\omega,\eta>=\tau(<<\omega,\eta>>)$ and denote its completion by
${\cal H}$.
\begin{lmma}
\label{ss}
$U$, defined by $U(\sum_{i=1}^{k}\xi_{i}\otimes q_{i})=\sum_{i=1}^{k}d\alpha_{(2)}(\xi_{i})(1\otimes q_{i})$, extends as a unitary representation of ${\cal Q}$ on ${\cal H}$(In the sense of Hilbert space).
\end{lmma}
{\it Proof:}\\
\begin{eqnarray*}
&&<U(\sum_{i=1}^{l}\xi_{i}\otimes q_{i}),U(\sum_{j=1}^{k}\xi_{j}^{'}\otimes q_{j}^{'})>\\
&=&\sum_{i,j}<d\alpha_{(2)}(\xi_{i})(1\otimes q_{i}),d\alpha_{(2)}(\xi_{j}^{'})(1\otimes q_{j}^{'})>\\
&=&\sum_{i,j}q_{i}^{*}<d\alpha_{(2)}(\xi_{i}),d\alpha_{(2)}(\xi_{j}^{'})>q_{j}^{'}\\
&=& \sum_{i,j}q_{i}^{*}(\tau\otimes id)<<d\alpha_{(2)}(\xi_{i}),d\alpha_{(2)}(\xi_{j}^{'})>>q_{j}^{'}\\
&=& \sum_{i,j} q^{*}_{i}(\tau\otimes id) \alpha(<<\xi_{i},\xi_{j}^{'}>>)q_{j}^{'}\\
&=&\sum_{i,j}q_{i}^{*}\tau <<\xi_{i},\xi_{j}^{'}>>q_{j}^{'}\\
&=&<\sum_{i}(\xi_{i}\otimes q_{i}),\sum_{j}(\xi_{j}^{'}\otimes q_{j}^{'})>.
\end{eqnarray*}
That $U$ has a dense range is clear as $d\alpha_{(2)}$ is a unitary. So $U$ extends to a unitary representation of ${\cal Q}$ on ${\cal H}$.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\
Recall that by lemma \ref{decomposition_forms}, $\Omega^{2}(C^{\infty}(M))=\Lambda^{2}(C^{\infty}(M))\oplus {\cal J}$. We have the following:
\begin{lmma}
$d\alpha_{(2)}$ leaves $\Lambda^{2}(C^{\infty}(M))$ invariant and thus induces an $\alpha$ equivariant representation from $\Lambda^{2}(C^{\infty}(M)$ to $\Lambda^{2}(C^{\infty}(M))\hat{\otimes} {\cal Q}$ , which is again denoted by $d\alpha_{(2)}$.
\end{lmma}
{\it Proof:}\\
 Denote $\Omega^{2}(C^{\infty}(M))$ by $E$ and let $p_{{\cal J}}$ be the projection onto the closed submodule ${\cal J}$. Also, denote by $p$ the projection onto the Hilbert space closure of ${\cal J}$, i.e. $\bar{{\cal J}^{<,>}}$ \\
It is easy to see that $U$ leaves $p{\cal H}$ invariant. Indeed as ${\cal J}^{{\cal A}}$ is
dense in the Hilbert module ${\cal J}$ (where ${\cal A}$ is the $\tau$ dense subalgebra
of $C^{\infty}(M)$ over which the action is algebraic), it is enough to show
that $d\alpha_{(2)}$ leaves ${\cal J}^{{\cal A}}$ invariant. Let
$\sum_{i=1}^{k}f_{i}\delta g_{i}\in \Omega^{1}({\cal A})_{u}$ be such that
$\sum_{i=1}^{k}f_{i}dg_{i}=0$ in $\Omega^{1}({\cal A})$, so that
$\pi_{(2)}(\delta(\sum_{i=1}^{k}f_{i}\delta g_{i}))=\sum_{i=1}^{k}df_{i}dg_{i}
\in {\cal J}^{{\cal A}}$.\\
\begin{eqnarray*}
&&d\alpha_{(2)}(\sum_{i=1}^{k}df_{i}dg_{i})\\
&=& \sum_{i=1}^{k}df_{i(1)}dg_{i(1)}\otimes f_{i(2)}g_{i(2)}\\
\end{eqnarray*}
Also observe that by the $\alpha$ equivariance of $d\alpha$ we have $d\alpha(\sum_{i=1}^{k}f_{i}dg_{i})=\sum_{i=1}^{k}f_{i(1)}dg_{i(1)}\otimes f_{i(2)}g_{i(2)}=0$\\
in $\Omega^{1}({\cal A})\otimes {\cal Q}_{0}\subset \Omega^{1}(C^{\infty}(M))\hat{\otimes}{\cal Q}$.\\
In other words $(\pi_{(1)}\otimes id)(\theta)=0$, where $\theta= \sum_{i}f_{i(1)}\delta g_{i(1)}\otimes f_{i(2)}g_{i(2)}\in \Omega^{1}({\cal A})_{u}\otimes {\cal Q}_{0}$.\\
Thus,
\begin{eqnarray*}
d\alpha_{(2)} (df_{i}dg_{i})&=&(\pi_{(2)}\otimes id)(\delta f_{i(1)}\delta g_{i(1)}\otimes f_{i(2)}g_{i(2)})\\
&=& (\pi_{(2)}\otimes id)(\delta\otimes id)(\theta),
\end{eqnarray*}
with $(\pi_{(1)}\otimes id)(\theta)=0$.\\
Hence it is in ${\cal J}^{{\cal A}}\otimes {\cal Q}_{o}$.\\

So, $U$ being a unitary representation of a CQG on a Hilbert space it also leaves $p^{\perp}{\cal H}$ invariant.\\
{\bf Claim}: $$\Omega^{2}(C^{\infty}(M))\cap p^{\perp}{\cal H}=p_{{\cal J}}^{\perp} E(= \Lambda^{2}(C^{\infty}(M))).$$
{\it Proof of the claim:}\\
Let $e^{'}\in\Omega^{2}(C^{\infty}(M))\cap p^{\perp}{\cal H}$.\\
Hence we have
\begin{eqnarray*}
&& <e^{'},p_{{\cal J}}e>=0 \ \forall \ e\in E ( p{\cal H}= \bar {{\cal J}}^{<,>})\\
&\Rightarrow& \tau<<e^{'},p_{{\cal J}}e>> =0 \ \forall \ e\in E.
\end{eqnarray*}
In particular $\tau<<e^{'},p_{{\cal J}}e^{'}>>=0\Rightarrow \tau<<p_{{\cal J}}e^{'},p_{{\cal J}}e^{'}>>=0\Rightarrow <<p_{{\cal J}}e^{'},p_{{\cal J}}e^{'}>>=0$  , since $\tau$ is faithful. So $p_{{\cal J}}e^{'}=0\Rightarrow e^{'}\in p_{{\cal J}}^{\perp}E$.\\
Conversely suppose $e\in p_{{\cal J}}^{\perp }E$. To show $e\in p^{\perp}{\cal H}$.\\
$<e,pe^{'}>=\tau<<e,p_{{\cal J}}e^{'}>>=0 (\forall e^{'}\in E)$\\
 and as $\bar{ p_{{\cal J}}E}^{<,>}=p{\cal H} $, $<e,ph>=0$ for all $h\in {\cal H}$
That implies $e\in p^{\perp}{\cal H}$.\\
Hence $p^{\perp}{\cal H}\cap E=p_{{\cal J}}^{\perp}E$.\\
On $p^{\perp}{\cal H}\cap E$, $U$ agrees with $d\alpha_{(2)}$. Hence the result of the lemma follows. \mbox{}\hfill $\sqare$\vspace{1ex}\\
\vspace{0.3in}\\
Similarly we can show
\begin{thm}
\label{lifting_rep}
The restriction of $d\alpha_{(k)}$ on $\Lambda^{k}(C^{\infty}(M))$ gives an $\alpha$ equivariant unitary representation for all $k=1,..,n$.
\end{thm}
We have already introduced the Hodge $*$ operator in the first section. Now we shall derive a characterization for $(*\otimes id):\Lambda^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q}\rightarrow \Lambda^{n-k}(C^{\infty}(M))\hat{\otimes}{\cal Q} $
 for all $k=1(1)n$.\\
\begin{lmma}
\label{star_char}
Let $\xi\in \Lambda^{n-k}(C^{\infty}(M))\hat{\otimes}{\cal Q}$ and $X\in \Lambda^{k}(C^{\infty}(M)) \hat {\otimes}{\cal Q}$. Then the following are equivalent:\\
(i) $\xi\wedge Y=<<X,Y>>(dvol\otimes 1_{{\cal Q}})$ for all $Y\in \Lambda^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q}$\\
(ii) $\xi=(*\otimes id)X.$
\end{lmma}
{\it Proof:}\\
$(i)\Rightarrow (ii)$:\\
Let $x\in M$. Choose a coordinate neighborhood $(U,x_{1},x_{2},....,x_{n})$ around $x$ in $M$.\\
Then we can write $\xi(x)=\sum_{|I|=n-k}dx_{I}(x)q_{I}(x)$ where $q_{I}\in C^{\infty}(M)\hat{\otimes}{\cal Q}$, such that $\{dx_{1}|_{x}, dx_{2}|_{x},..., dx_{n}|_{x}\}$ is an onb of $\Lambda^{1}(M)$ at $x$.\\
Also let $X(x)=\sum_{|J|=k}dx_{J}(x)Q_{J}(x)$ where $Q_{J}\in C^{\infty}(M)\hat{\otimes}{\cal Q}$.\\
Taking $Y$ such that $Y(x)= dx_{J^{'}}(x)1_{{\cal Q}}$, we have $$<<X,Y>>(x)(dvol(x)\otimes 1_{{\cal Q}})=dvol(x)Q_{J^{'}}(x).$$
Also $(\xi\wedge Y)(x)=dvol \ q_{I}(x)$.\\
Hence \begin{eqnarray*}&&(\xi \wedge Y)(x)= <<X,Y>>(x)dvol(x).1_{{\cal Q}}\\
&\Rightarrow& Q_{J^{'}}(x)= q_{I}(x)
\end{eqnarray*}
Since $x$ was an arbitrary point in $M$, we conclude that $\xi=(*\otimes id)X$. The other direction of the proof is trivial.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}
\vspace{0.4in}\\
$\Lambda^{n}(C^{\infty}(M))\hat{\otimes}{\cal Q}$ is generated by $dvol$ as a $C^{\infty}(M))\hat{\otimes}{\cal Q}$ module. So We must have $$d\alpha_{(n)}(dvol)= dvol.q$$ where $q\in (C^{\infty}(M)\hat\otimes {\cal Q})^{s.a.}$.\\
As $\alpha: C^{\infty}(M)_{\mathbb {R}}\rightarrow (C^{\infty}(M) \hat {\otimes}{\cal Q})^{s.a.}$, where $C^{\infty}(M)_{\mathbb {R}}=C^{\infty}(M)^{s.a.}$ consists of the $*$ algebra generated by real valued functions, it is easy to see that $d\alpha_{(n)}$ leaves $\Lambda^{n}(C^{\infty}(M)_{\mathbb {R}})$ invariant. Hence $q$ must belong to $(C^{\infty}(M)\hat {\otimes} {\cal Q})^{s.a.}$ i.e. $q=q^{*}$.\\

With this notation we have the following:
\begin{lmma}
\label{char}
We have $\forall \ k=1(1)n, q^{-1} d\alpha_{(n-k)}(*\omega)\wedge \eta= <<d\alpha_{(k)}(\omega),\eta>>(dvol\otimes 1_{{\cal Q}})\ \forall \ \eta\in \Lambda^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q}.$
\end{lmma}
{\it Proof:}\\
By Theorem \ref{lifting_rep} $d\alpha_{(k)}:\Lambda^{k}(C^{\infty}(M))\rightarrow \Lambda^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q}$ is $\alpha$ equivariant unitary for all $k=1(1)n$. So in particular $d\alpha_{(n)}$ is also $\alpha$ equivariant. Hence $$<<d\alpha_{(n)}(dvol),d\alpha_{(n)}(dvol)>>=\alpha<<(dvol),(dvol)>> \Rightarrow q^{2}=1\otimes 1_{{\cal Q}}.$$
Also \begin{eqnarray}
&&<<d\alpha_{(n)}(dvol),d\alpha_{(n)}(f \ dvol)>>=\alpha<<(dvol),(f \ dvol)>>\nonumber\\ &\Rightarrow& q\alpha(f)=\alpha(f)q \ \forall \ f\in C^{\infty}(M).
\end{eqnarray}
Now let $\eta=d\alpha_{(k)}(\omega^{'})(1\otimes q^{'})$. Then $q$ being invertible,\\
\begin{eqnarray*}
&& q^{-1}d\alpha_{(n-k)}(*\omega)\wedge \eta \\
&=&q^{-1}d\alpha_{(n-k)}(*\omega)\wedge d\alpha_{(k)}(\omega^{'})(1\otimes q^{'})\\
&=&q^{-1}\alpha<<\omega,\omega^{'}>>q(dvol\otimes q^{'})\\
&=&\alpha<<\omega,\omega^{'}>> (dvol\otimes q^{'}) \  (by \ (9))
\end{eqnarray*}
On the other hand from unitarity of $d\alpha_{(k)}$,\\
\begin{eqnarray*}
&& <<d\alpha_{(k)}(\omega),d\alpha_{(k)}(\omega^{'})(1\otimes q^{'})>>\\
&=&\alpha<<\omega,\omega^{'}>>(1\otimes q^{'}).
\end{eqnarray*}
So by replacing $\eta$ by finite sums of the type $\sum_{i}d\alpha^{(k)}(\omega_{i})(1\otimes q_{i})$, we can show that $\omega \in \Lambda^{k}(C^{\infty}(M))$ and $\eta\in  Sp \ d\alpha^{(k)}(\Lambda^{k}(C^{\infty}(M)))(1\otimes {\cal Q})$, $$d\alpha_{(n-k)}(*\omega)\wedge \eta=q<<d\alpha_{(k)}(\omega),\eta>>(dvol\otimes 1_{{\cal Q}}).$$
Now, since Sp $d\alpha_{(k)}(\Lambda^{k}(C^{\infty}(M)))(1\otimes {\cal Q})$ is dense in $\Lambda^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q}$, we get a sequence $\eta_{n}$ belonging to Sp $d\alpha_{(k)}(\Lambda^{k}(C^{\infty}(M)))(1\otimes {\cal Q})$ such that $\eta_{n}\rightarrow \eta$ in the Hilbert module $\Lambda^{k}(C^{\infty}(M))\hat{\otimes}{\cal Q})$.\\
But we have
 $$d\alpha_{(n-k)}(*\omega)\wedge \eta_{n}=q<<d\alpha_{(k)}(\omega),\eta_{n}>>(dvol\otimes 1_{{\cal Q}}).$$
 Hence the claim follows from the continuity of $<<,>>$ and $\wedge$ in the Hilbert module $\dot{\Lambda}(C^{\infty}(M))\hat{\otimes} {\cal Q}$.\\
  \mbox{}\hfill $\sqare$\vspace{1ex}\\
 Now combining \ref{star_char} and \ref{char} we immediately conclude the following:\\
 \begin{crlre}
 \label{char_Hodge}
$q^{-1}d\alpha_{(n-k)}(*\omega)=(*\otimes id)d\alpha_{(k)}(\omega).$
 \end{crlre}
 We are now ready to prove the main result of this section.\\

 \begin{thm}
 \label{main_thm}
 Let $\alpha:C(M)\rightarrow C(M)\hat {\otimes}{\cal Q}$ be a smooth action of a CQG with faithful Haar state, which is inner product preserving on a $\tau$-dense subalgebra ${\cal A}\subset C^{\infty}(M)$. Then $\alpha$ commutes with the geometric Laplacian ${\cal L}=-(*d)^{2}.$
 \end{thm}
 {\it Proof}:
 Let $\omega\in \Omega^{1}(C^{\infty}(M))$.\\

 Then

 \begin{eqnarray*}
  \alpha(*d*\omega)&=& q(*\otimes id)d\alpha(d*\omega)\\
  &=&q(*d\otimes id)\alpha(*\omega)\\
  &=& q.q (*d*\otimes id)d\alpha (\omega)\\
  &=& (*d*\otimes id)d\alpha(\omega).
  \end{eqnarray*}
  Hence for $\phi\in C^{\infty}(M)$,
  \begin{eqnarray*}
  \alpha(-*d*d \phi)&=& (-*d*\otimes id)d\alpha(d\phi)\\
  &=&(-*d*d\otimes id)\alpha(\phi).
  \end{eqnarray*}
 Hence $\alpha$ commutes with ${\cal L}$.\\
   \mbox{}\hfill $\sqare$\vspace{1ex}\\
 Combining the above theorem and the remarks in section 3.1 we get\\
 \begin{crlre}
 Suppose that ${\cal Q}$ is a CQG with faithful Haar state, $\alpha:C(M)\rightarrow C(M)\hat{\otimes}{\cal Q}$ is a smooth action and in case $M$ has a non trivial boundary, assume also that $\alpha({\cal C})\subset {\cal C}\hat{\otimes} {\cal Q}$ where ${\cal C}=\{f\in C^{\infty}(M);f|_{\partial M}=0\}$. Then $\alpha$ is isometric if and only if $\alpha$ is inner product preserving.
 \end{crlre}
 \subsection{ Averaging of the Riemannian metric with respect to  CQG action}
 We do not need the assumption of faithful Haar state  in the present subsection.
Fix as before a $\tau$ dense subalgebra ${\cal A}$ of $C^{\infty}(M)$ over which the
action $\alpha$ is algebraic. In this section our aim is to equip $M$ with a new
Riemannian structure with respect to which the action becomes isometric using
the averaging technique of classical differential geometry. Note that a priori
we do not assume the action to be isometric.\\
\begin{thm}
The following are equivalent:\\
(i)$M$ has a Riemannian structure such that $\alpha$ is inner product preserving.\\
(ii) There exists a bimodule morphism $d\alpha:\Omega^{1}(C^{\infty}(M))\rightarrow
\Omega^{1}(C^{\infty}(M))\hat{\otimes} {\cal Q} $ satisfying $d\alpha(df)= (d\otimes
id)\alpha(f)$.
\end{thm}
{\it Proof}:\\
$(i)\Rightarrow (ii)$:\\
Let $M$ has a Riemannian structure such that $\alpha$ is inner product preserving. Then by Theorem \ref{inner_prod_pres}, we have a bimodule morphism $d\alpha$. \\
$(ii)\Rightarrow (i)$:\\
Suppose that $d\alpha$ a well defined bimodule morphism as in the proposition.\\
We have a $\tau$ dense subalgebra ${\cal A}$ of $C^{\infty}(M)$ over which the action is algebraic.\\
For $\omega,\eta \in \Omega^{1}({\cal A})$, define\\
$<<\omega,\eta>>^{'}:=(id\otimes h)(id\otimes m)(id\otimes \kappa\otimes id)(\alpha\otimes id)<<d\alpha(\omega),d\alpha(\eta)>>$\\
As the action is algebraic over ${\cal A}$, we shall use Sweedler's notation to prove the following claims.\\
\vspace{0.2in}\\
{\bf Claim 1}: $<<d\phi,d\psi \ f>>^{'}= <<d\phi,d\psi>>^{'}f$ for $\phi,\psi,f \in {\cal A}$.\\
Let $\alpha(\phi) = \phi_{(1)}\otimes \phi_{(2)}$ and $\alpha(\psi) = \psi_{(1)}\otimes \psi_{(2)}$ and $\alpha(f) =f_{(1)}\otimes f_{(2)}$.\\
We denote $<<d\phi_{(1)},d\psi_{(1)}>>$ by $x$ and $\phi_{(2)}^{*}\psi_{(2)}$ by $y$.\\
Then \begin{eqnarray*}
&& <<d\phi,d\psi f>>^{'}\\
&=& (id \otimes h)(id \otimes m)(id \otimes \kappa \otimes id)(\alpha \otimes id)<<d\alpha(d\phi),d\alpha(d\psi f)>>\\
&=& (id \otimes h)(id \otimes m)(id \otimes \kappa \otimes id)(\alpha \otimes id)(xf_{(1)}\otimes yf_{(2)})\\
&=& (id \otimes h)(id \otimes m)(id \otimes \kappa \otimes id) (x_{(1)}f_{(1)(1)} \otimes x_{(2)}f_{(1)(2)} \otimes yf_{(2)})\\
&=& (id \otimes h)(x_{(1)}f_{(1)(1)} \otimes \kappa(x_{(2)}f_{(1)(2)})yf_{(2)})\\
&=& x_{(1)}f_{(1)(1)} h(f_{(2)}\kappa(f_{(1)(2)})\kappa(x_{(2)})y) ( by \ tracial \ property \ of \ h)\\
&=& x_{(1)}f_{(1)}h(f_{(2)(2)}\kappa (f_{(2)(1)})\kappa(x_{(2)})y)\\
&=& x_{(1)}f_{(1)}h(m(id\otimes \kappa)\Delta^{op}(f_{(2)})\kappa (x_{(2)})y)\\
&=& x_{(1)}f_{(1)}h(\epsilon(f_{(2)}).1.\kappa(x_{(2)})y)\\
&=& x_{(1)}(id\otimes \epsilon)\alpha(f)h(\kappa(x_{(2)})y)\\
&=& x_{(1)}fh(\kappa(x_{(2)})y).
\end{eqnarray*}
Also \begin{eqnarray*}
<<d\phi,d\psi>>^{'}f &=& [(id\otimes h)(id\otimes m)(id\otimes \kappa\otimes id)(\alpha \otimes id)<<d\alpha(d\phi),d\alpha(d\psi)>>]f\\
&=& [(id\otimes h) (id\otimes m)(id\otimes \kappa \otimes id)(x_{(1)}\otimes x_{(2)}\otimes y)]f\\
&=& x_{(1)}fh(\kappa (x_{(2)})y).
\end{eqnarray*}
Actually we can show more.\\
{\bf Claim 2}:We can extend the definition of $<<,>>^{'}$ for $\omega,\eta \in \Omega^{1}({\cal A})C^{\infty}(M)$ such that
\begin{eqnarray}
 \forall \ \ f\in C^{\infty}(M) \ and \ \phi, \psi \in {\cal A},
<<(d\phi),(d\psi)f>>^{'}=<<d\phi, d\psi>>^{'}f
 \end{eqnarray}
{\it proof}:\\
For $f\in C^{\infty}(M)$, define $<<(d\phi),(d\psi)f>>^{'}:= \ \tau -lim<<d\phi, d\psi \ f_{n}>>^{'}$, where $f_{n}\in {\cal A}$ with $\tau -lim \ f_{n}= f$.\\
Observe that $<<d\phi, d\psi \ f_{n}>>^{'}$ is $\tau$ Cauchy as
\begin{eqnarray*}
&&  <<d\phi, d\psi \ f_{n}>>^{'}- <<d\phi, d\psi \ f_{m}>>^{'}\\
&=& <<d\phi,d\psi>>^{'}(f_{n}- f_{m})
\end{eqnarray*}
So $<<d\phi, d\psi  f>>^{'}= \ \tau-lim<<d\phi, d\psi >>^{'}f_{n}= <<d\phi,
d\psi >>^{'}f$.\\
That proves the claim.\\
\vspace{0.2in}\\
{\bf Claim 3}:
\begin{eqnarray}
<<d\alpha(d\phi),d\alpha(d\psi)>>^{'}= \alpha(<<d\phi,d\psi>>^{'})
\end{eqnarray}
{\it Proof}:\\
\begin{eqnarray*}
&& <<d\alpha(d\phi),d\alpha (d\psi)>>^{'}\\
&=& (id\otimes h\otimes id)(id\otimes m\otimes id)(id\otimes \kappa\otimes id\otimes id)(\alpha\otimes id\otimes id)(x\otimes \Delta(y))\\
&=& (id\otimes h\otimes id)(id\otimes m\otimes id)(id\otimes \kappa\otimes id\otimes id)(x_{(1)}\otimes x_{(2)}\otimes y_{(1)}\otimes y_{(2)})\\
&=& (id\otimes h\otimes id)(x_{(1)}\otimes \kappa(x_{(2)})y_{(1)}\otimes y_{(2)})\\
&=& x_{(1)}\otimes h(\kappa(x_{(2)})y_{(1)})y_{(2)}.
\end{eqnarray*}
\vspace{0.1in}
On the other hand
\begin{eqnarray*}
\alpha(<<d\phi,d\psi>>^{'})&=& x_{(1)(1)}h(\kappa(x_{(2)})y)\otimes x_{(1)(2)}\\
&=& x_{(1)}\otimes x_{(2)(1)}h(\kappa(x_{(2)(2)})y)
\end{eqnarray*}
Hence it is enough to show that $h(\kappa(c)b_{(2)}) b_{(1)}=h(\kappa(b)c_{(1)})c_{(2)}$ where $c\in {\cal Q}_{0}$.\\
We make the transformation $T(a\otimes b)=\Delta(\kappa(a))(1\otimes b)$. \\
Then
\begin{eqnarray*}
&&(h\otimes id)T(a\otimes b)\\
&=& (h\otimes id)\Delta(\kappa(a))(1\otimes b)\\
&=& ((h\otimes id)\Delta(\kappa(a)))b\\
&=& h(\kappa(a))b\\
&=& (h\otimes id)(a\otimes b)\\
\end{eqnarray*}
Hence $h(b_{(2)}\kappa(c))b_{(1)}=(h\otimes id)T(b_{(2)}\kappa(c)\otimes b_{(1)})$.\\
So, by using traciality of $h$ it is enough to show that $T(b_{(2)}\kappa(c)\otimes b_{(1)})= c_{(1)}\kappa(b)\otimes c_{(2)}$.\\
\begin{eqnarray*}
&& T(b_{(2)}\kappa(c)\otimes b_{(1)})\\
&=& \Delta(\kappa(b_{(2)}\kappa(c)))(1\otimes b_{(1)})\\
&=& \Delta(c \kappa(b_{(2)}))(1\otimes b_{(1)})\\
&=& (c_{(1)}\otimes c_{(2)})[\kappa(b_{(2)(2)})\otimes \kappa(b_{(2)(1)})](1\otimes b_{(1)})\\
&=& (c_{(1)}\otimes c_{(2)})m_{23}(\kappa(b_{(2)(2)})\otimes \kappa(b_{(2)(1)})\otimes b_{(1)})\\
&=& (c_{(1)}\otimes c_{(2)})m_{23}(\kappa\otimes \kappa\otimes id)\sigma_{13}( b_{(1)}\otimes b_{(2)(1)}\otimes b_{(2)(2)})\\
&=& (c_{(1)}\otimes c_{(2)})m_{23}(\kappa\otimes \kappa\otimes id)\sigma_{13}(b_{(1)(1)}\otimes b_{(1)(2)}\otimes b_{(2)})\\
&=& (c_{(1)}\otimes c_{(2)})m_{23}(\kappa(b_{(2)}\otimes \kappa (b_{(1)(2)})\otimes b_{(1)(1)})\\
&=& (c_{(1)}\otimes c_{(2)})m_{23} \sigma_{13}(\kappa(b_{(2)})\otimes (id\otimes \kappa)\Delta(b_{(1)}))\\
&=& (c_{(1)}\otimes c_{(2)})(\kappa(b_{(2)})\otimes \epsilon(b_{(1)}).1_{{\cal Q}})\\
&=& (c_{(1)}\otimes c_{(2)})(\kappa\otimes \kappa)((b_{(2)})\otimes \epsilon(b_{(1)}).1_{{\cal Q}})\\
&=& (c_{(1)}\otimes c_{(2)})(\kappa\otimes \kappa)(\epsilon(b_{(1)})b_{(2)}\otimes 1_{{\cal Q}})\\
&=&  c_{(1)}\kappa(b)\otimes c_{(2)}
\end{eqnarray*}
Which proves the claim.\\
\vspace{0.2in} \\
Now we proceed to define a new Riemannian structure on the manifold so that the action $\alpha$ will be inner product preserving.\\
Let $m\in M$. $v,w\in T^{*}_{m}(M)$. $T^{*}_{m}(M)$ is isomorphic to $\mathbb{R}^{n}$ as a vector space.\\
\vspace{0.2in}\\
{\bf Claim}: Sp $\{ds(m): s\in{\cal A}\}$ coincides with $T^{*}_{m}(M)$.\\
{\it Proof}: \\
Choosing a coordinate neighborhood $U$ around $m$ and a set of coordinates
$x_{1},...,x_{n}$ we have $ds(m)= \sum_{i=1}^{n}\frac{\partial s}{\partial
x_{i}}(m)dx_{i}(m)$.\\
Pick any $\eta\in T^{*}_{m}(M)$ i.e. we have $\eta= \sum_{i=1}^{n}c_{i}dx_{i}(m)$ for some $c_{i}$'s in $\mathbb{R}$.\\
Choose any $f\in C^{\infty}(M)$ with $\frac{\partial f}{\partial x_{i}}(m)= c_{i}$.\\
For $f\in C^{\infty}(M)$, by $\tau$ density of ${\cal A}$ we have  a sequence
$s_{n}\in {\cal A}$ and an $n_{0}\in \mathbb{N}$ such that $$|\frac{\partial
s_{n}}{\partial x_{i}}(m)-\frac{\partial f}{\partial x_{i}}(m)|< \epsilon \
\forall \ n\geq n_{0}.$$
So Sp $\{ds(m); s\in {\cal A}\}$ is dense in $T^{*}_{m}(M)$. $T^{*}_{m}(M)$ being finite dimensional Sp $\{ds(m): s\in{\cal A}\}$ coincides with $T^{*}_{m}(M)$. Which proves the claim.  \\
We have $s_{1},\ldots,s_{n}\in {\cal A}$ such that $\{ ds_{1}(m),\ldots,ds_{n}(m)\}$
is a basis for $T^{*}_{m}(M)$. Hence there is a neighborhood $V$ of $m$ such
that $\{ ds_{1}(x),\ldots,ds_{n}(x)\}$ is a basis for $T^{*}_{x}(M)$ for all
$x\in V$. So corresponding to $v,w$ we have $\{c_{i},d_{i}:i=1(1)n\}$ so that
$v=\sum_{i=1}^{n}c_{i}ds_{i}(m)$ and $w=\sum_{i=1}^{n}d_{i}ds_{i}(m)$. Then we
define $<v,w>^{'}$ as:
$$<v,w>^{'}:= \sum_{i,j}\bar{c_{i}}d_{j}<<ds_{i},ds_{j}>>^{'}(m)$$
where $<<,>>^{'}$ is the new $C^{\infty}(M)$ valued inner product introduced earlier. We have to show that this definition is independent of the choice of the basis.\\
\begin{lmma}
Let $m\in M$ and $\omega \in \Omega^{1}({\cal A})$ such that $\omega= 0$ in a
neighborhood $U$ of $m$. Then $<<\omega, \eta>>^{'}= 0$ for all $\eta\in
\Omega^{1}({\cal A})$.
\end{lmma}
{\it proof}:\\
Let $V\subset U$ such that $V\subset \bar{V}\subset U$.\\
Choose $f\in C^{\infty}(M)_{\mathbb{R}}$ such that $supp(f)\subset \bar{V}$, $f\equiv 1$ on $V$ and $f\equiv 0$ outside $U$.\\
So we can write $\omega=(1-f)\omega$. Then
\begin{eqnarray*}
&&<<\omega,\eta>>^{'}(m)\\
&=& <<(1-f)\omega, \eta>>^{'}(m)\\
&=& <<\omega,\eta>>^{'}(m)(1-f)(m) \ (by \ (11))\\
&=& 0.
\end{eqnarray*}
\begin{crlre}
\label{ind}
Let $m\in M$ and $\omega=\omega^{'}, \ \eta=\eta^{'}$ in a neighborhood $U$ of
$m$. Then $<<\omega,\eta>>^{'}= <<\omega^{'},\eta^{'}>>^{'}$, $\forall
\omega,\omega^{'},\eta, \eta^{'}\in \Omega^{1}({\cal A})$.
\end{crlre}
{\it proof}:\\
Just observe that $<<\omega,\eta>>^{'}(m)- <<\omega^{'},\eta^{'}>>^{'}(m)=<<\omega-\omega^{'},\eta>>^{'}(m)+ <<\omega^{'},\eta-\eta^{'}>>(m)$.\\
Now apply the previous lemma.\\
\vspace{0.2in}\\
Now we can prove that our definition is indeed independent of choice of basis.\\
Let $s_{1},..., s_{n}$ and $s^{'}_{1},..., s^{'}_{n}$ in ${\cal A}$ be such that
$ds_{1}(m),..., ds_{n}(m)$ and $ds^{'}_{1}(m),..., ds^{'}_{n}(m)$ are two bases
for $T^{*}_{m}(M)$. then they are actually bases for $T^{*}_{x}(M)$ in a
neighborhood $U$ of $m$. So there are $\{f_{ij}:i,j= 1(1)n\}$ in $C^{\infty}(M)$
such that $$ds_{i}= \sum _{j=1}^{n}f_{ij}ds^{'}_{j}$$ on $U$ for all $i= 1(1)n$.
Hence by corollary \ref{ind}
\begin{eqnarray}
<<ds_{i},ds_{j}>>^{'}(m)=<<\sum_{k}f_{ik}ds^{'}_{k}, \sum_{l}f_{jl}ds^{'}_{l}>>^{'}(m)
\end{eqnarray}
Let $v=\sum_{i=1}^{n}c_{i}ds_{i}(m)=\sum_{i=1}^{n}c^{'}_{i}ds^{'}_{i}(m)$ and $w=\sum_{i=1}^{n}d_{i}ds_{i}(m)=\sum_{i=1}^{n}d^{'}_{i}ds^{'}_{i}(m)$. So by definition
\begin{eqnarray*}
<v,w>^{'}&=& \sum_{ij}\bar{c_{i}}d_{j}<<ds_{i},ds_{j}>>^{'}(m)\\
&=&
\sum_{ijkl}\bar{c_{i}}d_{j}\bar{f_{ik}}(m)f_{jl}(m)<<ds^{'}_{k},ds^{'}_{l}>>^{'}
(m) \ ( \ by \ (13))\\
&=& \sum_{kl}\bar{c_{k}}^{'}d_{l}^{'}<<ds_{k}^{'},ds_{l}^{'}>>^{'}(m)
\end{eqnarray*}
Hence we see that our definition is independent of choice of basis. Now using
(12) we see that with respect to this new Riemannian structure on the manifold,
$\alpha$ is inner product preserving. This completes the proof of the Theorem on
$\Omega^{1}({\cal A})$ and hence on $\Omega^{1}(C^{\infty}(M))$.\\
 \mbox{}\hfill $\sqare$\vspace{1ex}\\

\section{Lifting an action of a CQG to tubular neighborhood.}

Let $M$ be a compact, oriented Riemannain $n$-manifold without boundary,
embedded isometrically in some Euclidean space $\mathbb{R}^N$ such that it has a
trivial normal bundle. Let ${\cal Q}$ be a CQG which acts faithfully on $M$ as in
the sense mentioned earlier. Suppose that $\alpha:C(M)\rightarrow
C(M)\widehat{\otimes}{\cal Q}$ is the (co)-action which is smooth i.e.
$\alpha(C^\infty(M)\subseteq C^\infty(M,{\cal Q}))$ and $\alpha$ commutes with the
geometric Laplacian, say ${\cal L}.$ So, by Lemma \ref{dense_alg}, we have a
$\tau$-dense subalgebra $\beta_0$ of $C^\infty(M)$ over which $\alpha$ is
algebraic and $Sp\left(\alpha(\beta_0)(1\otimes{\cal Q}_0)\right)=\beta_0\otimes{\cal Q}_0.$

Now since $M$ is a manifold with a trivial normal bundle, by Lemma \ref{global_diff}, we have a global diffeomorphism
$$F:M\times B^{N-n}_\epsilon(0)\rightarrow {\cal N}_\epsilon M.$$ Where $B^{N-n}_\epsilon(0)$ is as in the Lemma \ref{global_diff}. So, by Lemma \ref{top alg_iso}, we have an isomorphism between the topological algebras $$\pi_F:C^\infty({\cal N}_\epsilon M)\rightarrow C^\infty(M\times B^{N-n}_\epsilon(0)).$$
We have $\alpha:\beta_0\rightarrow \beta_0\otimes{\cal Q}_0.$ So, define $$\widehat{\alpha}:\beta_0\otimes C^\infty(B^{N-n}_\epsilon(0))\rightarrow \beta_0\otimes C^\infty(B^{N-n}_\epsilon(0))\otimes{\cal Q}_0$$  by $ \widehat{\alpha}=\sigma_{23} \circ (\alpha \hat{\otimes} {\rm id})$($\sigma_{23}$ denotes the flip of the second and third tensor copies), which is well-defined and continuous by Proposition \ref{Ext}, and satisfies
$\widehat{\alpha}( f \otimes g)= f_{(1)} \otimes g \otimes f_{(2)}$ (Swedler notation).

Now we have $\pi_F:C^\infty({\cal N}_\epsilon M)\rightarrow C^\infty(M\timesB^{N-n}_\epsilon(0)),$ which implies that
$\pi_{F^{-1}}:C^\infty(M\timesB^{N-n}_\epsilon(0))\rightarrow C^\infty({\cal N}_\epsilon M).$ Hence
$$(\pi_{F^{-1}}\hat{\otimes} id_{\cal Q}):C^\infty(M\timesB^{N-n}_\epsilon(0))\widehat{\otimes}{\cal Q}\rightarrow C^\infty({\cal N}_\epsilon M)\widehat{\otimes}{\cal Q}.$$  So, defining
$$\Phi:=(\pi_{F^{-1}}\otimes id)\circ\widehat{\alpha}\circ\pi_F: C^\infty({\cal N}_\epsilon M)\rightarrow C^\infty({\cal N}_\epsilon M)\widehat{\otimes}{\cal Q},$$ we see that $\Phi$ is an action of ${\cal Q}$ on $C({\cal N}_{\epsilon}M)$ which is smooth and $\tau$-continuous.

Set ${\cal A}_0:=\pi_{F^{-1}}(\beta_0\otimes C^\infty(B^{N-n}_\epsilon(0))).$ Then ${\cal A}_0$ is a $\tau$-dense subalgebra of $C^\infty({\cal N}_{\epsilon}(M)).$ By construction, $\Phi$ is algebraic over ${\cal A}_0$ and moreover,
 $Sp~\Phi({\cal A}_0)(1\otimes{\cal Q}_0)={\cal A}_0\otimes{\cal Q}_0.$\\
 As the normal bundle of the manifold is trivial, we can choose smoothly varying basis for normal space at each point of the manifold. Let $y\in {\cal N}_{\epsilon}(M)$ and $\{e_{i}(y): i=1,\ldots,(N-n)\}$ be a basis for the normal space to the manifold at the point $\pi(y)$. And $u_{1}, u_{2},..., u_{N-n}$ be components of ${\cal U}(y):= (y-\pi(y))$ with respect to the basis $\{e_{i}(y): i=1,\ldots,(N-n)\}$ .

We introduce a coordinate system for the manifold ${\cal N}_{\epsilon}(M)$ as follows:

$$G:{\cal N}_{\epsilon}(M)\rightarrow M\times B_{\epsilon}^{N-n}(0)\stackrel{\xi\otimes
id}{\rightarrow}\mathbb{R}^N~
({\xi~{\rm is~ a~ coordinate~ map ~for}~ M}):$$
$$y\rightarrow (\pi(y),{\cal U}(y))\rightarrow( x_1,...x_n,u_1,...u_{N-n}  ).$$

Suppose that $\phi \in C^\infty({N}_{\epsilon}(M))$ such that $\phi(y) \equiv \Psi({\cal U}(y)).$ Then

\begin{eqnarray*}
&\phi\circ G^{-1}(x_1,...x_n,u_1,...u_{N-n})\\
&=\phi(F(\xi(x_1,...x_n),u_1,...u_{N-n}))\\
&=\Psi(u_1,...u_{N-n}).
\end{eqnarray*}
Therefore $\frac{\partial}{\partial x_i}\phi=0$ if $\phi\equiv\Psi({\cal U}(y)),$ which implies that

$$d\phi=\sum_{i=1}^{N-n}\frac{\partial\phi}{\partial u_i} du_i.$$

\begin{lmma}\label{cordinate system}
$$ << \frac{\partial}{\partial u_i}, \frac{\partial}{\partial x_j} >> = 0   $$ for all $i,j,~i=1,...N-n,~j=1,...n.$
\end{lmma}

{\it Proof:}\\
Without loss of generality let $y \in {\cal N}_{\epsilon}(M)$ be an interior point (for points on the boundary the proof will be similar) and $\phi\in C^\infty({{\cal N}}_{\epsilon}M)$ and $y\in {{\cal N}}_{\epsilon}M$  such that $G^{-1}(0,...0)=y.$ Let $e_i(y)=(e^1_i(y),...e^N_i(y))$ for all $i=1,...N-n.$ Then

\begin{eqnarray*}
&\frac{\partial}{\partial u_i}\phi(y)=\frac{d}{dt}|_{t=0} \phi(G^{-1}(0,...t,...0))~(\mbox{$t$ in $ith$ position})\\
&=\frac{d}{dt}|_{t=0} \phi(\xi^{-1}(0)+t e_i)\\
&=\sum_{j=1}^{N}e^j_i(y) \frac{\partial \phi}{\partial y_j}|_{y} ,\\
\end{eqnarray*}
where $y_j$'s are coordinate functions for $\mathbb{R}^N.$ Therefore we have
\begin{eqnarray*}
&\frac{\partial}{\partial u_i}|_{y}=\sum_{j=1}^{N}e^j_i(y)
\frac{\partial}{\partial y_j} .
\end{eqnarray*}
That is, $\frac{\partial}{\partial u_i}|_{y}$ is nothing but the vector
$\{e_{i}^{j}(y); j=1,\ldots,N \}$ under the canonical identification of $\mathbb
{R}^{N}$ with $T_{y}\mathbb{R}^{N}$. Hence $\forall \ y$ in a coordinate
neighborhood, $<\frac{\partial}{\partial u_i}|_{y},\frac{\partial}{\partial
x_k}|_{y}> = 0$. If $\frac{\partial}{\partial
x_k}=\sum_{j=1}^{N}\frac{\partial}{\partial y_j} f^j_k,$ then by definition,
$e^j_i$ and $f^j_k$ are orthogonal to each other in $\mathbb{R}^N.$ Hence in coordinate
neighborhood, $\langle \frac{\partial}{\partial u_i}, \frac{\partial}{\partial
x_k}      \rangle=0.$
 \mbox{}\hfill $\sqare$\vspace{1ex}

\begin{lmma}\label{inn-prod-pres}
$d\Phi$ preserves inner-product on ${\cal A}_0.$
\end{lmma}

{\it Proof:}

Observe that  elements of ${\cal A}_0$ are of the form $\sum_{i=1}^{k}\phi_i\otimes\psi_i,$ where $\phi_i(y)=\xi_i(\pi(y))$ and $\psi_i(y)=\eta_i({\cal U}(y)).$

If $\phi(y)\equiv \psi(\pi(y)),$ this implies that $(\pi_F\phi)(x,u_1,...u_{N-n})=\psi(x).$ Therefore $(\pi_F\phi)$ is a function on $M$ alone i.e. $\pi_F\phi\equiv\psi$ as an element of $C^\infty(M),$ so that we have $\Phi(\phi)=\alpha(\phi).$

If $\phi(y)\equiv\eta({\cal U}(y)),$ this implies that $\Phi(\phi)(y)=\eta({\cal U}(y))1_{\cal Q}.$ Therefore we have $\Phi(\phi)=\eta\otimes 1.$ Hence, if $\phi(y)\equiv\psi(\pi(y)),$ then $\Phi(\phi)=\alpha(\psi)$ and if $\phi(y)\equiv\eta({\cal U}(y)),$ then $\Phi(\phi)=\eta\otimes 1_{\cal Q}.$

Let $f_1,g_1,f_2,g_2\in{\cal A}_0,$ $f_i(y)\equiv\psi_i(\pi(y))$ and $g_i(y)\equiv\eta_i({\cal U}(y))$ for $i=1,2.$ Then

\begin{eqnarray*}
&\langle\innerl d\Phi(d(f_1g_1)), d\Phi (d(f_2g_2))                    \rangle\innerr\\
&=\langle\innerl (d\ot1)\Phi(f_1g_1), (d\otimes 1)\Phi(f_2g_2) \rangle\innerr\\
&=\langle\innerl (d\ot1)\alpha(\psi_1)(\eta_1\otimes 1), (d\otimes 1)\alpha(\psi_2)(\eta_2\otimes 1)\rangle\innerr\\
&=\langle\innerl  (d\otimes 1)\alpha(\psi_1)(\eta_1\otimes 1),(d\otimes 1)\alpha(\psi_2)(\eta_2\ot1)\rangle\innerr\\
&+\langle\innerl \alpha(\psi_1)(d\eta_1\ot1),\alpha(\psi_2)(d\eta_2\ot1)\rangle\innerr\\
\end{eqnarray*}
(which follows by Lemma \ref{cordinate system})

\begin{eqnarray*}
&=\langle\innerl  (d\otimes 1)\alpha(\psi_1)(\eta_1\otimes 1),(d\otimes 1)\alpha(\psi_2)(\eta_2\ot1)\rangle\innerr\\
&+\alpha(\psi_1)^*(\langle\innerl  d\eta_1,d\eta_2 \rangle\innerr\otimes 1)\alpha(\psi_2).
\end{eqnarray*}
On the other hand,

\begin{eqnarray*}
&\Phi(\innerl\innerl d(f_1g_1),d(f_2g_2)\innerr\innerr)\\
&=\alpha(\innerl\innerl d(\psi_1\eta_1), d(\psi_2\eta_2)\innerr\innerr)\\
&=\alpha(\innerl\innerl d\psi_1\eta_1+\psi_1 d\eta_1, d\psi_2\eta_2+\psi_2 d\eta_2 \innerr\innerr)\\
&=\alpha(\innerl\innerl  d\psi_1\eta_1,d\psi_2\eta_2\innerr\innerr)+\alpha(\innerl\innerl  \psi_1d\eta_1,\psi_2d\eta_2\innerr\innerr)\\
&=\alpha(\innerl\innerl d\psi_1,d\psi_2\innerr\innerr)(\eta_1\eta_2\ot1)+\alpha(\psi_1)^*(\innerl\innerl d\eta_1,d\eta_2\innerr\innerr\otimes 1_{\cal Q})\alpha(\psi_2),
\end{eqnarray*}
(since $\alpha$ commutes with ${\cal L}$ and $d\alpha$ preserves the inner-product)

\begin{eqnarray*}
&=\innerl\innerl (d\ot1)\alpha(\psi_1)(\eta_1\otimes 1),(d\otimes 1)\alpha(\psi_2)(\eta_2\otimes 1)\innerr\innerr\\
&+\alpha(\psi_1)^*(\innerl\innerl d\eta_1,d\eta_2\innerr\innerr\otimes 1_{\cal Q})\alpha(\psi_2).
\end{eqnarray*}

Therefore we have $$\innerl\innerl d\Phi(d\phi),d\Phi(d\psi)\innerr\innerr=\Phi{\innerl\innerl d\phi, d\psi\innerr\innerr}$$ for $\phi,\psi\in{\cal A}_0.$

So, pick $f,g\in{\cal A}_0.$ Then

\begin{eqnarray*}
\innerl\innerl  d\Phi(fd\phi), d\Phi(g d\psi)\innerr\innerr=\innerl\innerl \Phi(f)d\Phi(d\phi), \Phi(g)d\Phi(d\psi) \innerr\innerr;
\end{eqnarray*}

\begin{eqnarray*}
&\Phi(\innerl\innerl fd\phi, gd\psi \innerr\innerr)\\
&=\Phi(f)^*\Phi(\innerl\innerl d\phi, d\psi              \innerr\innerr)\Phi(g)\\
&=\innerl\innerl \Phi(f)d\Phi(d\phi), \Phi(g)d\Phi(d\psi) \innerr\innerr.
\end{eqnarray*}
Similar calculations hold for finite sums. Hence $d\Phi$ is inner-product preserving.\\

  \begin{crlre}
  \label{157}
   If  the Haar state of ${\cal Q}$ is faithful, then $\Phi$ commutes with the geometric Laplacian ${\cal L}_{\mathbb{R}^{N}}$ on ${\cal N}_{\epsilon}M$ which is same as $\sum_{i=1}^{N}\frac{\partial^{2}}{\partial y_{i}^{2}}$, where $y_{i}$'s are coordinate functions for $\mathbb{R}^{N}$.
  \end{crlre}
{\it Proof:}\\
It is easy to see from Lemma \ref{inn-prod-pres} that $\Phi$ satisfies the
conditions of Theorem \ref{main_thm} provided the Haar state of ${\cal Q}$ is
faithful. The conclusion now follows from Theorem \ref{main_thm}.  \mbox{}\hfill $\sqare$\vspace{1ex}

  \section{Main Results about rigidity of quantum isometry groups}

  \subsection{Non-existence of isometric action of genuine compact quantum groups on stably parallelizable manifolds}

  Let $\{y_{i}:i=1,..,N\}$ be the standard coordinates for $\mathbb{R}^{N}$.\\
  \begin{dfn}
  A twice continuously differentiable, complex-valued function $\Psi$ defined on an open,non empty set $\Omega\subset \mathbb{R}^{N}$  is said to be  harmonic on $\Omega$ if $${\cal L}_{\mathbb{R}^{N}} \Psi\equiv 0,$$
  where ${\cal L}_{\mathbb{R}^{N}}\equiv \sum_{i=1}^{N}\frac{\partial^{2}}{\partial y_{i}^{2}}$.\\
  \end{dfn}
  We note the following result whose proof is essentially the same as Theorem 3.3 of \cite{Rigidity}.\\

  \begin{lmma}
\label{noaction}
Let ${\cal C}$ be a unital commutative $C^{*}$ algebra and $X_{1},X_{2},..., X_{N}$ be self adjoint elements of ${\cal C}$ such that $\{1, X_1,\ldots,X_N, X_{i}X_{j}:1\leq i\leq j\leq N\}$
is a linearly independent set and ${\cal C}$ be the  unital $C^{*}$ algebra generated by $\{1,X_{1},X_{2},...,X_{N}\}$. Let ${\cal Q}$ be a compact quantum group
acting faithfully on ${\cal C}$ such that the action is affine i.e. maps the linear span of  $\{1,X_{1},X_{2},...,X_{N}\}$ into its algebraic tensor product with ${\cal Q}$. Then ${\cal Q}$ must be commutative
as a $C^{*}$ algebra,i.e. ${\cal Q}\cong C(G)$ for some compact group $G$.
\end{lmma}

  \begin{lmma}
\label{quad}
Let $W$ be a manifold (possibly with boundary) embedded in some $\mathbb{R}^{N}$ and $\{y_{i}\}$'s for $i=1(1)N$, be coordinate functions for $\mathbb{R}^{N}$. If $W$ has non empty interior
in $\mathbb{R}^{N}$, then $\{1,y_1,\ldots, y_N, y_{i}y_{j}; 1\leq i,j\leq N\}$ are linearly independent.
\end{lmma}
{\it Proof:}\\
If possible, assume that on $W$, we have  $c.1+\sum_i c_i y_i+\sum_{i\leq j}
c_{ij}y_{i}y_{j}=0$ for some $c, c_i,c_{ij}$'s.\\
Pick an interior point $y\in W$. Then at $y$, we can take partial derivatives in any direction.\\
Applying $\frac{\partial}{\partial y_{i}}|_{y}\frac{\partial}{\partial
y_{j}}|_{y}$ to $c.1+\sum_i c_i y_i+\sum c_{ij}y_{i}y_{j}=0$, we conclude that
$c_{ij}=0 \ \forall \ i \leq j$, and then applying $\frac{\partial}{\partial
y_{i}}|_{y}$ on $c+\sum_i c_i y_i=0$, we get $c_i=0$.\\
So $c=0$ too.\\
 \mbox{}\hfill $\sqare$\vspace{1ex} \\
  \begin{lmma}
  \label{affine}
  Let $\Phi$ be as in Corollary \ref{157} of section 5. Then $\Phi$ is affine i.e.$$\Phi(y_{i})=1\otimes q_{i}+ \sum_{j=1}^{N}y_{j}\otimes q_{ij}, \ for \ some \ q_{ij},q_{i}\in {\cal Q} $$ for all $i=1(1)N$.
  \end{lmma}
 {\it Proof:}\\
 As $\Phi$ commutes with the geometric Laplacian ${\cal L}_{\mathbb{R}^{N}}$ on ${\cal N}_{\epsilon}M$, for $\psi\in {\cal Q}^{*}$, ${\cal L}_{\mathbb{R}^{N}} \Phi_{\psi}(y_{i})=\Phi_{\psi}({\cal L}_{\mathbb{R}^{N}} y_{i})=0$ for all $i=1(1)N$ (here $\Phi_\psi=({\rm id \otimes \psi} \circ \Phi$).\\
 Similarly, as ${\cal L}_{{\mathbb{R}^N}}$ commutes with $\frac{\partial}{\partial y_{j}}$, we have
 \begin{eqnarray*}
 && {\cal L}_{\mathbb{R}^{N}} (id\otimes \psi)(\frac{\partial}{\partial y_{j}}\otimes id)\Phi(y_{i})\\
 &=& (id\otimes \psi) (\frac{\partial}{\partial y_{j}}\otimes id)\Phi({\cal L}_{\mathbb{R}^{N}} y_{i})\\
 &=& 0.
 \end{eqnarray*}
Hence $((id\otimes \psi)(\frac{\partial}{\partial y_{j}}\otimes id)\Phi(y_{i}))(y)$ is harmonic for all $\psi\in {\cal Q}^{*}$ and $y\in {\cal N}_{\epsilon}M$.\\
Let $D_{ij}(y)=((\frac{\partial}{\partial y_{j}}\otimes id)\Phi(y_{i}))(y)$.
Note that $E:=\Lambda^1(C^\infty({\cal N}_\epsilon M))$ is a free $C^\infty(M)$ Hilbert module with an orthonormal basis
$\{ dy_1,\ldots,dy_N\}$, and $d \Phi$ is an equivariant unitary representation of ${\cal Q}$ on it. But $D:=(( D_{ij} )) \in M_N(C^\infty({\cal N}_\epsilon M) \hat{\otimes} {\cal Q}) \cong {\cal L}(E \hat{\otimes} {\cal Q})$ is nothing but the matrix for the map $d \Phi$ with respect to the
 basis $\{ dy_1,\ldots,dy_N\}$. Hence $D$ must be a unitary element of $M_N(C^\infty({\cal N}_\epsilon M) \hat{\otimes} {\cal Q})$.\\

Now, let us pick $y_{0}$ in the interior of ${\cal N}_{\epsilon}M$ (which is non empty). Then the new ${\cal Q}$ valued matrix $G(y):=((G_{ij}(y)))=((D_{ij}(y)))((D_{ij}(y_{0})))^{-1}$ is unitary (since $((D_{ij}(y)))$ is so).\\
The unitariry of $(( G_{ij}(y)))$  for all $y$ implies $|\psi(G_{ij}(y))|\leq 1$.
Moreover, $\psi(G_{ii}(y_{0}))=1$.
Thus,  $f(y):=\psi(G_{ii}(y))$ is a harmonic function on an open connected set $Int ({\cal N}_{\epsilon}M)$ such that $|f|$ attains the  supremum at an interior point. Applying Corollary 1.11 of \cite{Harmonic} we conclude that $\psi(G_{ii}(y))=\psi(G_{ii}(y_{0}))=\psi(1)$ for all $\psi$, hence $G_{ii}(y)=1$ for all $i$. \\
$((G_{ij}(y)))$ being unitary for all $y$, we must have $G_{ij}(\cdot)=\delta_{ij}.1_{{\cal Q}}$.\\
That is, $((D_{ij}(y)))((D_{ij}(y_{0})))^{-1}=1_{M_n({\cal Q})}$.\\
So $((D_{ij}(y)))=((D_{ij}(y_{0})))$ for all $y\in {\cal N}_{\epsilon}(M)$.
It is now easy to see that  $\Phi$ is affine with $q_{ij}=D_{ij}(y_{0})$ \mbox{}\hfill $\sqare$\vspace{1ex}  \\

Now we are ready to state and prove the main result of this paper.\\
\begin{thm}
\label{main}
Let $M$ be a smooth, compact, oriented, connected Riemannian manifold without boundary which is stably parallelizable.
Let ${\cal Q}$ be a compact quantum group acting faithfully and isometrically on $M$. Then ${\cal Q}$ must be commutative, i.e. of the form $C(G)$ for some compact group $G$.
\end{thm}
{\it Proof:}\\
First apply  Lemma \ref{noaction}, Lemma \ref{quad} and Lemma \ref{affine}  to the action of the reduced quantum group ${\cal Q}_r$ (which has a faithful Haar state) to conclude that ${\cal Q}_r$ is commutative. But this implies the commutativity of ${\cal Q}_0$ and hence that of ${\cal Q}$, by the fact that ${\cal Q}_0$ is embedded as a dense subalgebra in both ${\cal Q}_r$ and ${\cal Q}$ (in their respective norms). \mbox{}\hfill $\sqare$\vspace{1ex}\\
In fact, making use of  the averaging trick of Subsection 4.3, we can restate the above theorem as follows:
\begin{thm}
Let ${\cal Q}$ be a CQG acting faithfully on $C(M)$, where $M$ is smooth, compact,
connected, oriented, stably parallelizable manifold and such that the action is
smooth and satisfies the condition (\ref{dalpha_eq}) of Theorem \ref{def_dalpha}
of Subsection 3.2. Then ${\cal Q}$ must be commutative as a $C^*$ algebra.
\end{thm}

\subsection{Quantum isometry group of Rieffel-deformation of stably parallelizable manifolds}
We shall now consider deformation of classical manifolds and quantum group actions on them. Let $M$ be as in  Theorem \ref{main} above and assume also that the group of Riemannian isometries $G=ISO(M)$ has a toral subgroup $T$ of rank at least two.   We can then consider the Rieffel-deformation $C(M)_\theta$ using the left action of $T$ on $M$ (see \cite{rieffel} for the definition and details of such deformation), indexed by skew-symmetric $n \times n$ matrices $\theta$, which is a continuous field of   possibly noncommutative $C^*$ algebras.

In a similar way, if  a compact group $K$ has an $n$-toral subgroup $T$, we can consider the Rieffel-Wang deformation  $C(K)_{\tilde{\theta}}$ (see \cite{rieffel_wang}, \cite{toral}) of
$C(K)$ by the action of the $2n$-dimensional torus $T \times T$ on $K$ given by $(z,w)g:=zgw^{-1}$, $z,w \in T, g \in K$, and where $\tilde{\theta}=\left( \begin{array}{cc} 0 & \theta \\
-\theta & 0 \\ \end{array} \right).$ This becomes a compact quantum group with the same coalgebra structure as $C(G)$.

We recall from \cite{qiso_comp} that there is a natural Laplacian on the deformed noncommutative manifold $C(M)_\theta$ and its quantum isometry group is isomorphic with a suitable Rieffel-Wang deformation of QISO of the commutative manifold, which is nothing but $C(ISO(M))$ by Theorem \ref{main}. This gives us the following:
\begin{thm}
With the above notations, the quantum isometry group of the noncommutative manifold $C^\infty(M)_\theta$  is  isomorphic with the Rieffel-Wang deformation $C(G)_{\tilde{\theta}}$, where $G=ISO(M)$ and $\tilde{\theta}=\theta \oplus (-\theta).$
\end{thm}

redpajama_set_name: RedPajamaArXiv