f(\mathcal{L}_{E_0})\in \Psi^{m,\mathcal{L}}_{1,0}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).
\end{equation}

In particular, for $f_s(\lambda):=(1+\lambda)^{\frac{s}{2}},$ the symbol of $\mathcal{M}_{s,E_0}=f_s(\mathcal{L}_{E_0})$ belongs to $ S^{s,\mathcal{L}}_{1,0}( (G\times \widehat{G})\otimes \textnormal{End}(E_0)) $ for all $s\in \mathbb{R}.$ In other words, we have

\begin{equation}\label{thepowersof:LE0}
\mathcal{M}_{s,E_0}:=(1+\mathcal{L}_{E_0})^{\frac{s}{2}}\in \Psi^{s,\mathcal{L}}_{1,0}((G\times \widehat{G})\otimes\textnormal{End}(E_0))
\end{equation}

for all $s\in \mathbb{R}.$

We first show the following result which is used to prove  Theorem <ref>.
The matrix-valued symbol of the vector-valued operator $ f(\mathcal{L}_{E_0})$ is given by

\begin{equation}
\sigma_{f(\mathcal{L}_{E_0})}(i,r,[\xi])\equiv  f(\widehat{\mathcal{L}}_{E_0})(i,r,\xi)=f(\widehat{\mathcal{L}}(\xi))\delta_{i,r},\,\,1\leq i,r\leq d_\tau,\,\,[\xi]\in \widehat{G},
\end{equation}
where $\delta_{i,r}$ is the Kronecker-Delta.

Keeping in mind Remark <ref>, let us observe that the $\textnormal{End}(E_0)$-valued symbol  of $f({\mathcal{L}}_{E_0}),$ is given by

\begin{equation*}
\end{equation*}
so that

\begin{equation*}
f(\widehat{\mathcal{L}}_{E_0}(\xi))= f(\widehat{\mathcal{L}}(\xi)) \otimes I_{ \textnormal{End}(E_0)}  \in \mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0)).
\end{equation*}
Now, notice that the matrix-valued symbol of $f({\mathcal{L}}_{E_0})$ can be computed as follows:

\begin{align*}
f(\widehat{\mathcal{L}}_{E_0})(i,r,\xi)&= (f(\widehat{\mathcal{L}}_{E_0})(i,r,\xi)_{u,v})_{u,v=1}^{d_\xi} =(e_{r,E_0}^{*}( f(\widehat{\mathcal{L}}_{E_0}(\xi))_{uv}e_{i,E_0}) )_{u,v=1}^{d_\xi}\\
&=(e_{r,E_0}^{*}( f(\widehat{\mathcal{L}}(\xi))_{uv}e_{i,E_0}) )_{u,v=1}^{d_\xi}=(f(\widehat{\mathcal{L}}(\xi))_{uv})_{u,v=1}^{d_\xi}\delta_{ir}\\
\end{align*}
Thus, we end the proof.

In view of Lemma <ref>, in order to prove that  $f(\mathcal{L}_E)\in \Psi^{m,\mathcal{L}}_{1,0}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),$ it is enough to check the validity of the following symbol inequalities

\begin{equation}\label{InI:ex}
\sup_{\substack{1\leq i,r\leq d_\tau\,  [\xi]\in  \widehat{G} }
} \Vert \widehat{ \mathcal{M}}(\xi)^{\gamma+|\alpha|-m} \Delta_{\xi}^{\alpha}[f(\widehat{\mathcal{L}}(\xi))]\delta_{i,r}\mathcal{M}(\xi)^{-\gamma} \Vert_{\textnormal{op}} <\infty,
\end{equation}
for any $\gamma\in \mathbb{R}.$ But it was proved in Theorem 8.17 of  [10] that $f(\mathcal{L})\in S^{m,\mathcal{L}}_{1,0}(G\times \widehat{G}),$ from which the validity of (<ref>) follows. The proof is complete.

§.§ Mapping properties for subelliptic operators on vector bundles

In this subsection we will discuss the mapping properties of subellipic pseudo-differential  operators. Let us recall that an operator $A:C^\infty(G, E_0) \rightarrow C^\infty(G, F_0)$ can be expressed, in view of (<ref>) as

\begin{align}\label{sumofoperators}
Af(x)= \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} A_{ir}f_i(x) e_{r, F_0},
\end{align}

where $A_{ir}f_i$ is defined on the complex valued function $f_i(x)= \langle f(x), e_{i, E_0}\rangle$ for every $1 \leq i \leq d_\tau, 1 \leq r \leq d_\omega.$
Let $\mathcal{F}_1(G)$ and $\mathcal{F}_2(G)$ be two normed linear spaces containing $C^\infty(G)$ as a dense subspace. Let  $\mathcal{F}_1(G, E_0)=\mathcal{F}_1(G)\otimes E_0$ and  $\mathcal{F}_2(G, F_0)= \mathcal{F}_2(G)\otimes F_0,$ be defined by the norms

\begin{equation}
\Vert  f_1\Vert_{\mathcal{F}_1(G, E_0)} = \Vert \Vert f_1(\cdot)\Vert_{E_0} \Vert_{\mathcal{F}_1(G)}\,\,\Vert  f_2\Vert_{\mathcal{F}_2(G, F_0)} = \Vert \Vert f_2(\cdot)\Vert_{F_0} \Vert_{\mathcal{F}_2(G)}.
\end{equation}

$A$ admits a  bounded extension from  $\mathcal{F}_1(G, E_0)$ to  $\mathcal{F}_2(G, F_0),$  if and only if, for all $1\leq i\leq d_\tau,$ and $1\leq r\leq d_\omega,$ $A_{ir}$ in (<ref>), admits a bounded extension  from $\mathcal{F}_1(G)$ to $\mathcal{F}_2(G).$

Let us first assume that for all $1 \leq i \leq d_\tau$ and $1 \leq r \leq d_\omega,$ $ A_{ir} $ admit a bounded extension for $\mathcal{F}_1(G)$ into $\mathcal{F}_2(G)$, that is, for all $g \in C^\infty(G),$

\begin{equation}\label{boundedenessir}
\Vert A_{ir} g \Vert_{\mathcal{F}_2(G)} \leq  C \Vert g \Vert_{\mathcal{F}_1(G)}, \quad \textnormal{for some}\,\, C>0,\,\forall \,1\leq i\leq d_{\tau},\,1\leq r\leq d_\omega.
\end{equation}

Now, for $f \in C^\infty(G, E_0)$ we have

\begin{align*}
&\Vert A f\Vert_{\mathcal{F}_2(G, F_0)} = \Vert \Vert Af(\cdot)\Vert_{F_0} \Vert_{\mathcal{F}_2(G)} = \Vert \Vert \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} A_{ir}f_i(\cdot) e_{r, F_0} \Vert_{F_0} \Vert_{\mathcal{F}_2(G)} \\& \leq \Vert \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} \Vert  A_{ir}f_i(x) e_{r, F_0} \Vert_{F_0} \Vert_{\mathcal{F}_2(G)}
=  \Vert \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} |A_{ir}f_i(\cdot)| \Vert   e_{r, F_0} \Vert_{F_0} \Vert_{\mathcal{F}_2(G)}.
\end{align*}
Using (<ref>), we have

\begin{align*}
\Vert A f\Vert_{\mathcal{F}_2(G, F_0)} &  \leq \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} \Vert  A_{ir}f_i  \Vert_{\mathcal{F}_2(G)} \leq   \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} C \Vert f_i  (\cdot)\Vert_{\mathcal{F}_1(G)}.
\end{align*}
Using the Cauchy-Schwarz inequality, we have

\begin{align*}
\Vert f_i  (\cdot)\Vert_{\mathcal{F}_1(G)}\leq  \Vert |\langle f(\cdot), e_{r, E_0} \rangle| \Vert_{\mathcal{F}_1(G)}\leq \Vert  \Vert f(\cdot) \Vert_{E_0} \Vert e_{r, E_0}\Vert_{E_0} \Vert_{\mathcal{F}_1(G)},
\end{align*}
and consequently,

\begin{align*}
\Vert A f\Vert_{\mathcal{F}_2(G, F_0)} &\leq  \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} C \Vert  \Vert f(\cdot) \Vert_{E_0}  \Vert_{\mathcal{F}_1(G)}= \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} C \Vert f  \Vert_{\mathcal{F}_1(G, E_0)}= C' \Vert f  \Vert_{\mathcal{F}_1(G, E_0)}.
\end{align*}

Therefore, by using the density of $C^\infty(G, E_0)$ we deduce that $A$ has a bounded extension from $\mathcal{F}_1(G, E_0)$ to $\mathcal{F}_2(G, F_0).$ For the converse part, let us assume that for $h\in C^\infty(G, E_0)$ we have
$$\Vert Ah\Vert_{\mathcal{F}_2(G, F_0)} \leq C_1\Vert h \Vert_{\mathcal{F}_1(G, E_0)}.$$

Then for $f \in \mathcal{F}_1(G)$ and for $1 \leq j \leq d_\tau,$ consider the function $f_{e_{j, E_0}}:G \rightarrow E_0$ defined by $f_{e_{j, E_0}}(x):= f(x) e_{j, E_0}.$ Then it is easy to see that $f_{e_{j, E_0}} \in \mathcal{F}_1(G, E_0).$ Now,

\begin{align*}
A f_{e_{j, E_0}}(x)= \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} A_{ir}(f_{e_{j, E_0}})_i(x) e_{r, F_0} = \sum_{r=1}^{d_\omega} A_{jr} f(x) e_{r, F_0},
\end{align*}

and so, for $1 \leq s \leq d_\omega,$ we obtain $\langle A f_{e_{j, E_0}}(x), e_{s, F_0}   \rangle= A_{js}f(x),$ for $x \in G.$

\begin{align*}
\Vert A_{js}f \Vert_{\mathcal{F}_2(G)}&= \Vert |\langle A f_{e_{j, E_0}}(\cdot), e_{s, F_0}   \rangle| \Vert_{\mathcal{F}_2(G)} \\
&\leq \Vert \Vert A f_{e_{j, E_0}}(\cdot) \Vert_{F_0} \Vert e_{s, F_0}\Vert_{F_0} \Vert_{\mathcal{F}_2(G)}  \\& = \Vert \Vert A f_{e_{j, E_0}}(\cdot) \Vert_{F_0}  \Vert_{\mathcal{F}_2(G)}\\
&=\Vert A f_{e_{j, E_0}} \Vert_{\mathcal{F}_2(G, F_0)}\\
&\leq C_1 \Vert  f_{e_{j, E_0}} \Vert_{\mathcal{F}_1(G, F_0)} \\&= C_1 \Vert  f e_{j, E_0} \Vert_{\mathcal{F}_1(G, E_0)} = C_1 \Vert \Vert f(\cdot) e_{j, E_0} \Vert_{E_0} \Vert_{\mathcal{F}_1(G)} = C_1 \Vert f \Vert_{\mathcal{F}_1(G)},
\end{align*}
for all $f \in C^\infty(G).$ Therefore, for each $1 \leq j \leq d_\tau$ and $1 \leq s \leq d_\omega,$ $A_{js}$ admits a bounded extension from $\mathcal{F}_1(G)$ into $F_{2}(G).$

In view of Lemma <ref> and Theorem <ref> we obtain the following subelliptic vector-valued Calderón-Vaillancourt theorem.

Let $G$ be a compact Lie group and let us consider the sub-Laplacian $\mathcal{L}=\mathcal{L}_X,$ where  $X=\{X_{i}\}_{i=1}^{k}$ is a system of vector fields satisfying the Hörmander condition of order $\kappa$.  For  $0\leq \delta< \rho\leq    1,$  (or $0\leq \delta\leq \rho\leq 1,$ and $\delta<1/\kappa$) let us consider a continuous linear operator $A:C^\infty(G, E_0)\rightarrow\mathscr{D}'(G, F_0)$ with symbol  $\sigma\in {S}^{0,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0))$. Then $A$ extends to a bounded operator from $L^2(G, E_0)$ to  $L^2(G, F_0),$ and

\begin{equation}
\Vert  A\Vert_{\mathscr{B}(L^2(G,E_0), L^2(G, F_0))}\leq C \Vert \sigma\Vert_{\ell, {S}^{0,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0)) },
\end{equation}
for $\ell \in \mathbb{N}$ large enough.  In particular, if $A$ maps continuously $C^{\infty}(G,E_0)^\tau$ into $C^{\infty}(G,F_0)^\omega,$ then $A$ extends to a bounded operator from $L^2(G, E_0)^{\tau}$ to  $L^2(G, F_0)^{\omega}.$

Again,  Lemma <ref> implies the following  $L^p$-theorem of boundedness for vector-valued subelliptic pseudo-differential operators as a consequence of Theorem <ref>.

Let $G$ be a compact Lie group and let us denote by $Q$ the Hausdorff
dimension of $G$ associated to the control distance associated to the sub-Laplacian $\mathcal{L}=\mathcal{L}_X,$ where  $X=\{X_{i}\}_{i=1}^{k}$ is a system of vector fields satisfying the Hörmander condition of order $\kappa$.  For  $0\leq \delta<\rho\leq 1,$ let us consider a continuous linear operator $A:C^\infty(G, E_0)\rightarrow\mathscr{D}'(G, F_0)$ with symbol  $\sigma\in {S}^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0))$, $m\geq 0$. Then $A$ extends to a bounded operator from $L^p(G, E_0)$ to $L^p(G, F_0)$ provided that

\begin{equation*}
m\geq m_p:= Q(1-\rho)\left|\frac{1}{p}-\frac{1}{2}\right|.
\end{equation*}
In particular,  if $A$ maps continuously $C^{\infty}(G,E_0)^\tau$ into $C^{\infty}(G,F_0)^\omega,$ then $A$ extends to a bounded operator from $L^p(G, E_0)^{\tau}$ to  $L^p(G, F_0)^{\omega}.$

Now, Theorem <ref> provides a vector-valued version of Corollary <ref>.

Let $A:C^{\infty}(G, E_0)\rightarrow \mathscr{D}'(G, F_0)$ be a continuous linear operator with symbol $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0)),$ $0\leq \delta< \rho\leq    1,$ (or $0\leq \delta\leq \rho\leq 1,$ $\delta<1/\kappa$). Then $A:L_{s}^{2, \mathcal{L}}(G, E_0)\rightarrow L_{s-m}^{2,\mathcal{L}}(G, F_0) $ extends to a bounded operator for all $s\in \mathbb{R}.$ In particular,  $A:L_{s}^{2, \mathcal{L}}(G, E_0)^\tau \rightarrow L_{s-m}^{2,\mathcal{L}}(G, F_0)^\omega $ extends to a bounded operator for all $s\in \mathbb{R}.$

From Definition  <ref> and  (<ref>) we have

\begin{align}
\Vert Af \Vert_{L_{s-m}^{2,\mathcal{L}}(G, F_0)} = \Vert \mathcal{M}_{s-m} A f\Vert_{L^2(G, F_0)},
\end{align}
where $\mathcal{M}_{\lambda}:=\textnormal{Op}\left(\widehat{\mathcal{M}}(\xi)^{\lambda} \otimes I_{ \textnormal{End}(E_0)}\right),$ $\lambda\in \mathbb{R}.$

Now observing that $\mathcal{M}_{s-m} A \mathcal{M}_{-s} \in \Psi^{0,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0)) $ and  then using the vector-valued Calderón-Vaillancourt theorem (Theorem <ref>) we obtain

\begin{align*}
\Vert Af \Vert_{L_{s-m}^{2,\mathcal{L}}(G, F_0)}&= \Vert \mathcal{M}_{s-m} A \mathcal{M}_{-s} \mathcal{M}_s f\Vert_{L^2(G, F_0)}\\& \leq C \Vert \mathcal{M}_s f\Vert_{L^2(G, E_0)}=\Vert f \Vert_{L_{s-m}^{2,\mathcal{L}}(G, E_0)}.
\end{align*}

This completes the proof of the theorem.

§.§ Quantisation on homogeneous vector bundles
In this subsection we obtain a quantisation formula on homogeneous vector bundles. We will follow the notation in (<ref>) for the inverse mapping $\varkappa_{\omega}^{-1}:C^\infty(G,F_0)^\omega\rightarrow\Gamma^\infty(F).$ In view that the following  diagram
\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma^\infty(E)$}  ; & \node(S) {$\Gamma^\infty(F)$}; \\
\node(R/I) {$C^\infty(G,E_0)^\tau$}; & \node (T) {$C^\infty(G,F_0)^\omega$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_{\tau}$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\omega$};
\draw[->] (R/I) -- (T) node[anchor=north] {$A$};
\end{tikzpicture}
\end{eqnarray}
commutes, we have the identity $\tilde{A}s=\varkappa_\omega^{-1} A\varkappa_\tau s,$ for any section $s\in \Gamma^\infty(E).$ Since

\begin{equation}   {
(A\varkappa_\tau s)(g)= \,\sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\textnormal{Tr}[\xi(g)\sigma_A(q,r,g,[\xi])\widehat{\varkappa_\tau  s}(q,[\xi])]e_{r,F_0}  ,   }
\end{equation}
when applying $\varkappa_\omega^{-1}$ in both sides of the previous identity, we obtain

\begin{align*}
\tilde{A}s(gK)&= (\varkappa_\omega^{-1} A\varkappa_\tau s)(gK)=[g,(A\varkappa_\tau s)(g)]:=(g, (A\varkappa_\tau s)(g))\cdot K\\
&=\left(g, \sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\textnormal{Tr}[\xi(g)\sigma_A(q,r,g,[\xi])\widehat{\varkappa_\tau  s}(q,[\xi])]e_{r,F_0} \right)\cdot K.
\end{align*}

The previous analysis  proves the following quantisation theorem.

Let $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ be a continuous linear operator and let $A: C^\infty(G,E_{0})^\tau\rightarrow C^\infty(G,F_{0})^\omega$ be the induced operator by the identifications $\Gamma^\infty(E)\cong C^\infty(G,E_{0})^\tau$ and $\Gamma^\infty(F)\cong C^\infty(G,F_{0})^\omega.$ Then we have

\begin{equation}\label{quantizationonhomogeneous2}
\boxed{     \tilde{A}s(gK)\equiv \left(g, \,\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{d_\tau}\sum_{r=1}^{d_\omega}\textnormal{Tr}[\xi(g)\sigma_A(i,r,g,[\xi])  \widehat{\varkappa_\tau  s}(i,[\xi])]   e_{r,F_0}\right)\cdot K}
\end{equation}
for every section $s\in \Gamma^\infty(E),$ where $\sigma_{A}$ is the matrix-valued symbol of $A.$

Let us recall that the symbol $\sigma_A,$ can be obtained from $A=\varkappa_\omega\tilde{A}\varkappa_\tau^{-1}$ via

\begin{equation}\boxed{
\sigma_{A}(i_0,r_0,x,\xi)= \xi(x)^{*}
e_{r_0,F_0}^{*}[A(\xi\otimes e_{i_0,E_0})(x)]   }
\end{equation}
in view of (<ref>). Because of  Theorem <ref>, we will introduce in the following definition the global matrix-valued symbol associated to a continuous linear operator between the spaces of sections $\Gamma^\infty(E)$ and $\Gamma^\infty(F).$

Let $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ be a continuous linear operator.
In view of Theorem <ref>,  we will refer to $\sigma_{\tilde{A}}:=\sigma_A(\cdot,\cdot,\cdot,\cdot)$ as the matrix-valued symbol of $\tilde{A}.$ Keeping in mind the defined identity
we will refer also to $\sigma_A$ the matrix-valued symbol of $\tilde{A}.$

§.§ $L^p$-estimates for Hörmander classes on homogeneous vector-bundles

Having defined a matrix-valued Fourier   transform for sections and proved a quantisation formula on homogeneous vector bundles,
let us now introduce the subelliptic Hörmander classes.

Let $G$ be a compact Lie group and let  $0\leq \delta,\rho\leq 1.$
Let us consider the sub-Laplacian $\mathcal{L}=\mathcal{L}_X,$ where  $X=\{X_{i}\}_{i=1}^{k}$ is a system of vector fields satisfying the Hörmander condition of order $\kappa$.

The subelliptic Hörmander class of order $m,$ and of type $(\rho,\delta),$ $\Psi^{m,\mathcal{L}}_{\rho,\delta}(E,F),$ is defined by

\begin{equation}\label{Subelliptichormanderclassesonvectorbundles}
\Psi^{m,\mathcal{L}}_{\rho,\delta}(E,F):=\{\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)|\,\,\,  \sigma_A\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0))  \} ,
\end{equation}
where the classes ${S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0))$ were introduced in Definition <ref>.

As a consequence of our analysis for vector-valued operators, and in view that the subelliptic calculus for vector-valued operators  is stable under compositions (Theorem <ref>) and adjoints  (Theorem <ref>), in the following theorem we establish the pseudo-differential calculus for subelliptic operators on homogeneous vector bundles. We will denote by $H$ a homogeneous vector bundle over $M=G/K,$ and $\Gamma^\infty(H)\cong C^\infty(G,H_0)^{\nu}.$

Let  $0\leq \delta<\rho\leq 1.$ If $\tilde{A} \in \Psi^{m_1,\mathcal{L}}_{\rho,\delta}(E,F)$ and $\tilde{B}\in \Psi^{m_2,\mathcal{L}}_{\rho,\delta}(F,H)$ then the composition operator $\tilde{B}\tilde{A}:=\tilde{B}\circ\tilde{ A}:\Gamma^\infty(E)\rightarrow \Gamma^\infty(H)$ belongs to the subelliptic class $\Psi^{m_1+m_2,\mathcal{L}}_{\rho,\delta}(E,H).$ The symbol of $\tilde{B}\tilde{A},$ $\sigma_{\tilde{B}\tilde{A}}\in\Sigma ((G\times \widehat{G})\otimes \textnormal{End}(E_0,H_0)),$ satisfies the asymptotic expansion,

\begin{equation*}
\sigma_{\tilde{B}\tilde{A}}(i, s, x,\xi) \sim \sum_{r=1}^{d_\omega}\sum_{|\alpha|= 0}^\infty(\Delta_{\xi}^\alpha\sigma_{\tilde{B}}(r, s, x,\xi))(\partial_{X}^{(\alpha)} \sigma_{\tilde{A}}(i, r, x,\xi)),
\end{equation*}
this means that, for every $N\in \mathbb{N},$

\begin{align*}
&\Delta_{\xi}^{\alpha_\ell}\partial_{X}^{(\beta)}\left(\sigma_{\tilde{B}\tilde{A}}(i, s, x,\xi) - \sum_{r=1}^{d_\omega}\sum_{|\alpha| \leq N }(\Delta_{\xi}^\alpha\sigma_{\tilde B}(r, s, x,\xi))(\partial_{X}^{(\alpha)} \sigma_{\tilde A}(i, r, x,\xi))  \right)\\
&\hspace{2cm}\in {S}^{m_1+m_2-(\rho-\delta)(N+1)-\rho\ell+\delta|\beta|,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,H_0)),
\end{align*}
for every difference operator $\Delta_{\xi}^{\alpha_\ell}$ of order $\ell\in\mathbb{N}_0.$ Moreover, under the indentifications $F_0^{*}\cong F_0$ and $E_0^{*}\cong E_0,$ the symbol of the adjoint $\tilde{A}^* \in \Psi^{m,\mathcal{L}}_{\rho,\delta}(F,E)$ of $\tilde{A},$   satisfies the asymptotic expansion,

\begin{equation*}
\sigma_{\tilde{A}^*}(i,s, x,\xi)\sim  \sum_{|\alpha|= 0}^\infty\Delta_{\xi}^\alpha\partial_{X}^{(\alpha)} \sigma_{\tilde A}(i,s, x, \xi)^*.
\end{equation*}

This means that, for every $N \in \mathbb{N}$,

\begin{equation*}
\Delta_{\xi}^{\alpha_\ell} \partial_X^{(\beta)} \left(\sigma_{\tilde A^*}(i,s, x,\xi) -  \sum_{|\alpha|\leq N }\Delta_{\xi}^\alpha\partial_{X}^{(\alpha)} \sigma_{\tilde A}( i,s,x, \xi)^* \right)
\end{equation*}
belongs to the symbol class ${S}^{m-(\rho-\delta)(N+1)-\rho\ell+\delta|\beta|,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(F_0,E_0)).$

The asymptotic expansions were proved in Theorems <ref> and <ref>.
In view of the following two commutative diagrams,

\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma^\infty(E)$}  ; & \node(S) {$\Gamma^\infty(F)$};  & \node(S') {$\Gamma^\infty(H)$}; \\
\node(R/I) {$C^\infty(G,E_0)^\tau$}; & \node (T) {$C^\infty(G,F_0)^\omega$};& \node (T') {$C^\infty(G,H_0)^\nu$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_\tau$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}$};
\draw[->] (S) -- (S') node[anchor=south] {$\tilde{B}$};
\draw[->] (S') -- (T') node[anchor=west] {$\varkappa_\nu$};
\draw[->] (T) -- (T') node[anchor=south] {$B$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\omega$};
\draw[->] (R/I) -- (T) node[anchor=south] {$A$};
\end{tikzpicture}
\end{eqnarray}

\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma(F^*)$}  ; & \node(S) {$\Gamma(E^*)$}; \\
\node(R/I) {$C^\infty(G,F_0)^\omega$}; & \node (T) {$C^\infty(G,E_0)^\tau$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_{\tau}$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}^*$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\omega$};
\draw[->] (R/I) -- (T) node[anchor=north] {$A^*$};
\end{tikzpicture}
\end{eqnarray}

we can conclude the proof because of the identities $\widetilde{AB}=\tilde{A}\tilde{B}$ and $\widetilde{A^*}=(\tilde{A})^{*},$ which implies that the induced vector-valued  operators by $\tilde{A}\tilde{B}$ and  $(\tilde{A})^{*}$ are indeed, the continuous linear operators $AB$ and $A^{*}.$

Now, we present following $L^p$-estimates of subelliptic pseudo-differential operators on homogeneous vector bundles over compact homogeneous manifolds.
For  $0\leq \delta< \rho\leq    1$ (or $0\leq \delta\leq \rho\leq 1,$ $\delta<1/\kappa$), any continuous linear operator $\tilde{A}\in  \Psi^{0,\mathcal{L}}_{\rho,\delta}(E,F)$ extends to a bounded operator from  $L^{2}(E)$ to  $L^{2}(F).$ Moreover, if
$$m\geq Q(1-\rho)\left|\frac{1}{2}-\frac{1}{p}\right|,$$
with $1<p<\infty,$ then $\tilde{A}\in  \Psi^{-m,\mathcal{L}}_{\rho,\delta}(E,F)$ with  $0\leq \delta< \rho\leq    1,$ extends to a bounded operator from  $L^{p}(E)$ into $L^{p}(F).$

In view that the following  diagram
\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma^\infty(E)$}  ; & \node(S) {$\Gamma^\infty(F)$}; \\
\node(R/I) {$C^\infty(G,E_0)^\tau$}; & \node (T) {$C^\infty(G,F_0)^\omega$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_{\tau}$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\omega$};
\draw[->] (R/I) -- (T) node[anchor=north] {$A$};
\end{tikzpicture}
\end{eqnarray}
commutes, and the fact that the arrows $\varkappa_\tau$ and $\varkappa_\omega$ admit  unitary extensions, $\tilde{A}$ admits a bounded extension from $L^{2}(E)$ to  $L^{2}(F),$ if and only if, $A$  extends to a bounded operator from $L^2(G, E_0)^\tau$ to  $L^2(G, F_0)^\omega.$ But, this is consequence of the vector-valued Calderón-Vaillancourt theorem (Theorem <ref>). Now consider $m\geq Q(1-\rho)\left|\frac{1}{2}-\frac{1}{p}\right|,$   $1<p<\infty,$ and  $\tilde{A}\in  \Psi^{-m,\mathcal{L}}_{\rho,\delta}(E,F),$  $0\leq \delta< \rho\leq    1.$ If $s\in \Gamma^\infty(E),$ then there exists a unique $f\in C^{\infty}(G,E_0)^{\tau}$ such that

\begin{equation*}
\Vert  s\Vert_{L^p(E)}=\Vert \varkappa_\tau^{-1}f \Vert_{L^{p}(G\times_\tau E_0)}=\Vert g \mapsto(g,f(g))\cdot K \Vert_{L^{p}(G\times_\tau E_0)}=\Vert f\Vert_{L^p(G,E_0)}.
\end{equation*}
Also, observe that

\begin{align*}
\Vert \tilde A s\Vert_{L^p(E)}
&=\Vert \varkappa_\tau^{-1}\tilde{A} f \Vert_{L^{p}(G\times_\omega F_0)}\\
&=\Vert g \mapsto(g,Af(g))\cdot K \Vert_{L^{p}(G\times_\omega F_0)}=\Vert Af\Vert_{L^p(G,F_0)}.
\end{align*}
In view of the Fefferman theorem (Theorem <ref>), we deduce the inequality
$$\Vert Af\Vert_{L^p(G,F_0)}\leq C\Vert f\Vert_{L^p(G,E_0)},$$
for some $C>0,$ from which we deduce the $L^p$-boundeness of $\tilde A.$ The proof is complete.

Now, we have the following Sobolev estimate.
For $0\leq \delta< \rho\leq    1$ (or $0\leq \delta\leq \rho\leq 1,$ $\delta<1/\kappa$), any continuous linear operator $\tilde{A} \in \Psi^{m,\mathcal{L}}_{\rho. \delta}(E, F),$ extends to a bounded operator from  $L_{s}^{2, \mathcal{L}}(E)$ into $ L_{s-m}^{2,\mathcal{L}}(F)$ for all $s\in \mathbb{R}.$

By Definition <ref> and noting that $\varkappa_\omega \tilde{A} \varkappa_{\tau}^{-1}=A$ and $\varkappa_\tau s=f,$  we have

\begin{align*}
\Vert \tilde{A} s\Vert_{L_{s-m}^{2,\mathcal{L}}(F)}&= \Vert \mathcal{M}_{s-m} \varkappa_\omega \tilde{A} \varkappa_{\tau}^{-1} \varkappa_\tau s \Vert_{L^2(G, F_0)}\\&= \Vert \mathcal{M}_{s-m} Af \Vert_{L^2(G, F_0)} =\Vert  Af \Vert_{L^{2, \mathcal{L}}_{s-m}(G, F_0)}.
\end{align*}

Now, as an application of Theorem <ref>, we get

\begin{align*}
\Vert \tilde{A} s\Vert_{L_{s-m}^{2,\mathcal{L}}(F)} &= \Vert  Af \Vert_{L^{2, \mathcal{L}}_{s-m}(G, F_0)}\leq C \Vert f \Vert_{L^{2, \mathcal{L}}_{s}(G, E_0)}\\&= C \Vert \varkappa_\tau s \Vert_{L^{2, \mathcal{L}}_{s}(G, E_0)} = \Vert s\Vert_{L_{s-m}^{2,\mathcal{L}}(F)}.
\end{align*}

This complete the proof.

§.§ $G$-invariant operators  and quantisation formula

We begin with the following known
lemma (see e.g. <cit.>). We present a proof for completeness.
Let $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ be a continuous linear operator and let $A: C^\infty(G,E_{0})^\tau\rightarrow C^\infty(G,F_{0})^\omega$ be the induced operator by the identifications $\Gamma^\infty(E)\cong C^\infty(G,E_{0})^\tau$ and $\Gamma^\infty(F)\cong C^\infty(G,F_{0})^\omega.$   The following conditions are equivalent.

(A). $\tilde{A}$ is $G$-invariant. This means that $\tilde{A}$ is invariant under the action of $G$ on sections: $\tilde{A}{\pi} (g)s={\pi} (g) \tilde{A}s,$ $s\in \Gamma^\infty(E),$ where $\pi(g) s$ denotes the action of $g \in G$ on $s$ given by  (<ref>) (see also equality (<ref>)).

(B).  ${A}$ is invariant under the regular action of $G$ on functions. This means that $A\tilde{\pi} (g)f= \tilde{\pi} (g)Af,$ $f\in C^\infty(G,E_0)^\tau,$ where $\tilde{\pi}$ is defined as  $\tilde{\pi}(g_0)f(g):=f(g_0^{-1}g),$ for all $ f\in C(G,E_0)^\tau$ and $ g, g_0 \in G.$ This also implies that $A$ is a left-invariant operator with right-convolution kernel given by
where $K_{A}\in C^\infty(G)\otimes_{\pi}\mathscr{D}'(E_{0}^*\otimes F_{0})$ is the Schwartz kernel of $A.$

Let us define $f=\varkappa_\tau s,$ $s\in \Gamma^\infty(E).$ Observe that, $\tilde{A}$ is $G$-invariant if and only if

\begin{align*}
\tilde{A}(\pi(g)s)=\pi(g)(\tilde{A}s) &\Longleftrightarrow \tilde{A}(\varkappa_\tau^{-1} \varkappa_\tau\pi(g)s)=\pi(g)(\tilde{A}\varkappa_\tau^{-1} \varkappa_\tau s)\\
&\Longleftrightarrow \varkappa_{\omega} \tilde{A}(\varkappa_\tau^{-1} \varkappa_\tau\pi(g)s)=\varkappa_{\omega}\pi(g)(\tilde{A}\varkappa_\tau^{-1} \varkappa_\tau s)\\
&\Longleftrightarrow \varkappa_{\omega} \tilde{A}(\varkappa_\tau^{-1} \tilde{\pi}(g)\varkappa_\tau s)=\tilde{\pi}(g)\varkappa_{\omega}(\tilde{A}\varkappa_\tau^{-1} \varkappa_\tau s)\\
&\Longleftrightarrow A\tilde{\pi}(g)f=\tilde{\pi}(g)A f,
\end{align*}
which is equivalent to the fact   that ${A}$ is invariant under the regular action of $G$ on functions. Now, in order to prove that $A$ as in (B) is left-invariant, let us use the Schwartz kernel theorem. So, if $K_{A}\in C^\infty(G)\otimes_{\pi}\mathscr{D}'(E_{0}^*\otimes F_{0})$ is the Schwartz kernel of $A,$ we have that

\begin{align*}
A\tilde{\pi}{(g)}f(x) &=   \int\limits_{G}K_A(x,y)\tilde{\pi}{(g)}f(y)dy= \int\limits_{G}K_A(x,y)f(g^{-1}y)dy\\
\end{align*}
This implies that, for every $g\in G,$ and $x,y\in G,$ $K_A(g^{-1}x,g^{-1}y)=K_A(x,y).$ In particular, for $g=x$ we have
So, if $k(z):=K_A(e_{G},z^{-1}),$ then $k(y^{-1}x)=K_A(x,y)$ and   we have that

$$ Af(x)=\int\limits_{G}k({y^{-1}x}) f(y)dy=(f\ast k)(x),$$
which proves that $A$ is left-invariant. On the other hand, if in the identity $K_A(g^{-1}x,g^{-1}y)=K_A(x,y)$ we take $g=y,$ we have $K_{A}(x,y)=K_A({y^{-1}x,e_{G}})$ and from the uniqueness of the right-convolution kernel we have $k(z)=K_A(z,e_{G}).$

Let $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ be a continuous linear operator and let $A: C^\infty(G,E_{0})^\tau\rightarrow C^\infty(G,F_{0})^\omega$ be the induced operator by the identifications $\Gamma^\infty(E)\cong C^\infty(G,E_{0})^\tau$ and $\Gamma^\infty(F)\cong C^\infty(G,F_{0})^\omega.$ If $\tilde{A}$ is invariant, then the matrix-valued symbol of $A$ is given by

\begin{equation*}
\sigma_A(i,r,[\xi]):=\mathscr{F}_{G}({k}_{ir})(\xi),\,(i,r)\in I_{d_\tau}\times I_{d_\omega}, \,[\xi]\in \widehat{G}.
\end{equation*}
Here, $k(z):=R_{A}(e_{G},z)$ is the right-convolution kernel of $A$ and ${k}_{ir}\in \mathscr{D}'(G)$ are defined by the decomposition of $k(z)$ in the induced basis of $\textnormal{End}(E_0,F_0)=E_{0}^{*}\otimes F_{0}.$ Indeed,

\begin{align*}
Af(x)&=\int\limits_{G}k(z)f(xz^{-1})dz=\sum_{i=1}^{d_\tau}\sum_{i'=1}^{d_\tau}\sum_{r=1}^{d_\omega}\int\limits_G k_{i'r}(z)f_{i}(xz^{-1})( e_{i',E_0}^{*}\otimes e_{r,F_0})(e_{i,E_0})\\
&=\sum_{i=1}^{d_\tau}\sum_{r=1}^{d_\omega}\int\limits_G k_{ir}(z)f_{i}(xz^{-1}) e_{r,F_0}\\
&=\sum_{i,r,[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(x)\mathscr{F}_{G}({k}_{ir})(\xi)    \widehat{f}(i,\xi)   ]e_{r,F_0}\\
&=:\sum_{i,r,[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(x)\sigma_A(i,r,\xi)(\xi)    \widehat{f}(i,\xi)   ]e_{r,F_0},
\end{align*}
for all $f\in C^{\infty}(G,E_0)^{\tau}.$

In view of (<ref>), we have

Let $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ be a $G$-invariant  operator and let $A: C^\infty(G,E_{0})^\tau\rightarrow C^\infty(G,F_{0})^\omega$ be the induced operator by the identifications $\Gamma^\infty(E)\cong C^\infty(G,E_{0})^\tau$ and $\Gamma^\infty(F)\cong C^\infty(G,F_{0})^\omega.$ Then we have

\begin{equation}\label{quantizationonhomogeneous3}
\tilde{A}s(gK)\equiv \left(g, \,\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{d_\tau}\sum_{r=1}^{d_\omega}\textnormal{Tr}[\xi(g)\sigma_A(i,r,[\xi])  \widehat{\varkappa_\tau s}(i,[\xi])] e_{r,F_0}  \right)\cdot K,
\end{equation}
for every section $s\in \Gamma^\infty(E),$ where $\sigma_{A}$ is the matrix-valued symbol of $A.$

§.§ Characterisation of Hörmander classes

In this subsection we characterise the algebra of pseudo-differential operators on homogeneous vector bundles over compact homogeneous manifolds by using the notion of global symbol.

Now, we record the formulation of Hörmander's pseudo-differential operators on closed manifolds (and so on compact Lie groups) by using local coordinate systems  (see [28] and  [46] for more details).

If $U$ is an open topological subset of $\mathbb{R}^n,$ we say that a symbol  $a:U\times \mathbb{R}^n\rightarrow \mathbb{C},$ belongs to the Hörmander symbol class $S^m_{\rho,\delta}(U\times \mathbb{R}^n),$ $0\leqslant \rho,\delta\leqslant 1,$ if for every compact subset $K\subset U,$ the inequality,

\begin{equation*}
|\partial_{x}^\beta\partial_{\xi}^\alpha a(x,\xi)|\leqslant C_{\alpha,\beta,K}(1+|\xi|)^{m-\rho|\alpha|+\delta|\beta|},
\end{equation*}
holds true uniformly in $x\in K$ and $\xi\in \mathbb{R}^n.$
Then, a continuous linear operator $A:C^\infty_0(U) \rightarrow C^\infty(U)$
is a pseudo-differential operator of order $m$ and of  $(\rho,\delta)$-type if there exists
a function $a\in S^m_{\rho,\delta}(U\times \mathbb{R}^n)$ such that

\begin{equation*}
Af(x)=\int\limits_{\mathbb{R}^n}e^{2\pi i x\cdot \xi}a(x,\xi)(\mathscr{F}_{\mathbb{R}^n}{f})(\xi)d\xi,
\end{equation*}
for all $f\in C^\infty_0(U),$ where

\begin{equation*}
(\mathscr{F}_{\mathbb{R}^n}{f})(\xi):=\int\limits_Ue^{-i2\pi x\cdot \xi}f(x)dx, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
\end{equation*}
is the  Euclidean Fourier transform of $f$ at $\xi\in \mathbb{R}^n.$ The class $S^m_{\rho,\delta}(U\times \mathbb{R}^n)$ on the phase space $U\times \mathbb{R}^n,$ is invariant under coordinate changes only if $\rho\geqslant   1-\delta,$ while a symbolic calculus (which means that symbol classes are closed by taking products, adjoints, parametrices, etc.) is only possible for $\delta<\rho$ and $\rho\geqslant   1-\delta.$  Now, we state the following definition.
Let $M$ be a $C^\infty$-closed manifold. A continuous linear operator $A:C^\infty(M)\rightarrow C^\infty(M) $ is a pseudo-differential operator of order $m$ and of $(\rho,\delta)$-type, $ \rho\geqslant   1-\delta, $ if for every local  coordinate patch $\omega: M_{\omega}\subset M\rightarrow U\subset \mathbb{R}^n,$
and for every $\phi,\psi\in C^\infty_0(U),$ the operator

\begin{equation*}
Tu:=\psi(\omega^{-1})^*A\omega^{*}(\phi u),\,\,u\in C^\infty(U),\footnote{As usually, $\omega^{*}$ and $(\omega^{-1})^*$ are the pullbacks induced by the maps $\omega$ and $\omega^{-1},$ respectively.}
\end{equation*}
is a pseudo-differential operator with symbol $a_T\in S^m_{\rho,\delta}(U\times \mathbb{R}^n).$ In this case we write that $A\in \Psi^m_{\rho,\delta}(M,\textnormal{loc}).$
It was proved in [37, 38] by using the Beals characterisation [6] that the Hörmander classes defined via localisations and defined by global symbols are equivalent, this means that

\begin{equation}\label{equivalenceofclasses}
\Psi^m_{\rho,\delta}(G\times \widehat{G})= \Psi^m_{\rho,\delta}(G,\textnormal{loc}),
\end{equation}
for $\delta<\rho$ and $\rho\geq 1-\delta.$

If $u\in C^\infty_0(\mathbb{R}^n,\mathbb{C}^{\ell}),$ the Fourier transform of $u$ is defined by taking the Fourier transform of its components. So, $A:\textnormal{Dom}(A)\subset C^\infty(M,\mathbb{C}^{\ell})\rightarrow C^\infty(M,\mathbb{C}^{r}), $ is a pseudo-differential operator in the class $\Psi^{m}_{\rho,\delta}(M;\mathbb{C}^\ell,\mathbb{C}^r,\textnormal{loc})$, if locally (with the notation in Definition <ref>), we can write

\begin{equation*}
Tu:=\psi(\omega^{-1})^*A\omega^{*}(\phi u),\,\,u\in C^\infty(U, \mathbb{C}^{\ell}),
\end{equation*}
and for some matrix of symbols $\sigma \in \mathbb{C}^{r\times \ell }(S^{m}_{\rho,\delta}(U\times \mathbb{R}^n)),$ we have

\begin{equation}
Tu(x)= \int\limits_{\mathbb{R}^n}e^{2\pi i y\cdot \xi}\sigma(y,\xi)(\mathscr{F}_{\mathbb{R}^n}u)(\xi)d\xi,\,\quad u\in C^{\infty}_0({U}, \mathbb{C}^{\ell}).
\end{equation}

Now, Definition <ref> can be extended to vector bundles as follows.

Let $E$ and $F$ be  vector bundles on $M$. A continuous linear operator  $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ belongs to the Hörmander class of order $m,$ and of type $(\rho,\delta)$  (in the sense of Hörmander), if for every $x\in M,$ there exists a neighborhood $U$ of $x\in M,$ a diffeomorphism of $U$ with an open subset  $\tilde{U}\subset \mathbb{R}^n,$ via  $\phi:\tilde{U}\rightarrow {U },$ and $C^\infty$-trivialisations $\Psi:E_{U}\rightarrow {U}\times \mathbb{C}^{\ell}$ and $\Phi:F_{U}\rightarrow {U}\times \mathbb{C}^{r},$ such that for every section $s\in \Gamma^\infty(E),$ and for every $\psi_1,\psi_2\in C^\infty_0(\tilde{U}),$ there exist $H\in C^{\infty}(U, \mathbb{C}^\ell)$ and  $G\in C^{\infty}(U, \mathbb{C}^r)$ with

\begin{equation}
\Psi s(\phi(y))=(\phi(y), H(\phi(y))),\,\,\Phi As(\phi(y))=(\phi(y), G(\phi(y)) ),\,y\in \tilde{U},
\end{equation}
and  some $\sigma \in \mathbb{C}^{r\times \ell }(S^{m}_{\rho,\delta}(\tilde{U}\times \mathbb{R}^n)),$ such that

\begin{equation*}
G(\phi(y))=\int\limits_{\mathbb{R}^n}e^{2\pi i y\cdot \xi}\psi_2(y)\sigma(y,\xi)\mathscr{F}_{\mathbb{R}^n}[ \psi_1\cdot H\circ \phi](\xi)d\xi,\,\quad y\in \tilde{U}.
\end{equation*}

For $\delta<\rho$ and $\rho\geq 1-\delta,$ we write that $A\in \Psi^m_{\rho,\delta}(E,F;\textnormal{loc}).$

Let $G$ be a compact Lie group and let  $0\leq \delta,\rho\leq 1.$ Let $E_0$ and $F_0$ be complex vector spaces with finite dimensions $d_\tau$ and $d_\omega,$ respectively.  If $m\in \mathbb{R},$ the Hörmander symbol class
$$S^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0)):=S^{m,\mathcal{L}_G}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))$$
of  order $m$ and of type $(\rho,\delta),$ consists of those functions $\sigma \in \Sigma(I_{d_\omega} \times I_{d_\tau} \times G\times \widehat{G}),$ satisfying the symbol inequalities

\begin{equation}\label{InI:L:L}
\tilde{p}_{\alpha,\beta,\rho,\delta,m, \gamma } (\sigma):= \sup_{\substack{1\leq i\leq d_\tau\\ 1\leq r\leq d_\omega \\ (x, [\xi])\in G\times \widehat{G} }
} \Vert \langle\xi\rangle ^{(\rho|\alpha|-\delta|\beta|-m)}\partial_{X}^{(\beta)} \Delta_{\xi}^{\alpha}\sigma(i,r, x,\xi) \Vert_{\textnormal{op}} <\infty.
\end{equation}

We denote by $\Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))$ the class of continuous linear operators $A:C^\infty(G,E_0)\rightarrow C^\infty(G,F_0)$ with symbols in the class $S^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0)).$

Observe that any operator $A\in \Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))$ mapping $C^{\infty}(G,E_0)^{\tau}$ into  $C^{\infty}(G,F_0)^{\omega}$ induces an operator $\tilde{A}$ in the class

\begin{equation}
\Psi^{m}_{\rho,\delta}(E,F):=\{\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)|\,\,\,  \sigma_A\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0, F_0))  \} .
\end{equation}
For $E=F$ let us denote $\Psi^{m}_{\rho,\delta}(E):= \Psi^{m}_{\rho,\delta}(E,E),$ and  $\Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$

Observe that Definition <ref> is obtained from Definition <ref> if a sub-Laplacian $\mathcal{L}$ is replaced by the Laplacian $\mathcal{L}_G$ on $G.$

The following theorem follows from the vector-valued subelliptic calculus and from the subelliptic calculus on homogeneous vector-bundles developed above.  We do not require the standard restriction $\rho\geq 1-\delta$ for the invariance under changes of coordinates that is usual in the Hörmander-calculus via localisations.

Let $0\leqslant \delta<\rho\leqslant 1.$  Then we have the following properties.

-The mapping $A\mapsto A^{*}:\Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))\rightarrow \Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(F_0^*,E_0^*))$ is a continuous linear mapping between Fréchet spaces and  the  symbol of $A^*,$ $\sigma_{A^*}$ satisfies the asymptotic expansion,

\begin{equation}
\sigma_{A^*}(i,s, x,\xi)\sim  \sum_{|\alpha|= 0}^\infty\Delta_{\xi}^\alpha\partial_{X}^{(\alpha)} \sigma_{A}(i,s, x, \xi)^*.
\end{equation}

This means that, for every $N \in \mathbb{N}$,

\begin{eqnarray*}
& \Delta_{\xi}^{\alpha_\ell} \partial_X^{(\beta)} \left(\sigma_{A^*}( i,s,x,\xi) -  \sum_{|\alpha|\leq N }\Delta_{\xi}^\alpha\partial_{X}^{(\alpha)} \sigma_{A}(i,s, x, \xi)^* \right)\\
&\in {S}^{m-(\rho-\delta)(N+1)-\rho\ell+\delta|\beta|}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(F_0^*,E_0^*)).
\end{eqnarray*}
In particular, under the identifications $E_{0}^{*}\cong E_0$ and $F_{0}^{*}\cong F_0,$ the same asymptotic expansion also applies if additionally
$$A:C^{\infty}(G,E_{0})^\tau\rightarrow C^{\infty}(G,F_{0})^\omega$$
is a continuous linear operator. This means that (<ref>) remains valid for the continuous linear operator $A^*:C^{\infty}(G,F_{0})^\omega\rightarrow C^{\infty}(G,E_{0})^\tau.$ In this case, if for any operator $A,$ we define the operator $\tilde{A}^*$ by making commutative the following diagram

\begin{eqnarray}\label{Diagram:adjoint}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma^\infty(F)$}  ; & \node(S) {$\Gamma^\infty(E)$}; \\
\node(R/I) {$C^\infty(G,F_0)^\tau$}; & \node (T) {$C^\infty(G,E_0)^\omega$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_{\tau}$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}^{*}$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\omega$};
\draw[->] (R/I) -- (T) node[anchor=north] {$A^{*}$};
\end{tikzpicture}
\end{eqnarray}
then the mapping
$\tilde{A}\mapsto \tilde{A}^* :\Psi^{m}_{\rho,\delta}(F,E)\rightarrow \Psi^{m}_{\rho,\delta}(F,E)$ is a continuous linear mapping between Fréchet spaces.

- The mapping $(A,B)\mapsto A\circ B: \Psi^{m_1}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))\times \Psi^{m_1}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(F_0,H_0))\rightarrow \Psi^{m_1+m_2}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,H_0))$ is a continuous bilinear mapping between Fréchet spaces, and the  symbol of $BA,$ $\sigma_{BA},$ satisfies the asymptotic expansion,

\begin{equation}\label{compositionAandB2}
\sigma_{BA}(i, s, x,\xi) \sim \sum_{r=1}^{d_\omega}\sum_{|\alpha|= 0}^\infty(\Delta_{\xi}^\alpha\sigma_{B}(r, s, x,\xi))(\partial_{X}^{(\alpha)} \sigma_{A}(i, r, x,\xi)),
\end{equation}
this means that, for every $N\in \mathbb{N},$

\begin{align*}
&\Delta_{\xi}^{\alpha_\ell}\partial_{X}^{(\beta)}\left(\sigma_{BA}(i, s, x,\xi) - \sum_{r=1}^{d_\omega}\sum_{|\alpha| \leq N }(\Delta_{\xi}^\alpha\sigma_{B}(r, s, x,\xi))(\partial_{X}^{(\alpha)} \sigma_{A}(i, r, x,\xi))  \right)\\
&\hspace{2cm}\in {S}^{m_1+m_2-(\rho-\delta)(N+1)-\rho\ell+\delta|\beta|}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,H_0)),
\end{align*}
for every difference operator $\Delta_{\xi}^{\alpha_\ell}$ of order $\ell\in\mathbb{N}_0.$ In particular, the same asymptotic expansion applies if additionally
$$A:C^{\infty}(G,E_{0})^\tau\rightarrow C^{\infty}(G,F_{0})^\omega,$$
$$B:C^{\infty}(G,F_{0})^\tau\rightarrow C^{\infty}(G,H_{0})^\nu$$
are continuous linear operators. This means that (<ref>) remains valid for the continuous linear operator $B\circ A:C^{\infty}(G,E_{0})^\tau\rightarrow C^{\infty}(G,H_{0})^\nu.$ In this case, $\tilde{A} \in \Psi^{m_1}_{\rho,\delta}(E,F)$ and $\tilde{B}\in \Psi^{m_2}_{\rho,\delta}(F,H),$ and the composition operator $\tilde{B}\tilde{A}:=\tilde{B}\circ\tilde{ A}:\Gamma^\infty(E)\rightarrow \Gamma^\infty(H)$ belongs to the  class $\Psi^{m_1+m_2}_{\rho,\delta}(E,H).$

If the vector bundles $E\rightarrow M$ and $F\rightarrow M$ are real-vector bundles, under the identifications $E\cong E^{*},$ and $F\cong F^{*},$ the vector bundles $E\rightarrow M$ and $E^{*}\rightarrow M,$ and $F\rightarrow M$ and $F\rightarrow M,$ are isomorphic vector bundles. Indeed,  since the base space $M$  is a (para)compact and Hausdorff topological space, then the real, finite-rank vector bundles $E$ (respectively $F$) and its dual $E^{*}$ (respectively $F^{*}$) are isomorphic as vector bundles. Indeed, we pick a metric $E$ on the bundle $(\cdot,\cdot)$ (i.e. a continuously varying metric on the fibers, which exists because the base space $M$ is compact), and map a vector $v$ to $(v,\cdot)$, which is an isomorphism.  However, in our case the spaces $E_{0}\cong \mathbb{C}^{d_\tau}$ and $F_{0}\cong \mathbb{C}^{d_\omega}$ are complex-vector spaces making the fibrations  $E\rightarrow M$ and $F\rightarrow M$ complex-vector bundles. In the setting of complex vector bundles, choosing a hermitian metric on a complex vector bundle $E$ yields an isomorphism $E\cong \overline{E}^{*}$, where $\overline{E}^{*}$ is the vector bundle whose fibers are the complex conjugates of the fibers of the dual $E^{*}.$ In consequence, for complex vector bundles the property $E\cong E^{*}$ does not hold. For example, the tautological line bundle over the Riemann sphere is not isomorphic to its dual. For details we refer to Milnor and Stacheff [32].

We note that in order to define the adjoint $\tilde{A}^{*}$ of an operator $\tilde{A}:\Gamma^\infty(E)\rightarrow \Gamma^{\infty}(F),$ we consider the corresponding operator $A: C^{\infty}(G, E_0)^{\tau}\rightarrow C^{\infty}(G, F_0)^{\omega}$ making the following
\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma^\infty(E)$}  ; & \node(S) {$\Gamma^\infty(F)$}; \\
\node(R/I) {$C^\infty(G,E_0)^\tau$}; & \node (T) {$C^\infty(G,F_0)^\omega$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_{\tau}$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\omega$};
\draw[->] (R/I) -- (T) node[anchor=north] {$A$};
\end{tikzpicture}
\end{eqnarray}
commutative. Then, we define the operator $A^{*}: C^{\infty}(G, F_0)^{\omega}\rightarrow C^{\infty}(G, E_0)^{\tau}$ under the identification of the vector-spaces $E_{0}\cong E_0^{*}$ and $F_0\cong F_0^{*}.$ And finally, the operator $\tilde{A}^{*}$ is defined by the commutative diagram (<ref>). The reason for following this construction of the adjoint operator instead of considering it as a continuous operator from $\Gamma^{\infty}(F^*)\rightarrow \Gamma^{\infty}(E^*)$ is that according to Remark <ref> we do not have the isomorphism of bundles  $E\cong E^{*},$ and  $F\cong F^{*},$ respectively.

To characterise the Hörmander classes defined by localisations it will be useful to localise operators on homogeneous vector-bundles. We present it as follows.

If $E\cong G\times_\tau E_0 $ and $F\cong G\times_\omega F_0$ are homogeneous vector bundles on $M,$ then we define the class

\begin{align*}
& \Psi^{m}_{\rho,\delta}(C^{\infty}(G,E_0)^{\tau},C^{\infty}(G,F_0)^\omega;\textnormal{loc}),
\end{align*}
by those continuous linear operators $A:C^{\infty}(G,E_0)^{\tau}\rightarrow C^{\infty}(G,F_0)^\omega,$ such that for every $g\in G,$ and for every   coordinate patch $\omega: U_{\omega}\subset G\rightarrow U\subset \mathbb{R}^n,$ $g\in U_{\omega},$
and for every $\phi,\psi\in C^\infty_0(U),$ the operator $T:C^\infty(U,E_0)\rightarrow C^\infty(F,E_0),$ given by

\begin{equation*}
Tu:=\psi(\omega^{-1})^*A\omega^{*}(\phi u),\,\,u\in C^\infty(U,E_0),
\end{equation*}
is a pseudo-differential operator with symbol $a_T\in \mathbb{C}^{d_\omega\times d_\tau } (S^m_{\rho,\delta}(U\times \mathbb{R}^n)).$ Summarising,
$$ A\in \Psi^{m}_{\rho,\delta}(C^{\infty}(G,E_0)^{\tau},C^{\infty}(G,F_0)^\omega;\textnormal{loc}), \textnormal{   if   } A\in \Psi^{m}_{\rho,\delta}(G;E_0,F_0,\textnormal{loc}).$$
Note that with the notation of Remark <ref>, $ C^\infty(G, E_0)^{\tau} \subset \textnormal{Dom}(A) .$

Now, we are ready to prove the main theorem of this subsection.

Let $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ be a continuous linear  operator and let $A: C^\infty(G,E_{0})^\tau\rightarrow C^\infty(G,F_{0})^\omega$ be the induced operator by the identifications $\Gamma^\infty(E)\cong C^\infty(G,E_{0})^\tau$ and $\Gamma^\infty(F)\cong C^\infty(G,F_{0})^\omega.$ If $0\leq \delta<\rho\leq 1,$ and $\rho\geq 1-\delta,$ the following statements are equivalent:

(A) $\tilde{A}\in \Psi^{m}_{\rho,\delta}(E,F;\textnormal{loc}).$

(B) $\tilde{A}\in \Psi^{m}_{\rho,\delta}(E,F).$

(C) ${A}\in \Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))\cap \mathscr{L}(C^\infty(G,E_0)^\tau,C^\infty(G,F_0)^\omega).$

(D) ${A}\in \Psi^{m}_{\rho,\delta}(C^{\infty}(G,E_0)^{\tau},C^{\infty}(G,F_0)^\omega;\textnormal{loc}).$

First, let us prove that $(C)$ and $(D)$ are equivalent. So, let us choose $A\in \Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))\cap \mathscr{L}(C^\infty(G,E_0)^\tau,C^\infty(G,F_0)^\omega).  $ From (<ref>), we can decompose $A$ as

\begin{align}\label{sumofoperators22}
Af= \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega} A_{ir}f_i e_{r, F_0},
\end{align}
where every $A_{ir}\in \Psi^{m}_{\rho,\delta}(G\times \widehat{G}).$ Using the notation in Definition  <ref>, we have that the operator $T$ defined by

\begin{equation*}
Tu:=\psi(\omega^{-1})^*A\omega^{*}(\phi u)= \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega}  \psi(\omega^{-1})^*A_{ir}\omega^{*}(\phi u_i)e_{r, F_0},
\end{equation*}
can be decomposed in the form

\begin{align*}
Tu:=\psi(\omega^{-1})^*A\omega^{*}(\phi u)= \sum_{i=1}^{d_\tau} \sum_{r=1}^{d_\omega}  T_{ir}u_{i}e_{r, F_0},
\end{align*}
where $T_{ir}:C^{\infty}(U)\rightarrow C^{\infty}(U)$ is defined by $T_{ir}f:=\psi(\omega^{-1})^*A_{ir}\omega^{*}(\phi f),$ $f\in C^{\infty}(U).$ In view of the equivalence of classes (<ref>),  $ A_{ir}\in \Psi^{m}_{\rho,\delta}(G\times \widehat{G})=\Psi^{m}_{\rho,\delta}(G;\textnormal{loc})$ for $\delta<\rho$ and $\rho\geq 1-\delta,$ and consequently, we conclude that $T$ is a pseudo-differential operator with a matrix symbol in the class $S^{m}_{\rho,\delta}(U\times \mathbb{R}^n).$ This analysis in components shows that

\begin{align*}
&{A}\in \Psi^{m}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0,F_0))\cap \mathscr{L}(C^\infty(G,E_0)^\tau,C^\infty(G,F_0)^\omega)\\
&\Longleftrightarrow \forall i,r,\,  {A}_{ir}\in \Psi^{m}_{\rho,\delta}(G\times \widehat{G})\\
&\Longleftrightarrow \forall i,r,\,  {A}_{ir}\in \Psi^{m}_{\rho,\delta}(G;\textnormal{loc})\\
&\Longleftrightarrow{A}\in \Psi^{m}_{\rho,\delta}(C^{\infty}(G,E_0)^{\tau},C^{\infty}(G,F_0)^\omega;\textnormal{loc}).
\end{align*}
Observe that  (B) and (C) are equivalent by Definition <ref>. Under that identifications $\Gamma^\infty(E)\cong C^{\infty}(G,E_0)^\tau$ and $\Gamma^\infty(F)\cong C^{\infty}(G,F_0)^\omega,$ the diagram  (<ref>) shows that (A) and (D) are equivalent. So, we conclude the proof.

§ GLOBAL FUNCTIONAL CALCULUS ON HOMOGENEOUS AND APPLICATIONS

In this section we establish the global functional calculus for the subelliptic Hörmander classes of operators $\Psi^{m,\mathcal{L}}_{\rho,\delta}(E):=\Psi^{m,\mathcal{L}}_{\rho,\delta}(E,E),$ where $E\cong G\times_{\tau}E_0$ is a homogeneous vector bundle.

We will start by establishing the global functional calculus for the vector-valued classes and later we will extend the functional calculus to the setting of vector bundles.

§.§ Vector-valued $\mathcal{L}$-elliptic pseudo-differential operators

Here, we will discuss the vector-valued $\mathcal{L}$-elliptic pseudo-differential operators. The main aim of this section is  to show the existence of parametrices for $\mathcal{L}$-elliptic operator in the vector valued setting.

Let us consider the endomorphisms-valued symbols

\begin{equation}
\sigma_{A_1}:G\times \widehat{G}\rightarrow \bigcup_{[\xi]\in \widehat{G}}\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0,F_0)),
\end{equation}

\begin{equation}
\sigma_{A_2}:G\times \widehat{G}\rightarrow \bigcup_{[\xi]\in \widehat{G}}\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(F_0,H_0)).
\end{equation}
Observe that

\begin{equation*}
\sigma_{A_1}\sigma_{A_2}:G\times \widehat{G}\rightarrow \bigcup_{[\xi]\in \widehat{G}}\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0,H_0)),
\end{equation*}
where, for any $(x,[\xi]),$ the $(u,v)$-entry with $1\leq u,v\leq d_\xi,$ is given by

\begin{equation*}
\end{equation*}
The difference operators $\Delta_\xi^{\alpha}$ for matrix-valued symbols induce difference operators for endomorphisms-valued symbols and we will check the consistency of the Leibniz rule. First step is to define

\begin{equation}
\Delta_\xi^\alpha[\sigma_1\sigma_2](x, \xi):=[ \Delta_\xi^\alpha(\sigma_1(x,\xi)\sigma_2(x,\xi))_{u,v}]_{u,v=1}^{d_\xi},
\end{equation}
where  the family of endomorphisms $ \Delta_\xi^\alpha(\sigma_1(x, \xi)\sigma_2(x,\xi))_{u,v}$ is defined via

\begin{equation}
( \Delta_\xi^\alpha(\sigma_1(x, \xi)\sigma_2(x,\xi))_{u,v})e_{i,E_0},e_{r,H_0})=\Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)\sigma_2(i,r,x,\xi))_{u,v},
\end{equation}
on the standard orthonormal bases that we fixed for $E_0$ and $H_0.$ Note that

\begin{align*}
\Delta_\xi^\alpha(\sigma_1(i,r,x,\xi)\sigma_2(i,r,x,\xi))_{u,v}=\sum_{w=1}^{d_\xi} \Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)_{uw}\sigma_2(i,r,x,\xi)_{wv}),
\end{align*}
and then

\begin{align*}
\Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)\sigma_2(i,r,x,\xi)) &=[\Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)\sigma_2(i,r,x,\xi))_{u,v}]_{u,v=1}^{d_\xi}\\
&=\left( \sum_{w=1}^{d_\xi} \Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)_{uw}\sigma_2(i,r,x,\xi))_{wv}\right)_{u,v=1}^{d_\xi}.
\end{align*}

Using the Leibniz rule in Remark <ref> for the matrix-valued symbols, we obtain

\begin{align*}
&\Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)\sigma_2(i,r,x,\xi))\\
&=\sum_{  |\alpha_1|,|\alpha_2|\leq |\alpha|\leq |\alpha_1|+|\alpha_2|  } \Delta_\xi^{\alpha_1}(\sigma_1(i,r,x, \xi))(\Delta_\xi^{\alpha_2}(\sigma_2(i,r,x, \xi)).
\end{align*}

\begin{align*}
& \Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)\sigma_2(i,r,x,\xi))_{uv}\\
&=\sum_{  |\alpha_1|,|\alpha_2|\leq |\alpha|\leq |\alpha_1|+|\alpha_2|  }\sum_{w=1}^{d_\xi} \Delta_\xi^{\alpha_1}(\sigma_1(i,r,x, \xi))_{uw}(\Delta_\xi^{\alpha_2}(\sigma_2(i,r,x, \xi))_{wv}.
\end{align*}
This analysis leads to the following identities

\begin{align*}
& ( \Delta_\xi^\alpha(\sigma_1(x, \xi)\sigma_2(x,\xi))_{u,v})e_{i,E_0},e_{r,H_0})=\Delta_\xi^\alpha(\sigma_1(i,r,x, \xi)\sigma_2(i,r,x,\xi))_{u,v}\\
&=\sum_{w=1}^{d_\xi} \Delta_\xi^{\alpha}(\sigma_1(i,r,x, \xi)_{uw}\sigma_2(i,r,x,[\xi])_{wv})\\
&=\sum_{  |\alpha_1|,|\alpha_2|\leq |\alpha|\leq |\alpha_1|+|\alpha_2|  }\sum_{w=1}^{d_\xi}
\Delta_\xi^{\alpha_1}(\sigma_1(i,r,x, \xi))_{uw}(\Delta_\xi^{\alpha_2}(\sigma_2(i,r,x, \xi))_{wv}\\
&=\sum_{  |\alpha_1|,|\alpha_2|\leq |\alpha|\leq |\alpha_1|+|\alpha_2|  } [\Delta_\xi^{\alpha_1}(\sigma_1(i,r,x, \xi))(\Delta_\xi^{\alpha_2}(\sigma_2(i,r,x, \xi))]_{uv}\\
&=\sum_{  |\alpha_1|,|\alpha_2|\leq |\alpha|\leq |\alpha_1|+|\alpha_2|  }( \Delta_\xi^{\alpha_1}(\sigma_1(x, \xi))(\Delta_\xi^{\alpha_2}\sigma_2(x,\xi))_{u,v}))e_{i,E_0},e_{r,H_0}),
\end{align*}
showing the Leibniz rule

\begin{align}\label{leibniz:End:symbol} \boxed{
\Delta_\xi^\alpha(\sigma_1(x,\xi)\sigma_2(x,\xi)=\sum_{  |\alpha_1|,|\alpha_2|\leq |\alpha|\leq |\alpha_1|+|\alpha_2|  } \Delta_\xi^{\alpha_1}(\sigma_1(x, \xi))(\Delta_\xi^{\alpha_2}\sigma_2(x,\xi))}
\end{align}

Let $m\in \mathbb{R},$ and let $0\leqslant \delta<\rho\leqslant 1.$  Let  $a=a(x,\xi)\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)) .$  Assume also that $a(x,\xi)\in \mathbb{C}^{d_\xi \times d_\xi}(\textnormal{End}(E_0))$ is invertible for every $(x,[\xi])\in G\times\widehat{G},$[which means that there exists  $b_0(x,\xi)\in \mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0)) ,$ such that $b_{0}(x,\xi)a(x,\xi)=a(x,\xi)b_0(x,\xi)=I,$ where $I=(\textnormal{id}_{E_0})_{i,j=1}^{d_\xi}$ is the identity element in  $\mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0)).$] and satisfies

\begin{equation}\label{Ies}
\sup_{(x,[\xi])\in G\times \widehat{G}}\Vert\widehat{\mathcal{M}}(\xi)^m\otimes a(x,\xi)^{-1} \Vert_{\textnormal{op}}<\infty.
\end{equation}

Then, $a^{-1}:=a(x,\xi)^{-1}\in {S}^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0) ).$

Let us estimate $\partial^{(\beta)}_{X_i}a^{-1}$ first. Suppose we have proved that

\begin{equation*}
I:= \sup_{|\beta|\leqslant \ell}\sup_{(x,[\xi])\in  G\times \widehat{G} }\Vert \widehat{ \mathcal{M}}(\xi)^{(-\delta|\beta|+m)}\otimes \partial_{X}^{(\beta)} (a(x,\xi)^{-1})\Vert_{\textnormal{op}} <\infty,
\end{equation*}

for some  $ \ell\in \mathbb{N}.$ We proceed by mathematical induction. Let us analyse the cases $|\tilde\beta|=\ell+1.$ If we write $\partial_{X}^{(\tilde\beta)}= \partial_{X}\partial_{X}^{(\beta)}$ where $|\beta|\leqslant \ell,$ then  $\partial^{\tilde\beta}_{X_i}a^{-1}=\partial_{X}\partial_{X}^{(\beta)}a^{-1}.$ From the identity
$$ a(x,\xi) \circ a(x,\xi)^{-1}=I_{ \mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0)),      }$$
we have

\begin{equation*}
a(x,\xi) \circ \partial_{X}\partial_{X}^{(\beta)}a^{-1}(x,\xi)=-\sum_{\beta_1+\beta_2=\beta+e_{j},|\beta_2|\leqslant  |\beta|}C_{\beta_1,\beta_2}(\partial_{X}^{(\beta_1)}a(x,\xi))\circ (\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)).
\end{equation*}

\begin{equation*}
\partial_{X}\partial_{X}^{(\beta)}a^{-1}(x,\xi)=-a(x,\xi)^{-1} \circ\left(\sum_{\beta_1+\beta_2=\beta+e_{j},|\beta_2|\leqslant  |\beta|}C_{\beta_1,\beta_2}(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \right).
\end{equation*}
We want to prove the estimate

\begin{equation*}
\sup_{(x,[\xi])\in G\times \widehat{G} }\Vert \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta|+1)+m)}\otimes \partial_{X}\partial_{X}^{(\beta)} a(x,\xi)^{-1}\Vert_{\textnormal{op}} <\infty.
\end{equation*}

For this, we only need to show that for every $\beta_1$ and $\beta_2$ such that $\beta_1+\beta_2=\beta+e_{j},|\beta_2|\leqslant  |\beta|,$

\begin{equation*}
\sup_{(x,[\xi])\in G\times \widehat{G} }\Vert \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta|+1)+m)} \otimes a(x,\xi)^{-1}\circ(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi))   \Vert_{\textnormal{op}} <\infty.
\end{equation*}

Observe that

\begin{align*}
&\widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta|+1)+m)} \otimes a(x,\xi)^{-1}\circ(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi))   \\
&=\widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|)+m)} \otimes a(x,\xi)^{-1}\circ(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)).
\end{align*}
So, by Theorem <ref> and (<ref>) we have

\begin{align*}
& \Vert \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta|+1)+m)} \otimes a(x,\xi)^{-1}\circ(\partial_{X}^{(\beta_1)} a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \Vert_{\textnormal{op}}\\
&= \Vert \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|)} \widehat{ \mathcal{M}}(\xi)^{ m } \otimes  a(x,\xi)^{-1}\circ(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \Vert_{\textnormal{op}}\\
&\leqslant \Vert \widehat{ \mathcal{M}}(\xi)^{ m }\otimes a(x,\xi)^{-1} \Vert_{\textnormal{op}}\Vert(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|)}\Vert_{\textnormal{op}}\\
&\lesssim  \Vert(\partial_{X}^{(\beta_1)} a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|)}\Vert_{\textnormal{op}}.
\end{align*}

Again, by using  Theorem <ref>, we have

\begin{align*}
& \Vert(\partial_{X}^{(\beta_1)}a(x,\xi))\circ(\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes  \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|)}\Vert_{\textnormal{op}}\\
&\leqslant \Vert(\partial_{X}^{(\beta_1)}a(x,\xi)) \otimes  \widehat{ \mathcal{M}}(\xi)^{(-m-\delta|\beta_1|)}\Vert_{\textnormal{op}} \\
&\hspace{4cm}\times\Vert\widehat{ \mathcal{M}}(\xi)^{(m+\delta|\beta_1|)} \otimes   (\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|))}  \Vert_{\textnormal{op}}\\
&\lesssim  \Vert\widehat{ \mathcal{M}}(\xi)^{(m+\delta|\beta_1|)} \otimes   (\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes  \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_1|+|\beta_2|))}  \Vert_{\textnormal{op}}\\
&=\Vert\widehat{ \mathcal{M}}(\xi)^{(m+\delta|\beta_1|)} \otimes   (\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_2|+m)}\widehat{ \mathcal{M}}(\xi)^{(-\delta|\beta_1|-m))}\Vert_{\textnormal{op}}\\
&\lesssim \Vert   (\partial_{X}^{(\beta_2)}a^{-1}(x,\xi)) \otimes \widehat{ \mathcal{M}}(\xi)^{(-\delta(|\beta_2|+m))}\Vert_{\textnormal{op}}\\
& \lesssim 1.
\end{align*}
A similar analysis using the Leibniz rule for difference operators can be used in order to estimate the differences $\Delta_\xi^{\alpha}a^{-1}.$ For this, we need the following two estimates,

\begin{equation}
\Vert \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)} \otimes a(x,\xi)^{-1}\circ [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)]\Vert_{\textnormal{op}}=O(1),
\end{equation}

\begin{equation}
\Vert  [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)]\circ a(x,\xi)^{-1} \otimes\widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)}\Vert_{\textnormal{op}}\\ =O(1).
\end{equation}

For the proof, let us use  (<ref>) as follows

\begin{align*}
\Vert \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)} \otimes &a(x,\xi)^{-1}\circ[\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)]\Vert_{\textnormal{op}}\\
&= \Vert \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|-m)}\widehat{\mathcal{M}}(\xi)^{ m } \otimes a(x,\xi)^{-1}\circ[\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)]\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{ m } \otimes a(x,\xi)^{-1}\circ[\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)] \otimes \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|-m)}\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{ m } \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}\Vert[\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)] \otimes \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|-m)}\Vert_{\textnormal{op}}\\
&= O(1).
\end{align*}
On the other hand,

\begin{align*}
\Vert  [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)]& \circ a(x,\xi)^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)}\Vert_{\textnormal{op}}\\
&= \Vert  [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)] \otimes   \widehat{\mathcal{M}}(\xi)^{- m }\widehat{\mathcal{M}}(\xi)^{ m } \otimes a(x,\xi)^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)}\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)} \otimes [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)] \otimes \widehat{\mathcal{M}}(\xi)^{- m } \widehat{\mathcal{M}}(\xi)^{ m } \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(\rho|\gamma|-\delta|\beta|)} \otimes [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)] \otimes \widehat{\mathcal{M}}(\xi)^{- m }\Vert_{\textnormal{op}} \Vert\widehat{\mathcal{M}}(\xi)^{ m } \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(-m+\rho|\gamma|-\delta|\beta|)} \otimes [\Delta_{\xi}^{\gamma}\partial_{X}^{(\beta)}a(x,\xi)]\Vert_{\textnormal{op}} \\
&= O(1).
\end{align*}

Now, we will estimate in $(m+\rho\ell)$ the subelliptic order for the differences  $\Delta_{q_{\ell}}a^{-1}$ where $\Delta_{q_{\ell}}$ denotes a difference operator of order $\ell$. To do so, we will use mathematical induction. The case $\ell=0$ holds true from the hypothesis of Theorem <ref>. To study the differences of higher order we will use the Leibniz rule (<ref>),

\begin{align*}
\Delta_{q_\ell}[ a_{1} \circ a_{2}](x,\xi) =\sum_{ |\gamma|,|\varepsilon|\leqslant \ell\leqslant |\gamma|+|\varepsilon| }C_{\varepsilon,\gamma}(\Delta_{q_\gamma}a_{1})(x,\xi) \circ (\Delta_{q_\varepsilon}a_{2})(x,\xi).
\end{align*}

From the identity,
$$a(x,\xi)\circ a(x,\xi)^{-1}=I_{ \mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))},$$
we deduce that

\begin{align*}
(\Delta_{q_1}a)(x,\xi) \circ & a(x,\xi)^{-1}+a(x,\xi) \circ (\Delta_{q_1}a^{-1})(x,\xi)\\
&=-\sum_{ 1= |\nu|,|\nu'| }C_{\nu,\nu'}(\Delta_{q_{(\nu)}}a)(x,\xi) \circ (\Delta_{q_{(\nu')}}a^{-1})(x,\xi),
\end{align*}
where $\Delta_{q_{(\nu)}}$ are differences operators of order $|\nu|=1$. Consequently

\begin{align*}
&  (\Delta_{q_1}a^{-1})(x,\xi)=-a(x,\xi)^{-1}\circ(\Delta_{q_1}a)(x,\xi)\circ a^{-1}(x,\xi)\\
&\hspace{2cm}-\sum_{ 1= |\nu|,|\nu'| }C_{\nu,\nu'}a(x,\xi)^{-1}\circ (\Delta_{q_{(\nu)}}a)(x,\xi) \circ (\Delta_{q_{(\nu')}}a^{-1})(x,\xi).
\end{align*}
The differences of higher order  $\Delta_{q_{\ell+1}}a^{-1},$ can be estimated e.g. using the difference operators of the form

Then, by applying the difference operator $\Delta_{q_{\ell}}$ to $(\Delta_{q_1}a^{-1})(x,\xi)$ we essentially obtain linear combinations of terms of the following kind as a consequence of the Leibniz rule:

* $\textnormal{IV}_{(1)}(x,\xi):=\Delta_{q_{\ell}}a(x,\xi)^{-1}\circ (\Delta_{q_1}a)(x,\xi)\circ a^{-1}(x,\xi)$

* $\textnormal{IV}_{(2)}(x,\xi):=a(x,\xi)^{-1}\circ \Delta_{q_{\ell+1}}a(x,\xi)\circ a^{-1}(x,\xi)$

* $\textnormal{IV}_{(3)}(x,\xi):=a(x,\xi)^{-1}\circ \Delta_{q_{1}}a(x,\xi)\circ \Delta_{q_{\ell}}a^{-1}(x,\xi)$

* $\textnormal{IV}_{(4)}(x,\xi):=\Delta_{q_{\ell_1}} a(x,\xi)^{-1}\circ(\Delta_{q_{\ell_2+1}}a)(x,\xi)\circ \Delta_{q_{\ell_3+1}}a^{-1}(x,\xi),$ $1\leqslant \ell_{1},\ell_2,\ell_3\leqslant \ell\leqslant \ell_1+\ell_2+\ell_3,$

* $\textnormal{V}_{(1)}(x,\xi):=\Delta_{q_{\ell}}a(x,\xi)^{-1}\circ (\Delta_{q_{(\nu) } }a)(x,\xi)\circ \Delta_{q_{(\nu')}}a^{-1}(x,\xi)$

* $\textnormal{V}_{(2)}(x,\xi):=a(x,\xi)^{-1}\circ \Delta_{q_{\ell+\nu}}a(x,\xi)\circ \Delta_{q_{\nu'}}a^{-1}(x,\xi)$

* $\textnormal{V}_{(3)}(x,\xi):=a(x,\xi)^{-1}\circ \Delta_{q_{(\nu)}}a(x,\xi)\circ \Delta_{q_{\ell+\nu'}}a^{-1}(x,\xi)$

* $\textnormal{V}_{(4)}(x,\xi):=\Delta_{q_{\ell_1}} a(x,\xi)^{-1}\circ(\Delta_{q_{\ell_2+\nu}}a)(x,\xi)\circ \Delta_{q_{\ell_3+\nu'}}a^{-1}(x,\xi),$ $1\leqslant \ell_{1},\ell_2,\ell_3\leqslant \ell\leqslant \ell_1+\ell_2+\ell_3,$

which we can estimate as follows. First, assume that for every $\ell\geqslant  1,$ $\ell\in \mathbb{N},$ we have

$$\sup_{(x,[\xi])}  \Vert\widehat{\mathcal{M}}(\xi)^{(m+\rho\ell)} \otimes (\Delta_{q_{\ell}}a^{-1})(x,\xi)\Vert_{\textnormal{op}}  <\infty. $$
We need to prove that

$$\sup_{(x,[\xi])}  \Vert\widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes (\Delta_{q_{\ell+1}}a^{-1})(x,\xi)\Vert_{\textnormal{op}}  <\infty. $$

For this, it is enough to prove that

$$ \sup_{(x,[\xi])}  \Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes \textnormal{IV}_{(i)}(x,\xi)\Vert_{\textnormal{op}},\, \sup_{(x,[\xi])}  \Vert \widehat{\mathcal{M}}(\xi)^{ (m+\rho(\ell+1))} \otimes  \textnormal{V}_{(j)}(x,\xi)\Vert_{\textnormal{op}} <\infty, $$
for all $1\leqslant i,j\leqslant 4.$ Next, let us omit the argument $(x,\xi)$ in order to simplify the notation.
So, we have

\begin{align*}
&\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes  \textnormal{IV}_{(1)}\Vert_{\textnormal{op}}+     \Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes \textnormal{IV}_{(2)}(x,\xi)\Vert_{\textnormal{op}}\\
&+   \Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes \textnormal{IV}_{(3)}(x,\xi)\Vert_{\textnormal{op}}+\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes \textnormal{V}_{(1)}(x,\xi)\Vert_{\textnormal{op}}\\
&\hspace{4cm}+\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes \textnormal{V}_{(2)}(x,\xi)\Vert_{\textnormal{op}}\\
& =\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes\Delta_{q_{\ell}}a^{-1}\circ (\Delta_{q_1}a)\circ a^{-1}\Vert_{\textnormal{op}}\\&\hspace{2cm}+     \Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+1}}a\circ a^{-1}\Vert_{\textnormal{op}}\\
&\hspace{2cm}+   \Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{1}}a\circ \Delta_{q_{\ell}}a^{-1}\Vert_{\textnormal{op}}\\
&\hspace{2cm}+\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes \Delta_{q_{\ell}}a^{-1}\circ (\Delta_{q_{(\nu) } }a)\circ \Delta_{q_{(\nu')}}a^{-1}\Vert_{\textnormal{op}}\\
&\hspace{2cm}+\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\circ \Delta_{q_{\nu'}}a^{-1}\Vert_{\textnormal{op}}\\
\end{align*}

\begin{align*}
\lesssim\Vert &\widehat{\mathcal{M}}(\xi)^{(m+\rho\ell)} \otimes \Delta_{q_{\ell}}a^{-1}\circ (\Delta_{q_1}a)\circ a^{-1} \otimes\widehat{\mathcal{M}}(\xi)^{{\rho}}\Vert_{\textnormal{op}}\\&\hspace{2cm}+     \Vert \widehat{\mathcal{M}}(\xi)^{ m } \otimes a^{-1}\circ \Delta_{q_{\ell+1}}a\circ a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \Vert_{\textnormal{op}}\\
&\hspace{2cm}+   \Vert \widehat{\mathcal{M}}(\xi)^{{\rho}} \otimes a^{-1}\circ \Delta_{q_{1}}a\circ \Delta_{q_{\ell}}a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(m+\rho\ell)}\Vert_{\textnormal{op}}\\
&\hspace{2cm}+\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho\ell)} \otimes \Delta_{q_{\ell}}a^{-1}\Vert_{\textnormal{op}}\Vert  (\Delta_{q_{(\nu) } }a)\circ \Delta_{q_{(\nu')}}a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{{\rho}}\Vert_{\textnormal{op}}\\
&\hspace{2cm}+\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\circ \Delta_{q_{\nu'}}a^{-1}\Vert_{\textnormal{op}}\\
\\
& \lesssim\Vert (\Delta_{q_1}a)\circ a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{{\rho}}\Vert_{\textnormal{op}}\\&\hspace{2cm}+      \Vert  \Delta_{q_{\ell+1}}a\circ a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \Vert_{\textnormal{op}}\\
&\hspace{2cm}+   \Vert \widehat{\mathcal{M}}(\xi)^{{\rho}} \otimes a^{-1}\circ \Delta_{q_{1}}a\Vert_{\textnormal{op}}\\
&\hspace{2cm}+\Vert  (\Delta_{q_{(\nu) } }a)\circ a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{{\rho}}\Vert_{\textnormal{op}}\\
&\hspace{2cm}+\Vert \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\Vert_{\textnormal{op}}\\
&\hspace{2cm}\lesssim 1,
\end{align*}

where we have used, and  the estimates

\begin{equation}\label{casobacano}
\Vert  (\Delta_{q_{(\nu) } }a)\circ \Delta_{q_{(\nu')}}a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{{\rho}}\Vert_{\textnormal{op}}\lesssim 1,
\end{equation}

\begin{equation}\label{casobacano2}
\Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes a^{-1} \circ \Delta_{q_{\ell+\nu}}a \circ \Delta_{q_{\nu'}}a^{-1}\Vert_{\textnormal{op}}\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \otimes a^{-1} \circ \Delta_{q_{\ell+\nu}}a\Vert_{\textnormal{op}}.
\end{equation}

Indeed, for the proof of (<ref>)  observe that from the induction hypothesis, we have

\begin{align*}
\Vert  (\Delta_{q_{(\nu) } }a)\circ &\Delta_{q_{(\nu')}}a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{{\rho}}\Vert_{\textnormal{op}}\\
&\lesssim   \Vert \widehat{\mathcal{M}}(\xi)^{(-m+\rho)} \otimes (\Delta_{q_{(\nu) } }a)\circ \Delta_{q_{(\nu')}}a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(m+\rho)} \widehat{\mathcal{M}}(\xi)^{-{\rho}} \Vert_{\textnormal{op}}   \\
&=   \Vert \widehat{\mathcal{M}}(\xi)^{(-m+\rho)} \otimes (\Delta_{q_{(\nu) } }a)\Vert_{\textnormal{op}} \Vert \Delta_{q_{(\nu')}}a^{-1} \otimes \widehat{\mathcal{M}}(\xi)^{(m+\rho)}\Vert_{\textnormal{op}} \Vert \widehat{\mathcal{M}}(\xi)^{-{\rho} } \Vert_{\textnormal{op}}   \\
&\lesssim 1.
\end{align*}
In order to prove (<ref>), observe that

\begin{align*}
& \Vert \widehat{\mathcal{M}}(\xi)^{(m+\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\circ \Delta_{q_{\nu'}}a^{-1}\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\circ \Delta_{q_{\nu'}}a^{-1} \otimes\widehat{\mathcal{M}}(\xi)^{ m }\Vert_{\textnormal{op}}\\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\Vert_{\textnormal{op}}\Vert \Delta_{q_{\nu'}}a^{-1} \otimes\widehat{\mathcal{M}}(\xi)^{(m+\rho)}\Vert_{\textnormal{op}}\Vert \widehat{\mathcal{M}}(\xi)^{-{\rho}}\Vert_{\textnormal{op}} \\
&\lesssim \Vert \widehat{\mathcal{M}}(\xi)^{(\rho(\ell+1))} \otimes a^{-1}\circ \Delta_{q_{\ell+\nu}}a\Vert_{\textnormal{op}}.
\end{align*}

A similar analysis can be used to study $\textnormal{IV}_{(4)}(x,\xi),\textnormal{V}_{(3)}(x,\xi)$ and $\textnormal{V}_{(4)}(x,\xi).$ Thus, we end the proof.

Let $\{m_j\}_{j=0}^{\infty}$ be a strictly decreasing sequence of real numbers such that $\lim_{j\rightarrow \infty}m_j=-\infty.$ Let $b_j=b_j(x,\xi)\in {S}^{m_j,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ be a sequence of symbols. Then, there exists $b\in {S}^{m_0,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ such that,

\begin{equation}
b\sim \sum_{j=0}^\infty b_j,
\end{equation}

in the sense that  for any $N\in \mathbb{N},$

\begin{equation}
b-\sum_{j=0}^{N}b_j\in {S}^{m_{N+1},\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).
\end{equation}

Let $\phi\in C^\infty(\mathbb{R}_0^+)$ be a smooth function such that $\phi\equiv 0,$ on the interval $[0,1/2],$ and such that $\phi\equiv 1,$ on  $[1,\infty).$ Note that $1-\phi$ is smooth and compactly supported. Let us consider the family of seminorms $ p_{\alpha,\beta,\rho,\delta,m,\textnormal{left}}(\cdot)$ as in Theorem <ref>.   In view of Lemma 6.4 of [10], if
$$\forall 1\leq i\leq d_\tau,\,\forall 1\leq r\leq d_\omega,\,  (i,r,x,[\xi])\mapsto b_{j}(i,r,x,[\xi])$$
denotes the matrix-valued symbol associated to $b_j,$ then we have the estimate

\begin{equation}
\exists C>0,\forall t\in (0,1),\,\,
\end{equation}

$$ p_{\alpha,\beta,\rho,\delta,m_0,\textnormal{left}}((i,r,x,[\xi])\mapsto b_{j}(i,r,x,[\xi])\phi (t\widehat{\mathcal{M}}(\xi)))$$

$$\lesssim_{m_0}  p_{\alpha,\beta,\rho,\delta,m_j,\textnormal{left}}\left((i,r,x,[\xi])\mapsto b_{j}(i,r,x,[\xi])\right)t^{m_0-m_j}.$$

In particular, if $t_j\rightarrow 0^+,$ is such that
$$p_{\alpha,\beta,\rho,\delta,m_j,\textnormal{left}}((i,r,x,[\xi])\mapsto b_{j}(i,r,x,[\xi]))t^{m_0-m_j}\leq 2^{-j},$$
then we define the symbol
$$\tilde{b}_j(i,r,x,[\xi])= {b}_j(i,r,x,[\xi]) \phi (t_j\widehat{\mathcal{M}}(\xi)).$$

$$\sum_{j=0}^{\infty}p_{\alpha,\beta,\rho,\delta,m_0,\textnormal{left}}((i,r,x,[\xi])\mapsto \tilde b_{j}(i,r,x,[\xi]))\leq \sum_{j=0}^\infty 2^{-j}<\infty,$$
and taking into account that the class $S^{m_0,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})$ is a Fréchet space endowed with  the family of seminorms $ p_{\alpha,\beta,\rho,\delta,m_0,\textnormal{left}},$ there exists $ b_{i,r}\in S^{m_0,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})$ such that $ b_{i,r}=\sum_{j=0}^\infty\tilde b_{j}(i,r,\cdot,\cdot)$ with the sum converging in the topology induced by the Fréchet structure in $S^{m_0,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}).$ Starting the summation at $j=N,$ the same proof above shows that $\sum_{j=N}\tilde b_{j}(i,r,\cdot,\cdot)\in S^{m_N,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}).$ Hence, the symbol defined by

$$b(i,r,x,[\xi])=\sum_{j=0}^\infty b_{ir}(x,\xi) ,$$
satisfies the condition of the theorem. Indeed, for any $N,$ we have that

\begin{align*}
b(i,r,x,[\xi])-&\sum_{j=0}^N b_j(i,r,x,\xi) \\
&= \sum_{j=0}^N b_j(i,r,x,\xi)(1-\phi)(t\widehat{\mathcal{M}}(\xi))+ \sum_{j=N+1}\tilde b_{j}(i,r,\cdot,\cdot).
\end{align*}
Note that the symbols $ b_j(i,r,x,\xi)(1-\phi)(t\widehat{\mathcal{M}}(\xi))$ are smoothing terms, namely these are symbols in $S^{-\infty,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}),$  since  $1-\phi$ is smooth and compactly supported. Because $\sum_{j=N+1}\tilde b_{j}(i,r,\cdot,\cdot)\in S^{m_{N+1},\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})$ we have that the difference $ b(i,r,x,[\xi])-\sum_{j=0}^N b_j(i,r,x,\xi) $ belongs to $ S^{m_{N+1},\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}).$ At the level of the vector-valued symbol $b:=b(x,\xi)$ induced by the matrix-valued symbol $(i,r,x,[\xi])\mapsto b(i,r,x,[\xi]),$ we have proved that

$$   b-\sum_{j=0}^{N}b_j\in {S}^{m_{N+1},\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$$
The proof is complete.

Now, we  present a technical result concerning   the existence of parametrices for $\mathcal{L}$-elliptic operators in the subelliptic calculus. We denote
$$ {\Psi}^{-\infty,\mathcal{L}}((G\times \widehat{G})\otimes \textnormal{End}(E_0)):=\bigcap_{m\in \mathbb{R}}{\Psi}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$$

Let $m\in \mathbb{R},$  $0\leqslant \delta<\rho\leqslant 1.$  Let  $a=a(x,\xi)\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$  Assume also that $a(x,\xi)\in \mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0))$ is invertible for every $(x,[\xi])\in G\times\widehat{G},$  and satisfies

\begin{equation}\label{Iesparametrix}
\sup_{(x,[\xi])\in G\times \widehat{G}} \Vert \widehat{\mathcal{M}}(\xi)^{m} \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}<\infty.
\end{equation}
Then, there exists $B\in {\Psi}^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)),$ such that $AB-I\in {\Psi}^{-\infty,\mathcal{L}}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)). $  Also, there exists  $B'\in {\Psi}^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)),$ such that $B'A-I\in {\Psi}^{-\infty,\mathcal{L}}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)). $

Let us follow the argument in Hörmander <cit.>.  Let $b_0(x,\xi)=a(x,\xi)^{-1},$ and let us consider its associated pseudo-differential operator $b_0(x,D).$ By the subelliptic calculus, we have that

$$ -r(x,D):=a(x,D)b_0(x,D)-I\in \textnormal{Op}( S^{-(\rho-\delta),\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})). $$

Define $b_k(x,D)=b_0(x,D)r(x,D)^{k} \in \textnormal{Op}(S^{-m-k(\rho-\delta),\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})),$ for any $k\in \mathbb{N}.$ By applying Lemma <ref>, consider a symbol $ b(x,\xi)$ satisfying the asymptotic expansion $b(x,\xi)\sim \sum_{k=0}^\infty b_k(x,\xi).$
Define $S_N(x,\xi)=\sum_{k=0}^{N-1} b_k(x,\xi).$ Note that

\begin{align*}
a(x,D)S_N(x,D)-I  &=a(x,D)\left(\sum_{k=0}^{N-1} b_0(x,D)r(x,D)^{k}\right)-I\\
& =a(x,D)b_0(x,D)\left(\sum_{k=0}^{N-1} r(x,D)^{k}\right)-I\\
&  =(I-r(x,D))\left(\sum_{k=0}^{N-1} r(x,D)^{k}\right)-I\\
&  =I-r(x,D)^{N}-I\\
&  =-r(x,D)^{N}\in \textnormal{Op}(S^{-N(\rho-\delta),\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})).
\end{align*}
Since $b(x,D)-S_{N}(x,D)\in  \textnormal{Op}(S^{-m-N(\rho-\delta),\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})),$ we have that

\begin{align*}
& a(x,D)b(x,D) -I\\
&= a(x,D)(b(x,D)-S_N(x,D))+  a(x,D)S_N(x,D)-I \\
&=a(x,D)(b(x,D)-S_N(x,D))-r(x,D)^{N}\in \textnormal{Op}(S^{-N(\rho-\delta),\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})),
\end{align*}
for all $N\in \mathbb{N}.$ Then $B=b(x,D)$ is such that $a(x,D)b(x,D) -I\in \textnormal{Op}(S^{-\infty,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})) .$ The same argument above can be applied to construct $B'=b'(x,D)\sim \sum_{k=0}^\infty b_k(x,D),$ by defining
$$b_k(x,D)=r(x,D)^{k}b_0(x,D) \in \textnormal{Op}(S^{-m-k(\rho-\delta),\mathcal{L}}_{\rho,\delta}(G\times \widehat{G})).$$

The proof of Proposition <ref> is complete.

§.§ $\mathcal{L}$-elliptic operators on homogeneous vector bundles

Now, we will prove the existence of parametrices for $\mathcal{L}$-elliptic operators  in the subelliptic calculus. We denote
$$ {\Psi}^{-\infty,\mathcal{L}}(E)=\bigcap_{m\in \mathbb{R}}{\Psi}^{m,\mathcal{L}}_{\rho,\delta}(E).$$

Let $m\in \mathbb{R},$ and let $0\leqslant \delta<\rho\leqslant 1.$  Let  $a=a(x,\xi)\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$  Assume also that $a(x,\xi)\in \mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0))$ is invertible for every $(x,[\xi])\in G\times\widehat{G},$ and satisfies

\begin{equation}\label{Iesparametrix2}
\sup_{(x,[\xi])\in G\times \widehat{G}} \Vert \widehat{\mathcal{M}}(\xi)^{m} \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}<\infty.
\end{equation}
Let $\tilde{A}\in \Psi^{m,\mathcal{L}}_{\rho,\delta}(E)$ be the operator associated to the symbol  $a$ via (<ref>). Then, there exists $\tilde{B}\in {\Psi}^{-m,\mathcal{L}}_{\rho,\delta}(E),$ such that $\tilde{A}\tilde{B}-I\in {\Psi}^{-\infty,\mathcal{L}}(E). $ Also there exists  $\tilde{B}'\in {\Psi}^{-m,\mathcal{L}}_{\rho,\delta}(E),$ such that $\tilde{B}'\tilde{A}-I\in {\Psi}^{-\infty,\mathcal{L}}(E). $

If $A$ and $B$ are the vector-valued pseudo-differential operators associated with $\tilde{A}$ and $\tilde{B},$ Proposition  <ref> shows that $AB-I,B'A-I\in \textnormal{Op}( {S}^{-\infty,\mathcal{L}}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))) $ which immediately implies that $\tilde{A}\tilde{B}-I,\tilde{B}'\tilde{A}-I\in {\Psi}^{-\infty,\mathcal{L}}(E). $ So, the proof is complete.

§.§ Fredholmness on homogeneous vector bundles
In this subsection, we discuss the Fredholmness of $\mathcal{L}$-elliptic pseudo-differential operators.

First, let us briefly recall the definition of Fredholm operators and index of an operator.
Let $X$ and $Y$ be two Banach spaces. A continuous linear operator $A:X \rightarrow Y$ is a Fredholm operator if its kernel $\textnormal{Ker}(A)$ is finite-dimensional and the range (the image) $\textnormal{Im}(A)$ is closed and of finite codimension. Then its index is defined as

$$\textnormal{Ind(A)}:=\textnormal{dim Ker}(A)-\textnormal{codim Im}(A),$$

or equivalently

$$\textnormal{Ind(A)}:=\textnormal{dim Ker}(A)-\textnormal{dim Ker}(A^*),$$
where $A^*:Y^{*}\rightarrow X^{*}$ is the adjoint of $A.$

The following criterion for a closed linear operator to be Fredholm is
usually attributed to Atkinson [4].

Let $A$ be a closed linear operator from a complex Banach space $X$ into a
complex Banach space $Y$ with dense domain $\mathcal{D}(A).$ Then $A$ is
Fredholm if and only if we can find a bounded linear operator
$B:Y\rightarrow X,$ a compact operator $K_{1}:X\rightarrow X$ and a
compact operator $K_2:Y\rightarrow Y$ such that $BA=I+K_1$ on $\mathcal{D}(A)$ and $AB=I+K_2$ on $Y$.

In the following theorem we will show that $\mathcal{L}$-ellipticity implies the Fredholmness.

Let us consider the identifications $E_{0}^{*}\cong E_0.$  Let $m\in \mathbb{R},$ and let $0\leqslant \delta<\rho\leqslant 1.$  Let  $a=a(x,\xi)\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$  Assume also that $a(x,\xi)\in \mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0))$ is invertible for every $(x,[\xi])\in G\times\widehat{G},$ and satisfies

\begin{equation*}
\sup_{(x,[\xi])\in G\times \widehat{G}} \Vert \widehat{\mathcal{M}}(\xi)^{m} \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}<\infty.
\end{equation*}

Let $A\in \Psi^{m,\mathcal{L}}_{\rho,\delta}((G \times \widehat{G}) \otimes \textnormal{End}(E_0))$ be the operator associated to the symbol  $a$. Then $A$ is a Fredholm operator from $L^{2, \mathcal{L}}_s(G, E_0)$ to $L^{2, \mathcal{L}}_{s-m}(G, E_0)$ for all $s \in \mathbb{R}.$
Moreover,  the kernels $\textnormal{Ker}(A),$ $\textnormal{Ker}(A^*)$ are in $C^\infty(G, E_0),$ so that $\textnormal{Ker}(A),$ $\textnormal{Ker}(A^*)$ and $\textnormal{Ind}(A)$ are independent on $s.$

In particular, $A$ is a Fredholm operator from $L^{2, \mathcal{L}}_s(G, E_0)^\tau$ to $L^{2, \mathcal{L}}_{s-m}(G, E_0)^\tau$ for all $s \in \mathbb{R}.$
Moreover, the kernels $\textnormal{Ker}(A),$ $\textnormal{Ker}(A^*)$ are in $C^\infty(G, E_0)^\tau,$ so that $\textnormal{Ker}(A),$ $\textnormal{Ker}(A^*)$ and $\textnormal{Ind}(A)$ are independent on $s.$

First, let us remark that any vector-valued operator $T$,  with symbol in the class ${S}^{-\infty,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)),$ is a compact operator from  $L^2(G)$ to itself,  and  from any subelliptic Sobolev space $L^{2,\mathcal{L}}_{s_1}(G,E_0)$ into $L^{2,\mathcal{L}}_{s_2}(G,E_0),$ for any pair $s_1,s_2\in \mathbb{R}.$

Indeed, its compactness on $L^2(G,E_0)$ can be deduced from Theorem    <ref> and the fact that any Hilbert-Schmidt operator is compact. Observe that the compactness of $T$ from $L^{2,\mathcal{L}}_{s_1}(G,E_0)$ into $L^{2,\mathcal{L}}_{s_2}(G,E_0),$ follows from the compactness of the operator $\mathcal{M}_{s_2,E_0} T \mathcal{M}_{-s_1,E_0} $ on $L^2(G,E_0).$

Now, we will show that  $A$ is a Fredholm operator from $L^{2, \mathcal{L}}_{s}(G, E_0)$ to $L^{2, \mathcal{L}}_{s-m}(G, E_0)$ for all $s \in \mathbb{R}.$ To prove this suppose that $B \in {S}^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ be a parametrix of the operator $A$ as in Proposition <ref>, that means, there exist operators $T$, $ T' \in {S}^{-\infty,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$  such that $BA=I-T$ and $AB=I-T'.$

Since the smoothing operators
$$T:L^{2, \mathcal{L}}_s(G, E_0) \rightarrow L^{2, \mathcal{L}}_s(G, E_0)$$
$$T':L^{2, \mathcal{L}}_{s-m}(G, E_0) \rightarrow L^{2, \mathcal{L}}_{s-m}(G, E_0)$$
are compact operators and $B: L^{2, \mathcal{L}}_{s-m}(G, E_0) \rightarrow L^{2, \mathcal{L}}_{s}(G, E_0)$ is a bounded linear operator (Theorem <ref>) we deduce, from Theorem <ref>, that $A$ is a Fredholm operator. This means, $\textnormal{Ker}(A)$ and   $ \textnormal{Ker}(A^*) $ are   subspaces  of finite dimension.

Now, we will prove that the index of $A$

$$ \textnormal{Ind}_s(A):=\textnormal{dim Ker}_s(A)-\textnormal{dim Ker}_s(A^*)  $$

is independent of $s\in \mathbb{R}.$
To show this, let us consider the space

$$\textnormal{Ker}_s(A)= \left\{ f \in L^{2, \mathcal{L}}_s(G, E_0): Af=0  \right\}.$$

We note that

$$ \textnormal{Ker}_s(A) \subset  \textnormal{Ker}_s(BA) =  \textnormal{Ker}_s(I-T) \subset C^\infty(G, E_0).$$

Indeed,  $T\in   {S}^{r,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ for any $r\in \mathbb{R},$ and then
$$T:L^{2, \mathcal{L}}_s(G, E_0) \rightarrow L^{2, \mathcal{L}}_{s-r}(G, E_0)$$
continuously for any $r\in \mathbb{R}.$ Now, if we  take $f\in \textnormal{Ker}_s(I-T),$  then it follows that $f\in L^{2, \mathcal{L}}_s(G, E_0),$ and $f=Tf\in L^{2, \mathcal{L}}_{s-r}(G, E_0), $ for any $r\in \mathbb{R}.$ This implies (see (<ref>))

\begin{equation*}
f\in  \bigcap_{t \in \mathbb{R}} L^{2, \mathcal{L}}_t(G, E_0)=C^\infty(G, E_0).
\end{equation*}

Similarly, in view of the identification $E_{0}^{*}\cong E_0$ we can see that $\textnormal{Ker}_{m-s}(A^*) \subset C^\infty(G, E_0)$ with the use of Theorem <ref>. Since these kernels are in $C^\infty(G, E_0)$, they do not depend on $s$, and thus the Fredholm index is independent of $s.$

Finally, using the fact $L^{2, \mathcal{L}}_s(G, E_0)^\tau$ is a closed subspace of $L^{2, \mathcal{L}}_s(G, E_0)$ with $\bigcap_{t \in \mathbb{R}} L^{2, \mathcal{L}}_t(G, E_0)^\tau=C^\infty(G, E_0)^\tau$ and Theorem <ref>, a similar argument can be made to establish that $A$ is a Fredholm operator from $L^{2, \mathcal{L}}_s(G, E_0)^\tau$ to $L^{2, \mathcal{L}}_{s-m}(G, E_0)^\tau$ for all $s \in \mathbb{R},$ and the kernels $\textnormal{Ker}(A),$ $\textnormal{Ker}(A^*)$ are in $C^\infty(G, E_0)^\tau,$ so that $\textnormal{Ker}(A),$ $\textnormal{Ker}(A^*)$ and $\textnormal{Ind}(A)$ are independent on $s.$

As a consequence of Theorem <ref> we have the following result in the context of homogeneous vector bundles.
Let $m\in \mathbb{R},$ and let $0\leqslant \delta<\rho\leqslant 1.$  Let  $a=a(x,\xi)\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$  Assume also that $a(x,\xi)\in \mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0))$ is invertible for every $(x,[\xi])\in G\times\widehat{G},$ and satisfies

\begin{equation}
\sup_{(x,[\xi])\in G\times \widehat{G}} \Vert \widehat{\mathcal{M}}(\xi)^{m} \otimes a(x,\xi)^{-1}\Vert_{\textnormal{op}}<\infty.
\end{equation}
Let $\tilde{A}\in \Psi^{m,\mathcal{L}}_{\rho,\delta}(E)$ be the operator associated to the symbol  $a$ via (<ref>). Then $\tilde{A}$ is a Fredholm operator from $L^{2, \mathcal{L}}_s(E)$ to $L^{2, \mathcal{L}}_{s-m}(E)$ for all $s \in \mathbb{R}.$

Moreover, the kernels $\textnormal{Ker}(\tilde{A}),$ $\textnormal{Ker}(\tilde{A}^*)$ are in $\Gamma^\infty(E),$ so that $\textnormal{Ker}(\tilde{A}),$ $\textnormal{Ker}(\tilde{A}^*)$ and $\textnormal{Ind}(\tilde{A})$ are independent on $s.$

In view of Remark <ref>,  the following  diagram
\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={10em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$L^{2, \mathcal{L}}_s(E)$}  ; & \node(S) {$L^{2, \mathcal{L}}_{s-m}(E)$}; \\
\node(R/I) {$L^{2, \mathcal{L}}_s(G, E_0)^\tau$}; & \node (T) {$L^{2, \mathcal{L}}_{s-m}(G, E_0)^\tau$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\varkappa_{\tau}$};
\draw[->] (R) -- (S) node[anchor=south] {$\tilde{A}$};
\draw[->] (S) -- (T) node[anchor=west] {$\varkappa_\tau$};
\draw[->] (R/I) -- (T) node[anchor=north] {$A$};
\end{tikzpicture}
\end{eqnarray}
commutes. Now, observe that  $\tilde{A}:L^{2, \mathcal{L}}_s(E) \rightarrow L^{2, \mathcal{L}}_{s-m}(E)$ is a Fredholm operator if and only $A:L^{2, \mathcal{L}}_s(G, E_0) \rightarrow L^{2, \mathcal{L}}_{s-m}(G, E_0)$ is a Fredholm operator. Indeed, using the relation,

$$AB-I= T \Leftrightarrow \varkappa_\tau A \varkappa_\tau^{-1} \varkappa_\tau B \varkappa_\tau^{-1}=I-\varkappa_\tau T \varkappa_\tau^{-1} \Leftrightarrow \tilde{A}\tilde{B}=I-\tilde{T}$$

along with Theorem <ref>, Theorem <ref> and the fact that $T$ is compact if and only if $\tilde{T}$ is compact,  the proof of theorem follows.

§.§ Parameter $\mathcal{L}$-ellipticity

To develop the subelliptic functional calculus we need a more wide notion of ellipticity, namely, the parameter ellipticity. Here, we will consider the case where $E_0=F_0,$ and every continuous linear operator $A$  on $C^\infty(G,E_0)$ will be considered with its $\textnormal{End}(E_0)$-valued symbol as in Definition <ref>.

Let $m>0,$ and let $0\leqslant \delta<\rho\leqslant 1.$
Let $\Lambda=\{\gamma(t):t\in J\footnote{where $J=[a,b],$ $-\infty<a\leqslant b<\infty,$ $J=[a,\infty),$ $J=(-\infty,b]$ or $J=(-\infty,\infty).$}\}$  be an analytic curve in the complex plane $\mathbb{C}.$ If $J$ is a finite interval we assume that $\Lambda$ is a closed curve. For  simplicity, if $J$ is an infinite interval we assume that $\Lambda$ is homotopy equivalent to the line $\Lambda_{i\mathbb{R}}:=\{iy:-\infty<y<\infty\}.$  Let  $a=a(x,\xi)\in {S}^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$  Assume also that
$$R_{\lambda}(x,\xi)^{-1}:=a(x,\xi)-\lambda \in \mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0)), \footnote{We have denoted $a(x,\xi)-\lambda:=a(x,\xi)-\lambda I_{ \mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))}$ to simplify the notation.}$$
is invertible   for every $(x,[\xi])\in G\times\widehat{G},$ and for all $\lambda \in  \Lambda.$ We say that $a$ is parameter $\mathcal{L}$-elliptic with respect to $\Lambda,$ if

\begin{equation}
\sup_{\lambda\in \Lambda}\sup_{(x,[\xi])\in G\times \widehat{G}}\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m}\otimes R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}<\infty.
\end{equation}

Observe that for $a=b=0,$ $J=\{0\},$ and for the trivial curve $\gamma(t)=0,$ that $a$ is parameter $\mathcal{L}$-elliptic with respect to $\Lambda=\{0\},$ is equivalent to say that $a$ is  $\mathcal{L}$-elliptic.

The following theorem classifies the matrix resolvent $R_{\lambda}(x,\xi)$ of a parameter $\mathcal{L}$-elliptic symbol $a.$
Let $m>0,$ and let $0\leqslant \delta<\rho\leqslant 1.$ If $a$ is parameter $\mathcal{L}$-elliptic with respect to $\Lambda,$ the following estimate

\begin{equation*}
\sup_{\lambda\in \Lambda}\sup_{(x,[\xi])\in G\times \widehat{G}}\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|}\otimes  \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}<\infty,
\end{equation*}
holds true for all $\alpha,\beta\in \mathbb{N}_0^n$ and $k\in \mathbb{N}_0.$

We will split the proof in the cases $|\lambda|\leqslant 1,$  and $|\lambda|> 1,$ where $\lambda\in \Lambda.$ It is possible however that one of these two cases could be trivial in the sense that $\Lambda_{1}:=\{\lambda\in \Lambda:|\lambda|\leqslant 1\}$ or $\Lambda_{1}^{c}:=\{\lambda\in \Lambda:|\lambda|> 1\}$ could be empty sets. In such a case the proof is self-contained in the situation that we will consider where we assume that $\Lambda_{1}$ and $\Lambda_{1}^c$ are not trivial sets.   For $|\lambda|\leqslant 1,$ observe that

\begin{align*}
&\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}\\
&=\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{-m(k+1)}\widehat{\mathcal{M}}(\xi)^{m(k+1)+\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}\\
&\leqslant \Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{-m(k+1)}\Vert_{\textnormal{op}}\\
&\hspace{5cm}\times\Vert\widehat{\mathcal{M}}(\xi)^{m(k+1)+\rho|\alpha|-\delta|\beta|} \otimes\partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}.
\end{align*}
We have

\begin{align*}
&\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{-m(k+1)}\Vert_{\textnormal{op}}\\
&= \Vert (|\lambda|^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi)^{-1}+I_{d_\xi})^{m(k+1)}\Vert_{\textnormal{op}}\leqslant \Vert |\lambda|^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi)^{-1}+I_{d_\xi}\Vert^{m(k+1)}_{\textnormal{op}}\\
&\leqslant \sup_{|\lambda|\in [0,1]}\sup_{1\leqslant j\leqslant d_\xi}(|\lambda|^{\frac{1}{m}}(1+\nu_{jj}(\xi)^2)^{-\frac{1}{2}}+1))^{k(m+1)}\\
\end{align*}
On the other hand, we can prove that

\begin{align*}
\Vert\widehat{\mathcal{M}}(\xi)^{m(k+1)+\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}=O(1).
\end{align*}
For $k=1,$ $\partial_{\lambda}R_{\lambda}(x,\xi)= R_{\lambda}(x,\xi)^{2}.$ This can be deduced from the Leibniz rule, indeed,

\begin{align*}
0=\partial_{\lambda}(R_{\lambda}(x,\xi) \circ (a(x,\xi)-\lambda))=(\partial_{\lambda}R_{\lambda}(x,\xi)) \circ (a(x,\xi)-\lambda)+ R_{\lambda}(x,\xi)(-1)
\end{align*}
implies that

\begin{align*}
\partial_{\lambda}(R_{\lambda}(x,\xi))\circ(a(x,\xi)-\lambda)= R_{\lambda}(x,\xi).
\end{align*}
Because $(a(x,\xi)-\lambda)=R_{\lambda}(x,\xi)^{-1}$ the identity for the first derivative of $R_\lambda,$ $\partial_{\lambda}R_{\lambda}$ it follows. So, from the chain rule we obtain that the term of higher order expanding the derivative   $ \partial_{\lambda}^kR_{\lambda} $ is $ R_{\lambda}^{k+1}.$ From Theorem   <ref> we deduce that $R_{\lambda}\in S^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$ The subelliptic calculus implies that $R_{\lambda}^{k+1}\in S^{-m(k+1),\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$ This fact, and the compactness of $\Lambda_1\subset \mathbb{C},$ provide us the uniform estimate

\begin{equation*}
\sup_{\lambda\in \Lambda_1}\sup_{(x,[\xi])\in G\times \widehat{G}}\Vert\widehat{\mathcal{M}}(\xi)^{m(k+1)+\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}<\infty.
\end{equation*}
Now, we will analyse the situation for $\lambda\in \Lambda_1^c.$ We will use induction over $k$ in order to prove that

\begin{equation*}
\sup_{\lambda\in \Lambda_1^c}\sup_{(x,[\xi])\in G\times \widehat{G}}\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}<\infty.
\end{equation*}

For $k=0$ notice that

\begin{align*}
&\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}\\
&=\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|}\otimes \partial_{X}^{(\beta)}  \Delta_{\xi}^{\alpha}(a(x,\xi)-\lambda)^{-1}\Vert_{\textnormal{op}},
\end{align*}
and denoting $\theta=\frac{1}{|\lambda|},$ $\omega=\frac{\lambda}{|\lambda|},$ we have

\begin{align*}
&\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}R_{\lambda}(x,\xi)\Vert_{\textnormal{op}}\\
&=\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m}|\lambda|^{-1}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}\\
&=\Vert (1+|\lambda|^{-\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}\\
&=\Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}\\
&=\Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m}\widehat{\mathcal{M}}(\xi)^{m+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}  \Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}\\
&\leqslant \Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m}\Vert_{\textnormal{op}}\Vert\widehat{\mathcal{M}}(\xi)^{m+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}.
\end{align*}
Because $ (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m} \in S^{0,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}) ,$ we have that the operator norm $ \Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m}\Vert_{\textnormal{op}}$ is uniformly bounded in $\theta\in [\varepsilon_0
,1]$ where $\varepsilon_{0}>0   $ is small enough and to be chosen later.  The same argument can be applied to the operator norm
$$ \Vert\widehat{\mathcal{M}}(\xi)^{m+\rho|\alpha|-\delta|\beta|}\otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}, $$

by using that $(\theta\times  a(x,\xi)-\omega)^{-1}\in S^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)), $ with $\theta\in [\varepsilon_0
,1]$  and with $\omega$ being an element of the complex circle. Now, let us analyse the case where $\theta\rightarrow 0^{+}.$ Note that in this case $|\omega|= 1.$ Consider the operator norm

$$\Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m}\widehat{\mathcal{M}}(\xi)^{m+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}  \Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}$$

when $|\alpha|=|\beta|=0.$ Then, we have that

\begin{align*}
&\lim_{\theta\rightarrow0^{+}} \Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m}\widehat{\mathcal{M}}(\xi)^{m} \otimes(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}\\
&\lesssim\lim_{\theta\rightarrow0^{+}} \Vert (\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}=|\omega|^{-1}=1.
\end{align*}
On the other hand, if $|\beta|\neq 0,$ when  $\theta\rightarrow 0^{+},\,|\omega|=1,$ the symbol $(\theta\times  a(x,\xi)-\omega)^{-1}$ tends to the constant operator $\omega^{-1}I,$ and the derivative   $ \partial_{X}^{(\beta)}  \Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}$ tends to the null operator with respect to the operator norm of matrices and uniformly in $[\xi]\in \widehat{G}$. Consequently, the term

$$\Vert (1+\theta^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-m}\widehat{\mathcal{M}}(\xi)^{m+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}  \Delta_{\xi}^{\alpha}(\theta\times  a(x,\xi)-\omega)^{-1}\Vert_{\textnormal{op}}$$

can be estimated from above by $O(1).$

The case  $k\geqslant 1$ for $\lambda\in \Lambda_1^c$ can be proved in an analogous way.
Combining Proposition <ref> and Theorem <ref> we obtain the following corollaries.

Let $m>0,$ and let $0\leqslant \delta<\rho\leqslant 1.$ Let  $a$ be a parameter $\mathcal{L}$-elliptic symbol with respect to $\Lambda.$ Then  there exists a parameter-dependent parametrix of $A-\lambda I,$ with symbol $a^{-\#}(x,\xi,\lambda)$ satisfying the estimates

\begin{equation*}
\sup_{\lambda\in \Lambda}\sup_{(x,[\xi])\in G\times \widehat{G}}\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|}\otimes \partial_{\lambda}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}a^{-\#}(x,\xi,\lambda)\Vert_{\textnormal{op}}<\infty,
\end{equation*}
for all $\alpha,\beta\in \mathbb{N}_0^n$ and $k\in \mathbb{N}_0.$

Let $m>0,$ and let $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)) $ where  $0\leqslant \delta<\rho\leqslant 1.$ Let us assume that $\Lambda$ is a subset of the $L^2$-resolvent set of $A,$ $\textnormal{Resolv}(A):=\mathbb{C}\setminus \textnormal{Spec}(A).$ Then $A-\lambda I$ is invertible on $\mathscr{D}'(G)$ and the symbol of the resolvent operator $\mathcal{R}_{\lambda}:=(A-\lambda I)^{-1},$ $\widehat{\mathcal{R}}_{\lambda}(x,\xi)$ belongs to $S^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)).$

§.§ Vector-valued subelliptic global functional calculus

Let $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ be a parameter $\mathcal{L}$-elliptic symbol  of order $m>0$ with respect to the sector $\Lambda\subset\mathbb{C}.$ For a pseudo-differential operator $A$ with symbol $a,$ let us define the operator $F(A)$  by the (Dunford-Riesz) complex functional calculus

\begin{equation}\label{F(A)}
F(A)=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(A-zI)^{-1}dz,
\end{equation}

(CI). $\Lambda_{\varepsilon}:=\Lambda\cup \{z:|z|\leqslant \varepsilon\},$ $\varepsilon>0,$ and $\Gamma=\partial \Lambda_\varepsilon\subset\textnormal{Resolv}(A)$ is a positively oriented curve in the complex plane $\mathbb{C}$.

(CII). $F$ is a holomorphic function in $\mathbb{C}\setminus \Lambda_{\varepsilon},$ and continuous on its closure.

(CIII). We will assume  decay of $F$ along $\partial \Lambda_\varepsilon$ in order that the operator (<ref>) will be densely defined on $C^\infty(G, E_0)$ in the strong sense of the topology on $L^2(G, E_0)$ (for instance, $|F(\lambda)|\leqslant C|\lambda|^s$ uniformly in $\lambda,$ for some $s<0$ will be enough).
Now, we will compute the  $\textnormal{End}(E_0)$-valued symbols for operators defined by this complex functional calculus.

Let $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ be a parameter $\mathcal{L}$-elliptic symbol  of order $m>0$ with respect to the sector $\Lambda\subset\mathbb{C}.$ Let $F(A):C^\infty(G, E_0)\rightarrow \mathscr{D}'(G, E_0)$ be the operator defined by the analytical functional calculus as in (<ref>). Under the assumptions $\textnormal{(CI)}$, $\textnormal{(CII)}$, and $\textnormal{(CIII)}$, the $\textnormal{End}(E_0)$-valued symbol of $F(A),$ $\sigma_{F(A)}(x,\xi)$ is given by,

\begin{equation*}
\sigma_{F(A)}(x,\xi)=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)\widehat{\mathcal{R}}_z(x,\xi)dz,
\end{equation*}
where $\mathcal{R}_z=(A-zI)^{-1}$ denotes the resolvent of $A,$ and $\widehat{\mathcal{R}}_z=\widehat{\mathcal{R}}_z(x,\xi)\in S^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0) ) $ is its symbol.

From Corollary <ref>, we have that  $\widehat{\mathcal{R}}_z\in S^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0) ) .$ Now, observe from (<ref>) that, for $i,r \in I_{d_\omega},$  we have

\begin{align*}
\sigma_{F(A)}(i, r, x,\xi)&=\xi(x)^* e_{r,E_0}^*[F(A)(\xi \otimes e_{i, E_0})(x)]\\&=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z) \xi(x)^* e_{r,E_0}^*[(A-zI)^{-1}(\xi \otimes e_{i, E_0})(x)]dz.  \end{align*}
We finish the proof by observing that $\widehat{\mathcal{R}}_z(i,r, x,\xi)=\xi(x)^* e_{r,E_0}^*[(A-zI)^{-1}(\xi \otimes e_{i, E_0})(x)],$  for every $z\in \textnormal{Resolv}(A).$

Assumption (CIII) will be clarified in the following theorem where we show that the subelliptic calculus on homogeneous vector bundles is stable under the action of the complex functional calculus.

Let $m>0,$ and let $0\leqslant \delta<\rho\leqslant 1.$ Let  $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ be a parameter $\mathcal{L}$-elliptic symbol with respect to $\Lambda.$ Let us assume that $F$ satisfies the  estimate $|F(\lambda)|\leqslant C|\lambda|^s$ uniformly in $\lambda,$ for some $s<0.$  Then  the symbol of $F(A),$  $\sigma_{F(A)}\in S^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)) $ admits an asymptotic expansion of the form

\begin{equation}\label{asymcomplex}
\sigma_{F(A)}(x,\xi)\sim
\sum_{N=0}^\infty\sigma_{{B}_{N}}(x,\xi),\,\,\,(x,[\xi])\in G\times \widehat{G},
\end{equation}
where $\sigma_{{B}_{N}}(x,\xi)\in {S}^{ms-(\rho-\delta)N,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ and

\begin{equation*}
\sigma_{{B}_{0}}(x,\xi)=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(a(x,\xi)-z)^{-1}dz\in {S}^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).
\end{equation*}

\begin{equation*}
\sigma_{F(A)}(x,\xi)\equiv -\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)a^{-\#}(x,\xi,\lambda)dz \textnormal{  mod  } {S}^{-\infty,\mathcal{L}}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),
\end{equation*}
where $a^{-\#}(x,\xi,\lambda)$ is the symbol of the parametrix to $A-\lambda I,$   in Corollary <ref>.

First, we need to prove that the condition $|F(\lambda)|\leqslant C|\lambda|^s$ uniformly in $\lambda,$ for some $s<0,$ is enough in order to guarantee that
\begin{equation*}
\sigma_{{B}_{0}}(x,\xi):=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(a(x,\xi)-z)^{-1}dz,
\end{equation*}
is a well defined matrix-symbol.
From Theorem <ref> we deduce that $(a(x,\xi)-z)^{-1}$ satisfies the estimate

\begin{equation*}
\Vert (|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m(k+1)}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{z}^k\partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}}<\infty.
\end{equation*}

Observe that

\begin{align*}
& =\Vert (|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{-m}(|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m} \otimes (a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}} \\
&\lesssim  \sup_{1\leqslant j\leqslant d_\xi}(|z|^{\frac{1}{m}}+(1+\nu_{jj}(\xi)^2)^{\frac{1}{2}})^{-m}\\&\leqslant |z|^{-1},
\end{align*}
and the condition $s<0$ implies

\begin{align*}
\left|\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(a(x,\xi)-z)^{-1}dz\right|\lesssim \oint\limits_{\partial \Lambda_\varepsilon}|z|^{-1+s}|dz|<\infty,
\end{align*}
uniformly in $(x,[\xi])\in G\times \widehat{G}.$ In order to check that $\sigma_{B_0}\in {S}^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ let us analyse the cases $-1<s<0$ and $s\leqslant -1$ separately. So, let us analyse first the situation of $-1<s<0.$ We observe that

\begin{align*}
&\Vert \widehat{\mathcal{M}}(\xi)^{-ms+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}\sigma_{B_0}(x,\xi)\Vert_{\textnormal{op}}\\
&\leqslant \frac{C}{2\pi }\oint\limits_{\partial \Lambda_\varepsilon} |z|^{s}\Vert      \widehat{\mathcal{M}}(\xi)^{-ms+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}} |dz|.
\end{align*}
Now, we will estimate the operator norm inside of the integral. Indeed, the identity

\begin{align*}
&\Vert      \widehat{\mathcal{M}}(\xi)^{-ms+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}}=\\
&\Vert (|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{-m}(|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{-ms+\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}}
\end{align*}
implies that

\begin{align*}
&\Vert      \widehat{\mathcal{M}}(\xi)^{-ms+\rho|\alpha|-\delta|\beta|}\otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}} \lesssim  \Vert (|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{-m}\widehat{\mathcal{M}}(\xi)^{-ms}\Vert_{\textnormal{op}}
\end{align*}
where we have used that

\begin{align*}
\sup_{z\in \partial \Lambda_\varepsilon}  \sup_{(x,\xi)} \Vert (|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{m}\widehat{\mathcal{M}}(\xi)^{\rho|\alpha|-\delta|\beta|} \otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}} <\infty.
\end{align*}

Consequently, by using that  $s<0,$ we deduce

\begin{align*}
& \frac{C}{2\pi }\oint\limits_{\partial \Lambda_\varepsilon} |z|^{s}\Vert      \widehat{\mathcal{M}}(\xi)^{ms+\rho|\alpha|-\delta|\beta|}\otimes \partial_{X}^{(\beta)}\Delta_{\xi}^{\alpha}(a(x,\xi)-z)^{-1}\Vert_{\textnormal{op}} |dz|\\
&\lesssim \frac{C}{2\pi }\oint\limits_{\partial \Lambda_\varepsilon} |z|^{s}\Vert (|z|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^{-m} \widehat{\mathcal{M}}(\xi)^{-ms}\Vert_{\textnormal{op}} |dz|\\
&= \frac{C}{2\pi }\oint\limits_{\partial \Lambda_\varepsilon} |z|^{s}\sup_{1\leqslant j\leqslant d_\xi} (|z|^{\frac{1}{m}}+(1+\nu_{jj}(\xi)^{2})^{\frac{1}{2}})^{-m}(1+\nu_{jj}(\xi)^{2})^{-\frac{ms}{2}}|dz|.
\end{align*}
To study the convergence of the last contour integral we only need to check the convergence of $\int_{1}^{\infty}r^s(r^{\frac{1}{m}}+\varkappa)^{-m}\varkappa^{-ms}dr,$ where $\varkappa>1$ in a parameter. The change of variable $r=\varkappa^{m}t$ implies that

\begin{align*}
\int\limits_{1}^{\infty}r^s(r^{\frac{1}{m}}+\varkappa)^{-m}\varkappa^{-ms}dr&=\int\limits_{\varkappa^{-m}}^{\infty}\varkappa^{ms}t^s(\varkappa t^{\frac{1}{m}}+\varkappa)^{-m}\varkappa^{-ms}\varkappa^mdt=\int\limits_{\varkappa^{-m}}^{\infty}t^s(t^{\frac{1}{m}}+1)^{-m}dt\\
&\lesssim \int\limits_{\varkappa^{-m}}^{1}t^sdt+\int\limits_{1}^{\infty}t^{-1+s}<\infty.
\end{align*}
Indeed, for $t\rightarrow\infty,$ $t^s(t^{\frac{1}{m}}+1)^{-m}\lesssim t^{-1+s}$ and we conclude the estimate because $\int\limits_{1}^{\infty} t^{-1+s'}dt<\infty,$ for all $s'<0.$ On the other hand, the condition $-1<s<0$ implies that

\begin{align*}
\int\limits_{\varkappa^{-m}}^{1}t^sdt=\frac{1}{1+s}-\frac{\varkappa^{-m(1+s)}}{1+s}=   O(1).
\end{align*}
In the case where $s\leqslant -1,$ we can find an analytic function $J(z)$ such that it is a holomorphic function in $\mathbb{C}\setminus \Lambda_{\varepsilon},$ and continuous on its closure and additionally satisfying that $F(\lambda)=J(\lambda)^{1+[-s]}.$[ $[-s]$ denotes  the integer part of $-s.$] In this case,  $J(A)$ defined by the complex functional calculus

\begin{equation}\label{G(A)}
J(A)=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}J(z)(A-zI)^{-1}dz,
\end{equation}

has symbol belonging to ${S}^{\frac{sm}{1+[-s]},\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ because $J$ satisfies the estimate $|J(\lambda)|\leqslant C|\lambda|^{\frac{s}{1+[-s]}},$ with $-1<\frac{s}{1+[-s]}<0.$
By observing that

\begin{align*}
\sigma_{F(A)}(x,\xi)&=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)\widehat{\mathcal{R}}_z(x,\xi)dz=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}J(z)^{1+[-s]}\widehat{\mathcal{R}}_z(x,\xi)dz\\
\end{align*}
and computing the symbol $\sigma_{J(A)^{1+[-s]}}(x,\xi)$ by iterating $1+[-s]$-times  the asymptotic formula for the composition in the vector-valued subelliptic calculus (see Theorem <ref>), we can see that the term with higher order in such expansion is $\sigma_{J(A)}(x,\xi)^{1+[-s]}\in {S}^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$ Consequently we have proved that $\sigma_{F(A)}(x,\xi)\in {S}^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$
This completes the proof for the first part of the theorem.
For the second part of the proof, let us denote by $a^{-\#}(x,\xi,\lambda)$  the symbol of the parametrix to $A-\lambda I,$   in Corollary <ref>. Let $P_{\lambda}=\textnormal{Op}(a^{-\#}(\cdot,\cdot,\lambda)).$ Because $\lambda\in \textnormal{Resolv}(A)$ for $\lambda\in \partial \Lambda_\varepsilon,$ $(A-\lambda)^{-1}-P_{\lambda}$ is a smoothing operator. Consequently, from Lemma <ref> we deduce that

\begin{align*}
& \sigma_{F(A)}(x,\xi)=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)\widehat{\mathcal{R}}_z(x,\xi)dz\\
&=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)a^{-\#}(x,\xi,z)dz-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(\widehat{\mathcal{R}}_z(x,\xi)-a^{-\#}(x,\xi,z))dz\\
&\equiv -\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)a^{-\#}(x,\xi,z)dz  \textnormal{  mod  } {S}^{-\infty,\mathcal{L}}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).
\end{align*}
The asymptotic expansion (<ref>) comes from the construction of the parametrix in the vector-valued subelliptic calculus (see Proposition <ref>).

§.§ Global functional calculus  on homogeneous vector bundles

Let $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0))$ be a parameter $\mathcal{L}$-elliptic symbol  of order $m>0$ with respect to the sector $\Lambda\subset\mathbb{C}.$ For a pseudo-differential operator $\tilde{A}$ with symbol $a,$ defined via (<ref>), let us define, again, as in the vector-valued case, the operator $F(\tilde{A})$  by using the (Dunford-Riesz) complex functional calculus:

\begin{equation}\label{F(A)'}
F(\tilde{A}):=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(\tilde{A}-zI)^{-1}dz,
\end{equation}

(CI)': $\Lambda_{\varepsilon}:=\Lambda\cup \{z:|z|\leqslant \varepsilon\},$ $\varepsilon>0,$ and $\Gamma=\partial \Lambda_\varepsilon\subset\textnormal{Resolv}(\tilde{A})$ is a positively oriented curve in the complex plane $\mathbb{C}$.

(CII)': $F$ is a holomorphic function in $\mathbb{C}\setminus \Lambda_{\varepsilon},$ and continuous on its closure.

(CIII)': We will assume  decay of $F$ along $\partial \Lambda_\varepsilon$ in order that the operator ${F(\tilde{A})}$ will be densely defined on $\Gamma^\infty(E)$ in the strong sense of the topology on $L^2(E).$  For instance, we require  $|F(\lambda)|\leqslant C|\lambda|^s$ uniformly in $\lambda,$ for some $s<0.$

Let $\tilde{A}\in \Psi^{m,\mathcal{L}}_{\rho,\delta}(E),$ and let $A$ be its associated vector-valued operator. Because $A$ and $\tilde{A}$ are unitarily equivalent, (because of  (<ref>) for $\tau=\omega$), and using the fact that the spectrum of densely defined linear operators is invariant under unitary transformations, we have that

\begin{equation*}
\textnormal{Resolv}(A)=\textnormal{Resolv}(\tilde{A}),\quad \textnormal{Spectrum}(A)= \textnormal{Spectrum}(\tilde{A}).
\end{equation*}
These equalities of real/complex sets imply that

\begin{equation*}
\widetilde{ [(A-\lambda I)^{-1}]}=(\tilde{A}-\lambda I)^{-1},
\end{equation*}
for any $\lambda \in  \textnormal{Resolv}(A).$ For the proof of the last statement observe that the identity

\begin{align}
\widetilde{ [(A-\lambda I)^{-1}]}=\varkappa_\tau^{-1} [(A-\lambda I)^{-1}]\varkappa_\tau.
\end{align}
implies that

\begin{align*}
\varkappa_\tau^{-1} [(A-\lambda I)^{-1}]\varkappa_\tau \circ  [(\tilde{  A} -\lambda I)]&= \varkappa_\tau^{-1} [(A-\lambda I)^{-1}] \varkappa_\tau \circ[(\varkappa_\tau^{-1}  A\varkappa_\tau -\lambda I)] \\
&=\varkappa_\tau^{-1} [(A-\lambda I)^{-1}]\varkappa_\tau \circ \varkappa_\tau^{-1} [(A-\lambda I)]\varkappa_\tau\\
&=\varkappa_\tau^{-1} [(A-\lambda I)^{-1}] \circ (A-\lambda I)\varkappa_\tau \\
\end{align*}

So, in view of Theorem <ref> and Remark <ref> we have:

Let $m>0,$ and let $0\leqslant \delta<\rho\leqslant 1.$ Let  $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ be a parameter $\mathcal{L}$-elliptic symbol with respect to $\Lambda.$ Let us assume that $F$ satisfies the  estimate $|F(\lambda)|\leqslant C|\lambda|^s$ uniformly in $\lambda,$ for some $s<0.$  Then  the symbol of $F(\tilde{A}),$  $\sigma_{F(\tilde{A})}\in S^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)) $ admits an asymptotic expansion of the form

\begin{equation}\label{asymcomplex2}
\sigma_{F(\tilde{A})}(x,\xi)=\sigma_{\widetilde{F(A)}}(x,\xi)\sim
\sum_{N=0}^\infty\sigma_{{B}_{N}}(x,\xi),\,\,\,(x,[\xi])\in G\times \widehat{G},
\end{equation}
where $\sigma_{{B}_{N}}\in {S}^{ms-(\rho-\delta)N,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ and

\begin{equation*}
\sigma_{{B}_{0}}(x,\xi)=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(a(x,\xi)-z)^{-1}dz\in {S}^{ms,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),
\end{equation*}
where $a$ is the $\textnormal{End}(E_0)$-valued symbol of the vector-valued operator $A$ associated with $\tilde{A}.$

Let us prove that $\widetilde{F(A)}=F(\tilde{A}).$ Indeed, Remark <ref> implies that

\begin{align*}
F(\tilde{A}):&=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(\tilde{A}-zI)^{-1}dz=-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)\varkappa_\tau^{-1} [(A-z I)^{-1}]\varkappa_\tau dz\\
&=\varkappa_\tau^{-1}\left(-\frac{1}{2\pi i}\oint\limits_{\partial \Lambda_\varepsilon}F(z)(A-z I)^{-1} dz\right)\varkappa_\tau=\varkappa_\tau^{-1}F(A)\varkappa_\tau=\widetilde{F(A)}.
\end{align*}
So, we can conclude that   $\sigma_{F(\tilde{A})}=\sigma_{\widetilde{F(A)}}:=\sigma_{{F(A)}}.$ Observe that the asymptotic expansion (<ref>) for the symbol $\sigma_{{F(A)}}$ was proved in Theorem <ref>. So, we end the proof.

§.§ Global functional calculus of the Laplacian
Let us consider the Laplacian $\mathcal{L}_G$ on $G.$  If we choose   $\mathbb{X}=\{X_1,\cdots X_n\}$ to be  an orthonormal basis of the Lie algebra $\mathfrak{g}$ of $G,$ then

The Laplacian $\mathcal{L}_G$ induces a vector-valued Laplacian $\mathcal{L}_{G,E_0}$ (see (<ref>)) that in principle maps $C^{\infty}(G,E_0)$ into $C^{\infty}(G,E_0).$ However, it was proved in <cit.>   that $\mathcal{L}_{G,E_0}$ maps  $C^{\infty}(G,E_0)^{\tau}$ into $C^{\infty}(G,E_0)^{\tau}.$ So, in view of the diagram

[every node/.style=midway]
[column sep=10em,between origins, row sep=4em] at (0,0)
(R) $\Gamma^\infty(E)$  ;     (S) $\Gamma^\infty(E)$;

(R/I) $C^\infty(G,E_0)^\tau$;     (T) $C^\infty(G,E_0)^\tau$;

[<-] (R/I) – (R) node[anchor=east]  $\varkappa_{\tau}$;
[->] (R) – (S) node[anchor=south] $\tilde{\mathcal{L}}_{G,E}$;
[->] (S) – (T) node[anchor=west] $\varkappa_\tau$;
[->] (R/I) – (T) node[anchor=north] $\mathcal{L}_{G,E_0}$;

$\mathcal{L}_{G,E_0}$ induces  a continuous linear operator $\tilde{\mathcal{L}}_{G,E}$ defined on the sections of the homogeneous vector-bundle $E.$ The operator $\tilde{\mathcal{L}}_{G,E}$  is $G$-invariant and it agrees with the Laplacian on the vector-bundle $E\rightarrow M$ (see <cit.> for details).

In this subsection we discuss the membership of the  operators defined by the functional calculus of ${\mathcal{L}}_{G,E}$  and $\tilde{\mathcal{L}}_{G,E}$ to the global (elliptic) Hörmander classes (see Definition <ref> and Remark <ref>).

Recall that  the vector-valued Laplacian is the operator defined via

\begin{equation}\label{Vec:sub:2}
\mathcal{L}_{G,E_0}:=\textnormal{Op}\left(\widehat{\mathcal{L}_G}(\xi) \otimes I_{ \textnormal{End}(E_0)}\right):C^{\infty}(G,E_0)\rightarrow C^{\infty}(G,E_0),
\end{equation}
In (<ref>), $\widehat{\mathcal{L}}_G$ denotes the matrix-valued symbol of the  Laplacian $\mathcal{L}_G.$
For $m\in \mathbb{R},$ let us consider the Euclidean class of symbols $S^{m}(\mathbb{R}^{+}_0),$ endowed with the standard countable family of seminorms

\begin{equation}
\Vert f \Vert_{d,\,S^{m}(\mathbb{R}^{+}_0)}:=\sup_{\ell\leq d}\sup_{\lambda\geq 0}(1+\lambda)^{-m+\ell}|\partial^\ell_{\lambda}f(\lambda)|,\,d\in \mathbb{N}_0.
\end{equation}

For any $f\in S^{m}(\mathbb{R}^{+}_0), $ we define

\begin{equation}\label{Vec:sub:f:2}
f( \mathcal{L}_{G, E_0}):=\textnormal{Op}\left(f(\widehat{\mathcal{L}}_G(\xi)) \otimes I_{ \textnormal{End}(E_0)}\right):C^{\infty}(G,E_0)\rightarrow C^{\infty}(G,E_0),
\end{equation}
with $f(\widehat{\mathcal{L}}_G(\xi)),$ $[\xi]\in \widehat{G},$ defined by the spectral calculus of matrices.
As a consequence of Theorem <ref> we have the following result.
For $f\in S^{\frac{m}{2}}(\mathbb{R}^{+}_0), $ $m\in \mathbb{R},$ we have

\begin{equation}
f(\mathcal{L}_{G,E_0})\in \Psi^{m}_{1,0}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).
\end{equation}

In particular for $f_s(\lambda):=(1+\lambda)^{\frac{s}{2}},$ the global symbol of ${M}_{s,E_0}=f_s(\mathcal{L}_{G,E_0})$ belongs to $ S^{s}_{1,0}( (G\times \widehat{G})\otimes \textnormal{End}(E_0)) $ for all $s\in \mathbb{R}.$ In other words, we have

\begin{equation}\label{thepowersof:LE0:2}
{M}_{s,E_0}:=(1+\mathcal{L}_{G,E_0})^{\frac{s}{2}}\in \Psi^{s}_{1,0}((G\times \widehat{G})\otimes\textnormal{End}(E_0))
\end{equation}

for all $s\in \mathbb{R}.$
As a consequence of Theorem <ref> and the previous corollary, we get the following result that guarantees the existence of matrix-valued symbols for the operators defined by the real functional calculus of the Laplacian on $E$.
For $f\in S^{\frac{m}{2}}(\mathbb{R}^{+}_0), $ $m\in \mathbb{R},$ we have

\begin{equation}
f(\tilde{\mathcal{L}}_{G,E})\in \Psi^{m}_{1,0}(E).
\end{equation}

In particular for $f_s(\lambda):=(1+\lambda)^{\frac{s}{2}},$ the global symbol of $\tilde{{M}}_{s,E}=f_s(\tilde{\mathcal{L}}_{G,E})$ belongs to $ S^{s}_{1,0}( E) $ for all $s\in \mathbb{R}.$ In other words, we have

\begin{equation}\label{thepowersof:LE0:3}
\tilde{M}_{s,E}:=(1+\tilde{\mathcal{L}}_{G,E})^{\frac{s}{2}}\in \Psi^{s}_{1,0}(E)
\end{equation}

for all $s\in \mathbb{R}.$

§.§ Vector-valued Gårding inequality

In this section we will prove the vector-valued Gårding inequality for the vector-valued subelliptic calculus.
We start our analysis with the following proposition.

Let $0\leqslant \delta<\rho\leqslant 1.$  Let  $a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0))$ be an $\mathcal{L}$-elliptic symbol where $m\geqslant 0$ and let us assume that $a$ is positive definite,  namely, that the symbol

$$ a(x,\xi):=(\sigma_A(x,\xi)_{u,v})_{1\leq v,u\leq d_\xi}\in \mathbb{C}^{d_\xi\times d_\xi}(\textnormal{End}(E_0)),$$
is self-adjoint and its spectrum is contained in $\mathbb{R}^{+}_0:=[0,\infty).$  Then $a$ is parameter-elliptic with respect to $\mathbb{R}_{-}:=\{z=x+i0:x<0\}\subset\mathbb{C}.$ Furthermore, for any number $s\in \mathbb{C},$

\begin{equation*}
\widehat{B}(x,\xi)\equiv a(x,\xi)^s:=\exp(s\log(a(x,\xi))),\,\,(x,[\xi])\in G\times \widehat{G},
\end{equation*}
defines a symbol $\widehat{B}\in S^{m\times\textnormal{Re}(s),\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$

Note that for any $(x,\xi)\in G\times \widehat{G},$ $a(x,\xi)$ is invertible. Then its spectrum is strictly contained in $\mathbb{R}^{+}:=(0,\infty).$ From the estimates

\begin{equation*}
\sup_{(x,[\xi])}\Vert\widehat{\mathcal{M}}(\xi)^{-m} \otimes a(x,\xi) \Vert_{\textnormal{op}}<\infty,\quad \sup_{(x,[\xi])}\Vert \widehat{\mathcal{M}}^m(\xi)\otimes a(x,\xi)^{-1} \Vert_{\textnormal{op}}<\infty,
\end{equation*}

we deduce that

\begin{equation*}
\sup_{x\in G}\Vert a(x,\xi) \Vert_{\textnormal{op}}\asymp\Vert  \widehat{\mathcal{M}}(\xi)^{m}\Vert_{\textnormal{op}}, \quad \sup_{x\in G}\Vert a(x,\xi)^{-1} \Vert_{\textnormal{op}}\asymp\Vert  \widehat{\mathcal{M}}(\xi)^{-m}\Vert_{\textnormal{op}}.
\end{equation*}
Consequently we have

\begin{equation*}
\Vert  \widehat{\mathcal{M}}(\xi)^{m}\Vert_{\textnormal{op}}^{-1} \sup_{x\in G}\Vert a(x,\xi) \Vert_{\textnormal{op}}\asymp 1, \quad \Vert  \widehat{\mathcal{M}}(\xi)^{-m}\Vert_{\textnormal{op}}^{-1} \sup_{x\in G}\Vert a(x,\xi)^{-1} \Vert_{\textnormal{op}}\asymp 1,
\end{equation*}
from which we deduce that

\begin{equation*}
\Vert  \widehat{\mathcal{M}}(\xi)^{-m}\Vert_{\textnormal{op}}^{-1}\textnormal{Spectrum}(a(x,\xi))\subset [c,C],
\end{equation*}
where $c,C>0$ are positive real numbers. For every $\lambda\in \mathbb{R}_{-}$ we have

\begin{align*}
&   \Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^m \otimes (a(x,\xi)-\lambda)^{-1}\Vert_{\textnormal{op}}\\
&\asymp \Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}(\widehat{\mathcal{M}}(\xi)^m-\lambda) \otimes (a(x,\xi)-\lambda)^{-1}\Vert_{\textnormal{op}} \\
&\lesssim \Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}\Vert_{\textnormal{op}} \Vert (\widehat{\mathcal{M}}(\xi)^m-\lambda)\otimes(a(x,\xi)-\lambda)^{-1}\Vert_{\textnormal{op}}\\
&\lesssim \Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}\Vert_{\textnormal{op}},
\end{align*}

where we have used that $\Vert (\widehat{\mathcal{M}}(\xi)^m-\lambda)\otimes(a(x,\xi)-\lambda)^{-1}\Vert_{\textnormal{op}}$ is uniformly bounded in $\lambda.$ Indeed, with $\theta=1/|\lambda|=-1/\lambda,$ and $\omega=-\lambda/|\lambda|=1,$ the previous operator norm can be written as

$$\Vert (\theta\widehat{\mathcal{M}}(\xi)^m+1)\otimes(\theta a(x,\xi)+1)^{-1}\Vert_{\textnormal{op}},\,\theta\in [0,1],$$

and the compactness of the unit interval allows us to estimate this operator norm uniformly in $[0,1].$ Indeed, note that the functional calculus, the positivity of the symbol $a(x,\xi),$  and its $\mathcal{L}$-ellipticity imply that for $\theta\in [0,1],$   $(\theta a(x,\xi)+1)^{-1}\in S^{-m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$ Because $(\theta \widehat{\mathcal{M}}(\xi)^m+1)\in S^{m,\mathcal{L}}_{1,0}(G\times \widehat{G})), $ then
$$  (\theta\widehat{\mathcal{M}}(\xi)^m+1)\otimes(\theta a(x,\xi)+1)^{-1}\in S^{0,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0),  $$

and then $\sup_{\theta\in [0,1]}\Vert (\theta\widehat{\mathcal{M}}(\xi)^m+1)\otimes(\theta a(x,\xi)+1)^{-1}\Vert_{\textnormal{op}}=O(1).$

Now, fixing again $\lambda\in \mathbb{R}_{-}$ observe that from the compactness of $[0,1/2]$ we deduce that

\begin{align*}
\sup_{0\leqslant |\lambda| \leqslant 1/2}\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}\Vert_{\textnormal{op}}&\asymp \sup_{0\leqslant |\lambda| \leqslant 1/2}\Vert \widehat{\mathcal{M}}(\xi)^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}\Vert_{\textnormal{op}} \\
%&\asymp \sup_{0\leqslant |\lambda| \leqslant 1/2}\Vert \widehat{\mathcal{M}}(\xi)^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}\Vert_{\textnormal{op}} \\
&\asymp \sup_{0\leqslant |\lambda| \leqslant 1/2}\Vert  (I_{d_\xi}-\lambda\widehat{\mathcal{M}}(\xi)^{-m})^{-1}\Vert_{\textnormal{op}} \lesssim 1,
\end{align*}

where (using the notation in (<ref>)) we have used the estimate

$$  0\leq |\lambda|(1+\nu_{ii}(\xi)^2)^{-\frac{m}{2}}\leq \frac{1}{2},  $$
for $\lambda<0$ with
$0\leq |\lambda|\leq 1/2,$ which implies

\begin{align*}
\Vert  (I_{d_\xi}-\lambda\widehat{\mathcal{M}}(\xi)^{-m})^{-1}\Vert_{\textnormal{op}}=\sup_{1\leq i\leq d_\xi}\left( 1+ |\lambda|(1+\nu_{ii}(\xi)^2)^{-\frac{m}{2}} \right)^{-1}\in [2/3,1].
\end{align*}

On the other hand, by writing $\theta=1/|\lambda|$ and $\omega=\lambda/|\lambda|,$ we have that

\begin{align*}
&  \sup_{ |\lambda|\geqslant 1/2 }\Vert (|\lambda|^{\frac{1}{m}}+\widehat{\mathcal{M}}(\xi))^m (\widehat{\mathcal{M}}(\xi)^m-\lambda)^{-1}\Vert_{\textnormal{op}}  \\
&=\sup_{ |\lambda|\geqslant 1/2 }\Vert (|\lambda|^{\frac{1}{m}}\widehat{\mathcal{M}}(\xi)^{-1}+I_{d_\xi})^m (I_{d_\xi}-\widehat{\mathcal{M}}(\xi)^{-m}\lambda)^{-1}\Vert_{\textnormal{op}}\\
&=\sup_{ |\lambda|\geqslant 1/2 }\Vert (\widehat{\mathcal{M}}(\xi)^{-1}+|\lambda|^{-\frac{1}{m}}I_{d_\xi})^m |\lambda|(I_{d_\xi}-\widehat{\mathcal{M}}(\xi)^{-m}\lambda)^{-1}\Vert_{\textnormal{op}}\\
&\leq \sup_{ |\theta|\leqslant 2 }\Vert (\widehat{\mathcal{M}}(\xi)^{-1}+\theta^{\frac{1}{m}}I_{d_\xi})^m (\theta-\omega\widehat{\mathcal{M}}(\xi)^{-m})^{-1}\Vert_{\textnormal{op}}.
\end{align*}
The previous operator norm is $O(1),$ by considering the compatness of the interval $[0,1/2],$ and the fact that $(\widehat{\mathcal{M}}(\xi)^{-1}+\theta^{\frac{1}{m}}I_{d_\xi})^m (\theta-\omega\widehat{\mathcal{M}}(\xi)^{-m})^{-1}\in S^{0,\mathcal{L}}_{1,0}(G\times \widehat{G}).$

So, we have proved that $a$ is parameter-elliptic with respect to $\mathbb{R}_{-}.$ To prove that $\widehat{B}=\widehat{B}(x,\xi)\in S^{m\times\textnormal{Re}(s),\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),$ we can observe that for $\textnormal{Re}(s)<0,$ we can apply Theorem <ref>. If  $\textnormal{Re}(s)\geqslant 0,$ we can find $k\in \mathbb{N}$ such that $\textnormal{Re}(s)-k<0$ and consequently from the spectral calculus of matrices we deduce that $a(x,\xi)^{\textnormal{Re}(s)-k}\in S^{m\times(\textnormal{Re}(s)-k),\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$ So, from the calculus we conclude that