$$a(x,\xi)^{s}=a(x,\xi)^{s-k}\circ a(x,\xi)^{k}\in S^{m\times\textnormal{Re}(s),\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$$
Thus the proof is complete.

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. Then
$\widehat{B}(x,\xi)\equiv a(x,\xi)^\frac{1}{2}:=\exp(\frac{1}{2}\log(a(x,\xi)))\in S^{\frac{m}{2},\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$

Now, let us assume that

\begin{equation*}
A(x,\xi):=\frac{1}{2}(a(x,\xi)+a(x,\xi)^{*}),\,(x,[\xi])\in G\times \widehat{G},\,\,a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),
\end{equation*}

\begin{align}\label{eqi}
\Vert\widehat{\mathcal{M}}(\xi)^{m} \otimes A(x,\xi)^{-1} \Vert_{\textnormal{op}}\leqslant C_{0}.
\end{align}
Observe that (<ref>) implies that

\begin{equation*}
\lambda(x,\xi):=\inf\{\tilde{\lambda}(x, \xi)^{-1}\}\leqslant C_{0},\,\,\,(x,[\xi])\in G\times \widehat{G},
\end{equation*}

where $\tilde{\lambda}(x,\xi)$ runs over the eigenvalues of the operator $\widehat{\mathcal{M}}(\xi)^{-m}\otimes A(x,\xi).$ So, $\lambda(x,\xi)^{-1}\geqslant \frac{1}{C_0}$ and consequently, in the sense of matrix of operators, we have

\begin{align*}
\widehat{\mathcal{M}}(\xi)^{-m} \otimes A(x,\xi)\geqslant\frac{1}{C_0}I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))}.
\end{align*}
This implies that

\begin{align*}
A(x,\xi)\geqslant\frac{1}{C_0}\widehat{\mathcal{M}}(\xi)^{m} \otimes I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))},
\end{align*}
and for $C_1\in(0, \frac{1}{C_0})$ we have that

\begin{align*}
A(x,\xi)-C_{1}  \widehat{\mathcal{M}}(\xi)^{m} \otimes I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))}\geqslant \left(\frac{1}{C_0}-C_1\right) \widehat{\mathcal{M}}(\xi)^{m} \otimes I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))}>0.
\end{align*}
If  $0\leqslant \delta<\rho\leqslant 1,$   from Corollary <ref>, we have that

\begin{align*}
q(x,\xi):=(A(x,\xi)-C_{1}  \widehat{\mathcal{M}}(\xi)^{m} \otimes I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))})^{\frac{1}{2}}\in  S^{\frac{m}{2},\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).
\end{align*}
From the symbolic calculus we obtain

\begin{align*}
q(x,\xi)q(x,\xi)^*= A(x,\xi)-C_{1}  \widehat{\mathcal{M}}(\xi)^{m} \otimes I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))}+r(x,\xi),\,\,
\end{align*}

where $  r= r(x,\xi)\in  S^{m-(\rho-\delta),\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)).$
Now, let us assume that $u\in C^\infty(G, E_0).$ Then we have, by setting
$$\mathcal{M}_{m,E_0}:=\textnormal{Op}(\widehat{\mathcal{M}}(\xi)^{m} \otimes I_{\mathbb{C}^{d_\xi\times d_\xi}( \textnormal{End}(E_0))}),$$

\begin{align*}
\textnormal{Re}(a(x,D)u,u)&=\frac{1}{2}((a(x,D)+a(x, \xi)^*))u,u)=(A(x,D)u,u)\\
&=C_{1}(\mathcal{M}_{m,E_0} u,u)+(q(x,D)q(x,D)^{*}u,u)-(r(x,D)u,u)\\
&\geqslant C_{1}\Vert u\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}-(r(x,D)u,u)\\
&= C_{1}\Vert u\Vert_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}^2-(\mathcal{M}_{-\frac{m-(\rho-\delta)}{2},E_0}r(x,D)u, \mathcal{M}_{\frac{m-(\rho-\delta)}{2},E_0}u).
\end{align*}
By using Remark <ref>, we have

\begin{align*}
(\mathcal{M}_{-\frac{m-(\rho-\delta)}{2},E_0}r(x,D)u, \mathcal{M}_{\frac{m-(\rho-\delta)}{2},E_0}u)&\leqslant \Vert \mathcal{M}_{-\frac{m-(\rho-\delta)}{2},E_0}r(x,D)u \Vert_{L^2(G, E_0)}\Vert u\Vert_{L^{2,\mathcal{L}  }_{\frac{m-(\rho-\delta)}{2}}(G, E_0)}\\
&= \Vert r(x,D)u \Vert_{L^{2,\mathcal{L}  }_{-\frac{m-(\rho-\delta)}{2}}(G, E_0)}\Vert u\Vert_{L^{2,\mathcal{L}  }_{\frac{m-(\rho-\delta)}{2}}(G, E_0)}\\
&\leqslant C\Vert u \Vert_{L^{2,\mathcal{L}  }_{\frac{m-(\rho-\delta)}{2}}(G, E_0)}\Vert u\Vert_{L^{2,\mathcal{L}  }_{\frac{m-(\rho-\delta)}{2}}(G, E_0)},
\end{align*}
where in the last line we have used the subelliptic Sobolev boundedness of $r(x,D)$ from $L^{2,\mathcal{L}  }_{\frac{m-(\rho-\delta)}{2}}(G, E_0)$ into $L^{2,\mathcal{L}  }_{-\frac{m-(\rho-\delta)}{2}}(G, E_0).$ Consequently, we deduce the lower bound

\begin{align*}
\textnormal{Re}(a(x,D)u,u) \geqslant C_{1}\Vert u\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}-C\Vert u\Vert_{L^{2,\mathcal{L}  }_{\frac{m-(\rho-\delta)}{2}}(G, E_0)}^2.
\end{align*}
If we assume for a moment that for every $\varepsilon>0,$ there exists $C_{\varepsilon}>0,$ such that

\begin{equation}\label{lemararo}
\Vert u\Vert_{{L}^{2,\mathcal{L}}_{\frac{m-(\rho-\delta)}{2}}(G, E_0)}^2\leqslant \varepsilon\Vert u\Vert_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}^2+C_{\varepsilon}\Vert u \Vert_{L^2(G, E_0)}^2,
\end{equation}
for $0<\varepsilon<C_{1}$ we have

\begin{align*}
\textnormal{Re}(a(x,D)u,u) \geqslant (C_{1}-\varepsilon)\Vert u\Vert_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}^2-C_{\varepsilon}\Vert u\Vert_{L^2(G, E_0)}^2.
\end{align*}
So, with the exception of the proof of (<ref>) we have deduced the following estimate which is the main result of this subsection.

Let $G$ be a compact Lie group. For $0\leqslant \delta<\rho\leqslant 1$, let $a(x,D):C^\infty(G, E_0)\rightarrow\mathscr{D}'(G, E_0)$ be an operator with symbol  $a\in {S}^{m,\mathcal{L}}_{\rho,\delta}(( G\times \widehat{G})\otimes \textnormal{End}(E_0))$, $m\in \mathbb{R}$. Let us assume that

\begin{equation*}
A(x,\xi):=\frac{1}{2}(a(x,\xi)+a(x,\xi)^{*}),\,(x,[\xi])\in G\times \widehat{G},\,\,a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),
\end{equation*}

\begin{align*}\label{garding}
\Vert\widehat{\mathcal{M}}(\xi)^{m} \otimes A(x,\xi)^{-1} \Vert_{\textnormal{op}}\leqslant C_{0}.
\end{align*}
Then, there exist $C_{1},C_{2}>0,$ such that the lower bound

\begin{align}
\textnormal{Re}(a(x,D)u,u) \geqslant C_1\Vert u\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}-C_2\Vert u\Vert_{L^2(G, E_0)}^2,
\end{align}
holds true for every $u\in C^\infty(G, E_0).$

In view of the analysis above, for the proof of Theorem <ref> we only need to prove (<ref>). However we will deduce it from the following more general lemma.
Let us assume that $s\geqslant t\geqslant 0$ or that $s,t<0.$ Then, for every $\varepsilon>0,$ there exists $C_\varepsilon>0$ such that

\begin{equation}\label{lemararo2}
\Vert u\Vert_{{L}^{2,\mathcal{L}}_{t}(G, E_0)}^2\leqslant \varepsilon\Vert u\Vert_{{L}^{2,\mathcal{L}}_{s}(G, E_0)}^2+C_{\varepsilon}\Vert u \Vert_{L^2(G, E_0)}^2,
\end{equation}
holds true for every $u\in C^\infty(G, E_0).$

Let $\varepsilon>0.$ Then, there exists $C_{\varepsilon}>0$ such that

\begin{equation*}
(1+\nu_{ii}(\xi)^2)^{t}-\varepsilon (1+\nu_{ii}(\xi)^2)^s\leqslant C_{\varepsilon},
\end{equation*}
uniformly in $[\xi]\in \widehat{G},$ where $\nu_{ii}(\xi)^2$ are eigenvalues of  the symbol of the sub-Laplacian $\mathcal{L}$ at $[\xi].$ Then (<ref>)  would follow from the Plancherel theorem. Indeed,

\begin{align*}
\Vert u\Vert_{{L}^{2,\mathcal{L}}_{t}(G, E_0)}^2& =\sum_{[\xi]\in \widehat{G};1\leq i_0\leq d_\tau}d_\xi\sum_{i,j=1}^{d_\xi} (1+\nu_{ii}(\xi)^2)^{t}|\widehat{u}_{ij}(i_0,\xi)|^{2}  \\
&\leqslant \sum_{[\xi]\in \widehat{G};1\leq i_0\leq d_\tau}d_\xi\sum_{i,j=1}^{d_\xi} (\varepsilon(1+\nu_{ii}(\xi)^2)^{s}+C_\varepsilon)|\widehat{u}_{ij}(i_0,\xi)|^{2}  \\
&= \varepsilon\Vert u\Vert_{{L}^{2,\mathcal{L}}_{s}(G, E_0)}^2+C_{\varepsilon}\Vert u \Vert_{L^2(G, E_0)}^2,
\end{align*}

completing the proof.

Let $G$ be a compact Lie group. Let $a(x,D):C^\infty(G, E_0)\rightarrow\mathscr{D}'(G, E_0)$ be an operator with symbol  $a\in {S}^{m,\mathcal{L}}_{\rho,\delta}(( G\times \widehat{G})\otimes \textnormal{End}(E_0))$, $m\in \mathbb{R}$ and let $0\leqslant \delta<\rho\leqslant 1$. Let us assume that

\begin{equation*}
a(x,\xi)\geqslant 0,\,(x,[\xi])\in G\times \widehat{G},
\end{equation*}

\begin{align*}\label{garding22}
\Vert\widehat{\mathcal{M}}(\xi)^{m} \otimes a(x,\xi)^{-1} \Vert_{\textnormal{op}}\leqslant C_{0}.
\end{align*}
Then, there exist $C_{1},C_{2}>0,$ such that the lower bound

\begin{align}
\textnormal{Re}(a(x,D)u,u) \geqslant C_1\Vert u\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}-C_2\Vert u\Vert_{L^2(G, E_0)}^2,
\end{align}
holds true for every $u\in C^\infty(G, E_0).$

§.§ Gårding inequality on homogeneous vector bundles
Now, the analysis above allows us to transfer the Gårding inequality from the vector-valued context to the setting of vector-bundles.

Let $E$ be a $G$-invariant vector bundle over $M=G/K,$ where $G$ is a compact Lie group. For $0\leqslant \delta<\rho\leqslant 1,$   let $\tilde{A}\in \Psi^{m,\mathcal{L}}_{\rho,\delta}(E)$ be an operator with symbol  $a\in {S}^{m,\mathcal{L}}_{\rho,\delta}(( G\times \widehat{G})\otimes \textnormal{End}(E_0))$, $m\in \mathbb{R}$. Let us assume that

\begin{equation*}
A(x,\xi):=\frac{1}{2}(a(x,\xi)+a(x,\xi)^{*}),\,(x,[\xi])\in G\times \widehat{G},\,\,a\in S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),
\end{equation*}

\begin{align*}\label{garding2}
\Vert\widehat{\mathcal{M}}(\xi)^{m} \otimes A(x,\xi)^{-1} \Vert_{\textnormal{op}}\leqslant C_{0}.
\end{align*}
Then, there exist $C_{1},C_{2}>0,$ such that the lower bound

\begin{align}
\textnormal{Re}(\tilde{A}s,s) \geqslant C_1\Vert s\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(E)}-C_2\Vert s\Vert_{L^2(E)}^2,
\end{align}
holds true for every $s\in \Gamma^\infty(E).$

Observe that, if $A:C^{\infty}(G,E_0)^{\tau}\rightarrow C^{\infty}(G,G_0)^{\omega} $ is the vector-valued operator associated to $\tilde{A},$ and $u=\varkappa_\tau s,$ $s\in \Gamma^\infty(E),$ then Theorem <ref> implies that

\begin{align*}
\textnormal{Re}(Au,u) &\geqslant C_1\Vert u\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(G, E_0)}-C_2\Vert u\Vert_{L^2(G, E_0)}^2\\
&= C_1\Vert s\Vert^2_{{L}^{2,\mathcal{L}}_{\frac{m}{2}}(E)}-C_2\Vert s\Vert_{L^2(E)}^2.
\end{align*}
Now, the proof is completed by observing that

\begin{align*}
\textnormal{Re}(\tilde{A}s,s)=\textnormal{Re}(\varkappa_{\tau }^{-1}{A}\varkappa_\tau \varkappa_\tau^{-1}u,\varkappa_{\tau}^{-1} u) =\textnormal{Re}({A}\varkappa_\tau \varkappa_\tau^{-1}u, u)=  \textnormal{Re}({A}u, u),
\end{align*}
where we have use that $\varkappa_\tau^{-1}$ is an isometry.

§.§ Global wellposedness  on homogeneous vector bundles

In this subsection we present some applications of the subelliptic pseudo-differential calculus developed in previous sections to study global solvability, global wellposedness and uniqueness of the solution of hyperbolic/parabolic PDE associated to subelliptic pseudo-differential operators on homogeneous vector bundles. As before we consider $E\cong G\times_{\tau}E_0.$

Specifically, let us consider the following Cauchy problem

\begin{equation}\label{PVI}(\textnormal{PVI}): \begin{cases}\frac{\partial u}{\partial t}=K(t,x,D)u+f ,& \text{ }u\in \mathscr{D}'((0,T)\times E),
\\u(0)=u_0, & \text{ } \end{cases}
\end{equation}
where the initial data $u_0\in L^2(E),$ $K(t):=K(t,x,D)\in C([0,T], S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)),$ $f\in L^2([0,T],L^2(E)) ,$ $m>0.$ In addition, we also assume that the associated operator $\textnormal{Re}(K(t))$ with operator $K(t)$ defined by

\begin{equation*}
\textnormal{Re}(K(t)):=\frac{1}{2}(K(t)+K(t)^*),\,\,0\leqslant t\leqslant T,\footnote{ This means that $A=K(t)$ is  strongly $\mathcal{L}$-elliptic.}
\end{equation*}
is $\mathcal{L}$-elliptic.

Under such assumptions we will prove the existence and uniqueness of a solution $u\in C^1([0,T],L^2(E))\cap C([0,T], L^{2, \mathcal{L}}_{m}(E)),$ $E\cong G\times_{\tau}E_0.$ We will start with the following energy estimate.

Let $K(t)=K(t,x,D)\in C([0,T], S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G}) \otimes \textnormal{End}(E_0)),$ $0\leqslant \delta<\rho\leqslant  1,$   be a subelliptic pseudo-differential operator of order $m>0,$ and let us assume that $\textnormal{Re}(K(t))$ is $\mathcal{L}$-elliptic, for every $t\in[0,T]$ with $T>0.$ If  $u\in C^1([0,T], L^2(E) )  \cap C([0,T],L^{2,\mathcal{L}}_m (E))$ is solution of the problem (<ref>) then there exist $C,C'>0,$ such that

\begin{equation} \label{GI-03}
\Vert u(t) \Vert_{L^2(E)}\leqslant   \left(C\Vert u(0) \Vert^2_{L^2(E)}+C'\int\limits_{0}^T \Vert (\partial_{t}-K(t_1))u(t_1) \Vert^2_{L^2(E)}dt_1 \right),
\end{equation}
holds for every $0\leqslant t\leqslant T.$

Moreover, we also have the estimate

\begin{equation}\label{Q*}
\Vert u(t) \Vert_{L^2(E)}\leqslant   \left(C\Vert u(T) \Vert^2_{L^2(E)}+C'\int\limits_{0}^T \Vert (\partial_{t}-K(\tau)^{*})u(t_1) \Vert^2_{L^2(E)}dt_1 \right).
\end{equation}

Let $u\in C^1([0,T], L^2(E) )  \cap C([0,T], L^{2,\mathcal{L}}_m (E)).$  We note  for further use that, by using the  embedding $L^{2, \mathcal{L}}_{m} \hookrightarrow L^{2, \mathcal{L}}_{\frac{m}{2}} $ we have   $u\in  C([0,T],L^{2, \mathcal{L}}_{\frac{m}{2}} (E)).$   So,
$u\in \textnormal{Dom}(\partial_{t_1}-K(t_1))$ for every $0\leqslant t_1 \leqslant T.$ In view of the embedding $L^{2, \mathcal{L}}_{m} \hookrightarrow L^2(E),$ we also have that  $u\in C([0,T], L^2(E) ).$  Let us define $f(t_1):=Q(t_1)u(t_1),$ $Q(t_1):=(\partial_{t_1}-K(t_1)),$ for every $0\leqslant t_1\leqslant T.$ Observe that

\begin{align*}
\frac{d}{dt}\Vert u(t) \Vert^2_{L^2(E)}&= \frac{d}{dt}\left(u(t),u(t)\right)_{L^2(E)}\\&=\left(\frac{d u(t)}{dt},u(t)\right)_{L^2(E)}+\left(u(t),\frac{d u(t)}{dt}\right)_{L^2(E)}\\
&=\left((K(t)+K(t)^{*})u(t),u(t)\right)_{L^2(E)}+2\textnormal{Re}(f(t), u(t))_{L^2(E)}\\
&=(2\textnormal{Re}K(t)u(t),u(t))_{L^2(E)}+2\textnormal{Re}(f(t), u(t))_{L^2(E)}.
\end{align*}
Now, from the subelliptic Gårding inequality on homogeneous vector bundles,

\begin{align}
\textnormal{Re}(-K(t)u(t),u(t)) \geqslant C_1\Vert u(t)\Vert_{L^{2, \mathcal{L}}_{\frac{m}{2}} (E)}-C_2\Vert u(t)\Vert_{L^2(E)}^2,
\end{align}
and  the parallelogram law, we have

\begin{align*}
2\textnormal{Re}(f(t), u(t))_{L^2(E)}&\leqslant 2\textnormal{Re}(f(t), u(t))_{L^2(E)}+\Vert f(t)\Vert_{L^2(E)}^2+\Vert u(t)\Vert_{L^2(E)}^2 \\
&= \Vert f(t)+u(t)\Vert^2_{L^2(E)}\leqslant \Vert f(t)+u(t)\Vert^2_{L^2(E)}+\Vert f(t)-u(t)\Vert^2_{L^2(E)} \\
&= 2\Vert f(t)\Vert^2_{L^2(E)}+2\Vert u(t)\Vert^2_{L^2(E)},
\end{align*}

and consequently

\begin{align*}
& \frac{d}{dt}\Vert u(t) \Vert^2_{L^2(E)}\\
&\leqslant 2\left(C_2\Vert u(t)\Vert_{L^2(E)}^2-C_1\Vert u(t)\Vert_{L^{2, \mathcal{L}}_{\frac{m}{2}} (E)}\right)+2\Vert f(t)\Vert^2_{L^2(E)}+2\Vert u(t)\Vert^2_{L^2(E)}.
\end{align*}
So, we have proved that

\begin{align*}
\frac{d}{dt}\Vert u(t) \Vert^2_{L^2(E)}\lesssim  \Vert f(t)\Vert^2_{L^2(E)}+\Vert u(t)\Vert^2_{L^2(E)}.
\end{align*}
By using Gronwall's Lemma we obtain the energy estimate

\begin{equation}
\Vert u(t) \Vert^2_{L^2(E)}\leqslant  \left(C\Vert u(0) \Vert_{L^2(E)}^2+C'\int\limits_{0}^T \Vert f(t_1) \Vert_{L^2(E)}^2dt_1 \right),
\end{equation}
for every $0\leqslant t\leqslant T,$ and $T>0.$

For the second part of the proof, we change the analysis above with $v(T-\cdot)$ instead of $v(\cdot),$ $f(T-\cdot)$ instead of $f(\cdot)$ and $Q^{*}=-\partial_{t}-K(t)^{*},$ (or equivalently $Q=\partial_{t}-K(t)$ ) instead of $Q^{*}=-\partial_{t}+K(t)^{*}$ (or equivalently $Q=\partial_{t}-K(t)$) using that $\textnormal{Re}(K(T-t)^*)=\textnormal{Re}(K(T-t))$ to deduce that

\begin{align*}
&\Vert u(T-t) \Vert^2_{L^2(E)}\\
&\leqslant \left(C\Vert u(T) \Vert^2_{L^2(E)}+C'\int\limits_{0}^{T} \Vert (-\partial_{t}+K(T-t_1)^{*})u(T-t_1) \Vert^2_{L^2(E)}dt_1 \right)\\
&=   \left(C\Vert u(T) \Vert^2_{L^2(E)}+C'\int\limits_{0}^T \Vert (-\partial_{t}-K(t_1')^{*})u(t_1') \Vert^2_{L^2(E)}dt_1' \right).
\end{align*}
So, we conclude the proof.

In the following theorem we prove the uniqueness and existence of the solution of the problem (<ref>) on the vector bundle $E\cong G\times_{\tau}E_0\rightarrow M$

Let $K(t)=K(t,x,D)\in C([0,T], S^{m,\mathcal{L}}_{\rho,\delta}((G\times \widehat{G})\otimes \textnormal{End}(E_0)),$ $0\leqslant \delta<\rho\leqslant  1,$   be a subelliptic pseudo-differential operator of order $m>0,$ and let us assume that $\textnormal{Re}(K(t))$ is $\mathcal{L}$-elliptic, for every $t\in[0,T]$ with $T>0.$ Let   $f\in  L^2([0,T],L^2(E))$, and let $u_0\in L^2([0,T],L^2(E)).$ Then there exists a unique $u\in C^1([0,T], L^2(E) )  \cap C([0,T],L^{2,\mathcal{L}}_m(E)) $ solving (<ref>). Moreover, $u$ satisfies the energy estimate

\begin{equation}
\Vert u(t) \Vert_{L^2(E)}\leqslant C\Vert u_0 \Vert_{L^2(E)}+C'\Vert f \Vert_{L^2([0,T],L^2(E))} ,
\end{equation}
for every $0\leqslant t\leqslant T.$

The energy estimate (<ref>) and the classical Picard iteration theorem implies the existence a solution $u$ of (<ref>). Now, in order to show the uniqueness of $u,$ let us assume that $v\in  C^1([0,T], L^2(E) )  \bigcap C([0,T],L^{2, \mathcal{L}}_m(E))$ is  another solution of the problem

\begin{equation*} \begin{cases}\frac{\partial v}{\partial t}=K(t,x,D)u+f ,& \text{ }v\in \mathscr{D}'((0,T)\times E),
\\v(0)=v_0 .& \text{ } \end{cases}
\end{equation*}
Then $\omega:=v-u\in C^1([0,T], L^2(E) )  \cap C([0,T],L^{2, \mathcal{L}}_m(E))$ solves the differential problem

\begin{equation*}
\begin{cases}
\frac{\partial \omega}{\partial t}=K(t,x,D)\omega,& \text{ }\omega\in \mathscr{D}'((0,T)\times E),
\\\omega(0)=0 .& \text{ }
\end{cases}
\end{equation*}

From Theorem <ref> it follows that $\Vert \omega(t)\Vert_{L^2(E)}=0,$ for all $0\leqslant t\leqslant T.$ Hence, from the continuity in $t$ of the functions we have that $u(t,x)=v(t,x)$ for all $t\in [0,T]$ and a.e. $x\in E.$

§  PSEUDO-DIFFERENTIAL OPERATORS ON COMPACT HOMOGENEOUS MANIFOLDS

The aim of this section is to use the general construction of the subelliptic pseudo-differential calculus developed in the previous sections to establish the properties of the subelliptic pseudo-differential calculus on the compact homogeneous manifolds $M=G/K$.  First, we  note the following observations:

* The space of smooth functions $ C^{\infty}(M)$ can be identified with

\begin{equation}\label{tauinvariantfunction1}
C^{\infty}(G)^K=\{\dot{f}\in C^{\infty}(G)\,\, |\forall g\in G,\,\forall k\in K,\,\,\dot{f}(gk)=\dot{f}(g) \}.
\end{equation}
Indeed, any function $f\in C^{\infty}(M),$ induces a unique function $\dot{f}\in C^{\infty}(G)^{K},$ via $\dot{f}(g):=f(gK),$ and viceversa.

* Let us take the trivial representation $\tau=\omega=1_{\widehat{K}}$ of $K,$ and $E_0=F_0=\mathbb{C}.$ Then, an element $[g,z]\in G\times_{1_{\widehat{K}} }\mathbb{C}$ has the form

$$ [g,z]:=(g,z)\cdot K=\{(gk,1_{\widehat{K}}(k)^{-1}z):k\in K\}\cong (gK,z)\in M\times \mathbb{C}.   $$

As a consequence, a section $s\in \Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})=\Gamma(M\times \mathbb{C}),$ is (a section on a trivial vector bundle $M\times \mathbb{C}\rightarrow M$) given by $s(x)=(x,{f}(x)),$ $x\in M,$ for some $f\in C^{\infty}(M).$ So, it is natural to identify $C^{\infty}(M)$ with $ \Gamma(G\times_{1_{\widehat{K}} }\mathbb{C} )$. Moreover, we also have the following isomorphisms
$$\Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})\cong C^\infty(G,\mathbb{C})^{ 1_{\widehat{K}}  }=:C^{\infty}(G)^{K}\cong  C^{\infty}(M).$$

Therefore, it makes sense to identify an operator
$$A:C^{\infty}(M)\rightarrow C^{\infty}(M),$$
with another operator
$$ \dot{A}: C^{\infty}(G)^{K}\rightarrow C^{\infty}(G)^{K}, $$
defined via $\dot{A}\dot{f}:=Af,$ $f\in C^{\infty}(M),$ and also
with another continuous and linear operator
$$\tilde{A}:\Gamma(G\times_{1_{\widehat{K}} } \mathbb{C})\rightarrow \Gamma(G\times_{1_{\widehat{K}} } \mathbb{C}),$$
defined by

$$   \tilde{A}s(x):=(x,Af(x)),\,s(x):=(x,f(x)),\,\,x\in M. $$

Based on these identifications we will construct the subelliptic pseudo-differential calculus on $M$ as a special case of the theory developed in previous sections.

§.§ The quantisation formula on compact homogeneous manifolds

It is easy to see  that every continuous linear operator $\mathbb{A}:C^\infty(G)\rightarrow C^\infty(G)$ admits a restriction $\mathbb{A}|_{C^\infty(G)^{K}}:C^\infty(G)^{K}\rightarrow C^\infty(G).$ However, the next theorem gives us a criterion in order that $\mathbb{A}({C^\infty(G)^{K}})\subset {C^\infty(G)^{K}}, $ or, in other words,  the restriction operator

$$\dot{A}\equiv \mathbb{A}|_{C^\infty(G)^{K}}:C^\infty(G)^{K}\rightarrow C^\infty(G)^{K} , $$
is well defined.  Indeed, we have the following characterisation obtained as a consequence of Theorem  <ref>.

Let $\mathbb{A}:C^\infty(G)\rightarrow  C^\infty(G),$ be a continuous linear operator. Then, for every $f\in C^\infty(G)^K$ we have $\mathbb{A}f\in  C^\infty(G)^{K}, $ if and only if, the Schwartz kernel of $\mathbb{A}$ satisfies the following identity

\begin{equation}
K_{\mathbb{A}}(g_1,g_2)=  K_{\mathbb{A}}(g_1k_1,g_2k_2),
\end{equation}
for all $k_1,k_2\in K,$ and all $g_1,g_2\in G.$

Let $\Omega^{k}(M):=\bigsqcup_{x\in M}\bigwedge^{k}(T^{*}_{x}M)$ be the space of differential $k$-forms on $M$ and  let us fix a $G$-invariant volume element $\Omega_M\in \Omega^{\textnormal{dim}(M)}(M),$ $\textnormal{dim}(M)=\textnormal{dim}(G)-\textnormal{dim}(K),$  such that, for $f \in C(G/K),$ (see <cit.>)

$$\int\limits_{M} f \Omega_M= \int\limits_G f(gK)\, dg .$$

Let us recall that for local coordinates $y=(y_1,\cdots,y_{\dim(M)})$ of $M,$ we have
$$\Omega_M(y)=\psi_{M}(y)dy_1\wedge\cdots \wedge dy_{\dim(M)},$$
for some smooth function $\psi_{M}\in C^{\infty}(M).$ We refer the reader to  [45] for the general aspects of the geometry of differential forms.

If $A:C^{\infty}(M)\rightarrow C^{\infty}(M),$ $C^{\infty}(M)\cong C^{\infty}(G)^K,$ is a continuous linear operator, the Schwartz kernel theorem provides $K_{A}\in \mathscr{D}'(M\times M)$ such that

\begin{equation}
\end{equation}
In view of Theorem <ref>, the kernel $K_{\mathbb{A}}(g,h):=K_{A}(gK,hK)$ is a well defined distribution in $\mathscr{D}'(G\times G),$ it is $K$-invariant in the sense that

\begin{equation}\label{Kinvarienceforkernel}
\forall k_1,\,k_2\in K,\,\, K_{\mathbb{A}}(g,h)=K_{\mathbb{A}}(gk_1,hk_2),
\end{equation}

and it defines the continuous linear operator

\begin{equation}
\mathbb{A}v(g):=\int\limits_{G}K_{\mathbb{A}}(g,h)v(h)dh,\,v\in C^{\infty}(G),
\end{equation}
with the property that  $\mathbb{A}|_{C^{\infty}(G)^K}=\dot{A}.$ By keeping the notation $\dot{f}(h):=f(hK),$ we have

\begin{equation}
Af(gK)=\int\limits_GK_{\mathbb{A}}(g,h)\dot{f}(h)dh=\int\limits_Gk_{g}(h^{-1}g)\dot{f}(h)dh=(\dot{f}\ast k_{g})(g),\,f\in C^{\infty}(M),
\end{equation}
where $k_{g}(z)\equiv R_{A}(g,z):=K_{\mathbb{A}}(g,gz^{-1})$ is the right-convolution kernel associated with $\mathbb{A}.$ So, the Fourier inversion formula gives

\begin{equation}
(\dot{f}\ast k_{g})(g)=\sum_{[\xi]\in \widehat{G}}d_\xi\textnormal{Tr}[\xi(g)\mathscr{F}_{G}[\dot{f}* k_{g}](\xi)].
\end{equation}
Now, using  the matrix-valued symbol of $\mathbb{A}$ we define

\begin{equation}
\sigma_{A}(g,\xi):= \sigma_{\mathbb{A}}(g,\xi)=\widehat{k}_{g}(\xi),\,\,[\xi]\in \widehat{G},
\end{equation}
the Fourier transform of the convolution leads to the following quantisation formula

\begin{equation}
Af(gK)= \sum_{[\xi]\in \widehat{G}}d_\xi\textnormal{Tr}[\xi(g)\sigma_A(g,\xi)\widehat{\dot{f}}(\xi)].
\end{equation}

Note that for any $k\in K,$ $1_{\widehat{K}}(k)=1.$ In view of Theorem <ref>, we have that

\begin{align*}
f\in  C^{\infty}(G)^K=C^{\infty}(G,\mathbb{C})^{1_{\widehat{K}}}
\end{align*}
if and only if
\begin{equation}
\widehat{f}(i,\xi)=\xi(k)^{*}\widehat{f}(i,\xi),
\end{equation}
for all $[\xi]\in \widehat{G},$ and all $k\in K.$ The proof is complete.

So, we have enough motivation to introduce the following definition.
Let $\Omega_M$ be a $G$-invariant volume form on $M.$ Let $A: C^\infty(M)\rightarrow C^\infty(M)$ be a continuous linear operator and let $K_{A}\in \mathscr{D}'(M\times M)$ be its Schwartz kernel, so that

\begin{equation}
Af(x)=\int\limits_MK_{A}(x,y)f(y)\Omega_M(y),\,\,f\in C^{\infty}(M).
\end{equation}

Let $K_{\mathbb{A}}\in \mathscr{D}'(G\times G)$ be the distribution defined via $K_{\mathbb{A}}(g_1,g_2):=K_A(g_1K,g_2K)$ and let us consider the operator

\begin{equation}
\mathbb{A}v(g_1):=\int\limits_{G}K_{\mathbb{A}}(g_1,g_2)v(g_2)dg_2,\,v\in C^{\infty}(G).
\end{equation}
The symbol of $A$ is defined to be the distribution

\begin{equation}\label{thesymbolofA:cM:CM}
\sigma_A(g,\xi)\equiv \sigma_{\mathbb{A}}(g,\xi):=\widehat{k}_{g}(\xi)=\int\limits_{G}k_{g}(z)\xi(z)^{*}dz,\,\,[\xi]\in \widehat{G},
\end{equation}
where $g\mapsto k_{g}:G\rightarrow \mathscr{D}'(G)$ is the right-convolution kernel of $\mathbb{A}.$ Because the Schwartz kernel $K_A$ depends on  the $G$-invariant volume form $\Omega_M,$ so are the kernel  $K_{\mathbb{A}}$ and the symbol $\sigma_{A}.$

The analysis above allows us to formulate the following version of  Theorem <ref> in the case of the trivial representation $\tau=\omega:=1_{\widehat{K}}.$
Let $A:  C^{\infty}(M)\rightarrow  C^{\infty}(M)$  be a continuous linear operator and let $\tilde{A}:\Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})\rightarrow\Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})$ be the induced operator by the identifications $C^{\infty}(M)\cong \Gamma(G\times_{1_{\widehat{K}} }\mathbb{C}).$  Then,  we have

\begin{equation}\label{quantizationonhomogeneous4}
\boxed{     \tilde{A}s(gK)=\left(gK, \,Af(gK)\right)}
\end{equation}
for any section $s=(x,f(x))\in \Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})=\Gamma(M\times \mathbb{C}),$  and we also have

\begin{equation}\label{quantisation2manifolds}
\boxed{    Af(gK)=\sum_{[\xi]\in \widehat{G}}\textnormal{Tr}[\xi(g)\sigma_A(g,[\xi])  \widehat{\dot{f}}([\xi])] ,\, f\in C^{\infty}(G)^{K}   }
\end{equation}
with $\dot{f}(h)=f(hK),$ $h\in G.$

§.§ The calculus on compact homogeneous manifolds

In this section we summarise the main properties of the subelliptic pseudo-differential calculus on $M=G/K.$ We start with the definition of the subelliptic Hörmander classes on $M$. Here, we consider a sub-Laplacian $\mathcal{L}=-(X_1^2+\cdots +X_k^2)$ on $G,$ where the system of vector fields $X=\{X_i\}_{i=1}^{k}$ satisfies the Hörmander condition.

Let $\Omega_M$ be a $G$-invariant volume form on $M.$ We say that  $\sigma_A\in S^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M), $ $0\leq \rho,\delta\leq 1,$ if the symbol $\sigma_{\mathbb{A}}$ in (<ref>) belongs to the subelliptic Hörmander class $S^{m,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}).$ We define

$$ \Psi^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M):=\{A:C^{\infty}(M)\rightarrow C^{\infty}(M):\sigma_A\in S^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)\}.   $$

Observe that

\begin{equation}\label{secondaruequivalence}
A\in  \Psi^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)\,\iff\,\tilde{A}\in  \Psi^{m,\mathcal{L}}_{\rho,\delta}(G\times_{1_{\widehat{K}} }\mathbb{C},\,G\times_{1_{\widehat{K}}}\mathbb{C}),
\end{equation}
with the notation of homogeneous vector-bundles in (<ref>).

The following theorem follows from the subelliptic calculus on homogeneous vector bundles developed in Section  <ref>.

Let $0\leqslant \delta<\rho\leqslant 1,$ let $\Omega_M$ be a $G$-invariant volume form on $M,$ and for any $m\in \mathbb{R},$ let $ \Psi^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)$ be the corresponding subelliptic Hörmander classes associated with $\Omega_M.$ Then we have the following properties.

- The mapping $A\mapsto A^{*}:\Psi^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)\rightarrow \Psi^{m,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)$ is a continuous linear mapping between Fréchet spaces and  the  symbol of $A^*,$ $\sigma_{A^*}(x,\xi)$ satisfies the asymptotic expansion,

\begin{equation*}
\sigma_{A^*}(g,\xi)\sim \sum_{|\alpha|= 0}^\infty\Delta_{\xi}^\alpha\partial_{X}^{(\alpha)} (\sigma_{A}(g,\xi)^{*}).
\end{equation*}
This means that, for every $N\in \mathbb{N},$ and all $\ell\in \mathbb{N},$

\begin{equation*}
\Small{ \Delta_{\xi}^{\alpha_\ell}\partial_{X}^{(\beta)}\left(\sigma_{A^*}(g,\xi)-\sum_{|\alpha|\leqslant N}\Delta_{\xi}^\alpha\partial_{X}^{(\alpha)} (\sigma_{A}(g,\xi)^{*}) \right)\in {S}^{m-(\rho-\delta)(N+1)-\rho\ell+\delta|\beta|,\mathcal{L}}_{\rho,\delta}(G\times\widehat{G}) },
\end{equation*}
where $|\alpha_\ell|=\ell.$

- The mapping $(A_1,A_2)\mapsto A_1\circ A_2: \Psi^{m_1,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)\times \Psi^{m_2,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)\rightarrow \Psi^{m_1+m_2,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)$ is a continuous bilinear mapping between Fréchet spaces. Moreover,  the symbol of $A=A_{1}\circ A_2,$ where each $A_j$ has symbol $\widehat{A}_j,$ with $j=1,2,$ satisfies the asymptotic expansion,

\begin{equation*}
\sigma_A(x,\xi)\sim \sum_{|\alpha|= 0}^\infty(\Delta_{\xi}^\alpha\widehat{A}_{1}(x,\xi))(\partial_{X}^{(\alpha)} \widehat{A}_2(x,\xi)),
\end{equation*}
this means that, for every $N\in \mathbb{N},$ and all $\ell \in\mathbb{N},$

\begin{align*}
&\Delta_{\xi}^{\alpha_\ell}\partial_{X}^{(\beta)}\left(\sigma_A(g,\xi)-\sum_{|\alpha|\leqslant N}  (\Delta_{\xi}^\alpha\widehat{A}_{1}(g,\xi))(\partial_{X}^{(\alpha)} \widehat{A}_2(g,\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}),
\end{align*}
for every  $\alpha_\ell \in \mathbb{N}_0^n$ with $|\alpha_\ell|=\ell.$

- For  $0\leqslant \delta< \rho\leqslant    1,$ (or for $0\leq \delta\leq \rho\leq 1,$ $\delta<1/\kappa$) let us consider a continuous linear operator $A:C^\infty(M)\rightarrow C^\infty(M)$ with symbol  $\sigma\in {S}^{0,\mathcal{L}}_{\rho,\delta}(M,\Omega_M)$. Then $A$ extends to a bounded operator from $L^2(M,\Omega_M)$ to  $L^2(M,\Omega_M).$

The asymptotic expansions for the product and the adjoint of subelliptic pseudo-differential operators follow from [10].  So, to finish the proof let us prove the $L^2$-boundedness for operators with symbols in the class ${S}^{0,\mathcal{L}}_{\rho,\delta}(M,\Omega_M).$ Observe that for $A\in \Psi^{0,\mathcal{L}}_{\rho,\delta}(M,\Omega_M),$ we have $\mathbb{A}\in \Psi^{0,\mathcal{L}}_{\rho,\delta}(G\times \widehat{G}),$ and then, for all $f\in C^\infty(M),$ we have

\begin{align*}
\Vert Af\Vert_{L^2(M,\Omega_M)}^{2}&=\int\limits_{M}|Af|^2\Omega_M=\int\limits_{G}|\dot{A}\dot{f}(g)|^2dg=\int\limits_{G}|\mathbb{A}\dot{f}(g)|^2dg\leq \Vert \mathbb{A}\Vert^2_{\mathscr{B}(L^2(G))}\Vert \dot{f}\Vert^2_{L^2(G)}\\
&=\Vert \mathbb{A}\Vert^2_{\mathscr{B}(L^2(G))}\Vert f\Vert^2_{L^2(M,\Omega_M)},
\end{align*}
having used the boundedness of $\mathbb{A}$ on $L^2(G)$ in view of the subelliptic Calderón-Vaillancourt Theorem <ref>.

We finish this section with the following characterisation of the Hörmander classes on $M$ defined by local coordinate systems.
We denote $\Psi^{m}_{\rho,\delta}(M,\Omega_M):=\Psi^{m,\mathcal{L}_G}_{\rho,\delta}(M,\Omega_M)$ the elliptic Hörmander classes defined by the Laplacian on $G,$ this means,
$$ A\in \Psi^{m}_{\rho,\delta}(M,\Omega_M)\,\iff\, \mathbb{A}\in \Psi^{m}_{\rho,\delta}(G\times \widehat{G}).$$

Let $m\in \mathbb{R},$ and let $0\leq \delta<\rho\leq 1,$ and $\rho\geq 1-\delta.$ Then,  the following facts are equivalent.

(A) $A\in \Psi^{m}_{\rho,\delta}(M,\Omega_M). $

(B) $A\in \Psi^{m}_{\rho,\delta}(M,\textnormal{loc}).$

Indeed, in view of Theorem <ref> and using the equivalence (<ref>) we have

\begin{align*}
A\in \Psi^{m}_{\rho,\delta}(M,\Omega_M)&\,\iff\,\tilde{A}\in  \Psi^{m}_{\rho,\delta}(G\times_{1_{\widehat{K}} }\mathbb{C},\,G\times_{1_{\widehat{K}}}\mathbb{C})\\
&\,\iff\,\tilde{A}\in  \Psi^{m}_{\rho,\delta}(G\times_{1_{\widehat{K}} }\mathbb{C},\,G\times_{1_{\widehat{K}}}\mathbb{C};\textnormal{loc})\\
&\,\iff\,{A}\in  \Psi^{m}_{\rho,\delta}(M;\textnormal{loc}),
\end{align*}
provided that $0\leq \delta<\rho\leq 1,$ and $\rho\geq 1-\delta.$ Indeed, the equivalence
$$\tilde{A}\in  \Psi^{m}_{\rho,\delta}(G\times_{1_{\widehat{K}} }\mathbb{C},\,G\times_{1_{\widehat{K}}}\mathbb{C};\textnormal{loc})\,\iff\,{A}\in  \Psi^{m}_{\rho,\delta}(M;\textnormal{loc})$$
is valid in view of the commutativity of the following diagram

[every node/.style=midway]
[column sep=10em,between origins, row sep=4em] at (0,0)
(R) $\Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})$  ;     (S) $\Gamma(G\times_{1_{\widehat{K}} }\mathbb{C})$;

(R/I) $C^\infty(M)$;     (T) $C^\infty(M)$;

[<-] (R/I) – (R) node[anchor=east]  $\varkappa_{1_{\widehat{K}}}$;
[->] (R) – (S) node[anchor=south] $\tilde{A}$;
[->] (S) – (T) node[anchor=west] $\varkappa_{1_{\widehat{K}}}$;
[->] (R/I) – (T) node[anchor=north] $A$;

provided that $0\leq \delta<\rho\leq 1,$ and $\rho\geq 1-\delta.$ Thus, the proof is complete.

§ GLOBAL SYMBOLS   ON  VECTOR BUNDLES OF DIFFERENTIAL FORMS

To illustrate the quantisation formula on homogeneous vector-bundles, in this section we compute the global symbol of the exterior derivative, its adjoint, the Hodge Laplacian and the Dirac operator. These are the operators of relevant interest in the theory of differential forms on $M.$ Notations and preliminaries are taken from Wallach [49] and from  <cit.>.

§.§ Differential forms on a homogeneous space

Let us consider $G$ to be connected and simply connected, and the quotient manifold $M$ to be orientable. We denote by $\mathfrak{g}$ and $\mathfrak{k}$ the Lie algebras of $G$ and $K,$ respectively. Because a $K$-invariant inner product on $\mathfrak{g}/\mathfrak{k}$ induces a $G$-invariant Riemannian metric on $G$ (see <cit.> for this and further details),  let us fix a $G$-invariant metric on $G,$ and let us extend it in the canonical way, to an Hermitian metric $\langle\cdot ,\cdot \rangle_{\Omega^p},$ on
$$\Omega^p(M):=\bigwedge^p T^{*}M\otimes \mathbb{C}, $$
that is, the  $p$-th exterior  power  of  the  complexified  cotangent bundle  of  $M$.
The space of (smooth) sections  of the homogeneous ($G$-invariant) vector bundle

\begin{equation}
\Omega^p(M)\rightarrow M,\,\,\,1\leq p\leq m:=\dim(M),
\end{equation}
is the set of $p$-differential forms on $M.$ So, a $p$-form $\omega\in \Gamma( \Omega^p(M))$ assigns to any point $x\in M,$ an element $\omega_x\in \bigwedge^{p}T_{x}^{*}M,$ of $C^{\infty}$-class  with respect to $x.$ In an arbitrary coordinate neighborhood $U$ of $M$, and $x_1,x_2,\cdots, x_m$ coordinate functions defined on $U,$ the family of vectors

$$  \frac{\partial}{\partial x_1}|_{x},\cdots,  \frac{\partial}{\partial x_m}|_{x} , $$

provides a basis of the tangent space $T_{x}M.$ The functions $x_i:U\rightarrow\mathbb{R}$ are smooth, and one can consider the differential $(dx_i)|_{x}: T_{x}U\rightarrow \mathbb{R},$ at $x,$ can be considered as an element of $T_{x}^{*}M,$ via

$$  dx_i\left(\frac{\partial}{\partial x_j}\right)=\delta_{ij}.  $$
In consequence,

$$ (dx_1)|_{x},\cdots,  (dx_m)|_{x},  $$
becomes a basis of the cotangent space $T_{x}^{*}M,$ and

\begin{equation}\label{defi:difffrom}
\omega_x=\sum_{I=(i_1,\cdots, i_p)}\omega_{I}(x)dx^{I}|_x,\,\,dx^{I}|_x:=dx_{i_1}|_x\wedge \cdots\wedge dx_{i_p}|_x,\,\omega_{I}\in C^{\infty}(M),
\end{equation}
where $1\leq i_1<\cdots <i_p\leq m.$ So, $\dim( \bigwedge^{p}T_{x}^{*}M)={{m}\choose{p}}=\textnormal{rank}(\Omega^p(M)).$

By considering the adjoint action of $K$ on $\mathfrak{g},$ we have the representation $\textnormal{Ad}|_{K}:K\rightarrow \textnormal{GL}(\mathfrak{g}).$ The fiber at $e_M=e_G K$ of the tangent bundle

\begin{equation}\label{tangentbundle}
TM\rightarrow M,
\end{equation}
that is $T_{e_M},$ can be identified with $\mathfrak{g}/\mathfrak{k}.$ There is a natural structure on (<ref>) making of it a homogeneous vector-bundle.
Indeed, let us consider the right action of $K$ on $G\times \mathfrak{g}/\mathfrak{k},$

\begin{equation}\label{equationofadjoint}
(G\times \mathfrak{g}/\mathfrak{k})\times K\rightarrow G\times \mathfrak{g}/\mathfrak{k},\,\quad(g,Y+\mathfrak{k})\cdot k=(gk,p_{\mathfrak{k}}\circ\textnormal{Ad}|_{K}(k^{-1})(Y+\mathfrak{k})),
\end{equation}
where $p_{\mathfrak{k}}:\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{k}$ is the quotient mapping. Let us record that the adjoint representation  $\textnormal{Ad}:G\rightarrow \textnormal{GL}(\mathfrak{g})$ is defined via

$$  \textnormal{Ad}(g):=(dC_{g})_{e_G}:\mathfrak{g}\rightarrow \mathfrak{g},\,\,\, C_{g}:G\rightarrow G,\,C_{g_1}(g_2):=g^{-1}_1g_2g_1,\,\,g_1,g_2\in G.  $$
In particular, in (<ref>) we have considered the restriction operator $\textnormal{Ad}|_{K}:K\rightarrow \textnormal{GL}(\mathfrak{g}).$
So, the fact that the fiber at $e_{G}K\in M,$ of the tangent bundle $TM\rightarrow M$ can be identified with $\mathfrak{g}/\mathfrak{k},$   implies that

\begin{equation}
TM\cong   G\times_{p_{\mathfrak{k}}\circ \textnormal{Ad}|_{K}}  \mathfrak{g}/\mathfrak{k}.
\end{equation}
Similarly, we have

\begin{equation}
T^*M\cong   G\times_{(p_{\mathfrak{k}}\circ\textnormal{Ad}|_{K})^{*} }  ( \mathfrak{g}/\mathfrak{k})^{*},
\end{equation}

with the representation $(p_{\mathfrak{k}}\circ\textnormal{Ad}|_{K})^{*}$ of $K$ given by

\begin{equation}
(\textnormal{Ad}|_{K})^{*}(k)(\eta)(Y+\mathfrak{k})=\eta(p_{\mathfrak{k}}\circ\textnormal{Ad}|_{K}(k^{-1})(Y+\mathfrak{k})  ).
\end{equation}

By considering the complexification $TM\otimes \mathbb{C}$ and $T^*M\otimes \mathbb{C},$ of the spaces $TM$ and $T^*M,$ respectively, we also have the following identifications for the space of $p$-forms,

\begin{equation}
\Omega^{p}(M)\cong G\times_{\bigwedge^{p}\textnormal{Ad}|_{K}}   \bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C}.
\end{equation}
Let us remark that the representation $\bigwedge^{p}\textnormal{Ad}|_{K}$ is defined via

\begin{align*}
& \bigwedge^{p}\textnormal{Ad}|_{K}(k)(e_{i_1}\wedge \cdots \wedge e_{i_p} )[Y_1+\mathfrak{k},\cdots,Y_p+\mathfrak{k}]\\
&:=e_{i_1}\wedge \cdots \wedge e_{i_p} [p_{\mathfrak{k}}\circ \textnormal{Ad}|_{K}(k^{-1})(Y_1+\mathfrak{k}),\cdots, p_{\mathfrak{k}}\circ \textnormal{Ad}|_{K}(k^{-1})(Y_1+\mathfrak{k})],\,\,k\in K,
\end{align*}
on canonical elements $e_{i_1}\wedge \cdots \wedge e_{i_p}$ of $\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C}.$
Because the previous identifications can be defined in terms of the adjoint action of $G$ into $\mathfrak{g},$ we allow the following notations

* $ TM\cong  G\times_{K}  \mathfrak{g}/\mathfrak{k}:= G\times_{p_{\mathfrak{k}}\circ \textnormal{Ad}|_{K}}  \mathfrak{g}/\mathfrak{k},$

* $ T^*M\cong  G\times_{K }  (\mathfrak{g}/\mathfrak{k})^{*}:= G\times_{(p_{\mathfrak{k}}\circ \textnormal{Ad}|_{K})^{*} }  (\mathfrak{g}/\mathfrak{k})^{*},$

* $ \Omega^{p}(M)\cong G\times_{K}   \bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C}:=    G\times_{\bigwedge^{p}\textnormal{Ad}|_{K}}   \bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C}.$
Finally, the family of sections $\Gamma(\Omega^{p}(M))$ of the homogeneous ($G$-invariant) vector-bundle $\Omega^{p}(M)\rightarrow M$ is the family of $p$-differential forms on $M,$ which admits the following identification

\begin{equation}
\Gamma(\Omega^{p}(M))\cong C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{\bigwedge^{p}\textnormal{Ad}|_{K}}=:C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}.
\end{equation}
In this terminology, a $p$-form $\omega\in \Gamma(\Omega^{p}(M)), $ is identified with a function from $G$ into $\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C}$ satisfying

\begin{equation}
\omega(gk)=\bigwedge^{p}\textnormal{Ad}|_{K}(k^{-1}) \omega(g).
\end{equation}
We also have the isomorphism of Hilbert spaces

\begin{equation}
L^2(\Omega^{p}(M))\cong L^2(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{\bigwedge^{p}\textnormal{Ad}|_{K}}=:L^2(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}.
\end{equation}

§.§ Fourier analysis associated to differential forms
Starting with a $K$-invariant inner product on $\mathfrak{g}/\mathfrak{k},$ we can identify $(\mathfrak{g}/\mathfrak{k})^{*},$ with the set of linear forms on $\mathfrak{g},$ vanishing in $\mathfrak{k}.$  Let $\mathfrak{m}$ be the orthogonal complement of $\mathfrak{k}$ in $\mathfrak{g}$ with respect to the the Killing form

\begin{equation*}
B(Y_1,Y_2):=\textnormal{Tr}(\textnormal{ad}(Y_1)\circ \textnormal{ad}(Y_2)).
\end{equation*}
Restricting    $-B$  to  $\mathfrak{m}$, we  can  define  a  $G$-invariant Riemannian  metric  on $M=G/K.$ Now, we choose an orthonormal basis

\begin{equation}\label{basisofg}
\{ Y_1,\cdots, Y_m, Y_{m+1}\cdots,Y_{n}    \},\,m=\dim(M),\,n=\dim(G),
\end{equation}
of $\mathfrak{g}$ with respect to the Killing form, in such a way that $\{ Y_1,\cdots, Y_m\}$ is a basis of $\mathfrak{m},$ and $\{ Y_{m+1}\cdots,Y_{n}    \}$ a basis of $\mathfrak{k}.$ A $p$-form $\omega \in  \Gamma(\Omega^{p}(M))\cong C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}, $ is determined by the system of $C^{\infty}$-functions

\begin{equation}
\omega(Y_{i_1},\cdots, Y_{i_p}),\,\,\,1\leq i_1<\cdots<i_p\leq m,
\end{equation}

\begin{equation}
\omega(g)=\sum_{I_p}\omega_{I_p}(g) \omega(Y_{i_1},\cdots, Y_{i_p}).
\end{equation}

Indeed, denoting $I_p:=(i_1,\cdots,i_p),$ and by considering the dual basis of (<ref>),

\begin{equation*}
\{ dY_1,\cdots, dY_m, dY_{m+1}\cdots,dY_{n}    \},\,dY_i(Y_j):=\delta_{i,j},
\end{equation*}

we have the unique representation

\begin{equation}\label{extensionintipicallemtnts}
\omega(g)=\sum_{I_p}\omega_{I_p}(g)dY_{i_1}\wedge\cdots\wedge dY_{i_p},\,\,
\end{equation}
from which one deduces

\begin{equation}
\omega(g)(Y_{i_1},\cdots, Y_{i_p})=\omega_{I_p}(g).
\end{equation}

Let  us consider a $p$-form $\omega \in  \Gamma(\Omega^{p}(M))\cong C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}. $ The Fourier transform of $\omega$ at $(I_p,[\xi])$ is given by

\begin{equation}
\widehat{\omega}(I_p,[\xi]):=\int\limits_G \omega_{I_p}(g)\xi(g)^{*}dg.
\end{equation}

§.§ Global symbol of the exterior derivative $d$

The exterior derivative $d_p:\Gamma(\Omega^{p}(M))\rightarrow \Gamma(\Omega^{p+1}(M)) $ acting on $p$-forms is defined in local coordinates on a $p$-form $\omega$ (see (<ref>)) by

\begin{equation}\label{defi:difffrom2}
d_p\omega_x=\sum_{i=1}^{m}\sum_{I=(i_1,\cdots, i_p)}\frac{\omega_{I}(x)}{\partial x_i}dx_{i}|_{x}\wedge dx^{I}|_{x},\,x\in M.
\end{equation}
As a section of the vector bundle $\Omega^{p+1}(M)\rightarrow M,$ $d_p\omega$ is defined via

\begin{equation}\label{defi:difffrom22}
d_p\omega=\sum_{i=1}^{m}\sum_{I=(i_1,\cdots, i_p)}\frac{\omega_{I}(x)}{\partial x_i}dx_{i}\wedge dx^{I}.
\end{equation}

As usual, let  $p_{M}:G\rightarrow M,$ $g\mapsto gK,$ $g\in G,$ be the natural projection. Let us define the function $\omega\mapsto \tilde{\omega},$

\begin{eqnarray}
\Gamma(\Omega^{p}(M))  \ni \omega\mapsto \tilde{\omega},\,&\tilde{\omega}(Y_1,\cdots, Y_p)=p_{M}^{*}\omega(Y_1,\cdots, Y_p)(g)\label{identificationof forms}\\
&:=\omega_{gK}(dp_{M}\circ dL_{g}|_{e_G}Y_1,\cdots,dp_{M}\circ dL_{g}|_{e_G}Y_p )\label{identificationof forms2}.
\end{eqnarray}
Observe that we have used the isomorphism between the set of left-invariant vector fields and the Lie algebra $\mathfrak{g}.$[Indeed, the mapping $TG\rightarrow\mathfrak{g},$ $Y\mapsto Y_{e_G},$ is an isomorphism, and the mapping

$$  Y_{g}:=dL_{g}|_{e_G}Y,\,\,Y\in \mathfrak{g},\,\,L_{g}:G\rightarrow G,\,L_{g}(h):=gh,\,g, h\in G, $$
associates to any vector $Y\in \mathfrak{g},$ a vector-field $g\mapsto Y_g$ in $TG$.] Observe that we also have done the  identification of  $(\mathfrak{g}/\mathfrak{k})^{*},$ with the set of linear forms on $\mathfrak{g},$ vanishing in $\mathfrak{k}$ which makes of $\tilde{\omega}$ in  (<ref>) defined via  (<ref>) a well defined function  in  $C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}.$

The exterior derivative $d_{p}$ induces an operator $\mathfrak{d}_{p}$ defined by the identification $\Gamma(\Omega^{p}(M))\cong C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}$ in Remark <ref>. It makes commutative the following diagram

\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={14em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma(\Omega^{p}(M))$}  ; & \node(S) {$\Gamma(\Omega^{p+1}(M))$}; \\
\node(R/I) {$C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}$}; & \node (T) {$C^{\infty}(G,\bigwedge^{p+1}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}.$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\cong$};
\draw[->] (R) -- (S) node[anchor=south] {$d_{p}$};
\draw[->] (S) -- (T) node[anchor=west] {$\cong$};
\draw[->] (R/I) -- (T) node[anchor=north] {$\mathfrak{d}_{p}$};
\end{tikzpicture}
\end{eqnarray}
More precisely,

\begin{equation}
\boxed{ \mathfrak{d}_{p}\tilde{\omega}=\widetilde{d_{p}\omega}  ,\,\,\omega\in \Omega^{p}(M) }
\end{equation}

Using the basis
\begin{equation}
\{ Y_1,\cdots, Y_m, Y_{m+1}\cdots,Y_{n}    \},\,m=\dim(M),\,n=\dim(G),
\end{equation}
of $\mathfrak{g}$ in (<ref>), one can express the operator $\mathfrak{d}_{p}$ by  (see <cit.>)

\begin{equation}
\mathfrak{d}_{p}\tilde{\omega}(Y_{i_1},Y_{i_2},\cdots ,Y_{i_{p+1}}):=\sum_{u=1}^{p+1}(-1)^{u-1}Y_{i_u}\tilde{\omega}(Y_{i_1},Y_{i_2},\cdots,Y_{i_u}^{\bullet} ,\cdots ,Y_{i_{p+1}}).
\end{equation}
The notation $Y_{i_u}^{\bullet}$ means that the term  $Y_{i_{u}}$ is omitted for any $1\leq i_{1}<\cdots < i_{p+1}\leq n.$ Let us consider the basis

\begin{equation}\label{basisiofthefiberat0}
\{dY^{I_p}:=dY_{i_1}\wedge\cdots\wedge dY_{i_{p+1}} \},\,\,\,I_{p}=(i_1,\cdots,i_p), \,1\leq i_1<\cdots < i_p\leq n,
\end{equation}
of $\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C}.$ We compute the symbol of the exterior derivative in the following theorem. We are going to use the notation $J_{p}=\{j_1,\cdots, j_p\},$ where $1\leq j_1<\cdots<j_{n}\leq n,$ for any $p\in \mathbb{N}.$
Let us consider  the Cartan decomposition $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{k}.$ Under the  identification of  $(\mathfrak{g}/\mathfrak{k})^{*},$ with the set of linear forms on $\mathfrak{g},$ vanishing in $\mathfrak{k},$[and in terms of the basis in (<ref>), where
\{ Y_1,\cdots, Y_m, Y_{m+1}\cdots,Y_{n}    \},\,m=\dim(M),\,n=\dim(G),
$ is  an orthonormal basis of $\mathfrak{g}$ with respect to the Killing form, in such a way that $\{ Y_1,\cdots, Y_m\}$ is a basis of $\mathfrak{m},$ and $\{ Y_{m+1}\cdots,Y_{n}    \}$ a basis of $\mathfrak{k}.$] the symbol of the exterior derivative $\mathfrak{d}_p$ is given by

\begin{equation}\label{symbol:of:d_p}
\sigma_{\mathfrak{d}_{p}}(I_{p},J_{p+1},[\xi])=\sum_{u=1}^{p+1}(-1)^{u-1}\textnormal{det}\left[\delta_{i_k,j_v}\right]_{ \begin{subarray}{l}  1\leq k\leq p\\
1\leq v\leq p+1,\,v\neq u
\end{subarray} }\sigma_{Y_{j_u}}(\xi),
\end{equation}
for any triple $(I_p,J_{p+1},[\xi]),$ with $[\xi]\in \widehat{G}.$

For any $\tilde{\omega}\in C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K} $ we have

\begin{equation}
\tilde{\omega}(g)=\sum_{I_p}\tilde{\omega}_{I_p}(g)dY_{i_1}\wedge\cdots\wedge Y_{i_p}=:\sum_{I_p}\omega_{I_p}(g)dY^{I_p},\,\,
\end{equation}
with components $\tilde{\omega}_{I_p}\in C^{\infty}(G).$ Let us choose a $(p+1)-$multi-index  $J_{p+1}=(j_{1},\cdots, j_{p+1}),$ $1\leq j_1<\cdots<j_{p+1}\leq n.$  Observing that
\begin{align*}
&  \mathfrak{d}_{p}\tilde{\omega}(g)(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p+1}})\\
&=\sum_{u=1}^{p+1}(-1)^{u-1}Y_{j_u}\tilde{\omega}(Y_{j_1},Y_{j_2},\cdots,Y_{j_u}^{\bullet} ,\cdots ,Y_{j_{p+1}})\\
&=\sum_{I_p}\sum_{u=1}^{p+1}(-1)^{u-1}Y_{j_u}\omega_{I_p}(g)dY^{I_p}(Y_{j_1},Y_{j_2},\cdots,Y_{j_u}^{\bullet} ,\cdots ,Y_{j_{p+1}})\\
&=\sum_{I_p}\sum_{u=1}^{p+1}(-1)^{u-1}Y_{j_u}\omega_{I_p}(g)\textnormal{det}[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p\\
1\leq v\leq p+1,\,v\neq u
\end{subarray} },
\end{align*}

and making use of the Fourier inversion formula, we have

\begin{align*}
& \mathfrak{d}_{p}\tilde{\omega}(g)(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p+1}})=\sum_{I_p}\sum_{u=1}^{p+1}(-1)^{u-1}Y_{j_u}\omega_{I_p}(g)\textnormal{det}[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p\\
1\leq v\leq p+1,\,v\neq u
\end{subarray} }\\
&=\sum_{[\xi]\in \widehat{G}}\sum_{I_p}d_\xi  \textnormal{Tr}[ \xi(g) \left(\sum_{u=1}^{p+1}(-1)^{u-1}\textnormal{det}[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p\\
1\leq v\leq p+1,\,v\neq u
\end{subarray} }Y_{j_u}\xi(e_G)\right)\widehat{\tilde{\omega}}_{I_p}(\xi)].
\end{align*}

Note that the terms $  \mathfrak{d}_{p}\tilde{\omega}(g)(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p+1}})$ are the coefficients of the function $\mathfrak{d}_{p}\tilde{\omega}(g)$ with respect to its expansion as a linear combination of elements of the basis $\{dY_{J_{p+1}}\},$ so that

\begin{align*}
& \mathfrak{d}_{p}\tilde{\omega}(g)=\\
&\sum_{[\xi]\in \widehat{G}}\sum_{I_p,J_{p+1}} d_\xi  \textnormal{Tr}[ \xi(g) \left(\sum_{u=1}^{p+1}(-1)^{u-1}\textnormal{det}[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p\\
1\leq v\leq p+1,\,v\neq u
\end{subarray} }Y_{j_u}\xi(e_G)\right)\widehat{\tilde{\omega}}_{I_p}(\xi)]dY_{J_{p+1}}.
\end{align*}

Since we want to  find the unique matrix-valued function
$$(I_{p},J_{p+1},[\xi])\mapsto \sigma_{\mathfrak{d}_{p}}(I_{p},J_{p+1},[\xi])$$

\begin{equation*}
\mathfrak{d}_p\tilde{\omega}(g)=\sum_{I_{p},J_{p+1},[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(g)\sigma_{\mathfrak{d}_p}(I_p,J_{p+1},\xi)    \widehat{\tilde{\omega}}(I_p,\xi)   ]dY^{J_{p+1}},
\end{equation*}

we deduce that

\begin{equation*}
\sigma_{\mathfrak{d}_{p}}(I_{p},J_{p+1},[\xi])=\sum_{u=1}^{p+1}(-1)^{u-1}\textnormal{det}[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p\\
1\leq v\leq p+1,\,v\neq u
\end{subarray} }Y_{j_u}\xi(e_G),
\end{equation*}
We conclude the proof by using  the identity $\sigma_{Y_{j_u}}(\xi)=Y_{j_u}\xi(e_G).$

§.§ The global symbol of $d^{*}$
The adjoint operator $d_{p}^{*}$ of the exterior derivative $d_{p}$ induces an operator $\mathfrak{d}_{p}^{*}$ defined by the identification $\Gamma(\Omega^{p}(M))\cong C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}$ in Remark <ref>. It makes commutative the following diagram

\begin{eqnarray}
\begin{tikzpicture}[every node/.style={midway}]
\matrix[column sep={14em,between origins}, row sep={4em}] at (0,0) {
\node(R) {$\Gamma(\Omega^{p+1}(M))$}  ; & \node(S) {$\Gamma(\Omega^{p}(M))$}; \\
\node(R/I) {$C^{\infty}(G,\bigwedge^{p+1}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}$}; & \node (T) {$C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}.$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\cong$};
\draw[->] (R) -- (S) node[anchor=south] {$d_{p}^*$};
\draw[->] (S) -- (T) node[anchor=west] {$\cong$};
\draw[->] (R/I) -- (T) node[anchor=north] {$\mathfrak{d}_{p}^*$};
\end{tikzpicture}
\end{eqnarray}
More precisely,

\begin{equation}
\boxed{ \mathfrak{d}_{p}^*\tilde{\omega}=\widetilde{d_{p}^*\omega}  ,\,\,\omega\in\Gamma( \Omega^{p+1}(M)) }
\end{equation}

Using the basis
\begin{equation}
\{ Y_1,\cdots, Y_m, Y_{m+1}\cdots,Y_{n}    \},\,m=\dim(M),\,n=\dim(G),
\end{equation}
of $\mathfrak{g}$ in (<ref>), one can express the operator $\mathfrak{d}_{p}^*$ by (see <cit.>)

\begin{equation}
\mathfrak{d}_{p}^*\tilde{\omega}(Y_{i_1},Y_{i_2},\cdots ,Y_{i_{p}}):=-\sum_{u=1}^{n}Y_{u}\tilde{\omega}(Y_u,Y_{i_1},Y_{i_2},\cdots ,Y_{i_{p}}).
\end{equation}

Let us consider the  Cartan decomposition $\mathfrak{g}=\mathfrak{m}\oplus \mathfrak{k}.$ Under the  identification of  $(\mathfrak{g}/\mathfrak{k})^{*},$ with the set of linear forms on $\mathfrak{g},$ vanishing in $\mathfrak{k},$ the symbol of $\mathfrak{d}_p^{*}$ is given by

\begin{equation}\label{symbol:of:d_p1}
\sigma_{\mathfrak{d}_{p}^*}(I_{p+1},J_{p},[\xi])=-\sum_{u=1}^{n}\sigma_{Y_{u}}(\xi) \det[\delta_{i_k,j_v}]_{ \begin{subarray}{l}  1\leq k\leq p+1\\
0\leq v\leq p,\,j_0= u
\end{subarray} },
\end{equation}
for any triple $(I_p,J_{p+1},[\xi]),$ with $[\xi]\in \widehat{G}.$

We want to  find the unique matrix-valued function
$$(I_{p+1},J_{p},[\xi])\mapsto \sigma_{\mathfrak{d}_{p}^*}(I_{p+1},J_{p},[\xi])$$
satisfying the quantisation formula

\begin{equation*}
\mathfrak{d}_p^*\tilde{\omega}(g)=\sum_{I_{p+1},J_{p},[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(g)\sigma_{\mathfrak{d}_p^*}(I_{p+1},J_{p},\xi)    \widehat{\tilde{\omega}}(I_{p+1},\xi)   ]dY^{J_{p}}.
\end{equation*}
By considering the expansion of the $p+1$-form
$\tilde{\omega}\in C^{\infty}(G,\bigwedge^{p+1}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K} $ we have

\begin{equation}
\tilde{\omega}(g)=\sum_{I_{p+1}}\omega_{I_{p+1}}(g)dY^{I_{p+1}}.\,\,
\end{equation}
By evaluating $\mathfrak{d}_p^*\tilde{\omega}(g)$ at $(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p}}),$ we obtain

\begin{align*}
\mathfrak{d}_p^*\tilde{\omega}(g)(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p}})
&=-\sum_{u=1}^{n}Y_{u}\tilde{\omega}(Y_u,Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p}})\\
&=-\sum_{I_{p+1}}\sum_{u=1}^{n}Y_{u}\omega_{I_{p+1}}(g)dY^{I_{p+1}}(Y_u,Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p}})\\
&=-\sum_{I_{p+1}}\sum_{u=1}^{n}Y_{u}{\omega}_{I_{p+1}}(g)\det[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p+1\\
0\leq v\leq p,\,j_0= u
\end{subarray} }.
\end{align*}

The Fourier inversion formula implies

\begin{align*}
&\mathfrak{d}_p^*\tilde{\omega}(g)(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{p}})\\
&=\sum_{[\xi]\in \widehat{G}} \sum_{I_{p+1}}d_\xi\textnormal{Tr}[\xi(g)\left(-\sum_{u=1}^{n}Y_{u}\xi(e_G) \det[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p+1\\
0\leq v\leq p,\,j_0= u
\end{subarray} } \right)\widehat{{\omega}}(I_{p+1},[\xi])].
\end{align*}

\begin{align*}
&=\sum_{[\xi]\in \widehat{G}} \sum_{I_{p+1},J_p}d_\xi\textnormal{Tr}[\xi(g)\left(-\sum_{u=1}^{n}Y_{u}\xi(e_G) \det[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p+1\\
0\leq v\leq p,\,j_0= u
\end{subarray} } \right)\widehat{{\omega}}(I_{p+1},[\xi])]dY_{J_p},
\end{align*}
and we deduce that

\begin{equation*}
\sigma_{\mathfrak{d}_{p}}(I_{p+1},J_{p},[\xi])=-\sum_{u=1}^{n}Y_{u}\xi(e_G) \det[dY_{i_k}(Y_{j_v})]_{ \begin{subarray}{l}  1\leq k\leq p+1\\
0\leq v\leq p,\,j_0= u
\end{subarray} }.
\end{equation*}
Thus, we conclude the proof.

§.§ The Dirac operator and the global symbol of the Hodge-Laplacian

Let us consider the total space $\Omega^{\bullet}(M)=\bigoplus_{p=0}^{m}\Omega^{p}(M)$ and the corresponding vector bundle

\begin{equation}
\Omega^{\bullet}(M)=\bigoplus_{p=0}^{m}\Omega^{p}(M)\rightarrow M,
\end{equation}

over $M.$ A section of the vector bundle $\Omega^{\bullet}(M)\rightarrow M$ is an element of the form

\begin{equation}
\omega=\bigoplus \omega^{p},\,\,\omega^{p}\in \Gamma(\Omega^p(M)),
\end{equation}
that is, a direct sum of $p$-differential forms over $M.$ The Dirac operator

\begin{equation}
\slashed{D}:=d+d^{*}:\Gamma(\Omega^\bullet(M))\rightarrow \Gamma(\Omega^\bullet(M))
\end{equation}
operates on a graded form $ \omega=\bigoplus_{p=0}^{m} \omega^{p},$ by

\begin{equation}
\slashed{D}\omega\equiv (d+d^{*})\omega:= d^{*}_0\omega^{1}\oplus(d_0\omega^{0}+d^{*}_1\omega^2)\oplus \cdots \oplus (d_{k-1}\omega^{k-1}+d^{*}_k\omega^{k+1})\oplus\cdots
\end{equation}

So, to compute the action of the Dirac operator on a graded form $\omega$, we may compute the components $d_{k-1}\omega^{k-1}+d^{*}_k\omega^{k+1}.$
The Fourier analysis above and Theorems  <ref> and <ref> imply

\begin{align*}
&\widetilde{\slashed{D}\omega}|_{\Omega^{k}(M)}:=  \widetilde{d_{k-1}\omega^{k-1}}+\widetilde{d^{*}_k\omega^{k+1}}\\ &=\sum_{I_{k-1},J_{k},[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(g)\sigma_{\mathfrak{d}_p}(I_{k-1},J_{k},\xi)    \widehat{\tilde{\omega}}^{k-1}(I_{k-1},\xi)   ]dY^{J_{k}}\\
&\,\,\,+\sum_{I_{k+1},J_{k},[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(g)\sigma_{\mathfrak{d}_{k}^*}(I_{k+1},J_{k},\xi)    \widehat{\tilde{\omega}}^{k+1}(I_{k+1},\xi)   ]dY^{J_{k}}\\
&=\sum_{\begin{subarray}{l} I_{k-1},I_{k+1}\\
J_{k},\,[\xi]\in \widehat{G}
\end{subarray}}d_{\xi}\textnormal{Tr}[\xi(g)\left(\sigma_{\mathfrak{d}_p}(I_{k-1},J_{k},\xi)    \widehat{\tilde{\omega}}^{k-1}(I_{k-1},\xi)+\sigma_{\mathfrak{d}_{k}^*}(I_{k+1},J_{k},\xi)    \widehat{\tilde{\omega}}^{k+1}(I_{k+1},\xi)    \right)  ]dY^{J_{k}}.
\end{align*}

Finally, if one considers the Hodge-Laplacian
$$ \slashed{D}^2_{k+1}=d_{k}d^{*}_k+d^{*}_{k+1}d_{k+1}:\Gamma(\Omega^{k+1}(M))\rightarrow \Gamma(\Omega^{k+1}(M)),$$
it was shown in <cit.> that

\begin{equation}
\boxed{ -\mathcal{L}_G\tilde\omega^{k+1}=\widetilde{\slashed{D}^2_{k+1}\omega^{k+1}}  ,\,\,\omega^{k+1}\in \Gamma( \Omega^{k}(M))  }
\end{equation}

restricting  the  Killing  form  sign  changed  to  $\mathfrak{m}$ in the Cartan decomposition $\mathfrak{g}=\mathfrak{m}\oplus\mathfrak{k}.$
By using the argument in the proof of Theorem <ref>, let us find the unique matrix-valued function
$$(I_{k+1},J_{k+1},[\xi])\mapsto \sigma_{\slashed{D}^2_{k+1}}(I_{k+1},J_{k+1},[\xi])$$
satisfying the quantisation formula

\begin{equation*}
-\mathcal{L}_G\tilde\omega^{k+1}(g)=\sum_{I_{k+1},J_{k+1},[\xi]\in \widehat{G}}d_{\xi}\textnormal{Tr}[\xi(g)\sigma_{\slashed{D}^2_{k+1}}(I_{k+1},J_{k+1},[\xi])   \widehat{\tilde{\omega}}^{k+1}(I_{k+1},\xi)   ]dY^{J_{k+1}}.
\end{equation*}
By considering the expansion of the $k+1$-form
$\tilde{\omega}\in C^{\infty}(G,\bigwedge^{k+1}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K} $ we have

\begin{equation}
\tilde{\omega}(g)=\sum_{I_{k+1}}\omega_{I_{k+1}}(g)dY^{I_{k+1}}.\,\,
\end{equation}
By evaluating $-\mathcal{L}_G\tilde\omega^{k+1}(g)$ at $(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{k+1}}),$ and using the Fourier inversion formula, we obtain

\begin{align*}
-\mathcal{L}_G\tilde\omega^{k+1}(g)(Y_{j_1},Y_{j_2},\cdots ,Y_{j_{k+1}})
&=\sum_{[\xi]\in \widehat{G}} \sum_{I_{p+1}}d_\xi\textnormal{Tr}[\xi(g)\left(-\lambda_{[\xi]}I_{d_\xi}\right)\widehat{{\omega}}(I_{k+1},[\xi])].
\end{align*}

\begin{align*}
&=\sum_{[\xi]\in \widehat{G}} \sum_{I_{k+1},J_{k+1}}d_\xi\textnormal{Tr}[\xi(g)\left(-\lambda_{[\xi]}I_{d_\xi}\right)\widehat{{\omega}}^{k+1}(I_{k+1},[\xi])]dY_{J_{k+1}},
\end{align*}
and we deduce that

\begin{equation*}
\sigma_{\slashed{D}^2_{k+1}}(I_{k+1},J_{k+1},[\xi])   =-\lambda_{[\xi]}I_{d_\xi},
\end{equation*}
as expected.

§.§.§ Acknowledgements
The authors thank Julio Delgado and  David Rottensteiner for discussions.

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§ APPENDIX VI: COMPACTNESS FOR VECTOR-VALUED SUBELLIPTIC  OPERATORS ON VECTOR BUNDLES
In this section we  put forward our discussion on compactness of subelliptic $G$-invariant operators. In the first step we state our result for subelliptic Hörmander class $S^{m, \mathcal{L}}_{\rho, \delta}(G \times \widehat{G})$ on  compact groups $G$ defined in Definition <ref>. Then we try to use these results to state and prove our results for subelliptic $G$-invariant operators on vector bundles of compact groups. A important result is Gohberg Lemma <cit.>. Gohberg Lemma was proved study for the torus and compact Lie groups with usual Hörmander symbols in <cit.> and [18] respectively.

We denote the smallest eigenvalue of positive matrix $P$ by  $\Vert P \Vert_{\textnormal{min}}.$

The following theorem is an analogue of Gohberg Lemma for subelliptic Hörmander classes $S^{0, \mathcal{L}}_{1,0}(G \times \widehat{G}).$

Let $A$ be a operator with matrix symbol $\hat{A}(x, \xi) \in S^{0, \mathcal{L}}_{1,0} (G\times \widehat{G}).$ Then for all compact operators $K$ on $L^2(G)$ we have the following inequality

\begin{align} \label{vis3.20}
\Vert A-K\Vert_{\mathscr{L}(L^2(G))} \geq d_{\textnormal{min}},
\end{align}

$$d_{\textnormal{min}} := \limsup_{\langle \xi \rangle \rightarrow \infty} \left\{\sup_{x \in G} \frac{\Vert \hat{A}(x, \xi) \hat{A}(x, \xi)^* \Vert_{\textnormal{min}}}{\Vert \hat{A}(x, \xi) \Vert_{\textnormal{op}}} \right\}.$$

§.§  Bott's Index Theorem

In order to study the index of subelliptic operators on homogeneous vector bundles we need to prove some technical results. We start with the following lemma.

Let $A:C^\infty(G,E_0)\rightarrow C^\infty(G,F_0)$ be a homogeneous differential operator. Then, $A$ is globally hypoelliptic, if  there exists  $\varrho>0,$ and a positive constant $C:=C_{G,A}>0,$ such that

\begin{equation}\label{hypo}
\Vert\widehat{Au}(r,\xi) \Vert_{\textnormal{HS}}\geq C\langle \xi\rangle^{\varrho} \Vert \widehat{u}(r,\xi) \Vert_{\textnormal{HS}},\quad u\in \mathscr{D}'(G,E_0),
\end{equation}
$\forall 1\leq i\leq d_\tau,$  $ \forall1\leq r\leq d_\omega,$ and for all but a finite number of $[\xi]\in \widehat{G}.$

Let $I$ be the set of equivalence classes in $\widehat{G}$ such that (<ref>) holds. Observe that for $s\in \mathbb{R},$

\begin{align*}
\Vert Au\Vert_{H^{s}(G,F_0)}^2:=\sum_{r=1}^{d_\omega}\sum_{[\xi]\in \widehat{G}}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2.
\end{align*}
By writing

\begin{align*}
\sum_{[\xi]\in \widehat{G}}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2&=\sum_{[\xi]\in \widehat{G}\setminus I}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2 +\sum_{[\xi]\in I}\langle \xi\rangle^{2s}d_\xi\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2
\end{align*}
and observing that for  $[\xi]\in I,$

\begin{equation*}
\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2 \geq C \langle \xi\rangle^{2\varrho(\xi)} \Vert \widehat{u}(i,\xi)\Vert_{\textnormal{HS}}^2, \forall 1\leq i\leq d_\tau,\forall1\leq r\leq d_\omega,
\end{equation*}
we deduce

\begin{align*}
&\sum_{[\xi]\in \widehat{G}\setminus I}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2 +\sum_{[\xi]\in I}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2 \\
&\gtrsim \sum_{[\xi]\in \widehat{G}\setminus I}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2+\sum_{[\xi]\in I}\sum_{i=1}^{d_\tau}d_\xi\langle \xi\rangle^{2\varrho+2s} \Vert \widehat{u}(i,\xi)\Vert_{\textnormal{HS}}^2\\
&\geqslant \sum_{[\xi]\in \widehat{G}\setminus I}d_\xi\langle \xi\rangle^{2s}\Vert \widehat{Au}(r,\xi)\Vert_{\textnormal{HS}}^2+\sum_{[\xi]\in I}\sum_{i=1}^{d_\tau}d_\xi\langle \xi\rangle^{2\varrho+2s} \Vert \widehat{u}(i,\xi)\Vert_{\textnormal{HS}}^2.
\end{align*}
Because $\widehat{G}\setminus I$ is a finite set, and

\begin{align*}
\Vert u\Vert_{H^{s+\varrho}(G,E_0)}^2&=\sum_{[\xi]\in I}\sum_{i=1}^{d_\tau}d_\xi\langle \xi\rangle^{2\varrho+2s} \Vert \widehat{u}(i,\xi)\Vert_{\textnormal{HS}}^2+\sum_{[\xi]\in \widehat{G}\setminus I}\sum_{i=1}^{d_\tau}d_\xi\langle \xi\rangle^{2\varrho+2s}\Vert \widehat{u}(i,\xi)\Vert_{\textnormal{HS}}^2\\
&\leq  \Vert Au\Vert_{H^{s}(G,F_0)}^2+\sum_{[\xi]\in \widehat{G}\setminus I}\sum_{i=1}^{d_\tau}d_\xi\langle \xi\rangle^{2\varrho+2s}\Vert \widehat{u}(i,\xi)\Vert_{\textnormal{HS}}^2\\
\end{align*}
the condition $Au\in H^{s}(G,F_0)$ implies that $u\in H^{s+\rho}(G,E_0).$ So, $Au\in C^\infty(G,F_0)$ implies that, for every $s\in \mathbb{R},$ $Au\in H^{s}(G,F_0)$ which also implies that for every $s\in \mathbb{R},$ $u\in H^{s+\rho}(G,E_0),$ which is equivalent to say that $u\in C^\infty(G,E_0).$ So, we have proved that if $A$ satisfies (<ref>), then $A$ is globally hypoelliptic.

§ APPENDIX I: CHARACTERISATION OF GLOBAL VECTOR-VALUED HÖRMANDER CLASSES

In this section we make the proof of Theorem <ref>.

In order to present the main result of this subsection, let us recall some notions of $K$-theory.

Let $S$ be a commutative semigroup written additively. We define $(K(S),j)$ to be the solution of the following universal problem: $K(S)$ is a semigroup homomorphism of $S$ into $K(S)$ such that if $N$ is a group and $\alpha:S\rightarrow N$ is a semigroup homomorphism, the there is a unique group homomorphism $\tilde{\alpha}:K(S)\rightarrow N$ so that $\tilde{\alpha}\circ j=\alpha,$ making commutative 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) {$K(S)$}  ; & \node(S) {$S$}; \\
\node(R/I) {$N$};\\
\draw[<-] (R/I) -- (R) node[anchor=east]  {$\tilde{\alpha}$};
\draw[<-] (R) -- (S) node[anchor=south] {$j$};
\draw[<-] (R/I) -- (S) node[anchor=south] {$\alpha\,\,$};
\end{tikzpicture}
\end{eqnarray}

Let $E(G)$ be the set of equivalence classes of (non-necessarily unitary) representations on $G.$ $E(G)$ is a semiring under the addition given by $[\omega]+[\eta]=[\omega\oplus \eta]$ and $[\omega]\cdot [\eta]=[\omega\otimes \eta],$ for  $[\omega],[\eta]\in E(G).$ Then, $R(G)=K(E(G))$ will called the ring of virtual representations.  Note that $\widehat{G}$ is a free basis for $R(G)$ over $\mathbb{Z}.$

Let us consider the inclusion $i:K\rightarrow G.$ Then $i$ induces a natural semiring homomorphism  $i^{*}:R(G)\rightarrow R(K),$ defined by $i^{*}[\omega]=[\omega\circ i].$ Let us define $i_{*}: R(K)\rightarrow R(G)$ by

\begin{equation}
i_{*}([\omega])=\sum_{[\xi]\in \widehat{G}}\langle i^{*}[\xi],[\omega] \rangle_K [\xi],\,\, \langle i^{*}[\xi],[\omega] \rangle_K:=\textnormal{dim}(\textnormal{Hom}_{K}(\mathbb{C}^{d_{\xi\circ i}},\mathbb{C}^{d_\omega})).
\end{equation}

Now, we prove that the Atiyah-Singer-Bott index theorem   remains valid for subelliptic operators.
Let $E=G\times_{\tau}E_0$ and $F=G\times_{\omega}F_0$ be two homogeneous vector bundles over $M.$
Let us assume that $\tilde{A}\in \Psi^{m,\mathcal{L}}_{\rho,\delta}(E,F)$ is  $\mathcal{L}$-elliptic and homogeneous.  Then $\tilde{A}$ admits a Fredholm extension from $L^2(E)$ into $L^2(F)$ and its index is given by

\begin{equation}
\textnormal{ind}(\tilde{A})=\textnormal{dim}(i_{*}([\tau]-[\omega])).
\end{equation}

* Let us observe that the space $\textnormal{Hom}_{K}(\mathbb{C}^{d_\xi},\mathbb{C})\neq \emptyset,$ if and only if $\xi$ is a type $I$ representation. A representation $\xi$ is of type $I,$ if $\xi$ has at least a non-zero invariant vector $v,$ with respect to $K,$ this means that $\xi(k)v=v,$ $v\neq 0,$ for all $k\in K.$ Let $\mathcal{B}_\xi$ be the space of these invariant vectors                                                                                and let $k_{\xi}:=\textnormal{dim}(\mathcal{B}_\xi).$ Moreover,

\begin{equation}
\end{equation}
Indeed, observe that

\begin{align*}
\textnormal{Hom}_K(\mathbb{C}^{d_\xi},\mathbb{C}) &=\{J\in \textnormal{End}(\mathbb{C}^{d_\xi},\mathbb{C}): \forall k\in K,\,J\xi(k)=J\}\\
&\cong \{J\in\mathbb{C}^{1\times d_\xi}: \forall k\in K,\,J\xi(k)=J\}.\\
&\cong \{J^{*}=\overline{J}^{\,t}\in\mathbb{C}^{d_\xi\times 1}: \forall k^{-1}\in K,\,\xi(k)^{*}J^{*}=J^{*}\}\\
&\cong \{J^{'}\in\mathbb{C}^{d_\xi\times 1}: \forall k\in K,\,\xi(k)J^{'}=J^{'}\}\\
&\cong \{J^{'}\in\textnormal{End}(\mathbb{C},\mathbb{C}^{d_\xi}): \forall k\in K,\,\xi(k)J^{'}=J^{'}\}.
\end{align*}
And, observing that $J^{'}\in\textnormal{End}(\mathbb{C},\mathbb{C}^{d_\xi})$ is uniquely determined by its value at $1\in \mathbb{C}$ (in view of $J'\lambda=\lambda J'1),$ $\lambda\in \mathbb{C}$),
$$\forall k\in K,\,\xi(k)J^{'}=J^{'},\,\textnormal{  if and only if, }\forall k\in K,\,\xi(k)J^{'}1=J^{'}1,$$
which implies that
$$\mathcal{B}_\xi=\textnormal{span}\{J'1:J'\in J^{'}\in\textnormal{End}(\mathbb{C},\mathbb{C}^{d_\xi}),\forall k\in K,\,\xi(k)J^{'}=J^{'}\,\},$$
and that $d_{K,\xi,1_{\widehat{K}}}=k_{\xi}.$

In terms of this notation, we will prove that every continuous linear operator $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ admits a representation of the form
(<ref>) where the matrix-valued function $\widehat{s}^{\,ij}:\widehat{G}\rightarrow \bigcup_{[\xi]\in \widehat{G}} \mathbb{C}^{d_\xi \times d_\xi}$ will be defined in terms of the Fourier transform of the section $s$ at $[\xi],$ (see (<ref>)) and

\begin{equation}\label{invariantdimension}
d_{K,\xi,\tau}:=\textnormal{dim}(\textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0))\equiv \textnormal{dim}(\textnormal{Hom}_K(\mathbb{C}^{d_\xi},\mathbb{C}^{d_\tau})).
\end{equation}

For the notation above see Theorem <ref>.   To be more precise, let us introduce the necessary notation which will be justified in Theorem <ref>. Our main tool is the understanding of the linear algebra of the $K$-invariant subspace $\textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0)$. So, let $\{ e_{i,d_\xi} \}_{i=1}^{d_\xi}$ be an orthonormal basis of the representation space $\mathbb{C}^{d_\xi}$ and $\{L_{j,[\xi]}\}_{j=1}^{d_{K,\xi,\tau}},$  be a basis of $\textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0).$
Because of the isomorphism
$$ \textnormal{Hom}(\mathbb{C}^{d_\xi},E_0)\cong (\mathbb{C}^{d_\xi})^{*}\otimes E_0,  $$
and the fact that $\textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0)\subset \textnormal{Hom}(\mathbb{C}^{d_\xi},E_0)$ we can assume that every mapping $L_{j,[\xi]}$ has the form,

\begin{equation*}
L_{j,[\xi]}=\tilde{e}_{j_1,d_\xi}^{*}\otimes \tilde{e}_{j_2,E_0},
\end{equation*}
where $\tilde{e}_{j_1,d_\xi}^{*}\otimes \tilde{e}_{j_2,E_0}$ is an orthonormal system in $\textnormal{Hom}(\mathbb{C}^{d_\xi},E_0)$ with respect to the inner product

\begin{align*}
(\tilde{e}_{j_1,d_\xi}^{*}\otimes \tilde{e}_{j_2,E_0},\tilde{e}_{ {j'_1},d_\xi}^{*}\otimes \tilde{e}_{ {j'_2},E_0})_{ \textnormal{Hom}(\mathbb{C}^{d_\xi},E_0) } &= (\tilde{e}_{j_1,d_\xi}^{*},\tilde{e}_{j'_1,d_\xi}^{*})_{(\mathbb{C}^{d_\xi})^{*}} (\tilde{e}_{j_2,E_0},\tilde{e}_{j'_2,E_0})_{E_0}
\\ &= (\tilde{e}_{j_1,d_\xi},\tilde{e}_{j'_1,d_\xi})_{\mathbb{C}^{d_\xi}} (\tilde{e}_{j_2,E_0},\tilde{e}_{j_2',E_0})_{E_0}
\\&=\delta_{(j_1,j'_1),(j_2,j_2')}.
\end{align*}
Then, the vectors,

\begin{equation*}
e_{i,d_{\xi}}\otimes L_{j,[\xi]},\,\quad 1\leq i\leq d_{\xi},\,1\leq j\leq d_{K,\xi,\tau},
\end{equation*}
provide a basis for $\mathbb{C}^{d_\xi}\otimes \textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0),$ and the family of functions $\alpha_{[\xi]}(e_{i,d_{\xi}}\otimes L_{j,[\xi]})$ provide an orthogonal basis of the subspace $\alpha_{[\xi]}(\mathbb{C}^{d_\xi}\otimes \textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0)).$ Indeed, this fact can be deduced from   the orthogonality of this family, which  can be proved as follows:

\begin{align*}
& (\alpha_{[\xi]}(e_{i,d_{\xi}}\otimes L_{j,[\xi]}),\alpha_{[\xi]}(e_{i',d_{\xi}}\otimes L_{j',[\xi]}))_{L^2(G,E_0)}\\&=\int\limits_{G}(L_{j,[\xi]}(\xi(g)^*e_{i,d_{\xi}} ),L_{j',[\xi]}(\xi(g)^*e_{i',d_{\xi}} ))_{E_0}dg\\
&=\int\limits_{G}(\tilde{e}_{j_1,d_\xi}^{*}\otimes \tilde{e}_{j_2,E_0}(\xi(g)^*e_{i,d_{\xi}} ),\tilde{e}_{j'_1,d_\xi}^{*}\otimes \tilde{e}_{j'_2,E_0}(\xi(g)^*e_{i',d_{\xi}} ))_{E_0}dg\\
&=\int\limits_{G} \tilde{e}_{j_1,d_\xi}^{*}(\xi(g)^*e_{i,d_{\xi}} ) \overline{   \tilde{e}_{j'_1,d_\xi}^{*}(\xi(g)^*e_{i',d_{\xi}} ) }   ( \tilde{e}_{j_2,E_0}, \tilde{e}_{j'_2,E_0})_{E_0}dg\\
&=\int\limits_{G} \tilde{e}_{j_1,d_\xi}^{*}(\xi(g)^*e_{i,d_{\xi}} ) \overline{   \tilde{e}_{j'_1,d_\xi}^{*}(\xi(g)^*e_{i',d_{\xi}} ) }   dg\,\delta_{j_2,j_2'}.
\end{align*}

Now, by observing that $\tilde{e}_{j_1,d_\xi}^{*}(\xi(g)^*e_{i,d_{\xi}} )=(\xi(g)^*e_{i,d_{\xi}},\tilde{e}_{j_1,d_\xi})_{\mathbb{C}^{d_\xi}}$ are the entries of the matrix representation $\xi(g)$ with respect to the basis $\{e_{i,d_{\xi}}\}$ and $\{\tilde{e}_{j,d_{\xi}}\},$ from the orthogonality of these entries[which is a consequence of the Peter-Weyl theorem.] we have that

\begin{align*}
(\alpha_{[\xi]}(e_{i,d_{\xi}}\otimes L_{j,[\xi]}),\alpha_{[\xi]}(e_{i',d_{\xi}}\otimes L_{j',[\xi]}))_{L^2(G,E_0)}=\frac{1}{d_\xi}\delta_{(i,j_1),(i',j_2)}\delta_{j_2,j_2'}.
\end{align*}
By following with the proof, we can identify $L_{j,[\xi]}\in \textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0)$ with a matrix of size $d_\tau\times d_\xi,$ by choosing the bases $\{e_{i,d_\xi}\}_{1\leq i\leq d_\xi}$ and $\{e_{i,E_0}\}_{1\leq i\leq d_\tau}$ of the representation spaces $\mathbb{C}^{d_\xi}$ and $E_0,$

\begin{equation*}
L_{j,[\xi]}:=[L_{j,[\xi]}^{\ell,\ell'}]_{1\leq \ell \leq d_\tau,\,1\leq \ell'\leq d_\xi }, \,\,\,L_{j,[\xi]}^{\ell,\ell'}:=(L_{j,[\xi]}e_{\ell',d_\xi},e_{\ell,E_0}).
\end{equation*}
In view of the Peter-Weyl theorem for sections (Theorem <ref>), the family of sections,

\begin{equation*}
\varkappa_{\tau}^{-1}\alpha_{[\xi]}(e_{i,d_{\xi}}\otimes L_{j,[\xi]})(gK):=\varkappa_{\tau}^{-1}L_{j,[\xi]}(\xi(g^{-1}) e_{i,d_{\xi}}  )=[g , L_{j,[\xi]}(\xi(g)^{-1}e_{i,d_{\xi}})],
\end{equation*}
where $1\leq i\leq d_{\xi},\,1\leq j\leq d_{K,\xi,\tau},$ provide a basis for the set of sections $\Gamma_{[\xi]}(E).$ This is because the mapping $\varkappa_{\tau}^{-1}$ is  unitary.  Using that $\widehat{s}(\xi)\in \Gamma_{[\xi]}(E), $ for every $s\in \Gamma^\infty(E),$ we can write in a unique way  the Fourier transform of $s$ at $[\xi]$ as

\begin{equation}\label{coefficientsofhats}
\widehat{s}(\xi)=\sum_{i=1}^{d_\xi}\sum_{j=1}^{d_{K,\xi,\tau}}s_{ij,[\xi]}[g , L_{j,[\xi]}(\xi(g)^{-1}e_{i,d_{\xi}})],
\end{equation}
and the Fourier inversion formula takes the form (compare with (<ref>))

\begin{align*}
s(gK)=&\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{d_\xi}\sum_{j=1}^{d_{K,\xi,\tau}}s_{ij,[\xi]}[g , L_{j,[\xi]}(\xi(g)^{-1}e_{i,d_{\xi}})]   .
\end{align*}

So, if $\tilde{A}:\Gamma^\infty(E)\rightarrow\Gamma^\infty(F)$ is a continuous linear operator and $A$ is its induced vector-valued operator, we have

\begin{align}\label{123456}
\tilde{A}s(gK)=\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{d_\xi}\sum_{j=1}^{d_{K,\xi,\tau}}s_{ij,[\xi]}\tilde{A}[g , L_{j,[\xi]}(\xi(g)^{-1}e_{i,d_{\xi}})],
\end{align}
where the right hand side of (<ref>) can be written as

\begin{equation*}
\tilde{A}[g , L_{j,[\xi]}(\xi(g)^{-1}e_{i,d_{\xi}})]=[g , AL_{j,[\xi]}(\xi(g)^{*}e_{i,d_{\xi}})].
\end{equation*}

\begin{align*}
\tilde{A}[g , L_{j,[\xi]}(\xi(g)^{-1}e_{i,d_{\xi}})]&=\varkappa_\omega^{-1}\varkappa_\omega\tilde{A}\varkappa_{\tau}^{-1}L_{j,[\xi]}(\xi(g^{-1}) e_{i,d_{\xi}}  )=\varkappa_\omega^{-1}AL_{j,[\xi]}(\xi(g^{-1}) e_{i,d_{\xi}}  )\\
&=:[g , AL_{j,[\xi]}(\xi(g)^{*}e_{i,d_{\xi}})].
\end{align*}

In order to compute this expression, let us denote by $\xi(g)=(\xi_{ip}(g))_{i,p=1}^{d_\xi},$ the matrix representation of $\xi(g)$ with respect to the basis $\{e_{i,d_\xi}\}_{1\leq i\leq d_\xi},$ and let $(L_{j,[\xi]}^{q,p})_{1\leq p\leq d_\xi,\,1\leq q\leq d_\tau}$ be the matrix of $L_{j,[\xi]}\in \textnormal{Hom}_K(\mathbb{C}^{d_\xi},E_0)$ with respect to the basis fixed above.    So, we have

\begin{align*}
AL_{j,[\xi]}(\xi(g)^{*}e_{i,d_{\xi}})&=AL_{j,[\xi]}(\sum_{p=1}^{d_\xi}\xi_{ip}(g^{-1})e_{p,d_{\xi}})=\sum_{p=1}^{d_\xi}A(    \xi_{ip}(g^{-1})L_{j,[\xi]}e_{p,d_\xi} )\\
&=\sum_{p=1}^{d_\xi}A\left(\xi_{ip}(g^{-1})\sum_{q=1}^{d_\tau}L_{j,[\xi]}^{q,p}e_{q,E_0} \right)\\
\end{align*}

By using the identity (<ref>) for the matrix-valued symbol of ${A},$ we have

\begin{align*}e_{r,F_{0}}^{*}(A(\xi_{ip}(g^{-1})e_{q,E_0})) &=e_{r,F_{0}}^{*}(A(\xi^{*}_{ip}(g)e_{q,E_0})) =(\xi(g)^{*}\sigma_{A}(q,r,g,[\xi^{*}]))_{ip} .
\end{align*}

The previous equality implies,

\begin{align*}
\tilde{A}s(gK)&=\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{d_\xi}\sum_{j=1}^{d_{K,\xi,\tau}}s_{ij,[\xi]}[g, \,\sum_{p=1}^{d_\xi}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}L_{j,[\xi]}^{q,p}e_{r,F_0}^{*}(A(\xi_{ip}(g^{-1})e_{q,E_0}))e_{r,F_0}    ] \\
&=\left[g, \,\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{d_\xi}\sum_{j=1}^{d_{K,\xi,\tau}}\sum_{p=1}^{d_\xi}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}s_{ij,[\xi]}L_{j,[\xi]}^{q,p}e_{r,F_0}^{*}(A(\xi_{ip}(g^{-1})e_{q,E_0}))e_{r,F_0}    \right]\\
&=\left[g, \,\sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\sum_{j=1}^{d_{K,\xi,\tau}}\sum_{i=1}^{d_\xi}\sum_{p=1}^{d_\xi}s_{ij,[\xi]}L_{j,[\xi]}^{q,p}(\xi(g)^{*}\sigma_A(q,r,g^{-1},[\xi^{*}]))_{ip}  e_{r,F_0}  \right].
\end{align*}

Defining the following auxiliary matrices in terms of the coefficients of the Fourier transform of $s$ at $[\xi^{*}]$ and in terms  of the entries of  the $K$-invariant homomorphism $L_{j,{[\xi]}},$

\begin{equation}\label{FouriercoefficientesofsectionsL}
\widehat{s}^{\,j,q}([\xi^{*}]):=(s^{[\xi],j,q}_{p,i})_{p,i=1}^{d_\xi},\,\,s^{[\xi],j,q}_{p,i}:=s_{ij,[\xi]}L_{j,[\xi]}^{q,p},
\end{equation}
we  can write

\begin{align*}
\tilde{A}s(gK)&=\left[g, \,\sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\sum_{j=1}^{d_{K,\xi,\tau}}\sum_{i=1}^{d_\xi}\sum_{p=1}^{d_\xi}s_{ij,[\xi]}L_{j,[\xi]}^{q,p}(\xi(g)^{*}\sigma_A(q,r,g,[\xi^{*}]))_{ip} e_{r,F_0}   \right]\\
&=\left[g, \,\sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\sum_{j=1}^{d_{K,\xi,\tau}}\textnormal{Tr}[ \widehat{s}^{\,j,q}([\xi^*])\xi^{*}(g)\sigma_A(q,r,g,[\xi^{*}]) ] e_{r,F_0}  \right].
\end{align*}
The change of variables $[\xi^{*}]\mapsto [\xi]$ on $\widehat{G}$ is a one-to-one correspondence of sets, and then we have

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

In view of the notation (<ref>) induced by the action of $K$ on $G\times F_0,$ and the properties of the trace,  we get

\begin{equation}\label{quantizationonhomogeneous}   {
\tilde{A}s(gK)\equiv \left(g, \,\sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\sum_{j=1}^{d_{K,\xi,\tau}}\textnormal{Tr}[\xi(g)\sigma_A(q,r,g,[\xi])  \widehat{s}^{\,j,q}([\xi])]e_{r,F_0}   \right)\cdot K   .}
\end{equation}

Let us simplify the formula <ref>. Let us record also the identity $A\varkappa_\tau=\varkappa_\omega\tilde{A}.$ Because

\begin{equation}   {
(\varkappa_\omega\tilde{A}s)(g)= \,\sum_{[\xi]\in \widehat{G}}\sum_{q=1}^{d_\tau}\sum_{r=1}^{d_\omega}\sum_{j=1}^{d_{K,\xi,\tau}}\textnormal{Tr}[\xi(g)\sigma_A(q,r,g,[\xi])  \widehat{s}^{\,j,q}([\xi])]e_{r,F_0}     ,}
\end{equation}
Observing that in the analysis above, we just can replace $A$ by the identity operator on $C^{\infty}()$

Consider a continuous linear operators
$$ A_{k_1,k_2}:\Gamma(\Omega^{k_1}(M))\rightarrow \Gamma(\Omega^{k_2}(M))$$
mapping $k_1$-forms into $k_2$-forms. Let us consider the identification $\Gamma(\Omega^{p}(M))\cong C^{\infty}(G,\bigwedge^{p}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}$ in Remark <ref>. Then, $A_{k_1,k_2}$ can be identified with an operator

\begin{equation}
\mathcal{A}_{k_1,k_2}: C^{\infty}(G,\bigwedge^{k_1}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K}\rightarrow  C^{\infty}(G,\bigwedge^{k_2}(\mathfrak{g}/\mathfrak{k})^{*}\otimes\mathbb{C})^{K},
\end{equation}
and to it we can associate its global symbol

\begin{equation}
\sigma_{A_{k_1,k_2}}:I_{{m}\choose{k_1}}\times I_{{m}\choose{k_1}}\times \widehat{G}\rightarrow\bigcup_{\ell\in \mathbb{N}}\mathbb{C}^{\ell \times \ell},
\end{equation}

$$ {I_{{m}\choose{k_j}}}:=\left\{  dY^{I_{s,p+1}}:1\leq s\leq {{m}\choose{k_j}} \right\},\,j=1,2,    $$
is an enumeration of the basis $\{  dY^{I_{k_j}} \},$ $j=1,2.$
The quantisation of the matrix-valued symbol $\sigma_{A_{k_1,k_2}}$ recover the operator $\mathcal{A}_{k_1,k_2}$ according to the formula,

\begin{equation}
\mathcal{A}_{k_1,k_2}\tilde\omega(g)=\sum_{[\xi]\in \widehat{G}}\sum_{i=1}^{ {{m}\choose{k_1}} }\sum_{r=1}^{ {{m}\choose{k_2}} }d_\xi\textnormal{Tr}[\xi(g)\sigma_{A_{k_1,k_2}}(i,r,[\xi])\widehat{\tilde\omega}(I_{i,k_1},[\xi])]dY^{I_{r,k_2}},
\end{equation}
with $\omega\in \Omega^{k_1}(M).$