Let $M=G/K$ with $G$ compact and let $(\tau,W_\tau)$ be an irreducible representation of $K$, let $E_\tau$ be the associated $G$-homogeneous vector bundle of $M$. Then, the space of $L^2$-sections of $E_\tau$ decomposes as $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau). $$

The proof follows by Peter-Weyl: $\Gamma^\infty(E_\tau)\simeq C^\infty(G;\tau):=\{f:G\to W_\tau:f(gk)=\tau(k)^{-1}f(g)\}$. For each $\pi\in\widehat G$, we have $V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau) \to C^\infty(G;\tau)$ given by $v\otimes L \mapsto (x\mapsto L(\pi(x^{-1}).v))$. This map is $G$-equivariant.

An other way, $C^\infty(G;\tau) = C^\infty(G,W_\tau)^K \simeq (C^\infty(G)\otimes W_\tau)^K$. Peter-Weyl theorem implies that $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} (V_\pi\otimes V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes (V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes\operatorname{Hom}_K(V_\pi,W_\tau). $$

When $M$ is symmetric (e.g. $M=S^3$) and $G$ semisimple, the Laplace operator acts as the Casimir element of $\mathfrak g$ (the Lie algebra of $G$). Furthermore, if $S^3=\operatorname{Spin}(4)/\operatorname{Spin}(3)$, the square of the Dirac operator $D^2$ coincides with the Casimir element plus $1/8 R$ ($R$ is the curvature curvature of $M$).

Let $M=G/K$ with $G$ compact and let $(\tau,W_\tau)$ be an irreducible representation of $K$, let $E_\tau$ be the associated $G$-homogeneous vector bundle of $M$. Then, the space of $L^2$-sections of $E_\tau$ decomposes as $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau). $$

The proof follows by Peter-Weyl: $\Gamma^\infty(E_\tau)\simeq C^\infty(G;\tau):=\{f:G\to W_\tau:f(gk)=\tau(k)^{-1}f(g)\}$. For each $\pi\in\widehat G$, we have $V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau) \to C^\infty(G;\tau)$ given by $v\otimes L \mapsto (x\mapsto L(\pi(x^{-1}).v))$. This map is $G$-equivariant.

An other way, $C^\infty(G;\tau) = C^\infty(G,W_\tau)^K \simeq (C^\infty(G)\otimes W_\tau)^K$. Peter-Weyl theorem implies that $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} (V_\pi\otimes V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes (V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes\operatorname{Hom}_K(V_\pi,W_\tau). $$

When $M$ is symmetric (e.g. $M=S^3$) and $G$ semisimple, the Laplace operator acts as the Casimir element of $\mathfrak g$ (the Lie algebra of $G$). Furthermore, the square of the Dirac operator $D^2$ coincides with the Casimir element plus $1/8$ the scalar curvature. See Section 3.5 of T. Friedrich, "Dirac operators in Riemannian geometry".

For example, let $S^{3}=\operatorname{Spin}(4)/\operatorname{Spin}(3)$, let $\tau$ be the spin representation, then $E_\tau$ is the spinor bundle. $\tau$ has highest weight $\frac12 \varepsilon_1$, thus $\operatorname{Hom}_K(V_\pi,W_\tau)\neq0$ if and only if $\pi$ has highest weight $\Lambda = \frac12((2k+1)\varepsilon_1\pm\varepsilon_2)$. The casimir element acts on $\pi_\Lambda$ as $\|\Lambda+\varepsilon_1\|-\|\varepsilon_1\|$ (see Wallach's book, Harmonic analysis on homogeneous vector bundles). Check that is equal to $$ \frac{(2k+1)(2k+5)+1}{4} $$ The sectional curvature is 6, then $D^2$ acts by $(k+3/2)^2$. Finally, the eigenvalues of $D$ are $\pm(k+3/2)$.

Let $M=G/K$ with $G$ compact and let $(\tau,W_\tau)$ be an irreducible representation of $K$, let $E_\tau$ be the associated $G$-homogeneous vector bundle of $M$. Then, the space of $L^2$-sections of $E_\tau$ decomposes as $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau). $$

The proof follows by Peter-Weyl: $\Gamma^\infty(E_\tau)\simeq C^\infty(G;\tau):=\{f:G\to W_\tau:f(gk)=\tau(k)^{-1}f(g)\}$. For each $\pi\in\widehat G$, we have $V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau) \to C^\infty(G;\tau)$ given by $v\otimes L \mapsto (x\mapsto L(\pi(x^{-1}).v))$. This map is $G$-equivariant. By Peter-Weyl, we have that the formula above.

When $M$ is symmetric (e.g. $M=S^3$) and $G$ semisimple, the Laplace operator acts as the Casimir element of $\mathfrak g$ (the Lie algebra of $G$). Furthermore, if $S^3=\operatorname{Spin}(4)/\operatorname{Spin}(3)$, the square of the Dirac operator $D^2$ coincides with the Casimir element plus $1/8 R$ ($R$ is the curvature curvature of $M$).

Let $M=G/K$ with $G$ compact and let $(\tau,W_\tau)$ be an irreducible representation of $K$, let $E_\tau$ be the associated $G$-homogeneous vector bundle of $M$. Then, the space of $L^2$-sections of $E_\tau$ decomposes as $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau). $$

The proof follows by Peter-Weyl: $\Gamma^\infty(E_\tau)\simeq C^\infty(G;\tau):=\{f:G\to W_\tau:f(gk)=\tau(k)^{-1}f(g)\}$. For each $\pi\in\widehat G$, we have $V_\pi\otimes \operatorname{Hom}_K(V_\pi,W_\tau) \to C^\infty(G;\tau)$ given by $v\otimes L \mapsto (x\mapsto L(\pi(x^{-1}).v))$. This map is $G$-equivariant.

An other way, $C^\infty(G;\tau) = C^\infty(G,W_\tau)^K \simeq (C^\infty(G)\otimes W_\tau)^K$. Peter-Weyl theorem implies that $$ L^2(E_\tau) = \sum_{\pi\in\widehat G} (V_\pi\otimes V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes (V_\pi^*\otimes W_\tau)^K = \sum_{\pi\in\widehat G} V_\pi\otimes\operatorname{Hom}_K(V_\pi,W_\tau). $$

When $M$ is symmetric (e.g. $M=S^3$) and $G$ semisimple, the Laplace operator acts as the Casimir element of $\mathfrak g$ (the Lie algebra of $G$). Furthermore, if $S^3=\operatorname{Spin}(4)/\operatorname{Spin}(3)$, the square of the Dirac operator $D^2$ coincides with the Casimir element plus $1/8 R$ ($R$ is the curvature curvature of $M$).