Peter-Weyl Theorem on the Sphere The Peter-Weyl theorem says that the matrix coefficients of the unitary irreps of a compact topological group $G$ form an orthonormal basis for $L^2(G)$. Similarly, spherical harmonics provide an orthonormal basis for $L^2(S^2)$, however the spherical harmonics are a basis for the irreps of $SO(3)$, not $S^2$, which has no Lie group structure (Lie Group Structure on the 2-Sphere: does the following argument hold?). I understand that the two spaces are intimately related since $SO(3)$ acts transitively on $S^2$, but precisely how does spherical harmonic decomposition relate to the Peter-Weyl theorem considering the space under question is not a topological group?
 A: Sorry for answering this old post. Since I also encountered this issue recently and no one has answered it yet, let me briefly summarize what I understand here.
You are right that Peter-Weyl theorem is only a statement for compact topological groups, in this case $SO(3)$. You are also right that $S^2$ is a homogeneous space for $SO(3)$, more precisely, $S^2 \cong SO(3) \big/ SO(2)$. Using these two facts we can establish a "Peter-Weyl theorem" on homogeneous spaces. In the following, let's use a more general notation, denoting $SO(3)$ by $G$, and $SO(2)$ by $H$.
Peter-Weyl theorem states that as a representation of $G \times G$, $L^2(G)$ can be decomposed into
$$L^2(G) \cong \bigoplus_{\lambda\in\hat G} V_\lambda \otimes V_\lambda^*,$$
where $\hat G$ denotes the set of irreducible representations (irreps) of $G$. Furthermore, for $(g_1,g_2) \in G \times G$ and $f \in L^2(G)$, $(g_1,g_2)$ acts on $f$ as
$$(g_1,g_2)\cdot f(g) = f(g_1^{-1}g g_2).$$
This implies that $L^2(G)$ can also be thought of as a representation of $G$, with $g_1 \in G$ acting on $f$ as $(g_1, e)$, where $e$ is the identity of $G$. This is known as the (left) regular representation of $G$. Thinking in this way, we see that each irrep $\lambda$ of $G$ appears with multiplicity $\dim(V_\lambda^*)$.
Now let's consider $L^2(G\big/H)$, where $G\big/H$ is the set of left cosets $\{gH| g\in G\}$. Then $L^2(G\big/H) = \{f \in L^2(G) | f(gh) = f(g), \forall g \in G \text{ and } h \in H\}$. Clearly $L^2(G\big/H)$ still forms a representation of $G$ on the left: for $g_1 \in G$, $f \in L^2(G\big/H)$, $g_1\cdot f(g) = f(g_1^{-1} g)$, and it can also be easily checked that $L^2(G\big/H)$ is closed under such action. However, $L^2(G\big/H)$ is a trivial representation of $H$ on the right. Therefore as a representation of $G$, $L^2(G\big/H)$ can be decomposed into
$$L^2(G\big/H) \cong \bigoplus_{\lambda\in\hat G} V_\lambda \otimes W_\lambda^*,$$
where $W_\lambda \subset V_\lambda$ is a trivial representation of $H$.
Finally, recall that irreps of $SO(3)$ can be labeled by spin $j$, and therefore
$$L^2(SO(3)) \cong \bigoplus_{j=0,\frac{1}{2},\cdots} V_j \otimes V_j^*,$$
where $V_j$ is an irrep of $SO(3)$ with spin $j$. $W_j \in V_j$ is a trivial representation of $SO(2)$ means that $W_j$ is spanned by the eigenvectors of $SO(2)$ with eigenvalue zero. Since we know $V_j$ has precisely one such eigenvector if and only if $j$ is an integer, therefore
$$L^2(S^2) \cong \bigoplus_{j=0,1,\cdots} V_j,$$
where $V_j$ corresponds to the spherical harmonics.
