Peter–Weyl decomposition of L2(SU(2))

Every element of $SU(2)$ has the unique form $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2=1$. This identifies it homeomorphically with the unit sphere in ${\mathbb{R}}^4$, which is compact and path connected: non-antipodal points are joined by normalizing their straight segment, and antipodal ones can be joined in two such segments through a perpendicular unit vector. The diagonal circle $T=\{\mathrm{diag}(z,z^{-1}):\mathrm{\mid }z\mathrm{\mid }=1\}$ is a maximal torus, since a matrix commuting with one of its elements having distinct eigenvalues must be diagonal; thus its centralizer is $T$, excluding any larger torus. Put $h=\mathrm{diag}(1,-1)$, $e=E_{12}$ and $f=E_{21}$. The matrices $ih,e-f,i(e+f)$ are a real basis of $\mathfrak{s}\mathfrak{u}(2)$ and a complex basis of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})$. Their brackets span the same real space, so the derived algebra is all of $\mathfrak{s}\mathfrak{u}(2)$. The relations $[h,e]=2e,[h,f]=-2f,[e,f]=h$ give the simple coroot $h$ and positive root character $z^2$. By [L4], characters of $T$ are exactly ${\chi }_n(z)=z^n$ for $n\in \mathbb{Z}$, since its exponential parameter has kernel $2\pi \mathbb{Z}$. The complexified differential satisfies $d{\chi }_n(h)=n$, and dominance is $n\ge 0$.

By [L2] and step 1.1 there is exactly one irreducible group representation $V(n)$ for each integer $n\ge 0$, and its differentiated highest weight is $n$. Its differentiated module is irreducible: if a complex subspace is invariant under $d\pi (\mathfrak{s}\mathfrak{u}(2))$, it is preserved by every $\exp (d\pi (X))=\pi (\exp X)$ by [L6]. The local exponential image generates the connected group $SU(2)$ of step 1.1, so the subspace is group-invariant and hence is either zero or all of $V(n)$. Complex linearity then makes it irreducible for ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{C})=\mathfrak{s}\mathfrak{u}(2{)}_{\mathbb{C}}$. Thus [L3] gives $d_n=\dim V(n)=n+1$. Choose invariant Hermitian forms and orthonormal bases using [L5] and [A1]. The dual of an irreducible finite-dimensional group representation is irreducible: the annihilator of a proper nonzero invariant subspace of the dual would be a proper nonzero invariant subspace of the original representation. Since double dual returns the original representation, duality permutes all irreducible classes bijectively.

For the representation ${\pi }_n$ on $V(n)$ define the linear coefficient map on the Hilbert tensor product by $C_n(v\otimes \phi )(g)=\sqrt{d_n}\phi ({\pi }_n(g^{-1})v)$. Under [L5], $C_n(v\otimes \phi )(a^{-1}gb)=\sqrt{d_n}\phi ({\pi }_n(b^{-1}){\pi }_n(g^{-1}){\pi }_n(a)v)=C_n({\pi }_n(a)v\otimes {\pi }_n^{\ast }(b)\phi )(g)$. Thus the left factor acts on $V(n)$ and the right factor on its dual, exactly as in the Example. For an orthonormal basis $e_i$ with dual basis $e_j^{\ast }$, these are the normalized matrix coefficients of the dual representation: $({\pi }_n^{\ast }(g)e_j^{\ast })(e_i)=e_j^{\ast }({\pi }_n(g^{-1})e_i)$. Hence [L1] proves that $C_n$ is an isometry and that its images for different $n$ are orthogonal.

By step 2.1 the dual representations occurring in step 3.1 exhaust the irreducible classes. Therefore [L1] says the union of the displayed orthonormal coefficient families is complete. The isometry on the algebraic direct sum extends to its Hilbert completion; its image is closed by completeness and dense by that orthonormal basis, hence is all of $L^2(SU(2))$. It is equivariant by step 3.1, proving the stated two-sided decomposition. On restriction to the left group each tensor product is $d_n$ copies of $V(n)$, so its multiplicity is $n+1$. At $n=0$ the representation is one-dimensional with trivial differential, hence trivial on the connected group, and its coefficient is the constant function $1$; at $n=1$ the block has dimension $4$ and left multiplicity $2$.