Compact Lie Groups, Maximal Tori, and Peter–Weyl Theory

A finite-dimensional Lie group is a Hausdorff topological group, and compactness is a topological property, so a compact Lie group is a compact Hausdorff group; by [L1] it therefore carries a unique left Haar probability measure $\mu$, which is right invariant and inversion invariant.

Since $G$ is compact, every left Haar measure on $G$ has finite positive total mass and can be normalized by dividing by that mass; compactness alone does not make an arbitrary Haar measure a probability measure. For the normalized measure, outer regularity also gives inner regularity on every Borel $E$: for $ϵ>0$ choose open $U\supseteq G\setminus E$ with $\mu (U)<\mu (G\setminus E)+ϵ$; then $C=G\setminus U$ is compact, $C\subseteq E$, and $\mu (C)>\mu (E)-ϵ$. Conversely, a regular Borel probability measure that is left invariant is a left Haar measure of total mass one in the sense of [L2]. Hence the uniqueness assertion of [L1] is exactly uniqueness among regular Borel probability measures that are left invariant.

By [L3] the measure $\mu$ of step 1.1 is bi-invariant satisfying all three invariance conditions, so a measure with the stated properties exists.

For uniqueness let $\nu$ be any regular Borel probability measure invariant under left and right translations and inversion. Then $\nu$ is a left Haar probability measure, so $\nu =\mu$ by the uniqueness in [L1], applied through step 1.2.

Existence and uniqueness are established, and the Axiom of Choice entered exactly through the uniqueness-and-existence statement [L1].

Let $E\subseteq G$ be Borel. Since $x\mapsto 1_E(hx)$ is the indicator of $h^{-1}E$, the left invariance in [L1] gives ${\int }_G{1}_E(hx)d\mu (x)=\mu (h^{-1}E)=\mu (E)$; since $x\mapsto 1_E(xh)$ is the indicator of $E{h}^{-1}$, the right invariance gives ${\int }_G{1}_E(xh)d\mu (x)=\mu (E{h}^{-1})=\mu (E)$; and since $x\mapsto 1_E(x^{-1})$ is the indicator of $E^{-1}$, inversion invariance gives ${\int }_G{1}_E(x^{-1})d\mu (x)=\mu (E^{-1})=\mu (E)$.

Let $g\ge 0$ be Borel. By [L2] choose an increasing sequence of nonnegative simple Borel functions $s_n\nearrow g$. Each $s_n$ is a finite linear combination of Borel indicators, so the three translation/inversion identities of step 1.1 and linearity of the simple integral give $\int s_n(hx)d\mu =\int s_n(xh)d\mu =\int s_n(x^{-1})d\mu =\int s_{n}d\mu$; the same sequences $s_n(h\cdot )$, $s_n(\cdot h)$, $s_n({\cdot }^{-1})$ increase to $g(h\cdot )$, $g(\cdot h)$, $g({\cdot }^{-1})$, so two applications of monotone convergence in [L2] identify all four nonnegative integrals.

For the conjugation identity write $\varphi (x)=hx{h}^{-1}$. Applying step 2.1 to the nonnegative Borel function $x\mapsto g(hx)$ in place of $g$ and then the left-translation identity gives $\int g(hx{h}^{-1})d\mu (x)=\int g(hx)d\mu (x)=\int g(x)d\mu (x)$ for nonnegative Borel $g$.

Now let $f$ be integrable complex-valued. The functions $f(h\cdot ),f(\cdot h),f(\varphi (\cdot )),f({\cdot }^{-1})$ are Borel because $x\mapsto hx$, $x\mapsto xh$, $\varphi$ and inversion are homeomorphisms of $G$, and they are integrable because the identities of steps 2.1 and 3.1 applied to $\mathrm{\mid }f\mathrm{\mid }$ show that each of these four functions has the finite integral $\int \mathrm{\mid }f\mathrm{\mid }d\mu$. Splitting $f$ into its four nonnegative parts by [L3] and applying the corresponding identity of steps 2.1 and 3.1 to each part, then recombining, gives the displayed chain of equalities. The Axiom of Choice entered only through [L1].

Fix a positive-definite Hermitian inner product $(\cdot ,\cdot {)}_0$ on $V$ and a basis of $V$, and for $v,w\in V$ define $⟨v,w⟩:={\int }_G(\pi (g)v,\pi (g)w{)}_{0}d\mu (g)$. The integrand is a finite sum of products of the continuous matrix entries of $\pi$ with the coordinates of $w$, hence continuous on $G$, and it is bounded because $G$ is compact; so by [L5] the integral exists in $\mathbb{C}$ and the definition is unambiguous.

The form $⟨\cdot ,\cdot ⟩$ is sesquilinear: for $a,b\in \mathbb{C}$ the identity $(\pi (g)(av+b{v}^{\mathrm{\prime }}),\pi (g)w{)}_0=a(\pi (g)v,\pi (g)w{)}_0+b(\pi (g)v^{\mathrm{\prime }},\pi (g)w{)}_0$ holds pointwise, and complex linearity of the integral [L5] turns it into $⟨av+b{v}^{\mathrm{\prime }},w⟩=a⟨v,w⟩+b⟨v^{\mathrm{\prime }},w⟩$; conjugate symmetry follows pointwise from conjugate symmetry of $(\cdot ,\cdot {)}_0$ together with reality of the integral of a real-valued function, and positive semidefiniteness follows because each integrand $(\pi (g)v,\pi (g)v{)}_0\ge 0$ and the integral of a nonnegative function is nonnegative.

If $v\mathrm{\ne }0$, the continuous function $\varphi (g):=(\pi (g)v,\pi (g)v{)}_0$ is nonnegative and satisfies $\varphi (e)=(v,v{)}_0>0$, so [L3] gives $⟨v,v⟩={\int }_G\varphi d\mu >0$; the form is therefore positive definite.

For every $h\in G$ and all $v,w\in V$ one has $⟨\pi (h)v,\pi (h)w⟩={\int }_G(\pi (g)\pi (h)v,\pi (g)\pi (h)w{)}_{0}d\mu (g)={\int }_G(\pi (gh)v,\pi (gh)w{)}_{0}d\mu (g)=⟨v,w⟩$, where the middle equality is the homomorphism property $\pi (g)\pi (h)=\pi (gh)$ of [L1] and the last equality is the right-translation invariance [L2] applied to $f(g)=(\pi (g)v,\pi (g)w{)}_0$.

By steps 2.1, 2.2 and 3.1 the form $⟨\cdot ,\cdot ⟩$ is a positive-definite Hermitian inner product preserved by every $\pi (h)$, so $\pi$ is unitary for it; the Axiom of Choice was used only through the existence and translation invariance of $\mu$ in [L2].

We argue by induction on $n=\dim V$, assuming the assertion for all representation spaces of dimension $<n$. If $n=0$ then $V$ is the empty direct sum of irreducibles by [L3]; otherwise $V$ has a nonzero subrepresentation, and choosing among the nonzero subrepresentations one of least positive dimension gives an irreducible subrepresentation $W$, since any nonzero proper subrepresentation of $W$ would be a nonzero subrepresentation of $V$ of strictly smaller positive dimension by [L3].

By [L1] fix a $G$-invariant positive-definite Hermitian inner product on $V$ and let $W^{\perp }$ be the orthogonal complement of $W$; then $W^{\perp }$ is a subrepresentation, because for $w^{\mathrm{\prime }}\in W^{\perp }$, $w\in W$ and $g\in G$ unitarity and invariance of $W$ give $⟨\pi (g)w^{\mathrm{\prime }},w⟩=⟨w^{\mathrm{\prime }},\pi (g{)}^{-1}w⟩$ with $\pi (g{)}^{-1}w=\pi (g^{-1})w\in W$, so $⟨\pi (g)w^{\mathrm{\prime }},w⟩=0$.

Since $W\mathrm{\ne }0$, [L3] gives $\dim W^{\perp }=\dim V-\dim W<\dim V$, so the inductive hypothesis applies to the subrepresentation $W^{\perp }$ and exhibits it as a direct sum of irreducible subrepresentations; adjoining the irreducible summand $W$ gives $V=W\oplus W^{\perp }$ as a direct sum of irreducibles by [L2]. The Axiom of Choice entered only through [L1].

For $A\in \mathrm{Hom}(V_{\sigma },V_{\pi })$ define $T(A):={\int }_G\pi (g)A\sigma (g{)}^{-1}d\mu (g)$, the entrywise integral of the continuous matrix-valued integrand; by [L4] this is a well-defined element of $\mathrm{Hom}(V_{\sigma },V_{\pi })$ and $A\mapsto T(A)$ is complex-linear.

For every $h\in G$ one has $T(A)\sigma (h)=\pi (h)T(A)$: indeed $T(A)\sigma (h)={\int }_G\pi (g)A\sigma (g{)}^{-1}\sigma (h)d\mu (g)={\int }_G\pi (g)A\sigma (g^{-1}h)d\mu (g)$, and substituting $g=h{g}^{\mathrm{\prime }}$ in the integral, which the left invariance [L3] permits, gives ${\int }_G\pi (h{g}^{\mathrm{\prime }})A\sigma (g^{\mathrm{\prime }-1})d\mu (g^{\mathrm{\prime }})=\pi (h)T(A)$.

In the special case $V_{\sigma }=V_{\pi }=V$ and $\sigma =\pi$, every $A\in \mathrm{End}(V)$ satisfies $\mathrm{tr}T(A)=\mathrm{tr}A$: by [L5] and [L4], $\mathrm{tr}T(A)={\int }_G\mathrm{tr}(\pi (g)A\pi (g{)}^{-1})d\mu (g)={\int }_G\mathrm{tr}Ad\mu (g)=\mathrm{tr}A,$ because $\mu$ is a probability measure.

Let $A=⟨\cdot ,e_l⟩f_j\in \mathrm{Hom}(V_{\sigma },V_{\pi })$. Then $A\sigma (g{)}^{-1}{e}_k=\overline{{\sigma }_{kl}(g)}{f}_j$ by [L6], and for all $i,k$, $⟨T(A)e_k,f_i⟩={\int }_G{\pi }_{ij}(g)\overline{{\sigma }_{kl}(g)}d\mu (g)\mathrm{.}$

If $\pi$ and $\sigma$ are inequivalent, then $T(A)=0$ for every $A$: by step 2.1 the map $T(A)$ intertwines the irreducible $\sigma$ with the irreducible $\pi$, so if it were nonzero it would be an isomorphism by Schur's lemma [L1], contradicting inequivalence.

If $V_{\sigma }=V_{\pi }=V$ and $\sigma =\pi$, then $T(A)=\lambda (A){\mathrm{id}}_V$ for every $A$: by step 2.1 the endomorphism $T(A)$ intertwines the irreducible representation $\pi$ with itself, so the intertwining endomorphisms form a division ring [L1]; if $T(A)\mathrm{\ne }0$, then $T(A)$ has an eigenvalue $\lambda$ by [L1] applied to the nonzero finite-dimensional complex space $V$, and $T(A)-\lambda {\mathrm{id}}_V$ is a non-injective intertwining endomorphism, hence $0$ in the division ring, so $T(A)=\lambda {\mathrm{id}}_V$ (the case $T(A)=0$ is the same statement with $\lambda =0$).

In the equal-representation case of the statement, use the fixed identification for which $V_{\sigma }=V_{\pi }=V$, $\sigma =\pi$, and $e_k=f_k$. Then the rank-one operator $A$ of step 2.3 is an endomorphism with $\mathrm{tr}A={\delta }_{jl}$. Combining steps 3.2 and 2.2 gives $T(A)=\lambda (A){\mathrm{id}}_V$ with $\lambda (A)=\mathrm{tr}A\mathrm{/}{d}_{\pi }={\delta }_{jl}\mathrm{/}{d}_{\pi }$, and hence $⟨T(A)e_k,f_i⟩={\delta }_{ik}{\delta }_{jl}\mathrm{/}{d}_{\pi }$.

In the general equivalent case fix an intertwining isomorphism $S:V_{\sigma }\to V_{\pi }$. The identity $\sigma (g{)}^{-1}{S}^{-1}=S^{-1}\pi (g{)}^{-1}$ shows that $T(A)S^{-1}={\int }_G\pi (g)(A{S}^{-1})\pi (g{)}^{-1}d\mu (g)$. Apply the equal-representation scalar and trace calculations of steps 3.2 and 2.2 to the endomorphism $A{S}^{-1}$ of $V_{\pi }$. They give $T(A)=S\mathrm{tr}(A{S}^{-1})\mathrm{/}{d}_{\pi }$. For the rank-one $A$ of step 2.3, the trace is $(S^{-1}{)}_{lj}$, so $⟨T(A)e_k,f_i⟩=S_{ik}(S^{-1}{)}_{lj}\mathrm{/}{d}_{\pi }$. No unitary normalization of $S$ is needed, and the formula holds for every such intertwiner.

Comparing step 2.3 with step 4.1 yields ${\int }_G{\pi }_{ij}(g)\overline{{\sigma }_{kl}(g)}d\mu (g)={\delta }_{ik}{\delta }_{jl}\mathrm{/}{d}_{\pi }$ in the equal-representation, aligned-basis case, and comparing step 2.3 with step 3.1 yields zero in the inequivalent case. Step 4.2 supplies the general equivalent case, so the alternatives exhaust all pairs. Irreducibility excludes the zero representation, hence $d_{\pi }>0$ and every division is defined. AC covers [L3] and the countable-choice terminology in [L6].

Expanding both characters by [L2] and using linearity of the integral [L3], $⟨{\chi }_{\pi },{\chi }_{\sigma }⟩={\int }_G{\chi }_{\pi }\overline{{\chi }_{\sigma }}d\mu ={\sum }_{i=1}^{d_{\pi }}{\sum }_{k=1}^{d_{\sigma }}{\int }_G{\pi }_{ii}(g)\overline{{\sigma }_{kk}(g)}d\mu (g)$, a finite sum of the orthogonality integrals of [L1].

If $\pi$ and $\sigma$ are inequivalent, every term in step 1.1 vanishes by the first case of [L1], so $⟨{\chi }_{\pi },{\chi }_{\sigma }⟩=0$. If $\pi =\sigma$, then the terms equal ${\delta }_{ik}{\delta }_{ik}\mathrm{/}{d}_{\pi }={\delta }_{ik}\mathrm{/}{d}_{\pi }$ by the second case of [L1] with $j=i$ and $l=k$, so $⟨{\chi }_{\pi },{\chi }_{\pi }⟩={\sum }_{i=1}^{d_{\pi }}1\mathrm{/}{d}_{\pi }=1$.

Conjugation invariance is [L2], so every character is a class function; combining with step 2.1, the characters of pairwise inequivalent irreducible unitary representations are orthonormal in $L^2(G,\mu )$. The Axiom of Choice entered only through the Haar-based supplier [L1].

Choose normalized Haar measure by [L8]. By [L5] fix an inner product $(\cdot ,\cdot {)}_0$ on $\mathfrak{g}$ and define $⟨X,Y⟩:={\int }_G(\mathrm{Ad}(g)X,\mathrm{Ad}(g)Y{)}_{0}d\mu (g)$ for $X,Y\in \mathfrak{g}$; the integrand is continuous in $g$ because $\mathrm{Ad}$ is smooth and $(\cdot ,\cdot {)}_0$ is bilinear in finite dimension, so it is integrable by [L6] and the definition is unambiguous.

The form $⟨\cdot ,\cdot ⟩$ is a positive-definite inner product: it is bilinear and symmetric because the integrand is, and if $X\mathrm{\ne }0$ then $g\mapsto (\mathrm{Ad}(g)X,\mathrm{Ad}(g)X{)}_0$ is continuous, nonnegative, and equal to $(X,X{)}_0>0$ at $g=e$, so its integral is positive by [L6], while $⟨X,X⟩\ge 0$ always. It is $\mathrm{Ad}$-invariant: for $h\in G$, writing $\mathrm{Ad}(g)\mathrm{Ad}(h)=\mathrm{Ad}(gh)$ and applying the right-translation invariance [L4] to the function $g\mapsto (\mathrm{Ad}(g)X,\mathrm{Ad}(g)Y{)}_0$ gives $⟨\mathrm{Ad}(h)X,\mathrm{Ad}(h)Y⟩={\int }_G(\mathrm{Ad}(gh)X,\mathrm{Ad}(gh)Y{)}_{0}d\mu (g)=⟨X,Y⟩$.

Define the metric $g_x(u,v):=⟨d{L}_{x^{-1}}u,d{L}_{x^{-1}}v⟩$ for $x\in G$ and $u,v\in T_{x}G$; it is a smooth positive-definite bundle metric, because left translation is a diffeomorphism and $⟨\cdot ,\cdot ⟩$ is a positive-definite inner product on the single vector space $\mathfrak{g}$ by step 2.1.

The metric $g$ is right-invariant: for $h\in G$, $x\in G$ and $u,v\in T_{x}G$, write $u=d{L}_x\xi$ and $v=d{L}_x\eta$. The identity in [L3] gives $g_{xh}(d{R}_{h}u,d{R}_{h}v)=⟨{\mathrm{Ad}}_{h^{-1}}\xi ,{\mathrm{Ad}}_{h^{-1}}\eta ⟩=⟨\xi ,\eta ⟩=g_x(u,v)$ by the $\mathrm{Ad}$-invariance of step 2.1. It is left-invariant by construction, so it is bi-invariant.

For any bi-invariant metric, conjugation $C_h=L_h{R}_{h^{-1}}$ is an isometry fixing $e$, so its inner product at $e$ is $\mathrm{Ad}$-invariant. For left-invariant vector fields $X,Y,Z$ the Levi-Civita connection of $g$ satisfies ${\mathrm{\nabla }}_{X}Y=\frac{1}{2}[X,Y]$: the Koszul identity $2⟨{\mathrm{\nabla }}_{X}Y,Z⟩=X⟨Y,Z⟩+Y⟨Z,X⟩-Z⟨X,Y⟩+⟨[X,Y],Z⟩-⟨[Y,Z],X⟩+⟨[Z,X],Y⟩$, which follows from symmetry and metric compatibility in [L1], reduces to $2⟨{\mathrm{\nabla }}_{X}Y,Z⟩=⟨[X,Y],Z⟩-⟨[Y,Z],X⟩+⟨[Z,X],Y⟩$ because $X⟨Y,Z⟩=Y⟨Z,X⟩=Z⟨X,Y⟩=0$ for left-invariant fields and a left-invariant metric; differentiating this $\mathrm{Ad}$-invariance along $g=\exp (tX)$ at $t=0$ and using [L7] gives the invariance identity $⟨[X,U],V⟩=-⟨U,[X,V]⟩$, which makes the last two terms cancel and yields $⟨{\mathrm{\nabla }}_{X}Y,Z⟩=⟨\frac{1}{2}[X,Y],Z⟩$; as $Z$ ranges over a basis of $\mathfrak{g}$ and $⟨\cdot ,\cdot ⟩$ is nondegenerate, ${\mathrm{\nabla }}_{X}Y=\frac{1}{2}[X,Y]$.