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

If $r=0$, both spaces are singletons, their probability measure is Dirac, all translations and inversion are identity, and the asserted integral is evaluation at the point. Now suppose $r\ge 1$. Every coset modulo ${\mathbb{Z}}^r$ has a unique representative in $Q$, by subtracting the coordinatewise integer floors. Define $\nu (B)={\lambda }_r(q^{-1}(B)\cap Q)$ for Borel $B\subseteq T^r$. The preimage is Borel and countable disjoint unions pull back to disjoint unions, so this is a Borel measure; [L2] gives $\nu (T^r)={\lambda }_r(Q)=1$.

Let $a\in {\mathbb{R}}^r$ and $E=q^{-1}(B)\cap Q$. Partition $E$ into the Borel sets $E_m=\{x\in E:x+a-m\in Q\}$, $m\in {\mathbb{Z}}^r$. Only finitely many are nonempty, since $x\in Q$ and $a$ is fixed. Unique representatives imply that the translates $E_m+a-m$ are pairwise disjoint and their union is exactly $q^{-1}(q(a)+B)\cap Q$: surjectivity follows by subtracting $q(a)$, and injectivity follows because two points of $Q$ differing by an integer vector coincide. Translation invariance and finite additivity give $\nu (q(a)+B)={\sum }_m{\lambda }_r(E_m+a-m)={\sum }_m{\lambda }_r(E_m)=\nu (B)$. Half-open faces cause no overlap or omitted boundary points.

To prove regularity, let $B$ be Borel and $E=q^{-1}(B)\cap Q$. By [L3], for each $\epsilon >0$ there is compact $C\subseteq E$ with ${\lambda }_r(E\setminus C)<\epsilon$. Then $q(C)$ is compact in $B$, and $\nu (B\setminus q(C))\le {\lambda }_r(E\setminus C)<\epsilon$. Applying this to $B^c$ gives a compact $D\subseteq B^c$ with $\nu (B^c\setminus D)<\epsilon$; the open set $T^r\setminus D$ contains $B$ and its excess over $B$ has measure less than $\epsilon$. Thus $\nu$ is both inner and outer regular.

The probability measure $\nu$ is compact finite and nonzero by step 1.1, left invariant by step 2.1, and regular by step 2.2. It is therefore a left Haar probability in the exact sense of [L1]. Apply the unique-left-Haar-probability clause of [L1]; it equals normalized Haar measure and in particular is also inversion invariant. We do not invoke uniqueness restricted to measures already known to be inversion invariant.

The integral formula holds for indicators of Borel sets by the definition of $\nu$ in step 1.1 and its identification in step 3.1. Finite linearity gives it for nonnegative simple functions. Increasing simple approximations, or equivalently the defining supremum for the nonnegative integral, give it for nonnegative Borel $f$. Applying this to the positive and negative parts of the real and imaginary parts proves the formula for integrable complex $f$; applying it first to $\mathrm{\mid }f\mathrm{\mid }$ verifies integrability on the cube. The singleton case was established separately in step 1.1.

The diagonal unitary matrices form a compact connected abelian subgroup of $U(n)$. A matrix commuting with every diagonal unitary matrix has zero $(i,j)$-entry for $i\mathrm{\ne }j$, by choosing diagonal phases whose $i$th and $j$th entries differ; hence the centralizer of $T_{U(n)}$ is itself, so it is maximal. For $n\ge 2$ the same entrywise argument uses determinant-one diagonal phases and shows that the centralizer of $T_{SU(n)}$ in $SU(n)$ is $T_{SU(n)}$; for $n=1$, $SU(1)$ is trivial. Thus both displayed tori are maximal.

The permutation matrices ${\pi }_{\sigma }$ are unitary and satisfy ${\pi }_{\sigma }\mathrm{diag}(z_1,\dots ,z_n){\pi }_{\sigma }^{-1}=\mathrm{diag}(z_{\sigma (1)},\dots ,z_{\sigma (n)})$, so they lie in the normalizer and induce $S_n$ in the Weyl group. For $SU(n)$ choose a diagonal unitary $d_{\sigma }$ with $\det d_{\sigma }=(\det {\pi }_{\sigma }{)}^{-1}$; then $d_{\sigma }{\pi }_{\sigma }\in SU(n)$ and, because $d_{\sigma }$ commutes with the diagonal torus, it induces the same coordinate permutation.

Conversely, let $g$ normalize the diagonal torus. Choose a regular element $t=\mathrm{diag}(z_1,\dots ,z_n)$ of the torus with pairwise distinct $z_j$. Then $gt{g}^{-1}\in T$, and the $z_j$ are the eigenvalues of $t$; since $gt{g}^{-1}$ has the same eigenvalues, and the eigenspaces of $t$ are the coordinate lines, $g$ permutes those lines up to scalars, hence equals a permutation matrix times a diagonal matrix. In $U(n)$ this says that the normalizer is generated by the torus and the permutation matrices. In $SU(n)$, if the induced permutation is $\sigma$, step 2.1 supplies the determinant-corrected representative $d_{\sigma }{\pi }_{\sigma }\in SU(n)$; multiplying $g$ by its inverse leaves a diagonal determinant-one matrix, so the normalizer is generated by $T_{SU(n)}$ and these corrected representatives. In either case the quotient is $S_n$.

Consequently the Weyl groups of $U(n)$ and $SU(n)$ are $S_n$ acting by coordinate permutation, in agreement with the root-system computation for types $A_{n-1}$.

$T$ is a torus by [L1]. If a connected abelian subgroup $S\supseteq T$ existed, then every element of $S$ would commute with every rotation about the $z$-axis, and a rotation commuting with all of them fixes the $z$-axis, hence is itself a rotation about the $z$-axis; so $S=T$ and $T$ is maximal.

The half-turn $s$ about the $x$-axis satisfies $st{s}^{-1}=t^{-1}$ for $t\in T$: conjugating a rotation about the $z$-axis by $s$ reverses its angle, so $s\in N_G(T)$ and its class in $W$ is nontrivial.

Conversely, if $g\in N_G(T)$ normalizes $T$, then $g$ preserves the axis of every nonidentity element of $T$, namely the $z$-axis as an unoriented line; hence $g$ either preserves or reverses the direction of the $z$-axis, and modulo $T$ the only two possibilities are the identity and the half-turn's coset.

Therefore $N_G(T)\mathrm{/}T\cong \mathbb{Z}\mathrm{/}2$, the nontrivial element acting by $t\mapsto t^{-1}$, i.e. by angle reversal; this agrees with the root-system computation for type $A_1$ with one positive root.

Substituting $\mathrm{\mid }W\mathrm{\mid }=2$ into [L2] and using that the class function is constant on Weyl orbits gives ${\int }_{SU(2)}fdg=\frac{1}{2}{\int }_{S^1}f(\mathrm{diag}(z,z^{-1}))J(\mathrm{diag}(z,z^{-1}))dz$.

By [L3] the single positive root satisfies $\alpha (\mathrm{diag}(z,z^{-1}))=z^2$, so $J(\mathrm{diag}(z,z^{-1}))=\mathrm{\mid }1-z^{-2}{\mathrm{\mid }}^2$, which is the displayed factor.

As a check, $f\equiv 1$ gives $\frac{1}{2}{\int }_{S^1}\mathrm{\mid }1-z^{-2}{\mathrm{\mid }}^{2}dz=\frac{1}{2}{\int }_{S^1}(2-z^2-z^{-2})dz=\frac{1}{2}\cdot 2=1$, consistent with the normalization of Haar measure.

$SU(2)$ is simply connected and semisimple with root system $A_1$, so $X^{\ast }(T_{SU(2)})=P=\mathbb{Z}\omega$ by [L1]; the characters are $\mathrm{diag}(z,z^{-1})\mapsto z^n$, with weight $n\omega$.

The kernel of $SU(2)\to SO(3)$ is $\{\pm I\}$. Since $-I=\mathrm{diag}(-1,-1)$ corresponds to $z=-1$ in $T_{SU(2)}$, the character of weight $n\omega$ takes the value $(-1{)}^n$ there, so it is trivial on the kernel exactly when $n$ is even.

By [L1] the intermediate lattice of $SO(3)$ is $X^{\ast }(T_{SO(3)})=\{\text{characters trivial on }\{\pm I\}\}=2\mathbb{Z}\omega =Q$, and the adjoint-form computation of [L1] gives the same answer.

Hence a dominant $SU(2)$ highest weight $n\omega$ descends to $SO(3)$ exactly when $n$ is even, i.e. exactly when $n\omega \in Q$; this is the explicit form of the finite central quotient obstruction.

By [L2] the simply connected form $G_{sc}$ itself has $X^{\ast }(T_{sc})=P$, corresponding under [L1] to the trivial central subgroup.

The adjoint form $G_{ad}=G_{sc}\mathrm{/}Z(G_{sc})$ has character lattice $Q$ by [L2]; by [L1] it corresponds to the full centre, and the annihilator of $Q$ in the finite dual pairing is exactly $Z(G_{sc})$.

For an intermediate lattice $Q⊊X⊊P$ the annihilator $C_X={\bigcap }_{\chi \in X}\ker \chi$ is a nontrivial proper central subgroup with $X^{\ast }(T_{sc}\mathrm{/}{C}_X)=X$, by [L1], and conversely every nontrivial proper central subgroup arises this way; so the intermediate quotients are exactly the intermediate lattices.

This yields the full menu of forms: the simply connected endpoint, the adjoint endpoint, and one marked quotient for each intermediate lattice, which is the sense in which compact semisimple groups are classified by root datum with the added lattice data.

Let $\pi :T^r\to U(V)$ be a finite-dimensional continuous irreducible representation. Since $T^r$ is abelian, every $\pi (t)$ commutes with every $\pi (s)$ and hence belongs to ${\mathrm{End}}_{T^r}(V)$; by [L4], every $\pi (t)$ is scalar. Thus every linear subspace of $V$ is invariant, so irreducibility and $V\mathrm{\ne }0$ force $\dim V=1$. Therefore $\pi$ is a character, and [L3] computes the characters: the quotient exponential ${\mathbb{R}}^r\to {\mathbb{R}}^r\mathrm{/}{\mathbb{Z}}^r$ has kernel ${\mathbb{Z}}^r$, so its allowed differentials are exactly $\lambda (X)=2\pi i{\sum }_j{n}_j{X}_j$ with $n_j\in \mathbb{Z}$. Thus the characters are exactly the $e_n$; distinct integer vectors have distinct differentials. Conversely each $e_n$ is a continuous unitary one-dimensional representation and hence irreducible.

Peter–Weyl [L2] therefore says exactly that the one-dimensional representations $e_n$, with their sole normalized matrix coefficient exactly $e_n$, form an orthonormal Hilbert basis of $L^2(T^r)$ and that the regular representation is their Hilbert direct sum weighted by dimension one. In the fixed left-action convention $L_y{e}_n=e_n(-y)e_n$, so the coefficient line has type $e_{-n}$; negation permutes ${\mathbb{Z}}^r$ and every character still occurs once.

For $r\ge 1$, [L1] gives the Fourier expansion and Parseval identity for precisely the basis identified in step 2.1. For $r=0$, the torus is a singleton, its normalized Haar measure has mass one at that point, and ${\mathbb{Z}}^0$ consists of the empty tuple alone. Its sole character is $e_0=1$ and $L^2(T^0)=\mathbb{C}$ with basis $1$; the expansion is $f=f(e)1$ and Parseval is $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=\mathrm{\mid }f(e){\mathrm{\mid }}^2$. Thus the zero-rank case is proved directly without applying [L1] outside its scope. The AC assumptions of the suppliers are covered by [A1].

With $g{e}_1=(a,-\bar{b})$ and $g{e}_2=(b,\bar{a})$, the matrix coefficients in the standard basis are ${\pi }_{11}=a$, ${\pi }_{21}=-\bar{b}$, ${\pi }_{12}=b$ and ${\pi }_{22}=\bar{a}$, by [L1]; these are exactly the four displayed coordinate functions.

The standard representation is irreducible. Indeed, for every unit vector $(u,v)\in {\mathbb{C}}^2$, the matrix $\left(\begin{array}{cc}u& -\bar{v}\\ v& \bar{u}\end{array}\right)$ lies in $SU(2)$ and sends $e_1$ to $(u,v)$; hence $SU(2)$ acts transitively on the unit sphere. Any nonzero invariant subspace therefore contains the whole unit sphere and equals ${\mathbb{C}}^2$. Since the representation has dimension two, [L2] applies with $d=2$: ${\int }_{SU(2)}{\pi }_{ij}\overline{{\pi }_{kl}}dg={\delta }_{ik}{\delta }_{jl}\mathrm{/}2$.

Reading off the four cases gives that each of $a,b,-\bar{b},\bar{a}$ has squared norm $\frac{1}{2}$ and that distinct coordinate functions are orthogonal, which is the assertion.

Give $G$ the discrete topology and singleton charts to ${\mathbb{R}}^0$. It is Hausdorff and second countable, its finite underlying space is compact, and multiplication and inversion are smooth in these zero-dimensional charts. Thus it is a compact Lie group. Since $G$ contains its identity, $\mathrm{\mid }G\mathrm{\mid }>0$ and $\mu (E)=\mathrm{\mid }E\mathrm{\mid }\mathrm{/}\mathrm{\mid }G\mathrm{\mid }$ is a probability. Every subset is open and compact, so this Borel measure is regular. Left translations, right translations and inversion permute $G$ and hence preserve cardinality and $\mu$. All hypotheses of [L1] hold, proving that $\mu$ is normalized Haar measure.

For a function $f$ on the finite group the Haar integral is therefore ${\int }_{G}fd\mu =\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{g\in G}f(g)$, and substituting this into the compact orthogonality relations of [L2] gives the displayed finite-group formula. If $\pi$ and $\sigma$ are equivalent they are the same chosen representative and use the same fixed orthonormal basis, so the delta case is licensed exactly. Otherwise the inequivalent case applies. For the trivial group the sole irreducible is one-dimensional and the formula is $1=1$.

Both groups are compact, connected and semisimple with root system $A_1$, and they are related by the finite central isogeny of [L1]; so the root system and the isogeny class coincide.

An isomorphism would carry the centre $\{\pm I\}$ of $SU(2)$ onto the centre of $SO(3)$, which is trivial by [L2]; as $\{\pm I\}$ has two elements and the trivial group has one, no isomorphism exists.

Equivalently, the two groups have different character lattices $P\mathrm{\ne }Q$ by [L3], and the fundamental weight of $SU(2)$ does not descend to $SO(3)$; the witness pair $(SU(2),SO(3))$ therefore refutes the claim that a shared root system determines a compact connected semisimple group up to isomorphism.

A reflection is an orthogonal matrix with determinant $-1$, so $r\mathrm{\notin }SO(2)=O(2{)}^0$.

Every torus contained in the identity component is a connected subgroup of $O(2)$ lying in $SO(2)$; by [L2] every connected subgroup of $O(2)$ lies in the kernel of the determinant, so no torus of the identity component contains $r$.

Hence $r$ is an element of the compact Lie group $O(2)$ that lies in no torus of the identity component, refuting the asserted statement and showing that connectedness of the ambient group is a necessary hypothesis for the torus-containment theorems.

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 $2$.