Complete Reducibility for Compact Groups — Examples

The set $S^1$ is a compact Hausdorff topological group. It contains $1$ and is closed under multiplication and inversion because $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $z^{-1}=\bar{z}$ with $\mathrm{\mid }\bar{z}\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$ by [F1]; associativity and the remaining group axioms are inherited from the field $\mathbb{C}$. In the coordinate plane, multiplication has the polynomial formula $(a,b,u,v)\mapsto (au-bv,av+bu)$ and inversion the formula $(a,b)\mapsto (a,-b)$ on $\mathrm{\mid }z\mathrm{\mid }=1$, so both operations are continuous on the product $S^1\times S^1$ respectively on $S^1$ by [F2]. Moreover $S^1$ is the preimage of $\{1\}$ under the continuous map $(a,b)\mapsto a^2+b^2$, hence closed in ${\mathbb{R}}^2$, and it is bounded because $a^2+b^2=1$; by Heine–Borel [F2] it is compact, and it is Hausdorff as a subspace of a metric space.

The map $\rho$ is a continuous finite-dimensional complex representation of $S^1$ on $V={\mathbb{C}}^2$, and $h_0$ is a Hermitian inner product on $V$. Indeed $\rho (z)\rho (w)=\mathrm{diag}(1,zw)=\rho (zw)$ and $\rho (1)=I$, each $\rho (z)=\mathrm{diag}(1,z)$ is invertible because $z\mathrm{\ne }0$, and $z\mapsto \rho (z)$ is continuous as a map into the finite-dimensional space $\mathrm{End}(V)$ because its matrix entries are continuous and all norms on that space are equivalent (All norms on a finite-dimensional complex normed space are equivalent). The form $h_0$ has the real symmetric matrix $\left(\begin{array}{cc}2& 1\\ 1& 3\end{array}\right)$, hence is conjugate-symmetric, and $h_0(v,v)=2\mathrm{\mid }{v}_1{\mathrm{\mid }}^2+2\mathrm{Re}(v_1\overline{v_2})+3\mathrm{\mid }{v}_2{\mathrm{\mid }}^2\ge \mathrm{\mid }{v}_1{\mathrm{\mid }}^2+2\mathrm{\mid }{v}_2{\mathrm{\mid }}^2>0$ whenever $v\mathrm{\ne }0$, because $2\mathrm{Re}(v_1\overline{v_2})\ge -(\mathrm{\mid }{v}_1{\mathrm{\mid }}^2+\mathrm{\mid }{v}_2{\mathrm{\mid }}^2)$; in particular $h_0(e_1,e_2)=1$.

The integrals of the characters vanish: ${\int }_{S^1}zd\mu (z)=0$ and ${\int }_{S^1}\bar{z}d\mu (z)=0$. Both characters are continuous and have modulus one, hence are integrable against the probability $\mu$. The map $j(z):=-z=(-1)\cdot z$ is a continuous self-map of $S^1$ with $j^{-1}(E)=(-1)E$, so $\mu (j^{-1}E)=\mu (E)$ by left invariance of [F3]: it is measure preserving, and [F5] gives $\int zd\mu =\int (-z)d\mu =-\int zd\mu$, hence $\int zd\mu =0$; replacing $z$ by $\bar{z}$, whose composite with $j$ is $-\bar{z}$, gives $\int \bar{z}d\mu =-\int \bar{z}d\mu =0$ in the same way.

For $z\in S^1$ one has $\rho (z)v=(v_1,z{v}_2)$ and $\mathrm{\mid }z{\mathrm{\mid }}^2=1$, so expanding $h_0$ in [F4] gives $h_0(\rho (z)v,\rho (z)w)=2v_1\overline{w_1}+\bar{z}{v}_1\overline{w_2}+z{v}_2\overline{w_1}+3\mathrm{\mid }z{\mathrm{\mid }}^2{v}_2\overline{w_2}=2v_1\overline{w_1}+3v_2\overline{w_2}+\bar{z}(v_1\overline{w_2})+z(v_2\overline{w_1})$ for all $v,w\in V$.

Integrating the expansion of step 3.1 and pulling out the constants $v_i\overline{w_j}$ by linearity of the integral [F5] yields $h(v,w)={\int }_{S^1}(2v_1\overline{w_1}+3v_2\overline{w_2}+\bar{z}{v}_1\overline{w_2}+z{v}_2\overline{w_1})d\mu (z)=2v_1\overline{w_1}+3v_2\overline{w_2}+(\int \bar{z}d\mu )v_1\overline{w_2}+(\int zd\mu )v_2\overline{w_1}=2v_1\overline{w_1}+3v_2\overline{w_2}$ by step 2.2.

Consequences. By step 4.1, $h(e_1,e_2)=0$, while $h_0(e_1,e_2)=1$ by step 2.1: the two weight lines are orthogonal for $h$ but not for $h_0$. Since $\rho (z)e_1=e_1$ and $\rho (z)e_2=z{e}_2$, the invariance failure is visible at $z=-1$: $h_0(\rho (-1)e_1,\rho (-1)e_2)=h_0(e_1,-e_2)=-h_0(e_1,e_2)=-1\mathrm{\ne }1=h_0(e_1,e_2)$, conjugate-linearity in the second variable producing the sign. The averaged form $h$ is positive definite and $S^1$-invariant by [F6], in agreement with the explicit formula of step 4.1.

$F$ is a compact Hausdorff topological group. Every subset of $F$ is open and closed in the discrete topology, and the product topology on $F\times F$ is discrete because its points are the basic open sets $\{g\}\times \{h\}$; hence the inversion $F\to F$ and the multiplication $F\times F\to F$ are continuous, being functions on discrete domains [F1]. Thus $F$ is a topological group; it is Hausdorff because distinct $g,h$ are separated by the disjoint open sets $\{g\}$ and $\{h\}$; and it is compact: given an open cover, enumerate $F=\{g_1,\dots ,g_n\}$ and choose a covering member $U_i\ni g_i$ for each $i$, which is a finite choice, and $U_1,\dots ,U_n$ is a finite subcover. [F1] 2.1 The normalized Haar probability $\mu$ exists on $F$ by [F2]. Every singleton is open, hence Borel [F3]. For every $g\in F$ the set $\{g\}=g\{e\}$ is a left translate of $\{e\}$, so left invariance gives $\mu (\{g\})=\mu (\{e\})$; writing $F=\{g_1,\dots ,g_n\}$ as a disjoint union of singletons and using finite additivity and $\mu (F)=1$ gives $1=\mu (F)={\sum }_{i=1}^n\mu (\{g_i\})=n\mu (\{e\})$, so $\mu (\{g\})=1\mathrm{/}\mathrm{\mid }F\mathrm{\mid }$ for every $g\in F$. [F2, F3, step 1.1] 3.1 Fix $v\in H$. The function $f_v(k):=\overline{{\chi }_{\sigma }(k)}\pi (k)v$ is constant on each singleton, with value $x_g:=\overline{{\chi }_{\sigma }(g)}\pi (g)v$ on $\{g\}$; its nonzero fibres are therefore unions of those singletons $g$ for which $x_g$ takes one fixed nonzero value, so $f_v={\sum }_j{y}_j{\mathbf{1}}_{B_j}$ for finitely many pairwise disjoint Borel sets $B_j$ and distinct nonzero $y_j\in H$ [F5]. Each $\mu (B_j)\le \mu (F)=1$ is finite, so $f_v$ is an integrable simple function and hence Bochner integrable, with Bochner integral equal to its simple integral; regrouping the singletons into the fibres and using finite additivity of $\mu$ gives ${\int }_F{f}_{v}d\mu ={\sum }_j\mu (B_j)y_j={\sum }_{g\in F}\mu (\{g\})x_g=\frac{1}{\mathrm{\mid }F\mathrm{\mid }}{\sum }_{g\in F}\overline{{\chi }_{\sigma }(g)}\pi (g)v$. [F3, F5, step 2.1] 4.1 Multiplying the identity of step 3.1 by $d_{\sigma }$ and comparing with the definition $P_{\sigma }v=d_{\sigma }{\int }_F{f}_{v}d\mu$ of [F4] yields $P_{\sigma }v=\frac{d_{\sigma }}{\mathrm{\mid }F\mathrm{\mid }}{\sum }_{g\in F}\overline{{\chi }_{\sigma }(g)}\pi (g)v$ for every $v\in H$, which is the displayed operator identity; the finite sum is independent of the enumeration by [F8]. [F4, F8, step 3.1] 5.1 The representation $\sigma$ and $\pi$ are strongly continuous, since every function on the discrete space $F$ is continuous, and $F$ is a compact Hausdorff group by step 1.1; so the A-page theorem [F6] applies and shows that this operator $P_{\sigma }$ is a bounded linear self-adjoint idempotent commuting with $\pi (F)$ whose range is exactly $H_{\sigma }$, and that for every irreducible $\tau$ inequivalent to $\sigma$ the corresponding projections satisfy $P_{\sigma }{P}_{\tau }=0=P_{\tau }{P}_{\sigma }$ and have orthogonal ranges. [F6, step 1.1, step 4.1] 6.1 Special cases of the formula of step 4.1. For the trivial representation ${\sigma }_0$ on $\mathbb{C}$ one has $d_{{\sigma }_0}=1$ and ${\chi }_{{\sigma }_0}\equiv 1$, so $P_{{\sigma }_0}=\frac{1}{\mathrm{\mid }F\mathrm{\mid }}{\sum }_{g\in F}\pi (g)$; the ${\sigma }_0$-copies are exactly the lines spanned by nonzero vectors fixed by $\pi (F)$ (a fixed vector spans a one-dimensional invariant subspace on which $\pi$ acts trivially, and conversely every ${\sigma }_0$-copy consists of fixed vectors); the fixed space is ${\bigcap }_{g\in F}\ker (\pi (g)-I)$, which is closed because each $\pi (g)-I$ is bounded, so its closed span of fixed lines is itself; by step 5.1 the range of $P_{{\sigma }_0}$ is exactly that fixed space. If $d_{\sigma }=1$, then on a $\sigma$-copy $\pi (g)$ acts as the scalar ${\chi }_{\sigma }(g)$ with $\mathrm{\mid }{\chi }_{\sigma }(g)\mathrm{\mid }=1$, so the formula gives $P_{\sigma }x=\frac{1}{\mathrm{\mid }F\mathrm{\mid }}{\sum }_{g\in F}\overline{{\chi }_{\sigma }(g)}{\chi }_{\sigma }(g)x=x$ for every $x$ in that copy, in agreement with the fixing property of step 5.1. For the one-element group $F=\{e\}$ irreducibility forces $d_{\sigma }=1$, because for $d_{\sigma }\ge 2$ a line in the finite-dimensional space $V_{\sigma }$ is a proper nontrivial closed $\sigma (e)$-invariant subspace, since $\sigma (e)=I$ and every finite-dimensional subspace is closed; then ${\chi }_{\sigma }(e)=1$, $\pi (e)=I$ and the formula gives $P_{\sigma }=I$, while $v=0$ gives $P_{\sigma }v=0$. [F4, F6, F7, step 5.1] 7.1 Explicit two-element group. Let $F=\mathbb{Z}\mathrm{/}2\mathbb{Z}=\{e,t\}$ and $H={\mathbb{C}}^2$ with orthonormal basis $e_1,e_2$, let $\pi (e)=I$, $\pi (t)=\mathrm{diag}(1,-1)$, and let ${\sigma }_0$ be the trivial representation and ${\sigma }_1$ the sign representation ${\sigma }_1(t)=-1$ on $\mathbb{C}$, both irreducible of degree one by [F7] and inequivalent because their characters differ at $t$. Applying the formula of step 4.1 ($\mathrm{\mid }F\mathrm{\mid }=2$, characters ${\chi }_{{\sigma }_0}(e)={\chi }_{{\sigma }_0}(t)=1$ and ${\chi }_{{\sigma }_1}(e)=1,{\chi }_{{\sigma }_1}(t)=-1$) gives $P_{{\sigma }_0}=\frac{1}{2}(I+\pi (t))=\mathrm{diag}(1,0)$ and $P_{{\sigma }_1}=\frac{1}{2}(I-\pi (t))=\mathrm{diag}(0,1)$; both matrices are self-adjoint idempotents, they are mutually orthogonal, and their ranges $\mathbb{C}{e}_1$ and $\mathbb{C}{e}_2$ are the trivial and sign copies inside $H$, so the ranges are orthogonal and span $H$.

The displayed table makes $C_2$ a group: $0$ is an identity by the first and third entries, $0+0=0$ and $1+1=0$ make every element its own inverse, the table is symmetric in its two arguments, and associativity holds because both $(x+y)+z$ and $x+(y+z)$ equal the sum of the three bits modulo $2$, as the eight triples of bits show. It follows that $K$ is a group under the coordinatewise operation: associativity is inherited coordinatewise from $C_2$, the constant function $e=0$ is an identity, and $x^{-1}=x$ because $x_n+x_n=0$ in every coordinate; moreover the function $z$ with $z(0)=1$ and $z(n)=0$ for $n\ge 1$ is not $e$, so $K\mathrm{\ne }\{e\}$.

Every subset of $C_2$ is open in the discrete topology, so every map with domain $C_2$, and every map with domain $C_2\times C_2$, is continuous; $C_2$ is compact as a finite space, and it is Hausdorff because $\{0\}$ and $\{1\}$ are disjoint open sets.

For a finite set $F\subseteq \mathbb{N}$ and a function $w:F\to C_2$ put $U_{F,w}:=\{y\in K:y_i=w_i\text{ for all }i\in F\}$. Each $U_{F,w}$ is ${\bigcap }_{i\in F}{\pi }_i^{-1}[\{w_i\}]$, a finite intersection of preimages of open sets, hence a basic product-open set and in particular open; conversely, if $W$ is open and $x\in W$, then by [F3] there is a basic product-open box $B={\prod }_n{V}_n$ with $x\in B\subseteq W$ and $V_n=C_2$ for all $n$ outside a finite set $F$, and then $x\in U_{F,x{\mathrm{\mid }}_F}\subseteq B\subseteq W$, because a point of $U_{F,x{\mathrm{\mid }}_F}$ has $i$-th coordinate $x_i\in V_i$ for $i\in F$ and an arbitrary coordinate of $C_2=V_n$ for $n\mathrm{\notin }F$. Also $U_{\mathrm{\varnothing },\mathrm{\varnothing }}=K$.

Since $V$ is finite dimensional it admits an ordered basis $e:n\to V$ of finite length, and then every $x\in V$ has exactly one coordinate list $\lambda :n\to \mathbb{C}$ with $x={\sum }_{i<n}{\lambda }_i{e}_i$. The assignment $\mathrm{\parallel }x\mathrm{\parallel }:={\sum }_{i<n}\mathrm{\mid }{\lambda }_i\mathrm{\mid }$ is a norm on $V$: definiteness is uniqueness of the coordinates, homogeneity is $\mathrm{\mid }\alpha {\lambda }_i\mathrm{\mid }=\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }{\lambda }_i\mathrm{\mid }$ applied termwise, and the triangle inequality follows from $\mathrm{\mid }{\lambda }_i+{\mu }_i\mathrm{\mid }\le \mathrm{\mid }{\lambda }_i\mathrm{\mid }+\mathrm{\mid }{\mu }_i\mathrm{\mid }$ and additivity of finite sums applied termwise. Since every linear map $V\to V$ is bounded for this norm, the operator norm makes $\mathcal{L}(V,V)$ a normed space with $\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$; and because all norms on this finite-dimensional complex space are equivalent, the topology induced on the subset $\mathrm{GL}(V)$ is the same for every choice of norm or of basis.

Let $T:V\to V$ be linear with $T^2={\mathrm{id}}_V$ and $T\mathrm{\ne }{\mathrm{id}}_V$. If $T+{\mathrm{id}}_V$ were injective, then it would be surjective, since an injective linear endomorphism of the finite-dimensional space $V$ is surjective; then $T-{\mathrm{id}}_V=(T-{\mathrm{id}}_V)(T+{\mathrm{id}}_V)(T+{\mathrm{id}}_V{)}^{-1}=0$ by $T^2={\mathrm{id}}_V$, that is $T={\mathrm{id}}_V$, contrary to hypothesis. Hence $T+{\mathrm{id}}_V$ is not injective, so there is $v\mathrm{\ne }0$ with $(T+{\mathrm{id}}_V)v=0$, that is $Tv=-v$.

Since $\rho$ is a group homomorphism, $\rho (e)=\rho (e\cdot e)=\rho (e{)}^2$, and multiplying by the inverse of the group element $\rho (e)$ gives $\rho (e)={\mathrm{id}}_V$; and $\rho (g^2)=\rho (g{)}^2$ for every $g\in K$.

$K$ is Hausdorff: if $x\mathrm{\ne }y$ in $K$, there is $n$ with $x_n\mathrm{\ne }{y}_n$, and then $F=\{n\}$ gives two points $w=x{\mathrm{\mid }}_F$, $w^{\mathrm{\prime }}=y{\mathrm{\mid }}_F$ of $C_2^F$ with $U_{F,w}\cap U_{F,w^{\mathrm{\prime }}}=\mathrm{\varnothing }$, while $x\in U_{F,w}$ and $y\in U_{F,w^{\mathrm{\prime }}}$ by step 1.3; these are disjoint open neighbourhoods.

$K$ is a topological group. For multiplication $m(x,y):=xy$ it suffices by [F3] to show that each component ${\pi }_n\circ m$ is continuous, and $({\pi }_n\circ m)(x,y)=x_n+y_n$. The map $K\times K\to C_2\times C_2$, $(x,y)\mapsto (x_n,y_n)$, is continuous because its components are ${\pi }_n\circ {\mathrm{p}\mathrm{r}}_1$ and ${\pi }_n\circ {\mathrm{p}\mathrm{r}}_2$, composites of continuous projections; the operation $C_2\times C_2\to C_2$ is continuous by step 1.2; so ${\pi }_n\circ m$ is a composite of continuous maps, hence continuous. Inversion is the identity map of $K$ by step 1.1 and therefore continuous.

The set $B:=\{T\in \mathcal{L}(V,V):\mathrm{\parallel }T-{\mathrm{id}}_V\mathrm{\parallel }<1\}$ is open in $\mathcal{L}(V,V)$ and contains ${\mathrm{id}}_V$, hence $B\cap \mathrm{GL}(V)$ is a neighbourhood of $\rho (e)={\mathrm{id}}_V$ in the subspace topology; by continuity of $\rho$ at $e$ there is an open neighbourhood $W$ of $e$ in $K$ with $\rho [W]\subseteq B$, and by step 1.3 applied to $W\ni e$ there is a finite $F\subseteq \mathbb{N}$ with $U_{F,0}\subseteq W$, where $0$ here denotes the zero function on $F$. By [F9] there is $k\in \mathbb{N}$ with $F\subseteq k$, and $U_{k,0}\subseteq U_{F,0}$ because a cylinder constrains more coordinates; hence $\mathrm{\parallel }\rho (g)-{\mathrm{id}}_V\mathrm{\parallel }<1$ for every $g\in U_{k,0}$.