A compact group with no faithful continuous finite-dimensional representation

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 $g\in U_{k,0}$.

Every $g\in K$ satisfies $g^2=e$ because each coordinate satisfies $x+x=0$ by step 1.1, and therefore $\rho (g{)}^2=\rho (g^2)=\rho (e)={\mathrm{id}}_V$ by step 1.6; consequently, whenever $\rho (g)\mathrm{\ne }{\mathrm{id}}_V$, step 1.5 applied to $T=\rho (g)$ provides $v\mathrm{\ne }0$ with $\rho (g)v=-v$.

Let $\mathcal{U}$ be an open cover of $K$ and call a pair $(F,w)$, with $F\subseteq \mathbb{N}$ finite and $w:F\to C_2$, bad when no finite subfamily of $\mathcal{U}$ covers $U_{F,w}$. If $w:k\to C_2$ is bad, then at least one of the two extensions $w_0,w_1:k+1\to C_2$, defined by $w_j{\mathrm{\mid }}_k=w$ and $w_j(k)=j$, is bad: indeed every $y\in U_{k,w}$ has $y(k)=0$ or $y(k)=1$, so $U_{k,w}=U_{k+1,w_0}\cup U_{k+1,w_1}$; if both cylinders on the right were covered by finite subfamilies of $\mathcal{U}$, their union, listed after one another, would be a finite subfamily covering $U_{k,w}$, contradicting badness.

Assume for contradiction that $\mathcal{U}$ is an open cover of $K$ with no finite subcover, so that the empty cylinder $U_{0,\mathrm{\varnothing }}=K$ is bad. Let $X$ be the set of all functions $w:D\to C_2$ whose domain $D\subseteq \mathbb{N}$ is finite, and define $\mathrm{\Phi }:X\to X$ by: if $dom(w)=k\in \mathbb{N}$, let $w_0,w_1$ be the two extensions of $w$ to $k+1$, and put $\mathrm{\Phi }(w):=w_0$ if $(k+1,w_0)$ is bad and $\mathrm{\Phi }(w):=w_1$ otherwise; if $dom(w)$ is not a natural number, put $\mathrm{\Phi }(w):=w$. By [F8] there is $h:\mathbb{N}\to X$ with $h(0)=\mathrm{\varnothing }$ and $h(n+1)=\mathrm{\Phi }(h(n))$. Induction on $n$ shows that $h(n)$ has domain $n$ and is bad: this holds at $n=0$ by the assumption, and if it holds at $n$ then step 2.5 produces a bad extension of $h(n)$ with domain $n+1$, which is exactly $\mathrm{\Phi }(h(n))=h(n+1)$. Since each $\mathrm{\Phi }(w)$ extends $w$, the functions $h(n)$ are coherent, and $x(i):=h(i+1)(i)$ defines a function $x:\mathbb{N}\to C_2$, that is, a point $x\in K$, with $x{\mathrm{\mid }}_n=h(n)$ for every $n$.

Every $g\in U_{k,0}$, for the $k$ of step 2.3, satisfies $\rho (g)={\mathrm{id}}_V$. Suppose $\rho (g)\mathrm{\ne }{\mathrm{id}}_V$; by step 2.4 there is $v\mathrm{\ne }0$ with $\rho (g)v=-v$, and then $2\mathrm{\parallel }v\mathrm{\parallel }=\mathrm{\parallel }\rho (g)v-v\mathrm{\parallel }=\mathrm{\parallel }(\rho (g)-{\mathrm{id}}_V)v\mathrm{\parallel }\le \mathrm{\parallel }\rho (g)-{\mathrm{id}}_V\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }<\mathrm{\parallel }v\mathrm{\parallel }$, using $\mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$ of step 1.4 and $\mathrm{\parallel }\rho (g)-{\mathrm{id}}_V\mathrm{\parallel }<1$ of step 2.3; but $2\mathrm{\parallel }v\mathrm{\parallel }<\mathrm{\parallel }v\mathrm{\parallel }$ is impossible because $\mathrm{\parallel }v\mathrm{\parallel }>0$ by the norm axioms of step 1.4. Hence $\rho (g)={\mathrm{id}}_V$ for every $g\in U_{k,0}$.

$K$ is compact. Let $\mathcal{U}$ be an open cover of $K$ and suppose it has no finite subcover. Steps 2.5 and 3.1 then produce a point $x\in K$ with $U_{n,x{\mathrm{\mid }}_n}$ bad for every $n\in \mathbb{N}$. Since $\mathcal{U}$ covers $K$ there is $U\in \mathcal{U}$ with $x\in U$, and since $U$ is open step 1.3 gives a finite $F\subseteq \mathbb{N}$ with $U_{F,x{\mathrm{\mid }}_F}\subseteq U$; by [F9] there is $m\in \mathbb{N}$ with $F\subseteq m$, so $U_{m,x{\mathrm{\mid }}_m}\subseteq U_{F,x{\mathrm{\mid }}_F}\subseteq U$, that is, the one-member finite subfamily $\{U\}$ of $\mathcal{U}$ covers the bad cylinder $U_{m,x{\mathrm{\mid }}_m}$ — a contradiction. Therefore every open cover of $K$ has a finite subcover, so $K$ is compact.

Every continuous finite-dimensional complex representation of $K$ has a nontrivial kernel. With $V,\rho ,k$ as above, step 3.2 gives $\rho (g)={\mathrm{id}}_V$ for all $g\in U_{k,0}$. The function $z:\mathbb{N}\to C_2$ with $z(k)=1$ and $z(i)=0$ for $i\mathrm{\ne }k$ lies in $U_{k,0}$ because it vanishes on all $i<k$, and $z\mathrm{\ne }e$ since $z(k)=1$; thus $\rho (z)={\mathrm{id}}_V$ with $z\mathrm{\ne }e$, and $\rho$ is not faithful, so the kernel of $\rho$ is nontrivial.

By steps 2.1, 2.2 and 4.1 the space $K={\prod }_{n\in \mathbb{N}}{C}_2$ is a compact Hausdorff topological group, and by step 4.2 every continuous finite-dimensional complex representation of $K$ has a nontrivial kernel, so no such representation is faithful. The entire argument uses only the two-element table, the definitions involved and recursion and induction on $\mathbb{N}$: no choice principle and in particular no Tychonoff theorem enters, so both claims are choice free.