Unitary Representations, Positive Type and GNS

If $\pi$ is strongly continuous, [A1] makes every mixed matrix coefficient continuous on $G$. In particular, every diagonal coefficient is continuous at $e$ on every total subset.

Now suppose the diagonal coefficients are continuous at $e$ for a total subset $S$. Fix $v\in S$, put $q(g)=⟨\pi (g)v,v⟩$, $r=\mathrm{\parallel }v{\mathrm{\parallel }}^2=q(e)$, and $z(g)=q(g)-r$. The homomorphism law gives $\pi (e)=I$. Using [A2] and [A4], $\mathrm{\parallel }\pi (g)v-v{\mathrm{\parallel }}^2=\mathrm{\parallel }\pi (g)v{\mathrm{\parallel }}^2-⟨\pi (g)v,v⟩-⟨v,\pi (g)v⟩+\mathrm{\parallel }v{\mathrm{\parallel }}^2=2r-q(g)-\overline{q(g)}=-(z(g)+\overline{z(g)})\mathrm{.}$ This is a nonnegative real number. By [A6] and [A7], $0\le \mathrm{\parallel }\pi (g)v-v{\mathrm{\parallel }}^2=\mid z(g)+\overline{z(g)}\mid \le \mathrm{\mid }z(g)\mathrm{\mid }+\mathrm{\mid }\overline{z(g)}\mathrm{\mid }=2\mathrm{\mid }z(g)\mathrm{\mid }\mathrm{.}$ Continuity of $q$ at $e$ in the metric of [A10] gives, for each $\epsilon >0$, a neighborhood of $e$ on which $2\mathrm{\mid }z(g)\mathrm{\mid }<{\epsilon }^2$. Then $\mathrm{\parallel }\pi (g)v-v{\mathrm{\parallel }}^2<{\epsilon }^2$, so [A5] and [A9] give $\mathrm{\parallel }\pi (g)v-v\mathrm{\parallel }<\epsilon$. No nonzero-vector hypothesis was used, so this also covers $v=0$.

If $w={\sum }_{j=1}^n{\lambda }_j{v}_j$ is a finite complex linear combination of elements of $S$, then linearity and [A5] give $\mathrm{\parallel }\pi (g)w-w\mathrm{\parallel }\le {\sum }_{j=1}^n\mathrm{\mid }{\lambda }_j\mathrm{\mid }\mathrm{\parallel }\pi (g)v_j-v_j\mathrm{\parallel }⟶0(g\to e)\mathrm{.}$ by step 1.2. Only finitely many orbit maps occur, so intersecting their neighborhoods proves convergence of the sum. If $n=0$, then $w=0$ and the orbit difference is identically zero.

For any $x\in H$ and $\epsilon >0$, totality supplies a finite-span vector $w$ with $\mathrm{\parallel }x-w\mathrm{\parallel }<\epsilon \mathrm{/}3$. Step 2.1 gives a neighborhood of $e$ on which $\mathrm{\parallel }\pi (g)w-w\mathrm{\parallel }<\epsilon \mathrm{/}3$. For such $g$, the triangle inequality and the isometry property in [A2] give $\mathrm{\parallel }\pi (g)x-x\mathrm{\parallel }\le \mathrm{\parallel }\pi (g)(x-w)\mathrm{\parallel }+\mathrm{\parallel }\pi (g)w-w\mathrm{\parallel }+\mathrm{\parallel }w-x\mathrm{\parallel }=2\mathrm{\parallel }x-w\mathrm{\parallel }+\mathrm{\parallel }\pi (g)w-w\mathrm{\parallel }<\epsilon \mathrm{.}$ This proves continuity at $e$ for every orbit map. If $S=\mathrm{\varnothing }$, totality means the only available finite sum is $0$ and still supplies the required approximation; step 2.1 and the same estimate apply (and force $H=\{0\}$).

Fix $x\in H$ and $g_0\in G$. As $g\to g_0$, continuity of multiplication in [A3] gives $h=g_0^{-1}g\to e$. The homomorphism law and the isometry $\pi (g_0)$ yield $\mathrm{\parallel }\pi (g)x-\pi (g_0)x\mathrm{\parallel }=\mathrm{\parallel }\pi (g_0)(\pi (h)x-x)\mathrm{\parallel }=\mathrm{\parallel }\pi (h)x-x\mathrm{\parallel }⟶0.$ by step 3.1. Thus every orbit map is norm-continuous on $G$, which is strong continuity.

If every matrix coefficient is continuous, its diagonal coefficients are continuous at $e$ on any total subset, so steps 1.2–4.1 prove strong continuity. Conversely step 1.1 proves that strong continuity implies both coefficient conditions. Hence all three conditions are equivalent.

AC implies Countable Choice by restricting a choice function to any countable subfamily, so [A3] supplies adjoints. If $S$ is a self-intertwiner of $\pi$, apply its relation at $g^{-1}$ and take adjoints; [A1] and [A3] give $\pi (g)S^{\ast }=S^{\ast }\pi (g)$. If $T:H\to K$ intertwines $\pi$ with $\rho$, the same operation gives $\pi (g)T^{\ast }=T^{\ast }\rho (g)$, so $T^{\ast }$ intertwines in the reverse direction.

Let $A=A^{\ast }$ be a bounded self-intertwiner of $\pi$. By [A9], $A$ is normal. If $\sigma (A)=\{\lambda \}$, the coordinate function on this singleton equals the constant $\lambda$; [A8] then gives $A=z(A)=\lambda I_H$. Hence a nonscalar $A$ has two distinct spectral points $\lambda ,\mu$. Choose disjoint nonempty relatively open neighborhoods $U,V$ of them in $\sigma (A)$. By [A7], $E_A(U)$ and $E_A(V)$ are nonzero; [A6] makes them orthogonal projections with $E_A(U)E_A(V)=E_A(\mathrm{\varnothing })=0$, so $P:=E_A(U)$ is nonzero and not $I_H$: if $P=I_H$, then $E_A(V)=P{E}_A(V)=0$, contrary to [A7].

Since $A$ belongs to the commutant and is self-adjoint, each $\pi (g)$ commutes with both $A$ and $A^{\ast }$. By [A4] and [A5], it commutes with $P={\mathbf{1}}_U(A)$. The range of the orthogonal projection $P$ is nonzero and proper, and is closed because $\mathrm{ran}P=\ker (I-P)$ for an idempotent bounded operator; commutation gives $\pi (g)\mathrm{ran}P\subseteq \mathrm{ran}P$, and applying the same inclusion to $g^{-1}$ gives equality. This contradicts [A2]. Therefore every bounded self-adjoint self-intertwiner of $\pi$ is scalar.

If $S$ is any bounded self-intertwiner, then by step 1.1 its adjoint also intertwines. The operators $A=(S+S^{\ast })\mathrm{/}2$ and $B=(S-S^{\ast })\mathrm{/}(2i)$ are self-adjoint self-intertwiners, so step 2.1 makes both scalar. Since $S=A+iB$, $S$ is scalar.

Let $T:H\to K$ be a nonzero bounded intertwiner. By step 1.1, $T^{\ast }$ intertwines in reverse, and $T^{\ast }T$ is a bounded self-intertwiner of $\pi$. Step 3.1 gives $T^{\ast }T=c{I}_H$ for some $c\in \mathbb{C}$.

The adjoint identities make $T^{\ast }T$ self-adjoint, hence $c$ is real. For any $x$, $\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=⟨Tx,Tx⟩=⟨x,T^{\ast }Tx⟩=⟨x,cx⟩=c\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by [A12]; since $T\mathrm{\ne }0$, this forces $c>0$. Let $s=\sqrt{c}>0$ by [A10] and set $U=s^{-1}T$. Then $\mathrm{\parallel }Ux{\mathrm{\parallel }}^2=c^{-1}\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ for every $x$, so the nonnegative norms are equal, $U$ is an isometry, and it still intertwines.

The range of $U$ is closed: if $U{x}_n$ converges, the isometry identity makes $(x_n)$ Cauchy, and completeness [A13] gives a limit whose image is the given range limit. Its range is nonzero because $H\mathrm{\ne }\{0\}$ and $U$ is an isometry. The intertwining identity and surjectivity of each $\pi (g)$ give $\rho (g)\mathrm{ran}U=U\pi (g)H=\mathrm{ran}U$, so the range is invariant. By irreducibility of $\rho$ it is all of $K$; thus $U$ is a unitary intertwiner and $\pi ,\rho$ are unitarily equivalent.

If the irreducible representations are inequivalent, a nonzero bounded intertwiner would produce the unitary equivalence in step 6.1, a contradiction. The self-intertwiner assertion is step 3.1.

Fix $g_0\in G$. By strong continuity, the orbit map for $\xi$ is continuous at $g_0$, so $\mathrm{\parallel }\pi (g)\xi -\pi (g_0)\xi \mathrm{\parallel }\to 0$ as $g\to g_0$.

Linearity in the first argument and Cauchy–Schwarz give $\mathrm{\mid }{c}_{\xi ,\eta }(g)-c_{\xi ,\eta }(g_0)\mathrm{\mid }=\mathrm{\mid }⟨\pi (g)\xi -\pi (g_0)\xi ,\eta ⟩\mathrm{\mid }\le \mathrm{\parallel }\pi (g)\xi -\pi (g_0)\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }\to 0$ by step 1.1. By [A3] this is continuity of the coefficient at the arbitrary point $g_0$, hence on $G$. This also holds when $\eta =0$.

Every complex-linear norm isometry $U:H\to H$ preserves the inner product. For $z=⟨u,v⟩$, expansion using [A4] gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2-\mathrm{\parallel }u-v{\mathrm{\parallel }}^2=4\mathrm{Re}z$ and $\mathrm{\parallel }u+iv{\mathrm{\parallel }}^2-\mathrm{\parallel }u-iv{\mathrm{\parallel }}^2=4\mathrm{Im}z$. Since $U$ is complex-linear and preserves norms, these identities give equality of the real and imaginary parts of $⟨Uu,Uv⟩$ and $⟨u,v⟩$. In particular, for every $h\in G$, $⟨\pi (h)u,v⟩=⟨u,\pi (h^{-1})v⟩,$ because $\pi (h^{-1})$ is the inverse of $\pi (h)$.

Cauchy–Schwarz and the isometry property give, for every $g\in G$, $\mathrm{\mid }{c}_{\xi ,\eta }(g)\mathrm{\mid }=\mathrm{\mid }⟨\pi (g)\xi ,\eta ⟩\mathrm{\mid }\le \mathrm{\parallel }\pi (g)\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }=\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }\mathrm{.}$ If either vector is zero, this also says directly that the coefficient is identically zero.

Fix $\epsilon >0$. By orbit continuity at $e$, choose an identity neighbourhood $U$ such that $\mathrm{\parallel }\pi (h)\xi -\xi \mathrm{\parallel }<\epsilon \mathrm{/}(1+\mathrm{\parallel }\eta \mathrm{\parallel })$ for every $h\in U$. If $(x,y)\in L_U$, then $h=x^{-1}y\in U$ and $y=xh$. The homomorphism law, linearity in the first argument, and [A5] give $\mathrm{\mid }{c}_{\xi ,\eta }(y)-c_{\xi ,\eta }(x)\mathrm{\mid }=\mathrm{\mid }⟨\pi (x)(\pi (h)\xi -\xi ),\eta ⟩\mathrm{\mid }\le \mathrm{\parallel }\pi (h)\xi -\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }<\epsilon \mathrm{.}$ The same $U$ works for every $x$, so this is uniform continuity for the left uniformity $L_U$.

Again fix $\epsilon >0$. Orbit continuity for $\eta$ gives an identity neighbourhood $V$ such that $\mathrm{\parallel }\pi (k)\eta -\eta \mathrm{\parallel }<\epsilon \mathrm{/}(1+\mathrm{\parallel }\xi \mathrm{\parallel })$ for every $k\in V$. By [A7], shrink to an identity neighbourhood $U$ such that $h\in U$ implies $h^{-1}\in V$. If $(x,y)\in R_U$, put $h=y{x}^{-1}\in U$, so $y=hx$. By step 1.1 and [A5], $\mathrm{\mid }{c}_{\xi ,\eta }(y)-c_{\xi ,\eta }(x)\mathrm{\mid }=\mathrm{\mid }⟨\pi (x)\xi ,\pi (h^{-1})\eta -\eta ⟩\mathrm{\mid }\le \mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\pi (h^{-1})\eta -\eta \mathrm{\parallel }<\epsilon \mathrm{.}$ This $U$ works for every $x$, proving uniform continuity for the right uniformity $R_U$.

Steps 1.3 and 2.1 give the entourage condition in [A9] for every metric entourage of $\mathbb{C}$, hence both asserted uniform continuities by [A10]. If $H=\{0\}$, or if either coefficient vector is zero, the function is identically zero and all conclusions hold. No commutativity, local compactness, Haar measure, or choice is used.

For any complex-linear isometry $U:H\to H$, expanding the squared norms using [A4] gives $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2-\mathrm{\parallel }x-y{\mathrm{\parallel }}^2=4\mathrm{Re}⟨x,y⟩$ and $\mathrm{\parallel }x+iy{\mathrm{\parallel }}^2-\mathrm{\parallel }x-iy{\mathrm{\parallel }}^2=4\mathrm{Im}⟨x,y⟩$. Since $U$ is linear and norm-preserving, the two differences are unchanged when $x,y$ are replaced by $Ux,Uy$. Their real and imaginary parts therefore agree, so $⟨Ux,Uy⟩=⟨x,y⟩$. In particular each $\pi (g)$ preserves the inner product.

Fix $n\ge 1$, elements $g_1,\dots ,g_n\in G$, and scalars $c_1,\dots ,c_n\in \mathbb{C}$, and put $v_i=\pi (g_i)\xi$. The homomorphism law and step 1.1 yield $\phi (g_i^{-1}{g}_j)=⟨\pi (g_i{)}^{-1}\pi (g_j)\xi ,\xi ⟩=⟨v_j,v_i⟩$. Hence the positive-semidefinite quadratic form from [A1] is ${\sum }_{i,j=1}^n\overline{c_i}{c}_j\phi (g_i^{-1}{g}_j)={\parallel {\sum }_{j=1}^n{c}_j{v}_j\parallel }^2\ge 0$. The calculation includes repeated group elements, zero coefficients, and $\xi =0$; when $\xi =0$ the value is zero.

By [A2], $\phi =c_{\xi ,\xi }$ is continuous. Step 2.1 proves its positive-type matrix test for every allowed finite list, so [A1] gives that $\phi$ is of positive type. The homomorphism law implies $\pi (e)=I$; therefore $\phi (e)=⟨\xi ,\xi ⟩=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$, including when $\xi =0$.

The sums defining $B_{\phi }(f,h)$ are finite because $f$ and $h$ have finite support, and the formula is linear in $f$ and conjugate-linear in $h$, hence sesquilinear with the convention in [A2].

If $f=0$, then $B_{\phi }(f,f)=0$; otherwise list its finite support as $g_1,\dots ,g_n$ and put $a_i=f(g_i)$. By [A1] and reindexing the finite sum, $B_{\phi }(f,f)={\sum }_{i,j}{a}_i\overline{a_j}\phi (g_j^{-1}{g}_i)={\sum }_{i,j}\overline{a_i}\phi (g_i^{-1}{g}_j)a_j\ge 0$. Thus every diagonal value is real and nonnegative.

For any $f,h$ and $z\in \mathbb{C}$, step 1.2 applied to $f+zh$ shows $B_{\phi }(f+zh,f+zh)\in \mathbb{R}$. Expanding by step 1.1, the diagonal terms are real and the cross term is $z{B}_{\phi }(h,f)+\bar{z}{B}_{\phi }(f,h)$; its being real for $z=1$ and $z=i$ implies $B_{\phi }(h,f)=\overline{B_{\phi }(f,h)}$. Hence the form is Hermitian.

Put $a=B_{\phi }(f,f)$, $b=B_{\phi }(h,h)$ and $c=B_{\phi }(f,h)$. By steps 1.2 and 2.1, $B_{\phi }(f+zh,f+zh)=a+2\mathrm{Re}(z\bar{c})+\mathrm{\mid }z{\mathrm{\mid }}^{2}b\ge 0$ for every $z\in \mathbb{C}$. If $b>0$, take $z=-c\mathrm{/}b$ to obtain $\mathrm{\mid }c{\mathrm{\mid }}^2\le ab$. If $b=0$ and $c\mathrm{\ne }0$, take $z=-tc\mathrm{/}\mathrm{\mid }c\mathrm{\mid }$ with $t>a\mathrm{/}(2\mathrm{\mid }c\mathrm{\mid })$; the displayed quantity is $a-2t\mathrm{\mid }c\mathrm{\mid }<0$, a contradiction. Thus $\mathrm{\mid }{B}_{\phi }(f,h){\mathrm{\mid }}^2\le B_{\phi }(f,f)B_{\phi }(h,h)$ in all cases.

By step 3.1, every $f\in N_{\phi }$ satisfies $B_{\phi }(f,h)=0$ for every $h$. This radical property makes $N_{\phi }$ a complex linear subspace, since sums of null vectors and scalar multiples remain null. Changing either representative by an element of $N_{\phi }$ leaves $B_{\phi }$ unchanged. The induced form on the quotient is positive definite: if $B_{\phi }([f],[f])=0$, then $f\in N_{\phi }$ and $[f]=0$. It is therefore an inner product under [A2], including the zero quotient when $\phi =0$.

For ${\delta }_x$ and ${\delta }_y$, only the summand with first index $x$ and second index $y$ survives, so $B_{\phi }({\delta }_x,{\delta }_y)=\phi (y^{-1}x)$.

For fixed $g\in G$, the support of $L_{g}f$ is $g\mathrm{supp}(f)$, so $L_g$ preserves finite support and is complex-linear. The group law gives $L_e=I$, $L_g{L}_h=L_{gh}$, and $L_g^{-1}=L_{g^{-1}}$.

For finitely supported $f,h$, [A2] gives $B_{\phi }(L_{g}f,L_{g}h)={\sum }_{a,b\in G}f(g^{-1}a)\overline{h(g^{-1}b)}\phi (b^{-1}a)\mathrm{.}$ Only finitely many terms are nonzero. Substitute $a=gx$ and $b=gy$; then $b^{-1}a=(gy{)}^{-1}(gx)=y^{-1}x$, so the sum becomes ${\sum }_{x,y\in G}f(x)\overline{h(y)}\phi (y^{-1}x)=B_{\phi }(f,h)$. Thus left translation preserves the whole form.

If $f,h\in N_{\phi }$, [A1] makes all four terms in $B_{\phi }(f+h,f+h)$ vanish, and sesquilinearity gives $B_{\phi }(\lambda f,\lambda f)=\mathrm{\mid }\lambda {\mathrm{\mid }}^2{B}_{\phi }(f,f)=0$ for every $\lambda \in \mathbb{C}$. Hence $N_{\phi }$ is a complex linear subspace. Step 2.1 implies $L_g{N}_{\phi }\subseteq N_{\phi }$; applying it to $g^{-1}$ and using step 1.1 gives $L_g{N}_{\phi }=N_{\phi }$. Therefore if $f-h\in N_{\phi }$, then $L_{g}f-L_{g}h=L_g(f-h)\in N_{\phi }$, so $[f]\mapsto [L_{g}f]$ is well-defined on the quotient. Its inverse is induced by $L_{g^{-1}}$, and the identities in step 1.1 descend to the quotient.

By [A13] and [A14], AC supplies Countable Choice. Hence [A7] gives the Hilbert completion $H_{\phi }$ of $Q_{\phi }$ with a dense linear isometry $\kappa :Q_{\phi }\to H_{\phi }$. By [A8], this completion is Banach. The quotient norm used here is the norm induced by its inner product, as in [A4]–[A5].

For $g\in G$, set $T_g[f]=[L_{g}f]$ on $Q_{\phi }$. It is well-defined by [A2] and complex-linear by the pointwise formula for $L_g$ and [A10]. For finitely supported $f,h$, reindex the finite sum in [A1] by $a=gx$ and $b=gy$; since $(gy{)}^{-1}(gx)=y^{-1}x$, this gives $B_{\phi }(L_{g}f,L_{g}h)=B_{\phi }(f,h)\mathrm{.}$ Thus $T_g$ preserves the quotient inner product and norm. The group laws for left translation give $T_e=I$, $T_g{T}_h=T_{gh}$, and $T_{g^{-1}}=T_g^{-1}$.