Unitary Representations, Positive Type and GNS — Examples

For $n\ge 1$, $g_i^{-1}{g}_j=e$ exactly when $g_i=g_j$; partitioning the finite list by its distinct values gives ${\sum }_{i,j=1}^n\overline{c_i}{c}_j{\delta }_e(g_i^{-1}{g}_j)={\sum }_{h\in \{g_1,\dots ,g_n\}}{\mid {\sum }_{j:g_j=h}{c}_j\mid }^2\ge 0$. This includes repeated group elements and zero coefficients.

Every orbit map $g\mapsto \sigma (g)v$ is continuous because its domain is discrete, so $\sigma$ is strongly continuous. Choose an orthonormal basis $(e_k{)}_{k<d}$ of $V$. Since $V\mathrm{\ne }\{0\}$, its dimension $d$ is nonzero; write $d=\mathrm{succ}(m)$. From $1=\mathrm{succ}(0)$ and the recursive addition law, induction gives $1+m=\mathrm{succ}(m)$, so $1\le d$ by the natural-order definition. Its canonical real scalar $d_{\mathbb{R}}$ is positive, and its image $d_{\mathbb{C}}$ in $\mathbb{C}$ is nonzero.

The function ${\delta }_e$ is continuous because $\mathrm{\Gamma }$ is discrete, and ${\delta }_e(e)=1$. Step 1.1 proves its finite-matrix test, so ${\delta }_e$ is of positive type.

For $k<d$ set ${\varphi }_k(g)=⟨\sigma (g)e_k,e_k⟩$. By [F5], each ${\varphi }_k$ is continuous and of positive type. If $\sigma (g)e_k={\sum }_{\mathrm{\ell }<d}{a}_{\mathrm{\ell }k}(g)e_{\mathrm{\ell }}$, orthonormality gives ${\varphi }_k(g)=a_{kk}(g)$; the trace definitions in [F7] consequently give $\mathrm{tr}(\sigma (g))={\sum }_{k<d}{\varphi }_k(g)$.

By [F8], $\sigma (e)=I_V$; its matrix in this basis is the $d$ by $d$ identity, so [F6, F7] give $\mathrm{tr}(\sigma (e))=d$. Hence ${\chi }_{\sigma }(e)=d_{\mathbb{C}}\mathrm{/}{d}_{\mathbb{C}}=1$ by the field embedding [F10]. Also ${\chi }_{\sigma }$ is continuous, since it is a function from a discrete domain.

For any finite test $(g_i,c_i{)}_{i=1}^n$ with $n\ge 1$, step 2.2 gives ${\sum }_{i,j=1}^n\overline{c_i}{c}_j{\chi }_{\sigma }(g_i^{-1}{g}_j)=d_{\mathbb{R}}^{-1}{\sum }_{k<d}({\sum }_{i,j=1}^n\overline{c_i}{c}_j{\varphi }_k(g_i^{-1}{g}_j))$. Each parenthesized value is a nonnegative real by [F5]; their finite sum is nonnegative, and $d_{\mathbb{R}}^{-1}>0$ by [F9]. The embedding in [F10] identifies this real nonnegative value with the displayed complex quadratic form, so [F1] shows that ${\chi }_{\sigma }$ is of positive type.

For finitely supported $f,h:\mathrm{\Gamma }\to \mathbb{C}$, [F2] gives $B_{{\delta }_e}(f,h)={\sum }_{x,y}f(x)\overline{h(y)}{\delta }_e(y^{-1}x)={\sum }_{x}f(x)\overline{h(x)}$. This is positive definite: if $f\mathrm{\ne }0$, a nonzero coordinate contributes a strictly positive squared modulus while every other term is nonnegative. Hence the GNS null space is $\{0\}$, and the quotient is this finite-support inner-product space.

Assume AC. By [F11], Countable Choice holds; [F12] gives a Hilbert completion $\widehat{E}$ of the inner-product space in step 3.2, with the finite-support functions embedded densely. On that dense subspace define $L_{g}f(x)=f(g^{-1}x)$. This is linear, and reindexing $x=gy$ gives $⟨L_{g}f,L_{g}h⟩={\sum }_{x}f(g^{-1}x)\overline{h(g^{-1}x)}={\sum }_{y}f(y)\overline{h(y)}=⟨f,h⟩$. Thus $L_g$ is bounded with norm bound one; by [F13] it extends uniquely to a linear contraction ${\lambda }_{\mathrm{\Gamma }}(g)$ on $\widehat{E}$. This is the completion stage in the GNS construction for ${\delta }_e$.

On the dense finite-support subspace, $L_g{L}_h=L_{gh}$ and $L_g{L}_{g^{-1}}=L_{g^{-1}}{L}_g=I$. The continuous extensions obey the same identities on $\widehat{E}$; since each extension and its inverse are contractions, each is an isometry and hence unitary. Therefore $g\mapsto {\lambda }_{\mathrm{\Gamma }}(g)$ is strongly continuous because $\mathrm{\Gamma }$ is discrete. The point masses $({\delta }_x{)}_{x\in \mathrm{\Gamma }}$ are orthonormal and their span is dense in $\widehat{E}$, so they form a complete orthonormal family. By [F14] the Fourier coefficient map $U:\widehat{E}\to {\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})$ is unitary and sends ${\delta }_x$ to the standard coordinate vector $e_x$. For each $g,x$, $U{\lambda }_{\mathrm{\Gamma }}(g){\delta }_x=e_{gx}$, which is the standard left translation of $e_x$; density and continuity imply that $U$ intertwines ${\lambda }_{\mathrm{\Gamma }}$ with the left regular representation on ${\mathrm{\ell }}^2(\mathrm{\Gamma })$. Finally ${\lambda }_{\mathrm{\Gamma }}(g){\delta }_e={\delta }_g$ and $⟨{\lambda }_{\mathrm{\Gamma }}(g){\delta }_e,{\delta }_e⟩=⟨{\delta }_g,{\delta }_e⟩={\delta }_e(g)$; the orbit spans a dense subspace, so ${\delta }_e$ is cyclic and this is its GNS representation.

The positivity and normalization proofs in steps 1.1–3.1 use no choice. AC is used only through Countable Choice in [F12]–[F14] for the Hilbert completion, bounded extensions, and the standard ${\mathrm{\ell }}^2$ coordinate identification. The extensions and Fourier coefficient map are unique, so the action and unitary equivalence require no further choice. All three claims are established.

The homomorphism law gives $\chi (e)=\chi (e{)}^2$. Since $\mathrm{\mid }\chi (e)\mathrm{\mid }=1$, the value is nonzero, so cancellation gives $\chi (e)=1$. Applying the homomorphism law to $g^{-1}g=e$ gives $\chi (g^{-1})=\chi (g{)}^{-1}=\overline{\chi (g)}$ by [A2]. [A1, A2] 2.1 For any $n\ge 1$, any $g_1,\dots ,g_n\in G$ (including repetitions), and any $c_1,\dots ,c_n\in \mathbb{C}$, the matrix quadratic form is ${\sum }_{i,j=1}^n\overline{c_i}\chi (g_i^{-1}{g}_j)c_j={\mid {\sum }_{j=1}^n{c}_j\chi (g_j)\mid }^2\ge 0.$ Thus the matrix is positive semidefinite. The given continuity of $\chi$ and $\chi (e)=1$ show $\phi \in P_1(G)$; zero coefficients are covered by the same identity. [A2, A6, step 1.1, algebra] 2.2 The pairing $⟨z,w⟩=z\bar{w}$ is the stated complex inner product, its induced norm is $\mathrm{\mid }z\mathrm{\mid }$, and the complex plane is complete; hence $\mathbb{C}$ is a complex Hilbert space by [A4]. For each $g$, multiplication by $\chi (g)$ is complex-linear by [A5] and is an isometry because $\mathrm{\mid }\chi (g)z\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$; multiplication by $\chi (g^{-1})$ is its inverse. The homomorphism law makes ${\pi }_{\chi }$ a representation. For fixed $z$ and $g_0$, $\mathrm{\parallel }{\pi }_{\chi }(g)z-{\pi }_{\chi }(g_0)z\mathrm{\parallel }=\mathrm{\mid }\chi (g)-\chi (g_0)\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }⟶0$ as $g\to g_0$, by continuity in [A3]. Thus it is strongly continuous. Since ${\pi }_{\chi }(e)1=1$, the orbit span contains $1$ and is all of $\mathbb{C}$; also $⟨{\pi }_{\chi }(g)1,1⟩=\chi (g)=\phi (g)$. [A3, A4, A5, step 1.1] 2.3 Define $L_{\chi }(f)={\sum }_{x}f(x)\chi (x)$ on finitely supported $f$. Using [A2] and step 1.1, $B_{\chi }(f,h)={\sum }_{x,y}f(x)\overline{h(y)}\chi (y{)}^{-1}\chi (x)=L_{\chi }(f)\overline{L_{\chi }(h)}\mathrm{.}$ Therefore $B_{\chi }(f,f)=\mathrm{\mid }{L}_{\chi }(f){\mathrm{\mid }}^2$ and $N_{\chi }=\ker L_{\chi }$. The map $[f]\mapsto L_{\chi }(f)$ is a well-defined linear isometry from the quotient onto $\mathbb{C}$: it is onto because $L_{\chi }(z{\delta }_e)=z$. For each $g$, $L_{\chi }({\delta }_g)=\chi (g)$ and $L_{\chi }(\chi (g){\delta }_e)=\chi (g)$, so injectivity on the quotient gives $[{\delta }_g]=\chi (g)[{\delta }_e]$. Left translation satisfies $L_{\chi }(L_{g}f)=\chi (g)L_{\chi }(f)$, in agreement with the scalar action ${\pi }_{\chi }(g)$ from [A7]. [A2, A6, A7, step 1.1, algebra] 3.1 By step 2.1, $\phi$ meets the input hypotheses of the GNS construction in [A7]. Its canonical triple is cyclic, strongly continuous, and has diagonal coefficient $\phi$. Step 2.2 gives the same properties and coefficient for $({\pi }_{\chi },\mathbb{C},1)$. The pointed uniqueness theorem in [A7] therefore gives the unique unitary intertwiner carrying the vector $1$ to the canonical GNS vector. [A5, A7, step 2.1, step 2.2] 4.1 The only choice used is AC, declared in the example, through the GNS completion/action and pointed uniqueness inputs in [A7]; [A8] identifies the precise reduction AC $\Rightarrow$ DC $\Rightarrow$ Countable Choice used for completion and bounded extensions. The finite matrix, scalar representation, and quotient calculations in steps 1.1–3.1 use no choice.

The additive real field gives the group laws for $(\mathbb{R},+)$, and addition and negation are continuous by the algebra of continuous real maps on a metric space, so $G$ is a topological group.

Let $h_0(s)=e^{-s^2\mathrm{/}4}$; evenness, reflection of the negative improper tail, and the mixed-integral convention give ${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }\mathrm{/}2$, so the substitution $s=2x$ yields ${\int }_0^{\mathrm{\infty }}{h}_0(s)ds=\sqrt{\pi }$.

For $R>0$, FTC gives ${\int }_0^{R}s{e}^{-s^2\mathrm{/}4}ds=2(1-e^{-R^2\mathrm{/}4})$, whose limit is $2$; both $h_0$ and $h_1(s):=\mathrm{\mid }s\mathrm{\mid }{e}^{-s^2\mathrm{/}4}$ are continuous and nonnegative on $[0,\mathrm{\infty })$, and their half-line improper integrals are respectively $\sqrt{\pi }$ and $2$.

For fixed $t$, $q_t(s):=e^{-s^2\mathrm{/}4}{e}^{its}$ has derivative $q_t^{\mathrm{\prime }}(s)=(-s\mathrm{/}2+it)q_t(s)$ by Euler's identity and the real product and chain rules; its continuous components are integrable on $[-n,n]$, and their Riemann and Lebesgue integrals agree, so componentwise FTC gives $q_t(n)-q_t(-n)={\int }_{-n}^n(-s\mathrm{/}2+it)q_t(s)ds$.

The formula $\mathrm{\mid }{e}^{its}\mathrm{\mid }=1$ makes multiplication by $e^{its}$ a well-defined complex-linear isometry on $L^2(\mathbb{R})$; the exponential addition law gives $\pi (t+u)=\pi (t)\pi (u)$ and $\pi (-t)$ is its inverse, so $\pi$ is a unitary representation.

For $f\in L^2$ and $t_n\to t$, the squared orbit difference has integrand $\mathrm{\mid }{e}^{i{t}_{n}s}-e^{its}{\mathrm{\mid }}^2\mathrm{\mid }f(s){\mathrm{\mid }}^2$, which converges pointwise to zero and is bounded by $4\mathrm{\mid }f(s){\mathrm{\mid }}^2$; dominated convergence gives $\mathrm{\parallel }\pi (t_n)f-\pi (t)f{\mathrm{\parallel }}_2\to 0$, hence $\pi$ is strongly continuous because $\mathbb{R}$ is metrizable.

The algebraic orbit span is linear, and its norm closure $H_0$ is a closed linear subspace: for $x,y$ in the closure, the metric-closure criterion approximates them by span elements within $\epsilon \mathrm{/}3$, whose sum is within $2\epsilon \mathrm{/}3$ of $x+y$; scalar multiples follow from norm homogeneity. Thus [A4] and the closed-subspace completeness theorem make $H_0$ a Hilbert space. The group law sends each orbit vector to another orbit vector, and continuity of $\pi (u)$ and its inverse shows $\pi (u)H_0=H_0$; by construction $\xi \in H_0$ is cyclic.

The half-line comparison turns the values in steps 1.2 and 1.3 into Lebesgue integrals; for either even function $h\in \{h_0,h_1\}$, splitting into positive and negative open half-lines and the null singleton gives ${\int }_{\mathbb{R}}h=2{\int }_{(0,\mathrm{\infty })}h$ by reflection invariance and integral additivity. Thus ${\int }_{\mathbb{R}}w=1$ and ${\int }_{\mathbb{R}}\mathrm{\mid }s\mathrm{\mid }w(s)ds=2\mathrm{/}\sqrt{\pi }<\mathrm{\infty }$.

The functions $s\mapsto e^{its}w(s)$ are measurable and have modulus $w(s)$, so they are integrable; for each fixed $s$, differentiation in $t$ gives ${\mathrm{\partial }}_t(e^{its}w(s))=is{e}^{its}w(s)$, whose modulus is bounded by the integrable majorant $\mathrm{\mid }s\mathrm{\mid }w(s)$. Hence differentiation under the integral sign gives $F^{\mathrm{\prime }}(t)=i{\int }_{\mathbb{R}}s{e}^{its}w(s)ds$ for $F(t):={\int }_{\mathbb{R}}{e}^{its}w(s)ds$.

The boundary terms in step 1.4 tend to zero, and dominated convergence passes the truncated integrals to their full-line integrals because $\mathrm{\mid }{q}_t\mathrm{\mid }=e^{-s^2\mathrm{/}4}$ and $\mathrm{\mid }s{q}_t\mathrm{\mid }=\mathrm{\mid }s\mathrm{\mid }{e}^{-s^2\mathrm{/}4}$ are integrable by step 2.1; therefore $\int s{q}_t(s)ds=2it\int q_t(s)ds$, or $\int s{e}^{its}w(s)ds=2itF(t)$.

Steps 3.1 and 3.2 give $F^{\mathrm{\prime }}(t)=-2tF(t)$, while step 2.1 gives $F(0)=1$; the product and chain rules show $(e^{t^2}F(t){)}^{\mathrm{\prime }}=0$, so applying the real FTC to each component on every compact interval yields $F(t)=e^{-t^2}$ for all $t\in \mathbb{R}$.

The norm identity $\mathrm{\parallel }\xi {\mathrm{\parallel }}_2^2=\int w=1$ holds, and the first-linear $L^2$ pairing gives $⟨\pi (t)\xi ,\xi ⟩=\int e^{its}w(s)ds=F(t)=e^{-t^2}$ by step 4.1; [A8] now gives normalized positive type and identifies $(\pi {\mathrm{\mid }}_{H_0},H_0,\xi )$ with the canonical GNS triple.

AC is propagated through Countable Choice exactly for the L² and measure-theoretic suppliers in [A1] and [A3], and is used directly by the canonical GNS construction and pointed uniqueness in [A8]; the Gaussian integral calculation and the phase representation use no further choice.

Addition on $\mathbb{R}$ is continuous because $\mathrm{\mid }(x+y)-(x_0+y_0)\mathrm{\mid }\le \mathrm{\mid }x-x_0\mathrm{\mid }+\mathrm{\mid }y-y_0\mathrm{\mid }$, and inversion $x\mapsto -x$ preserves distances; hence the usual additive group satisfies [A11]. The maps $t\mapsto t^4$ and $t\mapsto -t^4$ are polynomials and are continuous by [A6]. Composing with the continuous exponential by [A2] and [A7] proves that $f$ is continuous.

Write $t^4=(t^2{)}^2$. If $t^2=0$ then $t^4=0$; otherwise [A8] applied to $t^2$ gives $t^4>0$. Also $(-t{)}^4=t^4$ by finite multiplication, so $f$ is even. By [A3] and strict monotonicity in [A2], $0<f(t)=\exp (-t^4)\le \exp (0)=1$ for all $t$, and $f(0)=1$. Thus $f$ is real-valued, bounded by $1$ in modulus and normalized at the identity.

Put $a=\exp (-1\mathrm{/}16)$ and $b=\exp (-1)$. The matrix on the listed points is $M=\left(\begin{array}{ccc}1& a& b\\ a& 1& a\\ b& a& 1\end{array}\right)$, since $f$ is even. For $c=(1,-2,1)$, direct multiplication gives $c^{\ast }Mc=6-8a+2b$.

Applying [A4] at $x=-1\mathrm{/}16$ gives $a\ge 15\mathrm{/}16$, and at $x=1$ gives $2\le \exp (1)$. By [A10] and [A12], $2=1+1>0$. By [A3], $b=1\mathrm{/}\exp (1)>0$; if $\exp (1)=2$ then $b=1\mathrm{/}2$, while if $2<\exp (1)$ then [A5] gives $b<1\mathrm{/}2$. Hence $b\le 1\mathrm{/}2$, and step 2.1 yields $c^{\ast }Mc\le 6-8(15\mathrm{/}16)+2(1\mathrm{/}2)=-1\mathrm{/}2<0$. Thus $M$ is not positive semidefinite by [A1], so $f$ is not of positive type.