GNS representations with $\hat{G}$ and $C^*$-algebras

Question

Let $G$ be a commutative discrete group and $\mathcal{A} := l^1(G)$. Given a linear positive functional $\mu$ on $\mathcal{A}$, consider the GNS $(\mathcal{H}_\mu, \pi_\mu, \xi_\mu)$. Define $\mathcal{B} := \overline{\{\pi_\mu(f): f \in \mathcal{A}\}}$, which is a unital commutative $C^*$-algebra. $\mathcal{B}$ isomorphic to some $C(X)$ where $X$ is compact Hausdorff e.g. $X = \hat{\mathcal{B}}$ consider Gelfrands theorem. For any $x \in X$, we can define $\phi_x(f) := f(x)$, which is a character of $C(X)$. It's clear that $\phi_x \circ \pi_\mu \in \hat{\mathcal{A}}$(characters of $A$). I am a bit confused how $\phi_x \circ \pi_\mu$ gives a character of $\hat{G}$(group homomorphisms from $G$ into $\mathbb{T}$)? Is this what this means,consider $\delta_g \in l^1(G)$, define $$\chi(g) := \phi_x(\pi_\mu(\delta_g))$$It's immediately a group homomorphism because of the homomorphism properties on $\phi_x$ and $\pi_\mu$.Or is that wrong reasoning? Also I am trying to find an injective continuous $X$ into $\hat{G}$, I was thinking of using the map $x \mapsto \chi_x$. Any feed back will help.

Answer 1

For injectivity - Assume $\chi_x = \chi_y$. Then $$\chi_x(g) = \phi_x(\pi_\mu(\delta_g)) = \phi_y(\pi_\mu(\delta_g)) = \chi_y(g)\tag{1}$$ for all $g\in G$.

Now, Let $V\subseteq \mathcal{A}$ be the linear span of $\{\delta_g: g\in G\}$. Then $V$ is a dense subspace of $\mathcal{A}$ in the $l^1$-norm. For all $v\in V$ we have from $(1)$ by linearity of $\phi_x\circ\pi_\mu, \phi_y\circ\pi_\mu$ that $$\phi_x(\pi_\mu(v)) = \phi_y(\pi_\mu(v))\tag{2}$$ Since $\|\pi_\mu\|\le 1$ it follows by continuity of $\phi_x\circ\pi_\mu, \phi_y\circ\pi_\mu$ that $\phi_x(\pi_\mu(v))=\phi_y(\pi_\mu(v))$ for all $v\in \mathcal{A}$.

Let $W=\pi_\mu(\mathcal{A})\subseteq \mathcal{B}$. Since $(2)$ holds for all $v\in\mathcal{A}$, it follows that $$\phi_x(w)=\phi_y(w)\tag{3}$$ for all $w\in W$. Since $W$ is a dense subspace of $\mathcal{B}$ and $\phi_x, \phi_y$ are both continuous (being characters) it follows that $\phi_x=\phi_y$ on $\mathcal{B}$. Therefore they are the same character and $x=y$.

For continuity of $x\mapsto \chi_x$ here's a hint: the topology on $\hat{G}$ is the compact-open topology. A sub-base is given by taking a compact $C\subseteq G$ and an open $U\subseteq\Bbb{C}$ and considering the set $p(C,U)$ of all $f\in\hat{G}$ such that $f(C)\subseteq U$ open. Since $G$ is a discrete group, we get a reprieve. A subset of a discrete topological space is compact if and only if it is finite. If $C$ is finite, we can write $p(C,U)=\cap_{c\in C}\, p(\{s\}, U)$ which is a finite intersection (of open sets). Therefore a sub-base is given by $p(\{s\}, U)$ for all $s\in G$ and $U\subseteq\Bbb{C}$ open. The topology on $\hat{G}$ is the topology of pointwise convergence. I'll let you try to complete the proof that the map $x\mapsto \chi_x$ is contiuous from $X$ to $\hat{G}$.