Proposition 6.3.10.label The set $\OM(A)$ is locally compact with respect to $\s(A^{*},A)$-topology and compact if $A$ is unital.
Proof. Let $\OM'(A)=\OM(A)\cup\curl{0}$ then $\OM'(A)\suf S^{*}$ by Proposition 6.3.9. For each $x,y\in A$, the evaluation
\begin{align*}\Phi_{x,y}:A^{*}\to\com&&\om\mto \om(xy)-\om(x)\om(y)\end{align*}
is $\s(A^{*},A)$-continuous thus
\begin{align*}\OM'(A)=\caps{}{x,y\in A}\Phi_{x,y}\inv(\curl{0})=\caps{}{x,y\in A}\curl{\om\in \OM'(A):\om(xy)-\om(x)\om(y)=0}\end{align*}
which is closed. Thus $\OM'(A)$ is compact by Alaoglu’s Theorem. Since $\curl{0}$ is closed in $\OM'(A)$, $\OM(A)$ is an open subset of the compact space $\OM'(A)$ so it is locally compact. Suppose $A$ is unital then $0$ is isolated in $\OM'(A)$ because $\om(1)=1$ for every $\om\in \OM(A)$ so $\OM(A)$ is closed in $\OM'(A)$; hence compact.$\square$