so they are not in $\cl K.\big(L^2(G)\big)=\pi_0\big(C(G)\rtimes_\lambda G\big)$, but are in fact in its multiplier algebra. This is not a problem, as a state or representation on $C(G)\rtimes_\lambda G$ has a unique extension to its multiplier algebra, so it is fully determined on these elements. If one chose $C^*(U_G\cup T_{L^\infty(G)})$ as the field algebra instead of $C(G)\rtimes_\lambda G$, then this would contain many inappropriate representations, e.g. covariant representations for $\lambda:G\to\aut C(G)$ where the implementing unitaries are discontinuous w.r.t. $G$. Thus, our choice for the field algebra of a link remains as ${C(G)\rtimes_\lambda G}\cong\cl K.\big(L^2(G)\big)$. Clearly, as the momentum operators $P_X$ are unbounded, they cannot be in any C\*-algebra, but they are obtained from $U_G$ in the generalized Schrödinger representation.

We combine these C\*-algebras into the kinematic field algebra, which is $${\mathfrak A}_{\Lambda} :=
{\mathfrak F}_{\Lambda} \otimes \bigotimes_{\ell\in\Lambda^1}\big(C(G)\rtimes_\lambda G\big)$$ which is is well–defined as $\Lambda^1$ is finite, and the cross–norms are unique as all algebras in the entries are nuclear. (If $\Lambda^1$ is infinite, the tensor product ${\mathop{\bigotimes}\limits_{\ell\in\Lambda^1}\big(C(G)\rtimes_\lambda G\big)}$ is undefined, as $C(G)\rtimes_\lambda G$ is nonunital). Moreover, since $C(G)\rtimes_\lambda G\cong\cl K.\big(L^2(G)\big)$ and $\cl K.(\cl H._1)\otimes\cl K.(\cl H._2)\cong\cl K.(\cl H._1\otimes\cl H._2),$ it follows that $$\mathop{\bigotimes}\limits_{\ell\in\Lambda^1}\big(C(G)\rtimes_\lambda G\big)\cong\cl K.\big(
\mathop{\otimes}\limits_{\ell\in\Lambda^1}L^2(G)\big)\cong\cl K.(\cl L.)$$ as $\Lambda^1$ 