Let $G$ be a topological group, call $\mathcal{H}\in sGrp$ simplicial model of $G$ if there is a homomorphism of topological groups $|\mathcal{H}|\to G$ which is also a homotopy equivalence. In a perfect world, this map would be a homeomorphism, but it does not seem possible since $\mathcal{H}$ should be a Kan complex and hence $|\mathcal{H}|$ should be insanely large.
First, does this definition make sense? If I am not mistaken for any topological group $G$ we can make $S(G)$ into such a model using structure maps of $G$ and an isomorphism $S(G)\times S(G)\cong S(G\times G).$
Second, is there a way to obtain a manageable simplicial model of $S^1$ as a Lie group? Of course, it seems that we can take $S(S^1),$ but is there a way to make it with a countable number of simplices? For example, can we give the Kan complex $\operatorname{Ex}^\infty \mathscr{S}^1$ some kind of group structure making it a simplicial model for $S^1$? Here $\mathscr{S}^1=\Delta^1/\partial\Delta^1$ is the minimal simplicial circle.
Third, is there a way to make things better, if we work with semi-simplicial groups instead? For example, is it true that each Lie group $G$ admits a triangulation that gives rise to a semi-simplicial group $\mathcal{H}$ such that $|\mathcal{H}|\cong_{homeo} G?$