Let $G$ be a locally compact topological group (or more generally a localic group). Is there an infinity topos which classify principal $G$ bundles ?


More precisely, is there an $\infty$-topos $BG$ such that for every localic topos $\mathcal{L}$ the category of geometric morphism from $\mathcal{L}$ to $BG$ is equivalent to the category of $G$ principal bundle over $\mathcal{L}$, where a $G$-principal bundle over $\mathcal{L}$ is a locale $\mathcal{X}$ endowed with a $G$ action and an invariant map $p: \mathcal{X} \rightarrow \mathcal{L}$


such that:


1)$p$ is an open surjection.


2) The canonical map $\mathcal{X} \times G \rightarrow \mathcal{X} \times\_{\mathcal{L}} \mathcal{X}$ is an isomorphism.


I am especially interested in the cases where $G= \mathbb{U}$ (the group of complex number of module $1$) and $G=\mathbb{R}$.


Of course, if $G$ is pro-discrete, then the answer is yes: it suffice to consider the infinity topos associated to the $1$-topos of continuous $G$ set. In the general case, one should look for an infinity topos of spaces endowed with a $G$ action (up to homotopy), but my knowledge of homotopy theory is not enough to see if this trivially work/does not work or if it is a difficult question...



classifying $\infty$-toposes for topological/localic groups?
The projection map $p: \mathbf{R} \rightarrow \ast$ induces a fully faithful embedding of sheaf categories $p^{\ast}: Shv(\ast) \rightarrow Shv( \mathbf{R} )$. This is equally true for sheaves of sets, sheaves of spaces, or other variants. 
It follows that for any topos (or $\infty$-topos) $\mathcal{X}$, the ($\infty$-)category of geometric morphisms from $\ast$ to $\mathcal{X}$ embeds fully faithfully into the ($\infty$-)category of geometric morphisms from $\mathbf{R}$ into $\mathcal{X}$. As a consequence, there can't be a classifying topos (or $\infty$-topos) for $U(1)$-bundles in the sense you describe (the category of $U(1)$-bundles on a point does not embed fully faithfully in the category of $U(1)$-bundles on $\mathbf{R}$). The same counterexample works if you replace $U(1)$ by any topological group $G$ for which there exists a nonconstant continuous map $\mathbf{R} \rightarrow G$.


