Note: Gx Developing Pair

$$ \newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} $$

In the study of geometric structures on manifolds, there are multiple ways to work with the data of a $(G,X)$ structure. Two common ones are a $(G,X)$ atlas and a $(G,X)$ developing pair. In this short note we describe how to construct an atlas from a developing pair. To go the other way, see Goldman’s paper, Geometric Structures and Varieties of Representations or Goldman’s book Geometric Structures Chapter 5.

Let $M$ be a manifold equipped with a developing pair into $(G,X)$: that is, an immersion $\dev\colon \widetilde{M}\to X$ and a holonomy homomorphism $\hol\colon \pi_1(M)\to G$.

To construct a $(G,X)$ atlas for $M$ we can proceed as follows. About each $x\in M$ we can choose an open neighborhood $U_x$ satisfying the following properties:

• $U_x$ is evenly covered by $\pi\colon\widetilde{M}\to M$.
• At least one (and hence all) of the connected components of $\pi^{-1}(U_x)$ are small enough that $\dev$ restricts to a diffeomorphism onto its image.

Let $\mathcal{U}$ denote the open cover $\mathcal{U}={U_x}_{x\in M}$. We promote $\mathcal{U}$ to an $X$-atlas by making these each into a coordinate chart that takes values in $X$.
There’s some choice involved here, and the atlas constructed depends on the choices made, but all turn out to be $(G,X)$ atlases (and are all subsets of the same maximal atlas, and so determine the same smooth structure etc etc…) For each $U_x$, choose a particular preimage $\widetilde{U}_x$ under the universal covering map $\pi$, and denote by $\pi_x^{-1}$ the diffeomorphism $U_x\to\widetilde{U}_x$ inverting $\pi$. The chart associated to $U_x$ is then simply $\phi_x:=\dev\circ\pi_x^{-1}\colon U_x\to X$.

Now, we need to show that the atlas ${(U_x,\phi_x)}_{x\in X}$ is in fact a $(G,X)$ atlas, which involves showing that all transition maps are in $G$.

Let $x,y\in M$ be two points so that their corresponding neighborhoods $U_x$ and $U_y$ intersect. Then if $U=U_x\cap U_y$ we may look at two separate images of $U$ under its coordinate charts: $$\phi_x(U)=\dev\circ\pi_x^{-1}(U)$$ $$\phi_y(U)=\dev\circ\pi_y^{-1}(U)$$

There are two options here: either, we chose the inverses of $\pi$ in each case so that $\pi_x^{-1}(U)=\pi_y^{-1}(U)$, or we did not. In the first case, following these maps with $\dev$ makes $\phi_x(U)$ and $\phi_y(U)$ also coincide, and so they are trivially related by an element of $G$ (the identity). Otherwise, the sets $\pi_x^{-1}(U)$ and $\pi_y^{-1}(U)$ are disjoint ($U_x$ and $U_y$ were evenly covered neighborhoods) and are related to eachother in $\widetilde{M}$ by some deck transformation (they are each preimages of the same set $U\subset M$ under the universal covering). Thus we may write $\pi_x^{-1}(U)=\gamma.\pi_y^{-1}(U)$, where $\gamma.$ is the action of this element of the universal cover. Applying the developing map to both sides of this we see $$\dev\circ\pi_x^{-1}(U)=\dev(\gamma.\pi_y^{-1}(U))=\hol(\gamma).\dev\circ\pi_y^{-1}(U)$$ Where the last equality uses the interaction of $\dev$ and $\hol$. Since $\hol$ takes values in $G$ this tells us there is some element of $g$ such that $$\phi_x(U)=g.\phi_y(U)$$ and so our $X$-atlas is actually a $(G,X)$ atlas, as required.