Note: Gx Developing Pair

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 $\mathrm{dev}:\tilde{M}\to X$ and a holonomy homomorphism $\mathrm{hol}:{\pi }_1(M)\to G$.

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 $\pi :\tilde{M}\to M$.
• At least one (and hence all) of the connected components of ${\pi }^{-1}(U_x)$ are small enough that $\mathrm{dev}$ 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 ${\tilde{U}}_x$ under the universal covering map $\pi$, and denote by ${\pi }_x^{-1}$ the diffeomorphism $U_x\to {\tilde{U}}_x$ inverting $\pi$. The chart associated to $U_x$ is then simply ${\varphi }_x:=\mathrm{dev}\circ {\pi }_x^{-1}:U_x\to X$.

Now, we need to show that the atlas ${(U_x,{\varphi }_x)}_{x\in X}$ 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: $${\varphi }_x(U)=\mathrm{dev}\circ {\pi }_x^{-1}(U)$$ $${\varphi }_y(U)=\mathrm{dev}\circ {\pi }_y^{-1}(U)$$

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 $\mathrm{dev}$ makes ${\varphi }_x(U)$ and ${\varphi }_y(U)$ 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 $\tilde{M}$ 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 $$\mathrm{dev}\circ {\pi }_x^{-1}(U)=\mathrm{dev}(\gamma .{\pi }_y^{-1}(U))=\mathrm{hol}(\gamma ).\mathrm{dev}\circ {\pi }_y^{-1}(U)$$ Where the last equality uses the interaction of $\mathrm{dev}$ and $\mathrm{hol}$. Since $\mathrm{hol}$ takes values in $G$ this tells us there is some element of $g$ such that $${\varphi }_x(U)=g.{\varphi }_y(U)$$ and so our $X$-atlas is actually a $(G,X)$ atlas, as required.