From Developing Pairs to (G,X) Atlases

How a local diffeomorphism and holonomy map produce an atlas.

In the study of geometric structures on manifolds, there are multiple ways to work with the data of a $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} (G,X)$ structure. Two common ones are a $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} (G,X)$ atlas and a $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} (G,X)$ 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 $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} M$ be a manifold equipped with a developing pair into $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} (G,X)$: that is, an immersion $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} \dev\colon \widetilde{M}\to X$ and a holonomy homomorphism $\newcommand{\dev}{\operatorname{dev}} \newcommand{\hol}{\operatorname{hol}} \hol\colon \pi_1(M)\to G$.

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

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

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

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

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

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