# How to build an Atlas on the total space in bundle reconstruction

The idea behind a (differentiable) fibre bundle reconstruction is to obtain a differentiable fibre bundle from a minimal set of ingredients. For principal bundles, where the fibre $$F$$ and the structure group $$G$$ coincide, this set is composed by: $$M$$ (the base space), $$\{U_i \}$$ (an Atlas on $$M$$), $$F$$ (the fibre) and $$t_{ij}(p)$$ (the transition functions of the form $$U_i \cap U_j \rightarrow F$$). Then, from these ingredients, a differentiable fibre bundle can be constructed uniquely. Specifically, the the total space of such bundle, $$E$$, is obtained by constructing a topological sum of $$U_i \times F$$ and imposing an equivalence relation defined via $$t_{ij}$$. This is a standard procedure, which can be found in Nakahara (page 353) and Kobayashi & Nomizu (page 52), for instance. My question is really simple. $$E$$, by definition, needs to be a differentiable manifold. Hence, it needs to have an Atlas (manifold atlas) associated to it. My intuition is that it must be possible to construct such an Atlas explicitly from the given ingredients above in a natural way. However, I could not find a clear explanation of this construction anywhere. Any ideas on how such construction can be realised?

• Do you mean a bundle atlas or a manifold atlas? Jun 13, 2022 at 11:15
• I mean a manifold Atlas on E, thank you very much for the clarification, I will edit the question as well to avoid confusion Jun 13, 2022 at 13:01
• Also, can you make a bit more precise how far you came? E.g., do you know how the bundle charts look like? Jun 13, 2022 at 14:23
• The construction of the bundle charts is done as follows. First, we define the total space as $X/ \sim$, where $X = \cup_i U_i \times F$ and $U_i \times F \ni (p,f) \sim (q,g) \in U_j \times F$ iff $p=q$ and $g=t_{ij}(p) f$. Then, the projection map is $\pi : [(p,f)] \mapsto p$. The local trivialisations are of the form $\phi_i : (p,f) \mapsto [(p,f)]$ (so that $\phi : U_i \times f \rightarrow \pi^{-1}(U_i)$). Jun 14, 2022 at 9:17

Okay, following your comment, you know of the bijective maps $$\phi_i:\pi^{-1}(U_i)\to U_i\times F$$ (putting domain and codomain this way you would call it bundle chart). Note that $$U_i\times F$$ is a manifold, that is, it bears a topology and a differential structure and charts of product manifolds can be obtained by product charts.
Therewith you say that $$O\subset E$$ is open if for any bundle chart $$\phi_i:\pi^{-1}(U_i)\to U_i\times F$$ the set $$\phi_i\big(O\cap\pi^{-1}(U_i)\big)\subset U_i\times F$$ is open. "You pull the topology back through the bundle charts to make them homeomorphisms". That this is consistent over some $$U_i\cap U_j$$ is guaranteed by the equivalence relation used to obtain $$E$$ and the continuity of the $$t_{ij}$$.
Similarly, you pull back the differentiable structure back via the bundle charts. Given a smooth atlas $$\{(U_i,\varphi_i)\}_i$$ of $$M$$ (w.l.o.g. the bundle charts of $$E$$ and the manifold charts of $$M$$ share domains, otherwise trim them appropriately) and a smooth atlas $$\{(V_j,\psi_j)\}_j$$ of $$F$$ define the charts $$(W_{ij},\theta_{ij})$$ by \begin{align} W_{ij}&=\phi^{-1}(U_i\times V_j)\subset E\qquad\text{(this is open)},\\ \theta_{ij}&=(\varphi_i\times\psi_j)\circ\phi_i:~W_{ij}\to\mathbb{R}^n\times\mathbb{R}^r \end{align} ($$n=$$dim$$M$$, $$r=$$dim$$F$$). So in principle, you define to be a (manifold) chart domain precisely those sets which under a bundle chart are mapped to chart domains of the product charts of $$U_i\times F$$. That this is consistent over some $$U_i\times U_k\subset M$$ is now guaranteed by the smoothness of the $$t_{ik}$$. By some standard argumentation the differentiable structure does not depend on the choice of atlases of $$M$$ and $$F$$. Note that by construction the bundle charts are diffeomorphisms with this differentiable structure.
• Also, is it possible that you meant $\phi_i$ in all the instances where you wrote $\phi$? Jun 17, 2022 at 16:19