# Question on Collation as defined by Hirsch

On Hirsch I came across this:

"A differential structure $$\Phi$$ on $$M$$ is often obtain by collation of differential structures $$\Phi_i$$ on open sets $$U_i$$ covering $$M$$. This means that $$\left.\Phi_i\right|U_i\cap U_j= \left.\Phi_j\right|U_i\cap U_j,$$ for all $$i,j$$; and $$\Phi$$ is the unique differential structure on $$M$$ containing each $$\Phi_i$$ as a subset."

Collation is a term I've only come across in his textbook, and I cannot decipher what it means operationally. Also I do not understand the notation used in the equation he uses.

Specifically: what does it mean $$\Phi_i$$? Does he consider a family of differential structures? What does it mean to restrict a family of differential structures on $$U_i\cap U_j$$? Is that even possible given that the class of differential structures is immensely "big"?

• $\Phi_i$ is the collection of all coordinate systems on the open set $U_i$ and accordingly the formula above holds (note that the intersection of two open sets is open so a coordinate system lying in the intersection also lies in the ambient open sets).
– user555729
Apr 28, 2019 at 20:23

You know that a differential structure on $$M$$ is is a maximal differential atlas $$\Phi$$ and that each open $$U \subset M$$ inherits a differential structure $$\Phi \mid_U$$ from $$M$$ by taking all charts in $$\Phi$$ with domain contained in $$U$$.
Now let $$\{ U_i \}_{i \in I}$$ be an open covering of $$M$$. Assume that for each $$i$$ we are given a differential structure $$\Phi_i$$ on $$U_i$$. Then for any two $$i,j \in I$$ we get two differential structures $$\Phi_i \mid_{U_i \cap U_j}$$ on $$U_i \cap U_j \subset U_i$$ and $$\Phi_j \mid_{U_i \cap U_j}$$ on $$U_i \cap U_j \subset U_j$$. The assertion is that if for any two pairs $$i,j \in I$$ the differential structures $$\Phi_i \mid_{U_i \cap U_j}$$ and $$\Phi_j \mid_{U_i \cap U_j}$$ agree, then there exists a unique differential structure $$\Phi$$ on $$M$$ such that $$\Phi \mid_{U_i} = \Phi_i$$ for all $$i$$.