R-oriented vector bundle and orientation local system

Let $R$ be a commutative ring and let $\xi \to B$ be a metric rank-$n$ real vector bundle. Its $R$-orientation local system ${\mathcal{O}}_R(\xi )$ has stalk ${\mathcal{O}}_R(\xi {)}_b=H^n(D({\xi }_b),S({\xi }_b);R)\mathrm{.}$ On each bundle chart the fiber disk pairs identify with the standard disk pair. On overlaps, the linear transition functions induce units on its top relative cohomology; these units are locally constant and satisfy the cocycle law. They therefore define a rank-one local system, whose path transport is the corresponding composite of overlap units. Metric changes give the same system through the canonical radial pair isomorphisms.

The preceding disk-pair calculation identifies every stalk with a free rank-one $R$-module. An $R$-orientation is a section $o=(o_b)$ of ${\mathcal{O}}_R(\xi )$ such that every $o_b$ generates its stalk; equivalently it is a compatible locally constant family of fiber generators. The bundle is $R$-oriented when such a section is supplied.

For $R={\mathbb{F}}_2$, each one-dimensional stalk has exactly one nonzero generator and every transition automorphism fixes it, so every real vector bundle is canonically mod-two oriented. Over $\mathbb{Z}$, a loop whose monodromy sends a generator to its negative prevents an integral orientation: compatibility would require $1=-1$ in $\mathbb{Z}$.

For rank zero the stalk is $H^0(\ast ,\mathrm{\varnothing };R)=R$ and the unit gives the standard orientation. Empty bases have the unique empty section. The zero ring has its unique (zero) cyclic generator. Identity and constant paths act identically, reversed paths give inverse units, and no family of generators is selected in making the definition. It is choice-free.