An affine bundle has a global section?

Question

Let $X$ be a manifold. We say $\pi: Y \longrightarrow X$ is an rank $n$ affine bundle if there is an open cover $\{ U_\alpha \}$ of $X$ such that $ Y \big|_{U_\alpha} \cong U_\alpha \times \mathbb{R}^n $ and the transition function from $U_\alpha$ to $U_\beta$ is given by $$ (x,v) \mapsto (x, \rho_{\beta \alpha }(x) v + u_{ \beta \alpha} (x)) $$ satisfying the cocycle condition $ \rho_{\gamma \alpha} (x) = \rho_{\gamma \beta} (x) \rho_{\beta \alpha } (x) $ and $u_{\gamma \alpha}(x) = \rho_{\gamma \beta} (x) u_{\beta \alpha} (x) + u_{\gamma \beta}(x)$.

Wikipedia claims that an affine bundle has a global section so it can be identified with the vector bundle glued by the cocycles $\{ \rho_{\gamma \alpha} \}$ in a non-canonical way. How can we construct one exactly? Someone claimed that local sections exist so one can glue them to a global one by standard partition of unity argument. Since multiplying by constant doesn't make sense for affine bundle, I cannot see why this is obvious.

Answer 1

Multiplication by functions doesn’t in general make sense, but affine combinations of sections do.

Specifically, if $\sigma,\tau\in\Gamma(\pi)$ are sections and $f, g\in\mathrm C^\infty(X)$ are smooth functions on the base summing to unity, then the obvious definition of $f\sigma + g\tau$ is independent of trivialization if (!) $f(x)+g(x) = 1$. This defines affine combinations only for a finite number of terms, but you can of course also do that for an arbitrary one, as long as only a finite number of coefficients are non-zero at any given point.

Now choose a (locally finite) trivializing open cover of the bundle, a partition of unity corresponding to that cover and an arbitrary local section over every neighbourhood. Then the affine combination of these sections with coefficients given by this partition is a well-defined global section.