This has been asked here, though in the context of sheaves, and not for vector bundles (if it has been found elsewhere, I haven't found it).
From Wikipedia,
there is a bijection between vector bundle homomorphisms from $E$ to $F$ over $X$ and sections of $\text{Hom}(E, F)$ over $X$
Where $\text{Hom}(E, F)$ is the Hom-bundle, which has as fibre over each $x \in X$ the space of linear maps $E_x \to F_x$.
I am confused on two parts.
- Why can we even find nonzero global sections of the bundle here?
- What does the bijection look like?
Apparently it's an isomorphism of groups too, which I also don't see (but assume will follow once I understand the bijection itself).
I understand that a vector bundle homomorphism in this situation is just a continuous linear map between $E$ and $F$, which (loosely speaking) commutes with their respective projections. From the above definition of the Hom-bundle, I get that the projection $\pi: \text{Hom}(E,F) \to X$ is given by sending the pair $(\text{Hom}(E_x, F_x), x) \mapsto x$. Because of this, it doesn't make sense to me how we can have a nonzero global section, hence my first question.
Where am I misunderstanding the definitions here?
I have a very basic understanding of differential geometry and sheaf theory, so unless results from these areas are absolutely necessary, an answer that does o without them would be most appreciated 🙂
Best Answer
Every smooth vector bundle $E$ admits global sections. For example, let $U \subset M$ be a trivialising open set, then for every $\sigma \in \Gamma(U, E|_U)$ and every function $\rho : M \to \mathbb{R}$ with support contained in $U$, then $\rho\sigma$ is a global section of $E$, i.e. $\rho\sigma \in \Gamma(M, E)$.
If $\Phi : E \to F$ is a vector bundle homomorphism, then there is an associated section $\sigma : X \to \operatorname{Hom}(E, F)$ of $\pi$ given by $x \mapsto \Phi_x$. Conversely, If $\sigma : X \to \operatorname{Hom}(E, F)$ is a section of $\pi$, then we obtain a vector bundle homomorphism $\Phi : E \to F$ given by $e \mapsto \sigma(\pi(e))(e)$; that is, if $e \in E_x$, we have $e \mapsto \sigma(x)(e)$.