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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.

  1. Why can we even find nonzero global sections of the bundle here?
  2. 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 :)

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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 $e \mapsto \sigma(x)(e)$.

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  • $\begingroup$ What exactly is a trivialising set? $\endgroup$
    – mi.f.zh
    Dec 19, 2019 at 13:29
  • $\begingroup$ An open set $U$ such that $E|_U$ is trivial; such sets exist because $E$ is locally trivial. $\endgroup$ Dec 19, 2019 at 13:36
  • $\begingroup$ I'm not so familiar with what that means either - wikipedia mentions it several times, but never gives the definition. $\endgroup$
    – mi.f.zh
    Dec 19, 2019 at 13:50
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    $\begingroup$ It means for every $x \in X$, there is an open neighbourhood $U$ of $x$ such that $E|_U$ is isomorphic to a trivial bundle. This is part of the definition of a vector bundle. $\endgroup$ Dec 19, 2019 at 14:38
  • $\begingroup$ Got it - that makes sense! Thanks! $\endgroup$
    – mi.f.zh
    Dec 19, 2019 at 15:31

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