2

I start studying some complex geometry and I am lost about some facts about vector bundles.

Consider a smooth manifold $M$; and a holomorphic manifold $X$

I know that in the case of real smooth vector bundles over $M$ we have same canonical isomorphisms mirroring canonical constructions of linear algebra. For example,

  • $E^{*}\otimes F\ \simeq \operatorname{Hom}_{\mathbb{R}}(E,F)$
  • $(E^{*})^{*} \simeq E$
  • $\operatorname{Hom}_{\mathbb{R}}(E_1,F_1)\otimes \operatorname{Hom}_{\mathbb{R}}(E_2,F_2) \simeq \operatorname{Hom}_{\mathbb{R}}(E_1\otimes E_2; F_1 \otimes F_2)$

So in the case of smooth complex vector bundles over $M$; or in the case of holomorphic vector bundles over $X$ could we extend the same isomorphisms from linear algebra to vector bundles?

If this is the case how it is justified?

Is the lack of partitions of unity a problem for this to extend on smooth complex or holomorphic bundles? Because I no that for real smooth bundles one has the isomorphism $E \simeq E^{*}$ but this not true for the smooth complex vector bundles.

KReiser
  • 65,137
karhas
  • 675
  • 3
    These all follow from the corresponding canonical isomorphisms for vector spaces. So first make sure you understand them in that setting. After that, proving all of the above is relatively straightforward. – Deane Jun 02 '21 at 01:16
  • 1
    @Deane I think for the tensor product of sections, if $U$ is an open set over which the bundle is trivializable then everything is fine. But if $U$ is not such an open set, then in the case of holomorphic bundles and sections, I doubt this holds (because if I remember correctly, partitions of unity are involved, which obviously don't exist in the holomorphic case). Am I wrong? – peek-a-boo Jun 02 '21 at 04:33
  • 1
    For real and complex vector bundles, at least, the "fiberwise" constructions can be generalized using the notion of a smooth functor. – Kajelad Jun 02 '21 at 04:40

1 Answers1

1

Question: "In fact $\Gamma(∙,E)$ is a sheaf on the base space, c1.1 so are the isomorphisms extend to sheaf isomorphisms of $C^{\infty}(M)$-modules in case of real vector bundles? c1.2 and the same for complex vector bundles? c2 Are the isomorphisms extend to sheaf isomorphisms of $\mathcal{O}_X$ -modules in case of holomorphic vector bundles?"

Answer: If $(X, \mathcal{O})$ is a complex manifold there is an "equivalence of categories" between the category of finite rank holomorphic vector bundles on $X$ and the category of locally trivial $\mathcal{O}$-modules hence we may argue using locally free sheaves.

If $(X,\mathcal{O})$ is a locally ringed space and $E$ a finite rank locally trivial $\mathcal{O}$-modules on $X$ and $F$ any $\mathcal{O}$-module, there is a "canonical map"

$$\phi: E^*\otimes_{\mathcal{O}} F \rightarrow Hom_{\mathcal{O}}(E,F)$$

defined as follows: Let $U \subseteq X$ be an open set and let $F(U):=E^*(U) \otimes_{\mathcal{O}(U)}F(U)$ be the presheaf with associated sheaf $E^*\otimes F$.

There is a map

$$\phi(U): F(U) \rightarrow Hom_{\mathcal{O}(U)}(E(U), F(U))$$

defined by

$$\phi(U)(f \otimes w)(v):=f(v)w \in F(U).$$

The map $\phi$ is a map of presheaves and gives rise to a map

$$\phi^+: E^*\otimes F \rightarrow Hom_{\mathcal{O}}(E,F).$$

Since $E^*$ is finite rank locally trivial it follows $\phi^+$ is an isomorphism. You prove this by choosing a local trivialization of $E$.

Example: If $E:=A\{e_1,..,e_n\}$ is a free $A$-module with dual $E^*:=A\{x_1,..,x_n\}$ and $x_i:=e_i^*$ it follows

$$Hom_A(E,F) \cong F^n, E^*\otimes_A F \cong F^n$$

and the canonical map $E^*\otimes_A F \rightarrow Hom_A(E,F)$ induce an isomorphism. Let $\phi: E \rightarrow F$ with $\phi(\sum_j a_je_j):=\sum_ja_j\phi(e_j)$ and define

$$u:= \sum_i x_i \otimes \phi(e_i) \in E^* \otimes F.$$

It follows $u$ maps to $\phi$.

You may do similar constructions for all bundles mentioned above.

Example:

$$Hom(E_1,F_1)\otimes Hom(E_2,F_2)\cong (E_1)^*\otimes F_1 \otimes (E_2)^*\otimes F_2 \cong $$

$$E_1^*\otimes E_2^* \otimes F_1 \otimes F_2 \cong (E_1\otimes E_2)^*\otimes F_1\otimes F_2 \cong Hom(E_1\otimes E_2,F_1\otimes F_2).$$

As mentioned in the comments: Whenever you have a "canonical isomorphism" between vector spaces, such constructions carry over to locally free sheaves.

$$Hom_{\mathcal{O}}(E\otimes F,G) \cong Hom_{\mathcal{O}}(E, F^*\otimes G)$$

etc..

Question: "Is the lack of partitions of unity a problem for this to extend on smooth complex or holomorphic bundles?"

Answer: In the case of holomorphic bundles: By the above argument, no. There is for a real smooth manifold $M$ the "Serre-Swan theorem" saying there is an equivalence of categories between the category of finite rank real smooth vector bundles on $M$ and the category or finite rank projective $C^{\infty}(M)$-modules. Hence you may prove similar results using homological algebra. The formula

$$Hom_A(E\otimes F,G) \cong Hom_A(E,Hom_A(F,G)$$

holds for any $A$-modules $E,F,G$ with $E$ a (finite rank) projective $A$-modules. This proves the same result for vector bundles on $M$.

There is the "list of canonical isomorphisms in homological algebra and sheaf theory":

Stalks of exterior power

The "magic diagram" is cartesian

Prove $\phi$ to be isomorphism (an exercise in commutative algebra)

Form of basic open set of affine scheme: The intersection of two basic open sets.

hm2020
  • 1
  • Thank you for your response. My question was closed due to lack of focus so I had edited. I will make another question about the sheaf case, and linked it here. – karhas Jun 02 '21 at 20:24