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Given two holomorphic vector bundles $E,F \rightarrow X$ over a complex manifold $X$, an isomorphism is defined to be a holomorphic map $f: E \to F$ such that $f_x: E_x \to F_x$ is a linear isomorphism. (Griffiths and Harris p. 69-70). Now, one way to think of a holomorphic vector bundle is by fixing a complex vector bundle, say $E$, and adding a holomorphic structure. That is, a $\mathbb{C}$-linear operator

$$\bar{\partial}_{E}: \Omega^0(E) \to \Omega^{0,1}(E)$$

satisfying the Leibniz rule and the conditions $\bar{\partial}_{E}^2 = 0$. So a holomorphic bundle can be thought of as a pair $(E, \bar{\partial}_E)$.

My question is, in this frame work, what is the definition of isomorphism? We must have that if the holomorphic bundles $(E, \bar{\partial}_E)$ and $(F, \bar{\partial}_F)$ are isomorphic, then the complex bundles $E$ and $F$ must be isomorphic, but what is the condition on the holomorphic structures $\bar{\partial}_E$ and $\bar{\partial}_F$? They must be "the same" somehow, but I'm not sure what this conditions would be. Thanks for the help!

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  • $\begingroup$ Well, what do your D-bar operators act on? That is probably a good first question, because whatever they act on, they should act the same. $\endgroup$
    – Tabes Bridges
    May 26, 2017 at 17:06
  • $\begingroup$ @ Tabes Bridges Well, they both act on sections, but of different bundles. I think there should be some commutivity but I just can't convince myself of this. $\endgroup$
    – user46348
    May 26, 2017 at 21:03
  • $\begingroup$ Yes, different but isomorphic bundles. In particular, the space of sections of a given type is isomorphic over every open set by $s \mapsto f \circ s$ (and similarly for bundles of $E$ or $F$-valued forms), so any reasonable notion of isomorphism should allow you to differentiate and apply the isomorphism in either order. $\endgroup$
    – Tabes Bridges
    May 27, 2017 at 18:32

1 Answer 1

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$(E,\bar{\partial}_E)$ and $(F,\bar{\partial}_F)$ are isomorphic if there is a smooth isomorphism $f\colon E\to F$ such that $f\circ\bar{\partial}_E=\bar{\partial}_F\circ f$.

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  • $\begingroup$ It is worth noting what the left side your equation means, i.e. how $f:E \rightarrow F$ acts on $\Omega^{0,1}(E)$. The answer is: $\Omega^{0,1}(E)=C^\infty(\Lambda^{0,1} \otimes E)$, and in this tensor decomposition the map $f$ leaves the $\Lambda^{0,1}$-factor unchanged and acts only on $E$. Here, $C^\infty$ stands for "smooth sections". $\endgroup$
    – user505117
    Aug 16, 2022 at 16:46

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