I am at a loss as to several points made in A. Moroianu's "Lectures on Kähler Geometry", Chapter 9, section 9.2, from where I am trying to understand how to work with holomorphic vector bundles.

There, a pseudo-holomorphic structure on a merely complex vector bundle $\pi:E \rightarrow M$ on a complex manifold $M$ is defined to be an operator (or rather, a family of operators indexed after $(p,q)$) $\bar\partial: \Omega^{p,q}(E) \rightarrow \Omega^{p,q+1}(E)$ satifying the Leibniz rule, where $\Omega^{p,q}(E)$ is the vector bundle $\Lambda^{p,q}(M) \otimes E$. A section $\sigma$ of $E$ is called holomorphic if $\bar\partial \sigma = 0$.

Even though at the end of section 9.1 such an operator is defined explicitly on holomorphic vector bundles by using the usual operator $\bar\partial: \Lambda^{p,q}(M) \rightarrow \Lambda^{p,q+1}(M)$, from my understanding, since now we are just dealing with an arbitrary complex vector bundle, a pseudo-differential operator is just an arbitrary operator satisfying the Leibniz rule and it has nothing to do with the definition at the end of 9.1.

My problems come from trying to run through Lemma 9.1, the claim of which is that a complex bundle $E$ with a pseudo-holomorphic structure $\bar\partial$ is a holomorphic vector bundle (i.e. it admits a trivialization with holomorphic transition functions) if and only if each point has a neighborhood on which there are some holomorphic sections (in the sense that $\bar\partial \sigma = 0$ for $\bar \partial $ the pseudo-holomorphic structure) forming a basis on the fibers of $E$ over that neighborhood.

Supposing that $E$ is holomorphic, we are guaranteed a cover with trivializations $\psi_U: \pi^{-1}(U) \rightarrow U \times \mathbb{C}^k$ with $\psi_U^{-1} \circ \psi_V$ holomorphic on intersections. The claim is that, for $e_i$ the standard real basis of $\mathbb C^k$, we have that $\sigma_i (x):= \psi_U^{-1}(x, e_i)$ for $x \in U$ are holomorphic i.e. $\bar\partial \sigma_i = 0$. The problem is, I have no idea how to relate the pseudo-holomorphic structure $\bar\partial$ with the holomorphicity of the transition functions to show this. If it happened that $\bar\partial$ was not just any pseudo-holomorphic structure, but the one defined explicitly at the end of 9.1, then I would know that by definition $\bar\partial \sigma_i$ is in the trivialization $\psi_U$ equal to $(\bar\partial 0, \ldots, \bar\partial 0, \bar\partial 1, \bar\partial 0, \ldots, \bar\partial 0) = 0$ (where $0$ and $1$ are seen as constant functions on $U$) and this is independent of trivialization since the transition functions are holomorphic.

Conversely, supposing we have sections $\sigma_i$ with $\bar\partial \sigma_i = 0$ generating the fibers on some $U$, we can write $\sigma \in \Gamma(E,U)$ as $\sigma = \sum_i f_i \sigma_i$ with $f_i$ functions on $U$ (which must be smooth since $\sigma_i$ and $\sigma$ are smooth) and we define a trivialization on $U$ by $\psi_U(\sigma(x)):=(x, f_1(x), \ldots, f_k(x)) \in U \times \mathbb C^k$. If we have some other sections $\tilde\sigma_j$ on some $V$ intersecting $U$, we write $\sigma_i = \sum_j g_{ij} \tilde\sigma_j$ for $g_{ij}$ some (smooth) functions on $U \cap V$. Since $\bar\partial \sigma_i = 0$, and by the Leibniz rule we have $0 = \sum_j ( (\bar\partial g_{ij})\tilde\sigma_j + g_{ij} \bar\partial\tilde\sigma_j)= \sum_j (\bar\partial g_{ij})\tilde\sigma_j$ and since $\tilde\sigma_j$ form a basis we have $\bar\partial g_{ij} = 0$. However, this is again the pseudo-differential operator $\bar\partial$, not the "usual" $\bar\partial$ defined on forms on the complex manifold $M$.


1 Answer 1


For your first problem, I think that implicitly, whenever Moroianu talks about holomorphic bundles, he endows them with the induced (pseudo-)holomorphic structure as defined in 9.1. Otherwise you would deal with two different holomorphic bundles which have the same underlying complex bundle, which seems to be weird (actually I don't know if such a think exists).

However, this is again the pseudo-differential operator $\overline \partial$, not the "usual" $\overline \partial$ defined on forms on the complex manifold $M$.

That is not correct. For $\omega \in \Omega^{p,q}(M), \sigma \in \Omega^{r,s}(E)$, the Leibniz rule is $$\overline \partial (\omega \wedge \sigma) = \overline \partial\omega \wedge \sigma + (-1)^{p+q}\omega \wedge \overline \partial\sigma,$$ where $\overline \partial \omega$ is the standard antiholomorphic differential applied to $\omega$ (anything else would not make sense as $\omega$ is an ordinary differential form on $M$). So in your example, $$\overline \partial g_{ij} = 0$$ really means that the $g_{ij}$ are holomorphic functions $U \cap V \to \mathbb C$.


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