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I am reading an article in which there is the following statement:

Let $E\rightarrow X$ be a holomorphic vector bundle. The holomorphic sections of $E$ over a coordinate neighbourhood of $X$ are dense in the set of smooth sections of $E$.

I have some knowledge in complex geometry but I am not aware of this fact. For which topologies this fact holds? Does somebody knows a place in which I could find some similar statement with a proof? I already had a look in the textbooks of Huybrechts, Demailly and Griffiths-Harris but I did not see a similar statement yet.

I have another question that perhaps is related to the previous one (it is in the same article) but the statement hereafter is from me, not really from the author. Again, let $E\rightarrow X$ be a holomorphic vector bundle. Denote by $\mathcal{E}$ the sheaf of holomorphic sections of $E$. Then we have $\Gamma(U,E)=\mathcal{C}^\infty(U,\mathbb{C})\cdot\mathcal{E}(U)$ with $\Gamma(U,E)$ the sheaf of smooth sections of $E$, $\mathcal{C}^\infty(U,\mathbb{C})$ the sheaf of smooth $\mathbb{C}$-valued functions (seen as real smooth maps I presume) and the dot is just the usual multiplication of sections by functions.

Is this statement true? Does it hold for any holomorphic vector bundle? Where can I find a proof if it is true?

Thank you!

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Which article are you reading? – J.C. Ottem Apr 30 '11 at 23:28
It is the PhD thesis of M. Grützmann entitled "Courant Algebroids: Cohomology and Matched Pairs" (available on My first question comes from the first lines in the proof page 49 and my second comes from lemma 4.13 page 48. – Benjamin May 1 '11 at 8:25
up vote 5 down vote accepted

Dear Benjamin, the statement that holomorphic sections are dense in the smooth sections is false, already for the trivial bundle of rank one $E_1=X\times \mathbb C$ over $X=\mathbb C$. Indeed on any non-empty set $U \subset \mathbb C$ it is impossible to approximate the $C^{\infty}$ function $\bar z$ by holomorphic functions since the limit of a sequence of holomorphic functions on $U$ is a holomorphic function on $U$. [The limit is to be understood in the sense of uniform convergence on compact subsets of $U$. There is also an $L^2$- version stating that $L^2(U)\cap \mathcal O(U)$ is a Hilbert subspace of $L^2(U)$, so that a sequence of holomorphic functions converging only in the $L^2$ sense nevertheless has a holomorphic limit. ]

For your "another question", the answer is also "no":

Take $E_2=X\times \mathbb C^2$ , the trivial rank-2 bundle. Then you cannot write the section $(1,\bar z)\in C^{\infty}( U,E_2)$ as $(1,\bar z)=\phi (z) (f(z),g(z))$ with $f,g$ holomorphic and $\phi$ smooth , since $\phi g/\phi f=g/f$ is meromorphic while $\bar z/1=\bar z$ is not ( notice that $\phi$ never vanishes since $1=\phi f$)

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In the "other question", maybe Benjamin wants $\mathcal{C}^\infty(U,\mathbb{C})\cdot\mathcal{E}(U)$ to mean the set of functions which can be written as sums of products of smooth functions by holomorphic sections, and not just those products. – Mariano Suárez-Alvarez Apr 30 '11 at 23:11
Dear Mariano, in that case the answer is trivially "yes" on small open sets $U$: $(\phi, \psi)=\phi (1,0)+\psi (0,1)$. If however Benjamin wants global sections the answer is trivially "no": just take a holomorphic bundle whose only global section is zero! – Georges Elencwajg Apr 30 '11 at 23:23
@Georges Elencwajg: Thank you very much for your answers. Concerning the second question it was a local question and I really think the dot is the usual product of a section by a function here. Thank you! – Benjamin May 1 '11 at 5:51
@Mariano Suárez-Alvarez: In fact, you are right, this is the true question. The dot means "take linear combinations" here but I thought something else. And this is true: any smooth section over $U$ can be written as a $\mathcal{C}^\infty(U)$-linear combination of holomorphic sections, it is just the decomposition of the section in the canonical holomorphic frame. Moreover, if the section is holomorphic then the components will be holomorphic. – Benjamin May 2 '11 at 12:36

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