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$E \rightarrow M $ be a complex vector bundle (of real rank $2r$) with almost complex structure $J:E\rightarrow E \space\space\space(J^2 =-1)$ on it. $U\subset M$ be a trivial neighbourhood.

Does there exist a complex frame for $E \rightarrow M$ over U? By a complex frame, I mean a set of $r$ (real) linearly independent $E_U$-sections $\{X_1, ... , X_r\}$, such that $\{X_1, ... , X_r, JX_1, ... , JX_r\}$ forms a complete $E_U$-frame. (By "linearly independent", I mean they are linearly independent in the fibre above every point $x\in U$). It seems like such frames do exist as I recall seeing such frames in definition of complex connection and curvature matrices on complex vector bundles, which then lead to construction of Chern forms and classes.

I need such a frame to prove that existence of a $J$ map for $E \rightarrow M $ implies a reduction of structure group of E to $GL(r,C)$. I am stuck with this, even though I can prove the converse by explicitly constructing a complex frame using the reduction.

Edit: I am using the following definition for 'complex vector bundle': A real even ranked vector bundle with a J-map. Sorry for the confusion.

Michael Albanese
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Prajwal Samal
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    Given a real vector bundle, do you know how to find a local frame over $U$? – Michael Albanese Aug 26 '21 at 20:27
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    You mean a complex vector bundle of rank $r$, not $2r$. – Ted Shifrin Aug 26 '21 at 21:57
  • @TedShifrin Yes, I mean that the real rank is 2r. Sorry for that. – Prajwal Samal Aug 27 '21 at 03:14
  • I found an answer in https://math.stackexchange.com/questions/1817959/almost-complex-manifolds-are-orientable?rq=1 (the answer by @YYF). It was wrong of me to insist finding such a frame in a trivial neighbourhood. Such a frame, would be defined in a smaller neighbourhood, in general. @MichaelAlbanese I think this is what you are hinting at as well. – Prajwal Samal Aug 27 '21 at 04:05
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    That's not what I was hinting at. You can get a local frame on $U$ using local triviality. – Michael Albanese Aug 27 '21 at 14:43
  • @MichaelAlbanese $U$ is a trivial neighbourhood for $E\rightarrow M$ considered as a REAL vector bundle. Can a complex frame (as I described above) still be found, defined on all of $U$? – Prajwal Samal Aug 28 '21 at 05:22
  • In your question you said $U$ was a trivial neighbourhood for the complex vector bundle $E$, so my comments were regarding that question. Now you're saying $U$ is a trivial neighbourhood for $E$ as a real vector bundle, which is a different question. – Michael Albanese Aug 28 '21 at 13:45
  • Apologies for the confusion : ) – Prajwal Samal Aug 28 '21 at 14:19
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    If you have a complex vector bundle, why would you want to forget its complex structure? Trivialize it as a complex vector bundle. – Ted Shifrin Aug 28 '21 at 19:38

1 Answers1

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The definition of complex vector bundle you're using is non-standard. The standard definition mirrors the definition of a real vector bundle, but with $\mathbb{R}$ replaced by $\mathbb{C}$. Your definition is equivalent though, see this question.

To the question at hand: if $U \subset M$ is such that $E|_U$ is trivial as a real vector bundle, is it necessarily trivial as a complex vector bundle? The answer is no.

Example: Consider the complex vector bundle $E = TS^2\oplus\varepsilon_{\mathbb{C}}^1$ over $S^2$. The underlying real vector bundle is $TS^2\oplus\varepsilon_{\mathbb{R}}^2 \cong \varepsilon_{\mathbb{R}}^4$, i.e. it is trivial. On the other hand, the first Chern class of the complex vector bundle $E$ is $c_1(E) = c_1(TS^2\oplus\varepsilon_{\mathbb{C}}^1) = c_1(TS^2) \neq 0$, so it is not trivial. Therefore, by taking $U = S^2$, we have an example where $E|_U$ is trivial as a real vector bundle, but not as a complex vector bundle.

Regarding the reduction of structure group, it is true that if $E|_U$ is trivial as a real bundle, then there is $U' \subseteq U$ such that $E|_{U'}$ is trivial as a complex vector bundle, see the answer to the aforementioned question.

Michael Albanese
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