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Let $(M,g)$ be a compact complex manifold, and let $\mathcal{V}$ be an Hermitian holomorphic vector bundle over $M$, with $\nabla$ its Chern connection. If the curvature operator $\nabla^2 = 0$, then is it necessarily true that $\mathcal{V}$ is a trivial vector bundle. I guess this might be proved using Hirzebruch--Riemann--Roch, and the knowledge that a non-trivial vector bundle should have non-trivial Euler characteristic. However, it seems there should be a much simpler way to do this . . .

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  • $\begingroup$ Do you mean trivial as a $C^{\infty}$ complex vector bundle, or do you mean trivial as a holomorphic vector bundle? $\endgroup$
    – Malkoun
    Oct 3, 2019 at 15:47
  • $\begingroup$ I am a bit rusty on Chern connections, but from my experience, be very careful when you try to make general statements in complex geometry. There are often counterexamples. For example, you wrote that a non-trivial vector bundle should have non-trivial Euler characteristic. If I am not mistaken, the bundle $\mathcal{O}(1) \oplus \mathcal{O}$ over $P^2(\mathbb{C})$ has I think vanishing $c_2$, and therefore vanishing Euler characteristic, but it is not holomorphically trivial, since its $c_1$ is $1$. I am a bit rusty, but be careful with general statements! $\endgroup$
    – Malkoun
    Oct 3, 2019 at 16:04
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    $\begingroup$ No, there are flat bundles (i.e., ones coming from representations of $\pi_1(M)$) that are most definitely non-trivial. @Malkoun: I don't follow your suggestion, since holomorphic vector bundles on $\Bbb P^1$ holomorphically split as $\oplus \mathscr O(a_i)$. $\endgroup$ Oct 3, 2019 at 18:17
  • $\begingroup$ @TedShifrin, my bad. I am distinguishing between flat as smooth real vector bundles versus flat as holomorphic vector bundles. The bundle I have in mind is diffeomorphic to $S^2 \times \mathbb{R}^4$ as a real vector bundle, but after you put a complex structure, it becomes the holomorphic bundle $\mathcal{O}(1) + \mathcal{O}(1)$ over $P^1(\mathbb{C})$. I guess that comment was not relevant. $\endgroup$
    – Malkoun
    Oct 3, 2019 at 18:27
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    $\begingroup$ @Malkoun: So I should have thought of this, but since $TS^2\oplus \epsilon^1_{\Bbb R}$ is trivial, the real bundle $TS^2\oplus\epsilon^1_{\Bbb C}$ is trivial. But the first Chern class of the complex bundle is most definitely nontrivial. $\endgroup$ Oct 3, 2019 at 19:49

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