## New answers tagged vector-bundles

6

Since $H^2(S^6;\mathbb Z) = H^4(S^6;\mathbb Z) = 0$, we get $c_1(TS^6) = c_2(TS^6) = 0$. As for $c_3(TS^6)$, it equals the Euler class $e(TS^6)$, which is the Euler characteristic of $S^6$ (which equals $2$) times the generator of $H^6(S^6;\mathbb Z) = \mathbb Z$.

2

Yes, that is true. By "compact homogeneous space", I assume that you mean a smooth projective variety over an algebraically closed field $k$ (presumably $\mathbb{C}$ for you) that is homogeneous under an algebraic action of a group $k$-scheme $G$ (or holomorphic action of a complex Lie group). Let $\mathcal{F}$ be a locally free $\mathcal{O}_X$-module with ...

1

The curvature form of any bundle is closed by Bianchi identity, hence
your form $s\omega$ is necessarily closed. This implies that $s=const$, unless $dim_C M=1$, because $ds\wedge \omega=0$ implies $ds=0$ (multiplication by $\omega$ is injective on 1-forms).
As for the main question, projectively flat bundles are the same as flat PGL(n)-bundles, which is ...

1

Let $V_2$ be the image of $f:V\rightarrow W$; it is a subsheaf of $W$, hence locally free. Let $W_2$ be the quotient of $W/V_2$ by its torsion subsheaf, and let $W_1$ be the kernel of the projection $W\rightarrow W_2$. The induced map $V_2\rightarrow W_1$ has maximal rank, and you get exactly the situation described in the paper.

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