## New answers tagged vector-bundles

4

An obvious obstruction comes from topology: the Chern classes of your
bundle should be obtained from restriction of Hodge classes on an ambient variety. This is (more or less) enough to extend a smooth bundle $B$
from $Y$ to $X$. To be precise, you need the classifying map from $Y$ to the space $BU(r)$ to be extendable to a continuous map from $X\supset Y$ ...

5

Without hypotheses on $Y$ there is no hope to define such obstructions, already for line bundles. A natural hypothesis is to take for $Y$ a (smooth) ample divisor in $X$, of dimension $\geq 2$. In this paper, Fujita gives some cohomological conditions which imply that $E$ extends : $H^2(Y, \mathcal{E}nd(E)(-tY))=0$ for all $t\geq 1$ and $H^p(Y, E(tY))=0$ for ...

0

I don't know a reference either but you can proceed as follows...
In the vector space of numerical $\mathbb{R}$-divisors, $N^1(X)_{\mathbb{R}}$, being ample is an open condition. Take an open ball $B \subset N^1(X)_{\mathbb{R}}$ satisfying $L \in B$. Look at the map from $N^1(X)_{\mathbb{R}}$ to numerical $\mathbb{R}$-curves $N_1(X)_{\mathbb{R}}$ by taking ...

3

No: any finite set $S \subset M$ can be contained in the interior of an embedded closed disc in $M$, and cutting this out gives a manifold diffeomorphic to $M - \{x\}$. So if $M - S$ were parallelisable, $M-\{x\}$ would be too.

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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