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This semiring carries an enourmous amount of information about vector bundles on $\mathbb{P}^n$, including stuff we don't yet know. For example, you can read from it whether there are indecomposable v …

I don't think this is true, even for stable bundles. Assume that the genus of $X$ is at least 2. Take a point $p$ of $X$, and a nontrivial extension of ${\cal O}(p)$ by ${\cal O}$, this is indecomposa …

The only holomorphic subbudles of $T\mathbb P^3$ are the null-correlation bundles coming from symplectic forms in 4 variables (see for example http://www.math.ubc.ca/~reichst/nesting.pdf, Corollary 1. …

Swan has proved that taking global section gives an anti-equivalence between finitely generate projective $\Gamma^{\infty}(M)$-modules and $C^{\infty}$ vector bundles on $M$; this correspondence is fu …

No. When $E$ is a sum of two line bundles on $\mathbb P^2$, then $\mathrm{Ext}^1(E, E) = \mathrm H^1(E^\vee \otimes E) = 0$, but $\mathrm{Ext}^2(E, E) = \mathrm H^2(E^\vee \otimes E)$ is not necessari …

By Nori's theorem, the existence of non-trivial finite vector bundles on a reduced connected scheme $X$ of finite type over a perfect field $k$ with a rational point is equivalent to the fact that for …

The statement is true if $X$ is regular of dimension 2 (an in very few other cases, I would guess). Anyway, this certainly applies to your example. The point is that every locally free sheaf on $U$ h …

This is false as stated; for example, if $X$ is obtained from a projective geometrically connected smooth surface over a field $k$ by gluing two points together and $U$ is the complement of the singul …

The answer is positive. Let $P$ be the principal $\mathrm{GL}_n$-bundle associated with $E$; then the space of flags is the quotient $P/B$, where $B$ is the Borel subgroup of $\mathrm{GL}_n$ consistin …

Any curve of large enough degree will do. Set $F:= E'\otimes E^{\vee}$; if $d$ is a very large integer, then $\mathrm H^1(F(-d)) = 0$. Take any curve $C$ of degree $d$, and suppose that $E\mid_C$ and …

Complex curves with anti-holomorphic involutions correspond to real algebraic curves. Your involution has no fixed points, so your curve corresponds to a real curve $C$ of genus 0 with no real points …

The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$. In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory …

1. The argument seems to work fine in positive characteristic. 2. In Grothendieck's convention, the projectivization of vector bundle is defined with 1-dimensional quotients, and not 1-dimensional su …