7

Let $x_0,\ldots, x_n$ be homogeneous coordinates, then a section of
$H^0(\mathbb{P}^n,\mathcal{T}_{\mathbb{P}^n}(d))$ can be expanded as a sum
$$\sum_i f_i(x_0,\ldots, x_n) \frac{\partial}{\partial x_i}$$
where $f_i$ are homogenous polynomials of degree $d$. This follows more or less immediately from the Euler sequence
$$0\to \mathcal{O}\to \bigoplus_0^n \...

5

Let $Z_p \subset \operatorname{Gr}(k,n)$ be the subscheme parameterizing all subspaces parameterizing all $k$-planes containing $p$. Then $Z_p \cong \operatorname{Gr}(k-1,n-1)$ and the restriction of the tautological bundle to $Z_p$ splits as the sum of $\mathcal{O}$ and the tautological bundle $S'$ of $\operatorname{Gr}(k-1,n-1)$.
Indeed, let $V_1 \subset V$...

4

Theorem A in:
S. Kumar and M. S. Narasimhan. Picard group of the moduli spaces of G-bundles. Math. Ann., 308(1):155-173, 1997,
shows that when $G$ is a simple simply-connected connected complex affine algebraic group, $C$ is a complex smooth irreducible projective curve of genus at least 2, and $M$ is the moduli space of semistable principal $G$-bundles on $...

3

This is correct. Proof is by induction on the rank.
For rank one, this is obvious since you get a morphism from $\operatorname{Spec}R\to\operatorname{Pic} X$ which by your assumption is generically constant and thus constant.
For rank greater than one, you may twist by a large line bundle from $X$ and assume that the special member is globally generated and ...

2

Let $\text{G}$ be a group. Then maps
$$S\ \longrightarrow \ \text{BG}$$
from a test scheme $S$ is the same thing as a $\text{G}$-bundle $P\to S$ (or rather the groupoid of such).
Now let $X$ be another scheme with an action of $\text{G}$. Then to any $\text{G}$ bundle we can build its associated $X$ bundle:
$$(P\to S)\ \rightsquigarrow \ (P\times_\text{G}X\...

2

Because $L$ is ample, $E\otimes L^n$ is generated by global sections for $n\gg 0$, i.e., there is a surjective morphism $\mathscr O_X^{\oplus r} \to E\otimes L^n$, which implies that there is a surjective morphism $L^{\oplus r} \to E\otimes L^{n+1}$. As $L$ is ample, so is $L^{\oplus r}$ and then so is its quotient $E\otimes L^{n+1}$.

2

There have been recently various results revolving around this question. Let me quote a few:
$\bullet$ For any line bundle $L$ on $X$, the graded algebra $\mathrm{Ext}^*(L,L)$ is always graded-commutative. More generally, for any autoequivalence $\Phi$ of $\mathrm{D}^b(X)$, the graded algebra $\mathrm{Ext}^*(\Phi(\mathcal{O}_X),\Phi(\mathcal{O}_X))$ is ...

1

This is closely related to the resolution property [Tag 0F85], so it won't be true in complete generality. Indeed, if $\mathscr F$ is a coherent sheaf on $X = S$ that cannot be given as a quotient of a vector bundle $\mathscr E$ on $S$, then $Y = \mathbf{Spec}_X \operatorname{Sym}^* \mathscr F$ cannot be embedded into $\mathbf V_X(\mathscr E) = \mathbf{Spec}...

answered May 26 at 20:05

R. van Dobben de Bruyn

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1

This is an interesting question. Suppose that $ h = k-1$. Then the answer appears to be yes. Consider the determinant line bundle $L $ on $Gr(k,n) $; its fibre at $ H $ is $ \det H $.
Now let us pick some a point $ H \in X$. We have the short exact sequence
$$
0 \rightarrow \Gamma \rightarrow H \rightarrow H / \Gamma \rightarrow 0
$$
and thus $ \det \...

1

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 indecomposable and stable. Using the fact that the homomorphism ${\rm Ext}^1({\cal O(p), O}) = {\rm H}^1(X, {\cal O}(-p)) \to {\rm H}^1(X, {\cal O}) = {\rm Ext}^1({\cal O, ...

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