# Tag Info

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 $\mathbf V_X(\mathscr E) = \mathbf{Spec}... 1 This is an interesting question. Suppose that$ 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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