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2h
comment Lie algebra of holomorphic vector fields
Just to point out: even considered as a module over the infinite-dimensional algebra of holomorphic functions on the cotangent bundle, still the Lie algebra is not generated by $\mathfrak{g}$. In the case of the cotangent bundle of $\mathbb{P}^1$, the cokernel is one-dimensional, generated by the vector field of "scaling" on the quadric cone.
2h
comment Lie algebra of holomorphic vector fields
The cotangent bundle of $\mathbb{P}^1$ is a minimal desingularization of the affine quadric surface cone (with a single ordinary double point), e.g., the quotient of the affine plane by the $-1$-action. The Lie algebra is infinite-dimensional. In what terms do you want a description of this infinite-dimensional Lie algebra?
8h
comment Integral domains equal to intersection of their height one localizations
A one-dimensional, local, integral domain has this property. So there are many examples, e.g., the local ring of $k[x,y]/\langle y^2 - x^3 \rangle$ at the maximal ideal $\langle x,y\rangle$.
1d
comment Non trivial rank 2 holomorphic vector bundles in complex dimensions greater than or equal 2
Oops, complex tori don't work: $h^{0,1}$ is always nonzero.
1d
comment Non trivial rank 2 holomorphic vector bundles in complex dimensions greater than or equal 2
Aren't complex tori Oka manifolds?
Aug
27
revised Construction of coherent sheaf such that $\text{Proj}\,\text{Sym}\,(\mathcal{F}) = \text{Sym}^n X$
added 367 characters in body
Aug
27
answered Construction of coherent sheaf such that $\text{Proj}\,\text{Sym}\,(\mathcal{F}) = \text{Sym}^n X$
Aug
27
comment Construction of coherent sheaf such that $\text{Proj}\,\text{Sym}\,(\mathcal{F}) = \text{Sym}^n X$
I guess what you really need is a Poincare sheaf on $X\times \text{Pic}^nX$. Do you have a Poincare sheaf? If you have a zero-cycle of degree $1$ on $X$, then you have a Poincare sheaf.
Aug
27
comment Construction of coherent sheaf such that $\text{Proj}\,\text{Sym}\,(\mathcal{F}) = \text{Sym}^n X$
Every construction that I know requires choosing a rational point on $X$. Does your curve have a rational point? Does it at least have a zero-cycle of degree $1$?
Aug
27
comment Elementary examples on sheaf extension
Are you asking about $\text{Ext}^1_{\mathcal{O}_{C_V}}$, or are you asking about $\text{Ext}^1_{\mathcal{O}_{T*\mathbb{P}^n}}$? For the former, quite frequently the Ext group will be zero, e.g., when $V$ is any rational curve. However, for the latter, the group is nonzero already when $V$ is a line in $\mathbb{P}^2$.
Aug
26
comment Derived pullback of the coarse moduli morphism
Both exactness of $f_*$ and $\theta$ being isomorphic can be checked etale locally on $X$. Thus, assume that $\mathcal{X}$ is $[Y/G]$ with induced map $F:Y\to X$ and $|G|$ prime to the characteristic. Then $f_*f^*$ is the same as $(F_*F^*(-))^G$. Since $F$ is finite (hence affine), $F_*$ is exact. Since $|G|$ is prime to the characteristic $(-)^G$ is exact. Thus $f_*$ is exact. Because $f_*$ is exact, to check $\theta$ is isomorphic, it suffices to consider $\theta_{\mathcal{O}}$. This case follows from the definition of the categorical quotient of $Y$ by $G$.
Aug
26
comment Derived pullback of the coarse moduli morphism
Flatness of $f$ is a corollary of the local flatness criterion, cf. Theorem 23.1, p. 179 of H. Matsumura, Commutative Ring Theory, Cambridge U. Press. However, via the Chevalley-Shephard-Todd theorem, your hypotheses are extremely strong.
Aug
26
comment Derived pullback of the coarse moduli morphism
First of all, in the tame case, $Rf_*$ equals $f_*$. In the non-tame case, $Rf_*$ does not even map $D^b$ to $D^b$.
Aug
26
revised Derived pullback of the coarse moduli morphism
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Aug
26
revised Derived pullback of the coarse moduli morphism
added 1 character in body
Aug
26
answered Derived pullback of the coarse moduli morphism
Aug
25
answered The cohomology ring of a compact toric manifold
Aug
24
comment Annihilators of elements in symmetric algebras
@user43326. The map $M\to M\otimes_R M$ need not be injective. For instance, if $R$ is $\mathbb{Z}$ and $M$ is $\mathbb{Q}/\mathbb{Z}$, then the target module is the zero module, and the map is the zero map.
Aug
24
comment Is the elementary transformation along a curve decomposable?
Just to expand on Mohan's comment: if the kernel is $u:E\to \mathcal{O}_S^{\oplus 2}$, what is the image of the induced homomorphism $\bigwedge^2 u: \bigwedge^2 E \to \mathcal{O}_S$? If $E$ is a direct sum of invertible sheaves $F\oplus G$, what does this imply about $F\otimes_{\mathcal{O}_S}G$? What does this, further, imply about the pushforward of $c_1(A)$ versus $c_1(F)\cdot c_1(G)$?
Aug
24
answered Definition of Strongly Stable 0-cycle