bio | website | math.sunysb.edu/~jstarr/… |
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location | Stony Brook University, Stony Brook, NY USA | |
age | ||
visits | member for | 5 years, 7 months |
seen | 3 hours ago | |
stats | profile views | 4,347 |
May 11 |
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Why is the evaluation map of sheaves injective
The homomorphismm $\text{ev}$ is not necessarily injective if $S$ is reducible or nonreduced (I assume that your $X$ and your $S$ are the same). If $S$ is reducible and reduced, then a global section of $E$ is zero if and only if it is zero at the generic point. Thus, the stalk of $\text{ev}$ at the generic point is injective. A homomorphism of locally free sheaves (or even just torsion-free, coherent sheaves) on an integral scheme is injective. |
May 11 |
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Criterion for transverse boundary intersection of one-parameter family in $\overline M_{g,n}(X,\beta)$
I will try to find you a reference. Personally, I would handle it in two steps. First, the induced (1-)morphism from $B$ to the Artin stack $\mathfrak{M}_{0,n}$ of all $n$-pointed, at-worst-nodal curves pulls back the boundary divisor $\Delta$ to be $\underline{0}$, with multiplicity $1$. Second, because of convexity, the $1$-morphism $\overline{M}_{g,n}(X,\beta)\to \mathfrak{M}_{0,n}$ is smooth, so that the pullback of $\Delta$ is a Cartier divisor -- the usual boundary divisor. |
May 11 |
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Criterion for transverse boundary intersection of one-parameter family in $\overline M_{g,n}(X,\beta)$
For $g=0$, that is sufficient. For larger $g$, it depends on what you mean by "transverse to the boundary" in cases where "the boundary" is not actually a Cartier divisor in the moduli stack. |
May 6 |
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cup-length of the first Chern class of complex grassmannian
@RSQ. For the theorem abx states, you can look up "Hard Lefschetz Theorem" in Griffiths and Harris, among others. There are, of course, simpler proofs for the Grassmannian. I am sure that the description of the complex Grassmannian in Griffiths and Harris includes your result. In particular, $c_1^{2n-2}$ is computed as a Catalan number. |
May 5 |
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constructibility for pushforward
Okay, in that case, for every quasi-compact open $U$, there exists finite $n$ and an open $U_n \subseteq \mathbb{A}^n$ such that $U$ is the inverse image of $U_n$ under projection from $\mathbb{A}^{\mathbb{N}}$ to $\mathbb{A}^n$. There is a smooth base change theorem for constructible sheaves . . . |
May 5 |
answered | Are essentially smooth schemes noetherian? |
May 5 |
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constructibility for pushforward
With what transition maps? |
May 5 |
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Are essentially smooth schemes noetherian?
What about $k[x_0,x_1,x_2,x_3,\dots,x_n,\dots]/\langle x_0x_1-1,x_2^2-x_1,x_3^3-x_2,\dots,x_{n+1}^{n+1}-x_n,\dots \rangle$? This ring is the union of the subrings $R_n$ generated by $x_0,\dots,x_n$, i.e., $R\cong k[x_n,x_n^{-1}]$. Each transition map is etale. Yet the union is not Noetherian. |
May 5 |
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Do line bundles descend to coarse moduli spaces of Artin stacks with finite inertia?
You changed the question. |
May 4 |
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Do line bundles descend to coarse moduli spaces of Artin stacks with finite inertia?
The zeroth power descends :) |
May 4 |
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constructibility for pushforward
What precisely do you mean by $\mathbb{A}^{\mathbb{N}}$? |
May 4 |
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If the quotient of an algebraic space $X$ by a finite group is a scheme, is $X$ a scheme?
Chevalley's theorem is the statement, roughly, that $Y$ is affine if and only if $X$ is affine. The easy direction is: if $Y$ is affine, then also $X$ is affine. This is the direction you are interested in: every scheme is covered by open affines, you can consider the inverse image of these open affines in $X$, and then you apply the easy direction of Chevalley's Theorem. The hard direction is the one that Rydh generalizes: if $X$ is affine, then also $Y$ is affine. |
May 4 |
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If the quotient of an algebraic space $X$ by a finite group is a scheme, is $X$ a scheme?
This follows from the easy direction of Chevalley's theorem for algebraic spaces. For the hard direction, you can consult Knutson's "Algebraic Spaces" or the more recent proof by David Rydh in the not-necessarily-Noetherian setting. |
May 4 |
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Extending a section of a coherent sheaf and homomorphism
@Ron. The answer is still no, even with the edits. Let $\mathcal{F}$ be $\widetilde{A}\oplus \widetilde{A}$, let $\mathcal{G}$ be $\widetilde{A}$, let $\phi$ be projection onto the second factor, let $s$ be $1$ and let $s'$ be $(1/x,1)$. If you hope for something like this to be true, you typically allow modifications of $s$ and $s'$ by multiplying by powers of $f$, cf. Lemma 5.3 of Chapter II of Hartshorne's "Algebraic Geometry". |
May 4 |
answered | Extending a section of a coherent sheaf and homomorphism |
May 4 |
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Inverse problem of Chern Classes
Do you restrict the rank of the bundle (for fixed degree of the Chern class)? If so, I think the problem is difficult. Here is another issue: the Chern classes do not uniquely determine a complex vector bundle up to equivalence. To see this, consider the case when the base manifold is an odd-dimensional sphere, so that the Chern classes are all zero. Since homotopy groups with odd degree of the classifying space can be nontrivial, the Chern classes are not sufficient to characterize complex vector bundles. |
May 4 |
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Holomorphic extension of a section of a line bundle
"... I (implicitly, sorry for being imprecise) assume that $X$ is not a product." In that case, I think you should change the statement of your question. The basic issue persists: if you define $\mathcal{L}$ to be the invertible sheaf associated to the complex hypersurface $K$, then the tautological section $1$ on $X\setminus K$ is polar on $K$. In your special case, if $\mathcal{L}$ is the pullback of a holomorphic line bundle from the singular variety $\mathbb{C}^m/G$, then you should be able to use Hartog's theorem to extend $\sigma$ on $\mathbb{C}^m/G$, and then pull this back. |
May 4 |
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Holomorphic extension of a section of a line bundle
This already fails if $(X,g,\omega)$ is, say, $\mathbb{C}$ with its flat metric (complete and Kaehler), $\mathcal{L}$ is the trivial holomorphic line bundle $\mathbb{C}\times \mathbb{C}$, $K$ is a single point, say $\{0\}$, and $\sigma$ is a rational function like $1/z$. I realize that you write $m\geq 3$, but you can always repeat this with $X=\mathbb{C}\times T$, where $T$ is a complex torus of dimension $g\geq 2$, $K$ is $\{0\}\times T$, and $\sigma$ is $\text{pr}_{\mathbb{C}}^*(1/z)$. |
May 4 |
answered | Higher Fano varieties and Tsen's theorem |
May 4 |
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Algebraicity of the stack of coherent sheaves
I am afraid that I still do not understand your concern. |