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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.

13 votes

What does it mean that $[X]+[Y]=0$ in the Grothendieck ring of varieties?

If $k$ is an algebraically closed field of characteristic $0$, then the map $$X \mapsto N(X) := \#(\text{connected components of }X)$$ defined for smooth projective varieties extends to a ring homom …
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7 votes
Accepted

A funny factorization of the Jacobian coming from the lines on the Fermat cubic

The Jacobian matrix consists of coefficients of $t^3,t^2u,tu^2,u^3$ in the following $4$ partial derivatives $$\partial_p F(t,u,pt+ru,qt+su) = \partial_{y}F(t,u,pt+ru,qt+su)t,\\ \ldots\\ \partial_s F( …
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7 votes
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Deligne Pairing v.s. Weil Pairing on a Family of curves

Let $C$ be a smooth projective curve (say over $\mathbf{C}$ for simplicity). Given $L,M \in \mathrm{Pic}^0(C)[n]$, recall that the Weil pairing $e_n(L,M)$ is defined as follows. First of all, for an …
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6 votes
Accepted

Fujiki class $\mathcal C$ with a symplectic structure

If $X'$ is a Mukai flop of a compact hyper-Kähler manifold $X$, then $X'$ is in Fujiki class $\mathcal{C}$ and carries a holomorphic symplectic form $\sigma$. Taking the real part or the imaginary par …
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6 votes
2 answers
3k views

Algebraic varieties and UFD

Given an affine algebraic variety $V$ such that $\Gamma(V,\mathcal{O}_V)$ is a UFD, its sheaf of ring can be determined easily since one can show that: $$\Gamma(D(f_1) \cup \cdots \cup D(f_n),\mathca …
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4 votes

Lazarsfeld-Mukai bundles are stable on a K3 surface of picard number 1

Since the Lazarsfeld-Mukai bundle $E = E_{C,A}$ fits into the exact sequence $$0 \to H^0(C,A)^\vee \otimes \mathcal{O}_S \to E \xrightarrow{\phi} K_C(-A) \to 0,$$ the first Chern class of $E$ is $[ …
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4 votes
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Degeneration of Hodge-de Rham spectral sequence, exactness of a pairing and the trace morphism

First of all, Deligne-Illusie worked with relative Frobenius $F = F_{X/S}: X \to X'$, so the target of the composition should be $\Omega_{X'/S}^p$ instead of $\Omega_{X/S}^p$. Your question about why …
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4 votes
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Fibre product of complex analytic spaces - reference request

Apart from the reference given in the comment, you can also find a proof of the existence of the fiber product in Fischer's "Complex Analytic Geometry", Corollary 0.32. For direct products, a more st …
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4 votes
0 answers
216 views

Example of a non-algebraic singularity II

In an answer of this MO question, Frank Loray constructed an example of analytic singularity which is not algebraic. On the other hand, as I learned from one of Joël's comments in that question, Arti …
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3 votes
Accepted

Inequality on Kähler classes

Here is a simple proof using Theorem 1.6.1 in Lazarsfeld book, which is the following: Theorem (Demailly) If $H_1,\ldots,H_n$ are Kähler classes in a compact Kähler manifold of dimension $n$, then th …
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3 votes
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Demailly Campana Peternell Conjecture for isolated singularities

This is related to Mori's theorem through Grauert's ampleness criterion in Hartshorne's "Ample vector bundles" (Proposition 3.5). Let's assume that $M$ is projective and $\dim M \ge 2$. Let $\alpha : …
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3 votes

Is Kähler current class representable by semipositive forms?

This is an answer to your last question. In general we can't represent the class of a Kähler current by a semi-positive smooth form $\alpha$. Consider the blowup $\tilde{S} \to S$ of a compact Kähler …
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2 votes

Degree formalism for line bundles on Deligne-Mumford stacks

Let us do the example where $\mathscr{M}$ is a proper Deligne-Mumford moduli stack of varieties. Note that if $\mathscr{M}$ is one-dimensional, then the Cohen-Macauleyness of $\mathscr{M}$ is equival …
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2 votes

Is there a formula for the number of rational cuspidal curves in surfaces other than P^2?

Given a collection of isolated plane curve singularity type $\alpha = (\alpha_1,\ldots,\alpha_l)$, J. Li and Y.-J. Tzeng proved the existence of a polynomial $T_\alpha$ such that for any sufficiently …
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2 votes

About the isotriviality of pencils of plane curves

This is not an answer but rather a lengthy comment. A necessary condition for the pencil to be isotrivial is that a smooth member in that pencil has a non-trivial automorphism: By blowing-up the base …
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