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Algebraic varieties, stacks, sheaves, schemes, moduli spaces, complex geometry, quantum cohomology.
2
votes
A question on curves on effective divisors
That's not true. Curves are Cohen-Macaulay if and only if they don't have embedded primes (see Wikipedia). For example the curve
$$C = \{x=0, y^2 = 0\} \subset \mathbb P^3$$
is CM. But $C$ is not cont …
2
votes
Map of coherent sheaves inducing isomorphism on the stalks at the generic point
I think this can also be solved by taking the "coherent subsheaf generated by finitely many global sections of $i_* \mathcal{O}_U$". I'm not sure this is fundamentally different from Aknazar's solutio …
0
votes
Stable reduction for maps
An Invitation to Quantum Cohomology by Joachim Kock and Israel Vainsencher has a bit on this. First there is Example 2.2.2 on page 57, which is done in detail, and then there is a discussion in paragr …
6
votes
2
answers
469
views
Why is the scheme of isomorphisms of sheaves affine over the base?
Suppose $S$ a noetherian base scheme, $X \to S$ is projective and $F, G$ are coherent $\mathcal O_X$-modules. Then by EGA (7.7.8) and (7.7.9) there exists a scheme $H = \underline{\operatorname{Hom}}_ …
5
votes
Accepted
Why is the scheme of isomorphisms of sheaves affine over the base?
So I realized Jason Starr and Johan de Jong only claim that $H = \underline{\operatorname{Hom}}_S(F, G)$ is affine if $F$ and $G$ are locally free. In that case, if $U = \operatorname{Spec}(A) \subset …
2
votes
Accepted
Intersection pairing and birational morphisms
First note $\overline{f(C \cap V)} = f(C)$, since $f$ is closed and $C$ (and hence also $f(C)$) is irreducible. Also $f$ induces a birational map $C \to f(C)$, so $f_* [C] = [f(C)]$ where $[\cdot]$ de …
2
votes
1
answer
158
views
Irreducible components of a general singular fiber correspond to irreducible components of t...
I already asked this on math.SE, but didn't receive any response.
The following question arose when studying Hwang and Oguiso's Characteristic foliation on the discriminant hypersurface of a holomorph …
16
votes
0
answers
434
views
Can non-reduced fibers appear over a subset of codimension $\geq 2$?
I already asked this on math.stackexchange.com, but didn't receive an answer.
Suppose $f: X \to Y$ is a (possibly proper) morphism of complex manifolds (resp. smooth varieties) such that all fibers of …
2
votes
Accepted
Where are the critical points of a proper faithfully flat morphism
Take a point $x \in X$ and set $y = f(x) \in Y$. Let $X_y = f^{-1}(y)$ the "scheme-theoretic" fiber, i.e. the complex-analytic subspace of $X$ which is cut out the equation $f(x) = y$. Your question …
2
votes
Why is the normalization of a general fiber the general fiber of the normalization?
This is merely a recollection of the discussion in the comments, that I wrote for for myself. I think it is still incomplete, could anyone help me fill the gap at the end?
I will denote the normalizat …
2
votes
0
answers
145
views
Describing singular fibers of the lagrangian fibration $\mathcal M^s(0, [C], 1) \to |C|$
Let $S \to \mathbb P^2$ be a two-to-one cover branched over a sextic, i.e. $S$ is a K3-surface. Let $C \subset S$ be the preimage of a (smooth) quadric, so that by Hurwitz' formula, $g(C) = 5$. Accord …
13
votes
1
answer
757
views
J.-P. Serre: Duality of regular differentials on singular curves
I already asked this on math.stackexchange.com, but didn't get any responses. I hope it is appropriate here.
Let $X'$ be an irreducible singular algebraic curve over an algebraically closed field $k$, …
6
votes
J.-P. Serre: Duality of regular differentials on singular curves
In the algebraic setting, I solved the second case: Given $\omega \in \underline \Omega_Q' \setminus \underline \Omega_Q$, we want to find an $f \in \mathcal O_Q$, such that $\sum_P \operatorname{Res} …
4
votes
0
answers
218
views
Why does a deformation modify the complex structure *holomorphically*?
This is a question regarding Chapter 9.1 of Claire Voisin's book [1]
Let $\phi: \mathcal X \to B$ be a family of compact complex manifolds, that is a proper holomorphic submersion, with central fiber …
4
votes
0
answers
194
views
Is there Riemann-Roch without denominators for complex manifolds?
Let $X \subset Y$ be an inclusion of compact complex (possibly Kähler) manifolds. I'm wondering if "Riemann-Roch without Denominators" [1, Thm 15.3] holds in that situation. The statement is
For a ve …