10 votes

Complex vector bundles on compact complex manifolds

This is an explanation of my comment above, namely: "Complex vector bundles over a CW complex of dimension $\leq 4$ are classified by their Chern classes and rank. Moreover, every possible choice ...
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9 votes
Accepted

Examples of 6-manifolds without an almost complex structure

Turning comments into answer: An example of a closed 6-manifold not admitting an almost complex structure is $S^1 \times (SU(3)/SO(3))$. From the obstruction theory for lifting the map $M \to BSO(6)$ ...
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7 votes
Accepted

Difference between stabilizer and automorphism group of subvariety of an abelian variety

They have absolutely no reason to be equal. Consider the case where $A$ is the Jacobian of a genus 2 curve $C$, and $X=C$ embedded in $A$ by $x\mapsto [x]-[p]$ for some fixed point $p\in C$. Then $X$ ...
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  • 34.3k
4 votes

When is bijective map between closed point of varieties a morphism?

By Nakayama's lemma, the support of a coherent sheaf $\mathscr F$ on a Noetherian scheme $X$ is the set of $x \in X$ such that $\mathscr F \otimes_{\mathcal O_X} \kappa(x) \neq 0$ (as opposed to $\...
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4 votes
Accepted

Pull back a vector field

You are not pulling back vector fields. You are pulling back the vector bundle $T_B$ to be a bundle over $\mathcal{X}$. (See, e.g. https://en.wikipedia.org/wiki/Pullback_bundle for a description.) ...
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2 votes

When is a topological fiber bundle Zariski locally trivial?

Since you asked for "comments, suggestions" here are two thoughts off the top of my head: In Torsion Homologique et Sections Rationnelles by Grothendieck (Séminaire Chevalley, 1958), the ...
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  • 8,022
2 votes
Accepted

Pull-back of factor of automorphy

I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is ...
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  • 7,553
2 votes

When is bijective map between closed point of varieties a morphism?

It's important to distinguish two things. One is what it takes to uniquely specify a morphism, and the other is what it takes to construct a morphism. When we say a morphism is uniquely specified by ...
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  • 115k
1 vote

Can a non-Kähler complex manifold be rationally connected?

I am writing my comment as a question. I have certainly explained these examples before on MathOverflow, since they show that the Kollár-Miyaoka-Mori conjecture cannot hold beyond Fujiki class $\...
1 vote

Fundamental groups of primitive non-algebraic compact Kähler manifolds

I am not sure how one checks primitivity for a compact Kähler manifold, this seems to be too delicate property. Intuitively speaking, very non-algebraic Kähler manifolds have very simple fundamental ...
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1 vote

Holomorphic vector fields acting on Dolbeault cohomology

Klemyatin proved that this action is trivial if the corresponding ${\Bbb C}$-flow is compatible with some metric (hence can be extended to a compact torus action), https://arxiv.org/abs/1909.04075, (N....
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