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Complex geometry is the study of complex manifolds, complex algebraic varieties, complex analytic spaces, and, by extension, of almost complex structures. It is a part of differential geometry, algebraic geometry and analytic geometry.

4 votes
Accepted

Extending holomorphic forms

Here is a result which is sort of what you are asking: Theorem (Greb-Kebekus-Kovács-Peternell) Let $X$ be a complex quasi-projective variety of dimension $n$ and let $D$ be a $\mathbb Q$-divis …
Sándor Kovács's user avatar
5 votes
Accepted

Making Hironaka's theorem explicit for hypersurfaces

By now there are more tractable proofs of resolution of singularities than Hironaka's, so it no longer has to be a black box. A relatively elementary approach is described in Kollár's Lectures on Reso …
Sándor Kovács's user avatar
6 votes

Projective variety with no syzygies but not isomorphic to projective space

Here is an example for Torsten's Addendum: a line bundle with the required properties which is not semi-ample (no multiple is globally generated): Take an $\pi:X\to \mathbb P^n$ as in Torsten's examp …
Sándor Kovács's user avatar
2 votes

Uniformity of ampleness

You are essentially asking about a bound on the Seshadri constant of $A$ (on $X$). In fact, you are asking for something a little stronger. You can see the definition and basic properties of Seshadri …
Sándor Kovács's user avatar
8 votes
Accepted

Uniformity of ampleness

Remark: If $p=q$, then one can just work with $E_p$ instead of $2E_p$ and then at the end take $2k_0$ instead of $k_0$, so we may assume that $p\neq q$ and in particular that the exceptional diviso …
Sándor Kovács's user avatar
4 votes
Accepted

Canonical bundle of compactifications

perhaps I am misunderstanding what you ask, but the answer is no to both questions. Take an arbitrary projective variety $\overline X$, actually for simplicity let $\overline X$ be normal. Then it has …
Sándor Kovács's user avatar
4 votes

Sections of the canonical bundle of a blow-up

Lemma Let $\phi:Y\to X$ be a proper birational morphism of complex manifolds of dimension at least $2$. Let $E\subset Y$ be the exceptional locus of $\phi$. Note that $E$ is a Cartier divisor. Then fo …
Sándor Kovács's user avatar
5 votes

Smoothing subvarieties

I think the answer to your last question as posed is "no", but the first question may be actually quite difficult. #1 If you hadn't required $W$ to be a general complete intersection, then the answe …
Sándor Kovács's user avatar
5 votes

Is a smooth intersection of hypersurfaces equidimensional?

If the intersection $\cap_{i=1}^nV_i$ is irreducible, then it is equidimensional. Otherwise let $r<n$ be such that $\cap_{i=1}^rV_i$ is irreducible, but $A:=\cap_{i=1}^{r+1}V_i$ is not. If $\dim\cap_{ …
Sándor Kovács's user avatar
22 votes

flatness in complex analytic geometry

Mohammad, flatness is not simple, so you are not going to get a simple overall definition. On the other hand, in the example you mention in your comment, there is a simple criterion: Let $f:X\to …
Sándor Kovács's user avatar
8 votes

in the analytic category, finite morphisms are open maps?

Open Mapping Theorem [Grauert-Remmert: Coherent analytic sheaves, p.107] Let $X,Y$ be pure $d$-dimensional complex spaces and assume that $Y$ is locally irreducible. Then any holomorphic map $f:X\to Y …
Sándor Kovács's user avatar
6 votes

Question about Hodge number

Proposition. For compact complex manifolds of dimension $1$ (a.k.a. complex Riemann surfaces) the Hodge numbers are topological invariants. Proof. The Hodge numbers are determined by the genus, which …
Sándor Kovács's user avatar
5 votes
Accepted

Extension of Kollár's vanishing theorem to singular varieties?

If $X$ has rational singularities, then this vanishing (and the torsion-freeness as well) follows almost trivially. Let $g:Z\to X$ be a resolution of singularities. Then $Rg_*\omega_Z\simeq \omega_X$ …
Sándor Kovács's user avatar
6 votes
Accepted

When is a holomorphic submersion with isomorphic fibers locally trivial?

Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathb …
Sándor Kovács's user avatar
6 votes

Proper morphisms

No. Let $g:Y\to S$ be an arbitrary finite morphism, $Z\subset Y$ a closed subset such that $Z$ does not contain an entire fibre (for instance any closed point works if the degree of $g$ is larger than …
Sándor Kovács's user avatar

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