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Results tagged with complex-geometry
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user 4144
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.
7
votes
Accepted
Compact Riemann surfaces as holomorphically convex subsets of affine algebraic varieties
Perhaps I should turn my comment into an "answer".
Affine algebraic varieties over $\mathbb{C}$ are Stein spaces. That is,
they are already holomorphically convex, and points can be separated by glob …
- 35.2k
5
votes
Accepted
vanishing theorem in algebraic geometry
I guess I could have just told you in person, but anyway, yes your specific question has a positive answer. To see this, use Chow's theorem and resolution of singularities to find a birational map $\ …
- 35.2k
3
votes
Motivation behind spectral sequences
I suspect a previous comment of mine led to this question, so let me say a few words here. The basic problem is this: Suppose $(A^\bullet, F)$ is a filtered complex, then one wants to relate the (co …
- 35.2k
10
votes
Examples of compact complex manifolds for which the $dd^c$ lemma does not hold
A known consequence of the $dd^c$-lemma is the vanishing of Massey products, which are certain secondary cohomology operations, see Deligne, Griffiths, Morgan, Sullivan, Real homotopy theory of Kähler …
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11
votes
Are most Kähler manifolds non-projective?
I agree that in a sense it's harder to get your hands a non algebraic Kähler manifold, because you can't simply write an equation for one, but I would argue that there are plenty of them. You won't fi …
- 35.2k
5
votes
Accepted
complex algebraic morphisms as topological maps: every morphism is a topological fibration o...
I have also wondered about question (i) in the past, and fortunately the answer is yes.
Here is a reference:
Verdier, Stratification de Whitney et theoreme de Bertini-Sard, Invent 1976, Cor 5.1
The …
- 35.2k
14
votes
Newlander-Nirenberg theorem for general vector bundles
Let me supplement Johannes' answer a bit. In higher dimensions a holomorphic vector
bundle $V$ is equivalent to a $C^\infty$ bundle equipped with a Cauchy-Riemann
operator $\bar D$ (as explained in hi …
- 35.2k
5
votes
Complex line bundle over curves
As several people have pointed out, your example has degree $0$. Another way to
see this is to observe that given a section, the number of zeros minus poles in both hemispheres is a difference of two …
- 35.2k
3
votes
Compact holomorphic symplectic manifolds: what's the state of the art?
It's probably bad form to answer one's own question -- even the software has registered its disapproval -- but I thought I'd make a small update. Regarding classification in low dimensions, accordi …
- 35.2k
7
votes
Strong Kodaira vanishing
Perhaps I might add that the "strong Kodaira vanishing" holds more generally for smooth
projective toric varieties in any characteristic. This goes back to Danilov. This includes
your case 1 of course …
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3
votes
Accepted
A question on Steenbrink's paper, limit of Hodge structures
If $\pi$ is semistable, then Fujisawa, Limits of Hodge structures in several variables. Compositio (1999), does this. You might also look at some later papers by the same author for some refinements. …
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5
votes
Accepted
Admissible global residues on smooth variety with normal crossings divisor
Jason has given an essentially complete answer, which I'm just repeating it here, so that the question can be considered answered. (I think the sequence is OK, however. E.g. for $D=\{xy=0\}$ locally, …
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9
votes
Accepted
$h^{p,q} = h^{q,p}$ on complex smooth projective scheme
I remember seeing such a proof in an article by Messing, who attributed it to Gabber. Let $X$ be a smooth projective variety of dimension $n$ over a field of characteristic $0$.
Suppose that $p+q=i\l …
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4
votes
Accepted
Relations between Dolbeault cohomology and the corresponding $L^2$-cohomology
I don't know if that was a typo (did you want isomorphism, or did you really mean inclusion/injection?). In any case, the question in either interpretation has a negative answer. Let $X$ be a compact …
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9
votes
What can be the analogue of Frobenius in complex geometry?
I've been hesitating about whether or not to answer this question. Not because it's uninteresting, but because I think it's too interesting. I'm glad you didn't ask what the influence of Frobenius is …
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