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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.
29
votes
Accepted
Non-compact complex surfaces which are not Kähler
Following David Speyer's suggestion, let $X=\mathbb{C}^2-\{0\}/\lbrace(x,y)\mapsto (2x,2y)\rbrace$
be the standard Hopf surface. The image, $E$, of the $x$-axis is an elliptic curve.
Remove a point o …
25
votes
de Rham vs Dolbeault Cohomology
Although the main questions have been answered quite well, I would like to say a few words
about the first question "why would one bother...".
De Rham and Dolbeault cohomology are measuring differen …
24
votes
Accepted
Why are Stein manifolds/spaces the analog of affine varieties/schemes in algebraic geometry?
Also like affine varieties, we have:
Theorem. A complex manifold is Stein if and only if it embeds into some $\mathbb{C}^N$ as a closed complex submanifold.
For the "only if" direction, see Hörm …
22
votes
Accepted
Are "ample" and "positive" line bundle the same concept?
Yes. For ample implies positive, use the fact that $c_1(O(1))$ on projective space is the Kähler form of the Fubini-Study metric, and then restrict to $X$. For the converse, you need the Kodaira embed …
20
votes
Accepted
Mixed Hodge structure on sheaf cohomology of a variation of Hodge structures
The answer to your question is yes (provided the VHS is polarized). This is due to Morihiko Saito. It is implicitly contained his two long papers on (mixed) Hodge modules, and there is an explicit sta …
18
votes
Why do people think that abelian varieties are the hardest case for the Hodge conjecture?
The class of Abelian varieties is the simplest class of varieties where the Hodge conjecture is not known to be true. So naturally a certain amount of effort is directed toward them.
However, it's no …
15
votes
Hodge theory on complex spaces
The answer is that if $X$ is compact complex space of class $C$ in Fujiki's sense (i.e.
if it is dominated by compact Kaehler manifold) then $H^*(X)$ carries a natural
mixed Hodge structure. This is g …
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 …
14
votes
Accepted
Fundamental groups of normal complex quasi-projective varieties
Yes, sure. Take any polycyclic group which is not virtually nilpotent (e.g.
upper triangular matrices with entries in $GL_n(\mathbb{Z})$, $n\ge 2$). This cannot be the fundamental group of a normal q …
14
votes
Accepted
Variety without a compactification whose complement is smooth
Presumably, you want $\bar X$ to be smooth as well. Then there are many examples. Here is a simple one. Let $\bar Y$ be smooth projective curve of positive genus. Now remove at least two points to get …
13
votes
Accepted
$H^1(X,O_X)$, holomorphic $1$-forms, and $b_1(X)/2$ for normal $X$.
(Although I have pretty much "retired" from Mathoverflow, I will answer this, since the answer is nice but probably not all that well known.)
Theorem. If $X$ is complex normal projective variety, …
12
votes
Accepted
Can there exist Chow motives/motivic cohomology for compact Kähler manifolds?
Hi Mikhail,
I honestly don't have a good answer, but I'll share my thoughts on this and related things, since this is potentially quite interesting.
I don't see a problem in formally defining the C …
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 …
11
votes
Accepted
Do complex varieties have a dense open subset with residually finite fundamental group?
(I'm converting my comment to an answer.)
In SGA4 exp XI, Artin constructs a nonempty Zariski open $U\subset S$ which admits a sequence $U=U_n \to U_{n-1}\to\ldots $ which are topological fibrations …
10
votes
Intuition behind the Kodaira Vanishing Theorem?
Let me address the modified question (as corrected by Jason). There are many intuitions depending on which proof-style you want to use. Here are a few remarks.
The original proof of Kodaira uses th …