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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.
5
votes
Accepted
A corollary to Stone-Weierstrass theorem
In your case we can find a holomorphic function on the plane that uniformly approximates the
given continuous function .It is a consequence of the following .Suppose K is a compact measure zero subset …
7
votes
Accepted
Connectedness of boundary of a Stein domain
This is a consequence of the fact that any Stein manifold with complex dimension at least 2 has one end. This follows from Hartogs extension across compact sets in Stein manifolds. You can find a proo …
3
votes
Accepted
A cohomological variant of the second Riemann's extension theorem
For the first cohomology statement, you need the codimension to be at least three and for the first and second cohomology, codimension four. This theorem was proved by G Scheja in [1]. You can also fi …
2
votes
1-convex and holomorphically convex
The answer to your question is yes , 1-convex implies holomorphic convexity. This is Grauert's solution of the Levi problem.You can find a proof in the book of Fritzsche and Grauert titled From Holomo …
9
votes
$\partial \bar{\partial}$ lemma for contractible domains
If in addition you assume that your domain is pseudoconvex then by a theorem of
A.Aeppli what you want is true.The paper is titled :On the cohomology structure of Stein
manifolds 1965 (Proc.Conf.Comp …
1
vote
Reference request: Existence of plurisubharmonic potential for positive (1,1)-current on Ste...
The fact that you are looking for requires the current to be exact.On any Stein manifold for which the second cohomology group with integer coefficients has non trivial torsion free part,there are hol …
2
votes
Singularities of the Remmert reduction of a holomorphic convex manifold
The Cartan Remmert reduction can be any normal Stein space.By a theorem of Bingener and Flenner Arch Math (Basel)32(1979) pages 34-37 One can construct normal Stein spaces with unbounded embedding dim …
5
votes
Is hyperbolicity a Zariski open condition?
Yes,Hyperbolicity is an open condition .You can find it in Brody's paper in Trans Amer Math Soc vol235 1978 page 216 . A more general statement can be found in Kobayashi's book Hyperbolic Complex Spac …
2
votes
Extension of strictly plurisubharmonic functions on a Kähler manifold
If a complex manifold has a strictly plurisubharmonic function then it cannot contain
positive dimensional compact analytic sets.This is clearly a necessary condition.So you
might start considering yo …
4
votes
Accepted
The Levi form of the distance squared function in a non-positively curved Kaehler manifold
By the Hessian comparison theorem the square of the distance function on X is strictly convex.
On Kahler manifolds strictly convex functions are strictly plurisubharmonic .By Grauert's
solution of th …
3
votes
Accepted
Connectivity of complements of Stein opens
If you look at corollary 4.10 page 45 of the book of Banica and Stanasila titled Algebraic methods in the global theory of complex spaces,you will find a proof of the following .Any irreducible Stein …
13
votes
Accepted
Kähler manifold which is not algebraic
generic complex tori in complex dimension 2 or higher.
MR
5
votes
Accepted
Plurisubharmonic function and complete Kähler metric on certain Kähler manifold
Question 1: Plurisubharmonic functions extend across codimension 2 subvarieties . Let X be the complex projective plane blown up at one point and D be the exceptional divisor then any plurisubharmonic …
5
votes
Accepted
Monodromy representations and branched covers
For a modern treatment of the Grauert Remmert argument see Chapter 4 in the book Several Complex Variables vol 7 by Dethloff and Grauert . For an alternate proof using resolution of singularities see …
3
votes
Accepted
Normal Cones for Complex Spaces
Proposition 1.17 in the paper Complex Analytic Cones by Axelsson and Magnusson in
Math Ann 273 pages 601-628 answers your question.