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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.

3 votes

Closed analytic subvariety of $\mathbb C^n$ not defined by global holomorphic functions

A closed subset $X$ of $C^n$ satisfying your definition can be defined by global analytic functions. This follows from the solution of the "Second Cousin problem" in $C^n$, and is explained in any tex …
Alexandre Eremenko's user avatar
5 votes

intersection of holomorphic curve with hyperplane

EDITED. Suppose first that $f(C)$ is not contained in any hyperplane. Let us describe hyperplanes by equations $L(w)=a_1w_1+\ldots+a_nw_n-b=0$, where we can normalize so that $\sum |a_j|^2+b^2=1$. Now …
Alexandre Eremenko's user avatar
1 vote

A problem related to connectivity of analytic functions

No, it can be infinite. Start with the function $w=f(z)$ mapping the unit disc onto itself. Let $I_k$ be disjoint closed arcs on the image disc in the $w$ plane. Deform the image region by adding two …
Alexandre Eremenko's user avatar
1 vote

Riemann surface disconnected at infinity

Let $C$ be a complex line in $C^2$, say $y=0$. Project it on $x$-line, all properties are satisfied:-) If you really want "connected, but ONLY if one goes near the origin", take the set $\{(x,y): y^2 …
Alexandre Eremenko's user avatar
5 votes
Accepted

Is there an explicit formula for the modulus of an annulus given a parameterization of the i...

I think that there is no formula. The best one can do is to estimate. Here is a simpler problem of the same sort: suppose you have a parametrization of the boundary of a simply connected region, and s …
Alexandre Eremenko's user avatar
5 votes
Accepted

Extension of pluriharmonic functions

Your condition $\dim M>2$ does not save the situation: you can have many counterexamples with $M=M'\times C^n$ where $\dim M'=1$ and your functions are independent of the second variable. And in dimen …
Alexandre Eremenko's user avatar
2 votes
Accepted

Bounding the derivative of a holomorphic function on a disk by its absolute value

Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact. To obtain the estimate $|f'(z)|<1$, apply Cauc …
Alexandre Eremenko's user avatar
1 vote

Is there an underlying geometric reason for the rigidity of complex geometry?

To expand my comment, let me mention the wonderful book by R. Penrose, The road to reality. A complete guide to the laws of universe, A.A. Knopf, NY 2006. He discusses this under the title "Magic of c …
Alexandre Eremenko's user avatar
1 vote

Proof verification for a theorem about a harmonic function on the unit disc

This is a consequence of the Phragmen-Lindelof Principle, whose general formulation is the following: Let $D$ be a bounded region in the plane, and $\zeta_0\in\partial D$. Let $u$ be a subharmonic fun …
Alexandre Eremenko's user avatar
6 votes

An inf-sup estimate for holomorphic functions

This is not true: take $n=1$, $r=1$, $\eta(z)=e^{az},\; a>0,$ then $$\max_{z\in B(0,r)}|\eta(z)|=e^a,$$ while $$\min_{z\in B(0,1)}|\eta(z)|=e^{-a}.$$ Since $a>0$ is arbitrary, no $\kappa$ with require …
Alexandre Eremenko's user avatar
15 votes
Accepted

Interesting results for open Riemann surfaces

The results on open Riemann surfaces are not "rare". They are just well forgotten. I only list a few books which deal with open Riemann surfaces: MR0114911 (Zbl 0196.33801), MR0228671 (Zbl 0152.27401) …
Alexandre Eremenko's user avatar
1 vote

Higher dimensional analogue of Ahlfors covering surface theory

Yes, there exists such a generalization. It was done by Marie-Helene Schwartz. Formules apparentées à la formule de Gauss-Bonnet pour certaines applications d’une variété à n dimensions dans une autr …
Alexandre Eremenko's user avatar
3 votes

Examples of pluripolar sets

As you said, analytic sets (other than the whole space) are pluripolar. By playing with subharmnic functions of the form $\sum u_j$ you can arrange some more complicated sets which are neither open n …
Alexandre Eremenko's user avatar
3 votes
Accepted

Harmonic function osculating a given subharmonic function

The answer is no. The reason is that a subharmonic function does not have to be continuous. For example, there is a subharmonic function $u$ in the unit disk, $u(0)=1$ and $u(z_k)=0$ for a sequence $z …
Alexandre Eremenko's user avatar
3 votes

Which are the recommended books for an introductory study of complex manifolds?

MR2359489 Wells, Raymond O., Jr. Differential analysis on complex manifolds. With a new appendix by Oscar Garcia-Prada. Graduate Texts in Mathematics, 65. Springer, New York, 2008.

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