3
$\begingroup$

Here is the motivation of this question: $\mathbb C^n$ is already "local" in algebraic category. In other words, algebraic subvarieties of $\mathbb C^n$ are affine, so they are common zero locus of finitely many polynomials defined on $\mathbb C^n$.

However, $\mathbb C^n$ is not "local" enough for analytic varieties, since by definition (Griffiths & Harris, page 12), an analytic subvariety $X$ of $\mathbb C^n$ is that for each $x\in X$, there is an open neighborhood $U$ of $x$ in $\mathbb C^n$, such that $X\cap U$ is common zero loci of holomorphic functions $f_1,...,f_k$ defined on $U$.

Of course, globally defined holomorphic functions will do the job, for example, the graph of the entire function $z\mapsto e^z$ produces the simplest analytic subvariety of $\mathbb C^2$ that is not algebraic. However, is there an example of a closed analytic subvariety of $\mathbb C^n$ not defined by global holomorphic functions on $\mathbb C^n$?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I think that you need to be more careful with a definition of analytic subvariety, otherwise a zero set of a function holomorphic in some domain is an example. $\endgroup$
    – Oleg Eroshkin
    Commented May 4, 2020 at 0:34
  • $\begingroup$ @OlegEroshkin Thanks, I think the example I'm looking for is unbounded, so I added closedness assumption. $\endgroup$
    – AG learner
    Commented May 4, 2020 at 2:00

1 Answer 1

Reset to default
3
$\begingroup$

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 text on functions of several complex variables.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ Thank you. Vaninghing of $H^1(\mathbb C^n,\mathcal{O})$ implies gluing divisors on $\mathbb C^n$ has no obstruction. This solve the problem when $X$ is codimension one, but I don't see how the why general case follows from it. $\endgroup$
    – AG learner
    Commented May 4, 2020 at 15:13
  • 3
    $\begingroup$ @AGlearner You need Cartan's B theorem for arbitrary codimension. $\endgroup$
    – Oleg Eroshkin
    Commented May 4, 2020 at 15:46
  • $\begingroup$ @OlegEroshkin Sorry, it should be $H^1(X,\mathcal{O}_X^*)$ controls the divisor (instead of $H^1(X,\mathcal{O}_X)$ as I wrote). However, $\mathcal{O}_X^*$ is not a coherent sheaf anymore since it is not even a sheaf of $\mathcal{O}_X$-module, so I guess we cannot apply Cartan B here. But I should thank for your suggestion since it brought me to the fact that "$X$ being closed analytic subvariety of $\mathbb C^n$ implies $X$ is a stein space". $\endgroup$
    – AG learner
    Commented May 4, 2020 at 16:22
  • $\begingroup$ Dear @OlegEroshkin, if I apply Cartan's theorem A to the ideal sheaf $I_X$ on $\mathbb C^n$, there are finitely many global sections $s_1,...,s_k\in H^0(\mathbb C^n, I_X)$ which induce the finite generation $\oplus_{i=1}^k\mathcal{O}_{\mathbb C^n}\twoheadrightarrow I_X$. Therefore, as globally holomorphic functions, $\{s_1=0,...,s_k=0\}$ defines $X$. Do you think this is correct? Also, could you explain a little bit about your suggestion on applying Cartan B? Thanks! $\endgroup$
    – AG learner
    Commented May 7, 2020 at 2:45
  • 1
    $\begingroup$ That was my thinking. But, unless I am mistaken, you will get finite generation only over the arbitrary compact. I think, that over $\mathbb{C}^n$, the ideal sheaf may not be finitely generated. The idea: take a union of varieties, which required at least $k$ functions to define. By translating varieties, we may enforce that the union is locally finite. Then the union is an analytic set. To avoid such issues, you may consider an irreducible analytic set. $\endgroup$
    – Oleg Eroshkin
    Commented May 7, 2020 at 17:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .