Remmert-Stein theorem

For a complex analytic subvariety $V$ and $p\in V$ a regular point, let ${dim}_{p}V$ denote the complex dimension of $V$ near the point $p.$

Theorem (Remmert-Stein).

Let $U\mathrm{\subset }{\mathrm{C}}^n$ be a domain (http://planetmath.org/Domain2) and let $S$ be a complex analytic subvariety of $U$ of dimension Let $V$ be a complex analytic subvariety of $U\mathrm{\backslash }S$ such that ${\dim }_p\mathit{}V\mathrm{>}m$ for all regular points $p\mathrm{\in }V\mathrm{.}$ Then the closure of $V$ in $U$ is an analytic variety in $U\mathrm{.}$

The condition that ${dim}_{p}V>m$ for all regular $p$ is the same as saying that all the irreducible components of $V$ are of dimension strictly greater than $m.$ To show that the restriction on the dimension of $S$ is “sharp,” consider the following example where the dimension of $V$ equals the dimension of $S$. Let $(z,w)\in {ℂ}^2$ be our coordinates and let $V$ be defined by $w=e^{1/z}$ in ${ℂ}^2\setminus S,$ where $S$ is defined by $z=0.$ The closure of $V$ in ${ℂ}^2$ cannot possibly be analytic. To see this look for example at $W=\overline{V}\cap \{w=1\}.$ If $\overline{V}$ is analytic then $W$ ought to be a zero dimensional complex analytic set and thus a set of isolated points, but it has a limit point $(0,1)$ by Picard’s theorem.

Finally note that there are various generalizations of this theorem where the set $S$ need not be a variety, as long as it is of small enough dimension. Alternatively, if $V$ is of finite volume, we can weaken the restrictions on $S$ even further.

Title	Remmert-Stein theorem
Canonical name	RemmertSteinTheorem
Date of creation	2013-03-22 15:04:55
Last modified on	2013-03-22 15:04:55
Owner	jirka (4157)
Last modified by	jirka (4157)
Numerical id	9
Author	jirka (4157)
Entry type	Theorem
Classification	msc 32A60
Classification	msc 32C25
Synonym	Remmert-Stein extension theorem
Related topic	ChowsTheorem