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 $U$ ofdimension Let $V$ be a complex analytic subvariety of $U\mathrm{\backslash \backslash }S$ such that ${\dim }_p\mathit{}V\mathrm{>}m$ for allregular 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 irreduciblecomponents of $V$ are of dimension strictly greater than $m.$ To show that the restriction on the dimensionof $S$ 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 beanalytic. 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 dimensionalcomplex 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 therestrictions on $S$ even further.