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

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 $(z,w)\in{\mathbb{C}}^{2}$ be our coordinates and let $V$ be defined by $w=e^{1/z}$ in ${\mathbb{C}}^{2}\setminus S,$ where $S$ is defined by $z=0.$ The closure of $V$ in ${\mathbb{C}}^{2}$ 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.

References

• 1 Klaus Fritzsche, Hans Grauert. , Springer-Verlag, New York, New York, 2002.
• 2 Hassler Whitney. . Addison-Wesley, Philippines, 1972.
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