The condition that for all regular is the same as saying that all the irreducible components of are of dimension strictly greater than To show that the restriction on the dimension of is “sharp,” consider the following example where the dimension of equals the dimension of . Let be our coordinates and let be defined by in where is defined by The closure of in cannot possibly be analytic. To see this look for example at If is analytic then ought to be a zero dimensional complex analytic set and thus a set of isolated points, but it has a limit point by Picard’s theorem.
Finally note that there are various generalizations of this theorem where the set need not be a variety, as long as it is of small enough dimension. Alternatively, if is of finite volume, we can weaken the restrictions on even further.
- 1 Klaus Fritzsche, Hans Grauert. , Springer-Verlag, New York, New York, 2002.
- 2 Hassler Whitney. . Addison-Wesley, Philippines, 1972.
|Date of creation||2013-03-22 15:04:55|
|Last modified on||2013-03-22 15:04:55|
|Last modified by||jirka (4157)|
|Synonym||Remmert-Stein extension theorem|