Let $U \hookrightarrow X$ be an embedding of algebraic varieties such that $U$ is dense in $X$. Then any Zariski closed subset of $U$ is a trace of a Zariski closed subset of $X$.

It escapes me why a similar fact is not true for complex manifolds.

Again, let $U \hookrightarrow X$ be an embedding of complex manifolds such that $U$ is dense in $X$ and let $Z$ be an analytic subset of $X$. Then $Z$ is not necessarily an intersection $Z' \cap U$ where $Z'$ is an analytic subset of $X$. For example, if $X=\mathbb{P}^2$, $U=\mathbb{A}^2$ and $Z$ is the set defined by the equation $y=exp(x)$ then by Chow's theorem it cannot be a trace of an analytic subset of $\mathbb{P}^2$ since all such subsets are algebraic.

What is the reason that makes the extension of an analytic subset fail? Is it possible to formulate a general criterion?


1 Answer 1


The most useful general criterion for extending analytic subsets of an open subset of a manifold to the whole manifold is :

Theorem of Remmert-Stein
Let $X$ be an analytic space (for example a complex manifold), $Z\subset X$ a closed analytic subspace of dimension $\lt n$ and $A\subset U$ a purely $n$-dimensional closed analytic subset of the open set $U=X\setminus Z\subset X.$
Then the closure $\bar A\subset X$ is an analytic subspace of $X$.

So the "reason" for the failure of extension in your example is that your $Z$ is too big: it is a line of dimension one and your analytic subset , the graph of the exponential function, has dimension one as well, so that its closure in $\mathbb P^2 $ is no longer analytic.
However if you delete a point $Q\in \mathbb P^2 $ then any analytic curve $C\subset \mathbb P^2 \setminus \lbrace Q \rbrace$ will have analytic (even algebraic) closure $\bar C\subset \mathbb P^2$

In a comment below Dima asks about extending an analytic subset of $A\subset \mathbb A^2(\mathbb C)$ to $\mathbb P^2(\mathbb C)$, a situation where Remmert-Stein's theorem doesn't apply .
By Chow's theorem $\bar A \subset \mathbb P^2 $ is analytic iff it is algebraic, and in that case $A\subset \mathbb A^2$ is algebraic too, so that finally $A$ is extendable iff it is algebraic.
The simplest obstruction against algebraicity of $A$ might be that a line in $\mathbb A^2$ not contained in $A$ intersects $A$ only in finitely many points.
In Dima's example where $A$ is defined by $y=exp(x)$ , the line $y=1$ intersects $A$ in infinitely many points, which is an obstruction against $A$ being algebraic and thus explains why $A$ is not extendable to $\mathbb P^2$

  • $\begingroup$ Thank you for the answer. Yet there are many analytic subsets of $\mathbb{A}^2$ that do extend to $\mathbb{P}^2$. I wonder if there is some kind of obstruction in case of the set $y=exp(x)$ that vanishes for algebraic subsets of $\mathbb{A}^2$. $\endgroup$ Mar 8, 2012 at 10:51
  • $\begingroup$ Dear @Dima: I have written an edit about this question of obstruction. $\endgroup$ Mar 8, 2012 at 23:17
  • $\begingroup$ Dear Georges, thanks for the update. Your answer is tailored to this particular choice of X and Y. I was wondering about an obstruction in the general situation. I have so far found this article by Bishop: projecteuclid.org/… He proves that an analytic set can be extended if every point of $Y\setminus X$ has a neighbourhood $U$ such that $U \cap Z$ has a finite volume $\endgroup$ Mar 8, 2012 at 23:34
  • $\begingroup$ Dear @Dima: I had indeed a vague memory of having seen long ago a criterion for extendability in terms of volume but I didn't remember where. Bishop's article looks very interesting and it also proves a version of Remmert-Stein. Thanks a lot for that great link! $\endgroup$ Mar 9, 2012 at 0:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.