Say $Z$ is an anlatyic set in a complex manifold $M$. Is it true that irreducible components of $Z$ are just the closures in $Z$ of connected components of $Z\backslash sing(Z)$?

I need this for a paper I am reading. In general, can someone reccomend a good book about complex analytic geometry? I don't want to waste a lot of time figuring these things out without at least a good source for all these notions.

Thank you! Benny


1 Answer 1


[1] The answer to your question is yes.There is an important theorem on decomposition analytic subset into irreducible components:(Cf E.M.Chirka's Complex Analytic Sets chapter 5.4.Theorem1)

Let $A$ be an analytic subset of a complex manifold $\Omega$.Then:

(1)every irreducible component $A$ has the form $\overline S$,where $S$ is a connected component of $regA=A\backslash singA$;

(2)if $reg A=\bigcup_{j\in J}S_j$ is the decomposition into connected connected components ($J$ is finite or countable, $S_j\bigcap S_k=\varnothing$ for $j\not= k$),then $A=\bigcup_{j\in J}\overline {S_j}$,and this is the decomposition of $A$ into ieeducible components;

(3) the decomposition of $A$ into irreducible components is locally finite,i.e.for every compact $K\subset\Omega$ there is only a finite (or empty) set of indices $j\in J$ such that $K\bigcap\overline {S_j}$ is nonempty.

[2] Demailly's Complex Analytic and Differential Geometry is wonderful.I also recommend E.M.Chirka's Complex Analytic Sets,it's worth reading and full of a lot of classical theorems.

  • $\begingroup$ Thank you for your answer! In the case $Z$ is an algebraic set in $M=\mathbb{C}^{n}$, is it true that the irreducible components in the sense of zariski topology coincide with the irreducible components in the analytic sense? $\endgroup$ Commented Dec 11, 2019 at 19:34

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