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Let $M$ be a complex manifold, and $Z \subset M$ a closed real analytic subvariety. Suppose that the set of smooth points in $Z$ is complex analytic in $M$. Will it follow that $Z$ is complex analytic? I can deduce this statement from Remmert-Stein theorem when $2\dim_R S < \dim_R Z$, where $S$ is the singular set of $Z$. Also I can deduce this statement from Skoda-El Mir theorem when $S$ is pluripolar. I suspect that it should be true in bigger generality, maybe always.

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  • $\begingroup$ Dear Misha i believe there is a typo, it should be $2\rm{dim}_R S<\rm{dim}_R Z$ right? $\endgroup$ Commented Aug 24, 2022 at 10:04
  • $\begingroup$ Right! Thanks. I will correct it $\endgroup$ Commented Aug 26, 2022 at 21:09

2 Answers 2

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There is a paper by Hans-Jörg Reiffen called "Fastholomorphe Algebren" (1970) where he apparently proves something like this. Here is a link to the springer publication. I have not read the paper in detail so I can not really comment on the methods used there. Moreover, the paper is written entirely in german. The last highlighted statement in the appendix reads (translated to english by me):

"Let $G$ be a domain in $\mathbb{C}^N$ and $X\subseteq G$ a real analytic subset, which in the points of an open dense subset of $X$ is complex analytic, then $X$ is complex analytic."

EDIT: Below I have translated the statement 1' and 2' from the appendix of the mentioned paper. Apparently the above quoted statement is a corollary from these two facts, at least that is what the author of the paper states. In the paper the notation $R_{\omega}$ denotes the ring of complex-valued real analytic functions.

Satz 1'. Let $X_0$ be a germ of an irreducible real analytic set in $\mathbb{R}^N$. Then there exists a real analytic subgerm $S_0$ of $X_0$ with the following properties:

(a) $\mathrm{dim}X_0> \mathrm{dim}S_0$

(b) There exist representatives $X$ of $X_0$ and $S$ of $S_0$ in an open neighbourhood $U$ of $0$ such that for all $x\in X\setminus S$ the germ $X_x$ is non-singular and $\mathrm{dim}X_x= \mathrm{dim}S_0$.

(c) If $f_1,...,f_n$ is a generating system of $J_{\omega}(X_0):=\left\{f\in R_{\omega} : f\mid_{X_0}=0\right\}$, then the $f_1,....,f_n$ is a coherence basis on $X_0\setminus S_0$ (i.e. there exists an open neighbourhood $U$ of $0$ with representatives $X$ of $X_0$ and $S$ of $S_0$, such that the $f_1,....,f_n$ are defined on $U$ and for all $x\in X\setminus S$ the germs $(f_1)_x,...,(f_n)_x$ generate the ideal $J_{\omega}(X_x)$.)

Satz 2'. Let $X_0$ be a germ of an irreducible real analytic set in the origin of $\mathbb{C}^N$. Further assume that there exists a set $Y$ in a representative $X$ of $X_0$ with the following properties:

(a) $0$ is an accumulation point of $Y$.

(b) For $y\in Y$ one has $\mathrm{dim}X_y=\mathrm{dim}X_0$ and $X_y$ is complex analytic.

Then $X_0$ is complex analytic.

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Does this work?

Let $U\subset Z$ be the set of smooth points. This is open and dense in $Z$. Let $Z'\subset M$ be the smallest closed complex analytic variety containing $Z$. Let $U'$ be the set of smooth points of $Z'$. I believe that $U=Z\cap U'$ is not only open but also closed in $U'$, by some form of the uniqueness of analytic continuation. So $U'\backslash U$ is open in $Z'$ and disjoint from $Z$. Can you argue that the closed set $Z'\backslash (U'\backslash U)$ is complex analytic and therefore must be all of $Z$? If so, then $U=U'$ and therefore $Z=Z'$.

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  • $\begingroup$ I am not sure this could work; there is nothing prohibiting $Z'=M$, then $U'=M$ as well, your assumption $U=Z\cap U'$ will fail. however, it is not hard to find small-dimensional real analytic subvarieties which are Zariski dense. $\endgroup$ Commented Aug 30, 2022 at 17:50
  • $\begingroup$ How can $Z$ be Zariski dense without $U$ being Zariski dense? I confess that I am confused about local-versus-global issues. $\endgroup$ Commented Aug 30, 2022 at 18:57
  • $\begingroup$ Z is not a comples subvariety, an u is not closed; it is not har to find varieties which are not close, an the smallest complex subvariety containing them is the whole ambient variety; take, for instance, the graph of exponent in P2 $\endgroup$ Commented Sep 1, 2022 at 14:22

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