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My previous question on this topic had a negative answer, but Tom Goodwillie in the comments suggested a statement, which may be true, and even a strategy of how to prove it. I haven't been able to neither prove or refute it, and I abandoned the task, but now it seems that I need to know the answer again.

Let $U,V\subset \mathbb{C}^{n}$ be open and connected. Let $\Phi:U\to V$ be a holomorphic map.

Is it true that for every $x\in U$ there is an open connected neighborhood $W$ of $x$ and non-identically-zero holomorphic function $w$ on a (connected) neighbourhood of $\Phi(W)$, which vanishes at every critical value of $\Phi|_{W}$?

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  • $\begingroup$ I don't know the answer, but I would start by looking for literature about the singularity theory of holomorphic germs, such as "Equivalences between isolated hypersurface singularities" by Benson and Yau. $\endgroup$
    – Neil Strickland
    Commented Nov 8, 2017 at 9:51
  • $\begingroup$ Do you mean "vanishes at every critical value, AND nowhere else" ? (Otherwise take the zero function for $w$). $\endgroup$
    – Alexandre Eremenko
    Commented Nov 8, 2017 at 13:16
  • $\begingroup$ I think you want $w$ to be nowhere locally constant, otherwise you could take $w=0$ in any open neighborhood of $\Phi(W)$ and $w=1$ in some open set which does not intersect that open neighborhood. $\endgroup$
    – Ben McKay
    Commented Nov 8, 2017 at 15:26
  • $\begingroup$ You cannot expect $w$ to have no critical points. Think of the example $(x,y)\mapsto (u,v)=(x,y^3-3x^2y)$. The critical set is defined by $(y-x)(y+x)=0$ and its image is defined by $(v-2u^3)(v+2u^3)=0$. $\endgroup$
    – Tom Goodwillie
    Commented Nov 8, 2017 at 16:18
  • $\begingroup$ @BenMcKay Yes, you are right, or I could require $w$ to be holomorphic and non-zero on a connected neighborhood of $\Phi(W)$. $\endgroup$
    – erz
    Commented Nov 8, 2017 at 20:52

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If you mean that your function $w$ vanishes on the critical values and nowhere else, then the answer is no. Take $(y_1,y_2)=(x_1^2,x_1x_2)$. The Jacobian determinant is zero on the line $x_1=0$ but the image of this line is one point $(0,0)$. And the zero set of a holomorphic function cannot be one point.

If you mean that $w$ vanishes on the critical values, and perhaps somewhere else, then the answer is yes and trivial: take $w=0$.

EDIT. In the comments the question was modified as follows: does there exist a function $\not\equiv 0$ which is zero on the image of the critical set. To this the answer is yes when $n=2$ and no when $n\geq 3$.

Consider this map for $n=3$ $$x=u+w^2,\quad y=uv+w^2,\quad z=uve^v+w^2.$$ The Jacobian in zero in sufficiently small neighborhood of the origin of and only if $uvw=0$. Take the plane $w=0$. Its image is described parametrically as $$x(u,v)=u,\quad y(u,v)=uv,\quad z(u,v)=uve^v.$$ It is a famous result of W. Osgood that there is no non-zero analytic function satisfying $G(x(u,v),y(u,v),z(u,v))\equiv 0$.

Ch. Osgood, On functions of several complex variables, Trans. AMS 17 (1916), 1, 1-8, Theorem 1.

When $n=2$, the image of the critical set in a neighborhood of the origin is either a point or a curve. It is well-known and easy to prove that a parametrized curve is an analytic set.

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    $\begingroup$ I think you mean that your function vanishes on the critical values but does not vanish identically. $\endgroup$
    – Tom Goodwillie
    Commented Nov 8, 2017 at 13:49
  • $\begingroup$ Thank you for the answer, but like Tom Goodwillie says, the requirement is that the function vanishes on the critical values, but is not zero (identically). I hoped that it was clear in the question. $\endgroup$
    – erz
    Commented Nov 8, 2017 at 20:48
  • $\begingroup$ @Tom Goodwille: now I addressed this question too. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 9, 2017 at 22:59

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