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Let $f:\mathbb{CP}^n\to\mathbb{CP}^n$ be a holomorphic map; I am interested in what the subvariety of critical points could be.

More specifically, let $J=\{p\in \mathbb{CP}^n\ :\ \det\mathrm{Jac}(f)=0\}$:

  1. can $J$ be a smooth submanifold of $\mathbb{CP}^n$?
  2. can $J$ be an irreducible subvariety of $\mathbb{CP}^n$?
  3. is there a way to describe the subvarieties $J$ that can be obtained this way?

Obviously, for $n=1$, the answer to 1,2 is easy, because $J$ is a union of isolated points. Regarding 3, if we want a constructive answer (i.e., given $J$, we want to produce $f$) it is not immediate to answer to 3, already for $n=1$.

On a further level, I am also wondering if it can happen that $J$ is, e.g., smooth but $(\det\mathrm{Jac}(f))$ is not reduced (i.e., locally $\det\mathrm{Jac}(f)=(g)^k$ for some $g$ holomorphic and $k>1$).

I tried to write down the equations explicitly fixing the degree of $f$ (as a homogeneous polynomial on $\mathbb{C}^{n+1}$), but the result is not easily manageable, at least for me; I also tried to look at it from and algebraic viewpoint, noticing a link with the concept of socle of a Gorenstein algebra, but, as much as it was interesting, it did not give me any hint on how to exlpicitly describe $J$ (or $f$ given $J$, or given the equation I want for $J$).

I believe this kind of question could already appear somewhere in the literature, but it evaded my googleing efforts, up to now.

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  • $\begingroup$ For $n=1$, every point in the critical divisor has degree $\leq d$, where $d$ is the degree of the map, and the total degree of the critical divisor is $2d-2$, and any such divisor can occur. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 17, 2022 at 13:07

2 Answers 2

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For $n=2$, the locus $J$ is smooth and irreducible for a general $f$; i.e., these $f$ form a Zariski dense subset of the parameter space of such $f$. For $n\ge3$ and for general $f$, the locus $J$ will be (mildly) singular, irreducible, and of general type. See Theorems 14 and 15 in the linked paper. (The proof of Theorem 14, which proves general type, was shown to the authors by Jason Starr.)

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    $\begingroup$ Is it really singular if $n=3$? Neither Theorem 14 nor 15 seems to cover this. I would guss the singularities occur in codimension $4$ in $\mathbb P^n$, as they do for the space of all matrices with determinant zero. $\endgroup$
    – Will Sawin
    Commented Mar 17, 2022 at 11:40
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    $\begingroup$ Definitely it is singular for $n\geq 4$. I agree with Will Sawin that it appears to be smooth for $n=3$ and $f$ general. $\endgroup$
    – Jason Starr
    Commented Mar 17, 2022 at 11:48
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    $\begingroup$ Yes, sorry, agreed, it's $n\ge4$ where it's clear that the singularities appear. And while I'm adding this comment, I want to thank Jason again for taking the time to explain this material to me. $\endgroup$
    – Joe Silverman
    Commented Mar 17, 2022 at 13:02
  • $\begingroup$ @JasonStarr (or Joe Silverman ) could you be more precise on the reason why it should be clear that for $n\geq 4$ the locus $J$ will be singular (I am supposing that being non singular is an open condition so, if it is generically singular it will be always singular)? Sorry, but algebraic geometry is not really my piece of cake! $\endgroup$
    – Samuele
    Commented Mar 17, 2022 at 13:51
  • $\begingroup$ @Samuele. The singular locus of $\text{Zero}(\text{det}\ \text{Jac}(f))$ contains the locus where the Jacobian matrix has nullity at least $2$. For generic $g$, the locus where the Jacobian matrix has nullity at least $r$ has codimension $r^2$ in the projective space. So if $n\geq 4$, then the locus where the Jacobian matrix has nullity at least $2$ has codimension $4$ in $\mathbb{P}^n$, hence it is nonempty. In fact, by the Thom-Porteous formula, the cycle class of this codimension-$4$ locus can be computed (and it has been computed, and the computation is nonzero). $\endgroup$
    – Jason Starr
    Commented Mar 17, 2022 at 13:59
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For n=1, every point in the critical divisor has degree $≤d-1$, where $d$ is the degree of the map, and the total degree of the critical divisor is $2d−2$, and any such divisor can occur.

To state it without degree, the number $m$ of points (counting multiplicity) is even and multiplicity of each point is at most $m$, and this is a complete description of critical divisors.

This result is well-known and is easy to prove, but here is a recent reference:

https://www.math.ucdavis.edu/∼kapovich/EPR/covers.pdf

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