Timeline for Geometry of critical points of holomorphic maps in projective space
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2022 at 20:15 | vote | accept | Samuele | ||
| Mar 17, 2022 at 20:30 | comment | added | Samuele | @JasonStarr got it, thanks! For the case $n=3$, is it known if $J$ is generically smooth? | |
| Mar 17, 2022 at 13:59 | comment | added | Jason Starr | @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). | |
| Mar 17, 2022 at 13:51 | comment | added | Samuele | @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! | |
| Mar 17, 2022 at 13:02 | comment | added | Joe Silverman | 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. | |
| Mar 17, 2022 at 11:48 | comment | added | Jason Starr | 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. | |
| Mar 17, 2022 at 11:40 | comment | added | Will Sawin | 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. | |
| Mar 17, 2022 at 11:33 | history | answered | Joe Silverman | CC BY-SA 4.0 |