Timeline for irreducibility punctual Hilbert scheme of relative subschemes of length $2$
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 15, 2022 at 23:34 | vote | accept | Libli | ||
| Jan 15, 2022 at 18:17 | comment | added | Dori Bejleri | Maybe a better way to think of it is in terms of the monodromy action though. $p$ is a cyclic covering space so the branches are cyclically ordered. Then $S^2(X/S)$ consists of the diagonal, a component parametrizing two distinct points on adjacent branches of the covering space, and another component parametrizing two distinct points on non-adjacent branches. The non-diagonal components correspond to the two orbits of the $\mathbb{Z}/5\mathbb{Z}$ action on the set of $2$-element subsets of $\{1, \ldots, 5\}$ induced by the cyclic permutation. | |
| Jan 15, 2022 at 18:13 | comment | added | Dori Bejleri | @Libli yes that's right $S^2(X/S)$ is the disjoint union of 3 irreducible varieties which are each finite étale over $S$. One way to see this is that $X^2_S$ is the subvariety $x^5 = y^5$ inside $\mathbb{G}_m^2$ and the $\mathfrak{S}_2$ action swaps $x$ and $y$. | |
| Jan 15, 2022 at 9:47 | comment | added | Libli | Thanks a lot for your answer. I am very interested by your example. Just to be sure, the irreducible components of $S^2(X/S)$ are also connected components, aren't they? Otherwise it would contradict the comment I linked to in my question, since your morphism $p$ is generically smooth. Or do I miss something? | |
| Jan 15, 2022 at 5:30 | history | answered | Dori Bejleri | CC BY-SA 4.0 |