Timeline for moduli stack of double covers of $\mathbb{P}^1$ with one marked point
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 8, 2018 at 7:10 | comment | added | Will Sawin | @gmp I guess an alternate description is that it is the locus in $\mathbb A^3 - 0 $ times the space of pairs of a point $x \in \mathbb P^1$ and a section of $\mathcal O(1)$ at that point (i.e. literally a line bundle over $\mathbb P^1) where the sections satisfies $s^2=f(x)$ with $f$ the section of $\mathbb A^3-0$. | |
| Mar 8, 2018 at 7:07 | comment | added | Will Sawin | @gmp The double cover necessarily is a finite flat map to $\mathbb P^1$, hence the relative spectrum of a vector bundle, which must be rank two. This vector bundle admits an automorphism, the hypereliptic/Galois involution, with eigenvalues $1$ and $-1$. The eigenvalue $1$ part must be equal to the base ring, and has degree $0$. The eigenvalue $-1$ part must have degree $-1$ to make the arithmetic genus match up. I have written don the most general multiplication law for such a sum. I don't know a more concrete presentation than the relative spectrum here. | |
| Mar 8, 2018 at 0:53 | comment | added | gmp | I recently have been trying to write down the construction of the moduli space in detail following your explanations. The step I repeatedly struggled is the connection between double covers and the relative spectrum + multiplication mentioned in your answer. Could you point out the theory behind this relation? Just for clarification: the moduli stack of double covers is $M:=[\mathbb{A}^3-\{0\}/\mathbb{G}_m]$ and the moduli stack of double covers with one marked point is the universal family over $M$? Does the universal family possess a concrete presentation in this case? | |
| Mar 3, 2018 at 21:08 | vote | accept | gmp | ||
| Mar 3, 2018 at 21:03 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 1129 characters in body
|
| Mar 3, 2018 at 20:56 | comment | added | Will Sawin | @gmp But there is still exactly one stable curve, with no extra automorphisms or trickery, that corresponds to that blown-up curve. One has to check something more carefully at the level of functors and stacks but this has all been worked out. | |
| Mar 3, 2018 at 20:55 | comment | added | Will Sawin | @gmp Yes it's equivalent. Yes, $\mathbb P^2$ is the moduli space of degree $2$ divisors on $\mathbb P^1$, so doing it this way you don't have to deal with two marked branch points (which is not the right choice anyways). I was wrong about the $\mathbb Z/2$ quotient, it's actually a gerbe, I will edit to explain. The way the stability conditions are set up, you don't have to worry too much about that - though if another marked point was there it would require some attention. When the two branch points collide, that forms a node. When the node would be marked, you blow it up. | |
| Mar 3, 2018 at 20:19 | comment | added | gmp | Thanks a lot for your detailed answer. In deed i am interested in genus zero covers/ 2 branch points ( it is an equivalence due to Riemann-Hurwitz, isn't it?). Furthermore I had Kontsevich stability in mind. Am I getting it right that by exchanging $\mathbb{P}^1 \times \mathbb{P}^1$ with $\mathbb{P}^2$ I avoid having to deal with the two marked branch points globally on the whole stack? Could you explain in detail why the action of $\mathbb{Z}/2$ is trivial on $\mathbb{P}^2$? Last but not least, do I have to pay special attention to the case when all three points coincide? | |
| Mar 3, 2018 at 16:46 | history | answered | Will Sawin | CC BY-SA 3.0 |