Timeline for Embedding of symmetric square in Jacobian
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 12, 2023 at 9:51 | vote | accept | kindasorta | ||
| Sep 11, 2023 at 17:35 | comment | added | kindasorta | Though it should be said that from your description it isn't immediately obvious that the automorphism $\sigma$ should be algebraic | |
| Sep 11, 2023 at 17:31 | comment | added | kindasorta | Thank you very much for this clarification! Actually, I am interested in the case where the base is an arbitrary number field/ring of integers. How does the definition of "hyperelliptic" changes in this setting? | |
| Sep 11, 2023 at 17:16 | comment | added | David E Speyer | So having a degree $2$ map to $\mathbb{P}^1$ is the stronger property and (at least for $g \geq 2$ and $K$ algebraically closed), that is the definition of "hyperelliptic". | |
| Sep 11, 2023 at 17:15 | comment | added | David E Speyer | In the comments above, you refer to order $2$ automorphisms of $C$. The logical implications are (a) If $f:C \to \mathbb{P}^1$ is a degree $2$ rational map, then there is an order $2$ automorphism $\sigma : C \to C$ so that, for all $x \in C$, we have $f^{-1}(f(x)) = x+\sigma(x)$ (equality of divisors). (b) If $\sigma : C \to C$ is an order two automorphism, then there is an algebraic curve $C/\sigma$, and $C \to C/\sigma$ is degree $2$. However, the curve $C/\sigma$ may or may not be $\mathbb{P}^1$. (continued) | |
| Sep 11, 2023 at 16:53 | comment | added | David E Speyer | By definition (taking $g \geq 2$ and $K$ algebraically closed to avoid awkward cases), a curve $C$ is hyperelliptic if and only if there is a degree $2$ rational map $f : C \to \mathbb{P}^1$. | |
| Sep 11, 2023 at 16:48 | comment | added | kindasorta | So, what is the relation with hyperellipticity? | |
| Sep 11, 2023 at 16:35 | history | answered | David E Speyer | CC BY-SA 4.0 |