Timeline for Holomorphic covers pulling back the volume form to any integer multiple
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Oct 1, 2020 at 16:04 | history | bounty ended | CommunityBot | ||
| S Oct 1, 2020 at 16:04 | history | notice removed | CommunityBot | ||
| S Sep 23, 2020 at 14:23 | history | bounty started | CommunityBot | ||
| S Sep 23, 2020 at 14:23 | history | notice added | user164751 | Authoritative reference needed | |
| S Sep 12, 2020 at 11:01 | history | bounty ended | CommunityBot | ||
| S Sep 12, 2020 at 11:01 | history | notice removed | CommunityBot | ||
| S Sep 4, 2020 at 9:29 | history | bounty started | CommunityBot | ||
| S Sep 4, 2020 at 9:29 | history | notice added | user164740 | Authoritative reference needed | |
| Aug 31, 2020 at 11:03 | comment | added | Robert Bryant | @vrz: (This is what my actual comment above should have been; the remainder didn't post for some reason [probaby user-error].) No, that is not right, at least, not a power higher than 1. The point is that, for self-maps of a complex $n$-torus $M$ generated by the ring $R$ as above (when $R$ has rank larger than 1), the volume form can't necessarily be written as a higher power of a symplectic form, and, even when it can, these self-maps may not carry any symplectic form to an integer multiple of itself. | |
| Aug 24, 2020 at 22:25 | comment | added | Robert Bryant | @vrz: No, that is not right, at least, not a power higher than 1. | |
| Aug 24, 2020 at 20:18 | comment | added | user145520 | @RobertBryant for a complex torus, the volume form is an exterior power of the symplectic form so the determinant in question has to be a power of some integer. Isn't that right? | |
| Aug 24, 2020 at 16:00 | comment | added | Robert Bryant | If $M = \mathbb{C}^n/\Lambda$ where $\Lambda$ is a lattice, let $R\subset M_n(\mathbb{C})$ be the ring satisfying $R(\Lambda)\subseteq\Lambda$. Then $R$ is additively a free abelian group of rank at most $2n^2$, and $D(r) = \det_\mathbb{R}(r)$ for $r\in R$ is a nonnegative integer-valued form of degree $2n$ on $R$. Given any nonnegative integer $m$, one can choose $n$ and $\Lambda$ so that the multiplicative set $D(R)\subset\mathbb{Z}$ contains $\{1,2,\ldots, m\}$. One can even arrange that $M$ be a product of elliptic curves. Might there be a $\Lambda$ where $D(R)$ contains $\mathbb{Z}^+$? | |
| Aug 19, 2020 at 9:19 | comment | added | Robert Bryant | It's not true that the degrees of the self-maps of an elliptic curve are necessarily square integers; it depends on the elliptic curve. For example, for the square torus, the degree of a self-map can be the sum of any two square integers: Multiplication by a Gaussian integer $a+b\,i$ preserves the lattice of Gaussian integers $\mathbb{Z}[i]\subset\mathbb{C}$ and induces a self-map of degree $a^2+b^2$ on the square torus. More generally, for an elliptic curve that admits a complex multiplication, the degrees of its self-maps are the values of a quadratic form on an integer lattice of rank $2$. | |
| Aug 17, 2020 at 20:08 | comment | added | user145520 | @BenMcKay elliptic curves do have many maps but the self-maps multiply the volume form by a square integer. | |
| Aug 17, 2020 at 20:07 | comment | added | user145520 | @RobertBryant you are correct that there no covering maps for a simply-connected manifold. But I am looking for just one example. | |
| Aug 17, 2020 at 20:01 | comment | added | Ben McKay | Elliptic curves have many maps like these; take a rectangle in the plane, identify opposite sides, and map each complex number by an integer multiple. | |
| Aug 17, 2020 at 19:52 | comment | added | Robert Bryant | You can't mean literal 'covering maps'. Just take $M = \mathbb{P}^1$, which has no nontrivial covering maps of this kind. Maybe you mean 'branched covering maps'? Even in this case, it's not generally possible. | |
| Aug 17, 2020 at 19:44 | history | asked | user145520 | CC BY-SA 4.0 |