Timeline for Blow-up in family

Current License: CC BY-SA 3.0

9 events

when toggle format	what		by	license	comment
Jun 8, 2016 at 13:54	comment	added	Allen Knutson		[To see these facts, one looks for circle actions preserving the equations up to scale. The first one scales $Q$ while inverse scaling $s$; that preserves $X$ and $Y$. The second one scales $p$ while inverse scaling $r$; that preserves $X$ but not $Y$. Of the $4\times 2$ coordinate points on $\mathbb P^3\times\mathbb P^1$, only five are on $X$, which is why we get a pentagon. The line over e.g. the singular point $([0001],[01])$ has weights $(1,0)$ w.r.t. those two circle actions, obtained by dotting $[0001][01]$ with the vectors $[000-1][01],[10-10][00]$ corresponding to those actions.]
Jun 8, 2016 at 13:27	comment	added	Allen Knutson		For those who like toric pictures: this $X$ is a toric surface with moment pentagon a $2\times 2$ square with two corners cut off until their edges meet, i.e. the convex hull of $(1,0),(0,\pm 1),(-1,\pm 1)$. (Geometrically, blow up $\mathbb P^1\times \mathbb P^1$ at $(0,0),(0,\infty)$, then blow down the proper transform of $0\times\mathbb P^1$.) It has a $Z_2$-orbifold point over $(1,0)$. Its map to $T$ flattens this pentagon to a horizontal line, degenerating a degree $2$ $\mathbb P^1$ to a union of two. Unfortunately $Y$ is not torus-invariant. It of course goes through the orbifold point.
Jun 7, 2016 at 17:38	history	edited	Jason Starr	CC BY-SA 3.0	added 103 characters in body
Jun 7, 2016 at 13:46	history	edited	Allen Knutson	CC BY-SA 3.0	P^1 x P^3 was reversed
Jun 6, 2016 at 14:11	vote	accept	Giulio
Jun 6, 2016 at 14:11	vote	accept	Giulio
Jun 6, 2016 at 14:11
Jun 6, 2016 at 14:01	history	edited	Jason Starr	CC BY-SA 3.0	added 834 characters in body
S Jun 6, 2016 at 13:37	history	answered	Jason Starr	CC BY-SA 3.0
S Jun 6, 2016 at 13:37	history	made wiki			Post Made Community Wiki by Jason Starr