Talk:Linear system of divisors

Definition

I think the article should contain a precise definition of linear system of divisors. As it is now, the section "Definition by means of functions" defines only linear equivalence of divisors, and the following one already goes on defining the base locus of a system. I find this quite confusing. --F4wk3s (talk) 15:08, 21 February 2011 (UTC)

Linear equivalence

Just a note to anyone passing through who's feeling ambitious: it would be nice if this page said more about the classical notion of linear equivalence. At the moment the only definitive statement it makes about the subject is that linearly equivalent divisors have isomorphic line bundles. -- Walt Pohl 23:52, 7 August 2006 (UTC)

I've made a start. Charles Matthews 14:42, 8 August 2006 (UTC)

Note on the (Type I) family of conics

For reference, I’d like to elaborate in some detail on the family of conics that I’ve listed, as it is an archetypal example of an important type of family, viz, the conics defined by 4 points in general position (Type I), and it is frankly pretty, due to various symmetries and elegant placement of points. The below is too discursive to be included at WP (at Wikibooks it may fit), but is worth sharing, and may be a useful reference on gritty details of projective geometry.

The family is the pencil of conics passing through the points $(\pm 1,\pm 1).$

• The 3 degenerate conics are a pair of vertical lines ($y=\pm 1$), a pair of horizontal lines ($y=\pm 1$), and a pair of diagonal lines $y=\pm x$).
• The intersection point of these pairs of lines, corresponding to the apex of the cone, are at [1:0:0] (vertical), [0:1:0] (horizontal), [0:0:1] (origin).
• In all cases the center is at the origin.

Parametrization-wise:

• The simplest parametrization is $ax^{2}+(1-a)y^{2}=1,$ corresponding to affine (and, for $a=\infty$) projective combinations of the parallel vertical lines $x^{2}=1$ and the parallel horizontal lines $y^{2}=1,$ which places the degenerate conics at the standard points of $0,1,\infty$
• A more symmetric presentation is given by $(1+a)x^{2}+(1-a)y^{2}=2,$ where inverting a ($a\mapsto -a$) reverses x and y. This places the degenerate conics at $-1,1,\infty$

More subtly, I suggest the points $\infty ,2,1,\textstyle {\frac {1}{2}},0,\textstyle {\frac {-1}{2}},-1,-2,\infty$ as “representative points”. This can be justified and examined in some detail.

Properly, evenly spaced points in the sense of projective geometry would be $0,\pm 1,\infty ,\pm 1\pm {\sqrt {2}};$ these are $\tan(k\pi /8),$ and correspond geometrically to stereographic projection of those angles in the complex plane from i. (For anyone making an animation, using evenly spaced tan yields a smooth variation of the parameter a.) Concretely, these are approximately $0,.4,1,2.4,\infty$ (and negatives), to which $0,.5,1,2,\infty$ are close and considerably simpler. Note that the parameter slows down as it approaches 0, then speeds up after passing through 0.

However, the suggested points are more symmetric than the above suggests: any three points in the line are “evenly spaced” (due to the Möbius group being simply 3-transitive – any 3 points have dihedral symmetry), and starting from the standard 3 points $0,1,\infty ,$ the “midpoints” of the segments (corresponding to the fixed points under inverting any two of the points; the stabilizer of these 3 points is a very pretty subgroup of the modular group) are $-1,\textstyle {\frac {1}{2}},2,$ so the points $-1,0,\textstyle {\frac {1}{2}},1,2,\infty$ form a “hexagon” (there is a 6-cycle relating them), and correspond to equally space points stereographically projected from $\textstyle {\frac {1}{2}}+bi$ for some b – the parameter slows down as it approaches 1/2, then speeds up as it leaves.

Thus the $0,1/2,1,2$ placement of points corresponds, not to a regular octogon (as would be evenly spaced), but to taking 4 sides each of two regular hexagons and gluing them together (or taking a 60°/120° rhombus and doubling the number of sides) – the polygon is mostly regular, except for a bend at 0. In terms of smoothness, this corresponds to a hiccup in the 2nd derivative, and looks roughly like the “W” shape of $x^{4}-x^{2}$ – the parameter slows down as it approaches 1/2, speeds up towards 0, then slows back down going to $-1/2,$ then speeds back up. Due to sampling, rather than this speeding up and slowing down one instead has 5 points that are evenly spaced linearly ($-1,-1/2,0,1/2,1$), which is not even spacing projectively – the spacing should slow down approaching 0, then speed up; thus there is an awkward flatness.

In sum, the simplified placement of points yields a very minor hiccup at 0, but is otherwise quite elegant.

—Nils von Barth (nbarth) (talk) 02:38, 25 February 2010 (UTC)