Skip to main content
13 events
when toggle format what by license comment
Dec 14, 2021 at 17:48 comment added wlad @Anixx OK, but $\overline{\mathbb R} + i\overline{\mathbb R} \neq P(\mathbb C)$. In fact, I've never seen that before
Dec 14, 2021 at 15:25 comment added Anixx Well, what I do is considering pairs of $(u,v)$, $u,v\in\overline{\mathbb{R}}$ so this compactification has elements which cannot be represented in $a+bj$ basis, even if we allow $a,b\in\overline{\mathbb{R}}$.
Dec 14, 2021 at 15:21 comment added wlad Yes, but the difference between this and $P(\mathbb R^2)$ is not to do with the basis
Dec 14, 2021 at 15:16 comment added Anixx I use $\overline{\mathbb{R}}\times\overline{\mathbb{R}}$
Dec 14, 2021 at 15:15 comment added wlad @Anixx en.wikipedia.org/wiki/Projective_line_over_a_ring
Dec 14, 2021 at 15:11 comment added wlad @Anixx I don't understand what definition of compactification you're using. I'm using the definition of a projective line over a ring, which is basis independent
Dec 14, 2021 at 13:42 comment added Anixx The compactification with diagnal lines allows to easily generalize functions to this extended set in the diagonal basis.
Dec 14, 2021 at 13:32 comment added Anixx I would argue that the compactification with lines parallel and perpendicular to the real line is more natural for the complex numbers, and the compactification by adding diagonal lines is more natural for hyperbolic numbers.
Dec 14, 2021 at 13:31 comment added Anixx The thing is, different basises bring different compactifications. In one basis we add a family of oriented lines, parallel and pependiclar to the real line, in the other we add a family of oriented diagonal lines.
Dec 14, 2021 at 13:17 history edited wlad CC BY-SA 4.0
added 201 characters in body
Dec 14, 2021 at 13:07 history edited wlad CC BY-SA 4.0
added 233 characters in body
Dec 14, 2021 at 13:01 history edited wlad CC BY-SA 4.0
added 233 characters in body
Dec 14, 2021 at 12:55 history answered wlad CC BY-SA 4.0