Timeline for Representing split-complex numbers as intervals and related compactification
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 14, 2021 at 13:40 | comment | added | Anixx | @LSpice 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. 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 12:59 | comment | added | wlad | The idea of looking at this in terms of intervals is intriguing, but intervals usually correspond better to unordered pairs as other people have pointed out. Interesting idea though! | |
| Dec 14, 2021 at 12:57 | comment | added | wlad | There is a reason to consider a projective line over the split-complex numbers. The elements of this set represent oriented lines in the hyperbolic plane. Analogously, the projective line over the dual numbers represents oriented lines in the Euclidean plane. Over the complex numbers, oriented lines in the elliptic plane. Notice the three classic non-Euclidean geometries {hyperbolic, Euclidean, elliptic} correspond to the three planar number systems {split-complex, dual, complex}. | |
| Dec 14, 2021 at 12:55 | answer | added | wlad | timeline score: 2 | |
| Dec 14, 2021 at 12:30 | comment | added | Anixx | @LSpice for instance, if we are speaking about the infinite point at the end of the diagonal $(j+1)$, this point in one basis is $\infty+j\infty$, and in the other basis is $(0,\infty)$. Now, take logarithm of it. In the first basis it is impossible, in the second one we can write it with divergent integrals $\ln(0,\infty)=(-H-\gamma,H)$, where $H=\int_1^\infty\frac1x dx$. We clearly see that its real part is $-\gamma/2$, and full form is $-\gamma/2+jH+j\gamma/2$. | |
| Dec 14, 2021 at 11:53 | comment | added | Anixx | @LSpice Yes, it is an ordered pair. What do I gain is that the number $(2,\infty)$, for instance, cannot be written in the form $a+bj$, even if we allow $a$ and $b$ infinite, so this form of representation allows to make more complicated compactification than $a+bj$, $a,b\in\overline{\mathbb{R}}$. | |
| Dec 14, 2021 at 11:49 | comment | added | LSpice | So what does $(1, -1)$ mean as an interval? And, again, if it is just a pair, then what do you gain by writing an element of $\mathbb R^2$ in one basis $\{(1, 0), (0, 1)\}$ rather than another $\{(1/2, -1/2), (1/2, 1/2)\}$? | |
| Dec 14, 2021 at 11:45 | comment | added | Anixx | @LSpice $(u,v)=\frac{v+u}2+j\frac{v-u}2$. $-j=(1,-1)$ We can write $j=\pm1$,$-j=\mp1$ | |
| Dec 14, 2021 at 11:45 | comment | added | LSpice | Then what stands for $-j$? You refer to $(u, v)$ first as a pair and then as an interval. If you really meant $(u, v)$ to be just a pair, then I do not understand what you are proposing to gain by writing, say, $(2, \infty)$ instead of $\frac{\infty + 2}2 + j\frac{\infty - 2}2$; both seem equally meaningful, or meaningless. If you meant for it to be an interval, then it seems that you would have to write $(1, -1)$ for $j$; and what does that mean as an interval? | |
| Dec 14, 2021 at 11:44 | comment | added | Anixx | @LSpice no, $(-1,1)$ stands for $j$ only, see the formula in the post. | |
| Dec 14, 2021 at 11:42 | comment | added | LSpice | The interval $(-1, 1)$ stands for both $+j$ and $-j$, so its sum with itself could be $+2j$ or $-2j$, both represented by $(-2, 2)$, or $0$, represented by $(0, 0)$. | |
| Dec 14, 2021 at 11:42 | comment | added | Anixx | @LSpice $(-2,2)$. I do not see the problem. | |
| Dec 14, 2021 at 11:37 | comment | added | LSpice | One problem is that you seem to be representing a quotient of $\mathbb R^2$, not $\mathbb R^2$ itself. That is, your pair notation has no way of distinguishing $+j = \langle{-1}, 1\rangle$ from $-j = \langle1, -1\rangle$, although they are distinct elements of $\mathbb R^2$. For example, should the sum of the interval (not the ordered pair) $(-1, 1)$ with itself be $(0, 0) = \emptyset$ or $(-2, 2)$? | |
| Dec 14, 2021 at 11:12 | comment | added | Anixx | @Wojowu well, due to isomorphism, $f(u,v)=(f(u),f(v))$, so when $f$ is defined on $\overline{\mathbb{R}}$, it is defined on extended split-complex numbers as well. | |
| Dec 14, 2021 at 11:06 | comment | added | Wojowu | "we can extend the set of split-complex numbers and define operations on this extended set." How do you define the operations? | |
| Dec 14, 2021 at 10:08 | history | asked | Anixx | CC BY-SA 4.0 |