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doc: Another note about inversive coordinates

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Glen Whitney 2024-09-18 20:27:04 -07:00
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notes/inversive.md
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@ -10,7 +10,7 @@ These coordinates are of form $I=(c, r, x, y, z)$ where we think of $c$ as the c
| Plane p with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note $Q(I_p, I_p)$ is still −1. This plane is at distance $s$ from the origin. And note that these are _oriented_ planes. For example, $(-2, 0, -1/\sqrt3, -1/\sqrt3, -1/\sqrt3)$ and $(2, 0, 1/\sqrt3, 1/\sqrt3, 1/\sqrt3)$ represent planes that coincide in space, just the former has normal pointing away from the origin and the latter pointing toward it. | | Plane p with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note $Q(I_p, I_p)$ is still −1. This plane is at distance $s$ from the origin. And note that these are _oriented_ planes. For example, $(-2, 0, -1/\sqrt3, -1/\sqrt3, -1/\sqrt3)$ and $(2, 0, 1/\sqrt3, 1/\sqrt3, 1/\sqrt3)$ represent planes that coincide in space, just the former has normal pointing away from the origin and the latter pointing toward it. |
| Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. Because of this we might choose some other scaling of the inversive coordinates, say ... | | Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. Because of this we might choose some other scaling of the inversive coordinates, say ... |
| ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by the above case. | | ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by the above case. |
| P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | | | P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | Actually also works if $I$ is the coordinates of a point, in which case "lies on" simply means "coincides with". |
| Sphere/planes represented by I and J are tangent | If $I$ and $J$ have the same orientation where they touch, $Q(I,J) = -1$. If they have opposing orientations, $Q(I,J) = 1$. | For example, the $xy$ plane with normal $-e_z$, represented by $(0,0,0,0,1)$, is tangent with matching orientation to the unit sphere centered at $(0,0,1)$ with outward normals, represented by $(0,1,0,0,1)$. Accordingly, their $Q$-product is −1. | | Sphere/planes represented by I and J are tangent | If $I$ and $J$ have the same orientation where they touch, $Q(I,J) = -1$. If they have opposing orientations, $Q(I,J) = 1$. | For example, the $xy$ plane with normal $-e_z$, represented by $(0,0,0,0,1)$, is tangent with matching orientation to the unit sphere centered at $(0,0,1)$ with outward normals, represented by $(0,1,0,0,1)$. Accordingly, their $Q$-product is −1. |
| Sphere/planes represented by I and J intersect (respectively, don't intersect) | $\|Q(I,J)\| < (\text{resp. }>)\; 1$ | Follows from the angle formula, at least conceptually. | | Sphere/planes represented by I and J intersect (respectively, don't intersect) | $\|Q(I,J)\| < (\text{resp. }>)\; 1$ | Follows from the angle formula, at least conceptually. |
| P is center of sphere represented by I | Well, $Q(I_P, I)$ comes out to be $(\|P\|^2/r - r + \|P\|^2/r)/2 - \|P\|^2/r$ or just $-r/2$ . | Is it if and only if ?   No this probably doesn't work because center is not conformal quantity. | | P is center of sphere represented by I | Well, $Q(I_P, I)$ comes out to be $(\|P\|^2/r - r + \|P\|^2/r)/2 - \|P\|^2/r$ or just $-r/2$ . | Is it if and only if ?   No this probably doesn't work because center is not conformal quantity. |