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Engine prototype #13

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glen merged 133 commits from engine-proto into main 2024-10-21 03:18:48 +00:00
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notes/inversive.md
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@ -20,7 +20,7 @@ These coordinates are of form $I=(c, b, x, y, z)$ where we think of $c$ as the c
| Sphere centered on point $P$ through point $R$ | | Probably just calculate distance etc. | | Sphere centered on point $P$ through point $R$ | | Probably just calculate distance etc. |
| Plane rep'd by $I$ goes through center of sphere rep'd by $J$ | This is equivalent to the plane being perpendicular to the sphere: that is, $Q(I, J) = 0$. | | | Plane rep'd by $I$ goes through center of sphere rep'd by $J$ | This is equivalent to the plane being perpendicular to the sphere: that is, $Q(I, J) = 0$. | |
| Dihedral angle between planes or spheres rep by $I$ and $J$ | $\theta = \arccos(Q(I,J))$ | Aaron Fenyes points out: The angle between spheres in $S^3$ matches the angle between the planes they bound in $R^{(1,4)}$, which matches the angle between the spacelike vectors perpendicular to those planes. So we should have $Q(I,J) = \cos(\theta)$. Note that when the spheres do not intersect, we can interpret this as the "imaginary angle" between them, via $\cosh(t) = \cos(it)$. | | Dihedral angle between planes or spheres rep by $I$ and $J$ | $\theta = \arccos(Q(I,J))$ | Aaron Fenyes points out: The angle between spheres in $S^3$ matches the angle between the planes they bound in $R^{(1,4)}$, which matches the angle between the spacelike vectors perpendicular to those planes. So we should have $Q(I,J) = \cos(\theta)$. Note that when the spheres do not intersect, we can interpret this as the "imaginary angle" between them, via $\cosh(t) = \cos(it)$. |
| Points $R, P, S$ are collinear | Maybe just cross product of two differences is 0. Or, $R,P,S,\infty$ lie on a circle, or equivalently, $I_R,I_P,I_S,I_\infty$ span a plane (rather than a three-space). | $R,P,S$ lying on a line isn't a conformal property, but $R,P,S,\infty$ lying on a circle is. | | Points $R, P, S$ are collinear | Maybe just cross product of two differences is 0. Or, $R,P,S,\infty$ lie on a circle, or equivalently, $I_R,I_P,I_S,I_\infty$ span a plane (rather than a three-space). Or we can add two planes constrained to be perpendicular with one constrained to contain the origin, and all three points constrained to lie on both. But that's a lot of auxiliary entities and constraints... | $R,P,S$ lying on a line isn't a conformal property, but $R,P,S,\infty$ lying on a circle is. |
| Plane through noncollinear $R, P, S$ | Should be, just solve $Q(I, I_R) = 0$ etc. | | | Plane through noncollinear $R, P, S$ | Should be, just solve $Q(I, I_R) = 0$ etc. | |
| circle | Maybe concentric sphere and the containing plane? Note it is easy to constrain the relationship between those two: they must be perpendicular. | Defn: circle is intersection of two spheres. That does cover lines. But you lose the canonicalness | | circle | Maybe concentric sphere and the containing plane? Note it is easy to constrain the relationship between those two: they must be perpendicular. | Defn: circle is intersection of two spheres. That does cover lines. But you lose the canonicalness |
| line | Maybe two perpendicular containing planes? Maybe the plane perpendicular to the line and through origin, together with the point of the line on that plane? Or maybe just as a bag of collinear points? | The first is the limiting case of the possible circle rep, but it is not canonical. However, there is a distinguished "standard" choice we could make: always choose one plane to contain the origin and the line, and the other to be the perpendicular plane containing the line. That choice or Plücker coordinates might be the best we can do. If we use the standardized perpendicular planes choice, then adding a line would be equivalent to adding two planes and the two constraints that one contains the origin and the other is perpendicular to it. That doesn't seem so bad. The second convention (perpendicular plane through the origin and a point on it) appears to be canonical, but there doesn't seem to be a circle representation that tends to it in the limit. | | line | Maybe two perpendicular containing planes? Maybe the plane perpendicular to the line and through origin, together with the point of the line on that plane? Or maybe just as a bag of collinear points? | The first is the limiting case of the possible circle rep, but it is not canonical. However, there is a distinguished "standard" choice we could make: always choose one plane to contain the origin and the line, and the other to be the perpendicular plane containing the line. That choice or Plücker coordinates might be the best we can do. If we use the standardized perpendicular planes choice, then adding a line would be equivalent to adding two planes and the two constraints that one contains the origin and the other is perpendicular to it. That doesn't seem so bad. The second convention (perpendicular plane through the origin and a point on it) appears to be canonical, but there doesn't seem to be a circle representation that tends to it in the limit. |