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doc: another unicode pseudo-space

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Glen Whitney 2024-10-20 19:41:05 -07:00
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notes/inversive.md
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@ -16,7 +16,7 @@ These coordinates are of form $I=(c, b, x, y, z)$ where we think of $c$ as the c
| $P$ is center of sphere rep'd by $I$ | $Q(I, I_P) = -r/2$, where $1/r = 2Q(I_\infty, I)$ is the signed bend of the sphere, and $I_P$ is normalized in the standard way, which is to set $Q(I_\infty, I_P) = 1/2$ | This relationship is equivalent to both of the following. (1) The point $P$ has signed distance $-r$ from the sphere. (2) Inversion across the sphere maps $\infty$ to $P$. |
| Distance between points $P$ and $R$ is $d$ | $Q(I_P, I_R) = d^2/2$ | If $P$ and $R$ are represented by non-normalized vectors $V_P$ and $V_R$, the relation becomes $Q(V_P, V_R) = 2\,Q(V_P, I_\infty)\,Q(V_R, I_\infty)\,d^2$. This version of the relation makes it easier to see why $d$ goes to infinity as $P$ or $R$ approaches the point at infinity. |
| Signed distance between point rep'd by $V$ and sphere/plane rep'd by $I$ is $d$ | In general, $\frac{Q(I, V)}{2Q(I_\infty, V)} = Q(I_\infty, I)\,d^2 + d$. When $V$ is normalized in the usual way, this simplifies to $Q(I, V) = d^2/r + d$ for a sphere of radius $r$, and to $Q(I, V) = d$ for a plane. | We can use a Euclidean motion, represented linearly by a Lorentz transformation that fixes $I_\infty$, to put the point on the $z$ axis and put the nearest point on the sphere/plane at the origin with its normal pointing in the positive $z$ direction. Then the sphere/plane is represented by $I = (0, 1/r, 0, 0, -1)$, and the point can be represented by any multiple of $I_P = (d^2, 1, 0, 0, d)$, giving $Q(I, I_P) = d^2/2r + d.$ We turn this into a general expression by writing it in terms of Lorentz-invariant quantities and making it independent of the normalization of $I_P$. |
| Distance between sphere/planes rep by $I$ and $J$ | Note that for any two Euclidean-concentric spheres rep by $I$ and $J$ with radius $r$ and $s,$ $Q(I,J) = -\frac12\left(\frac rs + \frac sr\right)$ depends only on the ratio of $r$ and $s$. So this can't give something that determines the Euclidean distance between the two spheres, which presumably grows as the two spheres are blown up proportionally. For another example, for any two parallel planes, $Q(I,J) = \pm1$. | Alex had said: $Q(I,J)=\cosh(d/2)^2$ maybe where d is distance in usual hyperbolic metric. Or maybe $\cosh(d)$. That may be right depending on what's meant by the hyperbolic metric there, but it seems like it won't determine a reasonable Euclidean distance between planes, which should differ between different pairs of parallel planes. |
| Distance between sphere/planes rep by $I$ and $J$ | Note that for any two Euclidean-concentric spheres rep by $I$ and $J$ with radius $r$ and $s,$ $Q(I,J) = -\frac12\left(\frac rs + \frac sr\right)$ depends only on the ratio of $r$ and $s$. So this can't give something that determines the Euclidean distance between the two spheres, which presumably grows as the two spheres are blown up proportionally. For another example, for any two parallel planes, $Q(I,J) = \pm1$. | Alex had said: $Q(I,J)=\cosh(d/2)^2$ maybe where d is distance in usual hyperbolic metric. Or maybe $\cosh(d)$. That may be right depending on what's meant by the hyperbolic metric there, but it seems like it won't determine a reasonable Euclidean distance between planes, which should differ between different pairs of parallel planes. |
| 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, $Q(I, J) = 0$. | |
| 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 $\cosh(t) = \cos(it)$. |