From bb881f806c121f41cfe7f24a77dd2edbd4c7ae28 Mon Sep 17 00:00:00 2001 From: Glen Whitney Date: Sun, 20 Oct 2024 16:17:18 -0700 Subject: [PATCH] doc: Won't tex formula immediately followed by dash --- notes/inversive.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/notes/inversive.md b/notes/inversive.md index 0211f5e..df77846 100644 --- a/notes/inversive.md +++ b/notes/inversive.md @@ -11,8 +11,8 @@ These coordinates are of form $I=(c, b, x, y, z)$ where we think of $c$ as the c | Point $P$ with Euclidean coordinates $(x,y,z)$ | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. This gives us the freedom to choose a different normalization. For example, we could scale the representation shown here by $(\|P\|^2+1)^{-1}$, putting it on the sphere where the light cone intersects the plane where the first two coordinates sum to $1$. | | ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by (some normalization of) the above case. | | Point $P$ 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 $J$ intersect (respectively, don't intersect) | $\lvert Q(I,J)\rvert \le (\text{resp. }>)\; 1$ | Follows from the angle formula and the tangency condition, at least conceptually. One subtlety: parallel planes have $Q$-product $\pm 1$, because they intersect at infinity (and in fact, are "tangent" there)! | +| 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) | $\lvert Q(I,J)\rvert \le (\text{resp. }>)\; 1$ | Follows from the angle formula and the tangency condition, at least conceptually. One subtlety: parallel planes have $Q$ - product $\pm 1$, because they intersect at infinity (and in fact, are "tangent" there)! | | $P$ 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 $I_P$. |