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doc: Two more small typos

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Glen Whitney 2024-10-20 16:10:01 -07:00
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notes/inversive.md
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@ -9,9 +9,9 @@ These coordinates are of form $I=(c, b, x, y, z)$ where we think of $c$ as the c
| Sphere $s$ with radius $r>0$ centered on $P = (x,y,z)$ | $I_s = (\frac1{c}, \frac1{r}, \frac{x}{r}, \frac{y}{r}, \frac{z}{r})$ satisfying $Q(I_s,I_s) = -1$, i.e., $c = r/(\|P\|^2 - r^2)$. | Note that $1/c = \|P\|^2/r - r$, so there is no trouble if $\|P\| = r$; we just get first coordinate to be 0. Using the point representation $I_P$ from below, let's orient the sphere so that its normals point into the "positive side," where $Q(I_P, I_s) > 0$. The vector $I_s$ then represents a sphere with outward normals, while $-I_s$ represents one with inward normals. |
| Plane $p$ with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note that $Q(I_p, I_p)$ is still $−1$. We orient planes using the same convention we use for spheres. 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, which have their normals pointing away from and toward the origin, respectively. Note that the ray from $(sx, sy, sz) \in p$ in direction $(-x, -y, -z)$ is the ray perpendicular to the plane through the origin; since $(-x, -y, -z)$ is a unit vector, $(sx, sy, sz)$ and hence $p$ is at distance $s$ from the origin. These coordinates are essentially the limit of a sphere's coordinates as its radius goes to infinity, or equivalently, as its bend goes to 0. |
| 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. |
| ∞, 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 $J$ 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. |