glen/dyna3
1
0
Fork 0
Code Issues 14 Pull Requests 1 Actions Projects Releases 2 Wiki Activity

doc: fix more weird Unicode characters, probably from marktest editor

Browse Source
This commit is contained in:
Glen Whitney 2024-10-20 19:27:05 -07:00
parent bb881f806c
commit 47c2217bcf
1 changed files with 3 additions and 3 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
notes/inversive.md
View File

@ -11,16 +11,16 @@ 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 $J$ 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 $I_P$. |
| 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 $\cosh(t) = \cos(it)$. |
| 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). | $R,P,S$ 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. | |
| 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. The second appears to be canonical, but I don't see a circle rep that corresponds to it. |