TITLE: Cross Ratios of an irregular octahedron
QUESTION [1 upvotes]: An irregular octahedron projects on a plane and associated with each of the 6 vertices is a Cross Ratio. Are these related? Does an invariance relation exist among the six by arbitrary rotations of the octahedron?

REPLY [1 votes]: Relationship between cross ratios
Ignore the octahedron for a moment. Cross ratios are invariant under projective transformations of the plane. So without loss of generality you can choose your six points as
$$
A=\begin{pmatrix}1\\0\\0\end{pmatrix}\quad
B=\begin{pmatrix}0\\1\\0\end{pmatrix}\quad
C=\begin{pmatrix}0\\0\\1\end{pmatrix}\quad
D=\begin{pmatrix}1\\1\\1\end{pmatrix}\quad
E=\begin{pmatrix}u\\v\\1\end{pmatrix}\quad
F=\begin{pmatrix}x\\y\\1\end{pmatrix}
$$
As there are only four real degrees of freedom here, your six cross ratios cannot be independent. There have to be at least two conditions to relate them. You can compute the cross ratios from the above points. I'll assume $A$ is opposite $B$, $C$ opposite $D$ and $E$ opposite $F$.
\begin{align*}
a &= (C,D;E,F)_A = \frac{v(y-1)}{(v-1)y} \\
b &= (C,D;E,F)_B = \frac{u(x-1)}{(u-1)x} \\
c &= (A,B;E,F)_C = \frac{vx}{uy} \\
d &= (A,B;E,F)_D = \frac{(v-1)(x-1)}{(u-1)(y-1)} \\
e &= (A,B;C,D)_E = \frac{(u-1)v}{u(v-1)} \\
f &= (A,B;C,D)_F = \frac{(x-1)y}{x(y-1)}
\end{align*}
You may notice that there are only four terms occuring in these fractions, namely $u,(u-1),v,(v-1),x,(x-1),y,(y-1)$. One can write the exponents of each as vectors like this:
$$\begin{pmatrix}
0 & 0 & 1 & -1 & 0 & 0 & -1 & 1 \\
1 & -1 & 0 & 0 & -1 & 1 & 0 & 0 \\
-1 & 0 & 1 & 0 & 1 & 0 & -1 & 0 \\
0 & -1 & 0 & 1 & 0 & 1 & 0 & -1 \\
-1 & 1 & 1 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 1 & 1 & -1
\end{pmatrix}$$
This matrix has rank 4, and the left kernel is
$$\begin{pmatrix}
1 & -1 & 0 & 0 & -1 & 1 \\
0 & 0 & 1 & -1 & -1 & 1
\end{pmatrix}$$
so the relations between these numbers may be described as
$$\frac ab=\frac cd=\frac ef$$
If you choose different orders for the terms in your cross ratios, the relations may differ, but that's just minor details.
Quantity invariant under rotations
A general rotation of an object in three-space has three real degrees of freedom. But one of these degrees of freedom can be associated with a rotation aroun an axis that is perpendicular to the plane, so an orthogonal projection on that plane would merely rotate within the plane, without affecting the cross ratios. So the set of all possible cross ratios resulting from rotations of a single octahedron should result in a two-dimensional variety.
Let's look at a simpler situation first, namely a projectively invariant one. The figure in the plane can be obtained by projecting from a center or projection $P$ which would lie at infinity for the case of orthogonal projection. You don't really need to specify the plane onto which this gets projected, since as long as the plane doesn't contain the center of projection, any choice will lead to a projectively transformed version of the image you get for any other choice, which doesn't change cross ratios. Knowing the center of projection, the cross ratios could be computed directly from the spatial homogeneous coordinates like this:
$$a=(C,D;E,F)_{A,P}=\frac{[C,E,A,P][D,F,A,P]}{[C,F,A,P][D,E,A,P]}$$
and so on for $b,c,d,e,f$, using the same orders of points as above. The square brackets denote determinants. This setup is invariant under projective transformations of the space, so in general you may choose coordinates of five points. I'll choose
$$
A=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\quad
B=\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\quad
C=\begin{pmatrix}0\\0\\1\\0\end{pmatrix}\quad
D=\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\quad
E=\begin{pmatrix}1\\1\\1\\1\end{pmatrix}\quad
F=\begin{pmatrix}u\\v\\w\\1\end{pmatrix}\quad
P=\begin{pmatrix}x\\y\\z\\1\end{pmatrix}
$$
A change of $P$ corresponds to a change in projection, and since a rotation of the octahedron may be seen as an opposite rotation of the center of projection, that does include rotations. It also includes changes of perspective, which explains why we are looking at three parameters $x,y,z$ instead of the two parameters I mentioned earlier.
The claim now is that there should be a relationship between the shape of the octahedron (as expressed by $(u,v,w)$) and the cross ratios $(a,b,c,d,e,f)$ which does not involve the center of rotation $(x,y,z)$. Standard elimination techniques (e.g. using resultants) will indeed yield such a relationship, but it's highly implicit:
\begin{align*}
(v - 1)\,(-u + w)\,(uv + w)&\,ac \\
{}-u\,(uvw + uv - 2uw - 2vw + w^2 + w)&\,ae \\
{}+\,w\,(-u^2v - uv^2 + 2uvw + 2uv - uw - vw)&\,ce \\
{}-(v - 1)\,(-u + w)\,u&\,a^2 \\
{}+\,(-uv + w)^2&\,ace \\
{}+\,(-uv + w)^2&\,bce \\
{}-w\,(-v + w)\,u\,(u - 1)&\,e^2 \\
{}+\,(-v + w)\,(u - 1)\,(uv + w)&\,bce^2 \\
{}-(-v + w)\,v\,(u - 1)&\,b^2c^2e^2 \\
{}-v\,(uvw + uv - 2uw - 2vw + w^2 + w)&\,bc^2e \\
{}-w\,v\,(v - 1)\,(-u + w)&\,c^2 \\
{}+\,(-u^2v - uv^2 + 2uvw + 2uv - uw - vw)&\,abce \\
&=0
\end{align*}
I don't believe that it's possible to turn this relationship into a form where you'd have an expression that only depends on $u,v,w$ on one side of the equation and another expression that only depends on $a,b,c,e$ on the other side of the equation. If one could find such a form, one could use the value of either of these sides as a reprojection-invariant term characterizing the shape of the octahedron.
In case you want to toy around with these ideas, but don't want to adopt my choice of coordinate system, here is how you'd obtain the parameters $u,v,w,x,y,z$ from any generic situation (provided points are in general position):
\begin{align*}
u &= (A,D;E,F)_{B,C} & x &= (A,D;E,P)_{B,C} \\
v &= (B,D;E,F)_{A,C} & y &= (B,D;E,P)_{A,C} \\
w &= (C,D;E,F)_{A,B} & z &= (C,D;E,P)_{A,B}
\end{align*}
With this you could compute $(u,v,w)$ and confirm the above implicit relationship for any octahedron you choose. Restricting yourself to isometric rotations instead of arbitrary reprojections would mean restricting $P$ to a certain plane (namely the plane at infinity), which can be expressed at the coordinate level and which will lead to a second implicit relationship that I doubt to be any easier than the one above.
Regular octahedron
The considerations above assume points in general position, i.e. no four of them on a common plane. The situation of a regular octahedron is therefore poorly captured by this. In order to investigate relationships for the regular octahedron, I'd choose
$$
A=\begin{pmatrix}1\\0\\0\\1\end{pmatrix}\quad
B=\begin{pmatrix}-1\\0\\0\\1\end{pmatrix}\quad
C=\begin{pmatrix}0\\1\\0\\1\end{pmatrix}\quad
D=\begin{pmatrix}0\\-1\\0\\1\end{pmatrix}\quad
E=\begin{pmatrix}0\\0\\1\\1\end{pmatrix}\quad
F=\begin{pmatrix}0\\0\\-1\\1\end{pmatrix}\quad
P=\begin{pmatrix}x\\y\\z\\0\end{pmatrix}
$$
This leads to
\begin{align*}
a=b&=\frac{(x+y+z)(x-y-z)}{(x-y+z)(x+y-z)}\\
c=d&=\frac{(x+y+z)(-x+y-z)}{(-x+y+z)(x+y-z)}\\
e=f&=\frac{(x+y+z)(-x-y+z)}{(-x+y+z)(x-y+z)}
\end{align*}
So here $P$ is anywhere at infinity, and you really get rotations. One very simple relation can be immediately seen from the equations I just gave:
$$a=b$$
which implies $c=d$ and $e=f$ thanks to the equation from the first section of this answer. Eliminating variables using resultants (and factorizing to get rid of terms that correspond to degenerate positions for $P$) one can obtain a second condition which looks more complicated. I've written it down in two different forms:
\begin{align*}
  a^2c^2e^2
- 2(a^2ce + ac^2e + 2ace^2)
+ 8ace
+ (a^2 + c^2 + e^2)
- 2(ac + ae + ce) &= 0 \\
(a+c+e)^2 - 4(ac+ae+ce) + ace(ace-2(a+c+e)+8) &= 0
\end{align*}
The first one is using monomial symmetric polynomials, the second is built on elementary symmetric polynomials. If you are familiar with notation used for symmetric functions, you might write this as
\begin{align*}
m_{(2,2,2)} - 2m_{(2,1,1)} + 8m_{(1,1,1)} + m_{(2)} - 2m_{(1,1)} &= 0 \\
e_1^2 - 4e_2 +e_3(e_3-2e_1+8) &= 0
\end{align*}
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