Second post in the cubic curves series. We will show how cubic curves arise as the loci of triple cross ratios.
Third post: TBD.
$\newcommand\ol{\overline}$ $\newcommand\abs[1]{\left|#1 \right|}$ $\newcommand\cK{\mathcal K}$
Let’s say $A,B,C,D,P,Q$ are six points such that $$(A,C; B,D)_P = (A, C; B, D)_Q$$ where we use the shorthand $(A,C ; B,D)_P := (PA, PC; PB, PD)$. Then we must have that $A,B,C,D,P,Q$ lie on the same conic. An easy proof involves projecting the conic $(ABCPQ)$ into a circle, then the result follows by an easy angle relation.
How do you show the converse - that the equation $(A,C; B,D)_P = k$ defines a conic? We may be able to adapt the previous argument, but it will be informative to try to verify this computationally using complex numbers.
First, we have to establish $\sin \angle APB$ is in terms of complex numbers. Note that if $z=re^{i\theta}$, then $$\sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} = \frac{z - \ol z}{2i|z|}$$
Hence, $$\sin \angle APB = \frac{\frac{p-a}{p-b} - \frac{\ol{p-a}}{\ol{p-b}}}{2i\left|\frac{p-a}{p-b} \right|} $$ if you sign your angles clockwise (that is, $\angle APB$ is measuring how much we have to turn $PA$ to hit $PB$ clockwise).
Recall that
$$(A,C; B,D)_P := \frac{\sin \angle APB}{\sin \angle CPB} \big / \frac{\sin \angle APD}{\sin \angle CPD}.$$
We’ve cleverly chosen the signs here such that the absolute value part cancels out, because $$\frac{ \abs{\frac{p-a}{p-b}} }{ \abs{\frac{p-c}{p-b}} } \big/ \frac{ \abs{\frac{p-a}{p-d}} }{ \abs{\frac{p-c}{p-d}} } = 1$$ so really, we only need to grab the numerators. Thus, we write $$(A,C; B,D)_P = \frac{ \frac{p-a}{p-b} - \frac{\ol{p-a}}{\ol{p-b}} }{ \frac{p-c}{p-b} - \frac{\ol{p-c}}{\ol{p-b}} } \big / \frac{ \frac{p-a}{p-d} - \frac{\ol{p-a}}{\ol{p-d}} }{ \frac{p-c}{p-d} - \frac{\ol{p-c}}{\ol{p-d}} }.$$
The trick now is that we’re allowed to cross multiply each term (because the product of the denominators cancel out), and within each term e.g. $(p-a)\overline{(p-b)} - \overline{(p-a)}(p-b)$ the quadratic term cancels out, so each factor becomes linear. So the final expression is $$(A,C; B,D)_P = \frac{ (p-a)\overline{(p-b)} - \overline{(p-a)}(p-b) }{ (p-c)\overline{(p-b)} - \overline{(p-c)}(p-b) } \cdot \frac{ (p-c)\overline{(p-d)} - \overline{(p-c)}(p-d) }{ (p-a)\overline{(p-d)} - \overline{(p-a)}(p-d) }.$$
Hence, we have a quadratic rational function in $p$ and $\ol p$, so setting this to a constant gives a conic.
Presumably, by repeating this argument with the triple cross ratio $$(A,D; B,E; C, F)_P := \frac{\sin \angle APB}{\sin \angle CPB} \cdot \frac{\sin \angle CPD}{\sin \angle EPD}\cdot \frac{\sin \angle EPF}{\sin \angle APF}$$ if everything falls into place then this ends up being a cubic rational function in $p, \ol p$, so the constant loci ends up being a cubic curve. This paves the road to more natural characterizations of cubic curves in general.
Remark. One can also interpret perspective cross ratios as a regular cross ratio at infinity: $$(A,C; B,D)_P = (\infty_{PA}, \infty_{PC}; \infty_{PB}, \infty_{PD}) = \frac{(\tilde a - \tilde b)(\tilde c - \tilde d)}{(\tilde c - \tilde b)(\tilde a- \tilde d)}$$
where $\tilde a = \frac{p-a}{\ol{p-a}}$ and so on. One also notes that $\tilde a$ is naturally interpretable as the direction of the line $PA$.
You might know that 9 points define a cubic, but yet the sets $$\mathcal K_k = \{P: (A,D; B,E; C,F)_P = k\}$$ exactly partition the (projective) plane (minus points $A$ to $F$), which suggests that this is a single-parameter family of cubics. So, other than passing through $A,B,C,D,E,F$ there must be other constraints at work.
So what exactly are these other constraints? Let’s use the notation $[a,b]_p = (p-a)\ol{(p-b)} - (p-b)\ol{(p-a)}$. Then, we can write $$(A, D; B, E; C, F)_P = \frac{[a,b]_p \cdot [c,d]_p \cdot [e,f]_p}{[b,c]_p \cdot [d,e]_p \cdot [f,a]_p}.$$
The key observation we had was that $[a,b]_p$ was actually linear in $p$ and $\ol p$: $$[a,b]_p = p\ol{(a-b)} - \ol p (a-b) + a\ol b - b\ol a.$$ One might recognise $[a,b]_p = 0$ as precisely the condition that $a,b,p$ are collinear, since $$[a,b]_p = \det \begin{pmatrix} 1 & 1 & 1 \\ p & a & b \\ \ol p & \ol a & \ol b \end{pmatrix} = \det \begin{pmatrix}1 & 0 & 0 \\ p & p-a & p-b \\ \ol p & \ol {p-a} & \ol {p-b} \end{pmatrix}$$
Thus, the two sides end up also vanishing at the diagonal points $AB\cap DE, BC \cap EF, CD \cap FA$. So, we have a pencil of cubics passing through 9 points, which also happen to be the intersection of $\overline{AB} \cup \overline{CD} \cup \overline {EF}$ and $\overline{BC} \cup \overline {DE} \cup \overline {FA}$.
Furthermore, the group law gives that $A+B = D+E$, so $A-D = E-B = C-F$. However, we can argue this in reverse, so we can in fact pick such $A,B,C,D,E,F$ on a cubic such that the cubic is the loci of a constant perspective cross-ratio. (So the answer is yes, as long as you picked your points well!)
Note that any diagonal cubic passes through $$A,B,C,D,E,F, L = AB\cap DE, M = BC\cap EF, N = CD\cap FA$$ and so must be a linear combination of $\ol {ABL} \cup \ol {CDM} \cup \ol {EFN}$ and $\ol {BCM} \cup \ol {DEL} \cup \ol {FAM}$. Hence, as long as $P$ is not any of $\{L,M,N\}$, there is unique diagonal cubic passing through $A,B,C,D,E,F$ and $P$.
The corollary is that if $A-D=E-B = C-F$ on $\cK$, then $$(A, B; C, D; E, F)_P \text{ is constant for all }\cK$$ and so we can denote this constant value as $(A,B;C,D;E,F)_\cK$.
Another corollary is that if we had $C', F'$ such that $A-D = E-B = C'-F'$, then diagonal cubic passing through $A,B,C',D,E,F'$ and $P$ is $\cK$ itself. However, the exact value of the cross ratio might differ - we’ll see this come into play very soon.
Let’s suppose we had 6 real numbers $a,b,c,d,e,f$ satisfying
$$\frac{(a-b)(c-d)(e-f)}{(b-c)(d-e)(f-a)} = k.$$
Suppose $a,b,d,e$ were fixed, and $c$ was variable. Then it’s not hard to see (algebraically) that the map sending $c\mapsto f$ is projective (i.e. a fractional linear transform), and that this map must also send $d\mapsto a$ and $b\mapsto e$). In equation form, we would take $$\frac{(a-b)(c-d)(e-f)}{(b-c)(d-e)(f-a)} = k$$ $$\frac{(a-b)(c'-d)(e-f')}{(b-c')(d-e)(f'-a)} = k$$
Then dividing the two, we get $(b,d;c,c') = (e,a; f,f')$.
This points to the following question: if $$A-D = E-B = C-F$$ then must the map $C\mapsto F = C + (A-D)$ look like a projective map from $P$? The answer is no, because then this map also sends $D_P := (-P-D)$ to $A_P$ and $B_P$ to $E_P$, so because this map swaps two pairs of points it must be an involution, which would imply that $(A,D; E,B)_P = -1$, which need not be true.
This doesn’t contradict our earlier results, because it’s possible that $$(A, D; B, E; C, F)_P \neq (A, D; B,E; C', F')_P$$ and so the cross-ratio is not kept constant.
Okay, but maybe the value of the cross ratio is -1, which would suggest that the underlying map is an involution anyway. Let’s see if in this special case we can get what we want.
First, we note that the third intersection map “preserves” the cross ratio: if $t_P: A \mapsto P-A$, then $$(A, B; C, D; E, F)_P = (t(A), t(B); t(C), t(D); t(E), t(F))_P$$ and also we have $t(A) - t(D) = t(E) - t(B) = t(C) - t(F)$, so these are true for any point on $\cK$: $$(A, B; C, D; E, F)_\cK = (t(A), t(B); t(C), t(D); t(E), t(F))_\cK$$
If $2(A-D)=0$, then $t_A \circ t_D$ sends $A \leftrightarrow D, B\leftrightarrow E, C\leftrightarrow F$. Hence, $$(A, B; C, D; E, F)_P = (D, E; F, A; B, C)_P$$ but these must be reciprocals. (TODO: eliminate the case where this is 1.) We conclude that $$(A,B; C, D; E, F)_\cK = -1.$$
In this case, the map $C \mapsto F = C + (A-D)$ preserves the perspective cross ratio, and so must look like an involution from any point.