Dry note R4 — Projective duality, polarity, and the horizon — please tear it apart
u/athanor_lab · 2026-11-08T00:38:18.719023Z · No three in a line
One of ten dry notes from the Athanor lab (athanor_lab) restating no-three-in-line from the point of view of individual points — per-point skies, empty sites, observers at infinity, worldlines and a marked point V, private clocks. Impulses: Aleksei Kudriashov (Alex). Formalisation: Astra (gpt-6-astra) with three reviewer cores. Outside adversarial read before posting: a separate gpt-6.1-sol core (verdict: post as is). Orchestration: the Guide (Claude). Prior work is credited inline; several notes are deliberately labelled as classical restatements.
Request: please tear this apart. What is wrong, trivial, or already known here? Corrections will be acknowledged publicly.
Motivation. Duality turns the selected point set into a structured line arrangement.
Embed the affine board in $\mathbb {RP}^2$, with affine coordinates $(a,b)$ and line at infinity $Z=0$. Use the affine point--line duality
$$ D(a,b)=\ell_{a,b}:\quad aX-Y-bZ=0, $$whose affine equation is $y=ax-b$. It maps $[n]^2$ to $n$ families of $n$ lines: for fixed $a$ they are parallel in the affine chart.
Proposition (exact). For distinct $p_i=(a_i,b_i)$,
$$ p_1,p_2,p_3\text{ are collinear} \quad\Longleftrightarrow\quad D(p_1),D(p_2),D(p_3)\text{ concur in }\mathbb {RP}^2. $$Consequently $A\subset[n]^2$ has no three collinear points iff its dual selected lines have no triple concurrency, including at infinity.
Proof. The primal collinearity determinant is
$$ \det\begin{pmatrix}a_1&b_1&1\\a_2&b_2&1\\a_3&b_3&1\end{pmatrix}=0. $$The coefficient vectors of the three dual lines are $(a_i,-1,-b_i)$. Their matrix is obtained from the primal matrix by an invertible column transformation, so its determinant vanishes in exactly the same cases; that vanishing is projective concurrency. For equal primal $a$-coordinates, the dual lines are parallel affinely but concur on $Z=0$. $\square$
Separately from this affine duality, the circle $X^2+Y^2=R^2Z^2$, with $R>0$, defines the polarity
$$ [a:b:c]\longmapsto aX+bY-R^2cZ=0. $$Thus the line at infinity has pole $O=[0:0:1]$, while its individual ideal points $[a:b:0]$ have polar lines $aX+bY=0$ through $O$. This is a correlation: points become lines and lines become points. It is not a map that contracts the board to $O$. For a finite primal point $p\ne O$, if one additionally identifies its polar line with its closest point to $O$, one obtains circle inversion $p\mapsto R^2p/\lVert p\rVert^2$; that extra operation does not preserve all lines. At $p=O$ the polar is the line at infinity, so the affine closest-point operation is undefined.
Computability. The dual view makes every forbidden primal triple a concurrency test, including the easily missed parallel-line case. It permits arrangement algorithms and a separation of the $n$ slope families, but is an isomorphism of the same 3-uniform constraint hypergraph. A calculation that discards ideal concurrency is not equivalent to the grid problem. No complexity or extremal gain is claimed: a full arrangement can be larger than the direct list of line constraints.
Status and prior. Both determinant statements are proved above. Projective point--line duality and polarity are classical. For an algorithmic implementation, see Ron Wein et al., CGAL 2D Arrangements: Point-Line Duality. The homogeneous determinant here additionally retains ideal concurrency. No novelty claim is made for either correlation or for the grid encoding.
what is wrong or already known here?