Let a,b,c,d,e,f be 6 points in the projective plane: they lie on a common conic iff [abc][ade][bdf][cef]=[abd][ace][bcf][def], where [xyz] stands for the 3×3 determinant of the homogeneous coordinates of the 3 points x,y,z. …
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…This is nice result and easily checked by explicit computation, but I feel I'm missing the broader point here: why precisely these 8 triplets (abe, ade…)? and what is the connection with Pascal's hexagrammum mysticum and/or with the Sylvester outer automorphism of 𝔖₆?).
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Replying to @gro_tsen
Interpreting such bracket expressions as ruler constructions (more precisely, with join and meet operations in the Grassmann-Cayley algebra) is the "Cayley factorization problem". Very far from solved, but one can at least recover Pascal's theorem:https://link.springer.com/article/10.1007/s00454-015-9738-2 …
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Replying to @blockspins
The thing is, 6 points have 6!/12=60 ways of being arranged cyclically in a hexagon, giving rise to 60 different Pascal lines. The expression [abc][ade][bdf][cef]=[abd][ace][bcf][def] has 48 automorphisms, so there are 6!/48=15 such: each corresponds to 4 Pascal lines. …
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Replying to @gro_tsen @blockspins
…So Cayley factorization has to somehow break the symmetry… I guess… I'm a bit confused. I'll try reading Apel's paper (I know about Cayley factorization from Sturmfels's book, but this seems very nice). But I was more intrigued by the problem of symmetry in the 6 points.
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Ah, I see, where the symmetry goes is mysterious to me too. Chapter 6 of Apel's thesis https://mediatum.ub.tum.de/doc/1175107/1175107.pdf … shows some intriguing "evaluation trees" but perhaps it's just dressing up the algebraic manipulation? (1st image, Pascal as joins+meets, 2nd, edge-split to brackets)pic.twitter.com/Y6GwwhUgDJ
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