mr_e_man wrote:Suppose some vertex v' has that form. Then any adjacent vertex v'' is a reflection of v', say, across the j'th mirror. [...]
v'' = v' - (2v'•uj)uj
wendy wrote:The chords of a general polygon can be derived by replacing edges with a triangle of two consecutive edges.
wendy wrote:The correct vectors to use for finding distances is not the Dynkin matrix, but the Stott matrix.
⌈ 2 -1/π⌉
⌊-1/π 2 ⌋
⌈ 2 π-6⌉
⌊π-6 2 ⌋
wendy wrote:You're not using chords here, but some kind of actual distance. The spherical and hyperbolic cases are dealt with by using euclidean chords, rather than arc (ie true) distances.
mr_e_man wrote:I was able to prove that, for any (finite) Wythoffian n-polytope, if the edge lengths are algints, then the chord lengths are algints. [...]
But this doesn't work for infinite polytopes, even regular polygons. Consider a hyperbolic apeirogon, or a zigzag apeirogon ( /\/\/\/\/ ), or a Wythoffian polygon with an irrational angle between mirrors (so the vertices are dense on the circumcircle); given edge length 1, the chord lengths may vary in a continuum, so they're not necessarily algebraic numbers, much less algebraic integers.
wendy wrote:Of course you can have anything in a dynkin matrix, but the thing is that for *any* polygon, the shortchord must solve an algebraic equation. I am not sure how you are pulling values of pi, but all chords of the nature product_(i=1)(n-1) (cho(ic/n)) = n [c = angle of 1 cycle], and every chord is an algebraic integer that divides n.
wendy wrote:Have a look at http://www.os2fan2.com/files/maths.pdf
wendy wrote:Of course, all of the non-wythoffian ones that i have played with still use chords, for example, the snub figures can be thought of as being {3,sN}, wjere sM os a polygon whose chord-4 is the shortchord of P. This is equal to a uniform polytope when N=2, 3, and 6, for which {3,N} corresponds to {3,4}, {3,5} and {3,6} respectively. The shortchord here is s, solving s^3-3s^1 = a, for a = r2 and phi.
mr_e_man wrote:I considered the snub cube and snub dodecahedron with algint edges. It turns out that the corresponding omnitruncate (the hull of the snub and its mirror image) also has algint edges, so this reduces to the Wythoffian case. Thus these snubs have algint chords.
mr_e_man wrote:wendy wrote:Of course, all of the non-wythoffian ones that i have played with still use chords, for example, the snub figures can be thought of as being {3,sN}, wjere sM os a polygon whose chord-4 is the shortchord of P. This is equal to a uniform polytope when N=2, 3, and 6, for which {3,N} corresponds to {3,4}, {3,5} and {3,6} respectively. The shortchord here is s, solving s^3-3s^1 = a, for a = r2 and phi.
[...]
That is new to me. However, it gives only one or two chords, and only in uniform snubs. Here I'm considering all chords, in non-uniform snubs.
mr_e_man wrote:I'm having trouble proving that the snub tetrahedron has algint chords.
The chiral tetrahedral symmetry group is 'epecs' (even permutations, and even changes of sign, of Cartesian coordinates). If one vertex is (x,y,z), then the snub's other vertices are epecs(x,y,z). The squared chords (including edges) are
f₀ = 0
f₁ = 4(y² + z²)
f₂ = 4(z² + x²)
f₃ = 4(x² + y²)
g₀ = 2(x² + y² + z² - yz - zx - xy)
g₁ = 2(x² + y² + z² - yz + zx + xy)
g₂ = 2(x² + y² + z² + yz - zx + xy)
g₃ = 2(x² + y² + z² + yz + zx - xy)
The goal is to prove that, if f₃,g₀,g₂ are algints, then all f,g are algints. (I've succeeded in special cases, such as x=0, y=0, z=0, z=x, or z=-x.)
Q := RationalField();
R<x1,x2,x3> := PolynomialRing(Q,3);
f1 := 4*(x2^2 + x3^2);
f2 := 4*(x3^2 + x1^2);
f3 := 4*(x1^2 + x2^2);
g0 := 2*(x1^2 + x2^2 + x3^2 - x2*x3 - x3*x1 - x1*x2);
g1 := 2*(x1^2 + x2^2 + x3^2 - x2*x3 + x3*x1 + x1*x2);
g2 := 2*(x1^2 + x2^2 + x3^2 + x2*x3 - x3*x1 + x1*x2);
g3 := 2*(x1^2 + x2^2 + x3^2 + x2*x3 + x3*x1 - x1*x2);
L := [f1,f3,g0,g2];
S<y1,y2,y3,y4> := PolynomialRing(Q,4);
RelationIdeal(L,S);
Ideal of Polynomial ring of rank 4 over Rational Field
Order: Lexicographical
Variables: y1, y2, y3, y4
Homogeneous, Dimension >0
Basis:
[
y1^4 - y1^3*y2 - 2*y1^3*y3 - 2*y1^3*y4 + 2*y1^2*y2*y3 + 2*y1^2*y2*y4 +
2*y1^2*y3^2 + 2*y1^2*y4^2 - y1*y2^3 + 2*y1*y2^2*y3 + 2*y1*y2^2*y4 -
8*y1*y2*y3*y4 - 4*y1*y3^3 + 4*y1*y3^2*y4 + 4*y1*y3*y4^2 - 4*y1*y4^3 +
y2^4 - 2*y2^3*y3 - 2*y2^3*y4 + 2*y2^2*y3^2 + 2*y2^2*y4^2 - 4*y2*y3^3 +
4*y2*y3^2*y4 + 4*y2*y3*y4^2 - 4*y2*y4^3 + 4*y3^4 - 8*y3^2*y4^2 + 4*y4^4
]
mr_e_man wrote:sin((k+1)θ)/sin(θ) = sum0≤j≤k/2 (-1)j (k-j choose j) (2cosθ)k-2j
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