## Chords and algebraic integers

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### Chords and algebraic integers

A complex number z is called an algebraic integer when it satisfies a polynomial equation, zn + an-1zn-1 + ... + a2z2 + a1z + a0 = 0, where the leading coefficient is an=1, and the other coefficients ak are ordinary integers (sometimes called rational integers to distinguish from the more general notion). This implies that the minimal polynomial of z also has that form.

For example, φ = (1+√5)/2 is an algebraic integer because φ2 - φ - 1 = 0, and 6cos(π/7) is an algebraic integer because it satisfies the equation x3 - 3x2 - 18x + 27 = 0, but 1/√2 is not an algebraic integer because its minimal polynomial is either 2x2 - 1 (which doesn't have leading coefficient 1) or x2 - 1/2 (which doesn't have integer coefficients).

It can be proven that the sum, difference, or product of any two algints is another algint. (That is, the set of algebraic integers is a ring.) Also, if a complex number z satisfies a polynomial equation as above, where the leading coefficient is an=1, and the other coefficients ak are algints, then z is an algint (generally of much higher degree than n).

Here Wendy suggested that polytopes of certain types have algints as chord lengths. I would like to know which types.

I was able to prove that, for any (finite) Wythoffian n-polytope, if the edge lengths are algints, then the chord lengths are algints. Here's the proof.

The polytope's vertices are generated by reflections from a single vertex v. A reflection sends v to an adjacent vertex v - (2vu)u, where u is the unit vector perpendicular to the mirror; so the length of the edge connecting these two vertices is just 2vu.

Any vertex has the form v + sum1≤k≤n akuk, involving a combination of the mirror vectors u1, u2, ... , un, where the coefficients ak are algints. We prove this by induction: Suppose some vertex v' has that form. Then any adjacent vertex v'' is a reflection of v', say, across the j'th mirror. (Some would say "reflection in the mirror", which suggests a mere visual effect; I prefer "reflection across the mirror", which suggests an actual transformation, moving the point from one side to the other.)

v'' = v' - (2v'•uj)uj
= (v + sum1≤k≤n akuk) - (2(v + sum1≤k≤n akuk)•uj)uj
= v + sum1≤k≤n akuk - (2vuj)uj - sum1≤k≤n ak(2ukuj)uj

We know that 2vuj (being an edge length) is an algint, and 2ukuj = -2cos(θk,j) is an algint for any k, because the angle between the k'th and j'th mirrors is a rational angle (otherwise the symmetry group would be infinite). Therefore v'' has the same form as v'.

Now the chord vector between any two vertices has the form

(v + sum1≤k≤n akuk) - (v + sum1≤k≤n bkuk)
= sum1≤k≤n (ak - bk)uk
= sum1≤k≤n ckuk

where the coefficients ck are algints, so the chordsquare is

(sum1≤k≤n ckuk)2
= sum1≤k≤n sum1≤j≤n ckcjukuj
= sum1≤k≤n ck2 + 2 sum1≤k<j≤n ckcjukuj
= sum1≤k≤n ck2 + sum1≤k<j≤n ckcj(2ukuj)

which is clearly a combination of algints and thus itself an algint.

(In fact this shows that any chordsquare is in the ring generated by the edge lengths and doubled cosines of angles between mirrors.)

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.

And, as I noted in the other thread, J37 (elongated square gyrobicupola) has chordsquares involving 1/√2, so a finite unit-edged polyhedron with congruent vertices doesn't necessarily have algint chords.

I suspect that any finite isogonal polytope with algint edges (and maybe some non-degeneracy conditions) has algint chords. Can anyone prove this?
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### Re: Chords and algebraic integers

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.

But that is just a coincidence. For the (non-uniform, yet isogonal) snub tetrahedron, or the snub dihedra (antiprisms), the omnitruncate doesn't always have algint edges. 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.)
Last edited by mr_e_man on Mon Mar 07, 2022 2:39 pm, edited 1 time in total.
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### Re: Chords and algebraic integers

Here is a counter-example to my conjecture in its full generality.

Take the comb product of a doubly-wound digon {4/2} and another polygon {2n}. Abstractly, this is a 4×2n rectangular tiling of a torus. It is realized with the vertices of a 2n-gon prism.
Now consider its alternation, which is abstractly a rhombus tiling of a torus, and is realized with the vertices of an n-gon antiprism. In fact it has the same vertices and edges (ignoring multiplicity) as a regular skew 2n-gon. Its faces are degenerate "dart" tetragons, with angles 0,θ,0,-θ.
This degenerate isogonal polyhedron is flexible; given edge length 1, the chord lengths may vary in a continuum.

Notice that all of the edges at one vertex are contained in a plane.

Is it sufficient to require the edge vectors at a vertex to span a full n-dimensional space? Then is it true that a finite isogonal n-polytope with algint edges has algint chords?
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### Re: Chords and algebraic integers

And it is hardly supprising why i loathe mathematical proofs.

The chords of a general polygon can be derived by replacing edges with a triangle of two consecutive edges. For the cyclotomic numbers, you imagine c1 is some point on the unit circle, and c(n) is its nth power. The chords of a polygon are made by progressing 1 in directions of steps of c2, thus c0, c2, c4, c6, c8 ... represent a unit step along successive edges of a polygon. Since we want this to fall onto the real axis, we should seek to have pairs of values adding to 0, thus c0, c1+c(-1), c2+c0+c(-2), ,,, give rise to chords of the polygon that fall on the unit circle.

The above will work for any ratio, and it is fairly easy to find that (c1+c(-1)) is (c2+2c0+c-2) = ch(0)+ch(2). where ch(n) is the chord n of the figure. It goes that ch(2).ch(m) = ch(m+1)+ch(m-1). Of course ch(0)=0, ch(1)=1. For the horogon, ch(m)=m.

This is how the chord-span happens, It follows from this that only odd chords can appear on the real axis, the even chords will also appear if the number of edges of the polygon is odd.

For a polygon, Ch(p)=0, that is, you have gone around the polygon once. In terms of the cyclotomic equation, c(p)=-1, or c(1)^p+1=0. The latter polynomial has known factors, and the largest such factor for c(1)+c(-1) is a monic polynomial, whose solution is the shortchord of the polygon 2p. This polynomial has as an order half the euler totient of 2p, and the solutions represent the shortchords of the various stars, eg for Z12, the resulting equation is x^4-4x^2+1, the solutions being for 24, 24/5, 24/7, 24/11.

The numbers are carried around the branches of a group, the resulting symmetry having the group enclosing these numbers, which may not necessarily be a polygon group. For example, o5o7o has a group of Z5Z7, which is the spans of the pentagon and heptagon, but is is only half of the group Z35, which includes a further factor sqrt(phi.sqrt(5)).sqrt(7/ab) a, b are the chords of the heptagon.

The correct vectors to use for finding distances is not the Dynkin matrix, but the Stott matrix. The product of these gives rise to the Determinate of the dynkin matrix, times the identity, if leading fractions are ignored. These are the normalised forms of the matrices, every element belongs to some integer system, as determined above. Thus any vector comprised of integers of this system will have a norm, whose square is in this set. The eutactic star, which represents vertex-normals are in this span of vectors, and therefore every wythoff mirror-edge of edges in this number system is itself in this system, including every possible chord of these polytopes.

It works of course, for infinite and hyperbolic polytopes as well. In the case of the horogons, the radius would be infinite, byt if one replaces the '2' on the main diagonal by (2-e) for any vanishing e, the matrix is well defined and solvable. To this end it does not give the distances within the lattice, but the scaling factor of the lattice symmetry to keep the input edges. In reality, you can replace the stott matrix with a stott-vector, s_i, such that the element of \(S_ij = s_is_j\), The dot product of this vector and the input dynkin vector is the scaling factor direct.

It works equally well in hyperbolic space, the distances quoted are not measures of 'arc' but measures of 'chord'. Like the spherical case, we do not do 'spherical trignometry of arcs on the surface', but rather the inscribed polytope. In the hyperbolic case, we use the excribed polytope, whose edges are the lengths of the horocyclic arc spanning the two vertices. This is identical to the spheric case, since the horocycle is the enclosing euclidean metric. Apart from some tricky sign tricks in the squared values, it's pretty much the same as for polytopes.
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### Re: Chords and algebraic integers

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

Oops, that's not quite right. By "adjacent" I meant "connected by an edge". The two ends of an edge are related by some reflection, but it's not necessarily one of the n generating reflections, if v' ≠ v.

That doesn't matter, though. If we interpret "adjacent" to mean "related by a generating reflection", the proof still works. By the nature of generators, any vertex can be reached from v by stepping between "adjacent" vertices.

wendy wrote:The chords of a general polygon can be derived by replacing edges with a triangle of two consecutive edges.

Indeed, it's clear that any chord vector is a sum of edge vectors, because the polytope (any polytope, not just a polygon) is connected; any vertex can be reached from any other vertex by travelling along edges.
(That doesn't imply that the chord length is a sum of edge lengths, of course.)

wendy wrote:The correct vectors to use for finding distances is not the Dynkin matrix, but the Stott matrix.

I'm not using matrices at all. I'm using just the standard dot product, without reference to any particular coordinate system.

It looks like you're giving your own kind of proof for Wythoffian polytopes. I suppose it's good to have several different explanations of the same thing, in case one of them doesn't make sense to someone.

The problem with infinite polytopes is that the Dynkin matrix isn't necessarily algebraic. For example, it could be
Code: Select all
`⌈ 2  -1/π⌉⌊-1/π  2 ⌋`
or
Code: Select all
`⌈ 2  π-6⌉⌊π-6  2 ⌋`
And if it is algebraic, it's not necessarily integral. Yet, we've seen that the chordsquares are in the same number system as the edge lengths and the matrix elements, even if that system is not made of algebraic integers.

The interest now is in non-Wythoffian polytopes. Still all vertices are equivalent.
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### Re: Chords and algebraic integers

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.

Although the dynkin matrix for zero-curvature figures is zero, the actual matrices do indeed have a recriprocal.

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.

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.

But even if there is no ambiant symmetry, the motions of using polygonal struts (say as in the johnson figures) still invokes algebraic conditions.

Have a look at http://www.os2fan2.com/gloss/maths.pdf

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### Re: Chords and algebraic integers

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.

I agree that curved arc lengths are not appropriate to use in this context. Did I say something that suggested otherwise?

I used π just as an arbitrary non-algebraic number, not for a radian measure of an angle.

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.

Chords and algebraic integers 1.png (57.6 KiB) Viewed 2406 times

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.

You're talking about finite regular polygons. Nothing new there. But it is an interesting fact: The product of chordsquares (including edges) at one vertex of a unit-circumradius regular n-gon equals n*n. And the sum of chordsquares equals n+n.

wendy wrote:Have a look at http://www.os2fan2.com/files/maths.pdf

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### Re: Chords and algebraic integers

Here, I'll show that an isogonal pyritohedron with algint edges has algint chords. ("Pyritohedron" isn't really the right word, as it implies pentagonal faces and two vertex types....)

The pyritohedral symmetry group is 'epacs' (even permutations, and all changes of sign, of Cartesian coordinates). If one vertex is (x,y,z), then the other vertices are (±x,±y,±z), (±y,±z,±x), and (±z,±x,±y). Let's assume the polyhedron is convex, and 0<x<y<z, just to find what the edges should be.

Chords and algebraic integers 2.png (19.24 KiB) Viewed 2406 times

The vertices adjacent to (x,y,z) are (-x,y,z), (x,-y,z), (y,z,x), and (z,x,y). The corresponding edge vectors are

(x,y,z) - (-x,y,z) = (2x,0,0)
(x,y,z) - (x,-y,z) = (0,2y,0)
(x,y,z) - (y,z,x) = (x-y,y-z,z-x)
(x,y,z) - (z,x,y) = (x-z,y-x,z-y),

so one edge length is a=2x, one is b=2y, and the other is given by

c² = (x-y)² + (y-z)² + (z-x)²
= 2(x² + y² + z² - xy - xz - yz)
= 2(a²/4 + b²/4 - ab/4 + z² - z(a+b)/2).

Solving for z, we find that 2z is an algebraic integer, provided that a,b,c are:

(2z)² - (a+b)(2z) = 2c²-a²-b²+ab
z = (a+b + √[8c² - 3(a-b)²])/4.

Now the squared chords are

((x,y,z)-(±x,±y,±z))² = 2(1±1)x² + 2(1±1)y² + 2(1±1)z²
= (1±1)/2 a² + (1±1)/2 b² + (1±1)/2 (2z)²

and

((x,y,z)-(±y,±z,±x))² = ((x,y,z)-(±z,±x,±y))²
= (x±y)² + (y±z)² + (z±x)² = 2(x² + y² + z² ± xy ± xz ± yz)
= 2(a²/4 + b²/4 ± ab/4 + z² + z(±a±b)/2)
= (a²+b²±ab)/2 + (2z)²/2 + z(±a±b)
= (a²+b²±ab)/2 + ((a+b)(2z) + 2c²-a²-b²+ab)/2 + z(±a±b)
= (1±1)/2 ab + c² + ((1±1)/2 a + (1±1)/2 b) (2z)

which are all algints because (1±1)/2 is either 0 or 1.

The question is whether this happens in general. Of course the chords are algebraic numbers (with exceptions like infinite or flexible polytopes), but why should they be algebraic integers?
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### Re: Chords and algebraic integers

The correct URL for maths.pdf is http://www.os2fan2.com/gloss/maths.pdf or http://www.os2fan2.com/glossn/maths.pdf The first has th, the second has þ as nature intended.

If you use algebraic numbers in a monic polynomial, the outcome will be algebraic numbers. So the snub, which is \(x^3-3x-a = 0\) will return an algebraic number if a is algebraic. The order of the equation will be no more than three times the one that a solves. The proof is by inserting arrays into the multiplication trimex that the equations solve.
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### Re: Chords and algebraic integers

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.

Ah, now I see what you mean.
The verf of snid, considered as a planar polygon (rather than a spheric polygon or something else), has four edges of length 1, meeting at three equal angles, and one edge of length φ. Omitting the φ edge, we have an equilateral and equiangular figure (an incomplete polygon), so its chords have the same algebraic relations as a regular polygon:
c1 = 1,
c2 = s,
c3 = s^2 - 1,
c4 = s^3 - 2s.
(not s^3 - 3s)
And the 4th chord is just that missing edge, so s^3 - 2s = φ. (This chord s is between vertices of two triangles sharing an edge.)

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.
Last edited by mr_e_man on Tue Apr 05, 2022 12:12 am, edited 2 times in total.
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### Re: Chords and algebraic integers

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.

A snub dodecahedron is the alternation of some omnitruncated dodecahedron x5y3z. Suppose the latter is generated by a seed vertex v₀ and mirror vectors u₁,u₂,u₃; their dot products are

u₁•u₁ = u₂•u₂ = u₃•u₃ = 1,
u₁•u₂ = -φ/2, u₂•u₃ = -1/2, u₁•u₃ = 0,
v₀•u₁ = x/2, v₀•u₂ = y/2, v₀•u₃ = z/2.

v₁ = v₀ - (2v₀•u₁) u₁ = v₀ - x u₁,
v₂ = v₀ - (2v₀•u₂) u₂ = v₀ - y u₂,
v₃ = v₀ - (2v₀•u₃) u₃ = v₀ - z u₃,

so the omnitruncate's edge lengths are x,y,z. Note, we assume nothing about these lengths; they may be unequal, and they may not be algints.

For the snub, the vertex v₀ is deleted, and the other three are connected to form a triangle, with edge vectors v₁ - v₂, v₁ - v₃, v₂ - v₃. The squared edge lengths are

a² = (v₁ - v₂)² = x²(u₁•u₁) + y²(u₂•u₂) - xy(2u₁•u₂) = x² + y² + φxy,
b² = (v₂ - v₃)² = y² + z² + yz,
c² = (v₁ - v₃)² = x² + z² + 0.

These are assumed to be algints (but still not necessarily equal to each other). We'll solve for x,y,z in terms of a,b,c, and find that x,y,z must also be algints, after all.

The third equation shows that z is an algint if x is:

z = √[c² - x²]

Putting this in the second equation:

b² = y² + (c²-x²) + y √[c²-x²]
x² - y² + b² - c² = y √[c²-x²]
(x²-y²)² + 2(b²-c²)(x²-y²) + (b²-c²)² = y²(c²-x²)
x⁴ - x²y² + y⁴ + (2b²-2c²)x² + (c²-2b²)y² + (b²-c²)² = 0
x⁴ + (2b²-2c²-y²)x² + (y⁴+(c²-2b²)y²+(b²-c²)²) = 0

Since the leading coefficient is 1, this shows that x is an algint if y is. Applying the quadratic formula:

x² = (1/2) (y²-2b²+2c² + √[(y²-2b²+2c²)² - 4(y⁴+(c²-2b²)y²+(b²-c²)²)])
x² = (1/2) (y²-2b²+2c² + y √[4b²-3y²])
x⁴ = (1/2) (2c²y²-y⁴+2(b²-c²)² + y(y²-2b²+2c²) √[4b²-3y²])

(Here '√' means '±√'; it will eventually be squared, so the sign doesn't matter.) Working on the first equation before eliminating x:

a² - x² - y² = φxy
a⁴ - 2a²x² - 2a²y² + x⁴ + 2x²y² + y⁴ = φ²x²y²
x⁴ + (2y²-φ²y²-2a²)x² + (a²-y²)² = 0
(1/2) (2c²y²-y⁴+2(b²-c²)² + y(y²-2b²+2c²) √[4b²-3y²]) + (2y²-φ²y²-2a²) (1/2) (y²-2b²+2c² + y √[4b²-3y²]) + (a²-y²)² = 0
2c²y²-y⁴+2(b²-c²)² + (2y²-φ²y²-2a²)(y²-2b²+2c²) + 2(a²-y²)² = -y(y²-2b²+2c² + 2y²-φ²y²-2a²) √[4b²-3y²]
(3-φ²)y⁴ + 2(φ²(b²-c²) + 3c²-3a²-2b²)y² + 2(a²+b²-c²)² = -y((3-φ²)y² - 2(a²+b²-c²)) √[4b²-3y²]
[(3-φ²)y⁴ + 2(φ²(b²-c²) + 3c²-3a²-2b²)y² + 2(a²+b²-c²)²]² = y²[(3-φ²)y² - 2(a²+b²-c²)]² (4b²-3y²)

Finally, we have a polynomial equation in y, with algint coefficients. This is already a big equation, and it would get much worse if we expanded the squares, but we don't need to. We just need the leading coefficient, after subtracting the right side from the left side:

(3-φ²)²y⁸ - y²(3-φ²)²y⁴(-3y²)
= 4(3-φ²)²y⁸

Notice that the lower-degree terms have factors of 2, so, after squaring, everything has a factor of 4. Dividing this out, we get a leading coefficient (3-φ²)².

Until now, we haven't used any properties of φ. What we've done applies just as well to a snub cube or a snub tetrahedron; replace φ with √2 or 1.

For a snub dodecahedron, the leading coefficient is (3-φ²)² = (2-φ)² = (φ⁻²)². Multiplying by φ⁴, we get an algint polynomial equation with leading coefficient 1, which shows that y is an algint. (Thus x and z are also.)

For a snub cube, the leading coefficient is (3-(√2)²)² = 1, so again y is an algint.

But for a snub tetrahedron, the leading coefficient is (3-1²)² = 4, so y is not necessarily an algint.

We know that if the omnitruncate's edges x,y,z are algints, then its chords are algints (since it's Wythoffian). And obviously the snub's chords are a subset of the omnitruncate's chords. Therefore, any snub dodecahedron or snub cube, with algint edges a,b,c, also has algint chords.
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mr_e_man
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### Re: Chords and algebraic integers

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.

And now I see that the other chords, in a uniform snub, actually can be gotten from that without too much difficulty.

Take basis vectors e₁,e₂,e₃ as the edges between triangles at one vertex.

Chords and algebraic integers 3.png (31.84 KiB) Viewed 2188 times

As in the Wythoffian case, we'll find that any chord vector is a combination of the basis vectors with algint coefficients. Then the squared length of the chord will be a combination (with algint coefficients) of these dot products (which are also algints):

e₁•e₁ = e₂•e₂ = e₃•e₃ = 1,
2e₁•e₂ = 2e₂•e₃ = 1,
2e₁•e₃ = e₁² + e₃² - (e₁ - e₃)² = 1² + 1² - s²
= 2 - s².

Since any chord vector is a sum of edge vectors, it suffices to show that any edge vector is a combination of the basis vectors with algint coefficients.

Chords and algebraic integers 4.png (24.57 KiB) Viewed 2188 times

Notice that the chord between vertices C and F is parallel to the edge between D and E. Comparing their lengths, we get

C-F = (s²-1)(D-E)
= (s²-1)(D-A + A-E)
= (s²-1)(e₁ - e₂).

It follows that the edge C-A is

C-F + F-A = (s²-1)(e₁ - e₂) + e₃.

A snub polyhedron has mirror symmetry if you don't look too far. The edges around vertex C can be gotten from the edges around vertex A, by reflecting along the unit vector C-A:

D-C = D-A - [2(D-A)•(C-A)](C-A)
= e₁ - [2e₁•((s²-1)(e₁ - e₂) + e₃)] ((s²-1)(e₁ - e₂) + e₃)
= e₁ - [1] ((s²-1)(e₁ - e₂) + e₃)
= (2-s²)e₁ + (s²-1)e₂ - e₃,

H-C = E-A - [2(E-A)•(C-A)](C-A)
= e₂ - [2e₂•((s²-1)(e₁ - e₂) + e₃)] ((s²-1)(e₁ - e₂) + e₃)
= e₂ - [2-s²] ((s²-1)(e₁ - e₂) + e₃)
= (s⁴-3s²+2)e₁ - (s⁴-3s²+1)e₂ + (s²-2)e₃,

G-C = F-A - [2(F-A)•(C-A)](C-A)
= e₃ - [2e₃•((s²-1)(e₁ - e₂) + e₃)] ((s²-1)(e₁ - e₂) + e₃)
= e₃ - [-s⁴+2s²+1] ((s²-1)(e₁ - e₂) + e₃)
= (s⁶-3s⁴+s²+1)(e₁ - e₂) + (s⁴-2s²)e₃.

We can continue going around the pentagon, getting new edges by reflecting old edges. For example, the next vertex past C has edges

(2-s²)(G-C) + (s²-1)(H-C) - (D-C),

(s⁴-3s²+2)(G-C) - (s⁴-3s²+1)(H-C) + (s²-2)(D-C),

(s⁶-3s⁴+s²+1)((G-C) - (H-C)) + (s⁴-2s²)(D-C);

these can be written as combinations of e₁,e₂,e₃, but the coefficients are fairly large polynomials in s, up to 12th degree. There are no fractions in these polynomials, so they are algebraic integers.

And we can move to an adjacent pentagon, changing to a different basis:

e₁' = A-D = -(D-A) = -e₁,

e₂' = C-D = -(D-C) = (s²-2)e₁ + (1-s²)e₂ + e₃,

e₃' = H-D = H-C + C-D = (s⁴-2s²)e₁ - (s⁴-2s²)e₂ + (s²-1)e₃.

All of the edge vectors around that pentagon can be written as combinations of e₁',e₂',e₃', which in turn are combinations of e₁,e₂,e₃. The coefficients always remain in the algebraic ring generated by s.

You get the idea.
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mr_e_man
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### Re: Chords and algebraic integers

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.)

The squared edges f₃,g₀,g₂ are algebraically independent polynomials. This means there is no polynomial in 3 variables, h(u,v,w), such that h(f₃(x,y,z),g₀(x,y,z),g₂(x,y,z)) = 0 (for all x,y,z), except the trivial polynomial h(u,v,w) = 0 (for all u,v,w).

(This follows from the Jacobian criterion: The determinant of the matrix [∇f₃,∇g₀,∇g₂] is not 0.)

But any four polynomials in the three variables x,y,z are algebraically dependent. Take three of them to be the squared edges f₃,g₀,g₂, and take the fourth to be a squared chord, say f₁. There is a polynomial in four variables, h(t,u,v,w) ≠ 0, such that h(f₁(x,y,z),f₃(x,y,z),g₀(x,y,z),g₂(x,y,z)) = 0.

According to this MSE answer, we can find such h using the following Magma code:

Code: Select all
`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);`

Here's the output:

Code: Select all
`Ideal of Polynomial ring of rank 4 over Rational FieldOrder: LexicographicalVariables: y1, y2, y3, y4Homogeneous, Dimension >0Basis:[    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]`

That is,

f₁⁴ - f₁³f₃ - 2f₁³g₀ - 2f₁³g₂ + 2f₁²f₃g₀ + 2f₁²f₃g₂
+ 2f₁²g₀² + 2f₁²g₂² - f₁f₃³ + 2f₁f₃²g₀ + 2f₁f₃²g₂
- 8f₁f₃g₀g₂ - 4f₁g₀³ + 4f₁g₀²g₂ + 4f₁g₀g₂² - 4f₁g₂³
+ f₃⁴ - 2f₃³g₀ - 2f₃³g₂ + 2f₃²g₀² + 2f₃²g₂² - 4f₃g₀³
+ 4f₃g₀²g₂ + 4f₃g₀g₂² - 4f₃g₂³ + 4g₀⁴ - 8g₀²g₂² + 4g₂⁴
= 0;

f₁⁴ + (- f₃ - 2g₀ - 2g₂)f₁³ + (2f₃g₀ + 2f₃g₂ + 2g₀² + 2g₂²)f₁²
+ (- f₃³ + 2f₃²g₀ + 2f₃²g₂ - 8f₃g₀g₂ - 4g₀³ + 4g₀²g₂ + 4g₀g₂² - 4g₂³)f₁
+ (f₃⁴ - 2f₃³g₀ - 2f₃³g₂ + 2f₃²g₀² + 2f₃²g₂² - 4f₃g₀³
+ 4f₃g₀²g₂ + 4f₃g₀g₂² - 4f₃g₂³ + 4g₀⁴ - 8g₀²g₂² + 4g₂⁴)
= 0.

Since the leading coefficient is 1, this shows that f₁ is an algint if f₃,g₀,g₂ are algints.

Similarly, doing this for the other chords, we get

f₂⁴ + (f₃ - 4g₀ - 4g₂)f₂³ + (- 4f₃g₀ - 4f₃g₂ + 8g₀² + 8g₀g₂ + 8g₂²)f₂²
+ (f₃³ - 2f₃²g₀ - 2f₃²g₂ + 6f₃g₀² + 8f₃g₀g₂ + 6f₃g₂² - 8g₀³ - 8g₀²g₂ - 8g₀g₂² - 8g₂³)f₂
+ (f₃⁴ - 2f₃³g₀ - 2f₃³g₂ + 2f₃²g₀² + 4f₃²g₀g₂ + 2f₃²g₂² - 4f₃g₀³ - 4f₃g₀²g₂
- 4f₃g₀g₂² - 4f₃g₂³ + 4g₀⁴ + 8g₀²g₂² + 4g₂⁴)
= 0,

g₁⁴ + (- 2f₃ - 2g₂)g₁³ + (3f₃² - f₃g₀ - g₀² + 3g₂²)g₁²
+ (- 2f₃³ + f₃²g₀ + f₃g₀² + f₃g₀g₂ - 2g₂³)g₁
+ (f₃⁴ - f₃³g₀ - 2f₃³g₂ + f₃²g₀g₂ + 3f₃²g₂² - f₃g₀³
+ f₃g₀²g₂ - f₃g₀g₂² - 2f₃g₂³ + g₀⁴ - g₀²g₂² + g₂⁴)
= 0,

g₃⁴ + (- 2f₃ - 2g₀)g₃³ + (3f₃² - f₃g₂ + 3g₀² - g₂²)g₃²
+ (- 2f₃³ + f₃²g₂ + f₃g₀g₂ + f₃g₂² - 2g₀³)g₃
+ (f₃⁴ - 2f₃³g₀ - f₃³g₂ + 3f₃²g₀² + f₃²g₀g₂ - 2f₃g₀³
- f₃g₀²g₂ + f₃g₀g₂² - f₃g₂³ + g₀⁴ - g₀²g₂² + g₂⁴)
= 0.

So indeed all chords are algints if all edges are algints, in a snub tetrahedron.

(Here you can see a snub tetrahedron, as well as that pyritohedral thing, in the lowest images.)
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mr_e_man
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### Re: Chords and algebraic integers

A snub dihedron is an alternated xPy2z. This could be called a P-gon antiprism, but it's generally non-uniform and chiral. The top P-gon is twisted by some angle relative to the bottom P-gon, and the two are connected by scalene triangles.

There is another type of isogonal polyhedron, "between" a prism and an antiprism; it has trapezoids instead of triangles or rectangles. It can be written in segmentotope notation as xPy || yPx. It can be gotten from a prism x2Pw2z (where w = (y-x)/(2cos(π/(2P))), assuming x<y) by deleting some pairs of adjacent vertices.

I did the chord calculations for both of these types of polyhedron (getting algint results as expected), back in February, but I don't remember how exactly. I'm not sure anyone's interested anyway. I do remember using the following fact.

If θ is a rational angle, or more generally if 2cos(θ) is an algint, then sin(kθ)/sin(θ) is an algint, for any integer k. This is seen most easily using complex numbers. Of course sin(0) = 0 and sin(-kθ) = -sin(kθ), so assume k is positive. Let u = exp(iθ) = cosθ + i sinθ; this is an algint because u² - (2cosθ)u + 1 = 0. And 1/u satisfies the same equation. Then

sin(θ) = (u - 1/u)/(2i),
sin(kθ) = (uk - 1/uk)/(2i),

and, using the general algebraic identity ak - bk = (a - b) (ak-1 + ak-2b + ak-3b2 + ... + bk-1), we have

sin(kθ)/sin(θ) = (uk - 1/uk)/(u - 1/u)
= (uk-1 + uk-3 + uk-5 + ... + 1/uk-1),

which is a sum of algints, thus an algint.

Alternatively, to show that it's an algint without using complex numbers, you could derive the formula sin((k+1)θ)/sin(θ) = sum0≤j≤k/2 (-1)j (k-j choose j) (2cosθ)k-2j.
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mr_e_man
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### Re: Chords and algebraic integers

Note that sin(kθ)/sinθ is the k'th chord of a (unit-edge) regular polygon with central angle θ between a vertex and an edge midpoint. The shortchord is s = sin(2θ)/sinθ = 2cosθ.

mr_e_man wrote:sin((k+1)θ)/sin(θ) = sum0≤j≤k/2 (-1)j (k-j choose j) (2cosθ)k-2j

sin(0θ)/sinθ = 0
sin(1θ)/sinθ = (0 choose 0) = 1
sin(2θ)/sinθ = (1 choose 0)s = s
sin(3θ)/sinθ = (2 choose 0)s² - (1 choose 1) = s² - 1
sin(4θ)/sinθ = (3 choose 0)s³ - (2 choose 1)s = s³ - 2s
sin(5θ)/sinθ = (4 choose 0)s⁴ - (3 choose 1)s² + (2 choose 2) = s⁴ - 3s² + 1
sin(6θ)/sinθ = (5 choose 0)s⁵ - (4 choose 1)s³ + (3 choose 2)s = s⁵ - 4s³ + 3s
sin(7θ)/sinθ = (6 choose 0)s⁶ - (5 choose 1)s⁴ + (4 choose 2)s² - (3 choose 3) = s⁶ - 5s⁴ + 6s² - 1
sin(8θ)/sinθ = (7 choose 0)s⁷ - (6 choose 1)s⁵ + (5 choose 2)s³ - (4 choose 3)s = s⁷ - 6s⁵ + 10s³ - 4s
sin(9θ)/sinθ = (8 choose 0)s⁸ - (7 choose 1)s⁶ + (6 choose 2)s⁴ - (5 choose 3)s² + (4 choose 4) = s⁸ - 7s⁶ + 15s⁴ - 10s² + 1

I guess that explicit formula is new to Wendy, as the "Chord or Iso-Arithmetic" section of her document only gives a recursive formula, equivalent to this:

sin((k+1)θ)/sinθ = (2cosθ) sin(kθ)/sinθ - sin((k-1)θ)/sinθ
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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### Re: Chords and algebraic integers

The classification of 3D symmetry groups shows that these are the only types of convex isogonal polyhedron: Wythoffians, snubs & antiprisms, trapezoprisms, pyritosnub cube (including pyritosnub tetrahedron as a special case, with one edge length vanishing).
We've accounted for all of these now. Indeed the chords are always algebraic integers if the edges are.

Would anyone care to examine the chords of some of the known polychora (e.g. non-convex uniforms), and determine whether they're algints?
Or maybe there's some isogonal CRF polychoron involving J27, J34, J37, or J72-82 (as most of these have non-algint chords)?
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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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### Re: Chords and algebraic integers

I found another counter-example. It's conceptually similar to the previous one: In both cases, I took a skew polygonal loop, turned the loop into a torus somehow, and then tiled the torus to make a polyhedron.

Consider a rectangular trapezoprism. If one vertex is (x,y,z), then the other vertices are (±x,±y,z) and (±y,±x,-z). The squared edges and chords are

a² = 4x²
b² = 4y²
c² = 2x² - 4xy + 2y² + 4z²
d² = 2x² + 2y² + 4z² = c² + ab
e² = 2x² + 4xy + 2y² + 4z² = c² + 2ab
f² = 4x² + 4y² = a² + b².

Take the loop formed by the a edges and c edges. The shortchords here are d, so the triangles formed by every three consecutive vertices in this loop have edges a,c,d. This folded-up strip of triangles has some loose edges (namely the d edges); put another strip of triangles in the same position, and attach the d edges of the two strips, so that every edge connects exactly two faces.

The resulting polyhedron is isogonal, and rigid (if the symmetry is maintained). The chords are given by b = (d²-c²)/a, e² = 2d²-c², f² = a² + (d²-c²)²/a². So, if the edges a,c,d are algints, then the chords may not be algints (for example a=2, c=1, d=2; b=3/2).

And the edge vectors at a vertex span the 3D space. However, the polyhedron is still degenerate, with coincident edges, and coincident faces.

Chords and algebraic integers 5.png (62.68 KiB) Viewed 1631 times

(The middle part of the image shows the abstract structure, as a triangular tiling of a torus.
The lower part of the image is almost another counter-example. It's just a graph (just vertices and edges), not a polyhedron. Its edges are the slant edges of a square antiprism, and the diagonals of the top and bottom squares. Those may be algints, while the graph's chords (the antiprism's horizontal edges) are not algints, being edge lengths divided by √2.)
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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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### Re: Chords and algebraic integers

And here's a non-degenerate counter-example: a hexagonal stephanoid. Its convex hull is a hexagonal prism. Each face is a self-intersecting tetragon, with two adjacent vertices of one hexagon and two opposite vertices of the other hexagon. Each edge has one vertex in each hexagon.

Chords and algebraic integers 6.png (24.11 KiB) Viewed 1557 times

The hexagonal prism's edge lengths are a,b,h, and its chords are given by

c² = a² + b² + ab
d = a + b
A² = a² + h²
B² = b² + h²
C² = c² + h² = a² + b² + ab + h²
D² = d² + h² = a² + b² + 2ab + h².

The hexagonal stephanoid's edge lengths are A,B,C, and its chords are given by

a² = (C² - B²)² / (2C² - A² - B²)
b² = (C² - A²)² / (2C² - A² - B²)
h² = (A²B² - (C² - A² - B²)²) / (2C² - A² - B²).

Because of the division, these are not necessarily algints. (For example A=B=2, C=3; a²=5/2.)

So it seems that this algint phenomenon only applies to convex polytopes, or uniform polytopes, and algebraic conjugates and edge length variations of these.

Or maybe it has something to do with the topology; all of these counter-examples have Euler characteristic 0.

On the other hand, I just did some calculations for the uniform star polyhedra (without coplanar faces or coincident edges etc.), particularly the six snubs, using the same method as here. I found that they do have algint chords, even though they're not Wythoffian nor variations of convex polyhedra nor having Euler characteristic 2. (Well, in three of the six cases, the Euler characteristic is 2.)

Maybe it's because the snubs are "close enough" to being Wythoffian, unlike the stephanoid.
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ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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