On May 25, 2013, at 11:09 PM, Dan Asimov wrote:
Let's see. If a Platonic solid K can be embedded in R^3 with integer vertices, then the center will be a rational point, so by an integer expansion it, too, will be integer. So by an integer translation we can assume K has integer vertices and center.
Then the isometry group Isom(K) will be a subgroup of GL(3,Z), so in particular GL(3,Z) must have an element g of order 5. Then a primitive 5th root of unity must be an eigenvalue of g's matrix. But such roots of unity have a minimal polynomial equal to (x^5-1)/(x-1) = an integer polynomial of 4th degree, so can't be an eigenvalue of an integer 3x3 matrix.
Adam>A pentagon cannot be embedded on an integer lattice of any number of dimensions. Suppose it could. Let one of the vertices be at the origin, and the other two vectors represented by x, y.
Then cos^2(3pi/5) = (x.y)^2/((x.x)(y.y)) = a rational number.
But cos^2(3pi/5) is irrational.
I believe that the only regular polygons embeddable in Z^n are the triangle, square and hexagon. (I made an equivalent assumption to reduce the latest Project Euler problem to a number-theoretic bash. Unfortunately, it still takes 400 CPU-hours... :/)
Let's see if we can prove it. cos^2(pi - 2pi/n) being rational is equivalent to the real and imaginary parts of e^(2 pi i/n) being roots of quadratic equations. And that is equivalent to e^(2 pi i/n) having minimal (cyclotomic) polynomial of degree <= 2, i.e. phi(n) <= 2.
But the only n for which phi(n) <= 2 are 1, 2, 3, 4 and 6. QED.
Corollary: The only regular polytopes embeddable in Z^n (for any n) are:
Line segment {(0), (1)} Hexagon {(0,1,2),(0,2,1),{1,2,0},(2,1,0),(2,0,1),(1,0,2)} Simplex {(1,0,0,0,...,0), (0,1,0,0,...,0), ... (0,0,0,0,...,1)} Hypercube {(±1, ±1, ..., ±1)} Orthoplex {(±1,0,0,...,0), (0,±1,0,...,0), ... (0,0,0,...,±1)} 24-cell {(±2,0,0,0), (0,±2,0,0), (0,0,±2,0), (0,0,0,±2), (±1,±1,±1,±1)}
Sincerely,
Adam P. Goucher http://cp4space.wordpress.com
From: Warren D Smith
Sent: 05/26/13 07:23 PM To: math-fun@mailman.xmission.com Subject: [math-fun] Embedding with integer coordinates
If your thing has got two interpoint distances A,B such that (A/B)^2 is irrational, then its points cannot be be embedded in any finite dimension with integer coordinates.
Regular pentagon, (A/B)^2=2.6180339887498948482 = g+1 where g=golden ratio where B=side, A=2-apart vertex sep
icosahedron (including center as point) (A/B)^2 = sin(2*pi/5)^2 = (g/2)^2 + 1/4 =0.9045084971874737120
dodecahedron (ditto) (A/B)^2 = g^2 * (3/4) = 1.96352549156242113610 where A=radius, B=side
all easily seen to be irrational, so none can be embedded.
rwg> http://www.tweedledum.com/rwg/rectarith12.pdf (end) illustrates hexagons in R^3, followed immediately by
"How many dimensions before we see (regular) octagons, pentagons, dodecagons, ...? We won't! Even in an infinite dimensional grid, the only regular polygons of finite size are triangles, hexagons, and squares. We could probably prescribe the nth coordinate of the kth vertex of a regular pentagon, say, but infinitely many coordinates would be nonzero, resulting in infinite size. And it would only approach regularity as we consider higher and higher dimensions." --rwg
--Eight(?) years ago I gave Julian, who is watching just my end of this discussion, a hardcopy of http://www.tweedledum.com/rwg/rectarith12.pdf
http://s216.photobucket.com/user/sjmathcircle/media/SJMC%2004-30-08/P4300085... http://s216.photobucket.com/user/sjmathcircle/media/SJMC%2004-30-08/P4300084... Last night he offered [Theorem: The only regular polygons embeddable in Z^n are triangles, squares, and hexagons.] On Sun, May 26, 2013 at 11:04 AM, Julian Ziegler Hunts <julianj.zh@gmail.com
wrote:
The proof: if we have a regular n-gon, we have two equally long vectors making an angle of 2π/n. Then cos^2(2π/n)=(u.v)^2/((u.u)*(v.v))=(u.v)^2/(u.u)^2 is a rational square, so 2π/n is a rational angle with rational cosine, which only happens for multiples of π/2 and π/3. The proof of that is something that I looked up a while ago, maybe when we were doing radical denesting, and goes as follows: Suppose cos(pπ/q) is rational. Then, since the map x->2x^2-1 takes cos(t)->cos(2t) and rationals to rationals, cos(2^npπ/q) is rational. Also, denominator(2x^2-1) is at least half the square of denominator(x), and so is larger than denominator(x) when denominator(x) is larger than 2. Since cosine has period 2π, and there are only finitely many rationals in [0,2) with denominator at most q, we must have cos(2^mpπ/q)=cos(2^npπ/q) for some m<n, so they have the same denominator. Since cos(2^npπ/q) is the image of cos(2^mpπ/q) under n-m applications of x->2x^2-1, this means that their denominators are ≤2, and therefore that the denominator of cos(pπ/q) is ≤2. All rationals with denominator ≤2 are accounted for by cosines of multiples of π/2 and π/3, so those are the only rational angles with rational cosines. Note that a slight modification of the above shows that the only rational angles achievable in Q^n are multiples of π/4 and π/6.
[Later:...] you could mention the extensions to higher dimensions (e.g. all but two of the (convex) regular polychora [namely, the 120- and 600-cell] can be embedded with integer coordinates), or other extensions of the question. And you could also add something about >4 dimensions (all embeddable) or nonconvex regular polytopes.
Julian --rwg