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Topic: Trinity, yes; Pentagram, no (Read 687 times) |
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ecoist
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Trinity, yes; Pentagram, no
« on: Nov 18th, 2006, 10:51am » |
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One can embed an equilateral triangle in 3-space or higher so that all of its vertices have integer coordinates. However, a regular pentagon cannot be so embedded. Why?
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Eigenray
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Re: Trinity, yes; Pentagram, no
« Reply #1 on: Nov 18th, 2006, 6:01pm » |
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Because GL3(Q) contains no element of order 5:
| hidden: | Let v1, ..., v5 in Q3 be the vertices of a regular pentagon. The center of the pentagon is the average of these vectors, so has rational coordinates; WLOG we may therefore assume the center is 0.
Let u = v1 x v2, a rational vector orthogonal to v1 and v2, hence orthogonal to their Q-span, which is the plane of the pentagon (in Q3). Then {v1, v2, u} is a basis of Q3, and v3 is a Q-linear combination of v1, v2.
Therefore there is a linear transformation A in GL(Q3) with Av1=v2, Av2 = v3, and Au = u. Then A is a rotation of order 5 in the plane of the pentagon.
But the minimal polynomial divides both the characteristic polynomial of A, which is of order 3, and the polynomial x5-1 = (x-1)(x4+x3+x2+x+1). Since the latter factor is irreducible, the minimal polynomial could only be x-1, but that would mean A=I, a contradiction. |
On the other hand, GL4(Q) does contain elements of order 5. But are any of them rotation matrices?
More generally, GLm(Q) contains an element of order n = [prod] piai iff m > m(n) := [sum] phi(piai) or n=2 mod 4 and m> m(n/2) = m(n)-1. But when is there a rotation of order n?
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Barukh
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Re: Trinity, yes; Pentagram, no
« Reply #2 on: Nov 21st, 2006, 3:44am » |
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Eigenray presented some general results; in case of a pentagon the reason is much more prosaic.
In any number of dimensions, two parallel vectors with rational coordinates of their endpoints, have length ratio equal to a square of a rational number.
However, in a regular pentagon the diagonal is parallel to a side, but their ratio (golden) is not of that kind.
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Eigenray
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Re: Trinity, yes; Pentagram, no
« Reply #3 on: Nov 21st, 2006, 7:58pm » |
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Ah, I was confusing myself before. The dimension doesn't really matter. The point is that if v1,...,vn form a regular n-gon in some plane P, and a linear transformation T takes v1 to v2, and v2 to v3, then T must map P to itself, and restricts to a rotation of order n. And the only finite-order elements of GL2(Q) (or GL3(Q)) have order 1,2,3,4 or 6.
To generalize Barukh's argument, note that
|v1-v2| = 2|v1|sin(pi/n),
so sin2(pi/n) is rational. The usual argument is then that 4sin2(pi/n) is also an algebraic integer, so 4sin2(pi/n) is a (rational) integer, hence either 0, 1, 2, 3, or 4. Then n=1,2,3,4, or 6, and all are realized (though n=3,6 are only possible in dimension at least 3).
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