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Topic: Triangles in Z^n (Read 801 times) |
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Eigenray
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Triangles in Z^n
« on: Aug 30th, 2007, 7:14pm » |
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When can a triangle be embedded (up to similarity) in  n?
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Barukh
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Re: Triangles in Z^n
« Reply #1 on: Aug 31st, 2007, 11:27pm » |
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Let one of the vertices be put at the origin, and two others at lattice points A, B in n. If a, b are corresponding vectors from the origin to points A, B, then the area S of the triangle satisfies:
(2S)2 = a2b2 - (ab)2,
which is an integer. Therefore, if S is not a square root of an integer, the triangle cannot be embedded.
Of course, this is not suficient to answer the original question, even in general.
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| « Last Edit: Aug 31st, 2007, 11:28pm by Barukh » |
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Barukh
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Re: Triangles in Z^n
« Reply #2 on: Sep 1st, 2007, 5:08am » |
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Here’s another necessary condition: If a triangle with angles  ,  is embeddable in n, then both quantities tan2() and tan()/tan() are rational.
This condition, I think, is also sufficient.
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| « Last Edit: Sep 1st, 2007, 5:13am by Barukh » |
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Michael Dagg
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Re: Triangles in Z^n
« Reply #3 on: Sep 1st, 2007, 7:59am » |
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Good question. Obviously the squares of the ratios of the side lengths
will have to be rational. But that's not enough: a classic question asks,
"Is there an equilateral triangle in Z^n ?" ; here the ratios of the
side lengths are all equal to 1.
But, there are no equilateral triangles in Z^n by Pick's Theorem.
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| « Last Edit: Sep 1st, 2007, 8:05am by Michael Dagg » |
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Michael Dagg
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Eigenray
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Re: Triangles in Z^n
« Reply #4 on: Sep 1st, 2007, 10:24am » |
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on Aug 31st, 2007, 11:27pm, Barukh wrote:| Therefore, if S is not a square root of an integer, the triangle cannot be embedded. |
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But I am asking about similarity classes of triangles.
on Sep 1st, 2007, 5:08am, Barukh wrote:If a triangle with angles , is embeddable in n, then both quantities tan2() and tan()/tan() are rational.
This condition, I think, is also sufficient. |
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It is in fact sufficient for the triangle to be embeddable in some n.
But for each n, which triangles can be embedded in n? It may help to first consider the case n=2.
on Sep 1st, 2007, 7:59am, Michael_Dagg wrote:| But, there are no equilateral triangles in Z^n by Pick's Theorem. |
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Not in 2, but there are in all higher dimensions. In fact, a regular n-simplex can be embedded in n when:
n is even and n+1 is a square;
n  3 mod 4; or
n 1 mod 4 and n+1 is the sum of two squares.
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| « Last Edit: Sep 1st, 2007, 10:28am by Eigenray » |
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Barukh
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Re: Triangles in Z^n
« Reply #5 on: Sep 2nd, 2007, 5:20am » |
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on Sep 1st, 2007, 10:24am, Eigenray wrote:
It may help to first consider the case n=2. |
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This case is easy: tan(), tan() both rational.
Proof: Let tan() = p/q, tan() = r/s. Then the lattice points (0, rp), (ps, 0), (rq, 0) form a triangle similar to given.
The converse may be shown using formulas for tan of the sum of two angles.
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Michael Dagg
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Re: Triangles in Z^n
« Reply #6 on: Sep 2nd, 2007, 11:21am » |
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I meant Z^2 when I referred to Pick's Theorem.
I don't think I've ever seen your result about the n-simplex.
I guess I can imagine a proof (assuming "when" means "if", not "iff").
Is there a simple way to handle the case n=2? I mean, besides the
Pick's Theorem method I mentioned, which I've always thought was a
wonderful use of an unexpected tool...
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Michael Dagg
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Eigenray
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Re: Triangles in Z^n
« Reply #7 on: Sep 2nd, 2007, 12:03pm » |
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on Sep 2nd, 2007, 5:20am, Barukh wrote:
Proof: Let tan() = p/q, tan() = r/s. Then the lattice points (0, rp), (ps, 0), (rq, 0) form a triangle similar to given. |
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Now can you think of a way to generalize that same construction?
on Sep 2nd, 2007, 11:21am, Michael_Dagg wrote:I don't think I've ever seen your result about the n-simplex.
I guess I can imagine a proof (assuming "when" means "if", not "iff").
Is there a simple way to handle the case n=2? I mean, besides the
Pick's Theorem method I mentioned, which I've always thought was a
wonderful use of an unexpected tool... |
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It is in fact iff, and it's proved in:
I. J. Schoenberg, Regular simplices and quadratic forms, J. London Math. Soc., 12 (1937) 48-55.
The relation to quadratic forms is this:
Lemma: Let  = OP1...Pn and ' = OP1'...Pn' be two given n-dimensional simplices, and let
F(x) =  i,k <Pi, Pk> xixk, F(y) = i,k <Pi', Pk'> yiyk
be their corresponding quadratic forms. Let L() and L(') be the lattices generated by the sets of vectors {Pi} and {Pi'}. A multiple  L() can be inscribed in L(') iff the quadratic forms F(x) and F'(y) are rationally equivalent up to a factor.
However, the case n even is relatively simple: just consider the volume.
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Barukh
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Re: Triangles in Z^n
« Reply #8 on: Sep 4th, 2007, 10:18am » |
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on Sep 2nd, 2007, 12:03pm, Eigenray wrote:
Now can you think of a way to generalize that same construction? |
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In fact, I've read Beeson's paper...
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Barukh
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Re: Triangles in Z^n
« Reply #9 on: Sep 7th, 2007, 10:05am » |
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Here’s what I know.
The embeddability of a triangle with rational tan2() = p/q in n implies a non-trivial solution of the following system of Diophantine equations (indices run 1 ... n):
pq ai2 = hi2,
aihi = 0
where a, h are unknowns.
It then follows that if pq is represented as a sum of n squares, the corresponding angle is embeddable in n+1. Therefore, any triangle with rational tangent squares may be embedded in 5.
That’s sufficient but not necessary. It’s further shown that if triangle is embeddable in 4, then it’s embeddable also in 3. The trivial example of this is the equilateral triangle: for it, pq = 3, a sum of three squares, and embeddable in three dimensional space.
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| « Last Edit: Sep 7th, 2007, 10:06am by Barukh » |
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