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Jan 3, 2012 at 14:58 answer added Hugh Thomas timeline score: 1
Dec 12, 2011 at 13:24 comment added Bruce Westbury @Alexander: I was thinking of taking the "scissors congruence" group over the rationals. Then the volume is a group homomorphism to the reals. Then I had in mind to reinterpret this in terms of $K$-theory and the Borel regulator.
Dec 12, 2011 at 6:52 history edited Theo Johnson-Freyd CC BY-SA 3.0
Incorporated some comments
Dec 11, 2011 at 22:31 comment added so-called friend Don @Theo: I learned about it on math.stackexchange: Two proofs are at math.stackexchange.com/questions/79861/…
Dec 11, 2011 at 22:27 comment added Theo Johnson-Freyd @Anonymous: Well, that seems to answer the first question. I'm sure that "If theta is a rational multiple of π, then 2cos(θ) is an algebraic integer" is well-known, but it is not well-known to me — can you point me to somewhere to read more?
Dec 11, 2011 at 20:44 comment added Alexander Woo @Bruce: Could you explain the reason 'scissors congruence' might be related? Or is it just a hunch?
Dec 11, 2011 at 20:32 comment added Bruce Westbury Have you looked into "scissors congruence"?
Dec 11, 2011 at 20:19 comment added so-called friend Don It seems your pessimism is justified. If theta is a rational multiple of $\pi$, then $2\cos(\theta)$ is an algebraic integer. But if $\theta = \arctan(\sqrt{3}/5)$, then $2\cos(\theta) = 5/\sqrt{7}$.
Dec 11, 2011 at 14:43 comment added Ben Webster You correctly described the weight lattice; the root lattice is the intersection with that hyperplane rather than the projection. Unfortunately, that's the only part of your question I have much intelligent to say about.
Dec 11, 2011 at 7:57 history asked Theo Johnson-Freyd CC BY-SA 3.0