Euclidean Geometry

The ancient Greeks managed to prove quite a lot about mathematics in the guise of geometry. Euclid systematised much of this in his Elements (of geometry). This book started from five common notions and five postulates – collectively known as Euclid's axioms – and proceeded to prove a broad collection of results. His fifth postulate, which it is easy to believe even Euclid didn't like much (in the Elements, he avoided using it for as long as possible), stands out as the only axiom that is not intuitively obvious. Ultimately, it has been found to be the one that distinguishes the geometry of a flat surface from that of a curved surface.

Euclid's fifth postulate concerns a pair of lines which cross a third: if the internal angles they make with that third on a given side sum to less than half a turn, he postulates that – if extended indefinitely – they must meet on that same side of the given line. This postulate is commonly referred to as the parallel postulate because of its implications for parallel lines: it could equally be regarded as specifying the equivalence of any pair among the following notions of parallelness. Two lines may be deemed parallel if

If any two of these are equivalent, then so are the rest. It should be noted that further notions of parallelism are possible, that may be identified with one or more of the above, either individually or as a consequence of the equivalence of some pair of them. For example, two lines deemed parallel by at least some subset of the above satisfy:

With any of these notions of parallelism, Euclid's fifth postulate becomes equivalent to for any line L and point P not on L, there is exactly one line parallel to L which passes through P. There are many other axioms to which Euclid's fifth postulate is equivalent. One of the nicest is the assertion that any triangle can be scaled arbitrarily without changing either its angles or the ratios among the lengths of its sides. Euclid's fifth has also been shown equivalent to the Pythagorean theorem.

Some observations

A Euclidean space, E, is isomorphic to its tangent bundle, V, which thus serves a a space of differences between members of E. Let E's dimension be dim.

Obtaining a basis from a simplex.

We can use a simplex in E of side h, obtained from a function (1+dim:p:E) as (1+dim | p), to define, via a (dim:θ:(0,π/2], an orthonormal basis (dim:b:V) with (m,r) = (middle, distance)(1+dim|p) satisfying: pdim - m = r bdim-1; for each n in dim \ 1 (ie for n = 1, ..., dim-1), pn - m = r{ product(dim\n:sin2θ) bn-1 + sum(dim\n: i-> product(dim\ (1+i):sin2θ) bicos2θi) } and p0 - m = r{ -product(dim\1:sin2θ) b0 + sum(dim\1: i-> product(dim\ (1+i):sin2θ) bicos2θi) } (viewed as a special case via θ0 = π/2, whence cos2θ0 = -1, sin2θ0 = 0).

This yields, for n in dim-1,

p1+n - p_n_ = r product(dim\(1+n):sin2θ) {(1-cos2θ_n_)b_n_ - sin2θ_n_ b_n-1_} &implies; h = | p_1+n_ - p_n_ | = r product(dim\(1+n):sin2θ)|1-2cos2θ_n_ + 1|^1/2^ =r product(dim\(1+n):sin2θ) 2sinθ_n_, as 1-cos2θ = 2sin^2^θ,

whence, for n in dim-2,

1 = h/h = {2r sinθ_1+n_ product(dim\2+n:sin2θ)} / {2r sinθ_n_ product(dim\1+n:sin2θ)} &implies; sinθ_1+n_ = sinθ_n_sin2θ_1+n_ = 2sinθ_n_ sinθ_1+n_ cosθ_1+n_ &implies 1/2 = sinθ_n_ cosθ_1+n_

for n in dim-2, with

θ_0_ = π/2 Suppose that i ↦ i/(1+i) agrees with 2cos^2^θ on some natural number n (this is easy on n = 0 (empty) and true on n=1 since i in 1 implies i=0, for which i/(i+1) = 0 and 2cos^2^θ_0_ = 2cos^2^(π/2) = 0). Then 2sin^2^θ_n-1_ = 2 - 2cos^2^θ_n-1_ = 2-(n-1)/n = (1+n)/n, whence 2cos^2^θ_n_ = 1/(2sin^2^θ_n-1_) = n/(1+n), whence i↦i/(1+i) agrees with 2cos^2^θ on 1+n. From n=0, 1, we thus induce for any natural number n and obtain 2cos^2^θ_n_ = n/(1+n), cos2θ_n_ = 1- (2+n)/(1+n) = -1/(1+n), sin2θ_n_ = sqrt((2+n)/(1+n)) sqrt(n/(1+n)) = sqrt(n(2+n)) / (1+n). Thus p_1+n_ - p_n_ = r sqrt((1+dim)/dim) sqrt((1+n)/(2+n)) sqrt(2+n) / (1+n) { b_n_ sqrt(2+n) - b_n-1_ sqrt(n)} = r sqrt((1+dim)/dim) sqrt(1/(1+n)) {b_n_ sqrt(2+n) - b_n-1_ sqrt(n)} for n in dim-1, and p_n_ - m = r sqrt((1+dim)/dim) {b_n-1_ sqrt(n/(1+n)) - sum(dim\n: i↦ b_i_ / sqrt((2+i).(1+i)))} Thus h = |p_1+n_ - p_n_| = r sqrt(2(1+dim)/dim) and p_n_ = m + (h/sqrt(2)){b_n-1_ sqrt(n/(1+n)) - sum(dim\n: i↦ b_i_ / sqrt((2+i).(1+i)))} for n in dim. Now, p_0_ = m - (h/sqrt(2))sum(dim: i↦ b_i_/sqrt((2+i)(1+i))) so observe p_n_ - p_0_ = (h/sqrt(2)){ b_n-1_ sqrt(n/(1+n)) + sum(n: i↦ b_i_/sqrt((2+i)(1+i))) } = (h/sqrt(2)){ b_n-1_(sqrt(n/(1+n)) + 1/sqrt(n(1+n))) + sum(n-1: i↦ b_i_/sqrt((2+i)(1+i))) } = (h/sqrt(2)){ b_n-1_sqrt((1+n)/n) + sum(n-1: i↦ b_i_/sqrt((2+i)(1+i))) } whence b_n-1_ = sqrt(n/(1+n)){ (p_n_-p_0_)sqrt(2)/h - sum(n-1: i↦ b_i_/sqrt((2+i)(1+i))) } or b_n_ = sqrt((1+n)/(2+n)){ (p_1+n_-p_0_)sqrt(2)/h - sum(n: i↦ b_i_/sqrt((2+i)(1+i))) }. This last form serves to define b, given p, without reference to dim. End
Written by Eddy.