A more intelligent choice of ideal basis, involving Grassmann relations and proportional directions, reduces the 216-term degree-8 horror to a 96-term degree-7 horror; but further progress in this direction eluded me. However, following up an apparently improbable but actually rather astute observation by Lanco, reveals after all a perfectly civilised alternative formula for distance between parallel lines in 3-space; and suggests immediately a simple generalisation to cope with any parallel situation, subspaces X,Y intersecting only at infinity in a subspace of dimension n-1 : d(X,Y)^2 = ||<XºY>_{m-k-l-n}|| / ||<X•Y>_{k+l+n-2}|| ??!! For non-parallel lines L,M in 3-space, k = l = 2, m = 4, n = 0, and we have (as before) d(L,M)^2 = ||<LºM>_0|| / ||<L•M>_2|| ; or in terms of Pluecker coordinates, ( L_1 M^1 + L_2 M^2 + L_3 M^3 + L^1 M_1 + L^2 M_2 + L^3 M_3 )^2 ------------------------------------------------------------------------------------------------------------- ( (L^2 M^3 - L^3 M^2)^2 + (L^3 M^1 - L^1 M^3)^2 + (L^1 M^2 - L^2 M^1)^2 ) where L_i, L^i denote moment plane and direction vector resp, and (...)^2 denote squares rather than superscripts. If the lines are parallel, n = 1 and we have instead d(L,M)^2 = ||<LºM>_2||/||<L•M>_0|| ; or in terms of Pluecker coordinates, ( (L^1 M_2 + M^2 L_1 - L^2 M_1 - M^1 L_2)^2 + (L^2 M_3 + M^3 L_2 - L^3 M_2 - M^2 L_3)^2 + (L^3 M_1 + M^1 L_3 - L^1 M_3 - M^3 L_1)^2 ) -------------------------------------------------------------------------------- (L^3 M^3 + L^2 M^2 + L^1 M^1)^2 Proving this looked awkward, but assistance is at hand in the form of the earlier horror --- scaled to the same denominator, the difference of both parallel numerators reduces to zero modulo the ideal basis, QED. At the moment I have not extended my minimisation verifier to prove particular instances of k,l,m,n when n-1 > 0 --- this shouldn't be difficult --- but I'd rather prove the entire GA result directly! Jean Gallier's "Gram determinants" must presumably be equivalent to my earlier conjecture when n = 0 ; the revised conjecture suggests that the Fresnel theorem can also be generalised simply to parallel situations. Have I caught up with XX-th century theorems yet, I wonder? Or even managed to find an acceptably elegant expression for Henry Baker? Fred Lunnon On 3/28/11, Jean Gallier <jean@cis.upenn.edu> wrote:
There is a formula giving the distance between two disjoint affine subspaces involving Gram determinants.
Recall that if e_1, ..., e_m are vectors in R^n (m <= n) then Gram(e_1, .., e_m) = det(<e_i, e_j>), where <e_i, e_j> is the inner product of e_i and e_j.
If U_1 and U_2 are disjoint affine subspaces, let V_1 and V_2 be their linear directions, that is, unique vector spaces such that U_1 = a_1 + V_1 and U_2 = a_2 + V_2, for any a_1\in U_1 and any a_2\in U_2. Pick any basis e_1, ..., e_m of V_1 + V_2 , then the square of the distance d(U_1, U_2) between U_1 and U_2 is given by
d(U_1, U_2)^2 = Gram(a_1 - a_2, e_1, ..., e_m)/Gram(e_1, ..., e_m).
In the special case where U_1 and U_2 are skew lines (i.e., not parallel), we also have the formula
d(U_1, U_2)^2 = Gram(a - a', a - b, a' - b')/Gram(a - b, a' - b'),
where a, b are any two distinct points on U_1 and a', b' any two distinct points on U_2. The above formula can be found in Berger, Geometry I, Chapter 9, Section 2.
I found the more general formula in "Methodes Modernes en Geometrie", by Jean Fresnel, Part C, Section 1.4.2. The proof is not entirely trivial. See also Problem 7.13 of my "Geometric Methods and Applications", TAM 38, Springer-Verlag.
Best, -- Jean Gallier
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