On 12/16/09, Hans Havermann <pxp@rogers.com> wrote:
"Alice's adventures in algebra: Wonderland solved" in New Scientist, 16 December 2009, by Melanie Bayley:
http://www.newscientist.com/article/mg20427391.600-alices-adventures-in-alge...
So of course, when the discussion on quaternions has been written up, the title will have to be "The Mad Hatter's Tea-Party". In the meantime, perhaps I may move on to what prompted raising these matters in the first place: 2. Factorisation --- but not as we know it __________________________________________ In pursuit of my protracted investigations into the murkier recesses of Geometric Algebra, I had occasion recently to study Cl(2,2), with generators \x,\y,\v,\w (say) satisfying \x \x = \y \y = +1; \v \v = \w \w = -1; \x \y + \y \x = \x \v + \v \x = \x \w + \w \x = \y \v + \v \y = \y \w + \w \y = \v \w + \w \v = 0. One element of particular interest happens to be A == (a^2 + b^2) + b \x\y + a \x\v - a \y\w + b \v\w + \x\y\v\w, A^+ = (a^2 + b^2) - b \x\y - a \x\v + a \y\w - b \v\w + \x\y\v\w, where a,b represent real coefficients. We might verify that A is a versor, painstakingly checking that A^+ (c \x + d \y + e \v + f \w) A = vector; in fact for versors of grade < 6, it suffices that the terms are all even-grade or all odd-grade, and the magnitude scalar: so here we find ||A|| = (a^2 + 2a + 1+b^2) (a^2 - 2a + 1+b^2). Another theorem (Chasles) states that any versor is a product of vectors (with components in the scalar field). There are many ways in which a given versor may be factorised: for example when a = 0, A = \x \v (b\v - \w) (b\x + \y) = -\w \y (b\y - \x) (b\w + \v). Exercise: factorise the case a = 1, b = 0, A = 1 + \x\v - \y\w + \x\y\v\w. [Why should your suspicions be aroused by this apparently innocent case? Is it really a proper versor? Are there more cases like it?] It would be reassuring to have an explicit vector factorisation of A for general a,b. And given the simple, symmetric nature of A, we might hope for reasonably simple, symmetric factors. So far the best I have is (ready? deep breath!) A = ( a\x + b\y ) ( b\v + a\w ) ( -2ab(-a-b^2+a^2+a^3+ab^2)\x + a(a^4-a^2-4ab^2-b^4+b^2)\y - b(a^2+b^2+1)(-a-b^2+a^2+a^3+ab^2)\v - (a^6-2a^4b^2+a^4+2a^3b^2-a^2b^4+2ab^2+2b^4a-b^4-b^2)\w ) ( (a^4-2ab^2+b^2+a^2b^2)\x + b(-a-b^2+a^2+a^3+ab^2)\y + a(a^2+b^2)\v ), with RHS scaled up by scalar factor (a^6+2a^4b^2-a^4-2a^3b^2+a^2b^4-2ab^2-2b^4a+b^4+b^2)(a^2+b^2)^2 Can anybody improve on this? WFL