Mathematics desk
< December 29	<< Nov \| December \| Jan >>	December 31 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 30

Monstrous moonshine "numerology"

Leafing through Symmetry and the Monster by Mark Ronan last nite(I plan to buy the book later today)I saw that 1+4+9+...+529+24^2-70^2=0 is speculated to have importance for moonshine and the special theory of relativity since it says there are pairs of points in the Leech lattice that have timelike distance=zero between each other, provided, if I understand rightly, if one adjoins a 25th time dimension to the 24 space dimensions of the lattice so that u get a 25 dim. spacetime with signature of 24 ones and 1 negative one.(Of course the coordinates of the two points must differ by 70 in the time coordinate and by any permutation of 1,2,3,..24 in the 24 Leech lattice coordinates.)

My question is about this:Is the following numerology also important for moonshine? (don't know if in itself as just numbers if it's new or not, but even if it isn't new, i'm hoping it is important for moonshine):

Instead of summing squares of 1 thru 24, start from negative 7.

(-7)^2+(-6)^2+..+0+1+4+9+...24^2=7!=71^2-1. (Notice 71 is the largest prime dividing the order of the Monster, and it starts with minus Seven squared and equals Seven factorial.) Now 7!=72*70=7*8*9*10(starts with Seven again) and there is a wellknown identity for the product of four numbers in arithmetic progression, namely x(x+k)(x+2k)(x+3k)=(x^2+3kx+k^2)^2-k^4.

Also, I recall vaguely an AMM article that said generalizations beyond this don't exist, so that you shouldn't expect to be able to multiply 6 or 7 numbers in arithm. progression and get the difference of two squares. (The only exceptions I know of are 7!(=7!/1, giving either six or seven as you prefer)and 5!, giving five.You can tell a bit from the fect that 8!/2 isn't one less than a square.) The expansion of x(x+1)((x+2)(x+3)(x+4)(x+5) has coefficients 15,85,225,274,120. Perhaps some fiddling can get the product of six numbers in progression as the difference of two squares added to a cube minus a fourth power?

Thanks, Rich (talk) 00:40, 30 December 2007 (UTC)[reply]

Symmetry and the Monster? I went to a lecture by the guy who wrote that. I really didn't understand it. mattbuck (talk) 19:47, 30 December 2007 (UTC)[reply]