Let's consider the matrices defined below as O(p, q, r) for p=1, q = 0, r = 1. These are the matrices preserving the 2D form Q(x,y) = x^2: (1 0) (-1 0) ( ) and ( ) (c d) (c d) for all c, d in R. If d = 0, these are not invertible, so let's assume d <> 0. Then we have a group with 4 components depending on the signs of the matrix elements +-1 and d. From now on we will assume the notation O(p,q,r) denotes only invertible matrices preserving the corresponding quadratic form. The identity component has both these signs positive, so let's consider that. This is easily seen to be isomorphic to the group of 1D "linear" maps of form L(x) = dx + c with d > 0. I asked Mathematica to identify the matrices of O(1,1,1), i.e., those preserving the form x^2 - y^2 in 3-space. It says the (real) matrices look like ( (s)*cosh(t) sinh(t) 0) ((s s')sinh(t) (s')cosh(t) 0) ( g h k) for any reals k <> 0, g, h, t, and independent signs s, s' = +-1. where the variable signs are either (+,+) or (-,-). To be invertible, it's clear that we need k <> 0 as well. It appears that this group of matrices has 8 components corresponding to the independent signs of k, s, and s'. The identity component has all three signs positive. The uppper-left 2x2 shows that (the identity component of) the group O+(1,1) is a subgroup of (the identity component of) the O+(1,1,0), in fact a normal subgroup of it. I'm too lazy to figure out the quotient group right now. --Dan On Sep 3, 2014, at 9:54 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I've never been very comfortable with this notation for matrix groups; but I cannot see anything wrong with extending it to degenerate forms, apart from the fact that for some reason it's just not "standard".
So I will just define O(p, q, r) == real (or whatever) n x n matrices preserving (x_1)^2 + ... + (x_p)^2 - (x_(p+1))^2 - (x_(p+q))^2 , where n = p+q+r .
Now the usual representation of Euclidean 3-space isometries by homogeneous projective matrices becomes E(3) = PO(3, 0, 1) --- not as earlier botched --- and the Poincaré / Laguerre 3-space group would be PO(3, 1, 1) . Presumably this representation is well-known, though I have not seen it mentioned.
Speaking of botches --- I managed to gain a extra space dimension earlier, my muddle arising from Mob(2) acting on the unit 2-sphere in 3-space. Should have read (groan) Lorentz == 2-space Möbius = Mob(2) = PO(3, 1) , dimension 6 ; Poincaré == 3-space Laguerre = Lag(3) = PO(3, 1, 1) , dimension 10 .
WFL
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