(G, K) in which G is really a Lie group and K
Now the subgroup K of transformations i 0 that leave 0 invariant is clearly K = e -i , R; note that its Cal context or a psychological attribution to yet another person might constitute Elements in0 e duce the ordinary rotations on the unit disk. First, let us examine the prior building and note that every geometrical entity we met might be defined in terms of the Euclidean group SE(two). create K for the subgroup of rotations around a provided point, say o. If g = (R, x) is any element of G, the conjugate subgroup gKg -1 = (A, x - Ax), A K is the set of rotations around o + x. Now, the set gK = (A, x), A K remembers x and only x, so we can recover the Euclidean plane by considering the family members of all such cosets, which is, the set GK = gK, g G. Now when G can be a basic Lie group and K is really a closed subgroup, the smooth manifold M = GK comes having a all-natural transitive G-action, and K is however the subset of transformations which do not move the point K of M. This can be summarized by saying that M can be a G-homogeneous space. With this in mind, we can rephrase our key objective within this paper it truly is to show that some Klein pairs (G, K) allow for a construction of V1-like maps on the homogeneous space M = GK, along with a calculation of pinwheel densities in these V1-like maps. We shall retain M two-dimensional here, and stick to the 3 maximally symmetric spaces --the Euclidean plane, the round sphere and the hyperbolic plane. To recover the usual geometry of the round sphere S2 from a Klein pair, we want the group of rotations around the origin in R3 , that's, G = SO(3) = t AA = I3 and det(A) = 1, and the closed subgroup K = R 0 , R 0 1 SO(2)--of course K would be the group of rotations fixing (0, 0, 1). Let us now briefly give some particulars of how the hyperbolic plane can be defined from a Klein pair. Here G will be the group of linear transformations of C2 that have unit determinant and preserve the quadratic type (z, z ) |z|two - |z |2 , that's, G = SU(1, 1) = , C, ||two - ||two = 1 .Elements of G operate around the complicated plane C by way of conformal (but nonlinear) trans formations any element g = of G offers rise to a homography z g z = in the complicated plane. It's straightforward to see that the origin may be sent anywhere on the (open) unit disk, but nowhere outside. Now the subgroup K of transformations i 0 that leave 0 invariant is definitely K = e -i , R; note that its components in0 e duce the ordinary rotations with the unit disk. So the unit disk D in C comes using a Klein pair (G, K), and hunting for a G-invariant metric on D famously produces the negatively-curved Poincarmetric (see [43, 53] and also the Appendix for information). Recall that the formula for the square with the length of a vector tangent to D at (x, y) is summarized by ds two = (1 - (x 2 4 dx two + dy two .