Question

(Applied Agebra)

Edge Coloring Symmetry

6. To more formally describe the action of geometric symmetries on edge colorings, we assume: Each symmetry is a permutation of the set of edges. . An edge coloring is a function whose donain is the set of edges and codomain is the set of available colors (informally, each edge gets assigned a color). For example, f(A-f(C) = BLUE. f(B) = RED gives the middle coloring of the triangle in the lecture notes, using edge labels as in theprevious problem. Applying symmetry to an edge coloring must result in an edge coloring. If s is a symmetry and e is an edge coloring, then we define A(s, c) to be the functionk,where k(x)sfor each edge r. Informally, this says "the new coor k of the edge z is the old color e of the edge that was sent to z by the symmetry s. For example, using the coloring f above, Alpf) is the coloring with k(A) = f(p"(X)) = f(Z) BLUE which gives the example coloring on the right. Note we've used ρ--ρ2 Using this definition with the above coloring f, verify that (a) Alp, Ap.f))A(p. f) (b) A(p. A(r,f))-A(рт.f) (c) A(r, A(p, f))-A(pr,f).

Answer 1

Let sigma_Rotation symmetry and tau-Reflection symmetry of triangle then sigma^-1 = sigma^2, sigma^3 = I (identity), tau^-1 = I and Tau^2 = I a) A(sigma, A(sigma, f))(x) = A(sigma, f)(sigma^-1(x)) = A(sigma, f)(sigma^2(x)) = f(sigma^-1(sigma^2(x)) = f(sigma^2(x)) = A((tau sigma^2)^-1, f) a) = A(sigma^2)^-1 tau^-1, f)(x) [(xy)^-1 = y^-1 x^-1] = A(sigma tau, P)(x) rightarrow A(sigma, A(tau, f)) = a(sigma Tau, f) c)A(tau, A(sigma, f))(x) = A(sigma, f)(tau^-1 (x)) = A(sigma, f)(tau(x)) = f(sigma^-1(tau(x))) = f(sigma^2 tau(x)) = A((sigma^2 tau)^-1, f)(x) = A(tau^-1 sigma^2^-1, f)(x) = A = (tau sigma, f)(x) Rightarrow A(tau, A(sigma, f)) = A(tau sigma, f) ----(1) Let tau = Tef_alpha (reflection through alpha/ 2 angle) sigma = rot (rotation by triangle) then sigma^2 = rot_360 degree - beta (rotation by 360 degree- beta^0 angle) tau sigma = rot_alpha cdot ref_beta = ref_alpha - beta sigma^2 tau = rot_360 - beta^0 rot_alpha = ref_360 - beta + alpha(rot_alpha ref_beta = put_alpha + beta)