- #1
marton
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The following matrices are written in Matlab codes form.
The standard basis for so(3) is: L1 = [0 0 0; 0 0 -1; 0 1 0], L2 = [0 0 1; 0 0 0; -1 0 0], L3 = [0 -1 0; 1 0 0; 0 0 0]. Since [L1, L2] = L3, the structure constants of this Lie algebra are C(12, 1) = C(12, 2) = 0, C(12, 3) = 1. According to do Carmo and other textbooks, if M1, M2 and M3 is the basis for the left-invariant vector fields of A, where A is a member of SO(3), we have [Mi, Mj] = C(ij, k)Mk, where Mi = A * Li. In the above case, we have [M1, M2] = M3.
But, when I put A = [cos(t) -sin(t) 0; sin(t) cos(t) 0; 0 0 1], then M1 = [0 0 sin(t); 0 0 -cos(t); 0 -1 0], M2 = [0 0 cos(t); 0 0 sin(t); -1 0 0], and M3 = [-sin(t) -cos(t) 0; cos(t) -sin(t) 0; 0 0 0]. By straightforward calculation, it can be seen that [M1, M2] is not equal to M3.
I believe the textbooks could not be wrong , but my calculation is also correct. I am in confusion. Please help me.
The standard basis for so(3) is: L1 = [0 0 0; 0 0 -1; 0 1 0], L2 = [0 0 1; 0 0 0; -1 0 0], L3 = [0 -1 0; 1 0 0; 0 0 0]. Since [L1, L2] = L3, the structure constants of this Lie algebra are C(12, 1) = C(12, 2) = 0, C(12, 3) = 1. According to do Carmo and other textbooks, if M1, M2 and M3 is the basis for the left-invariant vector fields of A, where A is a member of SO(3), we have [Mi, Mj] = C(ij, k)Mk, where Mi = A * Li. In the above case, we have [M1, M2] = M3.
But, when I put A = [cos(t) -sin(t) 0; sin(t) cos(t) 0; 0 0 1], then M1 = [0 0 sin(t); 0 0 -cos(t); 0 -1 0], M2 = [0 0 cos(t); 0 0 sin(t); -1 0 0], and M3 = [-sin(t) -cos(t) 0; cos(t) -sin(t) 0; 0 0 0]. By straightforward calculation, it can be seen that [M1, M2] is not equal to M3.
I believe the textbooks could not be wrong , but my calculation is also correct. I am in confusion. Please help me.