structure constants of Lie algebra


by marton
Tags: algebra, constants, structure
marton
marton is offline
#1
Jul24-06, 07:49 PM
P: 4
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 text books, 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 text books could not be wrong , but my calculation is also correct. I am in confusion. Please help me.
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Doodle Bob
Doodle Bob is offline
#2
Jul26-06, 10:11 AM
P: 255
The key problem here is that, in general, to find the left-invariant vector fields of a matrix group at a point A not equal to the identity, one can not simply left-multiply A by the basis of the Lie algebra at the identity. The resulting matrices {A*L1, A*L2, A*L3} will not form a Lie algebra, since [A*L1, A*L2]=A*L1*A*L2-A*L2*A*L1 will not in general be an element of the linear span of {A*L1, A*L2, A*L3}.
marton
marton is offline
#3
Jul26-06, 09:08 PM
P: 4
Thanks a lot. I think i need read the text books more carefully.

Any suggestion for books on matrix group?

Doodle Bob
Doodle Bob is offline
#4
Jul27-06, 05:02 AM
P: 255

structure constants of Lie algebra


I seem to recall that I learned quite a lot about Matrix Lie groups via Spivak's Comprehensive Introduction to Differential Geometry. I think the pertinent sections are at the tail end of Volume I.


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