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Lie group multiplication and Lie algebra commutation

  1. Nov 25, 2015 #1
    I've heard it said that the commutation relations of the generators of a Lie algebra determine the multiplication laws of the Lie group elements.

    I would like to prove this statement for ##SO(3)##.

    I know that the commutation relations are ##[J_{i},J_{j}]=i\epsilon_{ijk}J_{k}##.

    Can you suggest a possible next step for showing how this can be used to determine the multiplication law for ##SO(3)##?
     
  2. jcsd
  3. Nov 25, 2015 #2
    I'm not used to Lie but I think the steps are : You could find a set of matrices that satisfy these commutation relation, $$J_k $$

    Then it builds a basis for the Lie algebra.

    By exponentiating we get elements of the group SO (3)

    Thus we need to compute $$exp (aJ_x+bJ_z)$$ for exemple.
     
  4. Nov 25, 2015 #3
    Does the multiplication law refer to the fact that if we multiply two elements ##e^{i\theta_{1}J_{1}}## and ##e^{i\theta_{2}J_{2}}## from the Lie group, we get an element which is also in the Lie group?

    Or does the group multiplication law refer to the fact that ##e^{i\theta_{1}J_{1}}e^{i\theta_{2}J_{2}}=e^{i(\theta_{1}J_{1}+\theta_{2}J_{2})}##?
     
  5. Nov 25, 2015 #4
    The last is in general not true see Baker Campbell Hausdorff formula
     
  6. Nov 27, 2015 #5
    Alright then, how would you define the multiplication law for ##SO(3)##?
     
  7. Nov 27, 2015 #6
    The operation could be matrix multiplication but the exponential comes from writing a rotation out of an infinitesimal one.

    Infinitesimal rotations are commutative but rotations are not.
     
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