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How to Determine Group from Commutation Relations?

  1. Jan 20, 2015 #1
    Is there a way to determine the group from the commutation relations?

    For example, the commutation relations:

    [itex][J_x,J_y]=i\sqrt{2} J_z [/itex]
    [itex][J_y,J_z]=\frac{i}{\sqrt{2}} J_x [/itex]
    [itex][J_z,J_x]=i\sqrt{2} J_y [/itex]

    is actually SO(3), as can be seen by redefining [itex]J'_x =\frac{1}{\sqrt{2}} J_x [/itex]: then [itex]J'_x [/itex], [itex]J_y [/itex] and [itex]J_z [/itex] have the SO(3) algebra. So the commutation relations above is SO(3). But how do we know that just by looking at it?

    When you start taking linear combinations of generators, including sometimes with complex coefficients as in [itex] J_x+iJ_y[/itex], how can you tell the resulting commutators are SO(3)?
  2. jcsd
  3. Jan 20, 2015 #2


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    You can see if the <algebra closes>, in other words, if you want to have a Lie algebra, you need to have some requirements there: Leibniz derivation + Jacobi identity. If they fail, then the generators need to be adjusted (by rescaling or by considering linear combinations of them). Then you'll get the Lie algebra. By exponentiation, you get the connected component of the Lie group which is at least locally isomorphic to the Lie group you're really after.
  4. Jan 20, 2015 #3
    Can Jacobi Identity and Leibniz derivation alone tell you what group it is?

    For example, take the SO(3) commutation relations. If you change [Jx,Jy]=iJz to [Jx,Jy]=-iJz and leave all other commutators the same, then I think you get something like SO(2,1) rather than SO(3). SO(2,1) would still obey things like the Jacobi identity.

    So if I have 3 generators and 3 commutation relations, how do I know what group these generators belong to when you can always rescale and take linear combinations?
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