How to Determine Group from Commutation Relations?

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SUMMARY

The discussion focuses on determining the group associated with specific commutation relations, particularly in the context of Lie algebras. The commutation relations provided, specifically [J_x,J_y]=i√2 J_z, [J_y,J_z]=i/√2 J_x, and [J_z,J_x]=i√2 J_y, correspond to the SO(3) algebra. The discussion emphasizes the importance of verifying if the algebra closes through the application of the Jacobi identity and Leibniz derivation. Adjustments to the generators, such as rescaling or linear combinations, can lead to different groups, exemplified by the transition from SO(3) to SO(2,1) when modifying the commutation relation [J_x,J_y].

PREREQUISITES
  • Understanding of Lie algebras and their properties
  • Familiarity with commutation relations in quantum mechanics
  • Knowledge of the Jacobi identity and Leibniz derivation
  • Basic concepts of group theory, particularly SO(n) groups
NEXT STEPS
  • Study the properties of SO(3) and its Lie algebra structure
  • Learn about the Jacobi identity in the context of Lie algebras
  • Explore the implications of rescaling generators in Lie algebras
  • Investigate the relationship between commutation relations and different Lie groups, such as SO(2,1)
USEFUL FOR

This discussion is beneficial for theoretical physicists, mathematicians specializing in algebra, and students studying quantum mechanics or group theory, particularly those interested in the classification of Lie groups and algebras.

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Is there a way to determine the group from the commutation relations?

For example, the commutation relations:

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

is actually SO(3), as can be seen by redefining J'_x =\frac{1}{\sqrt{2}} J_x: then J'_x, J_y and J_z 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 J_x+iJ_y, how can you tell the resulting commutators are SO(3)?
 
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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.
 
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dextercioby said:
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.

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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