Lorentz generators

  • Thread starter sshaep
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I know the representations of Lorentz generators M[tex]\mu\nu[/tex] as 4X4 matrices.

This matrices satisfy the commutation relation(Lie algebra of O(3,1))

However I think these 4X4 matrix representations are not unique.

Is there any other representations satisfying the commutation relation? 2X2 matrix or

another?
 

Ben Niehoff

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Yes, there are infinitely many representations of the Lorentz algebra. One trivial example is to form a set of 8x8 matrices by taking the Kronecker product of M with the 2x2 identity.
 

George Jones

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Is there any other representations satisfying the commutation relation? 2X2 matrix or another?
Because [itex]SL \left(2 , \mathbb{C} \right)[/itex] is the universal cover of the restricted Lorentz group, they have isomorphic Lie algebras.
 
Because [itex]SL \left(2 , \mathbb{C} \right)[/itex] is the universal cover of the restricted Lorentz group, they have isomorphic Lie algebras.
This means can I get the 2x2 representation of M[tex]\mu\nu[/tex] from SL(2,C)?

But how? What I know is [tex]\sigma_{\mu}A^{\mu} _{\phantom{\mu}\nu}}[/tex]=[tex]L\sigma_{\nu}L^{\dagger}[/tex]
 

samalkhaiat

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Take

[tex]
M^{0i} = \frac{i}{2} \sigma^{i}, \ \ M^{ij} = \frac{1}{2} \epsilon^{ijk}\sigma^{k}
[/tex]

Also, if you define

[tex]\omega_{i0} = 2 Re(\alpha_{i}), \ \ \omega_{ij} = 2 \epsilon^{ijk} Im(\alpha_{k})[/tex]

then

[tex]
1 + \frac{i}{2}\omega_{\mu\nu}M^{\mu\nu} = \left( \begin {array}{rr} (1 + \alpha_{3}) & (\alpha_{1} - i \alpha_{2}) \\ (\alpha_{1} + i \alpha_{2}) & (1 - \alpha_{3}) \end {array} \right)
[/tex]

represents elements of SL(2,C) infinitesimally close to the identity element.

regards

sam
 

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