# Lorentz generators

1. Mar 27, 2009

### sshaep

I know the representations of Lorentz generators M$$\mu\nu$$ 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?

2. Mar 27, 2009

### Ben Niehoff

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.

3. Mar 27, 2009

### George Jones

Staff Emeritus
Because $SL \left(2 , \mathbb{C} \right)$ is the universal cover of the restricted Lorentz group, they have isomorphic Lie algebras.

4. Mar 27, 2009

### sshaep

This means can I get the 2x2 representation of M$$\mu\nu$$ from SL(2,C)?

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

5. Mar 28, 2009

### samalkhaiat

Take

$$M^{0i} = \frac{i}{2} \sigma^{i}, \ \ M^{ij} = \frac{1}{2} \epsilon^{ijk}\sigma^{k}$$

Also, if you define

$$\omega_{i0} = 2 Re(\alpha_{i}), \ \ \omega_{ij} = 2 \epsilon^{ijk} Im(\alpha_{k})$$

then

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

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

regards

sam