Understanding Lorentz Group Generators: Derivation & Step in Eq 15

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VVS
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Hi,
I am trying to understand the derivation of the Lorentz generators but I am stuck.
I am reading this paper at the moment: http://arxiv.org/pdf/1103.0156.pdf
I don't understand the following step in equation 15 on page 3:
[itex]\omega^{\alpha}_{\beta}=g^{\alpha\mu}\omega_{\mu\beta}[/itex]
I don't understand how this can be true. I mean g is not the identity matrix.
 
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VVS said:
Hi,
I am trying to understand the derivation of the Lorentz generators but I am stuck.
I am reading this paper at the moment: http://arxiv.org/pdf/1103.0156.pdf
I don't understand the following step in equation 15 on page 3:
[itex]\omega^{\alpha}_{\beta}=g^{\alpha\mu}\omega_{\mu\beta}[/itex]
I don't understand how this can be true. I mean g is not the identity matrix.

That's sort of by definition. [itex]g^{\alpha \mu}[/itex] raises indices on a tensor and [itex]g_{\alpha \mu}[/itex] lowers them. Note that in general [itex]\omega^{\alpha}_\beta \neq \omega_{\alpha \beta}[/itex]

In inertial coordinates, [itex]g^{t t} = +1[/itex], [itex]g^{x x} = g^{y y} = g^{z z} = -1[/itex], and [itex]g^{i j} = 0[/itex] if [itex]i \neq j[/itex]. So in inertial coordinates, [itex]\omega^{\alpha}_{\beta} = \pm \omega_{\alpha \beta}[/itex].