How Do Lorentz Group Generators Comply with Their Commutation Relations?

[spookyfish]

The Lorentz group generators, in any representation, satisfy the commutation relation

<br /> [S^{\mu \nu}, S^{\rho \sigma}] = i \left( g^{\nu \rho}S^{\mu \sigma} -g^{\mu \rho} S^{\nu \sigma} -g^{\nu \sigma}S^{\mu \rho} +g^{\mu \sigma} S^{\nu \rho} \right)<br />

and the Lorentz transformation is

<br /> \Lambda=\exp(-i \omega_{\mu \nu} S^{\mu \nu}/2)<br />

My question is: is it possible to prove the formula for the generators (the first formula I wrote), from the definition of the Lorentz group

<br /> \Lambda^T g \Lambda =g<br />

 

[Bill_K]

Not quite what you're asking for, but Eq.(8)-(10) of http://hitoshi.berkeley.edu/232A/Lorentz.pdf gives a quick derivation.

 

[dextercioby]

For infinitesimal transformations you can linearize the exponential and get the particular value of the generators as 4X4 matrices (you get the generators in the fundamental representation of the Lorentz group). Then compute the commutator and represent it as a linear combination of the metric.

 

[Bill_K]

For infinitesimal transformations you can linearize the exponential and get the particular value of the generators as 4X4 matrices (you get the generators in the fundamental representation of the Lorentz group). Then compute the commutator and represent it as a linear combination of the metric.

Can you do this without writing out the 4 x 4 matrices explicitly?

 

[dextercioby]

Well, you get a particular form of M_mu nu in terms of metric tensor and delta mu nu.

 

[spookyfish]

Thanks for the replies. I have not seen any replies for a few days, and I thought I wouldn't get ones so I stopped looking.

Not quite what you're asking for, but Eq.(8)-(10) of http://hitoshi.berkeley.edu/232A/Lorentz.pdf gives a quick derivation.

Yes, but they derive it in the (1/2,1/2) representation, and I want a proof for an arbitrary representation.

For infinitesimal transformations you can linearize the exponential and get the particular value of the generators as 4X4 matrices (you get the generators in the fundamental representation of the Lorentz group). Then compute the commutator and represent it as a linear combination of the metric.

I was trying to do this. I found the following lecture notes:
http://www.damtp.cam.ac.uk/user/ho/GNotes.pdf
Eqs. (4.26)-(4.32) are exactly what I was looking for, but I wasn't able to follow all of the derivation. Perhaps people here can help.

In Eq. (4.27) I substitute
<br /> {{\Lambda_1}^\mu}_\nu={\delta^\mu}_\nu+{{\omega_1}^\mu}_\nu\\<br /> {{\Lambda_2}^\mu}_\nu={\delta^\mu}_\nu+{{\omega_2}^\mu}_\nu\\<br /> {({\Lambda_1}^{-1})^\mu}_\nu={\delta^\mu}_\nu-{{\omega_1}^\mu}_\nu\\<br /> {({\Lambda_2}^{-1})^\mu}_\nu={\delta^\mu}_\nu-{{\omega_2}^\mu}_\nu<br />
when I multiply them together as in the LHS of Eq. (4.27) I don't get the RHS.

Also, when I substitute Eq. (4.29) and the line after that, into Eq. (4.30), the multiplication of the 4 terms
<br /> U[{\Lambda_2}^{-1}]U[{\Lambda_1}^{-1}]U[\Lambda_2]U[\Lambda_1]<br />
does not give me the last line of Eq. (4.30).

Thanks for any help.

 

[Bill_K]

Yes, but they derive it in the (1/2,1/2) representation, and I want a proof for an arbitrary representation.

No, they only use the definition of the Lorentz group as the rotation group in Minkowski space.

 

[spookyfish]

No, they only use the definition of the Lorentz group as the rotation group in Minkowski space.

Any relation to my last question?