# Representations of the Poincare Group

I am concerned about generators of the Poincare Lie Group that is responsible for boosts and rotations (often labelled by $$J_{\nu \mu}$$).

$$J_{\nu \mu}$$ is often divided into a term responsible for the spin angular momentum ($$S_{\nu \mu}$$) and another term responsible for the orbital angular momentum ($$L_{\nu \mu}$$). Therefore,

$$J_{\nu \mu}$$ = $$L_{\nu \mu}$$ + $$S_{\nu \mu}$$

I am wondering why is the $$S_{0 \mu}$$ components equal to zero?

• This depends on a particular choice of coordinate that you are using (time is the $0$th coordinate). Then it all boils down to the fact that $O(1,3)$ contains the orthogonal subgroup $O(3)$ fixing the $0$th coordinate vector pointwise, hence, the 1st column vector of the corresponding group elements is $(1,0,....))$. On the level of the Lie algebra, this becomes $(0,...,0)$. – Moishe Kohan Dec 6 '19 at 13:24
• @MoisheKohan Can you elaborate a bit? You said that "This depends" so do you mean the 𝑆_(0𝜇) components need not be zero? – The First StyleBender Dec 6 '19 at 14:32
• Yes, for instance, if you, say, swap the 1st and 4th coordinates making the Lorentzian form $x_1^2+x_2^2+x_3^2 -x_4^2$. – Moishe Kohan Dec 6 '19 at 15:06