# Tag Info

## New answers tagged lie-algebra

2

This answer is mostly rephrasing Trimok's correct answer in other words. The super-Poincare group is supposed to be an extension of the Poincare group, which contains the Lorentz group and translations. We will complexify the Lorentz group. The Lie group $G:=SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (isomorhic to the double cover of) the complexified ...

2

It is maybe simpler to consider all the generators as representations of $SL(2,C)$, so, using spinor indices, you will have : $M^{\alpha \dot \alpha \beta \dot \beta}, P^{\beta \dot \beta}, Q_\alpha, \bar Q^\dot\beta$ Indices are raised and lowered with the Levi-Civita symbols $\epsilon_{\alpha \beta}, \epsilon^{\alpha \beta},\epsilon_{\dot \alpha \dot ... 3 Short intro to ladders As you say, they're ladder operators. Let's get rid of the annoying$\hbar$by setting it to one, and call them more systematically$L_{-1},L_0,L_1$instead of$L_-,L_z,L_+\$. Then, the commutation relations take the uniform form $$[L_n,L_m] = (m-n)L_{m+n}$$ If we had countably many of these, we'd have a Witt algebra, if there was ...

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