Question for ${\cal N}=1$ supersymmetry representations

Question

Please see this lecture note: https://arxiv.org/abs/1011.1491.

In section "2.2.5 Massless supermultiplet" the author defines a Casimir and says it is zero. How can we confirm it? We take the frame where $p^\mu = (E,0,0,E)$. The Casimirs are \begin{equation} C_1 = P^\mu P_\mu,\;\; \tilde C_2 = C_{\mu\nu}C^{\mu\nu}, \end{equation} where \begin{equation} B_\mu := W_\mu - \frac14\bar Q_{\dot\alpha}(\bar\sigma_\mu)^{\dot\alpha\beta}Q_\beta,\;\; C_{\mu\nu} := B_\mu P_\nu - B_\nu P_\mu. \end{equation}

For the first Casimir we can find easily $C_1 = -P_0P_0 + P_3P_3 = -E^2 + E^2 = 0$. How to calculate the second one and is it really zero?

I show my calculation below. For the second Casimir, \begin{equation} C_{01} = B_0P_1 - B_1P_0 = -EB_1,\;\; C_{02} = -EB_2,\;\; C_{03} = E(B_0 - B_3),\;\; C_{12} = 0,\;\; C_{13} = EB_1,\;\; C_{23} = EB_2. \end{equation} Then \begin{equation} \tilde C_2 = -2(C_{01}C^{01} + C_{02}C^{02} + C_{03}C^{03} + C_{13}C^{13} + C_{23}C^{23}) = -2E^2(B_0 - B_3)^2, \end{equation} where \begin{equation} B_0 - B_3 = W_0 - W_3 - \frac14\bar Q_{\dot\alpha}\bigg[ \begin{pmatrix} 1& 0\\ 0& 1 \end{pmatrix} + \begin{pmatrix} 1& 0\\ 0& -1 \end{pmatrix}\bigg]^{\dot\alpha\beta}Q_\beta = W_0 - W_3 - \frac12\bar Q_{\dot1}Q_1. \end{equation} How do we show this is zero?

Answer 1

We take the frame where $p^\mu = (E,0,0,E)$. The components of $W_\mu$ are \begin{align} W_0 &= -\epsilon_{0312}P^3M^{12} = EJ_3,\;\; W_1 = -\epsilon_{1023}P^0M^{23} - \epsilon_{1302}P^3M^{02} = E(-J_1 + K_2),\\ W_2 &= -\epsilon_{2031}P^0M^{31} - \epsilon_{2301}P^3M^{01} = E(-J_2 - K_1),\;\; W_3 = -\epsilon_{3012}P^0M^{12} = -EJ_3,\\ W_\mu &= E(J_3,-J_1+K_2,-J_2-K_1,-J_3). \end{align} In the above we used $J_i = (1/2)\epsilon_{ijk}M^{jk},\;\;K_i = M^{i0}$ and $\epsilon_{123} = \epsilon^{0123} = 1$. Their commutators are \begin{align} [W_1,W_2] &= E^2[-J_1 + K_2,-J_2 - K_1] = E^2(i\epsilon_{123}J^3 + (-i\epsilon_{123}J^3)) = 0,\\ [W_3,W_1] &= E^2[-J_3,-J_1 + K_2] = E^2(i\epsilon_{312}J^2 - i\epsilon_{321}K^1) = E^2i(J^2 + K^1) = -iEW_2,\\ [W_2,W_3] &= E^2[-J_2-K_1,-J_3] = E^2(i\epsilon_{231}J^1 - i\epsilon_{312}K^2) = iE^2(J^1 - K^2) = -iEW_1. \end{align} To define a new Casimir we introduce $C_{\mu\nu} = B_\mu P_\nu - B_\nu P_\mu$, where $B_\mu := W_\mu - (1/4)\bar Q_{\dot\alpha}(\bar\sigma_\mu)^{\dot\alpha\beta}Q_\beta$. \begin{equation} C_{01} = B_0P_1 - B_1P_0 = -EB_1,\;\; C_{02} = -EB_2,\;\; C_{03} = E(B_0 + B_3),\;\; C_{12} = 0,\;\; C_{13} = EB_1,\;\; C_{23} = EB_2. \end{equation} Then \begin{equation} \tilde C_2 = -2(C_{01}C^{01} + C_{02}C^{02} + C_{03}C^{03} + C_{13}C^{13} + C_{23}C^{23}) = -2E^2(B_0 + B_3)^2, \end{equation} where \begin{equation} B_0 + B_3 = W_0 + W_3 - \frac14\bar Q_{\dot\alpha}\bigg[ \begin{pmatrix} -1& 0\\ 0& -1 \end{pmatrix} - \begin{pmatrix} 1& 0\\ 0& -1 \end{pmatrix}\bigg]^{\dot\alpha\beta}Q_\beta = EJ_3 + (-EJ_3) + \frac12\bar Q_{\dot1}Q_1 = \frac12\bar Q_{\dot1}Q_1. \end{equation} By taking the square, \begin{equation} \tilde C_2 = -\frac12E^2\bar Q_{\dot1}Q_1\bar Q_{\dot1}Q_1 = -\frac12E^2\bar Q_{\dot1}(-\bar Q_{\dot1}Q_1 + 2(\sigma^\mu)_{1\dot1}P_\mu)Q_1, \end{equation} where \begin{equation} \sigma^\mu P_\mu = E\bigg[\begin{pmatrix} -1& 0\\ 0&-1 \end{pmatrix} + \begin{pmatrix} 1& 0\\ 0&-1 \end{pmatrix}\bigg] = E\begin{pmatrix}0& 0\\ 0&-2\end{pmatrix},\;\; \therefore (\sigma^\mu P_\mu)_{1\dot1} = 0. \end{equation} Then we conclude that $\tilde C_2 = 0$.

I think I could reach a satisfactory answer by myself. ($\star$ I used the metric signature $\eta^{\mu\nu} = (-1,+1,+1,+1)$ above unlike the text.) However, I think this was so complicated calculation. If you can give a proof in more simple way, please tell me.. (*´Д｀人)