Graded Jacobi and real-form verification for the extended super-Poincare algebra

The matrix commutators are antisymmetric and the odd-odd brackets are symmetric under interchange of the full generator labels: for the same-chirality bracket, ${ϵ}_{\beta \alpha }{Z}^{JI}=(-{ϵ}_{\alpha \beta })(-Z^{IJ})={ϵ}_{\alpha \beta }{Z}^{IJ}$, and the dotted case is identical. The mixed bracket is symmetric by its Kronecker delta and the graded-skew rule. Each listed bracket has parity equal to the sum of its input parities, so the bracket is degree-preserving and graded-skew.

For matrices $X,Y,Z$, both $X(YZ-ZY)-(YZ-ZY)X$ and $(XY-YX)Z-Z(XY-YX)+Y(XZ-ZX)-(XZ-ZX)Y$ expand to $XYZ-XZY-YZX+ZYX$, so the commutator on $\mathfrak{l}$ satisfies Jacobi. Write ${\rho }_V(A,B)=A\otimes I+I\otimes B$. Operators acting on different tensor factors commute, hence $[{\rho }_V(A,B),{\rho }_V(A^{\mathrm{\prime }},B^{\mathrm{\prime }})]=[A,A^{\mathrm{\prime }}]\otimes I+I\otimes [B,B^{\mathrm{\prime }}]={\rho }_V([(A,B),(A^{\mathrm{\prime }},B^{\mathrm{\prime }})])$. For the odd action, [F1] uses column entries: $[A,Q_{\alpha }^I]={\sum }_{\gamma }{A}^{\gamma }{}_{\alpha }{Q}_{\gamma }^I$. Consequently $[A,[A^{\mathrm{\prime }},Q_{\alpha }^I]]-[A^{\mathrm{\prime }},[A,Q_{\alpha }^I]]={\sum }_{\delta }(A{A}^{\mathrm{\prime }}-A^{\mathrm{\prime }}A{)}^{\delta }{}_{\alpha }{Q}_{\delta }^I=[[A,A^{\mathrm{\prime }}],Q_{\alpha }^I]$, and the identical calculation applies to the dotted action of $B$. In particular, for $A=E_{12}$, $A^{\mathrm{\prime }}=E_{21}$ and $q=Q_1^I$, one has $[A,q]=0$, $[A^{\mathrm{\prime }},q]=Q_2^I$, and both sides of this identity are $q$. The actions on $C_N$ and by $V$ are zero. Thus $\mathfrak{l}$ acts by representations on the abelian space $V\oplus C_N$ and on ${\mathfrak{g}}_{\bar{1}}$; expanding the even Jacobi identity leaves precisely these representation identities, while brackets internal to $V\oplus C_N$ vanish.

For two even inputs and one odd input, the Jacobi expression is $[\rho (L),\rho (L^{\mathrm{\prime }})]q-\rho ([L,L^{\mathrm{\prime }}])q$, which is zero by the fundamental-action identities in [F1]; if either even input lies in $V\oplus C_N$, its odd action is zero and the same expression vanishes. This checks all orderings of the two-even/one-odd case by graded skew-symmetry.

For one even input $L=(A,B)$ and two odd inputs, Jacobi is equivariance of their odd bracket. On a mixed pair, the two fundamental actions on $Q$ and $\bar{Q}$ give exactly $A\otimes I+I\otimes B$ on $T$, so the mixed bracket is equivariant. On an equal-chirality pair, writing $A=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)$ gives $A^{\mathsf{T}}ϵ+ϵA=0$ by direct multiplication; the identical calculation for $B$ preserves the dotted alternating form. The flavor tensors $Z^{IJ}$ and ${\bar{Z}}_{IJ}$ are Lorentz scalars, so both central brackets are equivariant. Even inputs from $V\oplus C_N$ act trivially on each side.

The given star squares to the identity on each generator. On the matrix-commutator bracket, entrywise conjugation preserves matrix multiplication and the minus signs in $(A,B{)}^{\ast }$ reverse the bracket; on the fundamental and tensor actions, star exchanges the two spinor factors and the corresponding minus sign gives $[x,y{]}^{\ast }=-[x^{\ast },y^{\ast }]$. On odd-odd brackets, star exchanges $Q$ with $\bar{Q}$, transposes the bispinor indices of $T$, and exchanges $Z$ with $\bar{Z}$; the real alternating-form coefficients and antisymmetric flavor labels give $[x,y{]}^{\ast }=[x^{\ast },y^{\ast }]$, as required for two odd inputs. Zero brackets remain zero. For homogeneous even-even, even-odd and odd-odd inputs, substituting these identities into $\tau =-\ast$, $i\ast$ and $-\ast$ respectively gives $\tau ([x,y])=[\tau x,\tau y]$; conjugate-linearity gives ${\tau }^2=I$. Therefore the fixed space is closed under the bracket. Every $x$ decomposes as $x_{+}+i{x}_{-}$ with $x_{+}=(x+\tau x)\mathrm{/}2$ and $x_{-}=(x-\tau x)\mathrm{/}(2i)$ fixed by $\tau$, so this fixed real algebra complexifies to the displayed graded algebra.

For three odd inputs, each inner bracket lies in $V\oplus C_N$. Both summands act trivially on the odd space, so each of the three terms in the odd-odd-odd super-Jacobi identity is zero. The four parity patterns checked in the preceding cases exhaust homogeneous triples, and multilinearity gives the identity for arbitrary triples.

The Pauli matrices satisfy ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=2{\delta }_{ij}{I}_2$ and $\mathrm{tr}({\sigma }_i)=0$. For $\mu =\nu =0$ the sigma identity is $2I_2=-2{\eta }^{00}{I}_2$; for one temporal and one spatial index the two terms cancel; for spatial $i,j$ their sum is $-({\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i)=-2{\delta }_{ij}{I}_2$. Also $(p_0{I}_2+p_i{\sigma }_i{)}^2=(p_0^2+\mathrm{\mid }\vec{p}{\mathrm{\mid }}^2)I_2+2p_0{p}_i{\sigma }_i$ and direct $2\times 2$ determinant expansion gives $\det (p_0{I}_2+p_i{\sigma }_i)=p_0^2-\mathrm{\mid }\vec{p}{\mathrm{\mid }}^2=-{\eta }^{\mu \nu }{p}_{\mu }{p}_{\nu }$.