M2 Worldvolume Clifford Projector

Choose an oriented $\eta$-orthonormal basis $(e_0,e_1,e_2)$ of $W$ with $\eta (e_0,e_0)=-1$ and $\eta (e_1,e_1)=\eta (e_2,e_2)=1$. By [F1] and [F2], the associated matrices satisfy ${\gamma }_0^2=-I$, ${\gamma }_1^2={\gamma }_2^2=I$, and distinct matrices anticommute. The normalized oriented volume element is ${\mathrm{\Gamma }}_W={\gamma }_0{\gamma }_1{\gamma }_2$ in this basis. Reordering the six factors requires three transpositions and contributes $-1$, while the three squares multiply to $-I$; hence ${\mathrm{\Gamma }}_W^2=I$. Moving any one tangent generator past the other two contributes $(-1{)}^2=+1$, so ${\mathrm{\Gamma }}_W$ commutes with each ${\gamma }_i$.

In the same orthonormal basis, $\sqrt{-\det g}=1$. For $i=0$, ${\epsilon }^{0jk}{\gamma }_{jk}=2{\gamma }_1{\gamma }_2=2{\gamma }^0{\mathrm{\Gamma }}_W$ because ${\gamma }^0=-{\gamma }_0$. For $i=1$, the left side is $-2{\gamma }_0{\gamma }_2$, and ${\gamma }^1{\mathrm{\Gamma }}_W={\gamma }_1{\gamma }_0{\gamma }_1{\gamma }_2=-{\gamma }_0{\gamma }_2$. For $i=2$, the left side is $2{\gamma }_0{\gamma }_1$, and ${\gamma }^2{\mathrm{\Gamma }}_W={\gamma }_2{\gamma }_0{\gamma }_1{\gamma }_2={\gamma }_0{\gamma }_1$. Thus the displayed identity holds in this basis. Both sides transform as vector densities under oriented changes of worldvolume coordinates, so the identity holds for every oriented basis; its coordinate-independent content uses the induced volume density.

The orthogonal complement $W^{\perp }$ is positive definite of dimension eight, so choose a unit vector $n\in W^{\perp }$. By [F1], $\rho (\bar{n}{)}^2=I$, hence $\rho (\bar{n}{)}^{-1}=\rho (\bar{n})$. By [F2], $\rho (\bar{n})$ anticommutes with each of the three tangent generators and therefore with their product ${\mathrm{\Gamma }}_W$. Thus conjugation by $\rho (\bar{n})$ sends ${\mathrm{\Gamma }}_W$ to $-{\mathrm{\Gamma }}_W$. Trace is invariant under conjugation, so $\mathrm{tr}{\mathrm{\Gamma }}_W=-\mathrm{tr}{\mathrm{\Gamma }}_W$, and the real trace is zero.

[step 1.1] gives ${\mathrm{\Gamma }}_W^2=I$, so direct multiplication yields $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$, and $P_{+}+P_{-}=I$. The eigenspace decomposition of an idempotent gives $\mathrm{rank}{P}_{\pm }=\mathrm{tr}{P}_{\pm }=(32\pm 0)\mathrm{/}2=16$, using [A1] and step 2.2. Thus the projections are complementary and each has rank $16$.