# Almost complex structure $J_i^j$ from covariantly constant spinor $\eta$

In the context of superstring compactification on a 6-manifold which admits a covariantly conserved spinor $$\eta$$, which we normalize so that $$\eta^{\dagger} \eta = 1$$, I am trying to show that the almost complex structure

$$J_m^{\,\,n} = -i \eta^{\dagger} \gamma_{m}^{\,\,n}\gamma \eta$$

satisfies $$J_m^{\,\,n} J_n^{\,\,p} = -\delta_m^p.$$

Here $$\gamma_{mn} = \gamma_{[mn]}$$ and $$\gamma$$ is the chirality operator. I'm aware that the Fierz rearrangement identity should be used,

$$\chi_\alpha \zeta \xi + \zeta_\alpha \xi \chi + \xi_{\alpha} \chi \zeta= 0$$

but I cannot show $$J^2 = -1$$, and would appreciate any help.

Attempt. Take $$\eta$$ to be real and to have chirality $$+1$$ so $$\gamma \eta = \eta$$. Now, with some spinor indices explicit, $$J_m^{\,\,n} J_n^{\,\,p} = - \eta_{\alpha} (\gamma_m^{\,\,n}\eta)^{\alpha} \eta(\gamma_{n}^{\,\,p}\eta)$$ $$= \eta_{\alpha}\eta^\alpha (\gamma_{n}^{\,\,p}\eta)(\gamma_m^{\,\,n}\eta)+ \eta_\alpha (\gamma_{n}^{\,\,p}\eta)^\alpha(\gamma_m^{\,\,n}\eta)\eta$$ $$= 1 \cdot (\gamma_{n}^{\,\,p}\eta)(\gamma_m^{\,\,n}\eta)+ \eta_\alpha (\gamma_{n}^{\,\,p}\eta)^\alpha(\gamma_m^{\,\,n}\eta)\eta.$$ Perhaps there is some gamma matrix algebra which would help me simplify this to $$-\delta_m^{\,\,p}$$.

(Ref: "Vacuum Configurations for Superstrings", Candelas et al. 1985)

• Question: are $\chi,\zeta,\xi$ are arbitrary spinors in your Fierz identity formula? If we take $\chi = \zeta = \xi = \eta$, it looks like the Fierz identity reduces to $3\eta_\alpha = 0$ which looks weird? – chichi Aug 16 '19 at 2:25

We have $$SO(6)=SU(4)$$ so a chiral spinor will be in the fundamental of $$SU(4)$$: $$\eta^I$$, $$I=1,...,4$$. You are imposing an additional restriction to $$\eta^{I}$$:

$$\bar\eta_{I}\eta^{I}=1$$

where $$\bar\eta_{I}=(\eta^{I})^*$$.

Than you propose that

$$J_m^n = -i \bar\eta_{I}(\sigma_{m}^{n})^I\,_{J}\eta^{J}$$

defines an almost complex structure. So you need to check that $$J_m^nJ_n^p=-\delta_m^p$$, which is the same as checking that

$$(\sigma_{n}^{p})^{(I}\,_{(J}(\sigma_{m}^{n})^{K)}\,_{L)}=\sigma_{[n}^{(I|M}\bar\sigma_{p]M(J|}\sigma_{[m}^{|K)N}\bar\sigma_{n]N|L)}=\delta^{I}_{J}\delta^{K}_{L}\delta_{m}^{p}$$

The indices $$IK$$ and $$JL$$ are symmetrized because they are hitting pairs of equal bosonic spinors. Note that the vectorial indices can be lowered and raised since we are dealing with $$SO(6)$$, and the metric $$\delta_{mn}$$ is a Kronecker delta.

In order to check this identity you are going to need to know the $$SO(6)$$ bi-spinor decomposition:

$$\delta_K^I\delta_L^J=\frac{1}{4}\sigma_m^{IJ}\bar\sigma^m_{KL} +\frac{1}{4}\frac{1}{3!\times 2}\sigma_{mnp}^{IJ}\bar\sigma^{mnp}_{KL}$$

Since $$\sigma_m^{IJ}=-\sigma_m^{JI}$$, we have that $$\sigma_{mnp}^{IJ}=+\sigma_{mnp}^{JI}$$, we have:

$$\delta_K^{(I}\delta_L^{J)}=\frac{1}{4}\frac{1}{3!\times 2} \sigma_{mnp}^{IJ}\sigma_{KL}^{mnp}$$

$$\delta_K^{[I}\delta_L^{J]}=\frac{1}{4}\sigma_m^{IJ}\bar\sigma^m_{KL}$$

Since $$(\sigma^{m}\sigma^{npq}\sigma^{m})=0$$ by the anti-commutation relations, we can use the equations above to show that:

$$\bar\sigma_{K(L}^{m}\bar\sigma_{I)J}^{m}=0\implies\varepsilon_{IJKL}=\frac{1}{2}\bar\sigma_{KL}^{m}\bar\sigma_{IJ}^{m}\implies \bar\sigma_{IJ}^{m}=\frac{1}{2}\varepsilon_{IJKL}\sigma^{KL}_m$$

And similarly

$$\varepsilon^{IJKL}=\frac{1}{2}\sigma^{KL}_{m}\sigma^{IJ}_{m}\implies \sigma^{IJ}_{m}=\frac{1}{2}\varepsilon^{IJKL}\bar\sigma_{KL}^m$$

This is all you need in order to prove that $$J_m^nJ_n^p=-\delta_m^p$$, just expand the four terms in $$\sigma_{[n}^{(I|M}\bar\sigma_{p]M(J|}\sigma_{[m}^{|K)N}\bar\sigma_{n]N|L)}$$ that comes from the antisymmetrization in $$np$$ and $$mn$$, and apply the identities above for each pair of sigma matrices with vectorial indices contracted.

Hint: at the end, because of the symmetrization in $$IK$$ and $$JL$$, most of the terms in $$(\sigma_{n}^{p})^{(I}\,_{(J}(\sigma_{m}^{n})^{K)}\,_{L)}$$ cancel, and the only term left is proportional to:

$$(\sigma_{p}\sigma_{m}+\sigma_{m}\sigma_{p})^{(I}\,_{(L}\delta_{J)}^{K)}$$

and in my conventions the anti-commutation relation between gamma matrices are

$$(\sigma_{m}\sigma_{n}+\sigma_{n}\sigma_{m})^{I}\,_{J}=-2\eta_{mn}\delta_{J}^{I}$$