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