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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)

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  • $\begingroup$ 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? $\endgroup$ – chichi Aug 16 at 2:25
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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 $\sigma_{mnp}^{IJ}=+\sigma_{mnp}^{JI}$, so we can antisymmetrize in $IJ$ and obtain:

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

Hitting $\varepsilon_{MNIJ}$ in both sides and the fact that $\varepsilon_{MNIJ}\sigma^{IJ}_m=2\bar\sigma_{MN}^{m}$, we also have:

$$ \varepsilon_{IJKL}=-\frac{1}{2}\bar\sigma_{KL}^{m}\bar\sigma_{IJ}^{m} $$

And similarly

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

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:

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

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