Apendix
Here we include the proof of Proposition 5.3 which says that the following identities hold:
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} \omega _{ak} = -\psi ^{-}_{ija}.{} \end{aligned}$$
(7.1)
$$\begin{aligned}{} & {} \bullet ~\psi ^{-}_{ijk} \omega _{ak} = \psi ^{+}_{ija}.{}\end{aligned}$$
(7.2)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} \omega _{jk} = 0 =\psi ^{-}_{ijk} \omega _{jk}.{}\end{aligned}$$
(7.3)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} \psi ^{+}_{abk} = \delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja} - \omega _{ia} \omega _{jb} +\omega _{ib} \omega _{ja}.{}\end{aligned}$$
(7.4)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} \psi ^{+}_{ajk} = 4 \delta _{ik}.\text { (contraction of the previous one)}{}\end{aligned}$$
(7.5)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} \psi ^{-}_{abk} = \delta _{ia} \omega _{jb}+\delta _{jb} \omega _{ia}-\delta _{ib} \omega _{ja}-\delta _{ja} \omega _{ib}.{}\end{aligned}$$
(7.6)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} \psi ^{-}_{ajk} = 4 \omega _{ia}.\text { (contraction of the previous one)}{}\end{aligned}$$
(7.7)
$$\begin{aligned}{} & {} \bullet ~\psi ^{-}_{ijk} \psi ^{-}_{abk} = \delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja} - \omega _{ia} \omega _{jb} +\omega _{ib} \omega _{ja}. (same as \psi ^{+}_{ijk} \psi ^{+}_{abk}){}\end{aligned}$$
(7.8)
$$\begin{aligned}{} & {} \bullet ~\psi ^{-}_{ijk} \psi ^{-}_{ajk} = 4 \delta _{ik}.\text { (contraction of the previous one)}{}\end{aligned}$$
(7.9)
$$\begin{aligned}{} & {} \bullet ~\omega _{ik} (\star \omega )_{abck} = \delta _{ia} \omega _{bc}+\delta _{ib} \omega _{ca}+\delta _{ic} \omega _{ab}.{}\end{aligned}$$
(7.10)
$$\begin{aligned}{} & {} \bullet ~\omega _{ik} (\star \omega )_{abik} = 4 \omega _{ab}.\text { (contraction of the previous one)}{}\end{aligned}$$
(7.11)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} (\star \omega )_{abck} =- \delta _{ia} \psi ^{+}_{jbc}-\delta _{ib} \psi ^{+}_{ajc}-\delta _{ic} \psi ^{+}_{abj}+\delta _{aj} \psi ^{+}_{ibc}+\delta _{bj} \psi ^{+}_{aic}+\delta _{cj} \psi ^{+}_{abi} - \omega _{ij} \psi ^{-}_{abc}.{}\end{aligned}$$
(7.12)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} (\star \omega )_{abck} = -\psi ^{-}_{ija} \omega _{bc}-\psi ^{-}_{ijb} \omega _{ca}-\psi ^{-}_{ijc} \omega _{ab}. (alternative expression to the previous one){}\end{aligned}$$
(7.13)
$$\begin{aligned}{} & {} \bullet ~\psi ^{+}_{ijk} (\star \omega )_{abjk} =2\psi ^{+}_{iab}.\text { (contraction of the previous one)}{}\end{aligned}$$
(7.14)
$$\begin{aligned}{} & {} \bullet ~\psi ^{-}_{ijk} (\star \omega )_{abck}=\psi ^{+}_{ija} \omega _{bc}+\psi ^{+}_{ijb} \omega _{ca}+\psi ^{+}_{ijc} \omega _{ab}.{}\end{aligned}$$
(7.15)
$$\begin{aligned}{} & {} \bullet ~\psi ^{-}_{ijk} (\star \omega )_{abjk} = 2\psi ^{-}_{iab}.\text { (contraction of the previous one)}{}\end{aligned}$$
(7.16)
$$\begin{aligned}{} & {} \bullet ~(\star \omega )_{ijkl} (\star \omega )_{abkl} = 2 \delta _{ia}\delta _{jb} - 2 \delta _{ib}\delta _{ja}+2 \omega _{ij} \omega _{ab}.{}\end{aligned}$$
(7.17)
$$\begin{aligned}{} & {} \bullet ~(\star \omega )_{ijkl} (\star \omega )_{ajkl} = 12 \delta _{ia}.\text { (contraction of the previous one)}{} \end{aligned}$$
(7.18)
Proof
We repeatedly use (5.1) along with \(G_2\)-contraction identities. For (7.1):
$$\begin{aligned} \psi ^{+}_{ijk} \omega _{ak}&= \sum _{k=1}^{6} \varphi _{ijk}(-\varphi _{0ak})\\&= -\sum _{k=0}^{6} \varphi _{ijk}\varphi _{0ak}\\&= -(\delta _{i0} \delta _{ja}-\delta _{ia} \delta _{j0} - \psi _{ij0a})\\&= \psi _{0ija}\\&= -\psi ^{-}_{ija}. \end{aligned}$$
Contracting both sides of (7.1) with \(w_{au}\) we get:
$$\begin{aligned} \psi ^{+}_{ijk} \omega _{ak} \omega _{au}&= -\psi ^{-}_{ija} \omega _{au}\\ \psi ^{+}_{ijk} \delta _{ku}&= \psi ^{-}_{ija} \omega _{ua}\\ \psi ^{+}_{iju}&=\psi ^{-}_{ija} \omega _{ua}, \end{aligned}$$
which gives us (7.2).
Contracting on (7.1) and (7.2) on j, a immediately gives (7.3).
Next, for (7.4):
$$\begin{aligned} \psi ^{+}_{ijk} \psi ^{+}_{abk}&= \sum _{k=1}^{6} \varphi _{ijk}\varphi _{abk}\\&= \sum _{p=0}^{6} \varphi _{ijk}\varphi _{abk}- \varphi _{ij0}\varphi _{ab0}\\&=(\delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja} - \psi _{ijab}) - \omega _{ij} \omega _{ab}\\&=(\delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja}) +( (\star \omega )_{ijab}- \omega _{ij} \omega _{ab}).\\&=\delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja} + \omega _{ja} \omega _{ib} +\omega _{bj} \omega _{ia}\text { (by Lemma }~??)\\&=\delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja} - \omega _{ia} \omega _{jb} +\omega _{ib} \omega _{ja}. \end{aligned}$$
Contracting (7.4) on j, b gives us (7.5):
$$\begin{aligned} \psi ^{+}_{ijk} \psi ^{+}_{ajk}&= 6\delta _{ia} -\delta _{ia} + \omega _{ij} \omega _{ja}\\&= 6\delta _{ia} -\delta _{ia} -\delta _{ia}\\&= 4\delta _{ia}. \end{aligned}$$
Next, for (7.6):
$$\begin{aligned} \psi ^{+}_{ijk} \psi ^{-}_{abk}&= \sum _{k=1}^{6} \varphi _{ijk} (-\psi _{0abk})\\&= -\sum _{k=0}^{6} \varphi _{ijk} \psi _{0abk}\\&= -(\delta _{i0} \varphi _{jab}+\delta _{ia} \varphi _{0jb}+\delta _{ib} \varphi _{0aj}-\delta _{0j} \varphi _{iab}-\delta _{aj} \varphi _{0ib}-\delta _{bj} \varphi _{0ai})\\&= \delta _{ia} \omega _{jb}+\delta _{ib} \omega _{aj}-\delta _{aj} \omega _{ib}-\delta _{bj} \omega _{ai}\\&= \delta _{ia} \omega _{jb}+\delta _{jb} \omega _{ia}-\delta _{ib} \omega _{ja}-\delta _{ja} \omega _{ib}. \end{aligned}$$
Contracting (7.6) on j, b will give us (7.7):
$$\begin{aligned} \psi ^{+}_{ijk} \psi ^{-}_{ajk}&= 6 \omega _{ia} - \omega _{ia} - \omega _{ia}\\&= 4\omega _{ia}. \end{aligned}$$
Next, we show that \(\psi ^{-}_{ijk} \psi ^{-}_{abk} = \psi ^{+}_{ijk} \psi ^{+}_{abk}\), which means that (7.8) and (7.4) are the same. Using (7.1), we have:
$$\begin{aligned} \psi ^{-}_{ijk} \psi ^{-}_{abk}&=\psi ^{+}_{ijs} \omega _{ks} \psi ^{+}_{abt} \omega _{kt}\\&= \psi ^{+}_{ijs} \psi ^{+}_{abt} \delta _{st}\\&=\psi ^{+}_{ijs} \psi ^{+}_{abs}. \end{aligned}$$
Thus, we also get (7.9):
$$\begin{aligned} \psi ^{-}_{ijk} \psi ^{-}_{ajk} = \psi ^{+}_{ijk} \psi ^{+}_{ajk} = 4 \delta _{ia}. \end{aligned}$$
Next, for (7.10):
$$\begin{aligned} \omega _{ik} (\star \omega )_{abck}&= \sum _{k=1}^6 (-\varphi _{0ik}) (-\psi _{abck})\\&= \sum _{k=0}^6 \varphi _{0ik} \psi _{abck}\\&= \delta _{0a} \varphi _{ibc}+\delta _{0b} \varphi _{aic}+\delta _{0c} \varphi _{abi}-\delta _{ai} \varphi _{0bc}-\delta _{bi} \varphi _{a0c}-\delta _{ci} \varphi _{ab0}\\&= \delta _{ia} \omega _{bc}+\delta _{ib} \omega _{ca}+\delta _{ic} \omega _{ab}. \end{aligned}$$
Contracting (7.10) on i, c yields (7.11):
$$\begin{aligned} \omega _{ik} (\star \omega )_{abik}&=\omega _{ba}+\omega _{ba}+6 \omega _{ab}\\&= 4 \omega _{ab}. \end{aligned}$$
For the next identity, there are two ways of computing the desired contractions yielding two different expressions (7.12) and (7.13). First, we use the usual way:
$$\begin{aligned} \psi ^{+}_{ijk} (\star \omega )_{abck}&= \sum _{k=1}^{6} \varphi _{ijk}(-\psi _{abck})\\&= -\sum _{k=0}^{6} \varphi _{ijk}\psi _{abck} + \varphi _{ij0}\psi _{abc0}\\&= -(\delta _{ia} \varphi _{jbc}+\delta _{ib} \varphi _{ajc}+\delta _{ic} \varphi _{abj}-\delta _{aj} \varphi _{ibc}-\delta _{bj} \varphi _{aic}-\delta _{cj} \varphi _{abi}) + (-\omega _{ij}) \psi ^{-}_{abc}\\&= - \delta _{ia} \psi ^{+}_{jbc}-\delta _{ib} \psi ^{+}_{ajc}-\delta _{ic} \psi ^{+}_{abj}+\delta _{aj} \psi ^{+}_{ibc}+\delta _{bj}\psi ^{+}_{aic}+\delta _{cj} \psi ^{+}_{abi} - \omega _{ij} \psi ^{-}_{abc}. \end{aligned}$$
Second, we can also use the previous results to get:
$$\begin{aligned} \psi ^{+}_{ijk} (\star \omega )_{abck}&= \psi ^{-}_{iju} \omega _{ku} (\star \omega )_{abck}\\&= -\psi ^{-}_{iju} \omega _{uk} (\star \omega )_{abck}\\&= -\psi ^{-}_{iju} (\delta _{au} \omega _{bc}+\delta _{bu} \omega _{ca}+\delta _{cu} \omega _{ab})\\&= -\psi ^{-}_{ija} \omega _{bc}-\psi ^{-}_{ijb} \omega _{ca}-\psi ^{-}_{ijc} \omega _{ab}. \end{aligned}$$
Note that both contractions of (7.12) and (7.13) on j, c yiled the same result (7.14):
$$\begin{aligned} \psi ^{+}_{ijk} (\star \omega )_{abjk}&= -\psi ^{+}_{abi}+ \psi ^{+}_{iba}+ \psi ^{+}_{aib}+6 \psi ^{+}_{abi} - \omega _{ij} \psi ^{-}_{abj}\\&= - \psi ^{+}_{iab}- \psi ^{+}_{iab}- \psi ^{+}_{iab}+6 \psi ^{+}_{iab} - \psi ^{+}_{abi}\\&=2 \psi ^{+}_{iab}, \end{aligned}$$
and
$$\begin{aligned} \psi ^{+}_{ijk} (\star \omega )_{abjk}&= -\psi ^{-}_{ija} \omega _{bj}-\psi ^{-}_{ijb} \omega _{ja}\\&= \psi ^{-}_{iaj} \omega _{bj}-\psi ^{-}_{ibj} \omega _{aj}\\&= \psi ^{+}_{iab} -\psi ^{+}_{iba}\\&=2 \psi ^{+}_{iab}. \end{aligned}$$
Since the second way of computing the contraction of \(\psi ^{+}\) and \(\star \omega \) gave us a nicer expression, we use it again for (7.15):
$$\begin{aligned} \psi ^{-}_{ijk} (\star \omega )_{abck}&= -\psi ^{+}_{iju} \omega _{ku} (\star \omega )_{abck}\\&= \psi ^{+}_{iju} \omega _{uk} (\star \omega )_{abck}\\&= \psi ^{+}_{iju} (\delta _{au} \omega _{bc}+\delta _{bu} \omega _{ca}+\delta _{cu} \omega _{ab})\\&= \psi ^{+}_{ija} \omega _{bc}+\psi ^{+}_{ijb} \omega _{ca}+\psi ^{+}_{ijc} \omega _{ab}. \end{aligned}$$
Contracting (7.15) on j, c yields (7.16):
$$\begin{aligned} \psi ^{-}_{ijk} (\star \omega )_{abjk}&= \psi ^{+}_{ija} \omega _{bj}+\psi ^{+}_{ijb} \omega _{ja}\\&= -\psi ^{+}_{iaj} \omega _{bj}+\psi ^{+}_{ibj} \omega _{aj}\\&= \psi ^{-}_{iab} - \psi ^{-}_{iba}\\&=2 \psi ^{-}_{iab}. \end{aligned}$$
Finally, we compute (7.17):
$$\begin{aligned} (\star \omega )_{ijkl} (\star \omega )_{abkl} =&\sum _{k,l=1}^6 \psi _{ijkl}\psi _{abkl}\\ =&\sum _{k,l=0}^6 \psi _{ijkl}\psi _{abkl}-\sum _{k=0}^6 \psi _{ijk0}\psi _{abk0}-\sum _{l=0}^6 \psi _{ij0l}\psi _{ab0l}\\ =&\sum _{k,l=0}^6 \psi _{ijkl}\psi _{abkl}-2\sum _{k=1}^6 \psi ^{-}_{ijk}\psi ^{-}_{abk}\\ =&(4 \delta _{ia}\delta _{jb} - 4 \delta _{ib}\delta _{ja} - 2\psi _{ijab})-2( \delta _{ia} \delta _{jb}-\delta _{ib} \delta _{ja} - \omega _{ia} \omega _{jb} +\omega _{ib} \omega _{ja})\\ =&2 \delta _{ia}\delta _{jb} - 2 \delta _{ib}\delta _{ja}+2 ((\star \omega )_{ijab} + \omega _{ia}\omega _{jb} + \omega _{aj}\omega _{ib})\\ =&2 \delta _{ia}\delta _{jb} - 2 \delta _{ib}\delta _{ja}+2 ((\omega _{ij} \omega _{ab} + \omega _{ja} \omega _{ib} +\omega _{bj} \omega _{ia}) + \omega _{ia}\omega _{jb} + \omega _{aj}\omega _{ib})\\ =&2 \delta _{ia}\delta _{jb} - 2 \delta _{ib}\delta _{ja}+2 \omega _{ij} \omega _{ab}. \end{aligned}$$
Contracting (7.17) on b, j gives us (7.18):
$$\begin{aligned} (\star \omega )_{ijkl} (\star \omega )_{ajkl} = 12 \delta _{ia} - 2 \delta _{ia}+2 \delta _{ia}= 12 \delta _{ia}.\square \end{aligned}$$
\(\square \)
