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Let's take a closed $G_2$ manifold $(M,\Phi)$. $\Phi$ denoting the three-form which defines the $G_2$ structure on $M$. Let's take a closed two form $\theta\in\Omega^2(M).$ Is \begin{align} d(\theta_{\Omega^2_7})\wedge (*\Phi)=0? \end{align}

$\Omega^2_7$ comes from the splitting of $\Omega^2$ into $7$ and $14$ dimensional subspaces. $\theta_{\Omega^2_7}$ is the part of $\theta$ which lies in the $7$-dimensional part.

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    $\begingroup$ Are you assuming that $\Phi$ is closed and co-closed? If so, then, yes, what you have written is an identity. If not, then it depends on the torsion of the $\mathrm{G}_2$-structure. $\endgroup$
    – Robert Bryant
    Commented Oct 3, 2023 at 9:15
  • $\begingroup$ Thanks, yes I see it now. All I have to use is $d(\theta_{\Omega^2_7})=-d(\theta_{\Omega^2_{14}})$ and $\theta_{\Omega^2_{14}}\wedge *\Phi=0.$ $\endgroup$
    – Partha
    Commented Oct 3, 2023 at 13:02
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    $\begingroup$ Yes, but you do have to at least assume that the $14$-piece of $\mathrm{d}({*}\Phi)$ vanishes in order to draw your conclusion. $\endgroup$
    – Robert Bryant
    Commented Oct 3, 2023 at 13:13

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