Decomposition of forms on a Spin$(7)$ manifold

Question

Let's take a $G_2$ manifold $(M,\Phi)$, then we get a Spin$(7)$ manifold by taking $(M\times\mathbb{R},\Psi:=\Phi\wedge dt+*_M\Phi),$ where $t$ is the coordinate in the $\mathbb{R}$-direction. $\Phi\in\Omega^3(M),\Psi\in\Omega^4(M\times\mathbb{R})$ determining the corresponding $G_2$ and Spin$(7)$ structures. Now these structures give us splitting of forms. E.g., \begin{align*} &\text{On } G_2: \Lambda^4=\Lambda^4_1\oplus \Lambda^4_7\oplus \Lambda^4_{27},\hspace{1 ex}\Lambda^3=\Lambda^3_1\oplus \Lambda^3_7\oplus \Lambda^3_{27},\\ &\text{On Spin}(7) : \Lambda^4=\Lambda^4_1\oplus \Lambda^4_7\oplus \Lambda^4_{27}\oplus \Lambda^4_{35} \end{align*} We start with a four form say, $\alpha\in\Omega^4(M\times\mathbb{R}).$ Now the component $\alpha_7\in\Omega^4_7(M\times\mathbb{R}).$ Moreover let's take $\alpha=\alpha_1+\alpha_2\wedge dt$, where $\alpha_1\in\Omega^4(M)$ and $\alpha_2\in\Omega^3(M)$.We can write $\alpha_7=\gamma+\beta\wedge dt$ (at least pointwise), where $\gamma\in\Omega^4(M)$ and $\beta\in\Omega^3(M)$. My question is what are the components of $\gamma$ and $\beta$ in $\Omega^4(M)$ and $\Omega^3(M)$ in terms of the splitting coming from the $G_2$ structure.

The question may seem a bit cryptic, I have done the part for a form in $\Omega^2_7$. Hope this clears things up. The known descriptions for $\Omega^4_7$ in the literature are not very calculation-friendly. That's where my difficulty arises.

On a $G_2$ manifold $(M,\Phi)$, $\Omega^2=\Omega^2_7\oplus \Omega^2_{14},$ and for a $2$-form $\xi,\xi_7=\frac{1}{3}\big(\xi+*(\xi\wedge\Phi)\big).$
On a Spin-$(7)$ manifold $(N,\Psi)$, $\Omega^2=\Omega^2_7\oplus \Omega^2_{21},$ and for a $2$-form $\xi,\xi_7=\frac{1}{4}\big(\xi+*(\xi\wedge\Psi)\big).$

\begin{align*} &\xi_{\Omega^2_7(M\times\mathbb{R})}\\ &=\frac{1}{4}\big(\xi+*(\xi\wedge(\Phi\wedge dt+*_M\Phi)\big)\\ &=\frac{1}{4}\big(\xi+*_M(\xi\wedge\Phi)+*(\xi\wedge*_M\Phi)\big)\\ &=\frac{3}{4} \xi_{\Omega^2_7(M)}+\frac{1}{4}*_M(\xi\wedge*_M\Phi)\wedge dt \end{align*}

Answer 1

Observe that, in a $\mathrm{Spin}(7)$-manifold, since $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$, the $\mathrm{Spin}(7)$ decomposition $\Lambda^4 = \Lambda^4_1\oplus \Lambda^4_7\oplus\Lambda^4_{27}\oplus\Lambda^4_{35}$ must refine the splitting $\Lambda^4=\Lambda^4_{+}\oplus\Lambda^4_{-}$ into self-dual and anti-self-dual parts. Since $\Lambda^4_1$ is self-dual, it follows that $\Lambda^4_{+} = \Lambda^4_1\oplus \Lambda^4_7\oplus\Lambda^4_{27}$ and $\Lambda^4_{-} = \Lambda^4_{35}$.

Thus, $\Lambda^4_7$ consists of self-dual forms, and hence, under the $G_2$-splitting $\alpha_7 = \beta\wedge \mathrm{d}t + \gamma$, since $G_2$ acts irreducibly on the $7$-dimensional representation, we must have $\beta\in\Lambda^3_7(M)$ and $\gamma = -{\ast}\beta\in \Lambda^4_7(M)$.