$\displaystyle(\mu_{3}+\mu_{4})\left\\{2(\mu_{3}+\mu_{4})\alpha_{1}+(\mu_{2}+\mu_{3})(\mu_{2}+\mu_{4})\right\\}$
$\displaystyle=0,$ (A.24)
$\displaystyle(\mu_{2}+\mu_{4})\left\\{2(\mu_{2}+\mu_{4})\alpha_{1}+(\mu_{2}+\mu_{3})(\mu_{3}+\mu_{4})\right\\}$
$\displaystyle=0,$ (A.25)
$\displaystyle(\mu_{2}+\mu_{3})\left\\{2(\mu_{2}+\mu_{3})\alpha_{1}+(\mu_{2}+\mu_{4})(\mu_{3}+\mu_{4})\right\\}$
$\displaystyle=0.$

Now, we further divide the situation.

  * (i)

Suppose that $\mu_{3}+\mu_{4}=0$. Then, by the same argument as above, we have
$\mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=0$ and $|\alpha|^{2}=1$, which is (10) with
$\alpha_{3},\alpha_{5},\alpha_{7}\neq 0$.

  * (ii)

If $\mu_{2}+\mu_{4}=0$, we have $\mu_{1}+\mu_{3}=0$. Then, (A.22) implies that

$2|\alpha|^{2}-2=-2\mu_{1}\mu_{2}=2\mu_{1}\mu_{2}.$

Hence, we obtain $|\alpha|^{2}=1$ and $\mu_{1}\mu_{2}=0$. If $\mu_{1}=0$ or
$\mu_{2}=-\mu_{4}=0$, $\sum_{j=1}^{4}\mu_{j}=0$ and
$\mu_{1}\leq\mu_{2}\leq\mu_{3}\leq\mu_{4}$ imply that
$\mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=0$. Hence, we obtain (10) with
$\alpha_{3},\alpha_{5},\alpha_{7}\neq 0$.

  * (iii)

Similarly, if $\mu_{2}+\mu_{3}=0$, we obtain (10) with
$\alpha_{3},\alpha_{5},\alpha_{7}\neq 0$.

  * (iv)

Now, suppose that $\mu_{2}+\mu_{3},\mu_{2}+\mu_{4},\mu_{3}+\mu_{4}\neq 0$.
Then, (A.23), (A.24) and (A.25) imply that

(A.26) $\displaystyle\alpha_{1}=$
$\displaystyle-\frac{(\mu_{2}+\mu_{3})(\mu_{2}+\mu_{4})}{2(\mu_{3}+\mu_{4})}$
$\displaystyle=$
$\displaystyle-\frac{(\mu_{2}+\mu_{3})(\mu_{3}+\mu_{4})}{2(\mu_{2}+\mu_{4})}=-\frac{(\mu_{2}+\mu_{4})(\mu_{3}+\mu_{4})}{2(\mu_{2}+\mu_{3})}.$

Then, as in the case (III)-(i)-(C), (A.26) implies that

(A.27) $\displaystyle(\mu_{2}+\mu_{3})(\mu_{2}-\mu_{3})(-\mu_{1}+\mu_{4})$
$\displaystyle=0,$ (A.28)
$\displaystyle(\mu_{2}+\mu_{4})(\mu_{2}-\mu_{4})(-\mu_{1}+\mu_{3})$
$\displaystyle=0.$

We patiently divide this situation.

    * (A)

Suppose that $\mu_{2}-\mu_{3}=0$ and $\mu_{2}-\mu_{4}=0$. Then, we have
$(\mu_{1},\mu_{2},\mu_{3},\mu_{4})=(-3\mu,\mu,\mu,\mu)$ for $\mu>0$. The
equations (A.22) and (A.26) imply that

$2|\alpha|^{2}-2=2\mu^{2},\quad\mbox{and}\quad\alpha_{1}=-\mu.$

Hence, we obtain (8) with $\alpha_{3},\alpha_{5},\alpha_{7}\neq 0$.

    * (B)

Similarly, if we suppose $\mu_{2}-\mu_{3}=-\mu_{1}+\mu_{3}=0$, we obtain (9)
with $\alpha_{3},\alpha_{5},\alpha_{7}\neq 0$. The equations
$-\mu_{1}+\mu_{4}=\mu_{2}-\mu_{4}=0$ and $-\mu_{1}+\mu_{4}=-\mu_{1}+\mu_{3}=0$
implies that $\mu_{1}=\mu_{2}=\mu_{3}=\mu_{4}=0$, which contradicts
$\mu_{2}+\mu_{3}\neq 0$.

Then, the proof is completed. ∎

Note that we can solve (A.15) with respect to $\alpha_{1}$ explicitly.

######  Lemma A.4.

The equation (A.15) holds if and only if

$\alpha_{1}=\sqrt{\frac{\sum_{j=1}^{4}\mu_{j}^{2}+4}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{-3\sqrt{3}\sum_{1\leq
i<j<k\leq
4}\mu_{i}\mu_{j}\mu_{k}}{\left(\sum_{j=1}^{4}\mu_{j}^{2}+4\right)^{3/2}}\right)+\frac{2{\ell}\pi}{3}\right)$

for ${\ell}=0,1,2$.

###### Proof.

We can obtain this by the Viète’s formula for a cubic equation. To apply this,
we have to show that $3\sqrt{3}\left|\sum_{1\leq i<j<k\leq
4}\mu_{i}\mu_{j}\mu_{k}\right|<\left|\sum_{j=1}^{4}\mu_{j}^{2}+4\right|^{3/2}.$
This follows from the following more general statement. ∎

######  Lemma A.5.

We have

$3\sqrt{3}\left|\sum_{1\leq i<j<k\leq
4}\mu_{i}\mu_{j}\mu_{k}\right|\leq\left|\sum_{j=1}^{4}\mu_{j}^{2}\right|^{3/2}.$

###### Proof.

Set $\nu_{1}=\mu_{2}+\mu_{3},\nu_{2}=\mu_{2}+\mu_{4}$ and
$\nu_{3}=\mu_{3}+\mu_{4}$. Then, by (A.17), we have $\sum_{1\leq i<j<k\leq
4}\mu_{i}\mu_{j}\mu_{k}=-\nu_{1}\nu_{2}\nu_{3}.$ We also compute

$\displaystyle\sum_{j=1}^{4}\mu_{j}^{2}=$
$\displaystyle(\mu_{2}+\mu_{3}+\mu_{4})^{2}+\mu_{2}^{2}+\mu_{3}^{2}+\mu_{4}^{2}$
$\displaystyle=$
$\displaystyle(\mu_{2}+\mu_{3})^{2}+(\mu_{2}+\mu_{4})^{2}+(\mu_{3}+\mu_{4})^{2}=\nu_{1}^{2}+\nu_{2}^{2}+\nu_{3}^{2}.$

By the inequality of arithmetic and geometric means, we have

$27|\nu_{1}\nu_{2}\nu_{3}|\leq(|\nu_{1}|+|\nu_{2}|+|\nu_{3}|)^{3}.$

We also have
$3(|\nu_{1}|^{2}+|\nu_{2}|^{2}+|\nu_{3}|^{2})-(|\nu_{1}|+|\nu_{2}|+|\nu_{3}|)^{2}=(|\nu_{1}|-|\nu_{2}|)^{2}+(|\nu_{2}|-|\nu_{3}|)^{2}+(|\nu_{3}|-|\nu_{1}|)^{2}\geq
0$. Hence, we obtain

$27|\nu_{1}\nu_{2}\nu_{3}|\leq\left(3(|\nu_{1}|^{2}+|\nu_{2}|^{2}+|\nu_{3}|^{2})\right)^{3/2}=3\sqrt{3}\left(|\nu_{1}|^{2}+|\nu_{2}|^{2}+|\nu_{3}|^{2}\right)^{3/2}.$

∎

Now, we show Proposition A.1.

###### Proof of Proposition A.1.

Since $\pi^{2}_{7}\left(F-*F^{3}/6\right)$, $*F^{4}/24$ and
$\pi^{4}_{7}(F^{2})$ are ${\text{\rm Spin}(7)}$-equivariant, we may assume
that $F$ is of the form of the right hand side of (2.12). Thus, we may show
that $1-*F^{4}/24=0$ or $\pi^{4}_{7}(F^{2})=0$ for all cases given in Lemma
A.3.

We now show that $1-*F^{4}/24=0$ for case (2). By Lemma A.2 (3), we have

$1-\frac{*F^{4}}{24}=1-|\alpha|^{4}-\mu_{1}\mu_{2}\mu_{3}\mu_{4}+I_{1}+I_{2},$

where

$\displaystyle I_{1}$
$\displaystyle=-|\alpha|^{2}(\mu_{1}\mu_{2}+\mu_{3}\mu_{4}),$ $\displaystyle
I_{2}$
$\displaystyle=-\alpha_{1}^{2}(\mu_{1}\mu_{3}+\mu_{2}\mu_{4}+\mu_{1}\mu_{4}+\mu_{2}\mu_{3})-\alpha_{1}\sum_{1\leq
i<j<k\leq 4}\mu_{i}\mu_{j}\mu_{k}$
$\displaystyle=-\alpha_{1}^{2}(\mu_{1}+\mu_{2})(\mu_{3}+\mu_{4})+\alpha_{1}(\mu_{2}+\mu_{3})(\mu_{2}+\mu_{4})(\mu_{3}+\mu_{4}),$

The last equality holds by (A.17). Then, we see that
$I_{1}=|\alpha|^{2}(2|\alpha|^{2}-2)$ and

$\displaystyle I_{2}=$
$\displaystyle\alpha_{1}^{2}(\mu_{3}+\mu_{4})^{2}+\alpha_{1}(\mu_{2}+\mu_{3})(\mu_{2}+\mu_{4})(\mu_{3}+\mu_{4})$
$\displaystyle=$
$\displaystyle-\frac{1}{4}(\mu_{2}+\mu_{3})^{2}(\mu_{2}+\mu_{4})^{2}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}(\mu_{2}(\mu_{2}+\mu_{3}+\mu_{4})+\mu_{3}\mu_{4})^{2}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}(-\mu_{1}\mu_{2}+\mu_{3}\mu_{4})^{2}.$

Hence, we have

$\displaystyle 1-\frac{*F^{4}}{24}$
$\displaystyle=1-|\alpha|^{4}-\mu_{1}\mu_{2}\mu_{3}\mu_{4}+|\alpha|^{2}(2|\alpha|^{2}-2)-\frac{1}{4}(-\mu_{1}\mu_{2}+\mu_{3}\mu_{4})^{2}$
$\displaystyle=(|\alpha|^{2}-1)^{2}-\frac{1}{4}(\mu_{1}\mu_{2}+\mu_{3}\mu_{4})^{2}=0.$

We can compute in the other cases similarly, and obtain the following Table 1.
By this table, the proof is completed.

Case | $1-*F^{4}/24$ | $\pi^{4}_{7}(F^{2})$
---|---|---
(1) | $1-(\alpha_{1}+\mu_{1})(\alpha_{1}+\mu_{2})(\alpha_{1}+\mu_{3})(\alpha_{1}+\mu_{4})$ | 0
(2) | 0 | $\neq 0$
(3)-(a) | $(\mu_{1}+\mu_{2})^{2}(\mu_{1}-\mu_{2})^{2}/4$ | 0
(3)-(b) | 0 | $\neq 0$
(4)-(a) | $(\mu_{1}+\mu_{2})^{2}(\mu_{1}-\mu_{2})^{2}/4$ | 0
(4)-(b) | 0 | $\neq 0$
(5) | 0 | $\neq 0$
(6) | 0 | $\neq 0$
(7)-(a) | 0 | 0
(7)-(b) | 0 | $\neq 0$
(7)-(c) | 0 | $\neq 0$
(8) | 0 | $\neq 0$
(9) | 0 | $\neq 0$
(10) | 0 | 0
Table 1.

∎

By Lemma A.3, we obtain the following simpler form of a 2-form
$F\in\Lambda^{2}W^{*}$ satisfying $\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and
$\pi^{4}_{7}(F^{2})=0$.

######  Corollary A.6.

If a 2-form $F\in\Lambda^{2}W^{*}$ satisfies
$\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and $\pi^{4}_{7}(F^{2})=0$, there
exists $h\in{\text{\rm Spin}(7)}$ such that

(A.29) $\displaystyle
h^{*}F=2\lambda^{2}(\alpha_{1}e^{1})+\mu_{1}e^{01}+\mu_{2}e^{23}+\mu_{3}e^{45}+\mu_{4}e^{67},$

where
$\alpha_{1},\mu_{j}\in{\mathbb{R}},\sum_{j=1}^{4}\mu_{j}=0,\mu_{1}\leq\mu_{2}\leq\mu_{3}\leq\mu_{4}$
and (A.15) holds.

###### Proof.

By Lemma A.3 and table 1, cases satisfying the assumption of this corollary
are (1),(3)-(a), (4)-(a), (7)-(a), (10). We only have to show that cases
(3)-(a), (4)-(a), (7)-(a) and (10) are reduced to (special cases of) (1) by
the ${\text{\rm Spin}(7)}$-action.

Suppose that (3)-(a) holds. Since $\alpha_{3}=\alpha_{7}=0$ and
$\mu_{2}+\mu_{4}=\mu_{1}+\mu_{3}=0$, we see that

$F_{7}=2\lambda^{2}(\alpha_{1}e^{1}+\alpha_{5}e^{5}),\qquad
F_{21}=\mu_{1}(e^{01}-e^{45})+\mu_{2}(e^{23}-e^{67}).$

Set
$\zeta_{0}=e^{0}-{\sqrt{-1}}e^{4},\zeta_{1}=e^{1}+{\sqrt{-1}}e^{5},\zeta_{2}=e^{2}+{\sqrt{-1}}e^{6},\zeta_{3}=e^{3}-{\sqrt{-1}}e^{7}$.
Then, it is straightforward to see that

$\displaystyle e^{01}-e^{45}$ $\displaystyle={\rm
Re}(\overline{\zeta_{0}}\wedge\zeta_{1}),$ $\displaystyle e^{23}-e^{67}$
$\displaystyle={\rm Im}(\zeta_{2}\wedge\overline{\zeta_{3}}),$ $\displaystyle
2\lambda^{2}(e^{1})+2{\sqrt{-1}}\lambda^{2}(e^{5})$
$\displaystyle=\zeta_{0}\wedge\zeta_{1}+\overline{\zeta_{2}\wedge\zeta_{3}}$

and

$\Phi=\frac{1}{2}\left(\frac{{\sqrt{-1}}}{2}\sum_{j=0}^{3}\zeta_{j}\wedge\overline{\zeta}_{j}\right)^{2}+{\rm
Re}(\zeta_{0}\wedge\zeta_{1}\wedge\zeta_{2}\wedge\zeta_{3}).$

Hence, for $\theta\in{\mathbb{R}}$, setting $h:W\to W$ by

$\left(h(\zeta_{0}),h(\zeta_{1}),h(\zeta_{2}),h(\zeta_{3})\right)=\left(e^{{\sqrt{-1}}\theta}\zeta_{0},\
e^{{\sqrt{-1}}\theta}\zeta_{1},\ e^{-{\sqrt{-1}}\theta}\zeta_{2},\
e^{-{\sqrt{-1}}\theta}\zeta_{3}\right),$

we see that $h\in{\text{\rm Spin}(7)}$, $h$ preserves
$e^{01}-e^{45},e^{23}-e^{67}$ and

$h^{*}\left(\lambda^{2}(e^{1})+{\sqrt{-1}}\lambda^{2}(e^{5})\right)=e^{2{\sqrt{-1}}\theta}\left(\lambda^{2}(e^{1})+{\sqrt{-1}}\lambda^{2}(e^{5})\right).$

This implies that $h$ rotates the plane ${\rm
span}\\{\lambda^{2}(e^{1}),\lambda^{2}(e^{5})\\}$ fixing $F_{21}$, and hence,
the case (3)-(a) is reduced to a special case of (1) by the ${\text{\rm
Spin}(7)}$-action.

Since we can permute $\\{\mu_{j}\\}$ by the ${\rm SU}(4)$-action, the case
(4)-(a) is reduced to the case (3)-(a). Hence, by the same argument as above,
the case (4)-(a) is reduced to a special case of (1) by the ${\text{\rm
Spin}(7)}$-action.

Suppose that (7)-(a) holds. Then,

$F_{7}=2\lambda^{2}(\alpha_{1}e^{1}+\alpha_{5}e^{5}+\alpha_{7}e^{7}),\qquad
F_{21}=\mu(-e^{01}-e^{23}+e^{45}+e^{67}).$

By the standard ${\rm SU}(2)$-action acting on ${\rm
span}\\{e^{4},e^{5},e^{6},e^{7}\\}$, which preserves $F_{21}$, we can reduce
the case (7)-(a) to the case of $\alpha_{7}=0$. Then, by the same argument as
above, the case (7)-(a) is reduced to a special case of (1) by the ${\text{\rm
Spin}(7)}$-action.

Suppose that (10) holds. Since $F_{21}=0$, there exists $h\in{\text{\rm
Spin}(7)}$ such that $h^{*}\alpha=e^{1}$. Hence, the case (10) is reduced to a
special case of (1) by the ${\text{\rm Spin}(7)}$-action. ∎

######  Remark A.7.

By Lemma A.4 and Corollary A.6, we see that for any solution
$F=F_{7}+F_{21}\in\Lambda^{2}_{7}W^{*}\oplus\Lambda^{2}_{21}W^{*}$ of
$\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and $\pi^{4}_{7}(F^{2})=0$, $F_{7}$ is
controlled by $F_{21}$. This fact is also reflected in Corollary A.11.

######  Theorem A.8.

For a 2-form $F\in\Lambda^{2}W^{*}$ define
$T_{F}:\Lambda^{2}W^{*}\rightarrow\Lambda^{2}_{7}W^{*}$ and
$S_{F}:\Lambda^{2}W^{*}\rightarrow\Lambda^{4}_{7}W^{*}$ by

$T_{F}(\beta)=\pi^{2}_{7}\left(\beta-*\left(\frac{F^{2}}{2}\wedge\beta\right)\right),\qquad
S_{F}(\beta)=2\pi^{4}_{7}\left(F\wedge\beta\right).$

  1. (1)

If $\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and $1-*F^{4}/24\neq 0$ (then
$\pi^{4}_{7}(F^{2})=0$ is satisfied by Proposition A.1), we have

$\ker T_{F}=(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*}).$

In other words, there exists an isomorphism
$P_{F}:\Lambda^{2}_{7}W^{*}\rightarrow\Lambda^{2}_{7}W^{*}$ such that

$T_{F}=P_{F}\circ\pi^{2}_{7}\circ((I+F^{\sharp})^{-1})^{*}=P_{F}\circ((I+F^{\sharp})^{-1})^{*}\circ\pi^{2}_{7,\Phi_{F}},$

where $\pi^{2}_{7,\Phi_{F}}$ is defined by (2.13) for $\sigma=I+F^{\sharp}$.

  2. (2)

If $\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and $\pi^{4}_{7}(F^{2})=0$, we have

$\ker T_{F}\cap\ker S_{F}=(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*}).$

In other words, there exists an isomorphism
$Q_{F}:\Lambda^{2}_{7}W^{*}\rightarrow{\rm Im\,}(T_{F},S_{F})$ such that

$(T_{F},S_{F})=Q_{F}\circ\pi^{2}_{7}\circ((I+F^{\sharp})^{-1})^{*}=Q_{F}\circ((I+F^{\sharp})^{-1})^{*}\circ\pi^{2}_{7,\Phi_{F}}.$

###### Proof.

Since $F$ satisfies $\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and
$\pi^{4}_{7}(F^{2})=0$ in all cases, we may assume that $F$ is of the form of
the right hand side of (A.29) and satisfies (A.15) by Corollary A.6.

First, we show that $\ker T_{F}\cap\ker
S_{F}\supset(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*})$ without any assumptions
on $1-*F^{4}/24$. Set

$C_{j}=\alpha_{1}+\mu_{j}\qquad j=1,\cdots,4.$

Then, we see that

$F=C_{1}e^{01}+C_{2}e^{23}+C_{3}e^{45}+C_{4}e^{67}.$

Set

$\xi=\sum_{1\leq i<j<k\leq 4}C_{i}C_{j}C_{k}-\sum_{j=1}^{4}C_{j}.$

Note that (A.15) is equivalent to

(A.30) $\displaystyle\xi=0.$

Indeed, since

$\displaystyle C_{i}C_{j}C_{k}$
$\displaystyle=\left(\alpha_{1}+\mu_{i}\right)\left(\alpha_{1}+\mu_{j}\right)\left(\alpha_{1}+\mu_{k}\right)$
$\displaystyle=\alpha_{1}^{3}+\left(\mu_{i}+\mu_{j}+\mu_{k}\right)\alpha_{1}^{2}+\left(\mu_{i}\mu_{k}+\mu_{j}\mu_{k}+\mu_{i}\mu_{j}\right)\alpha_{1}+\mu_{i}\mu_{j}\mu_{k}$

and

$\displaystyle\sum_{1\leq i<j<k\leq 4}\left(\mu_{i}+\mu_{j}+\mu_{k}\right)$
$\displaystyle=3\left(\mu_{1}+\mu_{2}+\mu_{3}+\mu_{4}\right)=0,$
$\displaystyle\sum_{1\leq i<j<k\leq
4}\left(\mu_{i}\mu_{k}+\mu_{j}\mu_{k}+\mu_{i}\mu_{j}\right)$
$\displaystyle=2\sum_{1\leq i<j\leq
4}\mu_{i}\mu_{j}=-\sum_{i=1}^{4}\mu_{i}^{2},$

we have

$\xi=4\alpha_{1}^{3}-\left(\sum_{i=1}^{4}\mu_{i}^{2}\right)\alpha_{1}+\sum_{1\leq
i<j<k\leq 4}\mu_{i}\mu_{j}\mu_{k}-4\alpha_{1},$

which is the left hand side of (A.15).

Fix $u,v\in V={\mathbb{R}}^{7}$ and set

$\tau=e^{0}\wedge\varphi\left(u,v,\cdot\right)-u^{b}\wedge v^{b}.$

By Lemma 2.4, $\Lambda^{2}_{21}W^{*}$ is spanned by elements of the form
$\tau$. Set

(A.31)
$T_{\mu}=\left\langle\left(I+F^{\sharp}\right)^{\ast}\tau-*\left(\dfrac{F^{2}}{2}\wedge\left(I+F^{\sharp}\right)^{\ast}\tau\right),2\lambda^{2}\left(e^{\mu}\right)\right\rangle$

for $\mu=1,\cdots,7$. Then, $\left(I+F^{\sharp}\right)^{\ast}\tau\in\ker
T_{F}$ if and only if $T_{\mu}=0$ for any $\mu=1,\cdots,7$. Using
$\left(I+F^{\sharp}\right)^{\ast}\alpha=\alpha-i\left(\alpha^{\sharp}\right)F$
for $\alpha\in W^{*}$, direct but lengthy computations show that

$\displaystyle T_{1}$
$\displaystyle=\xi(-C_{1}\varphi\left(e_{1},u,v\right)+C_{2}e^{23}\left(u,v\right)+C_{3}e^{45}\left(u,v\right)+C_{4}e^{67}\left(u,v\right)),$
$\displaystyle T_{\mu}$
$\displaystyle=-\xi*\varphi\left(e_{1},e_{\mu},u,v\right)$

for $\mu=2,\cdots,7$. Then, by (A.30), all of these vanish and hence, $\ker
T_{F}\supset(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*})$. Similarly, set

(A.32) $S_{\mu}=\left\langle
F\wedge\left(I+F^{\sharp}\right)^{\ast}\tau,2\lambda^{4}\left(e^{\mu}\right)\right\rangle$

for $\mu=1,\cdots,7$. Then, $\left(I+F^{\sharp}\right)^{\ast}\tau\in\ker
S_{F}$ if and only if $S_{\mu}=0$ for any $\mu=1,\cdots,7$. Direct but lengthy
computations show that

$\displaystyle S_{1}=0,\qquad
S_{\mu}=T_{\mu}=0\quad\mbox{for}\quad\mu=2,\cdots,7.$

Hence, we see that $\ker
S_{F}\supset(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*})$.

Next, we compute $\ker
T_{F}\cap\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{7}W^{*}$ and $\ker
T_{F}\cap\ker S_{F}\cap\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{7}W^{*}$.
We still do not assume that $*F^{4}/24\neq 1$. Since
$\Lambda^{2}W^{*}=\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{7}W^{*}\oplus\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{21}W^{*}$
and $\ker T_{F}\cap\ker S_{F}\supset(I+F^{\sharp})^{*}\Lambda^{2}_{21}W^{*}$,
we see that

$\displaystyle\ker T_{F}=$ $\displaystyle\left(\ker
T_{F}\cap\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{7}W^{*}\right)\oplus\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{21}W^{*},$
$\displaystyle\ker T_{F}\cap\ker S_{F}=$ $\displaystyle\left(\ker
T_{F}\cap\ker
S_{F}\cap(I+F^{\sharp})^{*}\Lambda^{2}_{7}W^{*}\right)\oplus\left(I+F^{\sharp}\right)^{\ast}\Lambda^{2}_{21}W^{*}.$

Fix $\eta=\sum^{7}_{\mu=1}\eta_{\mu}e^{\mu}$ for $\eta_{\mu}\in{\mathbb{R}}$.
Similarly to (A.31) and (A.32), set

$\displaystyle T^{\prime}_{\mu}$
$\displaystyle=\left\langle\left(I+F^{\sharp}\right)^{\ast}\left(2\lambda^{2}(\eta)\right)-*\left(\dfrac{F^{2}}{2}\wedge\left(I+F^{\sharp}\right)^{\ast}\left(2\lambda^{2}(\eta)\right)\right),2\lambda^{2}\left(e^{\mu}\right)\right\rangle,$
$\displaystyle S^{\prime}_{\mu}$ $\displaystyle=\left\langle
F\wedge\left(I+F^{\sharp}\right)^{\ast}\left(2\lambda^{2}(\eta)\right),2\lambda^{4}\left(e^{\mu}\right)\right\rangle$

for $\mu=1,\cdots,7$. Note that $\ker
T_{F}\cap(I+F^{\sharp})^{*}\Lambda^{2}_{7}W^{*}=\\{\,0\,\\}$ if and only if
$T^{\prime}_{\mu}=0$ for any $\mu=1,\cdots,7$ implies $\eta=0$. Similarly,
$\ker T_{F}\cap\ker
S_{F}\cap(I+F^{\sharp})^{*}\Lambda^{2}_{7}W^{*}=\\{\,0\,\\}$ if and only if
$T^{\prime}_{\mu}=S^{\prime}_{\mu}=0$ for any $\mu=1,\cdots,7$ implies
$\eta=0$. By direct but lengthy computations, we obtain

(A.33) $\displaystyle T_{1}^{\prime}$
$\displaystyle=4\eta_{1}\left(C_{1}C_{2}C_{3}C_{4}-\sum_{1\leq i<j\leq
4}C_{i}C_{j}+1\right),$ (A.34) $\displaystyle T_{2}^{\prime}$
$\displaystyle=4(C_{1}C_{2}-1)(C_{3}C_{4}-1)\eta_{2}+2\left\\{(C_{1}C_{2}-1)(C_{3}+C_{4})+(C_{1}+C_{2})(-C_{3}C_{4}+1)\right\\}\eta_{3},$
(A.35) $\displaystyle T_{3}^{\prime}$
$\displaystyle=-2\left\\{(C_{1}C_{2}-1)(C_{3}+C_{4})+(C_{1}+C_{2})(-C_{3}C_{4}+1)\right\\}\eta_{2}+4(C_{1}C_{2}-1)(C_{3}C_{4}-1)\eta_{3},$
(A.36) $\displaystyle T_{4}^{\prime}$
$\displaystyle=4(C_{1}C_{3}-1)(C_{2}C_{4}-1)\eta_{4}+2\left\\{(C_{1}C_{2}+1)(C_{3}-C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{5},$
(A.37) $\displaystyle T_{5}^{\prime}$
$\displaystyle=-2\left\\{(C_{1}C_{2}+1)(C_{3}-C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{4}+4(C_{1}C_{3}-1)(C_{2}C_{4}-1)\eta_{5},$
(A.38) $\displaystyle T_{6}^{\prime}$
$\displaystyle=4(C_{1}C_{4}-1)(C_{2}C_{3}-1)\eta_{6}+2\left\\{(C_{1}C_{2}+1)(-C_{3}+C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{7},$
(A.39) $\displaystyle T_{7}^{\prime}$
$\displaystyle=-2\left\\{(C_{1}C_{2}+1)(-C_{3}+C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{6}+4(C_{1}C_{4}-1)(C_{2}C_{3}-1)\eta_{7}.$

Note that we use (A.30) for the computation of $T_{1}^{\prime}$. Similarly,
direct but lengthy computations show that

$\displaystyle S_{1}^{\prime}$ $\displaystyle=0,$ (A.40) $\displaystyle
S_{2}^{\prime}$
$\displaystyle=-4(C_{1}+C_{2})(C_{3}+C_{4})\eta_{2}-2\left\\{(C_{1}C_{2}-1)(C_{3}+C_{4})+(C_{1}+C_{2})(-C_{3}C_{4}+1)\right\\}\eta_{3},$
(A.41) $\displaystyle S_{3}^{\prime}$
$\displaystyle=2\left\\{(C_{1}C_{2}-1)(C_{3}+C_{4})+(C_{1}+C_{2})(-C_{3}C_{4}+1)\right\\}\eta_{2}-4(C_{1}+C_{2})(C_{3}+C_{4})\eta_{3},$
(A.42) $\displaystyle S_{4}^{\prime}$
$\displaystyle=-4(C_{1}+C_{3})(C_{2}+C_{4})\eta_{4}-2\left\\{(C_{1}C_{2}+1)(C_{3}-C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{5},$
(A.43) $\displaystyle S_{5}^{\prime}$
$\displaystyle=2\left\\{(C_{1}C_{2}+1)(C_{3}-C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{4}-4(C_{1}+C_{3})(C_{2}+C_{4})\eta_{5},$
(A.44) $\displaystyle S_{6}^{\prime}$
$\displaystyle=-4(C_{1}+C_{4})(C_{2}+C_{3})\eta_{6}-2\left\\{(C_{1}C_{2}+1)(-C_{3}+C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{7},$
(A.45) $\displaystyle S_{7}^{\prime}$
$\displaystyle=2\left\\{(C_{1}C_{2}+1)(-C_{3}+C_{4})+(C_{1}-C_{2})(C_{3}C_{4}+1)\right\\}\eta_{6}-4(C_{1}+C_{4})(C_{2}+C_{3})\eta_{7}.$

First, we show the following.

######  Lemma A.9.

We have $C_{1}C_{2}C_{3}C_{4}-\sum_{1\leq i<j\leq 4}C_{i}C_{j}+1\neq 0$.

###### Proof.

If $C_{2}C_{3}+C_{2}C_{4}+C_{3}C_{4}-1\neq 0$, we have

$C_{1}=\frac{-C_{2}C_{3}C_{4}+C_{2}+C_{3}+C_{4}}{C_{2}C_{3}+C_{2}C_{4}+C_{3}C_{4}-1}$

by (A.30). Then,

$\displaystyle C_{1}C_{2}C_{3}C_{4}-\sum_{1\leq i<j\leq 4}C_{i}C_{j}+1$
$\displaystyle=$
$\displaystyle\frac{-(C_{2}C_{3}C_{4}-C_{2}-C_{3}-C_{4})^{2}-(C_{2}C_{3}+C_{2}C_{4}+C_{3}C_{4}-1)^{2}}{C_{2}C_{3}+C_{2}C_{4}+C_{3}C_{4}-1},$

which is nonzero since $C_{2}C_{3}+C_{2}C_{4}+C_{3}C_{4}-1\neq 0$.

Suppose that $C_{2}C_{3}+C_{2}C_{4}+C_{3}C_{4}-1=0$. Then, by (A.30), we have
$\xi=C_{2}C_{3}C_{4}-(C_{2}+C_{3}+C_{4})=0.$ Thus, we obtain

$\begin{pmatrix}C_{3}+C_{4}&C_{3}C_{4}-1\\\
-\left(C_{3}C_{4}-1\right)&C_{3}+C_{4}\end{pmatrix}\begin{pmatrix}C_{2}\\\
1\end{pmatrix}=\begin{pmatrix}0\\\ 0\end{pmatrix}.$

Since $(C_{3}+C_{4})^{2}+\left(C_{3}C_{4}-1\right)^{2}>0$, we have
$(C_{2},1)=(0,0)$, which is a contradiction. ∎

Hence, by (A.33), $T_{1}^{\prime}=0$ if and only if $\eta_{1}=0$. By (A.34)
and (A.35), $T_{2}^{\prime}=T_{3}^{\prime}=0$ implies $\eta_{2}=\eta_{3}=0$ if
and only if

(A.46) $\displaystyle(C_{1}C_{2}-1)(C_{3}C_{4}-1)$ $\displaystyle\neq
0\quad\mbox{or},$ (A.47)
$\displaystyle(C_{1}C_{2}-1)(C_{3}+C_{4})+(C_{1}+C_{2})(-C_{3}C_{4}+1)$
$\displaystyle\neq 0.$

Now, we show that $T_{2}^{\prime}=T_{3}^{\prime}=0$ implies
$\eta_{2}=\eta_{3}=0$ if $*F^{4}/24\neq 1$. To see this, suppose that the left
hand sides of (A.46) and (A.47) vanish. If $C_{1}C_{2}=1$, the assumption that
the left hand side of (A.47) vanishes implies that

$(C_{1}+C_{2})(-C_{3}C_{4}+1)=\frac{(C_{1}^{2}+1)(-C_{3}C_{4}+1)}{C_{1}}=0.$

Thus, we obtain $C_{3}C_{4}=1$. In particular, we have
$*F^{4}/24=C_{1}C_{2}C_{3}C_{4}=1$. Similarly, $C_{3}C_{4}=1$ implies that
$C_{1}C_{2}=1$ and $*F^{4}/24=C_{1}C_{2}C_{3}C_{4}=1$. Hence, we see that
$T_{2}^{\prime}=T_{3}^{\prime}=0$ implies $\eta_{2}=\eta_{3}=0$ if
$*F^{4}/24\neq 1$.

Similarly, we show that
$T_{2}^{\prime}=T_{3}^{\prime}=S_{2}^{\prime}=S_{3}^{\prime}=0$ implies
$\eta_{2}=\eta_{3}=0$ without any assumptions on $*F^{4}/24$. By (A.34),
(A.35), (A.40) and (A.41), this is the case if and only if one of the
following is nonzero.

$\displaystyle(C_{1}C_{2}-1)(C_{3}C_{4}-1),$
$\displaystyle(C_{1}+C_{2})(C_{3}+C_{4}),$
$\displaystyle(C_{1}C_{2}-1)(C_{3}+C_{4})+(C_{1}+C_{2})(-C_{3}C_{4}+1).$

To see that one of these is nonzero, suppose that all of these are 0. Then, by
the argument above, the first and the third equation imply that $C_{1}C_{2}=1$
and $C_{3}C_{4}=1$. Then,

$(C_{1}+C_{2})(C_{3}+C_{4})=\frac{(C_{1}^{2}+1)(C_{4}^{2}+1)}{C_{1}C_{3}}\neq
0,$

which is a contradiction.

We can discuss for $\eta_{4},\eta_{5},\eta_{6},\eta_{7}$ similarly and we see
that $\ker T_{F}=(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*})$ if
$1-*F^{4}/24\neq 0$ and $\ker T_{F}\cap\ker
S_{F}=(I+F^{\sharp})^{*}(\Lambda^{2}_{21}W^{*})$.

Finally, we prove the last statement of (1). As proved above, we see that

$P_{F}:=T_{F}\circ(I+F^{\sharp})^{*}|_{\Lambda^{2}_{7}W^{*}}:\Lambda^{2}_{7}W^{*}\rightarrow\Lambda^{2}_{7}W^{*}$

is an isomorphism if $*F^{4}/24\neq 1$. Then, for any
$\beta_{7}\in\Lambda^{2}_{7}W^{*}$ and $\beta_{21}\in\Lambda^{2}_{21}W^{*}$,
we see that

$\displaystyle\left(P_{F}\circ\pi^{2}_{7}\circ((I+F^{\sharp})^{-1})^{*}\right)\left((I+F^{\sharp})^{*}(\beta_{7})\right)$
$\displaystyle=P_{F}(\beta_{7})=T_{F}\left((I+F^{\sharp})^{*}(\beta_{7})\right),$
$\displaystyle\left(P_{F}\circ\pi^{2}_{7}\circ((I+F^{\sharp})^{-1})^{*}\right)\left((I+F^{\sharp})^{*}(\beta_{21})\right)$
$\displaystyle=0=T_{F}\left((I+F^{\sharp})^{*}(\beta_{21})\right).$

Hence, we obtain $T_{F}=P_{F}\circ\pi^{2}_{7}\circ((I+F^{\sharp})^{-1})^{*}$.

Similarly, for the last statement of (2), an isomorphism

$Q_{F}:=(T_{F},S_{F})\circ(I+F^{\sharp})^{*}|_{\Lambda^{2}_{7}W^{*}}:\Lambda^{2}_{7}W^{*}\rightarrow{\rm
Im\,}(T_{F},S_{F})$

satisfies the desired property. ∎

######  Remark A.10.

Since $C_{1}C_{2}C_{3}C_{4}-\sum_{1\leq i<j\leq 4}C_{i}C_{j}+1=1-\langle
F^{2},\Phi\rangle/2+*F^{4}/24$, we see that

$1-\frac{1}{2}\langle F^{2},\Phi\rangle+\frac{*F^{4}}{24}\neq 0$

for $F\in\Lambda^{2}W^{*}$ satisfying $\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$
and $\pi^{4}_{7}(F^{2})=0$ by Lemma A.9. We can also prove this by the
“mirror” of the Cayley equality in [10].

Finally, we give other applications of Corollary A.6. The following proves
Proposition 3.4.

######  Corollary A.11.

Suppose that $F\in\Lambda^{2}W^{*}$ satisfies
$\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and $\pi^{4}_{7}(F^{2})=0$. Set
$F_{7}=\pi^{2}_{7}(F)$ and $F_{21}=\pi^{2}_{21}(F)$.

  1. (1)

If $F_{21}=0$, we have $F_{7}=0$ or $|F_{7}|=2$.

  2. (2)

We have

$|F_{7}|\leq
2\sqrt{\frac{|F_{21}|^{2}+4}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{|F_{21}|^{3}}{(|F_{21}|^{2}+4)^{3/2}}\right)\right).$

###### Proof.

By Corollary A.6, we may assume that $F$ is of the form of the right hand side
of (A.29) satisfying (A.15).

If $F_{21}=0$, (A.15) implies that $(\alpha_{1}^{2}-1)\alpha_{1}=0$. Since
$|F_{7}|=2|\alpha_{1}|$, we obtain (1). Next, we show (2). Recall that
$|F_{21}|^{2}=\sum_{j=1}^{4}\mu_{j}^{2}$. By (A.15), we have

$\displaystyle\left(4\alpha_{1}^{2}-\sum_{j=1}^{4}\mu_{j}^{2}-4\right)|\alpha_{1}|\leq\left|4\alpha_{1}^{2}-\sum_{j=1}^{4}\mu_{j}^{2}-4\right||\alpha_{1}|=\left|\sum_{1\leq
i<j<k\leq 4}\mu_{i}\mu_{j}\mu_{k}\right|.$

Then, by Lemma A.5, we obtain

$4|\alpha_{1}|^{3}-(|F_{21}|^{2}+4)|\alpha_{1}|-\frac{\sqrt{3}}{9}|F_{21}|^{3}\leq
0.$

Thus, if we define a cubic polynomial $f(x)$ (with a parameter $\lambda\geq
0$) by

$f(x)=4x^{3}-\left(\lambda^{2}+4\right)x-\frac{\sqrt{3}}{9}\lambda^{3},$

$x=|\alpha_{1}|$ satisfies $f(x)\leq 0$ for $\lambda=|F_{21}|$. By the Viète’s
formula for a cubic equation, the largest solution of $f(x)=0$ is given by

$x=x_{0}:=\sqrt{\frac{\lambda^{2}+4}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{\lambda^{3}}{(\lambda^{2}+4)^{3/2}}\right)\right).$

Hence, we see that $f(x)\leq 0$ implies that $x\leq x_{0}$. This together with
the equation $|F_{7}|=2|\alpha_{1}|$ implies (2). ∎

The following is useful in Section 4.5.

######  Lemma A.12.

Suppose that $F\in\Lambda^{2}W^{*}$ satisfies
$\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$, $*F^{4}/24\neq 1$ and $F\neq 0$. Then,
if $\gamma\in\Lambda^{2}_{7}W^{*}$ satisfies

(A.48) $\displaystyle\pi^{4}_{1}(F\wedge\gamma)$ $\displaystyle=0,$ (A.49)
$\displaystyle\pi^{4}_{7}\left(\left(6F-*F^{3}\right)\wedge\gamma\right)$
$\displaystyle=0,$ (A.50)
$\displaystyle\pi^{4}_{35}\left(\left(6F+*F^{3}\right)\wedge\gamma\right)$
$\displaystyle=0,$

we have $\gamma=0$.

###### Proof.

By Corollary A.6, we may assume that $F$ is of the form of the right hand side
of (A.29) satisfying (A.15). Set $C_{j}=\alpha_{1}+\mu_{j}$. That is,

$F=C_{1}e^{01}+C_{2}e^{23}+C_{3}e^{45}+C_{4}e^{67}.$

Note that we have the equation (A.30). Set

$\gamma=2\lambda^{2}(\eta)=e^{0}\wedge\eta+i(\eta^{\sharp})\varphi,$

where $\eta=\sum_{j=1}^{7}\eta_{j}e^{j}\in V^{*}$. Since
$*\left(F\wedge\gamma\wedge\Phi\right)=3\langle F,\gamma\rangle$ and

$\displaystyle\langle F,\gamma\rangle$
$\displaystyle=C_{1}\eta_{1}+C_{2}\varphi(\eta,e_{2},e_{3})+C_{3}\varphi(\eta,e_{4},e_{5})+C_{4}\varphi(\eta,e_{6},e_{7})$
$\displaystyle=(C_{1}+C_{2}+C_{3}+C_{4})\eta_{1},$

(A.48) is equivalent to

(A.51) $\displaystyle(C_{1}+C_{2}+C_{3}+C_{4})\eta_{1}=0.$

Next, we rewrite (A.49) and (A.50). Since
$\pi^{2}_{7}\left(F-*F^{3}/6\right)=0$ and (2.11), we see that

$\pi^{4}_{7}\left(\left(F-\frac{*F^{3}}{6}\right)\wedge\gamma\right)=\frac{1}{2}\left(\left(F-\frac{*F^{3}}{6}\right)+*\left(F-\frac{*F^{3}}{6}\right)\right).$

Since $\Lambda^{4}_{35}W^{*}$ is the space of anti self dual 4-forms, we have

$\pi^{4}_{35}\left(\left(F+\frac{*F^{3}}{6}\right)\wedge\gamma\right)=\frac{1}{2}\left(\left(F+\frac{*F^{3}}{6}\right)-*\left(F+\frac{*F^{3}}{6}\right)\right).$

By a direct computation, we see that

$F\wedge\gamma\pm*(F\wedge\gamma)=\sum_{j=1}^{7}\eta_{j}\Theta_{j},$

where

$\displaystyle\Theta_{1}=$
$\displaystyle\left((C_{1}+C_{2})\pm(C_{3}+C_{4})\right)\left(e^{0123}\pm
e^{4567}\right)$
$\displaystyle+\left((C_{1}+C_{3})\pm(C_{2}+C_{4})\right)\left(e^{0145}\pm
e^{2367}\right)$
$\displaystyle+\left((C_{1}+C_{4})\pm(C_{2}+C_{3})\right)\left(e^{0167}\pm
e^{2345}\right),$ $\displaystyle\Theta_{2}=$ $\displaystyle(C_{1}\pm
C_{2})\left(e^{0146}-e^{0157}\pm e^{2346}\mp e^{2357}\right)$
$\displaystyle+(C_{3}\pm C_{4})\left(e^{0245}\pm e^{0267}-e^{1345}\mp
e^{1367}\right),$ $\displaystyle\Theta_{3}=$ $\displaystyle(C_{1}\pm
C_{2})\left(-e^{0147}-e^{0156}\mp e^{2347}\mp e^{2356}\right)$
$\displaystyle+(C_{3}\pm C_{4})\left(e^{0345}\pm e^{0367}+e^{1245}\pm
e^{1267}\right),$ $\displaystyle\Theta_{4}=$ $\displaystyle(C_{1}\pm
C_{3})\left(-e^{0126}+e^{0137}\mp e^{2456}\pm e^{3457}\right)$
$\displaystyle+(C_{2}\pm C_{4})\left(e^{0234}\pm e^{0467}-e^{1235}\mp
e^{1567}\right),$ $\displaystyle\Theta_{5}=$ $\displaystyle(C_{1}\pm
C_{3})\left(e^{0127}+e^{0136}\pm e^{2457}\pm e^{3456}\right)$
$\displaystyle+(C_{2}\pm C_{4})\left(e^{0235}\pm e^{0567}+e^{1234}\pm
e^{1467}\right),$ $\displaystyle\Theta_{6}=$ $\displaystyle(C_{1}\pm
C_{4})\left(e^{0124}-e^{0135}\pm e^{2467}\mp e^{3567}\right)$
$\displaystyle+(C_{2}\pm C_{3})\left(e^{0236}\pm e^{0456}-e^{1237}\mp
e^{1457}\right),$ $\displaystyle\Theta_{7}=$ $\displaystyle(C_{1}\pm
C_{4})\left(-e^{0125}-e^{0134}\mp e^{2567}\mp e^{3467}\right)$
$\displaystyle+(C_{2}\pm C_{3})\left(e^{0237}\pm e^{0457}+e^{1236}\pm
e^{1456}\right).$

Since

$\frac{*F^{3}}{6}=C_{2}C_{3}C_{4}e^{01}+C_{1}C_{3}C_{4}e^{23}+C_{1}C_{2}C_{4}e^{45}+C_{1}C_{2}C_{3}e^{67},$

(A.49) is equivalent to

(A.52)
$\displaystyle\begin{split}(C_{1}+C_{2})(C_{3}C_{4}-1)\eta_{a}&=(C_{3}+C_{4})(C_{1}C_{2}-1)\eta_{a}=0,\\\
(C_{1}+C_{3})(C_{2}C_{4}-1)\eta_{b}&=(C_{2}+C_{4})(C_{1}C_{3}-1)\eta_{b}=0,\\\
(C_{1}+C_{4})(C_{2}C_{3}-1)\eta_{c}&=(C_{2}+C_{3})(C_{1}C_{4}-1)\eta_{c}=0\end{split}$

for $a=2,3,b=4,5$ and $c=6,7$. Similarly, (A.50) is equivalent to

(A.53)
$\displaystyle\begin{split}\left\\{(C_{1}+C_{2})(C_{3}C_{4}+1)-(C_{3}+C_{4})(C_{1}C_{2}+1)\right\\}\eta_{1}&=0,\\\
\left\\{(C_{1}+C_{3})(C_{2}C_{4}+1)-(C_{2}+C_{4})(C_{1}C_{3}+1)\right\\}\eta_{1}&=0,\\\
\left\\{(C_{1}+C_{4})(C_{2}C_{3}+1)-(C_{2}+C_{3})(C_{1}C_{4}+1)\right\\}\eta_{1}&=0,\end{split}$
(A.54)
$\displaystyle\begin{split}(C_{1}-C_{2})(C_{3}C_{4}-1)\eta_{a}&=(C_{3}-C_{4})(C_{1}C_{2}-1)\eta_{a}=0,\\\
(C_{1}-C_{3})(C_{2}C_{4}-1)\eta_{b}&=(C_{2}-C_{4})(C_{1}C_{3}-1)\eta_{b}=0,\\\
(C_{1}-C_{4})(C_{2}C_{3}-1)\eta_{c}&=(C_{2}-C_{3})(C_{1}C_{4}-1)\eta_{c}=0\end{split}$

for $a=2,3,b=4,5$ and $c=6,7$. It is straightforward to see that all the
coefficients of $\eta_{1}$ in (A.51) and (A.53) vanish if and only if
$C_{1}=C_{2}=C_{3}=C_{4}=0$, which violates the assumption $F\neq 0$. Hence,
we obtain $\eta_{1}=0$.

It is also straightforward to see that all the coefficients of $\eta_{2}$ and
$\eta_{3}$ in (A.52) and (A.54) vanish if and only if
$C_{1}=C_{2}=C_{3}=C_{4}=0$ or $C_{1}C_{2}=C_{3}C_{4}=1$. Since
$*F^{4}/24=C_{1}C_{2}C_{3}C_{4}$, both cases are contrary to the assumptions.
Hence, we obtain $\eta_{2}=\eta_{3}=0$.

Similarly, we obtain $\eta_{4}=\eta_{5}=\eta_{6}=\eta_{7}=0$ and the proof is
completed. ∎

##  Appendix B Notation

We summarize the notation used in this paper. We use the following for a
manifold $X$ with a $G_{2}$\- or ${\text{\rm Spin}(7)}$-structure. Denote by
$g$ the associated Riemannian metric.

Notation |  | Meaning
---|---|---
$i(\,\cdot\,)$ |  | The interior product
$\Gamma(X,E)$ |  | The space of all smooth sections of a vector bundle $E\rightarrow X$
${\Omega}^{k}$ |  | ${\Omega}^{k}={\Omega}^{k}(X)=\Gamma(X,\Lambda^{k}T^{*}X)$
$S^{2}T^{*}X$ |  | The space of symmetric (0,2)-tensors on $X$
$S^{2}_{0}T^{*}X$ |  | The space of traceless symmetric (0,2)-tensors on $X$
${\mathcal{S}}^{2}$ |  | ${\mathcal{S}}^{2}=\Gamma(X,S^{2}T^{*}X)$
${\mathcal{S}}^{2}_{0}$ |  | ${\mathcal{S}}^{2}_{0}=\Gamma(X,S^{2}_{0}T^{*}X)$
$b^{k}$ |  | the $k$-th Betti number
$H^{k}_{dR}$ |  | the $k$-th de Rham cohomology
$H^{k}(\\#)$ |  | the $k$-th cohomology of a complex $(\\#)$
$Z^{1}$ |  | the space of closed 1-forms
$v^{\flat}\in T^{*}X$ |  |  $v^{\flat}=g(v,\,\cdot\,)$ for $v\in TX$
$\alpha^{\sharp}\in TX$ |  |  $\alpha=g(\alpha^{\sharp},\,\cdot\,)$ for $\alpha\in T^{*}X$
vol |  | The volume form induced from $g$
$\Lambda^{k}_{\ell}T^{*}X$ |  | The subspace of $\Lambda^{k}T^{*}X$ corresponding to
|  | an ${\ell}$-dimensional irreducible subrepresentation
${\Omega}^{k}_{\ell}$ |  | ${\Omega}^{k}_{\ell}=\Gamma(X,\Lambda^{k}_{\ell}T^{*}X)$
$\pi^{k}_{\ell}$ |  | The projection $\Lambda^{k}T^{*}X\rightarrow\Lambda^{k}_{\ell}T^{*}X$ or ${\Omega}^{k}\rightarrow{\Omega}^{k}_{\ell}$

Acknowledgements: The authors would like to thank an anonymous referee for the
careful reading of an earlier version of this paper and useful comments on it.

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We now use that $\tau>2u>Ct^{1/2-\delta/10}\geq C\cdot 2^{a(1/2-\delta/10)}$,
that $b\geq a(1/2-\delta/10)-5$, and that $a,b\leq C\log(\tau)$ to get that

$C\frac{\epsilon^{3/2}}{s^{2}}\sum_{a,b}2^{a-2b}+2^{2\delta^{2}a}2^{-b(1/2-\delta)}\leq
C\frac{\epsilon^{3/2}}{s^{2}}\sum_{a,b}2^{\delta a/4}+2^{-b/2+2\delta b}\leq
C\frac{\epsilon^{3/2}\tau^{\delta}}{s^{2}}$

####  8.5.5 $t<s/2,u>t^{1/2-\delta/10},\tau/2\leq
u\leq\tau-\tau^{1-0.6\delta}$

In this region, $u,\tau$ are comparable and $t<\tau^{2+\delta/2}$. Similarly
to Section (8.5.4), we note that $|\tau-u-u^{\prime}|>|\tau-u|/2$ and thus
obtain using Proposition 11 that

$|\partial\partial^{\beta}\phi|\leq
C\frac{t^{1/2}}{(\tau-u)^{1/2}}|\overline{\partial}\partial^{\beta}\phi|+C\frac{t^{1/2}}{(\tau-u)^{1/2}}|\overline{\partial}^{\prime}\partial^{\beta}\phi|.$

Thus, decomposing the null form relative to the frame adapted to
$\overline{\partial}^{\prime}$, we need to bound

$\displaystyle C$ $\displaystyle\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{t<s/2,u>t^{1/2-\delta/10},\tau/2\leq
u\leq\tau-\tau^{1-0.6\delta},|u^{\prime}|\leq
1}\left(\frac{\sqrt{t\tau}}{s^{2}}\right)|\overline{\partial}^{\prime}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|td\theta
dudtdu^{\prime}$ $\displaystyle\leq C\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{t<s/2,u>t^{1/2-\delta/10},\tau/2\leq
u\leq\tau-\tau^{1-0.6\delta},|u^{\prime}|\leq 1}$
$\displaystyle\qquad\left(\frac{t\tau^{1/2}}{s^{2}(\tau-u)^{1/2}}\right)\left(|\overline{\partial}^{\prime}\partial^{\alpha}\phi|||\overline{\partial}^{\prime}\partial^{\beta}\phi|+|\overline{\partial}^{\prime}\partial^{\alpha}\phi|||\overline{\partial}\partial^{\beta}\phi|\right)td\theta
dudtdu^{\prime}$ (64)

For the first term, we note that by Lemma 18, the $L^{2}$ norm of
$\overline{\partial}^{\prime}\partial^{\alpha}\phi$ and of
$\overline{\partial}^{\prime}\partial^{\beta}\phi$ is at most
$C\epsilon^{3/4}\tau^{-1}$ and we bound the prefactor pointwise and we use
that $t<\tau^{2+\delta/2}$ to get

$\displaystyle C\iint_{t<s/2,u>t^{1/2-\delta/10},\tau/2\leq
u\leq\tau-\tau^{1-0.6\delta},|u^{\prime}|\leq
1}\frac{t\tau^{1/2}}{s^{2}(\tau-u)^{1/2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|||\overline{\partial}^{\prime}\partial^{\beta}\phi|td\theta
dudtdu^{\prime}$ (65) $\displaystyle\leq
C\epsilon^{3/2}\frac{\tau^{2+\delta/2}\tau^{1/2}}{s^{2}\tau^{1/2-0.3\delta}}\tau^{-2}<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}}$

Now we bound the second term of (64). By Lemma 18, we know that the $L^{2}$
norm of $\overline{\partial}^{\prime}\partial^{\alpha}\phi$ is at most
$\epsilon^{3/4}\tau^{-1}$. We calculate the square of $L^{2}$ norm of the
remaining factor over the same region of integration (but not explicitly
writing it for ease of notation) using the pointwise bounds (and using that
$u$ and $\tau$ are comparable, and that $t<C\tau^{2+\delta}$)

$\displaystyle C\epsilon^{3/2}$
$\displaystyle\iint\frac{t^{2}\tau}{s^{4}(\tau-u)}\left(\frac{u^{\delta}}{t^{2}}\right)^{2-\delta^{2}}td\theta
dudtdu^{\prime}$
$\displaystyle\leq\frac{C\epsilon^{3/2}\tau^{1+2\delta}}{s^{4}}\iint\frac{t^{-1+2\delta^{2}}}{\tau-u}d\theta
dudtdu^{\prime}$
$\displaystyle\leq\frac{C\epsilon^{3/2}\tau^{1+2\delta}\log\tau}{s^{4}}\int_{\tau-2}^{C\tau^{2+\delta}}t^{-1+2\delta^{2}}dt\leq\frac{C\epsilon^{3/2}\tau^{1+2\delta+5\delta^{2}}\log\tau}{s^{4}}$

so the $L^{2}$ norm of the other factor is at most

$\frac{C\epsilon^{3/4}\tau^{1/2+2\delta}}{s^{2}}$

and the second term in (64) is bounded by

$\epsilon^{3/4}\tau^{-1}\frac{C\epsilon^{3/4}\tau^{1/2+2\delta}}{s^{2}}<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}}.$

####  8.5.6 $t<s/2,u>t^{1/2-\delta/10},\tau-u<\tau^{1-0.6\delta}$

In this region, $u$ and $\tau$ are comparable and $t<\tau^{2+\delta/2}$.
Decomposing the null form in a frame adapted to
$\overline{\partial}^{\prime}$, we use pointwise bounds for
$|\partial\partial^{\beta}\phi|$ to bound

$\displaystyle\sum_{|\alpha|+|\beta|\leq 3N_{0}/4+10}\iint_{|u^{\prime}|\leq
1,100<t<s/2,u>t^{1/2-\delta/10},\tau-u<\tau^{1-0.6\delta}}\frac{\sqrt{t\tau}}{s^{2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|td\theta
dudtdu^{\prime}$ $\displaystyle\leq C\epsilon^{3/4}\sum_{|\alpha|\leq
3N_{0}/4+10}\iint_{|u^{\prime}|\leq
1,100<t<s/2,u>t^{1/2-\delta/10},\tau-u<\tau^{1-0.6\delta}}\frac{\sqrt{t\tau}}{s^{2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|\left(\frac{u^{\delta}}{tu}\right)^{1-\delta^{2}}td\theta
dudtdu^{\prime}$ $\displaystyle\leq
C\epsilon^{3/4}\tau^{-1/2+\delta+\delta^{2}}s^{-2}\sum_{|\alpha|\leq
3N_{0}/4+10}\iint_{|u^{\prime}|\leq
1,100<t<s/2,u>t^{1/2-\delta/10},\tau-u<\tau^{1-0.6\delta}}t^{-1/2+\delta^{2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|td\theta
dudtdu^{\prime}.$

We now apply Cauchy-Schwarz and use Lemma 18 for the second factor to get that

$\displaystyle C\epsilon^{3/4}$
$\displaystyle\tau^{-1/2+\delta+\delta^{2}}s^{-2}\iint_{|u^{\prime}|\leq
1,100<t<s/2,u>t^{1/2-\delta/10},\tau-u<\tau^{1-0.6\delta}}t^{-1/2+\delta^{2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|td\theta
dudt$ $\displaystyle\leq
C\epsilon^{3/4}\tau^{-1/2+\delta+\delta^{2}}s^{-2}\left(\iint
t^{-1+2\delta^{2}}td\theta
dudt\right)^{1/2}\left(\iint|\overline{\partial}^{\prime}\partial^{\alpha}\phi|^{2}td\theta
dudt\right)^{1/2}$ $\displaystyle\leq
C\epsilon^{3/2}\tau^{-1/2+\delta+\delta^{2}}s^{-2}\left((\tau^{1-0.6\delta})\int_{100}^{\tau^{2+\delta/2}}t^{2\delta^{2}}dt\right)^{1/2}\tau^{-1}$
$\displaystyle\leq
C\epsilon^{3/2}\tau^{-1/2+\delta+\delta^{2}}s^{-2}\left(\tau^{1-0.6\delta+(2+\delta/2)(1+2\delta^{2})}\right)^{1/2}\tau^{-1}<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}},$

where we stopped writing the region of integration explicitly in some of the
terms above.

####  8.5.7 Case where $\tau<s^{1-10\delta}$, region where $t>s/2$

Here we use pointwise bootstrap assumptions to get that we need to bound

$\displaystyle C$ $\displaystyle\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-100>t>s/2,|u^{\prime}|\leq
10}\frac{\tau^{1/2}}{(s-t)^{3/2}}|\overline{\partial}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|(s-t)d\theta
dudtdu^{\prime}$ $\displaystyle\leq C\epsilon^{3/2}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-100>t>s/2,|u^{\prime}|\leq
10}\frac{\tau^{1/2}}{(s-t)^{3/2}}\left(\frac{u^{\delta}}{s^{2}}\right)^{1-\delta^{2}}\left(\frac{u^{\delta}}{us}\right)^{1-\delta^{2}}(s-t)d\theta
dudtdu^{\prime}$
$\displaystyle\leq\frac{C\epsilon^{3/2}}{s^{3-3\delta^{2}}}\int^{s-100}_{s/2}\frac{\tau^{1/2+2\delta+\delta^{2}}}{(s-t)^{1/2}}dt$
$\displaystyle\leq\frac{C\epsilon^{3/2}\tau^{1/2+2\delta+\delta^{2}}}{s^{5/2-3\delta^{2}}}\leq\frac{C\epsilon^{3/2}}{s^{2}}\leq
C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}}.$

####  8.5.8 Case where $\tau\geq s^{1-10\delta}$, region where $t\geq
s-\tau/4$

In this region, we have that $\tau$ and $u$ are of comparable size. We will
ignore that the derivatives of $\psi$ are only in certain directions and just
use the bound

$|\partial\psi|\leq\frac{1}{s-t}.$

Then we use pointwise bounds for $|\partial\partial^{\beta}\phi|$ and see that
we need to bound

$\displaystyle C$ $\displaystyle\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-100>t>s-\tau/2,|u^{\prime}|\leq
10}\frac{1}{s-t}|\overline{\partial}^{\prime}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|dxdt$
$\displaystyle\leq C\epsilon^{3/4}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-100>t>s-\tau/2,|u^{\prime}|\leq
10}\frac{1}{s-t}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|\left(\frac{u^{\delta}}{us}\right)^{1-\delta^{2}}dxdt$
$\displaystyle\leq
C\epsilon^{3/4}\frac{\tau^{\delta+\delta^{2}}s^{\delta^{2}}}{\tau
s}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-100>t>s-\tau/2,|u^{\prime}|\leq
10}\frac{1}{s-t}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|dxdt.$

We now apply Cauchy-Schwarz and use Lemma 18 for the second factor to get that

$\displaystyle\epsilon^{3/4}\frac{\tau^{\delta+\delta^{2}}s^{\delta^{2}}}{\tau
s}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-100>t>s-\tau/2,|u^{\prime}|\leq
10}\frac{1}{s-t}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|dxdt$
$\displaystyle\leq
C\epsilon^{3/4}\frac{\tau^{\delta+\delta^{2}}s^{\delta^{2}}}{\tau
s}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\left(\iint_{s-100>t>s-\tau/2,|u^{\prime}|\leq
10}\frac{1}{(s-t)^{2}}dxdt\right)^{1/2}\left(\iint|\overline{\partial}^{\prime}\partial^{\alpha}\phi|^{2}dxdt\right)^{1/2}$
$\displaystyle\leq
C\epsilon^{3/2}\frac{\tau^{\delta+\delta^{2}}s^{\delta^{2}}}{\tau
s}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\left(\int_{s-\tau/2}^{s-100}dt\right)^{1/2}\tau^{-1}$
$\displaystyle\leq
C\epsilon^{3/2}\frac{\tau^{\delta+\delta^{2}}s^{\delta^{2}}}{\tau
s}\tau^{1/2}\tau^{-1}<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}},$

where we note that we stopped writing the region of integration explicitly in
the above.

####  8.5.9 Case where $\tau\geq s^{1-10\delta}$, region where
$s/2<t<s-\tau/4$ and $u>\tau^{4/5}$

In this region, as in Section 8.5.8, we use pointwise bounds on
$|\partial\partial^{\beta}\phi|$, then Cauchy-Shwarz and Lemma 18 to obtain
the bound

$\displaystyle\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-\tau/4>t>s/2,u>\tau^{4/5},|u^{\prime}|\leq
10}\frac{\tau^{1/2}}{(s-t)^{3/2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|dxdt$
$\displaystyle\leq C\epsilon^{{3\over 4}}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\iint_{s-\tau/4>t>s/2,u>\tau^{4/5},|u^{\prime}|\leq
10}\frac{\tau^{1/2}}{(s-t)^{3/2}}\left(\frac{u^{\delta}}{us}\right)^{1-\delta^{2}}|\overline{\partial}^{\prime}\partial^{\alpha}\phi|dxdt$
$\displaystyle\leq C\epsilon^{{3\over 4}}\sum_{|\alpha|+|\beta|\leq
3N_{0}/4+10}\frac{\tau^{{1\over 2}}s^{\delta^{2}}}{s}(\tau^{{4\over
5}})^{-1+\delta+\delta^{2}}\left(\iint_{s-\tau/4>t>s/2,u>\tau^{4/5},|u^{\prime}|\leq
10}\frac{1}{(s-t)^{3}}dxdt\right)^{{1\over
2}}\left(\iint|\overline{\partial}^{\prime}\partial^{\alpha}\phi|^{2}dxdt\right)^{{1\over
2}}$ $\displaystyle\leq
C\epsilon^{3/2}\frac{s^{\delta^{2}}}{s}\tau^{-0.25}\left(\int_{s/2}^{s-\tau/4}\frac{1}{(s-t)}dt\right)^{1/2}\tau^{-4/5}<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}}.$

####  8.5.10 Case where $\tau\geq s^{1-10\delta}$, $s/2\leq t\leq s-\tau/4$,
and $u\leq t^{1/2-\delta/10}$

In this region we use the pointwise bound. We get that

$\displaystyle C\epsilon^{3/2}$ $\displaystyle\iint_{|u^{\prime}|\leq
1,s/2\leq t\leq s-\tau/4,u\leq
t^{1/2-\delta/10}}\left(\frac{\tau^{1/2}}{(s-t)^{3/2}}\right)\left(\frac{u^{\delta}}{tu}\right)^{1-\delta^{2}/100}\left(\frac{u^{\delta}}{t^{2}}\right)^{1-\delta^{2}/100}(s-t)d\theta
dudtdu^{\prime}$
$\displaystyle\leq\frac{C\epsilon^{3/2}\tau^{1/2}s^{0.03\delta^{2}}}{s^{3}}\iint_{|u^{\prime}|\leq
1,s/2\leq t\leq s-\tau/4,u\leq
t^{1/2-\delta/10}}(s-t)^{-1/2}u^{-1+2\delta+0.01\delta^{2}}d\theta
dudtdu^{\prime}$
$\displaystyle\leq\frac{C\epsilon^{3/2}\tau^{1/2}s^{0.03\delta^{2}}}{s^{3}}\int_{s/2}^{s-\tau/4}(s-t)^{-1/2}s^{(1/2-\delta/10)(2\delta+0.01\delta^{2})}dt$
$\displaystyle\leq\frac{C\epsilon^{3/2}\tau^{1/2}s^{0.03\delta^{2}}}{s^{3}}s^{1/2+(1/2-\delta/10)(2\delta+0.01\delta^{2})}dt<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}}.$

####  8.5.11 Case where $\tau\geq s^{1-10\delta}$, $s/2\leq t\leq s-\tau/4$,
and $t^{1/2-\delta/10}\leq u\leq\tau^{4/5}$

We obtain using Proposition 11 that

$|\partial\partial^{\beta}\phi|\leq
C\frac{(s-t)^{1/2}}{\tau^{1/2}}|\overline{\partial}\partial^{\beta}\phi|+C\frac{(s-t)^{1/2}}{\tau^{1/2}}|\overline{\partial}^{\prime}\partial^{\beta}\phi|\leq
Cs^{5\delta}(|\overline{\partial}\partial^{\beta}\phi|+|\overline{\partial}^{\prime}\partial^{\beta}\phi|).$

Thus, using pointwise bounds for
$|\overline{\partial}\partial^{\alpha}\phi|,|\overline{\partial}\partial^{\beta}\phi|$,
we obtain the bound

$\displaystyle\iint_{s/2\leq t\leq s-\tau/4,t^{1/2-\delta/10}\leq
u\leq\tau^{4/5},|u^{\prime}|\leq
1}\left(\frac{\tau^{1/2}}{(s-t)^{3/2}}\right)|\overline{\partial}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|dxdt$
$\displaystyle\qquad\leq Cs^{5\delta}\tau^{1/2}\iint_{s/2\leq t\leq
s-\tau/4,t^{1/2-\delta/10}\leq u\leq\tau^{4/5},|u^{\prime}|\leq
1}\frac{1}{(s-t)^{3/2}}\left(|\overline{\partial}\partial^{\alpha}\phi||\overline{\partial}\partial^{\beta}\phi|+|\overline{\partial}\partial^{\alpha}\phi||\overline{\partial}^{\prime}\partial^{\beta}\phi|\right)dxdt$
$\displaystyle\qquad\leq Cs^{5\delta}\tau^{1/2}\epsilon^{3/2}\iint_{s/2\leq
t\leq s-\tau/4,t^{1/2-\delta/10}\leq u\leq\tau^{4/5},|u^{\prime}|\leq
1}\frac{1}{(s-t)^{3/2}}\left(\frac{u^{\delta}}{s^{2}}\right)^{2-\delta^{2}}dxdt$
$\displaystyle\qquad\qquad+Cs^{5\delta}\tau^{1/2}\epsilon^{3/4}\iint_{s/2\leq
t\leq s-\tau/4,t^{1/2-\delta/10}\leq u\leq\tau^{4/5},|u^{\prime}|\leq
1}\frac{1}{(s-t)^{3/2}}\left(\frac{u^{\delta}}{s^{2}}\right)^{1-\delta^{2}/2}|\overline{\partial}^{\prime}\partial^{\beta}\phi|dxdt$
$\displaystyle\qquad\leq Cs^{5\delta}\tau^{1/2}\epsilon^{3/2}\iint_{s/2\leq
t\leq s-\tau/4,t^{1/2-\delta/10}\leq u\leq\tau^{4/5},|u^{\prime}|\leq
1}\frac{1}{(s-t)^{3/2}}\left(\frac{u^{\delta}}{s^{2}}\right)^{2-\delta^{2}}(s-t)d\theta
dudtdu^{\prime}$
$\displaystyle\qquad\qquad+Cs^{5\delta}\tau^{1/2}\epsilon^{3/4}\left(\iint_{s/2\leq
t\leq s-\tau/4,t^{1/2-\delta/10}\leq u\leq\tau^{4/5},|u^{\prime}|\leq
1}\frac{1}{(s-t)^{3}}\left(\frac{u^{\delta}}{s^{2}}\right)^{2-\delta^{2}}(s-t)d\theta
dudtdu^{\prime}\right)^{1/2}$
$\displaystyle\qquad\qquad\qquad\qquad\left(\iint_{s/2\leq t\leq
s-\tau/4,t^{1/2-\delta/10}\leq u\leq\tau^{4/5},|u^{\prime}|\leq
1}|\overline{\partial}^{\prime}\partial^{\beta}\phi|^{2}dxdt\right)^{1/2}.$

By Lemma 18, we know that the $L^{2}$ norm of
$\overline{\partial}^{\prime}\partial^{\beta}\phi$ is at most
$C\epsilon^{3/4}t^{-(1/2-\delta/10)}\leq C\epsilon^{3/4}s^{-(1/2-\delta/10)}$,
so

$\displaystyle\iint_{s/2\leq t\leq s-\tau/4,t^{1/2-\delta/10}\leq
u\leq\tau^{4/5},|u^{\prime}|\leq
1}\left(\frac{\tau^{1/2}}{(s-t)^{3/2}}\right)|\overline{\partial}\partial^{\alpha}\phi||\partial\partial^{\beta}\phi|dxdt$
$\displaystyle\qquad\leq
Cs^{5\delta}\tau^{1/2}\epsilon^{3/2}\frac{s^{2\delta^{2}}\tau^{4/5(1+2\delta)}}{s^{4}}\int_{s/2}^{s-\tau/4}(s-t)^{-1/2}dt$
$\displaystyle\qquad\qquad+Cs^{5\delta}\tau^{1/2}\epsilon^{3/2}\left(\frac{s^{2\delta^{2}}\tau^{4/5(1+2\delta)}}{s^{4}}\int_{s/2}^{s-\tau/4}(s-t)^{-2}dt\right)^{1/2}s^{-(1/2-\delta/10)}$
$\displaystyle\qquad\leq
Cs^{5\delta}\tau^{1/2}\epsilon^{3/2}\frac{s^{2\delta^{2}}\tau^{4/5(1+2\delta)}}{s^{4}}s^{1/2}+Cs^{5\delta}\tau^{1/2}\epsilon^{3/2}\left(\frac{s^{2\delta^{2}}\tau^{4/5(1+2\delta)-1}}{s^{4}}\right)^{1/2}s^{-(1/2-\delta/10)}$
$\displaystyle\qquad<C\epsilon^{3/2}\frac{\tau^{\delta}}{s^{2}}.$

####  8.5.12 Case where $\tau\geq s/{10}$

We need to bound (67), which comes out to bounding an integral over a backward
cone of thickness 1 given certain bounds on $\psi$ and certain bootstrap
assumptions on $\phi$ in this cone. We can perform a Lorentz boost to make it
so the tip of the cone is at a point where $s/100\leq\tau\leq s/20$, and the
same bootstrap assumptions hold (up to constant factors). The integral we need
to bound is new, but it is equivalent up to constant factors to the integral
we have already bounded in the previous subcases, so we are done.

###  8.6 Recovery of bootstrap assumption 1

This bootstrap assumption contains the pointwise estimates for $\phi$ and up
to ${3N_{0}\over 4}+1$ generic derivatives.

Proceeding as in Section 8.5 but using Proposition 5 instead of Proposition 6
gives us that

$|\partial\partial^{\alpha}\phi|(s,\tau,0,0)\leq{C\over(1+s)}\chi_{|\tau|\leq
5}\|\partial\partial^{\alpha}\phi\|_{H^{{3N_{0}\over
4}+10}(\Sigma_{0})}+C\sup_{|\beta|\leq
10}\sup_{\psi_{1}}\int_{0}^{s}\int_{\Sigma_{t}}|\partial^{\beta}\partial^{\alpha}(fg(d\phi,d\phi))||\partial\psi|dxdt.$

Now, we have that

${C\over(1+s)}\chi_{|\tau|\leq
5}\|\partial\partial^{\alpha}\phi\|_{H^{{3N_{0}\over
4}+10}(\Sigma_{0})}\leq{C\epsilon\over(1+s)}\chi_{\tau\leq 5},$

meaning that this term is appropriately controlled in order to recover the
bootstrap assumption. All that remains is to control the error integral given
by

$\sup_{|\beta|\leq
10}\sup_{\psi_{1}}\int_{0}^{s}\int_{\Sigma_{t}}|\partial^{\beta}\partial^{\alpha}(fg(d\phi,d\phi))||\partial\psi|dxdt.$
(66)

####  8.6.1 $s\leq 200$, $t\leq 100$, or $s-t\leq 100$

In the case that $s\leq 200$ or $s>200$ and $t\leq 100$, we use that

$|\partial\psi|\leq\frac{C}{1+s}\chi_{|u^{\prime}|\leq 1}$

and put both derivatives of $\phi$ in the integrand of (66) in $L^{2}$ and use
bootstrap assumption (7) to get the bound

$\frac{C\epsilon^{3/2}}{(1+s)(1+\tau)^{1-\delta}}.$

In the case that $s-t\leq 100$, we put $|\partial\psi|$ in $L^{2}$, put
whichever factor of $\phi$ gets hit by more derivatives in $L^{2}$ and bound
it using bootstrap assumptions (7), and put whichever factor of $\phi$ gets
hit by fewer derivatives in $L^{\infty}$ and bound it using bootstrap
assumptions (1). This then gives us an upper bound of

$C(C\epsilon^{3/4})\frac{C\epsilon^{3/4}}{(1+s)(1+\tau)^{1-\delta}}\leq\frac{C\epsilon^{3/2}}{(1+s)(1+\tau)^{1-\delta}}.$

Thus we only need to bound

$C\sup_{|\gamma|\leq
3N_{0}/4+10}\int_{100}^{s-100}\int_{\Sigma_{t}}\partial^{\gamma}g(d\phi,d\phi)|\partial\psi|dxdt.$
(67)

####  8.6.2 All other regions

Here, we use the upper bound of $\frac{C}{s-t}$ for $|\partial\psi|$. When
$t<s/2$, we note that

$\frac{1}{1+s-t}\leq
C\left(\frac{s}{\langle\tau\rangle}\right)\frac{\sqrt{t\langle\tau\rangle}}{s^{2}},$

pull the prefactor out of the integral, and then observe that we can just use
the bound on the integral we obtain in different regions in Sections 8.5.2
through 8.5.6. We note that this inequality uses that $|u^{\prime}|\leq 1$.

When $t\geq s/2$, we similarly note that

$\frac{1}{s-t}\leq
C\left(\frac{s}{\langle\tau\rangle}\right)\frac{\langle\tau\rangle^{1/2}}{(s-t)^{3/2}},$

pull the prefactor out of the integral, and then observe that we can just use
the bound on the integral we obtain in different regions in Sections 8.5.7
through 8.5.12. This completes the recovery of the bootstrap assumptions found
in Section 8.1 and, thus, completes the proof of Theorem 2.

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