Klainerman's Null Forms (A Question of Dimension) If $F$ has a particular form, then the wave equation $\square u = F(u,u')$ has a global solution for sufficiently small $C_0^\infty$ Cauchy data. Here $u'=(\partial_tu,\partial_1u,\dots,\partial_nu)$. Klainerman's null condition states that for $n=3$ the quadratic part of $F$ must be a bilinear form $Q(u',u')$ satisfying
$$
Q(\xi,\xi)=0 \qquad\text{whenever}\qquad \xi_0^2=\xi_1^2+\xi_2^2+\xi_3^2. \tag{1}
$$
The only such forms are linear combinations of the following seven null forms:
$$
Q_0(\xi,\eta)=\xi_0\eta_0-(\xi_1\eta_1+\xi_2\eta_2+\xi_3\eta_3)
$$
and
$$
Q_{ab}(\xi,\eta)=\xi_a\eta_b-\eta_a\xi_b, \qquad 0 \leq a < b \leq 3.
$$
The following facts are clear to me:


*

*The set of all bilinear forms on $\mathbb{R}^{1+3}$ may be identified with a 16-dimensional vector space.

*Those forms satisfying the null condition form a subspace.

*Both $Q_0$ and $Q_{ab}$ satisfy the null condition.

*The list $Q_0,Q_{01},\dots,Q_{23}$ is linearly independent.


But how does (1) imply that the subspace has dimension seven?
 A: Scratched this out over the weekend. Looking at components, write
$$
Q = \begin{bmatrix}
a_{00} & a_{01} & a_{02} & a_{03} \\
a_{10} & a_{11} & a_{12} & a_{13} \\
a_{20} & a_{21} & a_{22} & a_{23} \\
a_{30} & a_{31} & a_{32} & a_{33}
\end{bmatrix}
$$
Reduction. Assuming $\xi\neq0$ implies $\xi_0\neq0$ by (1). Dividing through by $|\xi_0|$ yields the equivalent condition
$$
Q(\xi,\xi)=0 \qquad\text{whenever}\qquad 1=\xi_1^2+\xi_2^2+\xi_3^2. \tag{2}
$$
Proceed investigating six cases by zeroing out components of $\xi$.
Case 1. Let $\xi_2=\xi_3=0$, which implies $\xi_1=\pm1$. Imposing $Q(\xi,\xi)=0$, we see that
$$
a_{00}+a_{11}\pm(a_{01}+a_{10})=0.
$$
Adding and subtracting gives the two relations
\begin{align}
a_{00}+a_{11} &= 0 \\
a_{01}+a_{10} &= 0.
\end{align}
Cases 2 and 3. By symmetry from case 1, setting $\xi_1=\xi_3=0$ and $\xi_1=\xi_2=0$ produces
\begin{align}
a_{00}+a_{22} &= 0 \\
a_{00}+a_{33} &= 0 \\
a_{02}+a_{20} &= 0 \\
a_{03}+a_{30} &= 0.
\end{align}
Case 4. Let $\xi_3=0$ and $\xi_1=\xi_2=\pm1/\sqrt{2}$. Imposing $Q(\xi,\xi)=0$ and applying the previous relations involving indices 1 and 2, we see that
$$
a_{21}+a_{12}=0.
$$
Cases 5 and 6. From case 4 and symmetry
\begin{align}
a_{31}+a_{13} &= 0 \\
a_{32}+a_{23} &= 0.
\end{align}
Therefore any bilinear form satisfying (1) must have the form
$$
Q = \begin{bmatrix}
a_{00} & a_{01} & a_{02} & a_{03} \\
-a_{01} & -a_{00} & a_{12} & a_{13} \\
-a_{02} & -a_{12} & -a_{00} & a_{23} \\
-a_{03} & -a_{13} & -a_{23} & -a_{00}
\end{bmatrix}.
$$
This can be written as a linear combination of the seven null forms. Since each null form satisfies (1) in general, so does every $Q$ of the above form.
